Reflection Theorems for Number Rings

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ig vraDdkn domain Dedekind a over Rings cohomol 6 Galois explicitize to theory Kummer of Extensions 5 groups Galois their and algebras Étale 4 cohomology Galois II Notation 3 xmlsfrtelyreader lay the for Examples 2 Introduction 1 Introduction I Contents . ui ig ...... rings . . . . resolvent ...... cubic . . . their . . . . and ...... rings . . . . Quartic ...... rings . . 6.5 . . . . Cubic ...... rings . . . 6.4 . . . Quadratic ...... 6.3 . . Discriminants ...... lattices . . . of . . 6.2 . Indices ...... 6.1 ...... symbol ...... Hilbert . . . . the . . . and . . . . . pairing . . . Tate . . . The ...... 5.1 ...... cohomology . Galois . . . . . at . . . . look . . fresh . . . . A ...... 4.6 . Torsors . . . automorphisms . . and . 4.5 . . Subextensions . . algebra étale . . 4.4 an Resolvents . of . group . Galois . 4.3 . The . . . algebras 4.2 . Étale . . 4.1 ...... equations quartic for Reflection 2.4 arXiv:2107.04727v1 ...... for . . . . . Reflection . equations . . . . . cubic . . . for . . . 2.3 . Reflection . . equations . . . quadratic . . . . . for 2.2 . . Reflection ...... [math.NT] . . 2.1 ...... Acknowledgements . . . paper the . . . of 1.5 . . . Outline ...... 1.4 Results . . . . 1.3 Methods . background 1.2 Historical 1.1 10 Jul 2021 2 eeto hoesfrnme rings number for theorems Reflection × n × n oe 11 ...... boxes vnO’Dorney Evan uy1,2021 13, July 1 35 ...... g 23 ogy 7 ...... 27 ...... 12 ...... 18 . . . 17 ...... 9 ...... 20 ...... 29 ...... 4 ...... 32 . 30 ...... 16 ...... 7 7 ...... 17 ...... 33 ...... 6 ...... 5 ...... 18 ...... 15 29 16 15 4 7 4 7 Cohomology of cyclic modules over a local field 41 7.1 Discriminants of Kummer and affine extensions ...... 42 7.2 TheShafarevichbasis ...... 44 7.3 ProofofTheorem7.1...... 49 7.4 Thetamecase ...... 51

III Composed varieties 51

8 Composed varieties 52 8.1 Examples ...... 53 8.2 Integral models; localization of orbit counts ...... 54 8.3 Fourier analysis of the local and global Tate pairings ...... 56

IV Reﬂection theorems: ﬁrst examples 60

9 Quadratic forms by superdiscriminant 61

10 Class groups: generalizations of the Scholz and Leopoldt reﬂection theorems 67 10.1Dualorders ...... 67 10.2 Dual orders are plentiful for quadratic extensions ...... 68 10.3 Localdualgeneralizedorders ...... 69 10.4 Relation to the Scholz reﬂection theorem ...... 71

V Reﬂection theorems: cubic rings 71

11 Cubic Ohno-Nakagawa 71 11.1 Abijectiveproofofthetamecase...... 74 11.2 Acomputationalproof...... 77 11.3 Abijectiveproofofthewildcase ...... 85

12 Non-natural weightings 90 12.1 Local weightings given by splitting types ...... 90 12.2 Discriminantreduction...... 91 12.3 Subringzetafunctions ...... 93 12.4 Invariant lattices at 2 ...... 96 12.5 Binary cubic forms over Z[1/N] ...... 98

VI Reﬂection theorems: quartic rings and related objects 100

13 Reﬂection for 2-adic quartic orders, and applications 100 13.1 Resultsonbinaryquarticforms ...... 103 13.2 The conductor property of the resolvent ring ...... 107

14 Tame quartic rings with non-split resolvent, by multijection 109 14.1 Invertibilityofidealsinorders...... 109 14.2 Self-duality of the count of quartic orders ...... 111

VII Counting quartic rings with prescribed resolvent 113

15 Introduction 113

2 16 The group H1 of quartic algebras with given resolvent 114

17 Reduced bases 115 17.1 The extender basis of a cubic resolventring ...... 119

18 Resolvent conditions 120 18.1 Transformation, and ring volumes in the white zone ...... 124 18.2 Fromringvolumestoringcounts ...... 125

19 The conic over K 127 19.1Diagonalconics...... O ...... 130 19.2 Thesolutionvolumeoftheconic ...... 132 19.3TheBrauerclass ...... 135 19.4Thesquareness ...... 136 19.5 ...... 138 N11 20 Boxgroups 141 20.1Signatures...... 141 20.2 Boxgroups in unramiﬁed splitting type ...... 142 20.3 Boxgroups in splitting type 13 ...... 145 20.4 Therecenteringlemma...... 149 20.5Charmedcosets...... 150 20.6Theprojectors ...... 152 20.7Notation...... 153

21 Ring volumes for ξ1′ 154 21.1Thesmearinglemma...... 154 21.2 The zones when is strongly active (black, plum, purple, blue, green, and red)...... 155 N11 21.3 The zones when 11 isweaklyactive(brownandyellow) ...... 166 21.4Thebeigezone ...... N ...... 171 21.5Orthogonality...... 173 21.6Smearedanswers ...... 176 21.7 The average value of a quadratic character on a box ...... 178

22 Ring volumes for ξ2′ 180 22.1 12 ...... 180 22.2 M ...... 181 M22 23 Further remarks on the code 184

VIII Unanswered questions 184

24 Doubly traced quartic rings 185

25 Reﬂection for 2 n n boxes 187 × ×

IX Appendices 187

A The Grothendieck-Witt ring and the proof of Lemma 19.11 188

3 B Examples of zone totals 189

Abstract The Ohno-Nakagawa reflection theorem is an unexpectedly simple identity relating the number of GL2Z-classes of binary cubic forms (equivalently, cubic rings) of two different discriminants D, −27D; it generalizes cubic reciprocity and the Scholz reflection theorem. In this paper, we provide a framework for generalizing this theorem using a global and local step. The global step uses Fourier analysis on the 1 adelic cohomology H (AK , M) of a finite Galois module, modeled after the celebrated Fourier analysis on AK used in Tate’s thesis. The local step is combinatorial, more elementary but much more mysterious. We establish reflection theorems for binary quadratic forms over number fields of class number 1, and for cubic and quartic rings over arbitrary number fields, as well as binary quartic forms over Z; the quartic results are conditional on some computational algebraic identities that are probabilistically true. Along the way, we find elegant new results on Igusa zeta functions of conics and the average value of a quadratic character over a box in a local field.

Part I Introduction

1 Introduction

1.1 Historical background In 1932, using the then-new machinery of class field theory, Scholz [49] proved that the class groups of the quadratic fields Q(√D) and Q(√ 3D), whose discriminants are in the ratio 3, have 3-ranks differing by at most 1. This is a remarkable− early example of a reflection theorem. A generalization− due to Leopoldt [31] relates different components of the p-torsion of the class group of a number field containing µp when decomposed under the Galois group of that field. Applications of such reflection theorems are far-ranging: for instance, Ellenberg and Venkatesh [21] use reflection theorems of Scholz type to prove upper bounds on ℓ-torsion in class groups of number fields, while Mihăilescu [34] uses Leopoldt’s generalization to simplify a step of his monumental proof of the Catalan conjecture that 8 and 9 are the only consecutive perfect powers. Through the years, numerous reflection principles for different generalizations of ideal class groups have come into print. A very general reflection theorem for Arakelov class groups is due by Gras [23]. A quite different direction of generalization was discovered by accident in 1997: The following relation was conjectured by Ohno [44] on the basis of numerical data and proved by Nakagawa [38], for which reason we will call it the Ohno-Nakagawa (O-N) reflection theorem:

Theorem 1.1 (Ohno–Nakagawa). For a nonzero integer D, let h(D) be the number of GL2(Z)-orbits of binary cubic forms f(x, y)= ax3 + bx2y + cxy2 + dy3 of discriminant D, each orbit weighted by the reciprocal of its number of symmetries (i.e. stabilizer in GL2(Z)). Let h3(D) be the number of such orbits f(x, y) such that the middle two coeﬃcients b,c are multiples of 3, weighted in the same way. Then for every nonzero integer D, we have the exact identity

3h(D), D> 0 h3( 27D)= (1) − ( h(D), D< 0.

By the well-known index-form parametrization (see 6.9 below), h(D) also counts the cubic rings of discriminant D over Z, weighted by the reciprocal of the order of the automorphism group. It turns out that h (D) counts those rings C for which 3 tr ξ for every ξ C. When D is a fundamental discriminant, the 3 | C/Z ∈ corresponding cubic extensions are closely related, via class ﬁeld theory, to the 3-class group of Q(√D) and we get back Scholz’s reﬂection theorem, as Nakagawa points out ([38], Remark 0.9).

4 Theorem 1.1 was quite unexpected, because GL2(Z)-orbits of binary cubics have been tabulated since Eisenstein without unearthing any striking patterns. Even the exact normalizations h(D), h3(D) had been in use for over two decades. They appear in the Shintani zeta functions

h( n) ζ±(s)= ± ns n 1 X≥ h3( 27n) ζˆ±(s)= ± , ns n 1 X≥ a family of Dirichlet series which play a prominent role in understanding the distribution of cubic number ﬁelds, similar to how the famous Riemann zeta function controls the distribution of primes. As Shintani proved as early as 1972 [52], the Shintani zeta functions satisfy a matrix functional equation (see Nakagawa [38], eq. (0.1))

+ ˆ+ ζ (1 s) 1 3s 2 4s 1 2 1 sin 2πs sin πs ζ (s) − =2− 3 − π− Γ s Γ(s) Γ s + (2) ζ−(1 s) − 6 6 3 sin πs sin 2πs ζˆ (s) − − The condition that 3 divide b and c is equivalent to requiring that the cubic form f is integer-matrix, that is, its corresponding symmetric trilinear form

b/3 c/3 ⑧⑧ ⑧⑧ ⑧⑧ ⑧ a b/3

c/3 d ⑧ ⑧⑧⑧ ⑧⑧ b/3 c/3 has integer entries. This condition arose in Shintani’s work by taking the dual lattice to Z4 under the pairing 1 1 (a,b,c,d), (a′,b′,c′, d′) = ad′ bc′ + cb′ da′, (3) h i − 3 3 − which plays a central role in proving the functional equation. However, as we will find, the pairing (3) does not figure in the proof of our reflection theorems, which indeed often relate lattices that are not dual under it. Using the functional equation, Shintani proved that the ζ± admit meromorphic continuations to the complex plane with simple poles at 1 and 5/6, inspiring him to conjecture that the number N (X) of cubic fields of positive or negative discriminant up to X has the shape ±

N (X)= a X + b X5/6 + o(X5/6) ± ± ± for suitable constants a and b . This conjecture was proven by Bhargava, Shankar, and Tsimerman [8] and independently by Taniguchi± and± Thorne [54]. Neither proof needs the Ohno-Nakagawa reﬂection theorem (Theorem 1.1), which appears in the notation of Shintani zeta functions in the succinct form

+ + ζˆ (s)= ζ−(s) and ζˆ−(s)=3ζ (s). (4)

Remark 1.2. In the earlier papers, the term “Ohno-Nakagawa identities” was used, referring to the pair (4). Our work conﬁrms the intuition that, despite the diﬀerent scalings, both identities are essentially one theorem.

1.2 Methods Several proofs of O-N are now in print ([38, 33, 43, 22]), all of which consist of two main steps:

5 • A “global” step that uses global class ﬁeld theory to understand cubic ﬁelds, equivalently GL2(Q)-orbits of cubic forms;

• A “local” step to count the rings in each cubic field, equivalently the GL2(Z)-orbits in each GL2(Q)- orbit, and put the result in a usable form. In this paper, the distinction between these steps will be formalized and clarified. For the global step, we take inspiration from Tate’s celebrated thesis [55], which uses Fourier analysis on the adeles to give illuminating new proofs of the functional equations for the Riemann ζ-function and various L-functions. Taniguchi and Thorne (see [53]) used Fourier analysis on the space of binary cubic forms over Fq to get the functional equation for the Shintani zeta function of forms satisfying local conditions at primes. Despite the similarities, their work is essentially independent from ours. We are also inspired by a remark due to Calegari in a paper of Cohen, Rubinstein-Salzedo, and Thorne ([12], Remark 1.6), pointing out that their reflection theorem counting dihedral fields of prime order can also be derived from a theorem of Greenberg and Wiles for the sizes of Selmer groups in Galois cohomology. We present a notion of composed variety, a scheme over the ring of integers K of a number field admitting an action of an algebraic group over .V Our guiding example is theO scheme of binary G OK V cubic forms of discriminant D with its action of = SL2. The term “composed” refers to the presence of a composition law on the orbits, which relate naturallyV to a Galois cohomology group H1(K,M). Our (global) reflection theorems can be stated as saying that two composed varieties (1), (2) have the same number V V of K -points, with a suitable weighting. Introducing a new technique of Fourier analysis on the adelic O 1 1 cohomology group H (AK ,M) = v′ H (Kv,M), based on Poitou-Tate duality, we present a generalized reflection engine (Theorems 8.12 and 8.13) that reduces global reflection theorems to local reflection theorems, Q (1) (2) that is, statements involving only the Kv -points of and for a single place v of K. A typical case is Theorem 11.2. O V V These local reflection theorems are approachable by elementary methods but can be difficult to prove. We present two kinds of proofs. The first is a bijective argument involving Bhargava’s self-balanced ideals that is very clean but has only been discovered at the “tame primes” (p ∤ 3 in the cubic case, p ∤ 2 in the quartic). The second is by explicitly computing the number of orders of given resolvent in a cubic or quartic algebra. We express it as a generating function in a number of variables depending on the splitting type of the resolvent. The generating function is rational, and local reflection can be written as an equality between two rational functions; but these functions are so complicated that the best approximation to a proof of the identity that we can find is a Monte Carlo proof, namely, substituting random values for the variables in some large finite field and verifying that the equality holds. The reader is invited to recheck this verification using the source code in Sage that will be made available with the final version of this paper.

1.3 Results We are able to prove O-N for binary cubic forms over all number fields K, verifying and extending the conjectures of Dioses [20, Conjecture 1.1]. However, we go further and ask whether every SL ( )-invariant 2 OK lattice within the space V (K) of binary cubic forms admits an O-N-style reflection theorem. Over Z, this question was answered affirmatively for each of the ten invariant lattices by Ohno and Taniguchi [45]. Over K , such lattices were classified by Osborne [47], and they differ from one another only at the primes dividingO 2 and 3. The lattices at 3 yield an elegant reflection theorem (Theorem 11.3) in which the condition 1 b,c t, where t is an ideal dividing 3 in K , reflects to b,c 3t− , the complementary divisor. At 2, the corresponding∈ reflection theorems still exist,O though they become∈ difficult to write explicitly: see Theorem 12.14. We also find a new reflection theorem (Theorem 9.3) counting binary quadratic forms, not by discriminant, but by a curious invariant: the product a(b2 4ac) of the discriminant and the leading coefficient. Over − Z, the reflection theorem (Theorem 9.5) has the potential to be proved simply using quadratic reciprocity, eschewing the machinery of Galois cohomology, though it seems unlikely that the theorem would have ever been discovered without it. Nakagawa has also conjectured [36] a reflection theorem for pairs of ternary quadratic forms, which parametrize quartic rings. The natural invariant to count by is the discriminant, but it is more natural from our perspective to subdivide further and ask for a reflection theorem for rings with fixed cubic resolvent,

6 which holds in the known cases [36, Theorem 1]. Here our global framework applies without change, but the local enumeration of orders in a quartic field presents formidable combinatorial difficulties, especially in the wildly ramified (2-adic) setting, which have been attacked in another work of Nakagawa [37]. Our methods have the potential to finish this work, but because we count by resolvent rather than discriminant, our answers do not directly match his. The process of proving local quartic O-N leads us down some fruitful routes that do not at first sight have any connection to reflection theorems or to the enumeration of quartic rings. These include new cases of the Igusa zeta functions of conics (Lemmas 19.9 and 19.10) and a result on the average value of a quadratic character on a box in a local field (Theorem 21.21). If quartic O-N holds true in all cases, it implies that the cubic resolvent ring (in the sense of Bhargava) of a maximal quartic order has a second natural characterization: it is the “conductor ring” for which the Galois-naturally attached extension K6/K3 is a ring class field (Theorem* 13.15).

1.4 Outline of the paper In Section 2, we state and give examples of the main global reflection theorems of the paper over Z, in a fashion that requires a minimum of prior knowledge, for the end of further diffusing interest in, and appreciation of, the beauty of number theory. In Part II, we lay out preliminary matter, much of which is closely related to results that have appeared in the literature but under different guises. It includes a simple characterization (Proposition 4.21) of Galois H1 in terms of étale algebras whose Galois group is a semidirect product. It also includes a theorem (Theorem 1 7.1) on the structure of H (K,M) in the case that K is local and M ∼= p (with any Galois structure), which will be invaluable in what follows. C In Part III, we lay out the framework of composed varieties, on which we perform the novel technique of Fourier analysis of the local and global Tate pairings to get our main local-to-global reflection engine (Theorems 8.12 and 8.13). The remainder of the paper will concern applications of this engine. In Part IV, we prove two relatively simple reflection theorems: one for quadratic forms (Theorem 9.3), and a version of the Scholz reflection principle for class groups of quadratic orders (Theorem 10.3). In Part V, we prove our extensions of Ohno-Nakagawa for cubic forms and rings. The quartic case is dealt with in Parts VI and VII: the first part dealing with the bijective methods, and the second with the (long) work of explicitly counting orders in each quartic algebra. The case of partially ramified cubic resolvent (splitting type 121) is still in progress, so we restrict our attention to the four tamely splitting types in the present version. We conclude the paper with some unanswered questions engendered by this research.

1.5 Acknowledgements For fruitful discussions, I would like to thank (in no particular order): Manjul Bhargava, Xiaoheng Jerry Wang, Fabian Gundlach, Levent Alpöge, Melanie Matchett Wood, Kiran Kedlaya, Alina Bucur, Benedict Gross, Sameera Vemulapalli, Brandon Alberts, Peter Sarnak, and Jack Thorne.

2 Examples for the lay reader

Fortunately for the non-specialist reader, the statements (though not the proofs) of the main results in this thesis can be stated in a way requiring little more than high-school algebra. We here present these statements and some examples to illustrate them.

2.1 Reflection for quadratic equations Definition 2.1. Let f(x)= ax2 +bx+c be a quadratic polynomial, where the coefficients a, b, c are integers. The superdiscriminant of f is the product I = a (b2 4ac) · − of the leading coefficient with the usual discriminant.

7 Lemma 2.2. If we replace x by x + t in a quadratic polynomial f, where t is a ﬁxed integer, then the superdiscriminant does not change. Proof. This can be veriﬁed by brute-force calculation, but the following method is more illuminating. The discriminant is classically related to the two roots of f,

b + √b2 4ac b √b2 4ac x = − − and x = − − − , 1 2a 2 2a through their diﬀerence:

2√b2 4ac √b2 4ac x x = − = − 1 − 2 2a a b2 4ac (x x )2 = − 1 − 2 a2 a3(x x )2 = a (b2 4ac)= I. 1 − 2 · −

If we replace x by x + t, then a does not change, and both roots x1, x2 are decreased by t, so their diﬀerence x x is unchanged. Therefore I is unchanged. 1 − 2 Deﬁnition 2.3. Call two quadratics f1, f2 equivalent if they are related by a translation f2(x)= f1(x + t). If I is a nonzero integer, let q(I) be the number of quadratics of superdiscriminant I, up to equivalence. Let + + q2(I), q (I), q2 (I) be the number of such quadratics that satisfy certain added conditions:

• For q2(I), we require that the middle coeﬃcient b be even. • For q+(I), we require that the roots be real, that is, that b2 4ac > 0. − + • For q2 (I), we impose both of the last two conditions. We are now ready to state a quadratic reﬂection theorem, the main result of this section. Theorem 2.4 (“Quadratic O-N”). For every nonzero integer n,

+ q2 (4n)= q(n) + q2(4n)=2q (n). Proof. The proof is not easy. See Theorem 9.5. It’s not hard to compute all quadratics of a fixed superdiscriminant I. The leading coefficient a must be a divisor of I (possibly negative), and there are only finitely many of these. Then, by replacing x by x + t where t is an integer nearest to b/(2a), we can assume that b lies in the window a < b a . We can try each of the integer values in this− window, checking whether −| | ≤ | | ab2 I c = − 4a2 comes out to an integer. Example 2.5. There are five quadratics of superdiscriminant 15:

+ + f(x) q q q2 q2 x2 + x 4 X 15− x2 + x− X X 15x2 x X X 15x2 +− 11x +2 X X 15x2 11x +2 X X − You might think we left out x2 x 4, but it is equivalent to another quadratic on the list: − − − x2 x 4= (x + 1)2 + (x + 1) 4. − − − − −

8 So we get the totals q(15) = 5 and q+(15) = 4. There are 18 quadratics of superdiscriminant 60:

+ + f(x) q q q2 q2 x2 15 X X X X x−2 15 X X −3x2−+2x 2 X X −3x2 2x − 2 X X −4x2 +− x −1 X −4x2 x − 1 X 15− x2 +2− x− X X X X 15x2 2x X X X X 15x2 +8− x +1 X X X X 15x2 8x +1 X X X X 60x2 +− x X X 60x2 x X X 60x2 +− 31x +4 X X 60x2 31x +4 X X 60x2 +− 41x +7 X X 60x2 41x +7 X X 60x2 +− 49x + 10 X X 60x2 49x + 10 X X − Counting carefully, we get

+ + q(60) = 18, q2(60) = 8, q (60) = 13, q2 (60) = 5. The equalities q+(60) = 5 = q(15) and q (60)=8=2 4=2q+(15) 2 2 · are instances of Theorem 2.4. From the same theorem, we derive, without computation, that

+ + q2 (240) = q(60) = 18 and q2(240) = 2q (60) = 26. This short investigation raises many questions. The superdiscriminant I = a(b2 4ac) does not seem to have been considered before. Is there an explicit formula for q(I)? Is there an elementary− proof of Theorem 2.4? See Example 9.6 for a connection to Gauss’s celebrated law of quadratic reciprocity.

2.2 Reflection for cubic equations Definition 2.6. For a cubic polynomial f(x)= ax3 + bx2 + cx + d, we define the discriminant to be disc f = a4(x x )2(x x )2(x x )2, (5) 1 − 2 1 − 3 2 − 3 where x1, x2, x3 are the roots. Explicitly, disc f = b2c2 4ac3 4b3d 27a2d2 + 18abcd. (6) − − − There are many transformations of a cubic polynomial that don’t change the discriminant. One is changing x to x + t, where t is a constant. Another is reversing the coefficients, 1 f(x)= ax3 + bx2 + cx + d x3f = dx3 + cx2 + bx + a. 7−→ x Both of these are special cases of the following construction.

9 Deﬁnition 2.7. Two cubic polynomials f1, f2 with integer coeﬃcients are equivalent if there is a matrix

p q r s whose determinant ps qr is 1 such that − ± px + q f (x) = (rx + s)3 f . 2 · 1 rx + s A matrix that makes f equivalent to itself, that is,

px + q f(x) = (rx + s)3 f , · rx + s is called a symmetry of f. The number of symmetries of f is denoted by s(f). Deﬁnition 2.8. If D is a nonzero integer, deﬁne h(D) to be the number of cubic polynomials

f(x)= ax3 + bx2 + cx + d of discriminant D, up to equivalence, each f counted not once but 1/s(f) times, where s(f) is the number of symmetries. Define h3(D) to be the number of cubics of discriminant D for which the middle two coefficients, b and c, are multiples of 3, up to equivalence, each f counted 1/s(f) times as before. We can now state the Ohno-Nakagawa reflection theorem that got this research project started:

Theorem 2.9 (Ohno-Nakagawa; Theorem 1.1). For every nonzero integer D,

3h(D), D> 0 h3( 27D)= − ( h(D), D< 0.

Proof. Several proofs are in print (see the Introduction). In this paper, we prove this theorem as a special case of Theorem 11.3.

Example 2.10. Take D =1. There is just one cubic with integer coeﬃcients and discriminant 1, namely

f(x)= x(x +1)= x2 + x.

The reader may balk at considering a quadratic polynomial as a “cubic” with leading coeﬃcient 0, but the polynomial can be replaced by any number of equivalent forms, for instance

x (x 1)3 f = x(x 1)(2x 1). − · x 1 − − − We will suppress this detail in subsequent examples. (A program for computing all cubics of a given discriminant is found in the attached ﬁle cubics.sage, based on an algorithm of Cremona [15, 16]). The cubic f has six symmetries, which is related to the fact that three linear factors can be permuted in 3!=6 ways. In terms of f(x)= x(x + 1), the symmetries are

1 0 1 1 0 1 1 1 1 0 0 1 , , , , , . 0 1 −0− 1 1 0 −1− 0 1 1 1 1 − − − − So h(1) = 1/6. Correspondingly, we look at cubics of discriminant 27. There are two: − f(x)= x2 + x +7 and f(x)= x3 +1.

10 Each admits two symmetries: the ﬁrst has

1 0 1 1 , , 0 1 −0− 1 and the second has 1 0 0 1 , . 0 1 1 0 So h( 27) = 1/2+1/2=1 and h ( 27) = 1/2. In particular, − 3 − h ( 27) = 3h (1), 3 − 3 in conformity with Theorem 2.9.

2.3 Reflection for 2 n n boxes × × Bhargava [4] studied 2 3 3 boxes as a visual representation for quartic rings, as cubic polynomials do for cubic rings. We think that× × reflection holds not only for 2 3 3 boxes but for 2 5 5, 2 7 7, and so on. We nearly prove the 2 3 3 case in this paper. We× are× quite far from proving× it× for the× lar×ger boxes. × × Definition 2.11. A box is a pair (A, B) of n n integer symmetric matrices. The resolvent of a box is the polynomial × f(x) = det(Ax B). − It is a polynomial in x, of degree at most n. If A is the identity matrix, the resolvent devolves into the standard characteristic polynomial.

Deﬁnition 2.12. Two boxes (A1,B1) and (A2,B2) are equivalent if there is an integer n n matrix X, 1 × whose inverse X− also has integer entries, such that

A2 = XA1X⊤ and B2 = XB1X⊤.

If (A2,B2) = (A1,B1) = (A, B) are the same pair, then X is called a symmetry of (A, B). The number of symmetries of (A, B) will be denoted by s(A, B). Conjecture 2.13 (“O-N for 2 n n boxes”). Let n be a positive odd integer. Let f be a polynomial of degree n with no multiple roots and× only× one real root. Denote by h(f) the number of 2 n n boxes with resolvent f, up to equivalence, each box weighted by the reciprocal of its number of symmetries.× × Denote by h2(f) the number of such boxes with even numbers along the main diagonals of A and B, weighted the same way. Then n 1 n 1 − h (2 − f)=2 2 h(f). (7) 2 · Remark 2.14. The condition that f have no multiple roots (even complex ones) is needed to ensure that there are only finitely many boxes with f as a resolvent. The condition that f have no more than one real root can be eliminated, but then we must impose conditions on the real behavior of the boxes that are difficult to state succinctly. Example 2.15. Take as resolvent f(x) = x3 x 1, the simplest irreducible cubic. It has one real root ξ 1.3247 and discriminant 23. There are two− boxes− with resolvent f, up to equivalence: ≈ − 0 0 1 0 1 0 0 1 0 1 0 1 − − − − − 0 1 0 , 1 0 1 , 1 0 1 , 0 1 1 .  1− 0 1   −0 1 −1   −0 1 −1   1 −1 −1  − − − − − − − − −         3 (These were computed from the balanced pairs ( R, 1) and ( R, ξ) in the number field R = Z[ξ]/(ξ ξ 1) corresponding to f.) Neither has any symmetriesO besides theO two trivial ones, the identity matrix− and− its negative, so 1 1 h(f)= + =1. 2 2

11 There are many boxes with resolvent 2f, but just one with even numbers all along the main diagonals of A and B, namely 0 01 01 0 0 2 0 , 10 0 .  1− 02   0 0 2  −     (This was computed from the unique quartic ring = Z with resolvent .) It too has only the trivial O ×OR OR symmetries, to h2(2f)=1/2, in accord with Conjecture 2.13.

2.4 Reflection for quartic equations There are also reflection theorems that appear when counting quartic polynomials. Definition 2.16. If f(x)= ax4 + bx3 + cx2 + dx + e is a quartic polynomial with integer coefficients, its resolvent is

g(y)= y3 cy2 + (bd 4ae)y +4ace b2e ad2; (8) − − − − equivalently, if f(x)= a(x x )(x x )(x x )(x x ), − 1 − 2 − 3 − 4 then g(y)= y a(x x + x x ) y a(x x + x x ) y a(x x + x x ) . − 1 2 3 4 − 1 3 2 4 − 1 4 2 3 Remark 2.17. Cubic resolvents of this type have been used since the 16th century as a step in solving quartic equations. For instance, it is well known that if f(x) factors as the product of two quadratics with integer coeﬃcients, then g(y) has a rational root (the converse is not true). Analogously to Deﬁnition 2.7, we put:

Deﬁnition 2.18. Two quartic polynomials f1, f2 with integer coeﬃcients are equivalent if there is a matrix

p q r s whose determinant ps qr is 1 such that − ± px + q f (x) = (rx + s)4 f . 2 · 1 rx + s A matrix that makes f equivalent to itself, that is,

px + q f(x) = (rx + s)4 f , · rx + s is called a symmetry of f. The number of symmetries of f is denoted by s(f). We have:

Lemma 2.19. (a) If two quartics f1, f2 are equivalent, then their resolvents g1, g2 are related by a translation g2(x)= g1(x + t) for some integer t. (b) A quartic and its resolvent have the same discriminant

disc f = disc g = b2c2d2 4ac3d2 4b3d3 +18abcd3 27a2d4 4b2c3e+16ac4e+18b3cde 80abc2−de 6ab−2d2e + 144a2cd2e− 27b4e2−+ 144ab2ce2 128a2c2e2 − 192a2bde2−+ 256a3e3. − − −

12 Proof. Exercise. As before, our reflection theorem will relate general quartics to quartics satisfying certain divisibility relations. Here the relations are quite peculiar: Definition 2.20. A quartic polynomial f(x)= ax4 + bx3 + cx2 + dx + e is called supereven if b, c, and e are multiples of 4 and d is a multiple of 8. 4 Not every quartic equivalent to a super-even quartic is itself supereven. (For instance, f1 = x +4 and 4 0 1 f2 =4x +1 are equivalent under the flip 1 0 , but f2 is not supereven.) We therefore make the following definition. Definition 2.21. Two quartic polynomials f1, f2 with integer coefficients are evenly equivalent if there is a matrix p q r s whose determinant ps qr is 1, and r is even, such that − ± px + q f (x) = (rx + s)4 f . 2 · 1 rx + s Such a matrix that makes f equivalent to itself, that is, px + q f(x) = (rx + s)4 f , · rx + s is called an even symmetry of f. The number of even symmetries of f is denoted by s2(f). Theorem 2.22 (“Quartic O-N”). Let g be an integer cubic with leading coefficient 1, no multiple roots, and odd discriminant. Denote by h(g) the number of quartics whose resolvent is g(y + t) for some t, up to equivalence and weighted by the reciprocal of the number of symmetries. Denote by h2(g) the number of supereven quartics whose resolvent is g(y+t) for some t, up to even equivalence and weighted by the reciprocal of the number of even symmetries. Define g2 by y g (y)=64g 2 4 Then: • If g has one real root, then 4h(g)= h2(g2). • If g has three real roots, then we subdivide + h(g)= h (g)+ h−(g)+ h±(g) where the respective terms count only quartic functions that are always positive, always negative, and have four real roots. We subdivide + h2(g)= h2 (g)+ h2−(g)+ h2±(g). Then:

2h(g)= h2±(g2) + + 4 h (g)+ h±(g) = h2 (g2)+ h2±(g2)

4h−(g)+ h±(g) = h2−(g2)+ h2±(g2) Also, denote by k(g) the number of integral 3 3 symmetric matrices of characteristic polynomial g. Then × + k(g)=24 h±(g) h (g) h−(g) . − − 13 Proof. See Theorem 13.11. Remark 2.23. We think that the hypothesis of odd discriminant is removable, but we have not yet finished the proof. Example 2.24. Let g(y)= y3 y 1. By techniques presented in Section 13.1, it is possible to transform the boxes found in example 2.15− into− binary quartic forms. We find that there is only one quartic with resolvent g, namely f(x)= x3 x 1 − − (which, as before, can be transformed by an equivalence to one with nonzero leading coefficient); and four supereven binary quartics with resolvent g (y)= y3 16y 64, namely 2 − − f(x)=4x3 + 12x2 +8x 4=4 (x + 1)3 (x + 1) 1 − − − f(x)= x4 +4x3 + 12x2 +8x − f(x)= x4 +8x 4 − − f(x)= x4 +4x3 4. − − All these have one pair of complex roots (as must occur for a resolvent with negative discriminant) and only the trivial symmetries 1 0 , so ± 0 1 1 1 h(g)= and h (g )=2=4 , 2 2 2 · 2 in accord with the first part of the theorem. Example 2.25. Consider g(y)= y3 2y2 3y +6=(y 2)(y + √3)(y √3), a cubic with three real roots. The quartics with resolvent g are − − − − f(x)= x(2x 1)(3x2 1), − − − which has four real roots, and f(x) = (x2 + x + 1)(x2 + 1), which has no real roots and is positive for all real x. Each has only the trivial symmetries, so

1 + 1 h±(g)= , h (g)= , h−(g)=0. 2 2 + (Note the discrepancy between h and h−.) Correspondingly, there are eight supereven binary quartics with resolvent g (y) = (y 8)(y +4√3)(y 4√3): 2 − − f(x)= 2x4 8x3 4x2 +8x = 2x(x + 2)(x2 +2x 2) − − − − − f(x)=4x3 4x2 16x 8=4(x + 1)(x2 2x 2) − − − − − f(x)=8x4 16x3 + 20x2 12x +4=4(x2 x + 1)(2x2 2x + 1) − − − − f(x)= x4 6x3 + 20x2 32x +32=(x2 4x + 8)(x2 2x + 4) − − − − f(x)= x4 +8x2 12 = (x2 2)(x2 6) − − − − − f(x)= 3x4 +8x2 4= (x2 2)(3x2 2) − − − − − f(x)= x4 +8x2 +12=(x2 + 2)(x2 + 6) f(x)=3x4 +8x2 +4=(x2 + 2)(3x2 + 2). Thus + h2±(g2)=2, h2 (g2)=2, h2−(g2)=0. This is in accord with the theorem, from which we also learn that

+ k(g)=48 h±(g) h (g) h−(g) =0, − − so g is not the characteristic polynomial of any integer 3 3 symmetric matrix, despite having three real roots (which is a necessary, but not a suﬃcient, condition).×

14 Example 2.26. Let g(y)= y3 y. Knowing that f(x)= x3 x is the only quartic with cubic resolvent f, − 0 1 − and it has four symmetries, the powers of 1− 0 , we get + 1 k(g)=24 h±(g) h (g) h−(g) = 24 0 0 =6. − − 4 − − So there are six symmetric matrices with characteristic polynomial y3 y. Indeed, they are the diagonal matrices with 1, 0, and 1 along the diagonal in any of the 3!=6 possible− orders. − 3 Notation

The following conventions will be observed in the remainder of the paper. We denote by N and N+, respectively, the sets of nonnegative and of positive integers. If P is a statement, then 1 P is true 1P = ( 0 P is false.

If S is a set, then 1S denotes the characteristic function 1S(x)= 1x S. An algebra will always be commutative and of finite rank over∈ a field, while a ring or order will be a finite-dimensional, torsion-free ring over a Dedekind domain, containing 1. An order need not be a domain. If a,b L are elements of a local or global field, a separable closure thereof, or a finite product of the ∈ preceding, we write a b to mean that b = ca for some c in the appropriate ring of integers L. If a b and b a, we say that a and b are| associates and write a b. Note that a and b may be zero-divisors.O | | ∼ If S is a finite set, we let Sym(S) denote the set of permutations of S; thus Sn = Sym( 1,...,n ). If S = T , and if g Sym(S), h Sym(T ) are elements, we say that g and h are conjugate{ if there} is a| bijection| | | between S∈ and T under∈ which they correspond. Likewise when we say that two subgroups G Sym(S), H Sym(T ) are conjugate. ⊆We will use the⊆ semicolon to separate the coordinates of an element of a product of rings. For instance, in Z Z, the nontrivial idempotents are (1; 0) and (0; 1). × If n is a positive integer, then ζn denotes a primitive nth root of unity in Q¯ , while ζ¯n denotes the nth root of unity 2 n 1 n ζ¯ = 1; ζ ; ζ ; ... ; ζ − Q¯ . n n n n ∈ Throughout the proofs of the local reflection theorems, we will fix a local field K, its valuation v = vK , its residue field kK of order q, and a uniformizer π = πK . The letter e will denote the absolute ramification index (e = vK (2) in the quadratic and quartic cases, vK (3) in the cubic). We let mK denote the maximal ideal, and likewise m ¯ be the maximal ideal of the ring ¯ of algebraic integers over K; note that m ¯ is K OK K not finitely generated. We also allow v = vK to be applied to elements of K¯ , the valuation being scaled so that its restriction to K has value group Z. We use the absolute value bars for the corresponding metric, whose normalization will be left undetermined. |•| If K is a local field, an m-pixel is a subset of an affine or projective space over K defined by requiring n O2 the coordinates to lie in specified congruence classes modulo π . For instance, in P ( K ), a 0-pixel is the 2 2n 2 O whole space, which is subdivided into (q + q + 1)q − -many n-pixels for each n 1. If R/K is a finite-dimensional, locally free algebra over a ring, we denote by RN≥=1 the subgroup of units of norm 1. The group operation is implicitly multiplication, so RN=1[n], for instance, denotes the nth roots of unity of norm 1.

15 Part II Galois cohomology

4 Étale algebras and their Galois groups

4.1 Étale algebras If K is a field, an étale algebra over K is a finite-dimensional separable commutative algebra over K, or equivalently, a finite product of finite separable extension fields of K. A treatment of étale algebras is found in Milne ([35], chapter 8): here we summarize this theory and prove a few auxiliary results that will be of use. An étale algebra L of rank n admits exactly n maps ι1,...,ιn (of K-algebras) to a fixed separable closure K¯ of K. We call these the coordinates of L; the set of them will be called Coord(L/K) or simply Coord(L). Together, the coordinates define an embedding of L into K¯ n, which we call the Minkowski embedding because it subsumes as a special case the embedding of a degree-n number field into Cn, which plays a major role in algebraic number theory, as in Delone-Faddeev [19]. For any element γ of the absolute Galois group GK , the composition γ ιi with any coordinate is also a coordinate ι , so we get a homomorphism φ = φ : G Sym(Coord(L)))◦ such that j L K →

γ(ι(x)) = (φγ ι)(x) for all x L, ι Coord(L). This gives a functor from étale K-algebras to GK -sets (sets with a GK -action), which is∈ denoted∈ in Milne’s terminology. A functor going the other way, which Milne calls , takes F A φ : GK Sn to → L = (x ,...,x ) K¯ n γ(x )= x γ G , i (9) { 1 n ∈ | i φγ (i) ∀ ∈ K ∀ } Proposition 4.1 ([35], Theorem 7.29). The functors and establish a bijection between F A • étale extensions L/K of degree n, up to isomorphism, and • G -sets of size n up to isomorphism; that is to say, homomorphisms φ : G S , up to conjugation K K → n in Sn. Moreover, the bijection respects base change, in the following way:

Proposition 4.2. Let K1/K be a ﬁeld extension, not necessarily algebraic, and let L/K be an étale extension of degree n. Then L1 = L K K1 is étale over K1, and the associated Galois representations φL/K , φL1/K1 are related by the commutative⊗ diagram

¯ •|K GK1 / GK (10)

φL1/K1 φL/K ∼ Sym(CoordK1 (L1)) Sym(CoordK (L))

Proof. That L1/K1 is étale is standard (see Milne [35], Prop. 8.10). For the second claim, consider the natural restriction map r : CoordK1 (L1) CoordK (L). It is injective, since a K1 linear map out of L1 is determined by its values on L; and since both→ sets have the same size, r is surjective and is hence an isomorphism of GK1 -sets (the GK1 -structure on CoordK (L) arising by restriction from the GK -structure).

We will use this proposition most frequently in the case that K is a global ﬁeld and K1 = Kv one of its completions. The resulting L1 is then the product Lv = w v Lw of the completions of L at the ∼ | places dividing v. Note the departure from the classical habit of studying the completion Lw at each place Q individually. The preservation of degrees, [L1 : K1] = [L : K] will be important for our applications.

16 4.2 The Galois group of an étale algebra Define the Galois group G(L/K) of an étale algebra to be the image of its associated Galois representation φ : GK Sym(Coord(L)). It transitively permutes the coordinates corresponding to each field factor. For example,→ if L is a quartic field, then G(L/K) is one of the five (up to conjugacy) transitive subgroups of , S4 which (to use the traditional names) are 4, 4, 4, 4, and 4. Galois groups in this sense are used in the tables of cubic and quartic fields in Delone-FaddeevS A D V [19] andCthe Number Field Database [28]. Note that the Galois group G(L/K) is defined whether or not L is a Galois extension. If it is, then the Galois group is simply transitive and coincides with the Galois group in the sense of Galois theory. Important for us will be two notions pertaining to the Galois group.

Deﬁnition 4.3. Let G be a subgroup. A G-extension of K is a degree-n étale algebra L with a choice ⊆ Sn of subgroup G′ Sym(Coord(L)) that is conjugate to G and contains G(L/K), plus a conjugacy class of ⊆ isomorphisms G′ = G: the conjugacy being in G, not in . The added data is called a G-structure on L. ∼ Sn Proposition 4.4. G-extensions L/K up to isomorphism are in bijection with homomorphisms φ : GK G, up to conjugation in G. → Proof. Immediate from Proposition 4.1.

Example 4.5. L = Q(ζ5) is a 4-extension (taking 4 = (1234) 4), indeed its Galois group is isomorphic to ; and L admits two distinctC -structures, asC thereh are twoi ⊆ ways S to identify with its image in , C4 C4 C4 S4 which are conjugate in 4 but not in 4. Likewise, L = Q Q Q Q admits six 4-structures, one for each embedding of into S, as its GaloisC group is trivial. × × × C C4 S4 4.3 Resolvents This will be an important notion.

Deﬁnition 4.6. Let G , H be subgroups and ρ : G H be a homomorphism. Then for every ⊆ Sn ⊆ Sm → G-extension L/K, the corresponding φL : GK G may be composed with ρ to yield a map φR : GK H, which deﬁnes an étale extension R/K of degree→m. This R is called the resolvent of L under the map→ρ.

Example 4.7. Since there is a surjective map ρ4,3 : 4 3, every quartic étale algebra L/K has a cubic resolvent R. This resolvent appears in Bhargava [4],S but→ it S is much older than that. It is generated by a formal root of the resolvent cubic that appears when a general quartic equation is to be solved by radicals.

Example 4.8. Likewise, the sign map can be viewed as a homomorphism sgn : n 2, attaching to every étale algebra L a quadratic resolvent T . If L = K[θ]/f(θ) is generated by a polynomialS → Sf, and if char K =2, then it is not hard to see that T = K[√disc f] where disc f is the polynomial discriminant. Note that6 T still exists even if char K =2. We have that T ∼= K K is split if and only if the Galois group G(L/K) is contained in the alternating group . × An Example 4.9. The dihedral group has an outer automorphism, because rotating a square in the plane D4 by 45◦ does not preserve the square but does preserve every symmetry of the square. This map ρ : D4 → D4 associates to each -algebra L a new -algebra L′, not in general isomorphic. This is the classical D4 D4 phenomenon of the mirror ﬁeld. For instance, if L = Q[ 1+ √2], then p L′ = Q 1+ √2+ 1 √2 = Q 2+2√ 1 . − − q q q

Both L and L′ have the same Galois closure, a -octic extension of Q. Likewise, the outer automorphism D4 of permits the association to each sextic étale algebra L/K a mirror sextic étale algebra L′. S6 Example 4.10. The Cayley embedding is an embedding of any group G into Sym(G), acting by left multiplication. The Cayley embedding ρ : ֒ attaches to every étale algebra L of degree n an algebra L˜ of Sn → Sn! degree n! with an n-torsor structure. This is none other than the n-closure of L, constructed by Bhargava in a quite diﬀerentS way in [4, Section 2]. S

17 More generally, for any G Sn, the Cayley embedding G ֒ Sym(G) allows one to associate to each G-extension L a G-torsor T , which⊆ we may call the G-closure of→L. The name “closure” is justiﬁed by the following observation: if G Sn is a transitive subgroup, then, since any transitive G-set is a quotient of the simply transitive one, we⊆ can embed L into T by Proposition 4.11 below. More generally, G-closures of ring extensions, not necessarily étale or even reduced, have been constructed and studied by Biesel [10, 11]. If ρ : G H is invertible, as in many of the above examples, then the map from G-extensions to H- extensions is→ also invertible: we say that the two extensions are mutual resolvents.

4.4 Subextensions and automorphisms The Galois group holds the answers to various natural questions about an étale algebra. The next two propositions are given without proof, since they follow immediately from the functorial character of the correspondence in Proposition 4.1

Proposition 4.11. The subextensions L′ L of an étale extension L/K, correspond to the equivalence relations on Coord(L) stable under permutation⊆ by G(L/K), under the bijection ∼

L′ = x L : ι(x)= ι′(x) whenever ι ι′ . ∼ 7→ { ∈ ∼ }

Remark 4.12. Note that if L is a Galois ﬁeld extension, the image of φL is a simply transitive subgroup , and identifying Coord(L) with , the stable equivalence relations are just right congruences modulo subgroups− of : so we recover the Galois− correspondence between subgroups and subﬁelds. − The Galois group is not a group of automorphisms of L. However, the automorphisms of L as a K-algebra can be described in terms of the Galois group readily.

Proposition 4.13. Let L be Minkowski-embedded by its coordinates ι1,...,ιn. Then the automorphism group Aut(L/K) is given by permutations of coordinates,

1 1 τπ(x1; ... ; xn)= xπ− (1); ... ; xπ− (n), for π in the centralizer C(Sn, G(L/K)) of the Galois group. (For H G groups, the centralizer C(G, H) of H in G is the subgroup of elements of G that commute with every⊆ element of H.) This provides a characterization, in terms of the Galois group, of rings having various kinds of automorphisms.

• Since S2 is abelian, any étale algebra L of rank 2 has a unique non-identity automorphism, the conjugation x¯ = tr x x. − • If L has rank 4, automorphisms τ of L of order 2 whose fixed algebra is of rank 2 are in bijection with D4-structures on L. Indeed, the conditions force τ to correspond to the permutation π = (12)(34) or one of its conjugates, and the centralizer of this permutation is D4. • Particularly relevant is the case that L has a complete set of automorphisms that permute the coordinates simply transitively: this is a generalization of a Galois field extension called a torsor. This case is sufficiently important to merit its own subsection.

4.5 Torsors Deﬁnition 4.14. Let G be a ﬁnite group. A G-torsor over K is an étale algebra L over K equipped with an action of G by automorphisms τg g G that permute the coordinates simply transitively, that is, such that L K¯ is isomorphic to { } ∈ ⊗K K¯ g G M∈ with G acting by right multiplication on the indices.

18 Proposition 4.15. Let G be a group of order n. An étale algebra L is a G-torsor if and only if it is a G-extension, where G is embedded into Sn by the Cayley embedding (G acting on itself by left multiplication). Moreover, there is a bijection between • G-torsor structures on L, up to conjugation in G, and • G-structures on L.

The bijection is given in the following way: there is a labeling ιg of the coordinates of L with the elements of G such that the Galois action is by left multiplication { }

g(ιh(x)) = ιφg h(x) (11) while the torsor action is by right multiplication

1 ιg(τh(x)) = ιgh− (x). (12) Proof. We first claim that the only elements of Sym(G) commuting with all right multiplications are left multiplications, and vice versa. If π : G G is a permutation commuting with left multiplications, then → π(g)= π(g id )= g π(id ), · G · G so π is a right multiplication. So the embedded images of G in Sym(G) given by left and right multiplication 1 (which are conjugate under the inversion permutation − Sym(G)) are centralizers of one another. It is • ∈ then clear that conjugates G′ of G in Sym(Coord(L)) that contain G(L/K) are in bijection with conjugates G′′ that commute with G(L/K). This establishes the first assertion. For the bijection of structures, if an embedding G = G′ Sym(Coord(L)) is given, then we can label the coordinates with elements of G so that ∼ ⊆ G acts on them by multiplication; then G′′ gets identified with G by the corresponding right action. The only ambiguity is in which embedding is labeled with the identity element; if this is changed, one computes that the resulting identification of G′′ with G is merely conjugated, so the map is well defined. The reverse map is constructed in exactly the same way. Here is another perspective on torsors.

Proposition 4.16. G-torsors over a field K, up to isomorphism, are determined by their field factor, a .Galois extension L /K equipped with an embedding Gal(L/K) ֒ G up to conjugation in G 1 → Proof. If T is a G-torsor, then since G permutes the coordinates simply transitively, all the coordinates have the same image; that is, the field factors of G are all isomorphic to a Galois extension L/K. The torsor operations fixing one field factor Li of T realize the Galois group Gal(L/K) as a subgroup of G; changing the field factor Li and/or the identification Li ∼= L corresponds to conjugating the map Gal(L/K) ֒ G by an element of G. → Conversely, suppose L and an embedding

Gal(L/K) ∼ H G −→ ⊆ are given. Let 1 = g1,...,gr be coset representatives for G/H. Then g2,...,gr must map any field factor L1 ∼= L isomorphically onto the remaining field factors L2,...,Lr, each Li occurring once. To finish specifying the G-action on T L L , it suffices to determine g for each g G. Factor gg = g h for some ∼= 1 r Li i j j 1,...,r , h H. Then×···× for each x L , g(g (x)) = g (h| (x)), and the∈ value of this is known because ∈ { } ∈ ∈ 1 i j the H-action on L1 is known. It is easy to see that we get one and only one consistent G-torsor action in this way. Because all field factors of a torsor are isomorphic, we will sometimes speak of “the” field factor of a torsor.

19 4.5.1 Torsors over étale algebras On occasion, we will speak of a G-torsor over L, where L is itself a product K K of ﬁelds. By this 1 ×···× r we simply mean a product T1 Tr where each Ti is a G-torsor over Ki. This case is without conceptual diﬃculty, and some theorems×···× on torsors will be found to extend readily to it, such as the following variant of the fundamental theorem of Galois theory:

Theorem 4.17. Let T be a G-torsor over an étale algebra L. For each subgroup H G, ⊆ (a) The ﬁxed algebra T H is uniformly of degree [G : H] over L (that is, of this same degree over each ﬁeld factor of L); (b) T is an H-torsor over T H , under the same action; (c) If H is normal, then T H is also a G/H-torsor over L, under the natural action. Proof. Adapt the relevant results from Galois theory.

4.6 A fresh look at Galois cohomology Galois cohomology is one of the basic tools in the development of class field theory. It is usually presented in a highly abstract fashion, but certain Galois cohomology groups, specifically H1(K,M) for finite M, have explicit meaning in terms of field extensions of M. It seems that this interpretation is well known but has not yet been written down fully, a gap that we fill in here. We begin by describing Galois modules.

Proposition 4.18 (a description of Galois modules). Let M be a finite abelian group, and let K be a field. Let M − denote the subset of elements of M of maximal order m, the exponent of M. The following objects are in bijection: (a) Galois module structures on M over K, that is, continuous homomorphisms φ : G Aut M; K → (b) (Aut M)-torsors T/K; ;c) (Aut M)-extensions L /K, where Aut M ֒ Sym M in the natural way) 0 → .d) (Aut M)-extensions L−/K, where Aut M ֒ Sym M − in the natural way) → Proof. For item (d) to make sense, we need that M − generates M; this follows easily from the classification of finite abelian groups. The bijections are immediate from Propositions 4.4 and 4.15.

We will denote M with its Galois-module structure coming from these bijections by Mφ, MT , or ML0 . Note that T , L0, and L− are mutual resolvents. Example 4.19. For example (and we will return to this case frequently), if we let M = be the smallest C3 group with nontrivial automorphism group: Aut M = . Then the Galois module structures on M are in ∼ C2 natural bijection with 2-torsors over K, that is, quadratic étale extensions T/K. If char K =2, these can C 2 6 be parametrized by Kummer theory as T = K[√D], D K×/ (K×) . The value D =1 corresponds to the split algebra T = K K and to the module M with trivial∈ action. We have an isomorphism × M = 0, √D, √D T ∼ { − } of GK -sets, and of Galois modules if the right-hand side is given the appropriate group structure with 0 as identity. 2 In particular, the Galois-module structures on form a group Hom(G , ) = K×/ (K×) : the group C3 K C2 ∼ operation can also be viewed as tensor product of one-dimensional F3-vector spaces with Galois action.

20 4.6.1 Galois cohomology Note that the zeroth cohomology group H0(K,M) has a ready parametrization:

0 Proposition 4.20. Let M = ML0 be a Galois module. The elements of H (K,M) are in bijection with the degree-1 ﬁeld factors of L0.

Proof. Proposition 4.18 establishes an isomorphism of GK -sets between the coordinates of L0 and the points of M. A degree-1 field factor corresponds to an orbit of GK on Coord(L0) of size 1, which corresponds exactly to a fixed point of GK on M. Deeper and more useful is a description of H1. For an abelian group M, let (M) = M ⋊ Aut M be the semidirect product under the natural action of Aut M on M. We can describeGA (M) more explicitly as the group of affine-linear transformations of M; that is, maps GA

a (x)= gx + t, g Aut M,t M g,t ∈ ∈ composed of an automorphism and a translation, the group operation being composition. In particular, we have an embedding .(M) ֒ Sym(M) GA → 1 Proposition 4.21 (a description of H ). Let M = Mφ = ML0 be a Galois module.

1 (a) Z (K,M) is in natural bijection with the set of continuous homomorphisms ψ : GK (M) such that the following triangle commutes: → GA

ψ GK / (M) (13) ❍❍ GA ❍❍ ❍❍ φ ❍❍ ❍$ Aut M

1 (b) H (K,M) is in natural bijection with the set of such ψ : GK (M) up to conjugation by M (M). → GA ⊆ GA (c) H1(K,M) is also in natural bijection with the set of (M)-extensions L/K (with respect to the embedding (M) ֒ Sym(M)) equipped with an isomorphismGA from their resolvent (Aut M)-torsor to T . GA → Proof. By the standard construction of group cohomology, Z1 is the group of continuous crossed homomorphisms Z1(K,M)= σ : G M σ(γδ)= σ(γ)+ φ(γ)σ(δ) . { K → | } Send each σ to the map

ψ : G (M) K → GA γ a . 7→ φ(γ),σ(γ) It is easy to see that the conditions for ψ to be a homomorphism are exactly those for σ to be a crossed homomorphism, establishing (a). For (b), we observe that adding a coboundary σ (γ) = γ(a) a to a a − crossed homomorphism σ is equivalent to post-conjugating the associated map ψ : GK (M) by a. As to (c), a (M)-extension carries the same information as a map ψ up to conjugation→ GA by the whole of (M). SpecifyingGA the isomorphism from the resolvent (Aut M)-torsor to T means that the map π ψ = GA ◦ φ : GK Aut(M) is known exactly, not just up to conjugation. Hence ψ is known up to conjugation by M. →

Remark 4.22. The zero cohomology class corresponds to the extension L0, with its structure given by the embedding Aut M ֒ (M). This can be seen to be the unique cohomology class whose corresponding (X)-extension has→ a GA ﬁeld factor of degree 1. GA

21 If K is a local field, a cohomology class α H1(K,M) is called unramified if it is represented by a cocycle α : Gal(K/K¯ ) M that factors through∈ the unramified Galois group Gal(Kur/K). The subgroup → 1 of unramified coclasses is denoted by Hur(K,M). If M itself is unramified (and we will never have to think about unramified cohomology in any other case), this is equivalent to the associated étale algebra L being unramified. 1 If X = MT is a Galois module and σ Z (K,M) is the Galois module corresponding to a (X)- extension L/K, we can also take the (X∈)-closure of L, a (X)-torsor E which fits into the followingGA diagram: GA GA E ❄ ⑧⑧ ❄❄ ⑧⑧ ❄❄ ⑧⑧ ❄ L T (14) ❄❄ ⑧ ❄❄ ⑧⑧ ❄ ⑧⑧ K

Because of the semidirect product structure of (X), we have E ∼= L K T . It is also worth tabulating the permutation representations of ﬁnite groups thatGA yield each of the étale⊗ algebras discussed here:

0 / M / (M ) / Aut(M) / 0 (15) ♦GAJj _ _ yields L♦♦♦ ♦♦ yields E yields T ♦♦♦ w♦♦ Sym(M) Sym( (M)) Sym(Aut(M)) GA 4.6.2 The Tate dual If M is a Galois module and the exponent m of M is not divisible by char K, then

M ′ = Hom(M,µm) is also a Galois module, called the Tate dual of M. The modules M and M ′ have the same order and are isomorphic as abstract groups, though not canonically; as Galois modules, they are frequently not isomorphic at all.

Example 4.23. If M = MK[√D] is one of the order-3 modules studied in Example 4.19, then the relevant µm is µ3 = MK[√ 3]. ∼ − Examining the Galois actions (here it helps to use the theory of GK -sets of size 2 presented in Knus and Tignol [30]), we see that

M ′ = MK[√ 3D]. − This explains the D 3D pattern in the Scholz reﬂection theorem and its generalizations, including cubic Ohno-Nakagawa. 7→ −

Example 4.24. A module M of underlying group 2 2 is always self-dual, regardless of what Galois- module structure is placed on it. This can be provedC by×C noting that M has a unique alternating bilinear form

B : M M µ × → 2 1, x =0,y =0, or x = y (x, y) 7→ 1, otherwise. ( −

Being unique, it is Galois-stable and induces an isomorphism M ′ ∼= M. Particularly notable for us are the cases when (M) is the full symmetric group Sym(M), for then every étale algebra L/K of degree M has a (unique)GA (M)-aﬃne structure. It is easy to see that there are only four such cases: | | GA

22 • M = 1 , (M) = { } GA ∼ S1 • M = Z/2Z, (M) = GA ∼ S2 • M = Z/3Z, (M) = ⋊ = GA ∼ C3 C2 ∼ S3 • M = Z/2Z Z/2Z, (M) = ( ) ⋊ = . × GA ∼ C2 ×C2 S3 ∼ S4 For degree exceeding 4, not every étale algebra arises from Galois cohomology, a restriction that plays out in the existing literature on reflection theorems. For instance, Cohen, Rubinstein-Salzedo, and Thorne [12] prove a reflection theorem in which one side counts p-dihedral fields of prime degree p 3. From our perspective, these correspond to cohomology classesD of an M = whose Galois action≥ is by 1. The Cp ± Tate dual of such an M can have Galois action by the full (Z/pZ)×, and indeed they count extensions of Galois group ( p) on the other side of the reflection theorem. This will appear inevitable in light of the motivations elucidatedGA C in Part III.

5 Extensions of Kummer theory to explicitize Galois cohomology

Now that Galois cohomology groups H1(K,M) have been parametrized by étale algebras, can invoke parametrizations of étale algebras by even more explicit objects. The most familiar instance of this is Kummer theory, an isomorphism 1 m H (K,µm) ∼= K×/(K×) coming from the long exact sequence associated to the Kummer sequence

m 0 µ K¯ × • K¯ × 0. −→ m −→ −→ −→ 1 m In favorable cases, the cohomology H (K,M) of other Galois modules M can be embedded into R×/(R×) for some ﬁnite extension R of K. We ﬁrst state the hypothesis we need:

Definition 5.1. Let M be a finite Galois module of exponent m over a field K, and let X be a Galois-stable generating set of M. We say that M equipped with X is a good module if the natural map of Galois modules

X = (Z/mZ) M → x X M∈ e x x 7→ is split, that is, its kernel admits a Galois-stable complementary direct summand M˜ . Such a direct summand is known as a good structure on M. Proposition 5.2. The following examples of a Galois module M with generating set X are good:

(a) M = , with any action, and X = M 0 . ∼ Cp \{ } (b) M = n , with any action preserving a basis X. ∼ Cm n 1 (c) M ∼= m− , n 2 with gcd(m,n), with an action that preserves a hyperbasis X, that is, a generating set ofCn elements≥ with sum 0.

Proof. (a) Here the Galois modules are representations of Fp× ∼= p 1 over Fp. Since the group and ﬁeld are of coprime order, complete reducibility holds: any subrepresentationC − is a direct summand. In fact, X is the regular representation, M is the tautological representation in which each λ F× acts by ∈ p multiplication by λ, and M˜ can be taken (uniquely in general) to be the product of all the other isotypical components of X.

(b) Here the natural map X M is an isomorphism, so M˜ = X. →

23 (c) Here the natural map X M is the quotient by the one-dimensional space →

ex . *x X + X∈ This space has a Galois-stable direct complement, namely the kernel M˜ of the linear functional

X F → 2 e 1. x → Proposition 5.3. Let M be a Galois module with a good structure (X, M˜ ), and let R be the resolvent algebra corresponding to the GK -set X. For any Galois module A with underlying group Z/mZ, there is a natural injection H1(K,M A) H1(R, A) ⊗ → as a direct summand. The cokernel is naturally isomorphic to

H1(K, (X/M˜ ) A). ⊗ Proof. We use the good structure M = M˜ ֒ X ∼ → to embed .(H1(K, M˜ A) ֒ H1(X A ⊗ → ⊗ Since M˜ is a direct summand, this is an injection with cokernel naturally isomorphic to H1(K, (X/M˜ ) A). It remains to construct an isomorphism ⊗

H1(X A) ∼ H1(R, A). ⊗ −→ If R decomposes as a product R = R R ∼ 1 ×···× s of ﬁeld factors corresponding to the orbits X = i Xi of GK on X, then X has a corresponding decomposition F s X = Xi i=1 M

Ri where Xi = ex : x Xi is none other than the induced module IndK Z/mZ. Its cohomology is computed by Shapiro’sh lemma:∈ i

s s s H1(K, X A) = H1(K, X A)= H1(K, IndRi A) = H1(R , A)= H1(R, A). ⊗ ∼ i ⊗ K ∼ i i=1 i=1 i=1 M M M This is the desired isomorphism. We can harness Kummer theory to parametrize cohomology of other modules as follows.

Theorem 5.4 (an extension of Kummer theory). Let M be a ﬁnite Galois module, and assume that m = exp M is not divisible by char K. Let GK act on the set M ′− of surjective characters χ : M ։ µm through its actions on M and µm, and let F be the étale algebra corresponding to this GK -set. (a) There is a natural group homomorphism

1 m Kum : H (K,M) F ×/(F ×) . →

24 (b) If M = is cyclic of prime order, then Kum is injective, F is naturally a (Z/pZ)×-torsor, and ∼ Cp p c im(Kum) = α F ×/(F ×) : τ (α)= α c (Z/pZ)× . ∈ c ∀ ∈ If p =3, then the image simpliﬁes to

3 im(Kum) = α F ×/(F ×) : N (α)=1 , ∈ F/K 3 and the ( 3) ∼= S3-extension L corresponding to a given α F ×/(F ×) of norm 1 can be described as follows:GA DeﬁneC a K-linear map ∈

κ : K K¯ 3 → √3 ξ trK¯ 2/K ξω δ , 7→ ω where √3 δ K¯ 2 is chosen to have norm 1, and ω ranges through the set ∈ (1; 1), (ζ ; ζ2); (ζ2; ζ ) { 3 3 3 3 } of cube roots of 1 in K¯ 2 of norm 1. Then

L = K + κ(F ).

(c) If M = , then Kum is injective and ∼ C2 ×C2 2 im(Kum) = α F ×/(F ×) : N (α)=1 . ∈ F/K 2 Moreover, the (M) ∼= S4-extension corresponding to a given α F ×/(F ×) of norm 1 can be described as follows:GA Deﬁne a K-linear map ∈

κ : K K¯ 4 → √ ξ trK¯ 3/K ξω δ , 7→ ω where √δ K¯ 3 is chosen to have norm 1, and ω ranges through the set ∈ (1;1;1), (1; 1; 1); ( 1; 1; 1); ( 1; 1; 1) { − − − − − − } of square roots of 1 in K¯ 3 of norm 1. Then

L = K + κ(F ).

Proof. If χ : M ։ µm is a surjective character, let Fχ be the fixed field of the stabilizer of χ; thus Fχ is the field factor of F corresponding to the GK -orbit of χ. If χ1,...,χℓ are orbit representatives, we can map

res χi 1 1 ∗ 1 m m H (K,M) H (Fχi ,M) H (Fχi ,µm) = Fχ× /(Fχ× ) = F ×/(F ×) . Q−→ Q−→ ∼ i i ∼ i i i Y Y Y This yields our map Kum. Alternatively, note that by Shapiro’s lemma,

1 1 K 1 H (Fχi ,µm) = H (K, Ind µm) = H (K, I), ∼ Fχi ∼ i i Y Y where K IM = IndF µm = µm, ։ χ:MMµm a Galois module under the action

g (aχ)i = g(cg 1(χ))χ = g(cχ(g ))χ . − • 25 Under this identiﬁcation, it is not hard to check that Kum = j , where j is the inclusion M ֒ IM given by ∗ → a (χ(a)) . 7→ χ Although j is injective (because the characters of maximal order m generate the group of all characters), it is not obvious whether j induces an injection on cohomology, nor what the image is. What makes the modules M in parts (b) and (c) tractable is that, in these cases, M 0 is a good generating set for M, so \{ } M is a direct summand of IM . In part (b), we can identify

(Z/pZ)× IM = Ind 1 Fp Fp µm ∼ { } ⊗ as a twist of the regular representation of (Z/pZ)× over Fp. Since Fp has a complete set of (p 1)st roots of unity, this representation splits completely into one-dimensional subrepresentations. The image− of j is p the eigenspace generated by (c)c Z/pZ , so Kum is injective and its image is the subspace of F ×/(F ×) cut ∈ × out by the same relations τc(x)= cx (where τc is the torsor operation on F , resp. the automorphism of IM , indexed by c) that cut out j(M) in IM . As to part (c), since Z/2Z Z/2Z has three surjective characters whose product is 1, we have I /M = µ × M ∼ 2 with the map η : IM µ2 given by multiplying the coordinates. Since µ2 also injects diagonally into IM , we easily get a direct sum→ decomposition, which shows that Kum is injective. As to the image, it is not hard to show that the diagram 2 1 1 F ×/(F ×) ∼ / H (F, ) ∼ / H (K, I) C2 qqq N cor qq qqqη xqq ∗ 2 1 K×/(K×) ∼ / H (K, C2) commutes, establishing the desired norm characterization of im(Kum). The formulas by radicals for the cubic and quartic algebras corresponding to a Kummer element follow easily by chasing through the Galois actions on the appropriate étale algebras. The quartic case is also considered by Knus and Tignol, where a closely related description of L is given ([30], Proposition 5.13).

1 m Remark 5.5. For general M, the map H (K,M) F ×/(F ×) may be made by the construction in Theorem → 5.4, but its image is hard to characterize, and it may not even be injective: for instance, when M = , ∼ C4 coclasses correspond to 4-extensions, and Kum conﬂates each extension with its mirror extension (compare Example 4.9). D Though it will not be used in the sequel, it is worth noting that Artin-Schreyer theory is amenable to the same treatment.

Theorem 5.6. Let Let M be a ﬁnite Galois module with underlying abelian group A of exponent m = p = char K. (a) There is a natural map AS : H1(K,M) F/℘(F ). →

(b) If A = , then AS is injective, F is naturally a (Z/pZ)×-torsor, and ∼ Cp

im(AS) = α F/℘(F ): τ (α)= cα c (Z/pZ)× . ∈ c ∀ ∈ (c) If p =2 and A = , then AS is injective and ∼ C2 ×C2 im(AS) = α F/℘(F ):tr (α)=0 . ∈ F/K

26 5.1 The Tate pairing and the Hilbert symbol Assume now that K is a local field. Our next step will be to understand the (local) Tate pairing, which is given by a cup product 1 1 2 , : H (K,M) H (K,M ′) H (K,µ ) = µ . h iT × → m ∼ m As we were able to parametrize the cohomology groups H1(K,M) in favorable cases, it should not come as a surprise that we can often describe the Tate pairing with similar explicitness. Recall the definitions of the Artin and Hilbert symbols. If M ∼= Z/mZ has trivial GK -action, then M ′ ∼= µm, and we have a Tate pairing , : H1(K, Z/mZ) H1(K,µ ) µ h iT × m → m 1 1 Now H (K, Z/mZ) = Hom(K, Z/mZ) parametrizes Z/mZ-torsors, while by Kummer theory, H (K,µm) = m ∼ ∼ K×/(K×) . The Tate pairing in this case is none other than the Artin symbol (or norm-residue symbol) φ (x) which attaches to a cyclic extension L, of degree dividing m, a mapping φ : K× Gal(L/K) µ L L → → m whose kernel is the norm group NL/K(L×) (see Neukirch [40], Prop. 7.2.13). If, in addition, µm K, then 1 m ⊆ H (K, Z/mZ) is also isomorphic to K×/ (K×) , and the Tate pairing is an alternating pairing m m , : K×/ K× K×/ K× µ h i × → m classically called the Hilbert symbol (or Hilbert pairing ). It is defined in terms of the Artin symbol by

a,b = φ m (a). (16) h i K[ √b] In particular, a,b = 1 if and only if a is the norm of an element of K[ m√b]. This can also be described in terms of theh splittingi of an appropriate Severi-Brauer variety; for instance, if m = 2, we have a,b = 1 exactly when the conic h i ax2 + by2 = z2 has a K-rational point. See also Serre ([51], §§XIV.1–2). (All identiﬁcations between pairings here are up to sign; the signs are not consistent in the literature and are totally irrelevant for this paper.) Pleasantly, for the types of M featured in Theorem 5.4, the Tate pairing can be expressed simply in terms of the Hilbert pairing. We extend the Hilbert pairing to étale algebras in the obvious way: if L = K K , then 1 ×···× s

(a1; ... ; as), (b1; ... ; bs) := a1,b1 as,bs . h iL h iK1 ·····h iKs m√ Note that if a is a norm from L[ b] to L, then a,b L =1, but the converse no longer holds. We then have the following: h i Theorem 5.7 (a formula for the local Tate pairing). Let K be a local ﬁeld. For M, F as in Theorem 5.4, let M ′ be the Tate dual of M, and let F ′ be the corresponding étale algebra, corresponding to the GK -set M − of elements of maximal order in M, just as F corresponds to M ′−. The Tate pairing

1 1 2 , : H (K,M) H (K,M ′) H (K,µ ) = h• •i × → m ∼ Cm can be described in terms of the Hilbert pairing in the following cases:

(a) If A = , then both F and F ′ embed naturally into F ′′ := F [µ ], and the Tate pairing is the restriction ∼ Cp p of the Hilbert pairing on F ′′.

2 (b) If A ∼= (Z/2Z) , then we have natural isomorphisms M ∼= M ′, F ∼= F ′, and the Tate pairing is the restriction of the Hilbert pairing on F .

Proof. In case (b), set F ′′ = F = F [µ2]. We will do the two cases largely in parallel. Let Surj(A, B) Hom(A, B) denote the set of surjections between two groups A, B. Note that if A, B ⊆ are Galois modules, then Surj(A, B) is a GK -set. Note that F ′′ is the étale algebra corresponding to the GK -set Z = Surj(M,µ ) Surj(µ , Z/mZ). p × p

27 There is an obvious map Z Surj(M,µ ) given by projection to the ﬁrst factor, which allows us to recover → p the identiﬁcation F ′′ = F [µm]. There is also a map of GK -sets

Ψ: Z Surj(M ′,µ ) → p which sends a pair (χ,u) (where χ : M ։ µ , u : µ ) to the unique surjective ψ : M ′ µ satisfying m m → p 1 ψ(χ)= u− (1), A = ∼ Cp ψ(χ)=1, A = . ( ∼ C2 ×C2

This allows us to embed F ′ into F ′′. It is worth noting that when A = , u carries no information and ∼ C2 ×C2 F ∼= F ′ ∼= F ′′. Let F1′′,...,Fℓ′′ be the ﬁeld factors of F ′′; each Fi′′ corresponds to an orbit GK (χi,ui) on Z. Let ψi = 1 1 Ψ(χ ,u ). Then for σ H (K,M), τ H (K,M ′), i i ∈ ∈ σ, τ = Kum(σ), Kum(τ) h iHilb h iHilb; F ′′ K K = invF ′′ χi resF σ ψi resF τ . i ∗ i′′ ∪ ∗ i′′ F Yi′′ K Since invF ′′ = invK corF (a standard fact), we have i ◦ i′′

K K K σ, τ Hilb = invK corF χi resF σ ψi resF τ h i i′′ ∗ i′′ ∪ ∗ i′′ F Xi′′ K K = invK corF ev (χi ψi) resF (σ τ) i′′ ◦ ⊗ i′′ ∪ F ∗ Xi′′ where ev : M M ′ µm is the evaluation map. We now apply the following lemma, which slightly generalizes results seen in⊗ the→ literature. Lemma 5.8. Let H G be a subgroup of ﬁnite index. Let X and Y be G-modules, and let f : X Y be a map that is H-linear⊆ (but not necessarily G-linear). Denote by f˜ the G-linear map →

1 f˜(x)= gfg− (x). gH G/H X∈ Let σ Hn(H, Y ). Then ∈ G G ˜ corH (f resH σ)= f σ. ∗ ∗ Proof. Since we are concerned with the equality of a pair of δ-functors, we can apply dimension shifting to assume that n =0. The proof is now straightforward.

Applying with G = GK , H = GF , and f = ev (χi ψi): M M ′ µm, we get i′′ ◦ ⊗ ⊗ →

1 1 σ, τ = inv g ev (χ g− ψ g− ) h iHilb K ◦ ◦ i ◦ ⊗ i ◦ F ′′ g GK /GF Xi ∈ X i′′ = inv ev (χ ψ ) (σ τ), K ◦ i,g ⊗ i,g ∪ F ′′ g GK /GF ∗ Xi ∈ X i′′

where χi,g = g(χi) and ψi,g = g(ψi) are given by the natural action. Now the outer sum runs over all GK -orbits of Z while the inner sum runs over the elements of each orbit, so we simply get

σ, τ = inv ev (χ Ψ(χ,u)) (σ τ). h iHilb K ◦ ⊗ ∪ (χ,u) Z ∗ X∈

28 Since the Tate pairing is given by σ, τ Tate = invK ev (σ τ), h i ∗ ∪ it remains to check that ev (χ Ψ(χ,u)) = ev ◦ ⊗ (χ,u) Z X∈ 2 as maps from M M ′ to µ . In the case M = , each term is actually equal to ev, and there are (p 1) 1 ⊗ m ∼ Cp − ≡ mod p terms. In the case M = , a direct veriﬁcation on a basis of M M ′ is not diﬃcult. ∼ C2 ×C2 ⊗ 6 Rings over a Dedekind domain

Thus far, we have been considering étale algebras L over a field K. We now suppose that K is the fraction field of a Dedekind domain K (not of characteristic 2), which for us will usually be a number field or a completion thereof, althoughO there is no need to be so restrictive. Our topic of study will be the subrings of L that are lattices of full rank over —the orders, to use the standard but unfortunately overloaded word. OK There is always a unique maximal order L, the integral closure of K in L. If L = L1 Lr is a product of field factors, we have = O . O ×···× OL OL1 ×···×OLr 6.1 Indices of lattices There is one piece of notation that we explain here to avoid confusion. If V is an n-dimensional vector space over K and A, B V are two full-rank lattices, we denote by the index [A : B] the unique fractional ideal c such that ⊆ ΛnA = cΛnB n as K -submodules of the top exterior power Λ V . Alternatively, if A B, then the classification theorem forO finitely generated modules over lets us write ⊇ OK A/B = /c /c , ∼ OK 1 ⊕···⊕OK r and the index equals c c c . 1 2 ··· r The index satisfies the following basic properties:

• [A : B][B : C] = [A : C]; • If V is a vector space over both K and a ﬁnite extension L, and A and B are two -sublattices, then OL [A : B]K = NL/K[A : B]L;

• If V = L is a K-algebra and α L×, then [A : αA]= N (α). ∈ L/K Despite the apparent abstractness of its deﬁnition, the index [A : B] is not hard to compute in particular cases: localizing at a prime ideal, we can assume K is a PID, and then it is the determinant of the matrix expressing any basis of B in terms of a basis of AO. If L is a K-algebra, L is an order, and a L is a fractional ideal, the index [ : a] is called the norm of a and will be denotedO ⊆ by N (a) or, when⊆ the context is clear, by N(a). NoteO the following basic properties: O

• If a = α is principal, then N (a)= NL/K(α). O O • If a and b are two -ideals and a is invertible, then N(ab)= N(a)N(b). This is easily derived from the theorem that an invertibleO ideal is locally principal (Lemma 14.2). It is false for two arbitrary -ideals. O • If K = Z or Zp, then for any L and , the norm N (a) of an integral ideal is the ideal generated by theO absolute norm /a . O O |O |

29 6.2 Discriminants As is standard, we define the discriminant ideal of an order in an étale algebra L to be the ideal d generated by the trace pairing O 2n τ : K O → O (17) (ξ , ξ ,...,ξ , η , η ,...,η ) det[tr ξ η ]n . 1 2 n 1 2 n 7→ i j i,j=1 The trace pairing is nondegenerate, that is, d = 0 (this is one equivalent definition of étale). The primes dividing d are those at which L is ramified and/or6 is nonmaximal. This notion is standard and widely used. O However, it does not quite extend the (also standard) notion of the discriminant of a Z-algebra over Z, which has a distinction between positive and negative discriminants. The Ohno-Nakagawa theorem involves this distinction prominently; Dioses [20] and Cohen–Rubinstein-Salzedo–Thorne [12] each frame their extensions of O-N in terms of an ad-hoc notion of discriminant that incorporates the splitting data of an order at the infinite primes. Here we explain the variant that we will use. Since the trace pairing τ is alternating in the ξ’s and also in the η’s, it can be viewed as a bilinear form n n on the rank-1 lattice Λ . Identifying Λ with a (fractional) ideal c of K (whose class is often called the Steinitz class of ), weO can write O O O τ(ξ)= Dξ2 2 n for some nonzero D c− . Had we rescaled the identification Λ c by λ K×, D would be multiplied 2 ∈ O →2 ∈ by λ . We call the pair (c,D), up to the equivalence (c,D) (λc, λ− D), the discriminant of and denote it by Disc . ∼ O ThereO is another perspective on the discriminant Disc . Let L˜ be the S -torsor corresponding to L, O n which comes with n embeddings κ1,...,κn : L L˜ freely permuted by the Sn-action (not to be confused with the n coordinates of L). Noting that, for any→ α L, ∈

tr(α)= κi(α), i X we can factor the trace pairing matrix: [tr ξ η ] = [κ (ξ )] [κ (η )] . i j i,j h i i,h · h j h,j Define τ0(ξ1,...,ξn) = det[κh(ξi)]i,h, so that τ(ξ ,...,ξ , η ,...,η )= τ (ξ ,...,ξ ) τ (η ,...,η ). 1 n 1 n 0 1 n · 0 1 n Now look more carefully at the map τ0. First, τ0 is alternating under permutations of the ξi’s, so it defines a linear map τ : c L.˜ 0 → Moreover, τ0 is alternating under postcomposition by the torsor action of Sn on L˜, which permutes the κh An freely. Thus the image of τ0 lies in the C2-torsor T2 = L˜ , which we call the discriminant torsor of L, and even more specifically in the ( 1)-eigenspace of the nontrivial element of C . By (the simplest case − 2 of) Kummer theory, we may write T2 = K[√D′]. If (ξ1,...,ξn) is any K-basis of L, so that ξ1 ξn corresponds to some nonzero element c c, then O ∧···∧ ∈ 2 2 2 2 Dc = τ(c,c)= τ0(c) = a√D′ = D′a . Thus T2 = K[√D]. We summarize this result in a proposition. Proposition 6.1. If L is an étale algebra over K of discriminant (c,D), then K[√D] is the discriminant torsor of L; that is, the diagram of Galois structure maps

φL GK / Sn ❈❈ ❈❈ ❈❈ sgn φK[√D] ❈❈ ! S2

30 commutes. There is notable integral structure on D as well. Lemma 6.2 (Stickelberger’s theorem over Dedekind domains). If (c,D) is the discriminant of an 2 2 1 order , then D t mod 4c− for some t c− . O ≡ ∈ Remark 6.3. When K = Z, Lemma 6.2 states that the discriminant of an order is congruent to 0 or 1 mod 4: a nontrivial and classicalO theorem due to Stickelberger. Our proof is a generalization of the most familiar one for Stickelberger’s theorem, due to Schur [50].

2 Proof. Since D c− , the conclusion can be checked locally at each prime dividing 2 in . We can thus ∈ OK assume that K is a DVR and in particular that c = (1). Now there is a simple tensor ξ1 ξn that corresponds toO the element 1 c. By deﬁnition, ∧···∧ ∈ √D = τ (ξ ,...,ξ ) = det[κ (ξ )] = sgn(σ) κ (ξ ) = ρ ρ,¯ (18) 0 1 n h i i,h π(i) i − π S i X∈ n Y where ρ = κπ(i)(ξi) π A i X∈ n Y lies in T by symmetry and ρ¯ is its conjugate. By construction, ρ is integral over , that is to say, ρ +ρ ¯ 2 OK and ρρ¯ lie in K . Now O D = (ρ ρ¯)2 = (ρ +ρ ¯)2 4ρρ¯ − − is the sum of a square and a multiple of 4 in . OK Remark 6.4. One can write (18) in the suggestive form

θ (1) θ (1) det[κ (ξ )] = det 1 2 h i i,h θ (ρ) θ (ρ) 1 2 where θ1,θ2 are the two automorphisms of T2. This equates discriminants of orders in L with those of orders in T2. Equalities of determinants of this sort reappear in Bhargava’s parametrizations of quartic and quintic rings and appear to be a common feature of many types of resolvent ﬁelds. We can now state the notion of discriminant as we would like to use it.

2 2 Definition 6.5. A discriminant over K is an equivalence class of pairs (c,D), with D c− and D t 2 1 O ∈ ≡ mod 4c for some t c− , up to the equivalence relation ∈ 2 (c,D) (λc, λ− D). ∼ If is an étale order, the discriminant Disc is defined as follows: Pick any representation φ :Λn c of theO Steinitz class as an ideal class; then DiscO is the unique pair (c,D) such that O → O det[tr ξ η ]n = D φ(ξ ξ ) φ(η η ). i j i,j=1 · 1 ∧···∧ n · 1 ∧···∧ n Note the following points. • The discriminant recovers the discriminant ideal via d = Dc2. • If L has degree 3, the discriminant also contains the splitting information of L at the infinite primes. Namely, for each real place ι of K, if ι(D) > 0 then L = R R R, while if ι(D) < 0 then L = R C. v ∼ × × v ∼ × • By a usual abuse of language, if L is an étale algebra over a number field K, its discriminant is the discriminant of the ring of integers over . OL OK • The discriminants over form a cancellative semigroup under the multiplication law OK

(c1,D1)(c2,D2) = (c1c2,D1D2).

31 • If K is a PID, then we can take c = K , and then the discriminants are simply nonzero elements D O congruent to a square mod 4, upO to multiplication by squares of units. ∈ OK • We will often denote a discriminant by a single letter, such as . When elements or ideals of K appear in discriminants, they are to be understood as follows: D O

D (D ) means ((1),D) (19) ∈ OK c2 (c K) means (c, 1). (20) ⊆ The seemingly counterintuitive convention (20) is motivated by the fact that, if c = (c) is principal, then (c, 1) is the same discriminant as ((1),c2). With these remarks in place, the reader should not have diﬃculty reading and proving the following relation:

Proposition 6.6. If ′ are two orders in an étale algebra L, then O ⊇ O 2 Disc ′ = [ : ′] Disc . O O O · O 6.3 Quadratic rings

We will spend a lot of time investigating the number of rings over K of given degree n and discriminant = (c,D). For quadratic rings, the problem has a complete answer:O D Proposition 6.7 (the parametrization of quadratic rings). Let K be a Dedekind domain of characteristic not 2. For every discriminant , there is a unique quadraticO étale order having discriminant . D OD D Proof. Note first that the theorem is true when K = K is a field: by Kummer theory, quadratic étale 2 O algebras over K are parametrized by K×/ (K×) , as are discriminants; and it is a simple matter to check that Disc K[√D]= D. We proceed to the general case. 2 2 1 For existence, let = (c,D) be given. By definition, D is congruent to a square t mod 4c , t c− . Consider the lattice D ∈ t + √D = cξ, ξ = L = K[√D]. O OK ⊕ 2 ∈ To prove that is an order in L, it is enough to verify that (cξ)(dξ) for any c, d c, and this follows from the computationO ∈ O ∈ t2 D ξ2 = ξ(t ξ¯)= tξ − − − 4 1 2 2 and the conditions t c− ,t D 4c− . Now suppose that∈ and− ∈are two orders with the same discriminant = (c,D). Their enclosing O1 O2 D K-algebras L1, L2 have the same discriminant D over K, and hence we can identify L1 = L2 = L. Now project each i along π : L L/K is an K -lattice ci in L/K, which is a one-dimensional K-vector space: O → 2 O indeed, we naturally have L/K = Λ L, and upon computation, we find that Disc i = (ci,D). Consequently ∼ 1 O c1 = c2 = c. Now, for each β c, the fiber π− (c) i is of the form βi + K for some βi. The element β β is integral over and∈ lies in K, hence in ∩ O. Thus = . O 1 − 2 OK OK O1 O2 If is any order in an étale algebra L/K (char K =2), the quadratic order B = Disc having the same O 6 O O discriminant as is called the quadratic resolvent ring of . It embeds into the discriminant torsor T2, in two conjugate ways.O Indeed, it is not hard to show that B Ois generated by the elements

ρ(ξ ,...,ξ )= κ (ξ ) T 1 n π(i) i ∈ 2 π A i X∈ n Y appearing in the proof of Lemma 6.2. Remark 6.8. The notion of a quadratic resolvent ring extends to characteristic 2, being always an order in the quadratic resolvent algebra constructed in Example 4.8. We omit the details.

32 6.4 Cubic rings Cubic and quartic rings have parametrizations, known as higher composition laws, linking them to certain forms over K and also to ideals in resolvent rings. The study of higher composition laws was inaugurated by BhargavaO in his celebrated series of papers ([2, 3, 4, 5]), although the gist of the parametrization of cubic rings goes back to work of F.W. Levi [32]. Later work by Deligne and by Wood [58, 60] has extended much of Bhargava’s work from Z to an arbitrary base scheme. In a previous paper [42], the author explained how a representative sample of these higher composition laws extend to the case when the base ring A is a Dedekind domain. In the present work, we will need a few more; fortunately, there are no added difficulties, and we will briefly run through the statements and the methods of proof. Theorem 6.9 (the parametrization of cubic rings). Let A be a Dedekind domain with field of fractions K, char A =3. 6 (a) Cubic rings over A, up to isomorphism, are in bijection with cubic maps O Φ: M Λ2M → between a two-dimensional A-lattice M and its own Steinitz class, up to isomorphism, in the obvious sense of a commutative square

M1 ∼ / M2 i

Φ1 Φ2 2 2 Λ M1 ∼ / Λ M2. det i The bijection sends a ring to the index form Φ: /A Λ2( /A) given by O O → O x x x2. 7→ ∧

(b) If is nondegenerate, that is, the corresponding cubic K-algebra L = K A is étale, then the map Φ Ois the restriction, under the Minkowski embedding, of the index form of⊗K¯ 3O, which is Φ: K¯ 3/K¯ K¯ 2/K¯ → (21) (x; y; z) (x y)(y z)(z x), 0 . 7→ − − − (c) Conversely, let L be a cubic étale algebra over K. If ¯ L/K is a lattice such that Φ sends ¯ into Λ2 ¯, then there is a unique cubic ring L such that,O ⊆ under the natural identiﬁcations, /AO= ¯. O O⊆ O O Proof. (a) The proof is quite elementary, involving merely solving for the coeﬃcients of the unknown multiplication table of . The case where A is a PID is due to Gross ([24], Section 2): the cubic ring having index form O f(xξ + yη)= ax3 + bx2y + cxy2 + dy3 has multiplication table

ξη = ad, ξ2 = ac + bξ aη,η2 = bd + dξ cη. (22) − − − − − For the general Dedekind case, see my [42], Theorem 7.1. It is also subsumed by Deligne’s work over an arbitrary base scheme; see Wood [58] and the references therein. (b) This follows from the fact that the index form respects base change. The index form of K¯ 3/K¯ is a Vandermonde determinant that can easily be written in the stated form. ¯ 2 ¯ (c) We have an integral cubic map Φ ¯ : Λ , which is the index form Φ of a unique cubic ring O O → O O O over K . But over K, Φ is isomorphic to the index form of L. Since L (as a cubic ring over K) is O determinedO by its index form, we obtain an identiﬁcation K L for which /A, the projection K ∼= of onto L/K, coincides with ¯. The uniqueness of isO⊗ obvious,O as must lie inO the integral closure O O O O L of K in L. O

33 In this paper we only deal with nondegenerate rings, that is, those of nonzero discriminant, or equivalently, those that lie in an étale K-algebra. Consequently, all index forms Φ that we will see are restrictions of (21). When cubic algebras are parametrized Kummer-theoretically, the resolvent map becomes very explicit and simple:

Proposition 6.10 (explicit Kummer theory for cubic algebras). Let R be a quadratic étale algebra over K (char K =3), and let 6 L = K + κ(R) be the cubic algebra of resolvent R′ = R K[µ3] (the Tate dual of R) corresponding to an element δ K× of norm 1 in Theorem 5.4(b), where ⊙ ∈

√3 ¯ 3 κ(ξ)= trK¯ 2/K ξω δ N=1 K ω (K¯ 2) [3] ∈ ∈ so κ maps R bijectively onto the traceless plane in L. Then the index form of L is given explicitly by

Φ: L/K Λ2(L/K) → (23) κ(ξ) 3√ 3δξ3 1, 7→ − ∧ where we identify 2 3 2 Λ L/K = Λ L = Λ R′ = R′/K = √ 3 R/K ∼ ∼ ∼ ∼ − · using the fact that R′ is the discriminant resolvent of L. Proof. Direct calculation, after reducing to the case K = K¯ .

Theorem 6.11 (self-balanced ideals in the cubic case). Let K be a Dedekind domain, char K = 3, and let R be a quadratic étale extension. A self-balanced triple inO R is a triple (B,I,δ) consisting6 of a quadratic order B R, a fractional ideal I of B, and a scalar δ (KB)× satisfying the conditions ⊆ ∈ δI3 B,N(I) = (t) is principal, and N(δ)t3 =1, (24) ⊆ 3 (a) Fix B and δ R× with N(δ) a cube t− . Then the mapping ∈ I = + κ(I) (25) 7→ O OK deﬁnes a bijection between • self-balanced triples of the form (B,I,δ), and • subrings L of the cubic algebra L = K + κ(R) corresponding to the Kummer element δ, such that isO⊆3-traced, that is, tr(ξ) 3 for every ξ L. O ∈ OK ∈ (b) Under this bijection, we have the discriminant relation

disc C = 27 disc B. (26) − Proof. The mapping κ defines a bijection between lattices I R and κ(I) L/K. The difficult part is showing that I fits into a self-balanced triple (B,I,δ) if and only⊆ if κ(I) is the⊆ projection of a 3-traced order . Note that if (B,I,δ) exists, it is unique, as the requirement [B : I] = (t) pins down B. O Rather than establish this equivalence directly, we will show that both conditions are equivalent to the symmetric trilinear form

β : I I I Λ2R × × → (α , α , α ) δα α α 1 2 3 → 1 2 3 1 2 taking values in t− Λ I. ·

34 In the case of self-balanced ideals, this was done over Z by Bhargava [2, Theorem 3]. Over a Dedekind domain, it follows from the parametrization of balanced triples of ideals over B [42, Theorem 5.3], after specializing to the case that all three ideals are identiﬁed with one ideal I. It also follows from the corresponding results over an arbitrary base in Wood [60, Theorem 1.4]. In the case of rings, we compute by Proposition 6.10 that β is the trilinear form attached to the index 1 2 form of κ(I). By Theorem 6.9(c), the diagonal restriction β(α, α, α) takes values in t− Λ (I) if and only if 1 2 κ(I) lifts to a ring . We wish to prove that β itself takes values in t− Λ (I) if and only if is 3-traced. O O Note that both conditions are local at the primes dividing 2 and 3, so we may assume that K is a DVR. 1 2 O With respect to a basis (ξ, η) of I and a generator of t− Λ (I), the index form of has the form O f(x, y)= ax3 + bx2y + cxy2 + dy3, a,...,d . ∈ OK If this is the diagonal restriction of β, then β itself can be represented as a 3-dimensional matrix

b/3 c/3 ⑧⑧ ⑧⑧ ⑧⑧ ⑧ a b/3 , c/3 d ⑧ ⑧⑧⑧ ⑧⑧ b/3 c/3 which is integral exactly when b,c 3 . Since the trace ideal of is generated by ∈ OK O tr(1) = 3, tr(ξ)= b, tr(η)= c − (by reference to the multiplication table (22)), this is also the condition for to be 3-traced, establishing the equivalence. O The discriminant relation (26) follows easily from the deﬁnition of κ.

6.5 Quartic rings and their cubic resolvent rings The basic method for parametrizing quartic orders is by means of cubic resolvent rings, introduced by Bhargava in [4] and developed by Wood in [58] and the author in [42].

Deﬁnition 6.12 ([42], Deﬁnition 8.1; also a special case of [58], p. 1069). Let A be a Dedekind domain, and let be a quartic algebra over A. A resolvent for (“numerical resolvent” in [42]) consists of a rank-2 A-latticeOY , an A-module isomorphism Θ:Λ2Y Λ3(O/A), and a quadratic map Φ: /A Y such that there is an identity of biquadratic maps → O O →

x y xy = Θ(Φ(x) Φ(y)) (27) ∧ ∧ ∧ from to Λ3( /A). O × O O We collect some basic facts about these resolvents.

Theorem 6.13 (the parametrization of quartic rings). The notion of resolvent for quartic rings has the following properties. (a) If X is a rank-3 A-lattice and Θ:Λ2Y Λ3X, Φ: X Y satisfy (27), then there is a unique (up to → → isomorphism) quartic ring equipped with an identiﬁcation /A = X making ( ,Y, Θ, Φ) a resolvent. O O ∼ O (b) There is a canonical (in particular, base-change-respecting) way to associate to a resolvent ( ,Y, Θ, Φ) O a cubic ring C and an identiﬁcation C/A ∼= X with the following property: For any element x and any lift y C of the element Φ(x) C/A, we have the equality ∈ O ∈ ∈ x x2 x3 = Θ(y y2). ∧ ∧ ∧

35 It satisﬁes Disc C = Disc . O (Here the discriminants are to be seen as quadratic resolvent rings, as in [42]; this implies the corresponding identity of discriminant ideals.) If is nondegenerate, then C is unique. O (c) Any quartic ring has at least one resolvent. O (d) If is maximal, the resolvent is unique (but need not be maximal). O (e) The number of resolvents of is the sum of the absolute norms of the divisors of the content of , O O the smallest ideal c such that = + c ′ for some order ′. O OK O O (f) Let (Y, Θ, Φ) be a resolvent of with associated cubic ring C, and let K = Frac A. If the corresponding O quartic K-algebra L = K A is étale, then the cubic K-algebra R = C A is none other than the cubic resolvent of L, as deﬁned⊗ O in Example 4.7. The maps Θ and Φ are⊗ theO restrictions, under the Minkowski embedding, of the unique resolvent of K¯ 4, which is K¯ 3 with the maps

Θ:Λ2(K¯ 3/K¯ ) Λ3(K¯ 4/K¯ ) → (28) (0;1;0) (0;0;1) (0;1;0;0) (0;0;1;0) (0;0;0;1) ∧ 7→ ∧ ∧ and Φ: K¯ 4/K¯ K¯ 3/K¯ → (29) (x; y; z; w) (xy + zw; xz + yw; xw + yz). 7→ (g) Conversely, let L be a quartic étale algebra over K and R its cubic resolvent. Let

Θ :Λ2R/K Λ3L/K, Φ : L/K R/K K → K → be the resolvent data of L as a (maximal) quartic ring over K. Suppose ¯ L/K, C¯ R/K are lattices such that O ⊆ ⊆

• Φ sends ¯ into C¯, K O • Θ maps Λ3 ¯ isomorphically onto Λ2C¯. K O Then there are unique quartic rings L, C R such that, under the natural identifications, /A = ¯, C/A = C¯, and C¯ is a resolventO ⊆ with the⊆ restrictions of Θ and Φ . O O K K Proof. (a) See [42], Theorem 8.3. (b) See [42], Theorems 8.7 and 8.8. (c) See [42], Corollary 8.6. (d) This is a special case of the following part. (e) See [42], Corollary 8.5. (f) By base-changing to K, we see that Y K = R/K is a resolvent for L. Since the resolvent is unique, ⊗A it suffices to show that the cubic resolvent R′ from Example 4.7 is a resolvent for L also. The maps Θ and Φ defined in the theorem statement are seen, by symmetry, to restrict to maps of the appropriate K-modules. The verification of (27) and of the fact that the multiplicative structure on R′ is the right one can be checked at the level of K¯ -algebras. (g) Letting X = ¯, Y = C¯ in part (a), we construct the desired and C. By comparison to the situation under base-changeO to K, we see that , C naturally injectO into L, R respectively. Uniqueness is O obvious, as must lie in the integral closure L. O O

36 In this paper we only deal with nondegenerate rings, that is, those of nonzero discriminant, or equivalently, those that lie in an étale K-algebra. Consequently, all resolvent maps Θ, Φ that we will see are restrictions of (28) and (29). When quartic algebras are parametrized Kummer-theoretically, the resolvent map becomes very explicit and simple: Proposition 6.14 (explicit Kummer theory for quartic algebras). Let R be a cubic étale algebra over K (char K =2), and let 6 L = K + κ(R) be the quartic algebra of resolvent R corresponding to an element δ K× of norm 1 in Theorem 5.4(c), where ∈ √ ¯ 4 κ(ξ)= trK¯ 3/K ξω δ N=1 K ω (K¯ 3) [2] ∈ ∈ so κ maps R bijectively onto the traceless hyperplane in L. Then the resolvent of L is given explicitly by Θ:Λ3(R) Λ3(L/K) → 1 (30) α β γ κ(α) κ(β) κ(γ) ∧ ∧ 7→ 16 N(δ) · ∧ ∧

Φ:pL/K R/K → (31) κ(ξ) 4δξ2 7→ Proof. Since the resolvent is unique (over a field, any étale extension has content 1), it suffices to prove that (30) and (31) define a resolvent. This can be done after extension to K¯ , and then it is enough to prove that (30) and (31) agree with the standard resolvent on K¯ 4, given in Theorem 6.13(f). As to (30), since both sides are alternating in α, β, and γ, it suffices to prove it in the case that α = (1;0;0), β = (0;1;0) γ = (0;0;1) form the standard basis of R = K¯ 3. Let δ = (δ(1),δ(2),δ(3)). Then

κ(α)= δ(1), δ(1), δ(1), δ(1) − − p p p p κ(β)= δ(2), δ(2), δ(2), δ(2) − − p p p p κ(γ)= δ(3), δ(3), δ(3), δ(3) − − p p p p and hence the wedge product of these diﬀers from the standard generator of Λ3(L/K) by a factor of 1 1 1 1 √δ(1) √δ(1) √δ(1) √δ(1) − − √δ(2) √δ(2) √δ(2) √δ(2) − − √ (3) √ (3) √ (3) √ (3) δ δ δ δ − − 11 1 1

1 1 1 1 = N(δ) − − 1 1 1 1 − − p 1 1 1 1 − −

= 16 N (δ). · The calculation for (31) is even more routine.p Remark 6.15. The datum Θ of a resolvent carries no information, in the following sense. It is unique up to scaling by c A×, and the resolvent data (X,Y, Θ, Φ) and (X,Y,cΘ, Φ) are isomorphic under multiplication 1 ∈ 2 by c− on X and by c− on Y . If A is a PID, indeed, neither X nor Y carries any information, and the entire data of the resolvent is encapsulated in Φ, a pair of 3 3 symmetric matrices over A (with formal × factors of 1/2 oﬀ the diagonal) deﬁned up to the natural action of GL3A GL2A. This establishes the close kinship with Bhargava’s parametrization of quartic rings in [4]. However,× it is useful to keep Θ around.

37 6.5.1 Traced resolvents Just as we found it natural to study not just binary cubic 1111-forms, but also 1331-forms and their analogue for each divisor of the ideal (3), so too we study not just quartic rings in general but those satisfying a natural condition at the primes dividing 2. Definition 6.16. Let A be a Dedekind domain, char A = 2, and let t be an ideal dividing (2) in A. A resolvent ( ,Y, Θ, Φ) over A is called t-traced if, for all x and6 y in A, the associated bilinear form O Φ(x + y) Φ(x) Φ(y) Φ(x, y)= − − 2 1 whose diagonal restriction is Φ(x, x)=Φ(x) takes values in 2− tY . If A is a PID, this is equivalent to saying 1 that the off-diagonal entries in the matrix representation of Φ, which a priori live in 2 A, actually belong to t 2 A. We say that A is t-traced if it admits a t-traced resolvent. Here are some facts about traced resolvents: Proposition 6.17. Let be a quartic ring over a Dedekind domain . O OK (a) is t-traced if and only if O (i) t2 tr x for all x ; | ∈ O (ii) x2 A + t for all x . ∈ O ∈ O 3 (b) If is not an order in the trivial algebra K[ε1,ε2,ε3]/(εiεj)i,j=1, the number of t-traced resolvents of Ois the sum of the absolute norms of the divisors of its t-traced content, which is the smallest ideal c O such that = A + c ′ and ′ is also t-traced. O O O 2 2 (c) If ( ,Y, Θ, Φ) is a t-traced resolvent with associated cubic ring C, then t ct(C), that is, C = A + t C′ O | 8 for some cubic ring C′. We call C′ a “reduced resolvent” of the t-traced ring A. Also, t disc A. | Proof. (a) Since both statements are local at the primes dividing 2, we can assume that is a DVR, OK and thus that t = (t) is principal. With respect to bases (1 = ξ0, ξ1, ξ2, ξ3) for and (1 = η0, η1, η2) for a resolvent C, the structure constants ck of the ring , defined by O ij O k ξiξj = cij ξk, Xk are determined by the entries of the resolvent

Φ = ([aij ], [bij ]) via the determinants

ij 1i=j +1k=ℓ aij akℓ λ =2 6 6 kℓ b b ij kℓ

and a set of formulas appearing in Bhargava [4, equation (21)] and over a Dedekind domain by the author [42, equation (12)]: cj = ελii ii − ik k jj cij = ελii (32) cj ck = ελjk ij − ik ii ci cj ck = ελij , ii − ij − ik ik where (i, j, k) denotes any permutation of (1, 2, 3) and ε = 1 its sign. (Here the nonappearance of k ± some of the individual cij on the left-hand side of (32) stems from the ambiguity of translating each ξ by , which does not change the matrix of Φ.) i OK Assume ﬁrst that Φ: / C/ is t-traced. Then O OK → OK ij 1i=j +1k=ℓ λ t 6 6 . (33) kℓ ∈ We then prove that the conditions (i) and (ii) must hold:

38 (i) The trace

1 2 3 tr(ξ1)= c11 + c12 + c13 12 23 3 = λ13 +2λ11 +4c13 0 mod t2, ≡ and likewise tr(ξ ), tr(ξ ) t2. 2 3 ∈ i 2 (ii) The coeﬃcients c11 of ξ1 satisfy:

c2 = λ11 t 11 13 ∈ 3 1 1 2 3 2 and likewise for c11; and then c11 t also, since the trace c11 + c12 + c13 = tr(ξ1) t t. So 2 ∈ ∈ ⊆ the desired relation ξ K + t holds when ξ = ξ1, indeed ξ = a1ξ1 for any a1 K . The same proof works for ξ ∈= Oa ξ orOξ = a ξ . Since the case ξ = a is trivial and∈ squaring O is 2 2 3 3 0 ∈ OK a Z-linear operation modulo 2, we get the result for all ξ . ∈ O Conversely, suppose that (i) and (ii) hold. We ﬁrst establish (33). We have • λ11 = c2 t 13 11 ∈ • λ23 = c2 c3 = tr ξ c1 2c3 t 11 12 − 13 1 − 11 − 13 ∈ • λ12 = c1 c2 c3 = tr ξ 2λ23 +4c3 t2. 13 11 − 12 − 13 1 − 11 13 ∈ ij Permuting the indices as needed, this accounts for all the λkℓ about which (33) makes a nontrivial assertion. ij Now we work from the λkℓ back to the resolvent ( , ). We may assume that C is nontrivial (the trivial rings, one for each Steinitz class, are plainly 2-tracedA B with ( , )=(0, 0).) Then, in the proof of A B [42], Theorem 8.4, the author established that there are vectors µij in a two-dimensional vector space V over K, unique up to GL2(V ), such that

µ µ = λij ω ij ∧ kℓ kℓ · for some ﬁxed generator ω Λ2V . (The proof uses the Plücker relations, which are a consequence of the associative law on .) This∈ V is none other than R/K, the resolvent module of the quartic algebra L = K, whichO admits the unique resolvent O⊗OK

Φ(a1ξ1 + a2ξ2 + a3ξ3)= aiaj µij i

The resolvents of were found to be exactly the lattices M containing the span M0 of the six µij , with the correct indexO

ij j k j k i j k [M : M0]= c = λkℓ = cii,cij ,cij cik,cii cij cik : i = j = k = i , i,j,k,ℓ − − − 6 6 6 the content ideal of . By inspection of (34) that M is t-traced if and only if it actually contains the O span M˜ 0 of the six vectors 1i=j µ˜ij = t− 6 µij . Condition (33) is interpreted as saying that the µ˜ µ˜ are still integer multiples of ω. Then the ij ∧ kℓ t-traced resolvents are the lattices M M˜ . The needed index ⊇ 0 ˜ ˜ij ˜c = [M : M0]= λkℓ i,j,k,ℓ is an integral ideal, so such M exists, ﬁnishing the proof of (a).

39 (b) It suffices to prove that ˜c is the t-traced content of . To see this, note that if = K + a ′ has Ok O O O content divisible by a, then the structure coefficients cij of ′ are obtained from those of by dividing ij ˜ij O O by a. This means that the λkℓ and λkℓ are divided by a, and so remain integral (indicating that ′ is also t-traced) exactly when a ˜c. O | (c) We can again reduce to the case that K is a DVR so has an K -basis. Recall that the index form of the resolvent C is given by O O O f(x, y) = 4 det( x + y) A B 1 ([4], Proposition 11; [42], Theorem 8.7). If and have off-diagonal entries in 2− t, it immediately follows that f is divisible by t2, so t2 ct(AC). ConsequentlyB disc = disc C, being quartic in the coefficients of f, is divisible by t8. | O Similar to Theorem 6.11, we have the following relation between 2-traced quartic rings and self-balanced ideals:

Theorem 6.18 (self-balanced ideals in the quartic setting). Let K be a Dedekind domain, char K = 2, and let R be a cubic étale extension. A self-balanced triple in R is aO triple (C,I,δ) consisting of a cubic6 order C R, a fractional ideal I of C, and a scalar δ (KC)× satisfying the conditions ⊆ ∈ δI2 C,N(I) = (t) is principal, and N(δ)t2 =1, (35) ⊆ 2 Fix an order C R and a scalar δ R× with N(δ) a square t− . Then the mapping ⊆ ∈ I = + κ(I) (36) 7→ O OK deﬁnes a bijection between • self-balanced triples of the form (C,I,δ), and • subrings L of the quartic algebra L = K + κ(R) corresponding to the Kummer element δ, such that isO2-traced ⊆ with reduced resolvent C. O Proof. The proof is very similar to that of 6.11, so we simply summarize the main points. The linear isomorphism κ establishes a bijection between lattices I R and κ(I) L/K. We wish to prove that (C,I,δ) is balanced if and only if κ(I) is the projection of⊆ a 2-traced order⊆ with reduced resolvent C. First note that either of these conditions uniquely speciﬁes

[C : I] = (t), the former by the balancing condition N(I) = (t), and the latter by the Θ-condition that have discriminant 256 disc C. O Once again, it is diﬃcult to proceed directly, and we instead prove that both conditions are equivalent to the bilinear map

Φ: I I R/K × → (α , α ) 4δα α 1 2 → 1 2 taking values in 4C/K. On the self-balanced ideals side, this follows from the parametrization of balanced pairs of ideals by 2 3 3 boxes performed over Z by Bhargava [3, Theorem 2] and over a general base by Wood [60, Theorem 1.4].× × On the quartic rings side, the diagonal restriction of Φ is precisely the resolvent of κ(I), by Proposition 6.14. That Φ(α, α) 4C/K for each α I expresses the one condition remaining for κ(I) to lift (by Theorem ∈ ∈ 6.13(g)) to a quartic ring with resolvent K +4C. Then, by deﬁnition, this resolvent is 2-traced exactly when Φ itself has image inO4C/K. O

40 7 Cohomology of cyclic modules over a local ﬁeld

Let M be a Galois module with underlying group over a local ﬁeld K Q (that is, a wild local ﬁeld of Cp ⊇ p characteristic 0). Denote by T and T ′, respectively, the (Z/pZ)×-torsors corresponding to the action of GK on M 0 and on \{ } Surj(M,µ )= M ′ 0 , p \{ } and denote by τc : T ′ T ′ the torsor operation corresponding to c (Z/pZ)× ∼= Aut M. By Theorem 5.4, Kummer theory gives→ an isomorphism ∈

1 p c H (K,M) = α T ′×/(T ′×) : τ (α)= α c (Z/pZ)× . (37) ∼ ∈ c ∀ ∈ Our objective in this section is to understand the group on the right: that is, to describe a basis of it (a generalization of the well-known Shafarevich basis for T ′×) and understand how the Tate pairing respects it. Much of our work parallels that of Del Corso and Dvornicich [13] and Nguyen-Quang-Do [41]. If σ H1(K,M), we let L = L be the ( )-extension of degree p coming from the aﬃne action of ∈ σ GA Cp GK on M, while we let E = Eσ be the associated ( p)-torsor. Owing to the semidirect product structure of ( ), we get a natural decomposition GA C GA Cp E = L T. ∼ ⊗K Using the division algorithm in Z, we let ℓ = ℓ(L)= ℓ(σ) and θ = θ(L)= θ(σ) the integers such that

v (disc L)= p(e ℓ)+ θ, 1 θ p 2. K − − ≤ ≤ −

We call ℓ the level, and θ the oﬀset, of the ( p)-extension L or of the coclass σ. Although these deﬁnitions appear strange, they allow us to state conciselyGA C the following theorem, which will be the main theorem of this section.

Theorem 7.1 (levels and offsets). Let M be a Galois module with underlying group p over a local field K with char K = p. C 6 (a) The level ℓ of a coclass determines its offset θ uniquely in the following way:

(i) If ℓ = e, then θ = vK (disc T ). (ii) If 0 ℓ

41 (f) For d 1, a neighborhood ≤ 1 [α] H (K,M): α T ′×, α 1 d ∈ ∈ | − |≤ is a level space i whose indexi is given by L

vK (β) p 1 log d vK (β) + − 1 , d dmin i = p p log π − − p 1 ≥  & | K | − '  e +1, d < dmin,

where  p/(p 1) d = p − . min | | 1 1 (g) For 0 i e, with respect to the Tate pairing between H (K,M) and H (K,M ′), ≤ ≤

i(M)⊥ = e i(M ′). L L − One corollary is suﬃciently important that we state it before starting the proof:

Corollary 7.2. For 0 i e, the characteristic function Li of the level space i has Fourier transform given by ≤ ≤ L e i Li = q − Le i. (38) − where q = k . | K | c Proof. Immediate from Theorem 7.1, parts (d) and (g).

7.1 Discriminants of Kummer and aﬃne extensions The starting point for our investigation of discriminants is as follows:

Theorem 7.3. Let K be a local ﬁeld with µp K, and let u K× be a minimal representative of a class p ⊆ ∈ p in K×/(K×) . The discriminant ideal of the associated Kummer extension L = K[√u] is given by

p p 1 p πK − peK · p 1 K vK (u 1) < Disc(L/K)= (u 1) − O − p 1  − pe−K  (1) vK (u 1) .  − ≥ p 1 −  Proof. One can ﬁnd an explicit basis for L and compute the discriminant. For details, see Del Corso and Dvornicich [13, Lemmas 5, 6, and 7]. O In this section, we will prove the following generalization:

Theorem 7.4. Let K be a local ﬁeld, and let M be a G -module with underlying group . Let T ′ be the K Cp (Z/pZ)×-torsor corresponding to the GK -set Surj(M,µm), and let T1′ be the ﬁeld factor of T ′. Let u T1′× 1 ∈ be a minimal representative for a class in (T1′×)ω parametrizing, via Theorem 5.4 a coclass σ H (K,M), and let L be the corresponding ( )-extension. Then ∈ GA Cp p p 1 p π − peT ′ · K v (u 1) < 1 p 1 T1′ T1′ Disc(L/K) = (u 1) − O − p 1 (39) T1′  − pe− O T1′  Disc(T/K) T vT (u 1) ,  · O 1′ 1′ − ≥ p 1 −  where T is the (Z/pZ)×-torsor corresponding to M.

p 1 Remark 7.5. Note that (u 1) − T is the extension of an ideal of K, since eT /K divides p 1. − O 1′ 1′ −

42 Proof. If L is not a field, then the image of GK in ( p) lies in a nontransitive subgroup (viewing ( p) as embedded in Sym( )). It is not hard to show thatGA C every nontransitive subgroup of ( ) hasGA a fixedC Cp GA Cp point. Moving this fixed point to 0, we get that σ = 0, u = 1, and L ∼= K T , in accord with the second case of the formula. × We may now assume that L is a field. Although the extension L/K need not be Galois, we have

T [µ ] L = T [µ , √p u], p ⊗K ∼ p a Kummer extension of T [µp]. Let E be a field factor of T [µp] containing T1′. Then E L = E[√p u] ⊗K ∼ as extensions of E. Note that [E : K] (p 1)2; in particular, [E : K] is prime to p. So u remains a minimal | − representative in E×, and since E and L must be linearly disjoint, E L = EL is a field unless u =1. So ⊗K p p 1 p πE − peK · p 1 E vK (u 1) < Disc(EL/E)= (u 1) − O − p 1  − pe−K  (1) vK (u 1) .  − ≥ p 1 − We must now relate Disc(EL/E) to Disc(L/K ). If v (u 1) pe /(p 1), then EL/E is unramified, so K − ≥ K − p ∤ eEL/K and L/K is unramified as well. In particular, L/K is Galois, so M is trivial, T is totally split, and the formula again holds. pe We are left with the case that v (u 1) < K . Here EL/E, and hence L/K, are totally ramified. We K − p 1 relate their discriminants by the following trick,− which also appears in Del Corso and Dvornicich [13]. An K -basis for L is given by O O p 1 1, πL,...,πL− . (40)

The same elements form an -basis for an order , but their EL-valuations are 0,e′,..., (p 1)e′, OE O ⊆ OEL − where e′ = eE/K . Divide each basis element by πE as many times as possible so that it remains integral. We get a new system of elements p 1 πL πL− 1, a1 ,..., ap 1 . (41) πE πE −

Since e′ is coprime to p, these elements have EL-valuations 0, 1,...,e′ 1 in some order and thus form an -basis for . We have − OE OEL

vE([ EL : ]) = a1 + + ap 1 O O ··· − [e′ + + (p 1)e′] [1 + + (p 1)] = ··· − − ··· − p (p 1)(e′ 1) = − − , 2 and hence

Disc(L/K) = Disc( / ) O OE = Disc( / ) [ : ]2 OEL OE · OEL O p p 1 p πE − (p 1)(e′ 1) = · π − − (u 1)p 1 · E · OE − − p e′(p 1) p π − = · E (u 1)p 1 · OE − − p (p 1) p π − = · K . (u 1)p 1 · OE − − Remark 7.6. Along the lines of the preceding argument, we can prove the following more general result on discriminants in extensions of coprime degree:

43 Proposition 7.7. Let L and M be two extensions of a local ﬁeld K with gcd([L : K], [M : K]) = 1. Then

fL/K (eL/K 1) π − Disc(L/K) = Disc(LM/M) M . · π K 7.2 The Shafarevich basis We start with the following exposition of the Shafarevich basis theorem. Although this theorem has appeared many times in the literature (see Del Corso and Dvornicich, [13], Proposition 6), we include a proof here by a method that will establish some important corollaries for us. Filter U = K× by the subgroups

i U = x × : x 1 mod π , i { ∈ OK ≡ } p and let U¯i be the projection of Ui onto U¯ := K×/(K×) . Note that U¯i =0 for i>peK /(p 1), as the Taylor p pe /(p 1) +1 − series for √x about 1 converges for x 1 mod π⌊ K − ⌋ . So ≡ ¯ ¯ ¯ ¯ ¯ U ∼= U/U0 Ui/Ui+1 (42) ⊕ pe T1′ 0 iMp 1 ≤ ≤ − as Fp-vector spaces, and we can produce a basis for U¯ by lifting a basis for each of the composition factors on the right-hand side.

Proposition 7.8 (the Shafarevich basis theorem). Let K Q be a local ﬁeld, and let i 0. The ⊇ p ≥ structure of U¯i/U¯i+1 is as follows: • If p 0

for any a K with p ∤ tr /Q (a). We call such a generator an intimate unit, and we let ∈ O OK p p/(p 1) d = p(ζ 1) = p − , min | p − | | | the distance of an intimate unit to 1.

• For all other i, we have U¯i/U¯i+1 =0. ¯ ¯ ¯ ¯ Proof. Note that U0/U1 =0 because U0/U1 = kK× has order prime to p. To compute Ui/Ui+1, where i 1, ∼ i i+1 ≥ we must see how many of the congruence classes 1+ xπ mod π (where x kK ) contain a pth power. p j ∈ Consider a general pth power u , 1 = u × . Write u =1+ yπ , π ∤ y. By the binomial theorem, 6 ∈ OK up 1+ pyπj + ypπpj mod π2j+eK , ≡ so eK = pj, j< p 1 , − p peK eK v(u 1)  p 1 , j = p 1 , − ≥ − −  eK  = j + eK , j> p 1 . − We now perform the needed analysis in each case:

44 p • If 0 peK/(p 1), then the pth powers of elements of the form 1+ xπ − surject onto the congruence classes, repeating− what we knew from the Taylor series. • Finally, if i = pe /(p 1), then we can only use powers up = (1+ πj)p where j = e /(p 1). We have K − K − (1 + yπj ) 1+ pyπj + ypπpj 1 + (yp cy)πi mod πi+1, ≡ ≡ − where p × c = (−p 1)j K . π − ∈ O So we must analyze the (clearly linear) map ℘ : k k given by ℘ (y)= yp cy. c K → K c − If c is not a (p 1)st power in k (or in , which amounts to the same thing by Hensel’s lemma), − K OK then ℘c is injective and hence surjective, so U¯i/U¯i+1 =0. Also, K has no nontrivial pth roots of unity, as u = ζ would yield a nontrivial element of ker ℘ (since v (ζ )= e /(p 1)). p c p p K − p 1 j If c = b − is a (p 1)st power in K , then y = b is an element of ker ℘c. Note that u = 1+ bπ − O p i+1 lifts to a nontrivial pth root of unity ζp, since u 1 mod π has (by the Taylor series again) a pth j+1 ≡ root that is 1 mod π . Note that ker ℘c has dimension only 1, since b is unique up to µp 1 = Fp×. ¯ ¯ − Consequently coker ℘c ∼= Ui/Ui+1 has dimension exactly 1. This does not tell us how to ﬁnd a generator for U¯i/U¯i+1. For this, put y = by′ so

j p p i p i+1 (1 + by′π ) 1+ b (y y)π 1+ b ℘(y) mod π , ≡ − ≡ p p where ℘(y) = y y is the usual Artin-Schreyer map. Since y and y are Galois conjugates over Fp, −p ¯ ¯ we have trkK /Fp (y y)=0, so if a kK is an element with nonzero trace to Fp, then Ui/Ui+1 is generated by − ∈ 1+ abpπi 1+ abcπi 1+ ap(ζ 1) mod πi+1. ≡ ≡ p − We draw two corollaries of the above method. Corollary 7.9. U¯ has dimension [K : Qp]+1+ 1µ K p⊆ with a basis consisting of πK and (arbitrary lifts of) the elements in the bases of U¯i/U¯i+1 in Proposition 7.8. We call this basis the Shafarevich basis for U. Remark 7.10. The dimension of U¯ follows also from the Euler-characteristic computation

0 2 H (K,µp) H (K,µp) [K:Q ] | || | = p− p H1(K,µ ) | p | Proof. Clearly U/¯ U¯ = is generated by π . The result follows from the composition series (42). 0 ∼ Cp K Corollary 7.11. An element x U¯ belongs to U¯ if and only if, in the expansion of x in the Shafarevich ∈ i basis, only the basis elements in Ui appear (to nonzero exponents).

Proof. Filtering U¯i by the U¯j , for j i, we see that a basis for U¯i is given by the portion of the Shafarevich basis coming from U¯ /U¯ for j ≥i. This is just the basis elements that lie in U . j j+1 ≥ i The following simple result is one I have not seen in the literature before:

45 p 1 Corollary 7.12. The cyclotomic extension Q [µ ] is isomorphic to Q [ −√ p] and has Kummer element p p p − p as a µp 1-torsor. − − p 1 p 1 Proof. Let K = Qp[ −√ p], with uniformizer πK = −√ p. In the notation of the intimate unit case of Proposition 7.8, we have−i = p, j =1, − p c = −p 1 =1, πK−

p 1 2 and we can take b = −√c =1. Accordingly, 1+ πK is congruent mod πK to a unique pth root of unity ζp. i 2 Note that ζp 1+ iπK mod πK . Since µp 1 acts on πK by multiplication, it must act on the powers of ζp ≡ − by ω. Hence K = Qp[µp] as µp 1-torsors. ∼ − In the rest of this section we will study how U¯ = U¯(K) behaves under field extension. We use the following notational conventions: p Elements u K×/(K×) are classified by their distance, by which we mean the closest distance of a representative from∈ 1: p d(u)= dK (u) = min uy 1 . y K | − | ∈ × Here the absolute value is the local one on K. (We could choose a normalization of this absolute value, but eK we prefer to express d(u) in terms of an undetermined πK and p = πK .) Note that d(u) 1, since u can always be taken to have nonnegative valuation. Also,| it| is easy| | to| see| that d(uv) max d(≤u), d(v) , so p ≤ { } d defines a norm on K×/(K×) . For ease in stating theorems involving distances, we note that an ideal (or even a fractional ideal) a of a local field K is uniquely determined by the largest absolute value of its elements, which we denote by a . We have | | a = x K : x a . { ∈ | | ≤ | |} p For any real d> 0, let B d denote the closed ball of radius d about 1 in K×/(K×) : ≤ p B d = u K×/(K×) : d(u) d ≤ { ∈ ≤ } and likewise for B

Lemma 7.13. If L/K is an extension of local ﬁelds whose degree n is prime to p, then the canonical map p p p from K×/(K×) to L×/(L×) is injective and preserves distance: that is, for every u K×/(K×) , ∈

dK (u)= dL(u).

n Proof. The injectivity follows from the fact that if u K, then NL/K = u and n is prime to p. It is obvious that d (u) d (u), so it suﬃces to∈ prove the opposite inequality. It’s easy to see that if L ≤ K x L, then ∈ O N (x) 1 x 1 . | L/K − | ≤ | − | p Let y L× achieve uy 1 = d (u). Then ∈ | − | L d (u)= uyp 1 N (uyp) 1 = u[L:K]N (y)p 1 d u[L:K] . L | − | ≥ | L/K − | | L/K − |≥ K But u is a power of u[L:K] up to pth powers, so d (u) d u[L:K] , completing the proof. K ≤ K Assume K Qp. We ﬁrst parametrize M itself. By (classical) Kummer theory, we have canonical isomorphisms ⊇

1 1 p 1 Hom(GK , Aut( p)) = H (GK , (Z/pZ)×) = H (GK ,µp 1) = K×/(K×) − C ∼ ∼ − ∼ where the isomorphism (Z/pZ)× = µp 1 is given by Teichmüller lift. (Note that we do not need to pick a ∼ − generator of (Z/pZ)× to do this.) We have the following:

46 Lemma 7.14. If M is the Galois module with underlying group p corresponding to the Kummer element p 1 C β K×/(K×) − , then the Tate dual M ′ has Kummer element β′ = p/β. ∈ − Proof. When M is trivial, the result was proved as Corollary 7.12. The lemma then follows by noting that if M1, M2 are cyclic Galois modules of order p with Kummer elements β1, β2, respectively, then Hom(M1,M2) is also cyclic of order p and has Kummer element β2/β1.

If T ′ has r ﬁeld factors (all necessarily isomorphic to one T1′), then, by Corollary 7.9,

p n +2r, µp T1′ dimFp T ′×/(T ′×) = ⊆ ( n + r otherwise.

p The group T ′×/(T ′×) is a representation of (Z/pZ)× over the ﬁeld Fp. Since Fp has the (p 1)st roots of unity, such a representation splits as a direct sum of 1-dimensional representations; there are−p 1 of these, and they are the powers of the standard representation −

ω : (Z/pZ)× GL (F ) → 1 p given by the obvious isomorphism. By Theorem 5.4, H1(K,M) is parametrized by the ω-isotypical compo- p nent of T ′×/(T ′×) , which we denote by Tω′× for brevity. We can reduce the problem from T ′ to T1′ in the following way:

Lemma 7.15. Let T ′ be a µ -torsor, t p 1. Let T ′ be the ﬁeld factor of T ′, and let r = [T ′ : K]. Then: t | − 1 1

(a) The subgroup fixing T1′ (as a set) is µr µp 1 = (Z/pZ)×, and T1′ is a µr-torsor; ⊆ − ∼ (b) Projection to T1′ defines an isomorphism Tω′× ∼= (T1′×)ω, where p c T ′× = α T ′×/(T ′×) : τ (α)= α c µ (43) ω ∈ c ∀ ∈ t p c (T ′×) = α T ′×/(T ′×) : τ (α)= α c µ . (44) 1 ω ∈ 1 1 c ∀ ∈ r (c) More generally, for any s with r s t, the orbit µ (T ′) consists of s/n field factors whose product T ′ is | | s 1 s a µs-torsor. Projection onto Ts′ and then onto T1′ defines isomorphisms

Tω′× ∼= (Ts′×)ω ∼= (T1′×)ω.

Remark 7.16. A result with much the same content, but in a slightly diﬀerent setting, is proved by Del Corso and Dvornicich ([13], Proposition 7).

Proof. The torsor action must permute the ﬁeld factors transitively; since µn is cyclic, a generator ζn must n/r cyclically permute them, and the stabilizer of T1 (as a set) is ζ = µr, proving(a). Since the action simply transitively permutes the coordinates of T1, T1 is a µr-torsor. If we know the T1-component α of T1× p | an α (T ×/(T ×) ) , then all the other components are uniquely determined by the eigenvector condition; ∈ ω k it is only necessary for α T × to behave properly under µr, namely that αT × (T1×)ω . | 1 1 ∈ This proves (b). Also, it is clear from our analysis that µs(T1) is a µs-torsor. Applying (b) to this torsor proves (c).

We now ﬁlter T1′× as above to discover its ω-component.

p 1 Proposition 7.17 (the Shafarevich basis for Tω′×). As above, let T ′ = K[ −√β] be the (Z/pZ)×-torsor corresponding to the Tate dual M ′ of a cyclic Galois module M of order p with Kummer element β p 1 ¯ p ¯ ∈ K×/(K×) − , and let T1′ be the ﬁeld factor of T ′. Filter U = T1′×/(T1′×) by the subgroups Ui as in the previous subsection. Since the µr-torsor action on T1′ preserves the valuation, each Ui is a subrepresentation p of T1′×/(T1′×) . Then:

(a) The ω-isotypical component of U/¯ U¯0 has dimension 1 if M ′ is trivial, 0 otherwise.

47 (b) The ω-isotypical component of U¯i/U¯i+1 has dimension • f if pe e v (β) T1′/Qp T1′/K K 0

Proof. Note that U/U = π is a copy of the trivial representation and that ω is trivial (as a representation 0 T1′ of µ ) exactly when r =1, that is, M is trivial. r ′ Our convention for the Kummer map is such that

p 1 p 1 τc( − β)=˜c − β.

p p (p 1)/r By a standard result in Kummer theory, the degree r is the least integer such that β = β1 − is a (p 1)/rth r − power, and T1′ = K[√β1] with r r τc( β1)=˜c β1.

Let f ′ = gcd vK (β1), r and e′ = r/f ′. We claimp that e′ andp f ′ are respectively the ramiﬁcation and inertia indices of T over K. By the Euclidean algorithm, we may choose integers g and h such that 1′ gv(β ) hr = f ′. (45) 1 − Construct the elements r g r e′ √β1 √β1 π′ = and u′ = . πh vK (β1)/f ′ K πK

Note that vK (π′)=1/e′, so eT /K e′. (46) 1′ ≥ On the other hand, vK (u′)=0, and u′ is an f ′th root of the unit

β1 β2 = . vK (β1) πK

Note that β2 is not an ℓth power for any prime ℓ f ′, as otherwise β would be a (p 1)ℓ/rth power, contradicting what we know about r. So the residue class| of u generates a degree-f extension− of k inside k ; in ′ ′ K T1′ particular, fT /K f ′. (47) 1′ ≥ Equality must hold in (46) and (47), so T1′ has uniformizer π′ and residue ﬁeld generator u′. Note that for c F×, ∈ p g e′ τc(π′)=˜c π′ and τc(u′)=˜c u′.

We now have what we need to compute the Galois action on U¯i/U¯i+1. By Proposition 7.8, the space U¯i/U¯i+1 is nonzero only for peT 0

48 i j gi+e′ j Thus the basis element 1+ π′ u′ generates a 1-dimensional µr-submodule of Ui/Ui+1 isomorphic to ω . Accordingly, we select the generic units satisfying

gi + e′j 1 mod r. ≡

Since r = e′f ′, there are exactly f ′ values of j satisfying this when

e′ gi 1, (48) | − and none otherwise. By (45), g is the multiplicative inverse of v(β1)/f ′ = e′v(β)/(p 1) mod e′, so we can rewrite (48) as − e′v(β) i mod e′, ≡ p 1 − as desired. As to the case that i = pe /(p 1) (the intimate units), we simply note that, by Proposition 7.4, we T1′ have − ¯ 1 (Ui)ω ∼= Hur(K,M), so (U¯ ) = H1 (K,M) = H0(K,M) . | i ω| | ur | | | ¯ (A direct computation of the torsor action on the intimate units is also possible; it turns out that Ui ∼= µm as µm 1-modules.) −

By complete reducibility, we can get a basis for U¯ω from the bases for its composition factors: Corollary 7.18. If M = , then H1(K,M) has dimension ∼ Cp 0 0 [K : Qp] + dimFp H (K,M) + dimFp H (K,M ′), with a basis consisting of appropriate lifts of the Shafarevich basis elements picked out by Proposition 7.17. In particular, we have proved Theorem 7.1(e).

7.3 Proof of Theorem 7.1 It now remains to recast the above results in terms of levels and oﬀsets and prove the remaining parts of Theorem 7.1. Let α (T1′×)ω, α being a minimal-distance element. We consider the possibilities for the leading factor in α∈ with respect to the Shafarevich basis; this determines α 1 by Corollary 7.9, and thence | − | d := vK (Disc(L/K)) and hence the level and oﬀset of α.

• If α is led by the uniformizer, then M ′ is trivial. From Theorem 7.4, we get d = peK + p 1, so ℓ(α)= 1 and θ(α)= 1. − − − • If α is led by a generic unit, then we have vT (α 1) = i, where i is an integer satisfying 1′ − pe e v (β) T1′/Qp T1′/K K 0

p 1

49 This proves (a) and (c). In the case that α is led by a generic unit of level ℓ, 0 ℓ

α 1 d. | − |≤

If d < dmin, then every α in this ball is a pth power (by Proposition 7.8, or simply by noting that the Taylor series for pth root converges on this ball), so the range of [α] is 1 = . If d d , then the intimate { } Le+1 ≥ min units are certainly included, so the range of [α] is at least e; it also includes all generic units whose levels ℓ satisfy L log d v (α 1) K − ≥ log π | K | pℓ v (β) v (β) log d − K +1+ K p 1 p 1 ≥ log π − − | K | pℓ v (β) log d v (β) − K 1 K . p 1 ≥ log π − − p 1 − | K | − Using the exchange x y x y x y , ⌊ ⌋≥ ⇐⇒ ⌊ ⌋≥⌈ ⌉ ⇐⇒ ≥⌈ ⌉ valid for all real numbers x and y, we can get this into a form solvable for ℓ:

pℓ v (β) log d v (β) − K 1 K p 1 ≥ log π − − p 1 − | K | − v (β) p 1 log d v (β) ℓ K + − 1 K . ≥ p p log π − − p 1 | K | − So the range of [α] is , where Li v (β) p 1 log d v (β) K + − 1 K , (49) p p log π − − p 1 & | K | − ' as claimed in (f). For (b), we note that as d decreases from 1 to dmin, the corresponding i in (49) hits every value from 0 to e, since the argument to the outer ceiling increases by jumps of (p 1)/p < 1. So each i is a subgroup. For (d), we note that = H0(K,M) , while for 1 i e, − L |L0| | | ≤ ≤ / = U¯ /U¯ Li+1 Li ∼ j j+1 has pf = q elements, where j is the unique value of v (α 1) for values of α having level i. T1′ Finally, we have claimed a relation (g) regarding how level− spaces interact with the Tate pairing. In the case of the Hilbert pairing, the result we need is as follows:

50 Lemma 7.19 (an explicit reciprocity law). Let K be a local field with µ K. If α, β × satisfy p ⊆ ∈ OK p/(p 1) α 1 β 1 < d = p − , | − | · | − | min | | then the Hilbert pairing α, β vanishes. h iK Proof. This is a consequence of the conductor-discriminant formula (see Neukirch [39], VII.11.9): For a Galois extension L/K, Disc(L/K)= f(χ)χ(1), χ Y where χ ranges over the irreducible characters of Gal(L/K). Here we apply the formula to L = K[√p α]. Scale α by pth powers to be as close to 1 as possible. If α =1 or α is an intimate unit, the Hilbert pairing clearly vanishes since L is unramified and β is a unit. So we can assume that L is a ramified extension of degree p. Then there are p-many characters on Gal(L/K), all of dimension 1. One is the trivial character, whose conductor is 1. The others all have the same conductor f, so

p 1 Disc(L/K)= f − .

By Theorem 7.4, we have p p 1 p π − Disc(L/K) · K , ∼ (α 1)p 1 − − so f is generated by any element f with

p/(p 1) p − π f = · K . | | α 1

− p/(p 1) p Note that dmin = p − is actually an attainable norm of an element of K, namely (ζp 1) . By the given inequality, β | | 1 mod f which implies that the Hilbert symbol − ≡ α, β = φ (β) h iK L/K vanishes.

If 0 i e, α i(M), and β e i(M), then it is easy to verify that the hypothesis of Lemma 7.19 ≤ ≤ ∈ L ∈ L − holds in each ﬁeld factor of T [µm], in which the Hilbert pairing is being computed. Hence

i(M)⊥ e i(M ′). L ⊇ L − However, since

eK 0 0 1 i(M) e i(M ′) = q H (K,M) H (K,M ′) = H (K,M) , |L | · |L − | · | | · | | | | equality must hold.

7.4 The tame case If K is a tame local field, that is, char k = p, the structure of H1(K,M) is well known. We put K 6 1 1 e =0, 1 = 0 , 0 = Hur(K,M), 1 = H (K,M) L− { } L L and observe that Theorem 7.1(c), (d), (e), (g) and Corollary 7.2 still hold. The wild function field case K = Fpr ((t)) admits a similar treatment, but now the number of levels is infinite. We do not address this case here.

51 Part III Composed varieties

8 Composed varieties

It has long been noted that orbits of certain algebraic group actions on varieties over a field K parametrize rings of low rank over K, which can also be identified with the cohomology of small Galois modules over K. The aim of this section is to explain all this in a level of generality suitable for our applications. Definition 8.1. Let K be a field and K¯ its separable closure. A composed variety over K is a quasi-projective variety V over K with an action of a quasi-projective algebraic group Γ over K such that:

(a) V has a K-rational point x0;

(b) the K¯ -points of V consist of just one orbit Γ(K¯ )x0;

(c) the point stabilizer M = StabΓ(K¯ ) x0 is a ﬁnite abelian subgroup. The term composed is derived from Gauss composition of binary quadratic forms and the “higher composition laws” of the work of Bhargava and others, from which we derive many of our examples. Proposition 8.2.

(a) Once a base orbit Γ(K)x0 is fixed, there is a natural injection (ψ : Γ(K) V (K) ֒ H1(K,M \ → by which the orbits Γ(K) V (K) parametrize some subset of the Galois cohomology group H1(K,M). \ (b) The Γ(K)-stabilizer of every x V (K) is canonically isomorphic to H0(K,M). ∈ Proof. (a) Let x V (K) be given. Since there is only one Γ(K¯ )-orbit, we can find γ Γ(K¯ ) such that γ(x )= x. For∈ any g Gal(K/K¯ ), g(γ) also takes x to x and so differs from γ by right-multiplication∈ 0 ∈ 0 by an element in Stab ¯ x = M. Define a cocycle σ : Gal(K/K¯ ) M by Γ(K) 0 x → 1 σ (g)= g(γ) γ− . x · It is routine to verify that

• σx satisﬁes the cocycle condition σx(gh) = σx(g) g(σx(h)) and hence deﬁnes an element of H1(K,M); ·

• If a diﬀerent γ is chosen, then σx changes by a coboundary; • If x is replaced by αx for some α Γ , the cocycle σ is unchanged; ∈ K x • If the basepoint x0 is replaced by αx0 for some α Γ(K), the cocycle σx is unchanged, up to 1 ∈ identifying M with StabΓ(K¯ )(αx0)= αMα− in the obvious way. (This is why we can ﬁx merely a base orbit instead of a basepoint.) So we get a map ψ : Γ(K) V (K) H1(K,M). \ →

We claim that ψ is injective. Suppose that x1, x2 V (K) map to equivalent cocycles σx1 , σx2 . Let γ Γ(K¯ ) be the associated transformation that maps∈ x to x . By right-multiplying γ by an element i ∈ 0 i 1 of M, as above, we can remove any coboundary discrepancy and assume that σx1 = σx2 on the nose. That is, for every g Gal(K/K¯ ), ∈ 1 1 g(γ ) γ− = g(γ ) γ− , 1 · 1 2 · 2 which can also be written as 1 1 g(γ2γ1− )= γ2γ1− . 1 Thus, γ2γ1− is Galois stable and hence deﬁned over K. It takes x1 to x2, establishing that these points lie in the same Γ(K)-orbit, as desired.

52 1 (b) If γ(x0)= x, then the Γ(K¯ )-stabilizer of x is of course γMγ− . We claim that the obvious map

1 M γMγ− → 1 µ γµγ− 7→ is an isomorphism of Galois modules. We compute, for g Gal(K/K¯ ), ∈ 1 1 1 1 1 g γµγ− = g(γ)g(µ)g(γ)− = γσx(g)g(µ)σx(g)− γ− = γg(µ)γ− ,

0 establishing the isomorphism. In particular, the Galois-stable points StabΓ(K) x0 = H (K,M) are the same at x as at x0. Note the crucial way that we used that M is abelian. By the same token, the identiﬁcation of stabilizers is independent of γ and is thus canonical. The base orbit is distinguished only insofar as it corresponds to the zero element 0 H1(K,M). Changing base orbits changes the parametrization minimally: ∈

1 Proposition 8.3. The parametrizations ψx0 , ψx1 : Γ(K) V (K) H (K,M) corresponding to two basepoints x , x V (K) differ only by translation: \ → 0 1 ∈ ψ (x)= ψ (x) ψ (x ), x1 x0 − x0 1 under the isomorphism between the stabilizers M established in the previous proposition. Proof. Routine calculation. While ψ is always injective, it need not be surjective, as we will see by examples in the following section. Definition 8.4. (a) A composed variety is full if ψ is surjective, that is, it includes a Γ(K)-orbit for every cohomology class in H1(K,M). (b) If K is a global field, a composed variety is Hasse if for every α H1(K,M), if the localization α H1(K ,M) at each place v lies in the image of the local parametrization∈ v ∈ v ψ : V (K ) Γ(K ) H1(K ,M), v v \ v → v then α also lies in the image of the global parametrization ψ.

8.1 Examples In this section, K is any field not of one of finitely many bad characteristics for which the exposition does not make sense. Example 8.5. The group Γ= G can act on the variety V = A1 0 , the punctured affine line, by m \{ } λ(x)= λn x. ·

There is a unique K¯ -orbit. The point stabilizer is µn, and the parametrization corresponding to this composed variety (choosing basepoint x0 =1) is none other than the Kummer map

n 1 K×/(K×) H (K,µ ). → n That V is full follows from Hilbert’s Theorem 90.

Example 8.6. Let V be the variety of binary cubic forms f over K with ﬁxed discriminant D0. This has an algebraic action of SL2, which is transitive over K¯ (essentially because PSL2 carries any three points of P1 to any other three), and there is a ready-at-hand basepoint D f (X, Y )= X2Y Y 3. 0 − 4

53 √ The point stabilizer M is isomorphic to Z/3Z, but twisted by the character of K( D); that is, M ∼= 0, √D, √D as sets with Galois action. Coupled with the appropriate higher composition law (Theorem {6.9), this− recovers} the parametrization of cubic étale algebras with ﬁxed quadratic resolvent by H1(K,M) in Proposition 4.18. To see that it is the same parametrization, note that a γ Γ(K/K¯ ) that takes f to f ∈ 0 is determined by where it sends the rational root [1 : 0] of f0, so the three γ’s are permuted by Gal(K/K¯ ) just like the three roots of f. In particular, V is full. Example 8.7. Continuing with the sequence of known ring parametrizations, we might study the variety V of pairs of ternary quadratic forms with ﬁxed discriminant D0. This has one orbit over K¯ under the action of the group Γ = SL2 SL3; unfortunately, the point stabilizer is isomorphic to the alternating group A4, which is not abelian. × So we narrow the group, which widens the ring of invariants and requires us to take a smaller V . We let Γ=SL3 alone act on pairs (A, B) of ternary quadratic forms, which preserves the resolvent

g(X, Y )= 4det(AX + BY ) , a binary cubic form. We let V be the variety of (A, B) for which g = g0 is a fixed separable polynomial. These parametrize quartic étale algebras L over K whose cubic resolvent R is fixed. There is a natural base orbit (A ,B ) whose associated L = K R has a linear factor. The point stabilizer M = Z/2Z Z/2Z, with 0 0 ∼ × ∼ × the three non-identity elements permuted by Gal(K/K¯ ) in the same manner as the three roots of g0. We have reconstructed the parametrization of quartic étale algebras with fixed cubic resolvent by H1(K,M) in Proposition 4.18. In particular, V is full.

Example 8.8. Alternatively, we can consider the space V of binary quartic forms whose invariants I = I0, J = J0 are ﬁxed. The orbits of this space have been found useful for parametrizing 2-Selmer elements of the 2 3 elliptic curve E : y = x + xI + J, because the point stabilizer is M ∼= Z/2Z Z/2Z with the Galois-module structure E[2]. This space V embeds into the space of the preceding example× via a map which we call the Wood embedding after its prominent role in Wood’s work [59]:

f (A, B) 7→ 1/2 a b/2 c/3 ax4 + bx3y + cx2y2 + dxy3 + ey4 1 , b/2 c/3 d/2 . 7→ 1/2 −  c/3 d/2 e      In general, V is not full. For instance, over K = R, if E has full 2-torsion, there are only three kinds of binary quartics over R with positive discriminant (positive definite, negative definite, and those with four real roots) which cover three of the four elements in H1(R, Z/2Z). Two of these three (positive definite, four real roots) form the subgroup of elements whose corresponding E-torsor z2 = f(x, y) is soluble at : these 1 ∞ are the ones we retain when studying Sel2 E. The fourth element of H (R, Z/2Z) yields étale algebras whose (A, B) has 1/2 A = 1 ,  1/2 a conic with no real points. However, over global fields, it ispossible to show that V is Hasse, using the Hasse-Minkowski theorem for conics. Remark 8.9. Because of the extreme flexibility afforded by general varieties, it is reasonable to suppose that any finite K-Galois module M appears as the point stabilizer of some full composed variety over K. However, we do not pursue this question here.

8.2 Integral models; localization of orbit counts

Let K be a number field and K its ring of integers. Let (V, Γ) be a composed variety, and let ( , ) be an integral model, that is, a pairO of a flat separated scheme and a flat algebraic group over actingV G on it, OK equipped with an identification of the generic fiber with (V, Γ). Then ( K ) ֒ Γ(K), and the Γ(K)-orbits on V (K) decompose into ( )-orbits. G O → G OK

54 Lemma 8.10 (localization of global class numbers). Let ( , ) be an integral model for a composed variety (V, Γ). For each place v, let V G w : ( ) ( ) C v G Ov \V Ov → be a function on the local orbits, which we call a local weighting. Suppose that: (i) (V, Γ) is Hasse. (ii) has class number one, that is, the natural localization embedding G ( Γ(K) ֒ ( ) Γ(K ( ) G OK \ → G Ov \ v v M is surjective.

(iii) For each place v, there are only ﬁnitely many orbits of ( v) on ( v). This ensures that the weighted local orbit counter G O V O

g : H1(K ,M) C v,wv v → α w (γx ) 7→ v α ( Kv )γ ( v) Γ(Kv) GsuchO that∈GXγxO \ ( ) α∈V Ov

takes ﬁnite values. (Here xα is a representative of the Γ(Kv)-orbit corresponding to α. If there is no

such orbit because V is not full, we take gv,wv (α)=0.)

(iv) For almost all v, ( v) ( v) consists of at most one orbit in each Γ(Kv)-orbit, and wv =1 identically. G O \V O Then the global integral points ( ) consist of ﬁnitely many ( )-orbits, and the global weighted orbit V OK G OK count can be expressed in terms of the gv,wv by

v wv(x) 1 h w := = gv,wv (α). (50) { v } 0 Stab ( K ) x H (K,M) 1 ( K )x ( K ) ( K ) Q G O α H (K,M) v G O ∈GXO \V O | | | | ∈ X Y Proof. Grouping the ( )-orbits into Γ(K)-orbits, it suﬃces to prove that for all α H1(K,M), G OK ∈

v wv(x) 1 = 0 gv,wv (α). (51) Stab ( K ) x H (K,M) ( K )x Γ(K)xα Q G O v G O X⊆ | | | | Y

If there is no xα, the left-hand side is zero by definition, and at least one of the gv,wv (α) is also zero since V is Hasse. So we fix an xα. The right-hand side of (51), which is finite by hypothesis (iv) since α is unramified almost everywhere, can be written as 1 w (γ x ), H0(K,M) v v α ( )γ v | | {G OXv v}v Y the sum being over systems of γ Γ(K ) such that γ x is -integral. Since has class number one, each v ∈ v v α Ov G such system glues uniquely to a global orbit ( K )γ,γ Γ(K), for which γxα is v-integral for all v, that is, -integral. Thus the right-hand side of (51)G O is now∈ transformed to O OK 1 w (γ x ). H0(K,M) v v α | | ( K )γ v γxG XO ( ) Y α∈V OK Now each γ corresponds to a term of the left-hand side of (51) under the map

( ) Γ(K) ( ) V (K) G OK \ → G OK \ ( )γ ( )γx . G OK 7→ G OK α

55 The ﬁber of each ( )x has size G OK H0(K,M) [StabΓ(K) x : Stab ( K ) x]= | | . G O Stab ( ) x | G OK | So we match up one term of the left-hand side, having value

wv(x)/ Stab ( ) x , | G OK | v Y 0 with H (K,M) / Stab ( K ) x -many elements on the right-hand side. In view of the outlying factor 1/ H0|(K,M) , this| | completesG O the| proof. | | 8.3 Fourier analysis of the local and global Tate pairings We now introduce the main innovative technique of this thesis: Fourier analysis of local and global Tate duality. In structure we are indebted to Tate’s celebrated thesis [55], in which he

1. constructs a perfect pairing on the additive group of a local ﬁeld K, taking values in the unit circle CN=1, and thus furnishing a notion of Fourier transform for C-valued L1 functions on K;

2. derives thereby a pairing and Fourier transform on the adele group AK of a global ﬁeld K;

3. proves that the discrete subgroup K AK is a self-dual lattice and that the Poisson summation formula ⊆ f(x)= fˆ(x) (52) x K x K X∈ X∈ holds for all f satisfying reasonable integrability conditions. In this paper, we work not with the additive group K but with a Galois cohomology group H1(K,M). The needed theoretical result is Poitou-Tate duality, a nine-term exact sequence of which the middle three terms are of main interest to us:

1 ′ 1 1 (ﬁnite kernel) H (K,M) H (K ,M) H (K,M ′)∨ (ﬁnite cokernel). → → v → → v M 1 1 This can be interpreted as saying that H (K,M) and H (K,M ′) (where M ′ = Hom(M,µ) is the Tate dual) map to dual lattices in the respective adelic cohomology groups

1 ′ 1 1 ′ 1 H (AK ,M)= H (Kv,M) and H (AK ,M ′)= H (Kv,M ′), v v M M which are mutually dual under the product of the local Tate pairings

α , β = α ,β µ. h{ v} { v}i h v vi ∈ v Y Here, for K a local ﬁeld, the local Tate pairing is given by the cup product

1 1 2 , : H (K,M) H (K,M ′) H (K,µ) = µ. h• •i × → ∼ It is well known that this pairing is perfect. (The Brauer group H2(K,µ) is usually described as being Q/Z but, having no need for a Galois action on it, we identify it with µ to avoid the need to write an exponential 1 in the Fourier transform.) Now, for any suﬃciently nice function f : H (AK ,M) C (locally constant and compactly supported is more than enough), we have Poisson summation →

f(α)= cM fˆ(β) α H1(K,M) β H1(K,M ) ∈ X ∈ X ′

56 1 for some constant cM which we think of as the covolume of H (K,M) as a lattice in the adelic cohomology. (In fact, by examining the preceding term in the Poitou-Tate sequence, H1(K,M) need not inject into 1 H (AK ,M), but maps in with finite kernel; but this subtlety can be absorbed into the constant cM .) We apply Poisson summation to the local orbit counters gv defined in the preceding subsection and get a very general reflection theorem. Definition 8.11. Let K be a local field. Let (V (1), Γ(1)) and (V (2), Γ(2)) be a pair of composed varieties over K whose associated point stabilizers M (1), M (2) are Tate duals of one another, and let ( (i), (i)) be an integral model of (V (i), Γ(i)). Two weightings on orbits V G

w(i) : ( ) ( ) C G OK \V OK → are called (mutually) dual with duality constant c Q if their local orbit counters gw(i) are mutual Fourier transforms: ∈ g(2) = c gˆ(1). (53) · where the Fourier transform is scaled by 1 fˆ(β)= f(α). H0(K,M) α H1 (K,M) ∈ X An equation of the form (53) is called a local reflection theorem. If the constant weightings w(i) = 1 are mutually dual, we say that the two integral models ( (i), (i)) are naturally dual. V G Theorem 8.12 (local-to-global reflection engine). Let K be a number field. Let (V (1), Γ(1)) and (V (2), Γ(2)) be a pair of composed varieties over K whose associated point stabilizers M (1), M (2) are Tate duals of one another. Let ( (i), (i)) be an integral model for each (V (i), Γ(i)), and let V G w(i) : (i)( ) (i)( ) C v G Ov \V Ov → be a local weighting on each integral model. Suppose that each integral model and local weighting satisfies the hypotheses of Lemma 8.10, and suppose that at each place v, the two integral models are dual with some duality constant c Q. Then the weighted global class numbers are in a simple ratio: v ∈

h (2) = cv h (1) . wv · wv v n o Y n o Proof. By Lemma 8.10, 1 h (i) = g (i) (α). wv H0(K,M (i)) v,wv α H1(K,M (i) ) v n o | | ∈ X Y At almost all v, each g (i) is supported on the unramiﬁed cohomology, and must be constant there because v,wv otherwise its Fourier transform would not be supported on the unramiﬁed cohomology. However, g (i) v,wv cannot be identically 0 because of the existance of a global basepoint. So for such v,

g (i) = 1 1 (i) and cv =1. v,wv Hur(K,M )

1 In particular, the product g (i) is a locally constant, compactly supported function on H (K, AK ), v v,wv which is more than enough for Poisson summation to be valid. Q 1 (i) Since the pairing between the adelic cohomology groups H (AK ,M ) is made by multiplying the local Tate pairings, a product of local factors has a Fourier transform with a corresponding product expansion:

\ g (1) = gˆ (1) = cv g (2) . v,wv v,wv · v,wv v v v v Y Y Y Y We then apply Poisson summation to get a formula for the ratio of the global weighted class numbers:

H0(K,M (1)) h (2) = | | cM (1) cv h (1) . wv H0(K,M (2)) · · wv v n o | | Y n o 57 This gives the desired identity, except for determining the scalar cM , which depends only on the Galois module (1) 1 M = M . This can be ascertained by applying Poisson summation to just one function f : H (AK ,M) C for which either side is nonzero. The easiest such f to think of is the characteristic function of a compact→ open box

X = Xv, v Y 1 with Xv = H (K,M) for almost all v. Such a speciﬁcation is often called a Selmer system, and the sum

1α X v v ∈ v∀ α H1(K,M) ∈ X is the order of the Selmer group Sel(X) of global cohomology classes obeying the speciﬁed local conditions. Poisson summation becomes a formula for the ratio Sel(X) / Sel(X⊥) as a product of local factors, commonly known as the Greenberg-Wiles formula. By appealing| to any| | of the known| proofs of the Greenberg- Wiles formula (see Darmon, Diamond, and Taylor [17, Theorem 2.19] or Jorza [29, Theorem 3.11]), we pin down the value 0 H (K,M ′) c = | |. M H0(K,M) | | At certain points in this paper, it will be to our advantage to consider multiple integral models at once. The following theorem has suﬃcient generality.

Theorem 8.13 (local-to-global reﬂection engine: general version). Let K be a number ﬁeld. Let (V (1), Γ(1)) and (V (2), Γ(2)) be a pair of composed varieties over K whose associated point stabilizers M (1), M (2) are Tate duals of one another. For each place v of K, let

(i) (i) (i) ( , )j : jv J Vjv G v ∈ v n o (i) (i) be a family of integral models for each (V indexed by some ﬁnite set Jv , and let

(i) (i) (i) w : ( v) ( v) C jv G O \V O → be a weighting on the orbits of each integral model. Similarly to Lemma 8.10 and Theorem 8.12, assume that (i) (V, Γ) is Hasse.

(ii) For each combination of indices j = (j ) , j J (i), the local integral models ( (i), (i)) glue together v v v v jv jv (i) (i) ∈ V G to form a global integral model ( j , j ). (Since the integral models are equipped with embeddings (i) (i) V G jv Vv , the gluing is seen to be unique; and its existence will be obvious in all the examples we consider.)V →

(iii) Each such (i) has class number one. Gj (i) (i) (iv) For each jv, there are only ﬁnitely many orbits of on , ensuring that the local orbit counter Gjv Vjv gjv,wjv takes ﬁnite values.

(v) For almost every v, the index set Jv = jv has just one element, with the corresponding integral model (i) { } (i) consisting of at most one orbit in each Γ(Kv)-orbit, and w =1 identically. Vjv jv (vi) At each v, we have a local reﬂection theorem

gˆjv ,wjv = gjv ,wjv . j J(1) j J(2) vX∈ v vX∈ v

58 (i) (i) (i) (i) Then the class numbers of the global integral models ( j , j ) with respect to the weightings wj = v wjv satisfy global reflection: V G Q h (1), w(1) = h (2), w(2) . Vj j Vj j j J(1) j J(2) ∈ Xv v ∈ Xv v Proof. Except for complexitiesQ of notation, the proof closelyQ follows the preceding one. The first five hypotheses ensure that each global integral model (i), (i) satisfies the hypotheses of Lemma 8.10, so its Vj Gj class number is representable as a sum over the lattice of glo bal points in adelic cohomology: 1 h (i), w(i) = g (α). Vj j H0(K,M (i)) jv ,wjv α H1 (K,M (i)) v | | ∈ X Y When we sum over all j, the contributions of each α factor to give 1 h (i), w(i) = g (α). Vj j H0(K,M (i)) jv,wjv j J(i) | | α H1(K,M (i) ) v j J(i) ∈ Xv v ∈ X Y vX∈ v Q But by the assumed local reflection identity, we have

g = g = g .  jv ,wjv   jv ,wjv  jv,wjv v (1) b v (1) b v (2) Y jvXJv Y jvXJv Y jvXJv  ∈   ∈  ∈ So we get the desired identity from Poisson summation. The scale factor was determined in proving the previous theorem.

Remark 8.14. Unlike in the previous theorem, we have not included duality constants cv, but the same eﬀect (i) can be obtained by taking the appropriate constant for the weighting wj .

8.3.1 Examples As one might guess, there are many pairs of composed varieties whose point stabilizers M (1), M (2) are Tate duals; and, given any integral models, it is usually possible to concoct weights w(i) that are mutually dual, thereby getting reflection theorems from Theorem 8.12. More noteworthy is when a pair of integral models are naturally dual at all finite places. Even more significant is if a group acts on a large variety Λ, leaving certain functions I on Λ invariant, such that every level set of I is an integralG model for a composed variety with natural duality. This is the case for O-N. We have found three families of naturally dual composed varieties of this sort: ΓΛ I Parametrizes M

λ t Quadratic forms, 2 2 GL2 2 a(b 4ac) ? 2 0 λ− ⊂ Sym (2) − C Cubic rings / Cubic forms, SL Discriminant 3-torsion in 2 Sym3(2) 3 quadratic rings C Pairs of ternary Quartic rings / SL3 quadratic forms, Cubic resolvent 2-torsion in 2 2 2 2 C ×C Sym (3)⊕ cubic rings These three representations will be considered in detail in Section 9, Part V, and Parts VI–VII, respectively. In each case, there is a local reflection that pairs two integral models of V over which look alike over K. OK Remark 8.15. In the latter two cases, the integral models are dual under an identification of V with its dual V ∗ (which are isomorphic, up to an outer automorphism of Γ=SL3 in the last case). But in the quadratic 2 case, V ∗ decomposes into K¯ -orbits according to a different invariant J = a/∆ , and the integral orbit counts are infinite, so the alignment with duals in the classical sense must be considered at least partly coincidental.

59 4 Closely related to the quartic rings example is the action of SL2 on binary quartic forms Sym (2). Here, 1 the orbits are parametrized by a subset of a cohomology group H (K,M) (M ∼= 2 2 as a group) cut out by a quadratic relation. Nevertheless, we will state some interesting reflection identitiesC ×C for these spaces in Section 13.1. More generally, we can consider the space Λ of pairs (A, B) of n-ary quadratic forms, on which Γ=SLn acts preserving a binary n-ic resolvent I = det(Ax By). − Although we do not consider it in this paper, preliminary investigations suggest that its integral models are naturally dual to one another for n odd, yielding a corresponding global reflection theorem (Conjecture 2.13). This composed variety figures prominently in the study of Selmer elements of hyperelliptic curves [6]. On the other hand, the following families of composed varieties do not admit natural duality:

• The action of Gm on the punctured affine line by multiplication by nth powers. The orbits do 1 n parametrize H (K,µm) = K×/(K×) . But over a local or global field, there are infinitely many integral orbits in each rational orbit. 2 2 • The action of SO2 (the group of rotations preserving the quadratic form x + xy + y ) on binary cubic forms of the shape f(x, y)= ax3 + bx2y + ( 3a + b)xy2 + ay3 − which is symmetric under the threefold shift x y, y x y. This representation is used by Bhargava and Shnidman [9] to parametrize cyclic7→cubic7→ rings, − that− is, those with an automorphism of order 3. The reason for failure of natural duality is quite simple. Within the representation over Z for p 1 mod3, take the composed variety where the discriminant is p2. The cohomology group p ≡ H1(Q , Z/3Z) is isomorphic to Z/3Z Z/3Z, and a function f on it may be written as a matrix p × f(0) f(α) f(2α) f(β) f(α + β) f(2α + β) f(2β) f(α +2β) f(2α +2β) in which the zero-element and the unramified cohomology α are marked off by dividers. h i The six ramified cohomology elements each have one integral orbit, corresponding to the maximal 3 order; the three unramified cohomology elements—the zero element for Qp, and the other two for the degree-3 unramified field extension in its two orientations—all have no integral orbits, because the three orders (x , x , x ) Z3 : x x mod p { 1 2 3 ∈ p i ≡ j } 3 are all asymmetric under the threefold automorphism of Zp. So we get a local orbit counter 0 0 0 1 1 1 1 1 1 whose Fourier transform 2 1 1 − − 0 0 0 0 0 0 has mixed signs and thus cannot be the local orbit counter of any composed variety. Similar obstruc- tions to natural duality have obtained in many of the composed varieties parametrizing rings with automorphisms found by Gundlach [25].

Part IV Reﬂection theorems: ﬁrst examples

The remainder of this thesis will be devoted to stating and proving explicit reﬂection theorems for various objects of interest.

60 9 Quadratic forms by superdiscriminant

We begin with the simplest Galois module M ∼= Z/2Z. There are many full composed varieties whose point stabilizer is of order 2, and the one we take is, to say the least, one of the more unexpected. The group GL2 acts on the space

V = Sym2(2) = ax2 + bxy + cy2 : a,b,c G { ∈ a} of binary quadratic forms in the natural way. Let Γ be the algebraic subgroup, deﬁned over Z, of elements of a peculiar form: u t 2 : u Gm,t Ga . 0 u− ∈ ∈ Abstractly, this group is a certain semidirect product of Ga by Gm. As is not too hard to verify, the restriction of Λ to Γ has a single polynomial invariant, the superdiscriminant

I := aD = a(b2 4ac). − Because of the asymmetry between x and y, there is no harm in writing forms in V inhomogeneously as f(x)= ax2 + bx + c, as was done in Section 2. Then the variety V (I)= f V : I(f)= I { ∈ } is full composed. We take the basepoint 1 f = Ix2 + 0 4I 2 2 of discriminant 1. Then the rational orbits are parametrized by D = b 4ac K×/(K×) consistent with the parametrization of their splitting fields via Kummer theory. − ∈ Remark 9.1. The group Γ is not reductive, that is, does fit into the classical Dynkin-diagram parametrization for Lie groups. Non-reductive groups are decidedly in the minority within the whole context of using orbits to parametrize arithmetic objects, but they have occurred before: Altuğ, Shankar, Varma, and Wilson [1] count D4-fields using orbits of pairs of ternary quadratic forms under a certain nonreductive subgroup of GL SL . 2 × 3 Now we introduce integral models. Suppose K K is a PID with field of fractions K. If τ K divides 2, then O ⊆ ∈ O V = ax2 + bx + c : a,c , τ b τ { ∈ OK | } is a GL2( K )-invariant lattice in V . For any I K , we can take ( , ) = (Vτ (I), Γ( K )) as an integral model forOV (I). For it to have any integral points,∈ O we must have τ 2 VI. G O Our first local reflection theorem says that each of these integral| models has a natural dual.

Theorem 9.2 (“Local Quadratic O-N”). Let K be a non-archimedean local ﬁeld, char K = 2. For I, τ elements dividing 2, the integral models 6 ∈ OK 4 1 Vτ (I) and V2τ − 4τ − I are naturally dual with scale factor the absolute norm N(τ) = K /τ K . In other words, the local orbit counters are related by |O O |

4 gˆVτ (I) = N(τ) gV 1 (4τ − I). (54) · 2τ−

Proof. We prove this result by explicitly computing the local orbit counter gVτ(I) , which sends each [D] 2 1 ∈ K×/(K×) = H (K, Z/2Z) to the number of cosets [γ] Γ( ) Γ(K) such that γv V (I)( ), where ∼ ∈ OK \ 0 ∈ τ OK v0 is an arbitrary vector in V (K) with I(v0)= I and D(v0)= D. Let t = v(τ) and e = v(2); we have e> 0 exactly when K is 2-adic, and 0 t e. ≤ ≤

61 A coset [γ] is speciﬁed by two pieces of information. First is the valuation v(u) of the diagonal elements; this is equivalent to specifying v(D) and v(a), where, as is natural we set

γv = ax2 + bxy + cy2 and D = b2 4ac. 0 − Second, we specify t modulo the appropriate integral sublattice. If (as we may assume) v(u)=0, then t is deﬁned modulo 1, which is the same as specifying b modulo 2a. So the problem of computing gVτ (I) devolves onto computing how many b τ , up to translation by 2a, yield an integral value for ∈ OK b2 D c = − ; 4a that is, we must solve the quadratic congruence

b2 D mod 4a. (55) ≡ The answer, in general, depends on how close D is to being a square in K. So we will express our answer in terms of the level spaces introduced in Theorem 7.1. Here the level of a coclass [α], α K×, is deﬁned in terms of the discriminant of K[√α], which, by Theorem 7.3, can be computed from the minimal∈ distance α 1 , over all rescalings of α by squares. The level spaces thus correspond to the natural ﬁltration of | − | 2 K×/(K×) by neighborhoods of 1:

2 K×/(K×) , i = 1 2 2e 2i − i = [α] × /( × ) : α 1+ π − , 0 i e L  { ∈ OK OK ∈ OK } ≤ ≤  1 , i = e +1. { } Let L be the characteristic function of . By Corollary 7.2, the Fourier transform of each L is a scalar i Li i multiple of Le i. We now claim− that, if we ﬁx v(I) and v(a) (and hence v(D)), then the contribution of all solutions of (55) to gτ,I can be expressed as a linear combination of the Li. The basic idea, which will be a recurring one, is to 2 group the solutions into families that have a constant number of solutions over some subset S K×/(K×) . The subset S will be called the support of the family, and the number of solutions for each ⊆D S will be called the thickness of the family. ∈ If v(D) v(4a), then (55) simpliﬁes to 4a b2, that is, ≥ | 1 v(b) e + v(a) . ≥ 2 Since we are counting values of b modulo 2a, the number of solutions is simply

(e+v(a)) e 1 v(a) 1 v(a) q −⌈ − 2 ⌉ = q⌊ 2 ⌋.

We get a family with this thickness, supported on either 0 or 1 0 according as v(D) is even or odd. If v(D) < v(4a), then b2 must be actually able to cancelL L at− least\ L the leading term of D to get any v(D) 1 v(D) solutions. In particular, v(D) must be even. Let D˜ = D/π , and let ˜b = b/π 2 , so ˜b must be a unit satisfying 4a ˜b2 D˜ mod . (56) ≡ D Let m = v(4a/D). If m 2e +1, a unit is a square modulo πm only if it is a square outright, so we get a ≥ 2 family supported just on the trivial class 1 K×/(K×) . Otherwise, we have 1 m 2e, and the support ∈ ˜ ≤ ≤ is L m/2 . The corresponding thicknesses are easy to compute. The b satisfying (56) form a ﬁber of the group⌈ homomorphism⌉ 2 v(2a) 1 v(D) × v(4a) v(D) × φ = : /π − 2 /π − , • OK → OK

62 and the cokernel of this homomorphism has size [ : ], so the thickness is L0 Li

v(2a) 1 v(D) × /π − 2 OK ker φ = [ 0 : i] | | L L · v(4a) v(D) K /π × O − 1 v(2a) 1 v(D) 1 q − 2 − q = [ 0 : i] L L · 1 1 qv(4a) v(D) − q − 1 v(D) e = [ : ] q 2 − L0 Li · 1 v(D) e+ m/2 v(a)/2 q 2 − ⌈ ⌉ = q⌊ ⌋, 1 m 2e = 1 v(D) ≤ ≤ 2q 2 , m 2e +1. ( ≥ We have not mentioned the condition b (τ), because it is equivalent to v(D) 2v(τ), and eliminates some families, leaving the others intact. ∈ ≥ By way of illustration, we tabulate the contributions to gτ,I in the example where e = 2. It is already easy to check many examples of Theorem 9.2.

v(a); v(D) 01234 5 6 7 8 ↓ → 0 L2 L1 L0 L 1 L0 L0 L 1 L0 L0 − − − − 1 L3 L1 L0 L 1 L0 L0 L 1 L0 L0 − − − − 2 L3 qL2 qL1 qL0 q (L 1 L0) qL0 − − 3 L3 qL3 qL1 qL0 q (L 1 L0) qL0 2 2 − − 2 4 L3 qL3 q L2 q L1 q L0

We have shown the subdivision of the table into three zones given by the inequalities: • Zone I: v(D) v(4a) ≥ • Zone II: v(a) < v(D) v(4a) ≤ • Zone III: v(a) > v(D). (A more general deﬁnition of a zone will be given later.) In general, the shapes of these zones, together with the needed condition v(D) 2t, will look as follows: ≥ 2t 2e v✲(D) 0 ❅ Zone I 2t ❅ ❅ Zone II ❅ ❅ ❅ Zone III❅ ❅ ❅ ❅ ❅ v(a) ❄ ❅

The feature to be noted is that, under the transformation t e t, the shape of Zone II is flipped about a diagonal line and Zones I and III are interchanged. This will7→be the− basis for our proof of Theorem 9.2; but there will be irregularities owing to the floor functions in the formulas and the fact that 1 0, instead of L− \ L 1, appears as a support. L− There are two ways to finish the proof. One is to establish a bijection of families, as outlined in the previous paragraph, so that Li and Le i are interchanged as supports and all the thicknesses correspond appropriately. Such an approach will be− used for cubic O-N in Section 11.3. The other is to verify the local reflection computationally, by means of a generating function. We choose the second, admittedly less elegant, method, mainly because it shows, in a context simple enough to be worked by hand, transformations that we will relegate to a computer in the succeeding sections.

63 Let n F (Z)= gτ,πn Z , n 0 X≥ 2 a formal power series whose coeﬃcients are functions of D K×/(K×) . We write F = F + F + F , ∈ I II III where FX is the contribution coming from Zone X in the preceding analysis. Writing i = v(a) and d = v(D), we proceed to compute

i/2 i+d q⌊ ⌋L0Z , d even FI = q i/2 (L L )Zi+d, d odd i 0 d 2e+i+1 ( ⌊ ⌋ 1 0 X≥ ≥ X − − q i/2 L Zi+d, d even = Z2e ⌊ ⌋ 0 q i/2 (L L )Zi+d, d odd. i 0 d i+1 ( ⌊ ⌋ 1 0 X≥ ≥X − − Splitting i =2i + i , where 0 i 1, and likewise d =2d + d , we get f p ≤ p ≤ f p 1 2e if d d+2if +ip if 2df +2if +ip+1 FI = Z q ( 1) L0Z + q L 1Z  − −  ip=0 i 0 d 2i +i +1 d i X Xf ≥ ≥ Xf p Xf ≥ f 1  if ip 4if +2ip+1 if 4if +ip+1  2e q ( 1) L0Z q Z L 1 = Z − + − 1+ Z 1 Z2 ip=0 if 0 X X≥ − 1 ip 2ip ip+1 2e ( 1) Z L0 Z L 1 = Z − + − (1 + Z)(1 qZ4) (1 Z2)(1 qZ4) i =0 Xp − − − 2 2e (1 Z ) Z(1 + Z) = Z − L0 + L 1 (1 + Z)(1 qZ4) (1 Z2)(1 qZ4) − − − − Z2e(1 Z) Z2e+1 = − L0 + L 1. 1 qZ4 (1 Z)(1 qZ4) − − − − For Zone II, which appears only when e> 0, the most sensible way to evaluate the sum

i d e+ 2 2 i+d FII = q ⌊ ⌋− Le+ i d Z 2 − 2 i 0 i d

i d is to group terms with the same level Lj. We have j = e + 2 2 , so the values of i and j determine d. The condition d 2t reduces to i 2(j + t e); the other condition− i d

i/2 i+2(e+ i 2j) q⌊ ⌋Lj Z ⌊ 2 ⌋− i max 0,2(j+t e) ≥ {X − } 1 if 4if +ip+2e 2j = q Z − Lj

ip=0 i max 0,j+t e X f ≥ X{ − } 1 i 4 if 2e 2j = Z p qZ Z − L     j ip=0 i max 0,j+t e X f ≥ X{ − }   max 0,j+t e  4 { − } (1 + Z) qZ 2e 2j = Z − L . 1 qZ4 · j −

64 For j =0 and j = e, since ip can only take one of its two values, the initial factor 1+ Z is to be replaced by Z and 1 respectively. Finally, Zone III presents no particular diﬃculties:

d/2 i+d FIII = 2q Le+1Z d 2t i d+1 dX≥even ≥X Z2d+1 =2 qd/2 L · 1 Z · e+1 d 2t − dX≥even 2qsZ4t+1 = L . (1 Z)(1 qZ4) e+1 − − Summing up, we get for e 1 (the case e =0 can be handled similarly) ≥

F = FI + FII + FIII max 0,j+t e 2e+1 2e 2e+1 4 { − } Z Z (1 Z) Z (1 + Z) qZ 2e 2j = L 1 + − L0 + L0 + Z − Lj (1 Z)(1 qZ4) − 1 qZ4 1 qZ4 1 qZ4 · 1 j e 1 − − − − ≤X≤ − − t qZ4 2qsZ4t+1 + L + L 1 qZ4 e (1 Z)(1 qZ4) e+1 − − − max 0,j+t e 2e+1 2e 4 { − } Z Z (1 + Z) qZ 2e 2j = L 1 + L0 + Z − Lj (1 Z)(1 qZ4) − 1 qZ4 1 qZ4 · 1 j e 1 − − − ≤X≤ − − t t qZ4 2Z qZ4 + L + L . 1 qZ4 e (1 Z)(1 qZ4) e+1 − − −

Now the evident symmetry between the coeﬃcients of Lj and Le j , when the transformation t e t is made, establishes the theorem. − 7→ − Inserting this into the machinery of Part III produces global reﬂection theorems:

Theorem 9.3 (“Quadratic O-N”). Let K be a number ﬁeld of class number 1. Then for any I, τ K with τ 2, ∈ O | 1 N (τ) 1 = | K/Q | r2(K) StabΓ( K )f 2 4 StabΓ( K )f f Γ( K ) Vτ (I)( K ) | O | f Γ( K ) V2τ 1 (4τ − I)( K ) | O | disc f>∈ 0OatX every\ realO place ∈ O \ X− O where r2(K) is the number of complex places of K.

4 1 Proof. We verify the hypotheses of Lemma 8.10 on the integral models Vτ (I) and V2τ − (4τ − I): (i) V (I) is Hasse because it is full, as previously noted.

(ii) To check that Γ has class number 1, it suﬃces to check the factors Gm and Ga of which Γ is a semidirect product. The former of these has the same class number as K, explaining the restriction in the theorem statement.

(iii) The ﬁniteness of the local orbit counter follows from the formulas for it computed in the previous theorem.

(iv) Finally, at almost all places, we plug in e = t =0 to get F = L0, establishing the needed convergence.

Now we need the local reﬂection itself. We keep track of the constants cv accrued: • If v ∤ 2 , the integral models are naturally dual with constant c =1. ∞ v

65 • If v 2, the integral models are naturally dual with constant cv = [ v : τ v ]. Multiplying over all v 2 and| using that τ 2 yields a factor O O | | c = [ : τ ]= N τ . v OK OK | K/Q | v 2 Y| • If v is real, the integral models are no longer naturally dual at v. We place the non-natural weighting

(2) w = 10

that picks out α H1(K, Z/2Z) that vanish at v, that is, forms with positive discriminant at v. This ∈ (1) is the Fourier transform of w =1, so cv =1. • Finally, if v is complex, then the integral models are certainly naturally dual at v, because H1 = 1. | | However, the scaling of the Fourier transform by 1/ H0(C,M) =1/2 requires that we take c =1/2. | | v Multiplying these constants gives the constant claimed.

Remark 9.4. The condition that be a PID can be dropped, but then Γ no longer has class number 1, OK and each side of the theorem becomes a sum of orbit counts on Cl( K )-many global integral models that locally look alike. We do not spell out the details here. We wonder whetherO such a method works in general to circumvent the class-number-1 hypothesis in Theorem 8.12. We conclude by specializing further to the case K = Q. We replace Γ(Z) by its index-2 subgroup, the group Z of translations. This merely doubles all orbit counts, and it acts freely on quadratics with nonzero discriminant, so we can suppress all mention of stabilizers for the following charmingly simple statement, also featured in Section 2:

Theorem 9.5 (“Quadratic O-N”). If n is a nonzero integer, let q(n) be the number of integer quadratic polynomials f(x)= ax2 + bx + c with a(b2 4ac)= n, − + + up to the trivial change x x + t (t Z). Let q2(n), q (n), and q2 (n), respectively, be the number of these 7→ ∈ + + f such that 2 b (for q2), such that the roots of f are real (for q ), or which satisfy both conditions (for q2 ). Then for all nonzero| integers n,

+ q2 (4n)= q(n) + q2(4n)=2q (n).

Example 9.6. Looking at n = p1p3, where p1 1 (mod 4) and p3 3 (mod 4) are primes, the counts ≡ ≡ 2 involve certain Legendre symbols. For instance, the combination a = p3, b 4ac = p1 is feasible if and only if the congruence − b2 p mod 4p ≡ 1 3 has a solution, which happens exactly when p1 =1. Working out all cases, we ﬁnd that p3 p p q+(p p )=5+ 1 and q (4p p )=10+2 3 . 1 3 p 2 1 3 p 3 1 Thus our reﬂection theorem recovers the quadratic reciprocity law

p p 1 = 3 . p p 3 1 We wonder: does there exist a proof of Theorem 9.5 using no tools more advanced than quadratic reciprocity?

66 10 Class groups: generalizations of the Scholz and Leopoldt reﬂec- tion theorems

We now return to the consideration with which we began: reflection theorems for class groups. Scholz [49] proved a relation between the 3-torsion in the class groups of Q(√D) and Q(√ 3D). Leopoldt [31] − significantly generalized this result. We here present a generalization of Leopoldt’s result to orders in (Fp)- extensions, where exact formulas (as opposed to bounds) can often be obtained. We will not use composedGA varieties; instead, we will use Poisson summation in the form of the Greenberg-Wiles formula to get reflection theorems. Let T/K be a (Z/pZ)×-torsor. If T is a Galois-invariant K -order, then (Z/pZ)× acts on the class group Cl( ). The p-primary partOCl( ⊆ ) is broken up into eigenspaces,O one for each character χ : O O p (Z/pZ)× µp 1 Zp×. There is a distinguished character χ = ω lifting the reduction map modulo p (the → − ⊆ Teichmüller lift). We will concern ourselves with the ω-part Cl( )p,ω. (The remaining parts are related to the ω-parts of the class groups of other torsors.) We look at theOp-torsion, or equivalently the p-cotorsion:

Cl( )[p] = (Cl( )/p Cl( )) . O ω ∼ O O ω 10.1 Dual orders

We now develop a condition on two orders 1 T , 2 T ′ that will suﬃce to produce a reﬂection theorem between their class groups. First, a simpleO lemma:⊆ O ⊆

Lemma 10.1. Let L be an étale algebra over a number field K, let L be an order, and let M/L be a G-torsor. The following conditions are equivalent: O ⊆ (a) M is a ring class algebra for ; that is, the global Artin map O ψ = ψ : (L, m) G M/L Mi/Li I → i Y factors through Cl( ), where m is an admissible modulus for M/L, (L, m) is the group of O ⊆ OK I invertible fractional ideals of L prime to m, and the product runs through all field factors Mi of M, with Li being the corresponding field factor of L; (b) For every valuation v of K, the local Artin map

φ = φ : L× G Mv /Lv Mu/Lw v → u w v Y| |

vanishes on ×. Ov Proof. By local-global compatibility, the global Artin map can be described idelically as the product of the local ones. Indeed, (L, m) embeds into A×/ × , and I L v∤m OLv Q ψ = φ : A× G. M/L Mv /Lv L → v Y Now the idele-theoretic description of Cl( ) is O

Cl( )= A×/ L× × . O L · Ov v ! Y

Since ψ always vanishes on the principal ideles L×, it factors through Cl( ) if and only if it vanishes M/L O on each v× (v a place of K), where it reduces to the product φMv /Lv of the local Artin maps at the primes dividingOv, as desired.

67 There is an analogue for narrow ring class algebras: here φMv /Lv is required to vanish on v× for v finite only. O This motivates the following definitions. Definition 10.2.

(a) Let K be a local ﬁeld, T/K a (Z/pZ)×-torsor and T ′ its Tate dual. Two (Z/pZ)×-invariant orders T , T ′ are called dual if the ω-parts of the multiplicative groups, ( ×) and ( ×) , are O1 ⊆ O2 ⊆ O1 ω O2 ω orthogonal complements under the Hilbert pairing, which as we know is perfect between (T ×)ω and (T ′×)ω.

(b) Let K be a global ﬁeld, T/K a (Z/pZ)×-torsor and T ′ its Tate dual. Two orders 1 T , 2 T ′ are called dual if the completions , are dual for all primes q of K. O ⊆ O ⊆ O1,q O2,q A dual pair yields a reﬂection theorem, as follows.

Theorem 10.3. Let T , T ′ be dual orders. Then O1 ⊆ O2 ⊆ + 1 Cl ( )[p] p T is totally split × | O1 ω| = |O2,ℓ,ω| (57) 1 1T is totally split Cl( )[p] p T ′ is totally split · p 2 ω of p | O | v YK Proof of Theorem 10.3. The maps ψ : Cl( ) Z/pZ are the Artin maps of ring class algebras E/T of O1 → . Now (Z/pZ)× acts both on maps ψ and algebras E, and it is easy to see that the ψ belonging to the O1 ω-component correspond to E that are symmetric, that is, are (Fp)-torsors with resolvent subtorsor T . So we get an injection of groups GA

i : Hom(Cl( ), Z/pZ) H1(K,M ). O1 → T

By Lemma 10.1, the image of i is a Selmer group SelX (K,MT ), where the local conditions Xv are given by

X = the whole of H1(K ,M ) if v v v T |∞ X = ⊥ = , v ﬁnite v O1,v,ω O2,v,ω where the second equality uses the duality of and and the Kummer parametrization O1 O2 1 H (Kv,MT ) ∼= Tω′×. By the exact same argument, the dual Selmer system

X =0 if v v |∞ X = ⊥ = , v finite v O2,v,ω O1,v,ω + has Selmer group naturally identified with Hom(Cl ( 2), Z/pZ). (Of course, the distinction between wide and narrow class groups is only relevant if p =2, a caseO which we will exclude in the next section.) To finish, we apply the Greenberg-Wiles formula, as mentioned in the end of the proof of Theorem 8.12, and use that H0(M ) is either p or 1 according as T is totally split. | T | 10.2 Dual orders are plentiful for quadratic extensions

It’s not hard to show that the maximal orders T , T ′ are dual at primes ℓ ∤ p. At p, however, it is not obvious how one might find a pair of dual orders,O orO whether such orders exist. However, there is a case in which this is manageable, and it specializes to the Scholz reflection principle in the case p =3. Let p be an odd prime. We will assume that our base field K contains the element

1 2π ρ = ζ + ζ− = 2cos . p p p p

68 (Note that ρ = 1, so this assumption always holds when p = 3.) This entails in particular that K(ζ ) is 3 − p an extension of K of degree at most 2, being K(√D) where

1 2 2 D = (ζ ζ− ) = ρ 4. p − p p − 1/(p 1) We note that D is a unit locally at all ﬁnite primes q except those dividing p, in which case D q = p q − = 1/p | | | | dmin,q. (p 1)/2 If Q = K[√a] is an étale quadratic algebra, we may form the (Z/pZ)×-torsor T = Q − , with p the unique possible torsor action. Q is a µ2-torsor, and Q×/(Q×) is the direct sum of two components: p N=1 N=1 p Q×0 = K×/(K×) , and Q× = Q /(Q ) which parametrizes (M)-extensions whose resolvent torsor ω ∼ ω GA is T . Due to the splitting of T , these are in fact Dp-extensions, where Dp is the dihedral group (the permutation group that the symmetries of a regular p-gon induce on its vertices). (p 1)/2 The Tate dual T ′ is Q′ − , a product of copies of the quadratic algebra Q′ = K[√Da].

10.3 Local dual generalized orders Suppose our base field K has a distinguished subring of integers , a Dedekind domain with field of OK fractions K. If Q is an order over K , denote by ( ) the projection of × onto (Q×)ω, quotienting O ⊆ O O R O O p out by both pth powers and the eigenspace corresponding to the trivial character (namely ( × )/( × ) ). OK OK If K is local, we call a pair of orders Q, ′ Q′ dual if the associated unit class subgroups O ⊆ O ⊆ ( ) (Q×) , ( ′) (Q′×) are orthogonal complements. For example, it is not hard to prove that if R O ⊆ ω R O ⊆ ω char kK = p, the maximal orders in Q and Q′ are dual to one another. We pose the question of whether any order in6 Q admits a dual order. The answer is no, because ( ) can be as small as 1 but cannot be as R O { } big as (Q×)ω, being always contained in ( Q×)ω. This is essentially the only obstruction, and we remedy it by introducing a notion of generalized order.O Definition 10.4. If Q is a quadratic étale algebra over a field K, in which a Dedekind domain is fixed OK as a ring of integers, a generalized order in Q is a finitely generated K -subalgebra spanning Q over K and closed under the conjugation automorphism of Q. O O

If K is local, then as soon as contains an element of Q with negative valuation, even with respect to only oneO of the valuations on Q (ifOQ is split), then taking conjugates and powers shows that contains all elements of Q. Thus the only generalized orders in this case are that = Q or is an order inO the ordinary O O i sense, that is, a subring of Q that spans Q. Letting Q = K [ξ], these orders have the form = K [π ξ] for i 0. O O O O O ≥ In general, a generalized order over a Dedekind domain K is specified by a collection ( q)q of orders O O 1 1 O in the completions K , almost all maximal; and particular has the form [q− ,..., q− ] where is an q O1 1 r O1 order in Q and the qi are finitely many primes of K, at which 1 can be taken maximal. Class groups of generalizedO orders over number fields are not hard to study:O in the foregoing notation, we have that Cl( ) = Cl( )/ q ,..., q is formed by quotienting out by the classes of the relevant primes. O ∼ O1 h 1 ri Lemma 10.5. If K is a local field and is a quadratic generalized order, then ( ) is a level space O ⊆ OQ R O in Qω× (in the sense of Theorem 7.1). Moreover, all level spaces arise in this way.

Proof. In the tame case that char kK = p, there are at most three level spaces, and it is easy to identify the generalized orders to which they correspond:6

Q× = (Q) ω R ( )× = ( ) OQ ω R OQ 1 = ( ), any ( . { } R O O OQ 0 The last holds because any x × is the product of x0 K×, which maps into the ω -component, and an x 1 mod π which is necessarily∈ O a pth power. ∈ 1 ≡ K OQ In the wild case we use similar methods. Since p =2, we may write Q = QK[ βQ], where vK (βQ) is 0 6 (p 1)/2O (p 1)/2 or 1. The Kummer element β corresponding to the torsor T = Q − is β = βQp− .

69 The generalized order Q has unit class subgroup

(Q)= 1. R L− The remaining orders can be described as

j v(β)/2 = π − β , Oj OK h p i where j, the valuation of a generator, ranges over the nonnegative elements of Z (if β 1) or Z +1/2 (if β π). A unit in such an order is of the form ∼ ∼ j v(β)/2 u = a 1+ bπ − β , a × ,b ∈ OK ∈ OK p Since the factor a belongs to the ω0-component, it can be ignored. The range of [u] H1(K,M), by Theorem ∈ 7.1(f), is i, where L (p 1)(j 1) pe − − , j +1 p ≤ p 1 i =  pe− e +1, j> +1.  p 1 − It is easy to see that all i (0 i e +1 ) are attained thereby. ≤ ≤  Proposition 10.6. Every generalized order in a quadratic extension Q/K has a (not necessarily unique) dual order in the reﬂection extension Q′. Proof. Follows immediately from Lemma 10.5 and Theorem 7.1(g). In the tame case, we evidently have the dual pairs

,Q OQ ←→ OQ′ ←→ O for any ( . In the wild case, things are only a bit more involved: O OQ′ Proposition 10.7. Let , be the orders in Q and Q′ as parametrized in the proof of Lemma 10.5. OQ,j OQ′,j′ A dual to Q is any Q ,j for which O ′ ′ pe j > +1. p 1 − pe For 0 j p 1 , a dual to Q,j is Q′,j′ where ≤ ≤ − O O pe j′ = +1 j. p 1 − − Proof. The only slightly nontrivial step is to show that, in the second case, the corresponding level indices

(p 1)(j 1) (p 1)(j′ 1) i = − − and i′ = − − p p have sum e. But after noting that the arguments to the two ceilings have sum 1/p mod 1, the summation becomes easy. ≡ The method of proof of Theorem 10.3 applies without change to generalized orders and yields the following.

1 Theorem 10.8. Let p 3 be a prime, let K be a global ﬁeld with ζ + ζ− K, and let Let Q, ≥ p p ∈ O1 ⊆ Q′ be dual generalized quadratic orders. Then O2 ⊆ 1 Q=K K Cl( 1)[p]ω p ∼ × 2×,ℓ,ω | O | = 1 1|O | (58) Q K K Qp=K K Cl( )[p] p ′∼= · ∼ × 2 ω × of p | O | v YK

70 10.4 Relation to the Scholz reﬂection theorem

Example 10.9. Let p =3, K = Q, T = Q[√D], and T ′ = Q[√ 3D], where D is a fundamental discriminant − with 3 ∤ D. Construct a pair of dual orders , ′ by speciﬁcation at each prime ℓ of Z as follows: O O • If ℓ = 3, , we take and ′ to be maximal at ℓ, contributing nothing to the product in Theorem 10.8.6 ∞ O O • If ℓ = 3, using Proposition 10.7, we see that the orders Q [√D] and Q [√ 27D] are dual, as are 3 3 − Q [√9D] and Q [√ 3D]. The ﬁrst contributes 1 to the product, and the second contributes 3. 3 3 − The prime ℓ = does not enter into the construction of the dual orders, but it introduces a factor ∞ H0(R,M ) that depends on the sign of D. Finally, note that all of the class group Cl( ) of a quadratic | D | O order belongs to the ω-eigenspace, the 1-eigenspace being Cl(Z)=0. So we get an equality, which was also noticed by Nakagawa ([38], Theorem 0.5): Corollary 10.10. If D 0, 1 mod 4 is an integer, write Cl(D) for the class group of the quadratic ring ≡ over Z having discriminant D. Let D be a fundamental discriminant not divisible by 3. Then

1D= 3 1D=1+1D>0 Cl( 27D)[3] = Cl(D)[3] 3 − − (59) | − | | | · 1D= 3 1D=1+1D>0 1 Cl( 3D)[3] = Cl(9D)[3] 3 − − − . (60) | − | | | · Both equations are generalizations of the Scholz reﬂection principle, which states that for D =1, 3, 6 − 1 ε Cl( 3D)[3] = Cl(D)[3] 3 D>0− | − | | | · where ε 0, 1 . This theorem shows that ε can be explained by the size of the kernel of either of the maps ∈{ } Cl(9D)/ Cl(9D)3 Cl(D)/ Cl(D)3 or Cl( 27D)/ Cl( 27D)3 Cl( 3D)/ Cl( 3D)3. (61) → − − → − − It also shows that exactly one of the maps (61) is an isomorphism, the other having kernel of size 3—a theorem, perhaps, that has not appeared in the literature yet?

Part V Reﬂection theorems: cubic rings

11 Cubic Ohno-Nakagawa

The space V (K) of binary cubic forms over a local or global field K can have many integral models. Let V be the lattice of binary cubic forms with trivial Steinitz class; these can be written as OK 3 2 2 3 V = ax + bx y + cxy + dy : a,b,c,d K , OK ∈ O and we abbreviate the form ax3 + bx2y + cxy2 + dy3 to (a,b,c,d). A theorem of Osborne classifies all lattices 1 L V K that are GL2( K )-invariant and primitive, in the sense that p− L * V ( K ) for all finite primes p of ⊆ O: O O OK Theorem 11.1 (Osborne [47], Theorem 2). A primitive GL2( K )-invariant lattice in V ( K ) is determined by any combination of the primitive GL ( )-invariant latticesO in the completions V ( O ), which are: 2 OK,p OK,p (a) If p 3, the lattices Λ = (a,b,c,d): b c 0 mod pi , for 0 i v (3); | p,i { ≡ ≡ } ≤ ≤ p (b) If p 2 and N (p)=2, the five lattices | K/Q Λ = V ( ), p,1 OK,p Λ = (a,b,c,d) V ( ): a + b + d a + c + d 0 mod p p,2 { ∈ OK,p ≡ ≡ } Λ = (a,b,c,d) V ( ): a + b + c b + c + d 0 mod p p,3 { ∈ OK,p ≡ ≡ } Λ = (a,b,c,d) V ( ): b + c 0 mod p p,4 { ∈ OK,p ≡ } Λ = (a,b,c,d) V ( ): a d b + c mod p , p,5 { ∈ OK,p ≡ ≡ }

71 (c) For all other p, the maximal lattice V ( ) only. OK,p From the perspective of algebraic geometry, if p 2, the latter four lattices are not true integral models, | because they lose their SL2-invariance as soon as we extend scalars so that the residue ﬁeld has more than

2 elements. By contrast, if p 3, the SL2-invariance of the space Lpi can be established purely formally. This integral model, which we will| call the space of pi-traced forms, will be the subject of our main reﬂection theorem in this part. Although Osborne deals only with the case of Γ( ), his method generalizes easily to the lattice OK 3 2 2 3 1 2 V ( , a)= ax + bx y + cxy + dy : a a,b ,c a− , d a− OK { ∈ ∈ OK ∈ ∈ } that pops up when considering the maps Φ: M Λ2M → that appear in the higher composition law Theorem 6.9. Here the relevant action of

a11 a12 j i Γ( K , a) = Aut K ( K α)= GL2(K): aij a − O O O ⊕ a21 a22 ∈ ∈ is nontrivial on both M and Λ2M, thus aﬀecting V ( , a) via a twisted action OK

a11 a12 1 . Φ (x, y)= Φ(a11x + a21y,a12x + a22y). (62) a21 a22 a a a a 11 22 − 12 21 (Compare [14], p. 142 and [57], Theorem 1.2.) The twist by the determinant does not affect invariance of lattices but renders the action faithful, while otherwise scalar matrices that are cube roots of unity would act trivially. We sidestep this issue entirely by restricting the action to the group SL2, which preserves the 2 discriminant D a− of the form. The corresponding ring has discriminant (a,D). ∈ For instance, over K = Z there are ten primitive invariant lattices, comprising five types at 2 and two types at 3. The O-N-like reflection theorems relating all the types at 2 were computed by Ohno and Taniguchi [45] and will be considered later in this paper (Section 12.4). While the behavior at 2 admits only mild generalization, being based on the combinatorics of the finitely many cubic forms over F2, the behavior at 3 is robust. We begin by making some definitions needed to track the behavior of cubic forms and rings at primes dividing 3. If is a ring of finite rank over a Dedekind domain , define its trace ideal tr( ) to be the image of the O OK O trace map tr / K : K . Note that tr( ) is an ideal of K and, since 1 has trace n = deg( / K ), it is a divisorO ofO theO ideal → O (n). In particular,O if is a DVR,O this notion∈ is O uninteresting unless O Ohas OK OK residue characteristic dividing n. Let t be an ideal of K dividing (n). We say that the ring is t-traced if tr( ) t. O O OBy⊆ Theorem 6.9, we can parametrize cubic orders by their Steinitz class a and index form O Φ(xξ + yη) = (ax3 + bx2y + cxy2 + dy3)(ξ η) ∧ 1 2 relative to a decomposition = K K ξ aη, where a a, b K , c a− , and d a− . Then a short computation using the multiplicationO O ⊕ table O from⊕ Theorem∈ 6.9 shows∈ O that,∈ if (1,ξ,η) is a∈ normal basis, then tr(ξ)= b and tr(η)= c, so tr( ) = 3, b, ac . Thus the based t-traced rings over K are parametrized by the rank-−4 lattice of cubic formsO h i O

3 2 2 3 1 2 ( ) := ax + bx y + cxy + dy : a a,b t,c ta− , d a− , Va,t OK { ∈ ∈ ∈ ∈ } on which GL( a) acts by the twisted action (62). For instance, if K = Z, a = (1), and t = (3), this is the lattice of integer-matrixO⊕ cubic forms considered in the introduction.O Our goal in this section is to prove a generalization for all number ﬁelds K and spaces V ( , a, t). OK Theorem 11.2 (“Local cubic O-N”). Let K be a nonarchimedean local ﬁeld, char K = 3. Let V (D) be 6 the composed variety of binary cubic forms of discriminant D, under the action of the group Γ = SL2. If α K× and τ 3 in , let (D) be the integral model of V (K)(D) consisting of forms of the shape ∈ | OK Vα,τ 3 2 1 2 2 3 f(x, y)= aαx + bτx y + cα− τxy + dα− y ,

72 together with its natural action of = SL( α ). Then the integral models Gα OK ⊕ OK 6 ( (D), SL ) and 1 ( 27τ − D), SL , (63) V1,τ 2OK V1,3τ − − 2OK and consequently (V (D), ) and V 3 1 ( 27D), 3 (64) α,τ Gα ατ − ,3τ − − Gατ − are naturally dual with duality constant NK/Q(τ)= K /τ K . |O O | The two formulations are easily seen to be equivalent. The first one is the one we will prove, but the second one has the needed form of a local reflection theorem to apply at each place to get the following global reflection theorem:

Theorem 11.3 (O-N for traced cubic rings). Let

3 2 2 3 1 2 ( ) := f(x, y)= ax + bx y + cxy + dy : a a,b t,c ta− , d a− , Va,t OK { ∈ ∈ ∈ ∈ } a representation of := SL( a). Ga OK ⊕ Note that a,t is the integral model of (V (K), Γ(K)) parametrizing t-traced cubic rings over K with Steinitz V 2 2 O class a. For D t a− , deﬁne the class number ∈ 1 1 ha,t(D)= = . Stab Φ AutK Φ a a,t( K ) | | Disc =(a,D) | O| ∈Gdisc\VX Φ=DO t-tracedOX Then we have the global reﬂection theorem

2 # v :D (K×) 3 { |∞ ∈ v } 3 1 ha,t(D)= hat− ,3t− ( 27D). (65) NK/Q(t) · −

Proof. Use Theorem 11.2 at each ﬁnite place. At the inﬁnite places, the two integral models are necessarily 1 0 naturally dual because H (R,MD)=0; but the duality constant depends on H , which depends on the sign of D at each real place, as desired.

Observe that taking K = Q, a =1, t =1 recovers Ohno-Nakagawa (Theorem 1.1). This also yields the extra functional equation for the Shintani zeta functions (see Corollary 11.8 below). We can rewrite our results in terms of Shintani zeta functions.

Definition 11.4. Let K be a number field. If / K is a cubic ring of nonzero discriminant, the signature O O 2 σ( ) of is the Kummer element α (K R)×/((K R)×) corresponding to the quadratic resolvent O O ∈ ⊗Q ⊗Q of . That is, it takes the value α = +1 or 1 at each real place v of K according as = R R R or O v − Ov ∼ × × R C, and α =1 at each complex place. × v 2 Definition 11.5. Given a number field K, a signature σ (K Q R)×/((K Q R)×) , an ideal class [a] Cl(K), and an ideal t 3, we define the Shintani zeta function∈ ⊗ ⊗ ∈ |

1 s ξK,σ,[a],t(s)= NK/Q(discK L)− AutK ( ) XO | O | where the sum ranges over all cubic orders over K having signature σ, Steinitz class a, and trace ideal contained in t. We also deﬁne the Shintani zetaO functionO with unrestricted Steinitz class

ξK,σ,t(s)= ξK,σ,[a],t(s). [a] Cl(K) ∈X Remark 11.6. Confusingly, it is traditional to denote Shintani zeta functions by the Greek letter xi.

73 Remark 11.7. By Minkowski’s theorem on the finite count of number fields with bounded degree and discrim- s inant, each term n− has a finite coefficient, so the Shintani zeta function at least makes sense as a formal Dirichlet series. It generalizes the Shintani zeta functions for rings over Z mentioned in the introduction. Datskovsky and Wright [18] study an adelic version of the Shintani zeta function; they show that ξK,σ,(1) and ξK,σ,(3) are entire meromorphic with at most simple poles at s = 1 and s = 5/6, satisfying an explicit functional equation. We surmise that the same method will prove the same for ξK,σ,[a],t. However, we do not consider the analytic properties here. Then we have the following corollary, which generalizes Conjecture 1.1 of Dioses [20]. Corollary 11.8 (the extra functional equation for Shintani zeta functions). Let K be a number 2 field, σ (K R)×/((K R)×) a signature, [a] Cl(K) an ideal class, and t 3 an ideal. Then the ∈ ⊗Q ⊗Q ∈ | Shintani zeta function ξK,σ,[a],t(s) satisfies an extra functional equation

# v :σ =1 +3[K:Q]s 3 { |∞ v } ξ (s)= ξ 3 1 (s), (66) K,σ,[a],t 1+6s K, σ,[at− ],3t− NK/Q(t) − Hence, summing over all a, # v :σ =1 +3[K:Q]s 3 { |∞ v } ξ (s)= ξ 1 (s), (67) K,σ,t 1+6s K, σ,3t− NK/Q(t) − 3 2 Proof. Fix a and t. Sum Theorem 11.3 over all D t a− of signature σ, weighting each D by ∈ 2 s NK/Q Da − , the norm of the discriminant of the associated cubic rings. Then the left-hand side of the summed equality 2 6 matches that of (66). The right-hand side involves rings with discriminant ideal 27Da t− , so a compensatory factor of s N Da2 − 3[K:Q]s K/Q = 2 6 s N (t)6s NK/Q (27Da t− )− K/Q must be added to the right-hand side to pull out the desired Shintani zeta function. In the succeeding subsections, we present three approaches to the local duality (Theorem 11.2). First, we present a short conceptual proof in the special case that char kK = 3, a “tame” case. Second, we explicitly compute the local orbit counters for a computational proof. Third,6 we organize the local orbits into families for a more conceptual general proof.

11.1 A bijective proof of the tame case

Proof of the tame case of Theorem 11.2. Fix D K . Let T = K[√D] be the corresponding quadratic ∈ O i i algebra, and T ′ = K[√ 3D]. For brevity we will write H (T ) for the cohomology H (K,MT ) of the − i corresponding order-3 Galois module, and H (T ′) likewise. Denote by f(σ), for σ H1(T ), the number of orders of discriminant D in the corresponding cubic ∈ 1 algebra Lσ; and likewise, denote by f ′(τ), for τ H (T ′), the number of orders of discriminant 3D in Lτ . ˆ ∈ − Our task is to prove that f ′ = f. We note that if 3 is a square in K× , then T = T ′ and f = f ′. Note that f is even: σ and σ are parametrized− by the same cubicO algebra with opposite orientations of its resolvent. The Fourier transform− of an even, rational-valued function on a 3-torsion group H1(T ) is again even and rational-valued. So far, so good. Our method will be ﬁrst to prove the duality at 0: that is, that

f ′(0) = fˆ(0) (68)

f(0) = f ′(0). (69) ˆ 1 Let us explain how this implies that f ′ = f. We computeb H (T ) using the self-orthogonality of unramiﬁed cohomology: | | 1 1 ur 1 ur 0 0 H (T ) = H (T ) H (T ′) = H (T ) H (T ′) . | | | | · | | | | · | | So there are basically three cases:

74 1 1 (a) If neither D nor 3D is a square in K , then H (T ) = H (T ′) = 0, and (68) trivially implies that − v ∼ ∼ f ′ = fˆ.

1 1 (b) If one of D, 3D is a square, then H (T ) and H (T ′) are one-dimensional F3-vector spaces. The space of even functions− on each is 2-dimensional, and

g (g(0), gˆ(0)) 7→ g′ (g (0),g′(0)) 7→ ′ are corresponding systems of coordinates on them.b Consequently, the two equations (68) and (69) together imply that fˆ = f ′.

1 (c) Finally, if D and 3D are both squares, then f is a function on the two-dimensional F3-space H (T ) = 1 − 1 ∼ H (T ′) which we would like to prove self-dual. Note that H (T ) has four subspaces W1,...,W4 of dimension 1. Consider the following basis for the ﬁve-dimensional space of even functions on H1(T ): 1 1 1 f1 = W1 ,...,f4 = W4 ,f5 = 0.

Note that f1,...,f4 are self-dual (the Tate pairing is alternating, so any one-dimensional subspace is isotropic), while f is not: indeed fˆ (0) = f (0). Thus if (68) holds, then f is a linear combination of 5 5 6 5 f1,...,f4 only and hence fˆ = f. We have now reduced the theorem to a pair of identities, (68) and (69). By symmetry, it suﬃces to prove (69), which may be written ? ˆ 1 f(0) = f(0) = 0 f(τ). (70) H (T ′) τ H1(T ) | | ∈X ′ The proof is clean and bijective. The left-hand side of (70) counts orders of discriminant D in the split algebra K T . These can be straightforwardly parametrized as × +0 a, OK × where a is a multiplicatively closed lattice in T , that is, an invertible ideal in some quadratic order T . (Here we use that, in a quadratic algebra, any lattice a is an invertible ideal with respect to its endomorphismO ⊆ ring End a. This fails for higher-degree algebras, which we will encounter later.) The sum on the right-hand side of (70) counts all cubic orders of discriminant 3D. Any cubic order C of discriminant 3D can be assigned an ideal in T as follows. Let L be the fraction− algebra of C. By Theorem 5.4, we have− the description

3 3 L = K + ξ√δ + ξ¯ δ¯ ξ T ′ { | ∈ } N=1 N=1 3 p for some δ T ′ / T ′ ; and so, since 3 is invertible in , ∈ OK

3 C = + ξ√3 δ + ξ¯ δ¯ ξ c (71) OK { | ∈ } p for some lattice c in T ′. Now by Theorem 6.11, we get that ( , c,δ) is a self-balanced ideal, that is, OD δc3 ,N(c) = (t) is principal, and N(δ)t3 =1, (72) ⊆ OD

Now c need not be invertible in D; but let := End c. Note that = D/π2i for some positive integer i. Then form the shadow O O O O δc3 a = . πi Since the norm is multiplicative on invertible ideals, the properties of c in (72) can be recast as properties of a: a End a and N (a)=1. (73) ⊆ OD

75 The ﬁrst of these says that a is multiplicatively closed, and the second that the ring C = K +0 a corresponding to a has discriminant D. O × It remains to show that there are exactly H0(T ) cubic orders C corresponding to each shadow a satisfying (73), weighting each C by the number of isomorphic| | copies of C in its fraction algebra L = KC. First, we simply count C up to isomorphism. This is the same as counting the self-balanced ideals ( , c,δ) up to the equivalence relation OD 3 c λc, δ λ− δ (λ T ′×) (74) 7→ 7→ ∈ Note the slight subtlety in this step: we would like to deﬁne an isomorphism between the associated cubic

3 algebras Lδ, Lλ− δ by 3 3 ξ√3 δ + ξ¯ δ¯ λξ√3 λ 3δ + λ¯ξ¯ λ¯ 3δ,¯ 7→ − − which works, but only after potentially rescalingp the cube roots onp the right-hand side by a suitable 3rd root of unity in K so that their product is N(λ)t = N(λ)√3 δ√3 δ¯. Let a = α be the given shadow, where = D/π2i is its ring of invertibility. Clearly c must be invertible withO regard to . The possible (c,δ)OmayO be found by ﬁxing c = and taking δ = πiαε where O O ε × is constrained by the requirement that N(δ) be a cube. If, without loss of generality, we scale α so ∈ O i 3 that N(π α) is a cube, then the admissible values are ε × : N(ε) (K×) . Now in the equivalence { ∈ O ∈ } 3 relation 74, the multipliers λ preserving c = are λ ×, so we must consider ε up to × . Since 3 3 O ∈ O O N : ×/ × K× / K× is surjective, the number of distinct ε, which is the number of nonisomorphic C, isO simplyO → O O 3 ×/ × O O . 3 × / × OK OK

Next, we weight each C by the number of isomorphic copies of C in L = KC. The automorphisms of C are given by 3 3 ξ√3 δ + ξ¯ δ¯ ωξ√3 δ +¯ωξ¯ δ¯ 7→ 3 where ω T ′ satisﬁes ω = ωω¯ =1. But choicesp ω ﬁx C, sop we must quotient out by those, and the number of∈ isomorphicO copies is ∈ O N=1 [3] × [3] OT ′ = OT ′ , N=1[3] [3] |O | |O× | and the total number of cubic orders we seek is the product

3 ×/ × × [3] O O OT ′ . (75) 3 · [3] × / × × OK OK |O |

To maneuver this into the required form, ﬁrst note that

3 3 ×/ × T× / T× O O = O ′ O ′ ×[3] × [3] T′ |O | O by the Snake Lemma, since × / × is ﬁnite; so (75) takes the form OT ′ O 3 T× / T× O ′ O ′ , 3 × / × OK OK which we can now compute directly to equal

76 11.2 A computational proof We now turn our attention to the wild case of Theorem 11.2. We present two proofs, one computational, one more conceptual.

11.2.1 Trace ideals of maximal orders

Our first step is to compute tr( L) for the maximal orders of all cubic extensions L. The answer is delightfully simple. O Proposition 11.9. Let L/K be a cubic étale algebra over a 3-adic field. The trace ideal of the maximal order of L is e ℓ(L) tr( )= (3)+ m − OL K where ℓ(L) is the level. Remark 11.10. Hyodo ([27], equation (1–4); see also Xia and Zhukov [61]) proves a theorem like this one for an invariant he calls the depth of a ramified extension. The depth is in fact closely related to the level and offset. Proof. We first dispose of the case that L is not totally ramified, that is, has splitting type 111, 12, 3, or 2 1 1, by noting that in all these cases tr( L) = (1) and ℓ(L)= e. O 2 Now let L = K[πL] be a totally ramified extension. We have tr( L) = (3, tr πL, tr πL). There are two cases. O

2 1 In this case we assume vK (tr πL) vK (tr πL), including the case that both are inﬁnite. In this case it h i ≥ 2 is possible to adjust πL by a multiple of πL so as to make the trace vanish. Therefore we may assume that tr πL =0, so the Eisenstein minimal polynomial of πL is a depressed cubic,

φ(x)= x3 + ux + v,

with v (u) 1 and v (v)=1. We have tr(x2)=2u, so tr( )=(3,u). K ≥ K OK disc(L) = disc φ =4u3 27v2. − 2 Now vK (27v )=3e +2 is not a multiple of 3, so the two terms have unequal valuation. If vK (u) e, 3 ≤ℓ then 4u dominates so L has level ℓ = e vK (u) and oﬀset θ = 0; and the trace ideal is (u) = mK . 2 − If vK (u) > e, then 27v dominates so vK (disc L)=3e +2. This is the case L = K[√3 πK ] of a uniformizer radical extension.− The level is 1, the oﬀset is 1 and the trace ideal is (3). − − 2 2 2 In this case we assume vK (tr πL) < vK (tr πL). Note that vK (tr πK πL) vK (tr πL), so it is possible h i 2 ≤ to adjust πL by a multiple of πK πL to produce an element ρ such that tr ρ = 0 and vL(ρ)=2. The minimal polynomial of ρ is a depressed cubic

φ(x)= x3 + ux + v,

with vK (u) 1 and vK (v)=2. Now ρ does not generate all of L; instead, an K -basis of L is (1,ρ2/π ,ρ)≥so tr( )=(3,u/π ). O O O K OL K disc φ 4u3 27v2 disc(L)= 2 = −2 . πK πK

2 Now vK (27v )=3e+4 is not a multiple of 3, so the two terms in the numerator have unequal valuation. 3 If vK (u) e +1, then 4u dominates so L has level ℓ = e vK (u)+1 and oﬀset θ =1; and the trace ≤ ℓ 2 − ideal is (u/πK) = mK. If vK (u) > e +1, then 27v dominates and we have a uniformizer radical extension again. −

The foregoing proof has a corollary on the structure of totally ramiﬁed cubic extensions which will be important to us.

77 Corollary 11.11. Let L/K be a totally ramified extension. (a) If θ(L)=0, then L has a traceless uniformizer. (b) If θ(L)=1, then L has a traceless element of valuation 2. Proof. Simply note that Case 1 occurs only when θ = 0 or 1, and Case 2 occurs only when θ = 1 or 1. h i − h i − Problem 11.12. Find an analogue of Proposition 11.9 for (Z/pZ)-extensions of p-adic fields, p 5. GA ≥ 11.2.2 The subring zeta function If L/K is a cubic 3-adic algebra and 0 t e, let ≤ ≤ v (disc )/2 1/2 g(L,t)= z K O Z[[z ]] ∈ orders L O⊆ tr( X) mt O ⊆ K t be the generating function of mK-traced orders in L. This is related to the subring zeta function (see Section 12.3). The factor of 1/2 in the exponent is used (quite arbitrarily) to make a factor of z correspond to 1/2 passing to a subring of index mK . Note that g(L,t) z Z[[z]] or Z[[z]] according as T is ramified or not. ∈ t Note that if t e ℓ(L), then every order in L is automatically mK -traced, so g(L,t)= g(L, 0) is simply the generating function≤ − for all orders that was computed by Datskovsky and Wright. We will proceed to compute g(L,t) for all L and t. We begin by tabulating the possible splitting types for a cubic algebra L: σ(L) ℓ(L) θ(L) vK (Disc L) 111 e 0 0 12 e 0 0 3 e 0 0 121 e 1 1 13 0 ℓ

t 3t 3e gˆ(L,t)= q z − 2 g(L,e t). − Proof. The proof proceeds by writing g(L,t) as a linear combination of characteristic functions of level spaces. Let g(T,ℓ,t)= g(L,t) for any L of resolvent torsor T and level ℓ, where

1 if T ′ is split e ℓ ℓmin = − ≥ ≥ ( 0 otherwise. By Theorem 7.1(d), such L exists, and by Lemma 11.13, the series g(T,L,t) is independent of which L of this level we choose, with one exception: if T ∼= K K is split and ℓ =0, then L could have splitting type 111 or 3. We resolve the ambiguity as follows: give×g(K K, 0,t) the value of g(L,t) when σ(L)=3, and introduce a symbol g(K K, 1,t) with the value g(K ×K K,t) for splitting type 111. Correspondingly, deﬁne the level space × − × × = 0 , Le+1 { }

78 in spite of the fact that cubic algebras of splitting types 3 and 111 both have level e. Then in all cases, if we set e +1 if T is split ℓmax = ( e otherwise then we have 1 1 g(L,t)= L g(T,ℓ,t)= L ℓ (g(T,ℓ,t) g(T,ℓ 1,t)) ∈Lℓ\Lℓ+1 ∈L − − ℓ ℓ ℓ ℓ ℓ ℓ min≤X≤ max min≤X≤ max where g(T,ℓ 1,t)=0 (76) min − 1ˆ 1 Now ℓ = cℓ e ℓ , where L L − 1 3 ℓ = e +1= ℓmax c = |Lℓ| = qe ℓ 0 ℓ e ℓ H0(T )  − e ≤ ≤ | |  3q ℓ = 1= ℓmin. − Thus it suffices to prove that  t 3t 3e c (g(T,ℓ,t) g(T,ℓ 1,t)) = q z − 2 (g(T ′,e ℓ,t) g(T ′,e ℓ 1,t)) (77) ℓ − − − − − − for ℓmin ℓ ℓmax. It is easy to verify that flipping T T ′, ℓ e ℓ, t e t transforms (77) to an equivalent≤ equation,≤ so we assume that ℓ> 0 and ℓ + t 7→e, which7→ cuts− down7→ the number− of cases. We now enumerate the cases of (77), which by Lemma≥ 11.13, depend on the splitting types and offsets of the fields L appearing. Recall that when the level ℓ (0 ℓ e) and resolvent torsor T of a cubic algebra L are known, the offset can be determined by the congruence≤ ≤

3ℓ + θ = v (disc L) v (β ) mod2 K ≡ K T where βT is a Kummer element for T (here we simply have vK (βT ) vK (disc L) mod2). Now since β β = 3 (up to squares), we have ≡ T T ′ −

v (β )+ v (β′ ) e mod 2. K T K T ≡ From this, we ﬁnd that each oﬀset in (77) determines the other three, even without knowing T , and there are only six cases:

σ(T ) ℓ σ(LT,ℓ) σ(LT,ℓ+1) σ(LT ,e ℓ) σ(LT ,e ℓ+1) ′ − ′ − any 0 <ℓ

The dashes for ℓ = ℓmin indicate that the corresponding term was declared zero in (76). We now write each term of (77) in terms of sσ,θ(n) using Lemma 11.13. Thanks to our assumptions that ℓ > 0 and ℓ + t e, the terms on the right side involve only sσ,θ(n) for n < 0. In this case, we can replace n by 0 because≥ all orders in L automatically satisfy the trace condition. The corresponding generating function was computed by Datskovsky and Wright and will soon be recovered by us (see (94))

3 1 s = s1 ,θ(0) = , 0 (1 z)(1 qz3) − − independent of θ. The theorem is now reduced to the following lemma.

79 Lemma 11.15. For t 0, ≥ 3 3 s1 ,θ=0(n) z2s1 ,θ=1(n 1) = qnz3n(1 z2)s (78) − − − 0 13,θ=1 13,θ=0 n 3n s (n) zs (n 1) = q z (1 z)s0 (79) − 3 − − s3(n) z2s1 ,θ=1(n 1) = qnz3n(1 z)s (80) − − − 0 12 2 13,θ=1 n 3n s (n) z s (n 1) = q z s0 (81) 2 − 3 − s1 1(n) zs1 ,θ=0(n 1) = qnz3ns (82) − − 0 s111(n) s3(n)=3qnz3n+1s . (83) − 0 This lemma can be viewed as a set of coupled diﬀerence equations for computing the sσ,θ(t).

11.2.3 Counting traced cubic orders: the proofs of Lemma 11.13 and Lemma 11.15

t It now remains to count the mK -traced orders in each cubic algebra L. These are sublattices of L containing 1 and satisfying (a) the ring condition, that is, closure under multiplication, and (b) the trace conditionO that t each of their elements has trace in mK. We ﬁrst simplify these two conditions. If L is a sublattice containing 1, the quotient L/ ∼= ( L/ K )/( / K ) is a ﬁnite group generatedO ⊆ as O an -module by two elements. We may writeO O O O O O OK / = ( /mi )ξ ( /mj )η, (84) OL O OK K ⊕ OK K i j where ξ and η are generators such that L and have K -bases (1,ξ,η) and (1, πK ξ, πK η). By symmetry we may assume that i j. By varyingO the basisO(1,ξ,ηO), we get all lattices of “index (i, j)” in the sense that (84) holds. (This is≥ an example of a reduced basis, which we will use moreO systematically in the quartic case.) Note that = + mi + mj η, (85) O OK K OL K from which it is easy to see that two bases (1,ξ,η) and (1, ξ′, η′) yield the same if and only if O j i i j η uη′ mod m − , some u ( /m − )×. (86) ≡ K OL ∈ OK K i Note that ξ is irrelevant. Also note that when i = j there is a single lattice, the content ring K + mK L. Having constructed all lattices of index (i, j), we test whether they are rings using TheoremO 6.9(c): O is a ring if and only if its index form is integral. Let the index form of L be O

Φ (xξ + yη) = (ax3 + bx2y + cxy2 + dy3)(ξ η). L ∧ Then i j 2i j i j 2j i i j Φ (xπ ξ + yπ η) = (aπ − + bπ + cπ + dπ − )(π ξ π η). (87) L K K K K K K K ∧ K

Free✟ zone✟ ✟✟ ✟✟ ✟ Root zone ✟ ✟ i

Figure 1: Two zones for the indices (i, j) for a candidate subring of OL

80 From this we can deduce that there are two kinds of pairs (i, j). If j i 2j, then every lattice with ≤ ≤ i 2j indices (i, j) is a ring: we say that (i, j) is in the free zone. If i> 2j, the ring condition is that πK− d, that is, | i 2j Φ (η) 0 mod m − . (88) OL ≡ K Because η must in this sense be a root of Φ, we call this range of (i, j) values the root zone. Note that if L has splitting type (3), then Φ has no roots even mod mK and hence no orders in the root zone. For the remaining splitting types, the root-zone orders can be subdivided according to which root of Φ—which “1” in the splitting type—η reduces to mod mK. e ℓ+n Finally we must test our orders for the trace condition tr mK− , where ℓ = e vK (tr L) = max ℓ(L), 0 . We may assume that 0 n ℓ. The trace ideal ofO an ⊆ order (85) can be computed− byO { } ≤ ≤ tr = tr( + mi + mj η)= (3)+ mℓ+i + mj tr η. O OK K OL K K K Now if n>i, the trace condition is impossible; if n j, it is automatic; and if j

11.2.4 Traced orders in the free zone In this subsubsection we evaluate

FZ v ([ : ]) s (n)= z K OL O .

L free-zone orders, O⊂Otr( ) Xmn tr( ) O ⊆ K OL

In the free zone, we know that the lattices of index (i, j) are parametrized by elements η L/ K not i j 1 ∈ O O divisible by mK up to the equivalence relation (86), and all are orders. We get q − − (q + 1) orders if i > j, just 1 if i = j; and all of these satisfy the trace condition if n j. It remains to test them on the trace condition (89) under the hypothesis that j

FZ i n i+j i j 1 i+j i+j s (n)= q − z + (q + 1)q − − z + z . (90) j

RZ,η¯ v ([ : ]) s 0 (n)= z K OL O

L η¯0-orders, tr(O⊂O) Xmn tr( ) O ⊆ K OL

In each of the various cases that we will encounter, we will choose a basis (1, ξ0, η0) such that η0 reduces to η¯ mod + m . Then the lattices belonging to this root are parametrized by elements η = xξ + yη 0 OK K OL 0 0 where mK x; so mK ∤ y and we can scale so that y =1. Consequently we can take η = x′πK ξ0 + η0 where x′ | i j 1 runs over the residue classes mod mK− − . The ﬁrst case is that of the simple root, the “1” with no exponent that appears in the splitting types 2 111, 12, and 1 1. In this case Hensel’s lemma tells us that η¯0 can be lifted to an element η0 with Φ(η0)=0,

81 and we can identify it explicitly: L = K T splits and η = (1; 0). Note that tr(η )=1 and ℓ = e. We can × 0 0 complete to a basis (1, ξ0, η0) with tr(ξ0)=0. The index form of L has the form

Φ(xξ + yη ) = (ax3 + bx2y + cxy2)(ξ η ) 0 0 0 ∧ 0 2 where mK ∤ c, since the root is simple. When plugging in a value η = x′πK ξ0 + η0 with y = 1, the cxy i 2j 1 j term will dominate so the ring condition is mK− − x: we get q rings. But tr(η)=1 so the trace condition cannot be fulﬁlled unless n j, in which case it is vacuous.| Thus we get a root subring generating function ≤ sSR(n)= qj zi+j . (91) j n iX≥2j ≥ 2 3 Now assume that η¯0 is the multiple root of one of the splitting types 1 1 and 1 : that is, in a basis (1, ξ0, η0) with η0 lifting η¯0, the index form is

Φ (xξ + yη ) = (ax3 + bx2y + cxy2 + dy3)(ξ η ) L 0 0 0 ∧ 0 with m c and m d. A now-standard trick shows that m2 ∤ d, that is, v (d)=1: if not, then applying K | K | K K the formula (87) with i = 0 and j = 1 would show that (1, ξ0, (η0 + u)/πK) is a basis of a ring for some − 2 u K : that is, L would not be the maximal order in L. So mK ∤ d, and the root η¯0 mod mK has no lift ∈ O 2 O to mod mK. This shows that the only rings belonging to this root occur for i =2j +1, the very edge of the root zone, where the ring condition (88) is mod mK only (and is therefore automatically satisﬁed). There i j 1 j are q − − = q rings. The trace condition can be expressed in terms of the index form using the fact, previously mentioned, that tr(ξ )= b and tr(η )= c. 0 − 0 Thus tr(η) = tr(xπK ξ0 + η0) = c πK bx. We may assume that ℓ = e min vK(b), vK (c) > 0, as when ℓ =0 we have n =0 and no trace condition.− There are two cases. − { }

1 If vK (b) vK (c), then vK (tr(η)) = vK (c)= e ℓ no matter what x we pick, and so the trace condition h i is unsatisﬁable≥ for all n > j. −

2 But if vK (b) < vK (c), then the equation c πK bx =0 has a solution x0, and the solutions to the trace h i condition − c πK bx n j − − e ℓ 0 mod mK πK− ≡ i n 2j+1 n are x = x + u for v (u) n j 1 (since v (b)= e ℓ). There are q − = q − solutions. 0 K ≥ − − K − We must now determine for which algebras L Cases 1 and 2 occur. For splitting type 13, if we take 2 h i h i our basis (1, ξ0, η0)=(1, πL, πL), they match up exactly with Cases 1 and 2 in the proof of Proposition 11.9 and therefore correspond to the oﬀsets θ = 0 and θ = 1, respectively.h i Forh i splitting type 121, we have mK ∤ b (since otherwise Φ would have a triple root) and again Case 2 occurs. Since θ =1 in this case too, we can divide up the generating functions by θ-value rather than splittingh i type:

sMR,θ=0(n)= qj z3j+1 (92) j n X≥ MR,θ=1 j 3j+1 2j+1 n 3j+1 s (n)= q z + q − z . (93) n 1 j n − j

sMR(0) = qj z3j+1. (94) j 0 X≥

82 11.2.6 Putting it together The subring generating functions sσ,θ(n) are now derived by summing the free-zone and root-zone contributions: s111(n)= sFZ(n)+3sSR(n) s12(n)= sFZ(n)+ sSR(n) s3(n)= sFZ(n) (95) 2 s1 1(n)= sFZ(n)+ sSR(n)+ sMR,θ=1(n) 3 s1 ,θ(n)= sFZ(n)+ sMR,θ(n). The proof of Lemma 11.13 is now complete.

Proof of Lemma 11.15. It is possible to prove Lemma 11.15 in an automated fashion by summing the doubly geometric series (90), (91), (92), and (93), solving the linear recurrences (78)–(83), and checking that the resulting rational functions agree for both even and odd n. We here present a more illuminating method, which does not attempt to sum all the series but simply manipulates their terms—that is, it is very nearly a bijective proof. 3 3 We begin with a simpliﬁcation of the generating functions s1 ,θ=0 and s1 ,θ=1. Lemma 11.16.

3 s1 ,θ=0(n)= qazb (96) a 0 ≥ b max X3a, 3a+3n ≥ { 2 } 3 s1 ,θ=1(n)= qazb (97) a 0 ≥3a+3n 1 b max 3Xa, − ≥ { 2 } Proof. We have

3 s1 ,θ=0(n)= sFZ(n)+ sMR,θ=0(n) i n i+j i j 1 i+j i+j j 3j+1 = q − z + (q + 1)q − − z + z + q z . j

13,θ=0 i n i+j i j i+j i j 1 i+j j 3j+1 s (n)= q − z + q − z + q − − z + q z . j

MR,θ=1 MR,θ=0 2j+1 n 3j+1 a b s (n) s (n)= q − z = q z . − n 1 − j

83 ✏ a ✏✏ b =3✏a ✏ ✏✏ 3a +3n θ ✑✑ b = − ✑ 2 ✑ ✑ ✑ b

Figure 2: Terms qazb appearing in the count of traced orders of a totally ramiﬁed cubic ﬁeld

3 In other words, s1 ,θ(n) can be viewed as a sum of terms qazb for integer points (a,b) in a certain region : R In proving (78) and (79), we observe that changing n and θ moves only the left side of the region , while multiplication by powers of z moves the whole region to the right. The diﬀerences appearing in (78)R and (79) have the pleasant property that the composite of these two transformations causes the left sides to coincide, causing cancellation everywhere but along the top side b =3a. In symbols:

3 3 s1 ,θ=0(n) z2s1 ,θ=1(n 1) = qazb qazb − − − a 0 a 0 ≥ ≥ b max X3a, 3a+3n b max 3Xa+2, 3a+3n ≥ { 2 } ≥ { 2 } = qazb a n 3a Xb≥3a+1 ≤ ≤ = (1+ z) qaz3a a n X≥ (1 + z)qnz3n = 1 qz3 − 3 = qnz3n(1 z2)s1 (0). − This proves (78), and the proof of (79) is analogous. As for equations (80)–(83), the linear combination (78) (80) reduces to a formula for sMR,θ=0, − 3 sMR,θ=0(n)= qnz3n+1(1 z)s1 (0), − while the linear combinations (81) (80), (79) (82), and (83) all reduce to a formula for sSR: − − 3 sSR(n)= qnz3n+1s1 (0).

These formulas are easily derived from the series formulations (92) and (91). Also easily derived from the foregoing is a formula for the number of traced orders in a given cubic algebra, in other words, for the coeﬃcient of a single power zb in one of the series g(L,t): Theorem 11.17 (the traced subring zeta function). Let L be a cubic algebra over a 3-adic ﬁeld K with d0 discriminant disc L = mK . Let d and t be integers satisfying the necessary restrictions d d , d d mod 2, 0 t e . ≥ 0 ≡ 0 ≤ ≤ K t d Then the number g(L,d,t) of mK-traced orders of discriminant mK in L is a linear combination of the three

84 functions

r 0, d< 3t 13 q 1 d g (d0,d,t)= − , r = t +1, 3t d 6t d q 1  3 − ≤ ≤ − 0 d d0 −  − +1, d 6t d0 6 ≥ − r  0, d< 3t 3 q 1 d g (d0 =0,d,t)= − , r = t +1, 3t d 6t q 1  3 − ≤ ≤ −  d 2 d +1, d 6t 2 − 6 ≥ r t SR q q  t, d< 6t g (d0 =0 or 1,d,t)= − , r = q 1 d t +1, d 6t − ( 6 − ≥ in a manner dependent on the splitting type of L:

3 • σ(L)=13: g = g1

3 • σ(L)=121: g = g1 + gSR • σ(L)=3: g = g3 • σ(L)=12: g = g3 + gSR • σ(L) = 111: g = g3 + g3SR.

Remark 11.18. When t =0 (and here we need no longer assume that char kK =3), we recover the formulas for orders in a cubic ﬁeld computed by Datskovsky and Wright and written more explicitly by Nakagawa and the author.

11.3 A bijective proof of the wild case Ideally we would desire a bijective proof of global reflection identities such as Theorem 11.3. This is still beyond reach. However, using the machinery of Galois cohomology and Poitou-Tate duality, we reduced this theorem to a local result, Theorem 11.14. In this section we will prove this theorem using a bijective method. It is not obvious what “bijective” means when trying to prove that two local weightings are Fourier 1 transforms of each other. Recall that in Theorem 7.1, we constructed level spaces H (T ) 0 ) 1 ) ) with the property that, for i, ⊇ L L ··· Le i⊥ = e i. (98) L L − As in the preceding proof, we extend this notation slightly. If e = 0 , which happens exactly when 1 L 6 { } T = K K is split, then let e+1 = 0 ; and dually, if 0 = H (T ), which happens exactly when T ′ is split, ∼ × 1 L { } L 6 then let 1 = H (T ). Then e e+1 consists of the unramified field extension of K, and 1 0 consists L− L \ L L− \ L of the uniformizer radical extensions (UREs) K[√3 π], the cubic extensions of maximal discriminant-valuation 3e +2. We have 1 11 1 e1 \e+1 = 1 and [1 =3q e+1 . L 3 L− L− L We also let

1 T ′ ∼= K K 1 ℓmin = − × min = ℓmin = H (T ), ( 0 otherwise, L L

e +1 T ∼= K K ℓmax = × max = ℓmax = 0 . ( e otherwise, L L { }

The level spaces i and their associated characteristic functions Li will be central to our proof. Our strategy is as follows: we groupL all cubic rings whose resolvent torsor is T into families with the following properties: F

85 • All rings in a family have the same discriminant and trace ideal.

• All rings in a family are contained in étale algebras L belonging to some level space i; this i is called the support supp(F ) of the family. L L F • Each L i has the same number of orders in the family; this number is called the thickness th( ) of the family.∈ L F The proof of Theorem 11.14 will then consist in exhibiting an involution between the families of support i and e i which aﬀects their thicknesses, discriminants, and trace ideals in such a manner that they contributeL L − equally to both sides of the theorem. The remainder of this section will be spent in carrying this out. Lemma 11.19. Let T/K be a quadratic torsor and let n = v (disc T ) 0, 1 . T K ∈{ } Then the cubic orders whose resolvent torsor is T can be partitioned into families n,k indexed by the pairs of integers (n, k) satisfying the conditions F n 0 k , n n mod 2, ≤ ≤ 3 ≡ T j k with the following properties: n min k,e 1. The rings in have discriminant ideal (π ) and trace ideal (π { }). Fn,k 2. The support and thickness of each n,k depend on which of three zones the pair (n, k) belongs to, as follows: F Zone (n, k) supp( n,k) th( n,k) n F F I 0 k< H0(T ) ≤ 6 Lmax | | n n n e n (99) 3 k II k + e 2k+ n q⌊ ⌋− 6 ≤ ≤ 3 − 6 2 L − ⌊ 3 ⌋ n n j e k n n k III + < k q 3 − 3 − 6 2 ≤ 3 Lmin ⌊ ⌋ j k j k The relative positions of these zones follow a pattern like that in quadratic O-N, up to O(1) discrepancies in the indices: t e ✲k 0 ❅ Zone III t ❅ ❅ Zone II ❅ ❅ ❅ Zone❅ I ❅ ❅ ❅ ❅ n k ❄ ❅ 3 − Proof of Theorem 11.14. Once Lemma 11.19 is proved, we can prove Theorem 11.14 quite simply by sending the family = to ′ = , where F Fn,k F Fn′,k′ n′ = n +3e 6t − n k′ = k + e t. 3 − − nj k n If the original satisﬁed the bounds t k , then it is easy to see that t′ k′ where t′ = e t, F ≤ ≤ 3 ≤ ≤ 3 − and likewise n nT mod 2 implies n′ nT ′ . Thus ′ is a family of rings of resolvent torsor T ′ whose trace ≡ e t ≡ F ideal is contained in (π − ). It is not hard to see that ′ lies in zone III, II, or I according as lies in zone I, II, or III. We leave it to the reader to check the neededF identities F

supp( ′) = supp( )⊥ F F supp( ) th( ′)= | F | th( ). F H0(T ) · F | |

86 Remark 11.20. Zone I, which is supported on max = 0 , consists precisely of those rings whose structure uses in an essential way that L is split, that is,L has more{ } than one ﬁeld factor. Zone II has the approximate shape of a band of constant width, n n e k + ; 6 ≤ ≤ 6 2 but the dependency on the value of n modulo 3 attests to a waviness of the boundary between zones II and III that cannot be avoided. Proof of Lemma 11.19. We now begin to enumerate all the orders in every cubic K-algebra L and arranging n0 k0 them into families. Let disc L = (π ) and tr( L) = (π ), and let θ = n0 3k0. As our investigations of the structure of cubic ﬁeldsO have found, we haveOθ 0, 1, 2 , the case θ =2−corresponding to the URE. ∈{ } Let [1, ξ0, η0] be a basis for L. We can arrange so that η0 is traceless and tr(ξ0) is a generator for the trace ideal tr( ). Any order CO L then has a unique basis of the form OK ⊆ i j [1, ξ = π ξ0 + uη0, η = π η0] (100)

j where i and j are nonnegative integers and u ranges over a system of coset representatives in K /π K . (Note the departure from the reduced basis used in the preceding subsection.) Such a C has discriminantO O valuation n = v(disc C)= n0 +2i +2j and trace ideal i min e,k0+i (3, tr(π ξ0)) = π { }.

Let k = k0 +i. With one exception, namely when L is a URE (which case we will handle later), we will place such a ring C into the family n,k. We must now compute the sizes of the families we have thus constructed. Whether or not a lattice FC with a basis (100) is actually a ring is determined by the integrality of its index form. The following lemma reduces the number of coefficients to be checked from four to two. Lemma 11.21. Let ξ, η be integral elements of a nondegenerate cubic algebra L over the field of ∈ OL fractions K of a Dedekind domain K such that the sublattice C = K 1,ξ,η is of full rank. If the outer coefficients O O h i

Φ L (ξ) Φ L (η) ΦC (ξ)= O and ΦC (η)= O π[ L:C] [ : C] O OL of the index form of C are integral, then the entire index form of C is integral and C is a ring.

Proof. If the whole index form of C is integral, then there is a ring C′ = 1, ξ′, η′ with the same index form h i (with respect to its basis) as C. Using the identity of index forms for L over K, we can embed C′ into L with ξ′ = ξ + u, η′ = η + v for some u, v K. But since ξ′,ξ,η′, η are integral elements and is integrally ∈ OK closed, we have u, v K so C′ = C. So it suffices to prove∈ O that the index form is integral. This is a local statement, so we may assume that is a DVR. Passing to a finite extension, we may assume that L = K K K is totally split. Let OK ∼ × × ξ = (a1; a2; a3) and η = (b1; b2; b3). Then 1 1 1 D = [ : C] = det a a a OL  1 2 3 b1 b2 b3 We are given that the outer coefficients   1 1 c = (a a )(a a )(a a ) and c = (b b )(b b )(b b ) 0 D 1 − 2 2 − 3 3 − 1 3 D 1 − 2 2 − 3 3 − 1 of the index form of C are integral. We wish to prove that the same applies to the two middle coefficients. By symmetry, we can consider just the x2y-coefficient

3 1 c = (a a )(a a )(b b ) . 1 D i − i+1 i+1 − i+2 i+2 − i "i=1 # X

87 (Here, and for the rest of the proof, indices are modulo 3.) It is easy to verify that 3 c = a +2a a + (a a )(a a )(b b ). 1 − 1 2 − 3 D 1 − 2 2 − 3 3 − 1 So it is enough to show that 1 d = (a a )(a a )(b b ) 1 D 1 − 2 2 − 3 3 − 1 is integral, or more generally any of the three 1 d = (a a )(a a )(b b ). i D i − i+1 i+1 − i+2 i+2 − i But we see that 2 a0a3 = d1d2d3.

Since a0 and a3 have nonnegative valuation, the three di cannot all have negative valuation, completing the proof. Remark 11.22. The hypothesis that L be nondegenerate is likely nonessential. We can now resume the proof of Lemma 11.19. Let the index form of be OL 3 2 2 3 Φ (xξ0 + yη0)= ax + bx y + cxy + dy . OL Of the coeﬃcients of the index form of C, we focus on the outer coeﬃcients,

3i 2i i 2 3 aπ + bπ u + cπ u + du 2j i Φ (ξ)= and Φ (η)= dπ − . C πi+j C

The latter coeﬃcient is the simpler one, depending only on i and j. Due to the tracelessness of η0, we have 3 c. We must then have π2 ∤ d, or else would be nonmaximal, and the discriminant analysis of | OL Proposition 11.9 shows that vK (d)= θ. Thus the condition ΦC (η) K comes out to 2j i + θ 0, which simpliﬁes to n 3k. ∈ O − ≥ (Incidentally,≥ when k e, the relation n 3k expresses an important relation between the trace ideal of a ring and its discriminant,≤ generalizing the≥ observation that an integer-matrix cubic form has discriminant divisible by 27.)

There thus remains the ξ-condition ΦC(ξ) K , which informally states that ξ is a root of Φ L modulo i+j ∈ O O π . Now Φ L is a homogeneous binary form, and it is natural to consider its roots on the projective lines 1 m O P ( K /π ); the factorization over the field K /m, for instance, gives the splitting type σ( L). However, in ourO situation there is a distinguished pointO on this projective line, at least for m 0. This is the motivation for the calculations to be undertaken now. O Suppose first that we are in zone I, that is, n > 6k, which translates into j > 2i + θ. Then the ΦC (ξ) condition j 2i i 2 2i 3 3i π − a + buπ− + cu π− + du π− | i is clearly dominated by a non-integral last term unless u is of the form π u′, in which case it simplifies to

j 2i 2 3 π − a + bu′ + cu′ + du′ . | j 2i In other words, ξ0 + u′η0 must be a root of Φ L modulo m − . The condition j 2i>θ rules out any O 2 − contribution from a multiple root modulo m, which never lifts to mod m (or else L would be nonmaximal). So the only roots that contribute are the simple roots occurring if L has splittingO type 111, 12, or 121. There are H0(T ) simple roots, and none of them are traceless (to be explicit, they are at (1; 0) for each | | decomposition L = K T into a linear and a quadratic algebra). By Hensel’s lemma, each simple root has ∼ × 0 j 2i a unique lift to any modulus. So the solutions u′ form a union of H (T ) congruence classes modulo m − . i j i | | Since u′ = u/π is deﬁned modulo m − , there are H0(T ) qi = H0(T ) qk | | · | | ·

88 rings for each pair (i, j). This completes the construction of the families n,k in zone I. Now suppose that (n, k) is in zone II or III, still assuming that L is notF a URE: we have 3k n 6k ≤ ≤ or, and the (i, j) coordinates, n i 2j + θ and j 2i + 0 θ. (101) ≤ ≤ 2 −

We claim that there are rings for this pair (i, j) if and only if L e 2k+ n , which may be also written as ∈ L − 3 n 2 ⌊ ⌋ k0 2k −3 or in (i, j) coordinates as ≤ − j 2i +1 θ. (102) ≤ − 3 Assume ﬁrst that k0 > 0, that is, L has splitting type 1 . Then the structure of L, and the fact that ξ is a depth element (a generator of the trace ideal) imply that vL(ξ)=2 θ, and hence that vK (a)=1 θ. Now it is easy to show that − −

v(Φ (ξ)) = v(aπ3i + buπ2i + cu2πi + du3) = min 3i +1 θ, 3v(u)+ θ OL { − } i+1 θ because the sum is dominated by its ﬁrst term if π − u and by its last term otherwise. So a necessary condition for there to be rings is that i + j 3i +1 θ, which| is equivalent to (102). If this condition holds, then the ξ-condition simply becomes ≤ − i + j θ v(u) − (103) ≥ 3 and we get i+j θ 2j+θ i n j − − k q −⌈ 3 ⌉ = q⌊ 3 ⌋ = q⌊ 3 − ⌋ solutions. If L has one of the other splitting types, then our task is simpliﬁed by the facts that k0 = 0 and n0 = θ 0, 1 . Note that the second inequality of (101) implies (102), so we are only trying to prove that ∈{ } 3 there are solutions in this case. If (103) does not hold, then the du term dominates in Φ L (ξ) and we do not get a solution. If (103) holds, we leave it to the reader to check the inequalities that implyO (even without i+j knowing anything about a, b, and c) that π divides each term of Φ L (ξ). So we get the same number of solutions as in the preceding case. O Lastly, we must address the exceptional case that L = K[√3 π] is a URE. We take ξ = √3 π and η = (√3 π)2, which are both traceless; and we have the explicit index form

Φ (xξ + yη)= x3 πy3. OL − This resembles the index form for a ramiﬁed L with θ =1, and analogously to that case, we compute that rings appear only for the pairs (i, j) with i 2j +1 and j 2i, (104) ≤ ≤ 2j+θ i − each such (i, j) yielding q⌊ 3 ⌋ solutions. Now we come to the least satisfying part of the bijection. In the absence of a distinguishing k (since all these rings are (3)-traced), we we simply have to place these rings into the families n,k of zone III so that the discriminant valuations and thicknesses match up. There is a unique choice: F

n = n0 +2i +2j =3e +2i +2j +2 n 2j +1 i k = − . 3 − 3 j k We leave it to the reader that this establishes a bijection between the pairs (i, j) in the region (104) with the pairs (n, k) in zone III. The discriminant and thickness are correct by construction, completing the proof.

n Remark 11.23. This lemma also yields a second proof of the number of t-traced rings of discriminant mK in a cubic algebra (Theorem 11.17). Remark 11.24. The method of the above proof can also be adapted to the tame case.

89 12 Non-natural weightings

1 Now that we know that the integral models Vt, V3t− of binary cubic forms are naturally dual, we can further look for duals for non-natural weightings. This has applications to counting cubic rings satisfying local conditions. We restrict ourselves to primes not dividing 3 . ∞ For simplicity we work over Z, though the techniques extend. Denote by MD the group Z/3Z with Galois action given by the quadratic character corresponding to Q(√D). Denote by V (R) the space of binary cubic forms over a ring R. As in Section 8, if W : V ( ) C is a locally constant weighting invariant under GL ( ), we denote OAQ → 2 OAQ by h(D, W ) the number of GL2Z-classes of binary cubic forms over Z of discriminant D, each form Φ weighted by W (Φ) . Stab Φ | GL2Z | If W = p Wp is a product of local weightings, then our local-to-global reflection engine (Theorems 8.12 and 8.13) produces identities relating different h(D, W ), if we can find a dual for each Wp. Q 12.1 Local weightings given by splitting types Let σ 111, 12, 3, 121, 13, 0 be one of the six splitting types a binary cubic form can have at a prime. Let ∈{ } T (σ)= T (σ): V (Z ) 0, 1 p p →{ } be the selector that takes the value 1 on binary cubic forms of splitting type σ. Then the associated weighted local orbit counter g (T (σ)) : H1 (Q ,M ) N D p p D → attaches to each cubic algebra L of discriminant K(√D) its number of orders of discriminant D and splitting type σ. There is another construction of interest to us. If a Q, then the varieties ∈ 2 VZ(D = D0) and VZ(D = a D0) (105) do not in general look alike. However, their base-changes to Q are isomorphic, being related by any g ∈ GL2(Q) of determinant a. Hence the two varieties (105) can be viewed as two integral models for VQ(D = D0). Coupled with Theorem 8.13, this viewpoint is very flexible. We denote by Zp the transformation that applies

1/p 1 to the vectors of an integral model of V and conjugates accordingly. Observe that Q G n 2n gD Tp(σ)Zp = gDp− (Tp(σ)) , and similarly for global class numbers. It is not hard to see that h(D, W ) is meaningful for any W in the Q-algebra generated by the Tp(σ)’s and the Zp’s for all p. Over Z , we still have Z , and we sometimes omit the subscript, as every Z with ℓ = p has no effect on p p ℓ 6 the integral model. We define Z = Zπ for integral models over a general local field similarly. Lemma 12.1. Let K be a local field, char k =3, and let D 0 . Then the weightings K 6 ∈ OK \{ } T (13)Z and 2 T (111) T (3) · − are dual with duality constant 1; that is, the associated local orbit counters satisfy

3 gˆ 3π2D(1 )=2gD(111) gD(3). (106) − −

90 Proof. The right-hand side of (106) can be written as

0 H (MD) 10 1 1 , | | · − Hur so it suﬃces to show that the left-hand side is the Fourier transform of this, namely

1 1 1 = 1 1 . − Hur Hram Look at binary cubic forms f(x, y) of splitting type 13 and discriminant 3π2D. Changing coordinates, we can assume − f(x, y)= ax3 + bx2y + cxy2 + dy3 x3 mod π. ≡ Then note that disc f 27a2d2 27d2 mod π3, so the only way that f can have discriminant 3π2D ≡ − ≡ − − is if π2 ∤ d. Then f is an Eisenstein polynomial, the index form of a maximal order in a totally ramiﬁed

1 1 extension L. Hence the weighting counting such f is Hram , as desired. Plugging this, together with the natural duality of Theorem 11.2 at the other primes, into Theorem 8.12 yields results such as the following: Theorem 12.2. Let p Z be a prime, p =3. For all integers D such that p ∤ D, ∈ 6 1 h ( 27p2D,T (13))=2h(D,T (111)) h(D,T (3)) (107) c 3 − p p − p ∞ 2 3 c h(p D,Tp(1 ))=2h3( 27D,Tp(111)) h3( 27D,Tp(3)) (108) ∞ − − − where c =3 for D> 0, c =1 for D< 0. ∞ ∞ 12.2 Discriminant reduction This can be used to improve a step that often occurs in arithmetic statistics, namely the production of discriminant-reducing identities that express the number of forms with certain non-squarefree discriminant in terms of lower discriminants. For N a positive integer, let h(D, RN ) be the number of classes of binary cubic forms of discriminant D, each weighted not only by the reciprocal of its number of automorphisms but also by its number of roots in P1(Z/NZ). Equivalently, consider the natural congruence subgroup a b GΓ0(N)= GL (Z): b 0 mod N , c d ∈ 2 ≡ 0 and let h(D, RN ) be the number of GΓ (N)-orbits of cubic 111N-forms (integral forms with a marked root) of discriminant D over Z, each weighted by the reciprocal of its stabilizer in GΓ0(N). (If N > 1, it is easy to prove that these stabilizers are trivial.) If 3 ∤ N, denote by h3(D, Rp) the analogous weighted count of 133N-forms. For a prime N = p, we have

2 h(D, Rp)=3h (D,Tp(111)) + h (D,Tp(12)) + 2h D,Tp(1 1) + h (D,Tp(111)) + (p + 1)h (D,Tp(0)) . This enables us to state succinctly the following theorem. Theorem 12.3 (discriminant reduction). Let p =3 be a prime, and D an integer divisible by p2. Then 6 D D D 1 27D 27D h(D)= h , R + h h , R + 2h − ,T (111) h − ,T (3) , p2 p p4 − p4 p c 3 p2 p − 3 p2 p ∞ where c =3 for D> 0, c =1 for D< 0. ∞ ∞ Proof. The cubic rings C counted by the left-hand side can be divided into maximal and nonmaximal at p. If C is maximal at p, then C Z Zp is the ring of integers of a totally tamely ramiﬁed cubic extension of Qp and p2 D. By Theorem 12.2,⊗ such rings are counted by the last term. k If C is nonmaximal at p, then C sits with p-power index inside an overring C′. By considering C′ + pC, we can take an inclusion C C′ of one of the following forms: ⊂

91 2 • C has index p in a C′ = C1 of discriminant D/p . Here the index form of C1 must have a marked root modulo p so that the transformation d Φ (x, y)= ax3 + bx2y + cxy2 + dy3 Φ (x, y)= p2ax3 + pbx2y + cxy2 + y3 C1 7−→ C p

2 keeps the form integral. This accounts for the term h(D/p , Rp).

2 4 2 • C has index p in a C′ = C2 of discriminant D/p with C2/C ∼= (Z/pZ) . This requires that the index form of C have content divisible by p; we have 1 Φ = C . C p 2

This accounts for the term h(D/p4).

Observe that C2 is unique if it exists. A choice of C1 corresponds to a choice of multiple root of ΦC , which is unique if ΦC is nonzero modulo p. Thus, the only chance of overcounting occurs when a C admits both a C2 and one or more C1’s. The C1’s are all the subrings of index p in C2 and thus correspond to the roots of ΦC2 modulo p. So we subtract 1 (more precisely, 1/ Aut C2 ) for each root of a form ΦC2 counted in the 4 4 | | term h(D/p ). That is, we subtract h(D/p , Rp), yielding the claimed total.

max More generally, we can reduce at multiple primes at once. Let Tp : V ( AQ ) Z be the selector for rings maximal at p. O →

Theorem 12.4 (discriminant reduction). Let q = p1 pr be a squarefree integer, 3 ∤ q. If D is a nonzero integer divisible by q2, then for any t < q, ···

D max h(D)= h 2 4 , Tp Rp (1 Rp) + q=q q q q2q3 −  1 2 3 p q1 p q2 p q3 qX1 t Y| Y| Y| ≤   1 27D 1 1 + h3 − , p2 D(Rp 1) p2 D(1 Rp) 2 2 k k c q=q q q  q1q3 − −  ∞ 1 2 3 p q1 p q3 qX1>t Y| Y|   and

27D max h3( 27D)= h3 −2 4 , Tp Rp (1 Rp) + − q=q q q  q2q3 −  1 2 3 p q1 p q2 p q3 qX1 t Y| Y| Y| ≤   D 1 1 + c h3 , p2 D(Rp 1) p2 D(1 Rp) . ∞ 2 2 k k q=q q q q1q3 − −  1 2 3 p q1 p q3 qX1>t Y| Y|   Remark 12.5. If we take t = √q, we ﬁnd that all discriminants appearing are at most 27D/q. − Proof. Since a ring of discriminant D is maximal or nonmaximal at each of the primes dividing q, we have

h(D)= h D, T max(13) T nonmax ,  p p  q q =q p q p q 1X2′ Y| 1 Y| 2′   max 3 max 3 nonmax max where Tp (1 )= Tp Tp(1 ) and Tp =1 Tp (as is natural). We transform each term in one of two ways, depending on· whether q t. − 1 ≤

92 nonmax 2 If q1 t, we simply replace each Tp by ZpRp + Zp (1 Rp) by the method of the preceding theorem, which respects≤ local conditions at other primes. We get a sum−

h D, T max(13) T nonmax  p p  p q p q Y| 1 Y| 2′   = h D, T max(13) (Z R + Z2(1 R ))  p p p p − p  p q p q Y| 1 Y| 2′   D = h , T max R (1 R ) . q2q4 p p − p  q =q q 2 3 p q p q p q 2′ X2 3 Y| 1 Y| 2 Y| 3   max 2 If q1 > t, we reﬂect. A dual of Tp , when restricted to discriminants D that are divisible by p , 1 nonmax max is p2 DZp(Rp 1) by Lemma 12.1. Hence a dual of Tp = 1 Tp on the same discriminants is 1 k − − 1+ p2 DZp(1 Rp). Applying the reﬂection theorem, k −

h D, T max(13) T nonmax  p p  p q p q Y| 1 Y| 2′   1 1 1 = h3 27D, p2 DZp(Rp 1) 1+ p2 DZp(1 Rp) c − k − · k −  ∞ p q1 p q′ Y| Y| 2   1 27D 1 1 = h3 − 2 2 , p2 D(Rp 1) p2 D(1 Rp) . c  q1q3 k − k −  ∞ q′ =q q p q1 p q3 2X2 3 Y| Y|   Summing over q1 yields the ﬁrst identity. The second is proved in the same way.

12.3 Subring zeta functions Fix a local ﬁeld K. Instead of restricting ourselves to local weightings taking values in N or C, consider the following (generalized) local weighting:

η : H1(K,M ) Z((Z)) D D → k L g 2k (L) Z . 7→ π D · k Z X∈ Note that the sum is a Laurent series since the discriminant π2kD is nonintegral for k sufficiently negative. This ηD is, up to renormalizing, (a local factor of) the subring zeta function that plays a central role in the study of Shintani zeta functions in works such as Datskovsky and Wright [18] and Nakagawa [38]. In like manner we can define k η (σ, L)= g 2k (σ, L) Z Z((Z)), D π D · ∈ k Z X∈ a partial subring zeta function that picks out the subrings of splitting type σ. We can even define

n n η (σZ )= η 2n (σ)= Z η (σ), D π− D · D which explains the use of the same symbol Z for the discriminant-shift operator and the formal variable in the subring zeta function. A formula for the subring zeta function ηdisc L(L), without splitting-type selector, is computed by Dats- kovsky and Wright [18] and put into a more explicit form by Nakagawa:

93 s Theorem 12.6 (Datskovsky–Wright; Nakagawa). The subring zeta function ηdisc L(L)(Z = p− ) is given by F η (L)= disc L (1 Z)(1 qZ3) − − where F is a polynomial depending on the splitting type σ( ) as follows. OL

σχ F 111 (1+ Z)2 12 1+ Z2 (109) 3 1 Z + Z2 121− 1+ Z 13 1

Proof. Although Datskovsky–Wright [18] and Nakagawa ([38], Lemma 3.2; see Lemma 3.6 for the simpliﬁ- cation method) work only with cubic extensions of Q, their method applies to this case. Alternatively, it follows immediately from Theorem 11.17.

Remark 12.7. When working over the field K = Q, the local subring zeta functions at each prime p form an Euler product expansion of the Dirichlet series ζ (s) L ζ(2s)ζ(3s 1), ζL(2s) − where ζ and ζL are the Riemann and Dedekind zeta functions, respectively. In principle, Theorem 11.17 allow us to write each Shintani zeta function ξK,σ,t in Definition 11.5 as an infinite sum of Euler products, one factor for each cubic étale algebra L/K, as was done for t = (1) by Datskovsky and Wright ([18]; see also [56]).

We turn to the computation of ηdisc L(σ, L), in which subrings are ﬁltered by splitting type. Happily the answers are not too hard to deduce from Theorem 12.6. The easiest cases are σ = 111, 12, and 3, which only occur in maximal orders: thus for these three values of σ,

1 if σ( L)= σ ηdisc L(σ, L)= O ( 0 otherwise.

2 Next we compute ηdisc L(1 1,L). Lemma 12.8. G η (121,L)= disc L 1 Z − where G is a polynomial depending on the splitting type σχ as follows.

σχ G 111 3Z 12 Z (110) 3 0 121 1 13 0

Proof. A ring C of splitting type 121 has its corresponding cubic form (in suitable coordinates) congruent to a multiple of xy2 modulo p. Write

3 2 2 3 ΦC (x, y)= pax + pbx y + cxy + pdy where p ∤ c. The ring C is non-maximal iﬀ p a, in which it is of index p in a unique overring C′ whose cubic form | a Φ (x, y)= x3 + bx2y + pcxy2 + p2dy3 C′ p

94 has a distinguished simple root, here at [0 : 1]. Conversely, a ring with a distinguished simple root has one subring of index p and splitting type 121. Now the only non-maximal rings whose cubic forms have a simple root are themselves of splitting type 121: thus the number of these is constant at index p,p2,.... The value of the constant is the number of simple roots of the maximal order. Thus we have the desired claim.

As for the subrings of splitting type 0, they are + πC for each subring C, and so OK

ηD(0,L)= ηD/π2 (L)= ZηD(L)

3 for all D and L. Finally, ηD(1 ,L) can be computed by subtracting oﬀ all the other splitting types from ηD(L). We do not need the explicit value in this case. Before computing the Fourier transforms of the ηD(σ, L), it is helpful to compute them for some simpler local weightings.

2 3 1 Lemma 12.9. Let p =3 be a prime. For σ 111, 12, 3, 1 1, 1 and D K×, let tD(σ): H (MD) N be the local weighting given6 by ∈{ } ∈ → 1 σ(L)= σ tD(L) := ( 0 otherwise.

Let TD be the Q-linear span of the tD(σ). Then the Fourier transform yields an isomorphism

: T 3D TD (111) − → given explicitly by b

(a) (t 3D(111) + t 3D(12) + t 3D(3)) = (t 3D(111) + t 3D(12) + t 3D(3)) − − − − − − 2 2 (b) t 3D(1 1) = tD(1 1) − b 3 (c) (2t 3D(111) t 3D(3)) = tD(1 ) − b − − 3 (d) t 3D(1 ) =2tD(111) tD(3) − −b

tD(12) q 1 mod3 (e) (t 3D(12))b = ≡ − t (111) + t (3) q 2 mod3 ( D D ≡ where q = k . b | K | Proof. We may assume that D = D0 K is a fundamental discriminant. Parts (a) and (b) reduce to the 1 ∈ O self-orthogonality of H for D × and D π × , respectively. Parts (c) and (d) were proved after D,ur ∈ OK ∈ OK showing that both sides of Lemma 12.1 concern only D × . Finally, part (e) reduces to part (a), using ∈ OK the fact that 3 × is a square if and only if q 1 mod 3. ∈ OK ≡ 2 Theorem 12.10. The weightings ηD, ηD(111), ηD(12), ηD(3), ηD(1 1) span TD((Z)) := TD Q Q((Z)). Their Fourier transforms are given by ⊗

(a)η ˆ 3D = ηD; − (b) If q 1 mod 3, then ≡

(η 3D(111) + η 3D(3)) = ηD(111) + ηD(3) and η 3D(12) = ηD(12); − − −

(c) If q 2 mod 3, then b b ≡

(η 3D(111) + η 3D(3)) = ηD(12) and η 3D(12) = ηD(111) + ηD(3); − − −

1 3 2 (d) (2η 3D(111) η 3D(3)) = (1 Z)(1b qZ )ηD +(Z 1)(b ηD(111)+ηD(12)+ηD(3))+Z (2ηD(111) − − − Z − − − − η (3)) (1 + Z)2η (121) ; D − D b 95 2 1 3 (e)η ˆ 3D(1 1) = (1 Z)(1 qZ )ηD + (Z 1)(ηD(111) + ηD(12) + ηD(3)) + Z(Z 1)(2ηD(111) − 1+Z − − − − − η (3)) Z(1 + Z)η (121) . D − D Proof. We may assume that D = D0 K is a fundamental discriminant. We need the following result, which is useful in its own right: ∈ O

Lemma 12.11. Let D = disc L be the discriminant of a maximal cubic order over a local field K, char kK = 2k O 6 3. Write D = π D0 where D0 is the associated fundamental discriminant. Then 1 σ( )=13 k = OL ( 0 otherwise. Proof. If L is unramified, then π ∤ D and k =0. If σ(L)=121, then D = disc L = disc(K K( D )) = D . × 0 0 3 2 So we are left with the case that σ(L)=1 . Note that L/K isp tamely ramified so D = π D′, π ∤ D′. We need 3 2 to prove that D′ is a discriminant, which is only nontrivial when char kK =2. Let f(x)= ax +πbx +πcx+πd be an Eisenstein polynomial for L. Then 2 2 2 3 2 3 2 2 2 D′ = π b c 4πac 4π b d 27a d + 18πabcd (ad πbc) mod 4, − − − ≡ − so D′ = D0.

2 This allows us to change the scaling in Lemma 12.8 (on ηD(1 1)) and Theorem 12.6 (on ηD) from ηdisc L 3 to ηD, multiplying the weighting by Z in the case that σ( L)=1 . Also, trivially ηD(σ) = tD(σ) for unramiﬁed σ. So we have ﬁve equations O

ηD(111) = tD(111)

ηD(12) = tD(12)

ηD(3) = tD(3) 3t (111) + t (12) + t (121) η (121) = D D D D 1 Z − (1 + Z)2t (111)+ (1 + Z2)t (12) + (1 Z + Z2)t (3)+(1+ Z)t (121)+ Zt (13) η = D D − D D D . D (1 Z)(1 qZ3) − − We see that there is an invertible transition matrix between the sets t (111),t (12),t (3),t (121),t and η (111), η (12), η (3), η (121), η { D D D D D} { D D D D D} is invertible (we cannot call these “bases,” because tD(σ) can vanish for certain D). Accordingly, we can rewrite Lemma 12.9 in terms of the η’s and get the identities claimed in the theorem.

Problem 12.12. Can the rich structure found in this section be carried out, to some extent, when char kK = 3? For instance, Vˆ (F3), the space of integral 1331-forms modulo 3993-forms, has six GL2(F3)-orbits, the analogues of splitting types, and it is natural to wonder whether the Fourier transform relates them to the six splitting types on V (F3).

12.4 Invariant lattices at 2 The foregoing investigation also allows us to extend the work of Ohno and Taniguchi [45] on extending O-N to counting binary cubic forms satisfying certain congruence conditions at 2. Assume that q = 2. Recall from Section 11 the ﬁve primitive invariant lattices Λ V ( ) of binary cubic forms: i ⊆ OK Λ = f(x, y)= ax3 + bx2y + cxy2 + dy3 : a,b,c,d 1 { ∈ OK } Λ = f Λ : σ(f) 0, 12 2 { ∈ 1 ∈{ }} Λ = f Λ : σ(f) 0, 111, 3 3 { ∈ 1 ∈{ }} Λ = f Λ : σ(f) 0, 111, 12, 13 4 { ∈ 1 ∈{ }} Λ = f Λ : σ(f) 0, 111 5 { ∈ 1 ∈{ }}

96 Let

λ (D): H1(K,M ) Z((Z)) i D → k L g 2k (Λ ,L) Z 7→ π D i · k Z X∈ be the analogue of the subring zeta function counting only those rings whose corresponding form is in Λi. Each λi(D) is a linear combination of the appropriate ησ,D, and the transition matrix is again invertible. (No reason is known, beyond pure coincidence, why the number of invariant lattices over Z2 should equal 5, the number of independent ησ,D.) So we get: Corollary 12.13 (Local O-N for general invariant lattices; cf. [45], Theorems 1.2, 1.3 and 1.4). Let K be an unramiﬁed extension of Q2. The Fourier dual of each λi( 27D) lies again in the span of the λi(D), explicitly: − λˆ ( 27D)= λ (D) 1 − 1 λˆ ( 27D)= λ (D) 2 − 3 λˆ ( 27D)= λ (D) 3 − 2 1 λˆ ( 27D)= ( 8Z3 +6Z2 Z)λ (D)+(2Z 1)λ (D) + ( 8Z2 +4Z)λ (D) 4 − 3Z − − 1 − 2 − 3 + ( 4Z2 + 1)λ (D)+(16Z2 1)λ (D) − 4 − 5 1 λˆ ( 27D)= ( 2Z3 +3Z2 Z)λ (D)+(2Z 1)λ (D) + ( 2Z2 + Z)λ (D) 5 − 3Z − − 1 − 2 − 3 + ( Z2 + 1)λ (D)+(4Z2 1)λ (D) − 4 − 5 Proof. The λi(D) are simply the ηD(σ) in disguise: for instance,

λ2(D)= ηD(0) + ηD(12), and so on. Rewriting the results of Theorem 12.10 in terms of the λi and plugging in q = 2 proves the theorem. The five equalities in this corollary can be viewed as local reflection theorems in the sense of Theorem 8.13, relating different integral models of the same composed variety on each side. We can now get global results in great generality. Theorem 12.14 (O-N for general invariant lattices). Let K be a number field. Let a be an ideal of K, and let Λ be a = SL( a)-invariant lattice of full rank in the space V (K) of binary cubic forms. Let t Ga OK ⊕ be the trace ideal of Λ, that is, the unique t 3 such that Λ = (p) at every prime p 3 (after identifying | p ∼ V(1,t) | a with SL2, which we can do after localization). For D K×, denote by hΛ(D) the number of a-orbits of Gbinary cubic forms of discriminant D, each orbit weighted∈ by the reciprocal of its stabilizer. G Then:

1. If Λ is of type Λ1, Λ2, or Λ3 at every prime whose residue ﬁeld is F2, then there is a lattice Λ∗ invariant under 3 such that we have the global reﬂection theorem: for each D K×, Gat− ∈ 2 # v :D (K×) 3 { |∞ ∈ v } hΛ(D)= hΛ∗ ( 27D). (112) N /Z(t) − OK

2. In general, there is a family Λ1∗,..., Λm∗ of lattices, all equal away from 2, each Λj∗ invariant under a 3 group asj t− ,... where sj is an ideal having nonzero valuation only at primes dividing 2, and a global reﬂectionG theorem of the shape

2 # v :D (K×) m 3 { |∞ ∈ v } hΛ(D)= cj hΛ ( 27D), N (t) j∗ − K /Z j=1 O X where the c Q depend only on Λ. i ∈

97 Proof. Theorem 11.1 limits the lattices we must consider. If Λ is of type Λ1 at each prime dividing 2, we have Λ= c a,t( K ). But since V O 2 c c = 1 c , ⊕ ∼ ⊕ there is an isomorphism of integral models

(c , ) = 2 , 2 . Va,t Ga ∼ Vac− ,t Gac− So we can take Λ∗ = c 3 1 , Vat− ,3t− and the desired reflection theorem follows from Theorem 11.3. The remaining cases can be solved with a bit of fiddling at 2. Let Λ¯ Λ be the lattice that sits over Λ ⊇ at each p 2 as Λ sits over the relevant Λ , 1 i 5, and let Λ¯ ∗ be the corresponding reflection lattice. All | 1 i ≤ ≤ Λi∗ will look like Λ∗ away from 2. k Locally at each p 2, we construct a collection of Λj,∗ p, 1 j mp as follows: for each term Z λi(D) in Corollary 12.13, take| the lattice ≤ ≤ πk Λ = Λ , j,∗ p 1 i 2k 1 for which π− -orbits of discriminant D correspond to SL2-orbits of discriminant D/π in Λi. Then Corollary 12.13 appearsG as a local reflection theorem

gˆΛ (D) = cj,p gΛ (D). p · j,∗ p j=1 X for the integral models Λp and Λj,∗ p. (Strictly speaking, these are not truly integral models, inasmuch as the SL2-invariance of Λ2,..., Λ5 is not given by an algebraic integrality; but we get the same local and global class numbers by going up to a lattice of type Λ1 and imposing non-natural weights to pick out the appropriate splitting types.) We then apply Theorem 8.13 to these local reﬂection theorems and, as usual, the ones relating Λ¯ and Λ¯ ∗ away from 2. The resulting reﬂection theorem involves the global integral models given by gluing the Λj,∗ p at each p 2 in all possible ways. When Λ is of type Λ1, Λ2, Λ3 at each p 2, there is only one Λ∗, and s, which arises| from the Z-operators in Corollary 12.13, disappears. |

12.5 Binary cubic forms over Z[1/N] For N a squarefree integer, it is natural to ask what happens if we invert finitely many primes and count binary cubic forms of discriminant D =0 over Z[1/N], up to the action of the relevant group SL2(Z[1/N]). There are still only finitely many for each6 degree, owing to Hermite’s theorem on the finiteness of the number of number fields with prescribed degree and set of ramified primes. Note that D matters only up to multiplication by the squares in Z[1/N]×; hence we can restrict our attention to D Z that are fundamental at each prime p N. (If p = 2, this means that p2 ∤ D. If p = 2, this means that∈D 1 mod4 or D 8, 12 mod 16. However,| we allow6 D to be non-fundamental at primes not dividing N.) ≡ ≡ We do not have O-N for forms over Z[1/N] in the same formulation as over Z. Nevertheless, the other side of the reflection theorem is noticeably not too complicated. Theorem 12.15. Let N be a squarefree integer. For 0 = D Z[1/N], let h (D) be the number of SL (Z[1/N])-orbits of integral binary cubic forms 6 ∈ Z[1/N] 2 over Z[1/N], each weighted by the reciprocal of its stabilizer in SL2(Z[1/N]). If 3 ∤ N, define h3,Z[1/N](D) to be the same count, counting only 1331-forms (that is, forms whose middle two coefficients belong to the ideal 3Z[1/N] ( Z[1/N]). Now let D be a discriminant that is fundamental at all primes dividing N. For each p N , let | ∞ 2 3 if D Z× ∈ p c = D,p  1 if D is a non-square unit modulo p,  1/2 if p D. |  98 and let cD,N = cD,p. ∞ p N |Y∞ Then:

(a) If 3 ∤ N, then

c 27D,N h (D)= − ∞ h ( 27D, R ) Z[1/N] 3 · 3 − N h3,Z[1/N]( 27D)= cD,N h(D, RN ). − ∞ · (b) If 3 N, then | hZ[1/N]( 27D)=3cD,N h3(D, RN ). − ∞ · Proof. We take the same composed variety (V, Γ) of binary cubic forms as before. However, we take integral models ( (i), G(i)) that are not even over the same ring of integers (i)! V OQ On the left-hand side, we take the scheme (1) of binary cubic 111N- or 133N-forms of discriminant V D or 27D. This does not admit an algebraic action of SL2Z, but it does admit an algebraic action of G(1) =− GΓ0(N). On the right-hand side, we take the scheme (2) of binary cubic forms over Z[1/N] of discriminant D, or V (2) 1331-forms as appropriate, with the natural action of G = SL2 over Z[1/N]. It is evident that the global class numbers of these integral models match the quantities studied in the theorem. The checking of most of the conditions of Theorem 8.12 is routine, so we content ourselves with checking the local duality. When p ∤ N, the integral model is identical to that used for O-N, so we already have the needed duality with an appropriate duality constant cD,p or c 27D,p. This includes the infinite prime, at which the duality − constant cD, tracks the sign of D as in O-N. When p ∞N, the computation of the local class numbers is not difficult: | • As to (1), we look for forms of discriminant D with a marked root modulo p. We first observe that forms correspondingV to nonzero cohomology classes are not counted, because they either have splitting type

– (3), and have no roots modulo p, or – (13), and have discriminant non-fundamental at p, by Lemma 12.11.

(1) 1 So g : H (Zp,M ′) N is a scalar multiple of 10, nonzero because the split ring Zp Zp[(D +√D)/2] has an index form with→ a root. ×

• As to (2), since the completion of Z[1/N] at p is Q , the local orbit counter counts cosets in V p SL (Q ) SL (Q ) that keep a certain form f “integral” over Q . There is obviously only one such 2 p \ 2 p p coset, regardless of the cohomology class of f, so g(2) : H1(Z ,M) N is identically 1. p → It remains only to compute the duality constant.

• If Qp[√D] is split, then there are three roots of f(x, y) = xy(x + y) to mark, but they all wind up equivalent. So g(1) = 1 , but because of the H0 =3 in the scaling of the Fourier transform, we need 0 | | to insert a factor of cD,p =3.

• If Q [√D] is inert, there is only one root to mark, and H0 =1, so c =1. p | | D,p

• If Qp[√D] is ramified, then we can mark either the single or the double root modulo p. These are non-equivalent GΓ0(p)(Z )-orbits inside the same SL (Q )-orbit, so g(1) = 2 1 and H0 = 1, so p 2 p · 0 | | cD,p =1/2. (If we modified the theorem by counting 111N-forms whose third coefficient c is coprime to N, another family stable under GΓ0(N), then we would be forced to mark the simple root, and this factor of 1/2 would disappear.)

99 Multiplying the duality constants obtained completes the proof.

Problem 12.16. Does the integral model of 111N-forms of non-fundamental discriminant D have a natural dual? The ﬁrst step in answering this is to check whether the Fourier transform of its local orbit counter takes nonnegative values.

Part VI Reﬂection theorems: quartic rings and related objects

13 Reﬂection for 2-adic quartic orders, and applications

Analogously to the cubic case, the reflection theorem that we state and prove is going to swap t-traced and 1 2t− -traced orders. It is not the most general reflection theorem that one can try to state: see Section 24 below. Fix a nondegenerate cubic ring C over a Dedekind domain . We can also fix a basis OK C = ξ aη, OK ⊕ OK ⊕ making the index form ΦC (xξ + yη) a cubic in the a,1 of Theorem 11.3. However, none of our work will depend on this basis. V We can then look at the scheme of pairs of ternary quadratic forms Vt,C ( , ): a ⇒ a, A B OK × OK × OK × that is, pairs of 3 3 symmetric matrices with entries in the ideals × 1 1 1 1 1 (1) 2− t 2− ta− a 2− ta 2− t 1 1 1 (1) 2− ta− , a 2− t  2   1  a− a−     satisfying the four equations det( x + y)=Φ (x, y) A B C asserting that ( , ) parametrizes a quartic ring L that is t-traced with reduced resolvent C. This t,C , together with theA naturalB action of the group = GL( a), is an integral form of the composedV G OK × OK × variety (V, GL3) of pairs of ternary quadratic forms over K. We assert that the integral models

and 1 Vt,C V2t− ,C are naturally dual at all ﬁnite places.

Notation 13.1. Here and in the sequel, we use the label “Theorem*” to denote a theorem proved with the following caveats:

(1) Resolvents that are wildly ramiﬁed at a 2-adic place are excluded. (2) In general, the theorem depends on a Monte Carlo veriﬁcation of a rational algebraic identity (as we will explain). However:

(3) The results for K/Q2 unramiﬁed (e.g. the cases over Z) are known unconditionally. (4) The results for the reduced resolvent being maximal (C = ) are also known unconditionally. t OR

100 Theorem* 13.2 (“Local Quartic O-N”). Let K be a nonarchimedean local field and C an order in an étale algebra R that is not wildly ramified over K. For τ a divisor of 2, let be the integral model Vτ,C parametrizing (τ)-traced orders with reduced resolvent C. Then and 1 are naturally dual with Vτ,C V2τ − ,C duality constant q2vK (τ); in order words, the associated local orbit counters g : H1(K,M ) N t,C R → satisfy the local reflection theorem 2 gˆ = /τ g 1 . t,C |OK OK | · 2t− ,C At the infinite places, we no longer have natural duality. (We did not have this problem in the cubic case because H1(R,M) is trivial for M odd.) Therefore, call a quartic algebra| |L/K over a number field nowhere totally complexified (ntc) if there is no 1 2 real place p of K such that Lp ∼= C C. This is equivalent to the cohomology element σL H (K, (Z/2Z) ) × 1 ∈ being trivial at all infinite places. Then the local specifications 0 for ntc quartic algebras and 1 for all quartic algebras are mutually dual, provided that one inserts the correct{ } scale factor. Theorem* 13.3 (“Quartic O-N”). Let K be a number field. Let C be an order in a cubic K-algebra R, and let t K be an ideal such that t (2). Let h(C, t) count the number of t-traced quartic rings with reduced resolvent⊆ O C, respectively, each weighted| by the reciprocal of its number of resolvent-preserving automorphisms. Let hntc(C, t) count the subset of the foregoing that are ntc, weighted in the same way. Then

2 N(t) ntc 1 h(C, t)= r h (C, 2t− ), 2 ∞ · where r is the number of real places of K over which R is not totally real plus twice the number of complex places of∞K. (1) (2) 1 Proof. We apply Theorem 8.12 to the composed varieties = t,C and = 2t− ,C just deﬁned, with (i) V V V V the following local weightings wv : (i) • At ﬁnite v ∤ 2, we take wv =1, which are mutually dual with duality constant 1 by Theorem 14.6.

(i) • At v = p 2, we take wv =1, which are mutually dual with duality constant | N(p)2vp(t) by Theorem* 13.2.

(i) • At complex v, we take wv =1, which are mutually dual with duality constant 1 1 fˆ = = . v H0(K ,M ) 4 | v R | (i) • At real v for which R = R C, we take wv =1, which are mutually dual with duality constant v ∼ × 1 1 fˆ = = . v H0(K ,M ) 2 | v R | (1) (2) 1 • At real v for which Rv ∼= R R R, we take wv = 1 and wv = 0, the selector for rings that are not totally complex at v. The× corresponding× duality constant is

1 H (Kv,MR) 4 0 = =1. H (Kv,MR) 4 | | The product of all duality constants is thus 2 2vp(t) 1 1 N(t) N(p) = r , · 4 · 2 2 ∞ p 2 Kv =C Rv =R C Y| Y∼ Y∼ × as desired.

101 Although we have been counting quartic rings by resolvent, the corresponding result where we count by discriminant follows quickly. We present the reﬂection theorem in two forms, one dealing with 2 3 3 symmetric boxes, the other with quartic rings (which are in bijection only in the case of rings of content× ×1): Theorem 13.4. Let K be a number ﬁeld. Denote by ( ) the space of pairs of ternary quadratic forms Vt,a OK ( , ): a ⇒ a, A B OK × OK × OK × that are t-traced in the sense that the entries belong to the ideals

1 1 1 1 1 (1) 2− t 2− ta− a 2− ta 2− t 1 1 1 (1) 2− ta− , a 2− t .  1    a− (1)     It has a natural action of the group

= SL( a) SL( a) Ga OK ⊕ OK ⊕ × OK ⊕ that preserves discriminant. Denote by h ( ) the number of orbits of pairs of ternary quadratic forms t D Gt,a having discriminant = (a,D), each orbit weighted by the reciprocal of the order of its stabilizer in a. Denote by hntc( ) theD number of such orbits (weighted in the same way) which are ntc, in the sense thatG at t D each real place of K, the conics and have a common point in RP2. Then for all discriminants prime to 2, A B D 2 8 N(t) ntc 8 ht(t )= h 1 (256t− ), r 2t− D 2 ∞ · D where r is the number of real places of K at which D< 0 plus twice the number of complex places of K. ∞ Proof. We sum the preceding theorem over all cubic rings C of discriminant , weighting each C by the reciprocal of the number of orientation-preserving automorphisms of C, whichD is the stabilizer of the corresponding form in SL( K a). It is easy to see that each orbit is counted the number of times it appears in the theorem. BecauseO the⊕ reduced discriminant is prime to 2 (a needed condition to avoid involving wildly ramiﬁed resolvents in the sum), we can state the theorem unconditionally.

Theorem 13.5. Let K be a number field, and let = (a,D) be a discriminant. Denote by h◦( ) the number D t D of t-traced quartic rings over K having discriminant , each weighted by 1/ AutK ( ) . Denote by ,ntc O O D O | O | ht◦ ( ) the number of such that are ntc, weighted in the same way. Then for all discriminants prime to 2, D D 2 8 N(t) ,ntc 8 h◦(t )= h◦ 1 (256t− ), (113) t r 2t− D 2 ∞ · D where r is the number of real places of K at which D< 0 plus twice the number of complex places of K. ∞ Proof. In the previous theorem, we studied h (t8 ), which can be interpreted as the number of quartic t D rings equipped with a resolvent C and an orientation, that is, an identification Λ4 = a for which O 8 O ∼ the discriminant is t . Every quartic ring admits two orientations (there are K× -many identifications 4 a D |O | 1 t8 Λ ∼= , but all but one and its negative yield a D scaled by a different square of a unit). So 2 ht( ) is theO number of resolvents ( ,C, Θ, Φ) of discriminant t8 , up to isomorphism, each weighted by the reciprocalDD of its number of automorphisms.O D 1 8 Let ht (t ) be the number of quartic rings of discriminant (a,D) with t-traced content 1, weighted by 1/ Aut D. This is related to If is an ntc quarticO ring of discriminant t8 having some t-traced content | O| O 3D8 6 8 c, then = K + c ′, where ′ has t-traced content 1, discriminant (ac− t ,D) = c− t , and the same automorphismO O groupO as . ThusO D O 1 8 1 8 6 h (t )= h (t c− ). 2 t D t D c3 Da2 X| On the other hand, the number of resolvents of depends on the t-traced content c (Proposition 6.17(b)): it is O σ1(c)= NK/Q(d). d c X|

102 These are resolvents as maps out of (as pointed out in [42], end of Section 8), which is the correct manner of counting to make O 8 1 6 8 h◦(t )= σ (c)h (c− t ). t D 1 t D c3 Da2 X| We can now write ht◦ in terms of ht:

13.1 Results on binary quartic forms We can also derive a reﬂection theorem about binary quartic forms, which correspond (via a completely general construction for binary n-ic forms) to a certain subclass of quartic rings. This subclass was identiﬁed explicitly by Wood: Theorem 13.6 ([59], Theorem 1.1). There is a natural, discriminant preserving bijection between the set of GL2(Z)-equivalence classes of binary quartic forms and the set of isomorphism classes of pairs (Q, C) where Q is a quartic ring and C is a monogenized cubic resolvent of Q (where isomorphisms are required to preserve the generator of C modulo Z). Proof. Regarding pairs (Q, C) of quartic rings as pairs (A, B) of 3 3 symmetric matrices via Bhargava’s parametrization, we send a form Φ(x, y)= ax4 + bx3y + cx2y2 + dxy×3 + ey4 to 1/2 a b/2 (A0,B)= 1 , b/2 c d/2 . 1/2 −   d/2 e      The distinguished generator arises because the resolvent form g(x, y) = 4 det(A x By) is monic, since 0 − det A0 =1/4. Further details will be found in [59]. To apply this theorem, we need to know the number of automorphisms of the quartic ring corresponding to a given form: Lemma 13.7. In this bijection, the group of resolvent-preserving automorphisms of a quartic ring is in natural isomorphism with the stabilizer (in PGL2(Z)) of the corresponding form. Proof. The conclusion follows easily from the method of proof of the preceding theorem. By [59], Theorem 2.5, we can choose bases for Q and C so that the corresponding pair of ternary quadratic forms, has the form (A0,B) above. A resolvent-preserving automorphism is a change of variables h SL3(Z) that preserves ˜ ∈ both A0 and B. By [59], Lemma 3.2, h = εA0 (h) lies in the image of the map ε : GL (Z) SL (Z) A0 2 → 3 a2 ab b2 a b 1 2ac ad + bc 2bd c d 7→ ad bc   − c2 cd d2   By [59], Theorem 3.1, h preserves B if and only if h˜ preserves the binary quartic form f. Moreover, it is easy to see that ker ε = 1. This constructs the desired isomorphism. A0 ±

103 We ﬁx a monic binary cubic form g(x, y) and let C = Z[ξ] be the corresponding monogenized cubic ring, with generator ξ. Then in the notation of Theorem* 13.3, 1 1 h C, (1) = = . AutC Q StabPGL Z f quartic rings Q | | binary quartics f(x,y) | 2 | with resolventX C, with resolventX g, up to C-isom up to PGL2Z

The quantity hntc C, (2) appearing on the opposite side of Theorem* 13.3 is not so straightforward to interpret. Here we are counting pairs (A, B) of integer symmetric matrices with det(Ax By)= g (with a certain condition at ), so we need to classify integer symmetric matrices A with det A =1− . There are, up to similarity, two: ∞

Lemma 13.8. Every integer symmetric matrix A with det A =1 is similar to

1 1 A1 = 1 or I = 1 . 1 −   1     Proof. Let A be an integral symmetric matrix of determinant 1. Look at the corresponding conic defined C by x⊤Ax =0. Note that for each rational prime p 2, , we have p ∤ 4 = det , so has good reduction 6∈ { ∞} C C to F : by the Chevalley-Warning theorem, has an F -point and hence a Q -point. Then, by Hilbert p C p p reciprocity, there are only two possibilities for the isomorphism type of over Q: C • If has an R-point, then also has a Q -point and hence (by the Hasse principle) a Q-point. By a C C 2 GL3Z-transformation, we set this point to [1 : 0 : 0], the tangent line there to go through [0 : 1 : 0], and then A must take the form 1 ± 1 a ,  1 −a b  ±   which one easily sees is similar to A1. • If has no R-points, then A is positive (or negative) definite. It is well known that the only positive definiteC integral unimodular matrix of any rank is the identity.

ntc Hence h C, (2) decomposes into the pairs (A, B) of “type A1” and of “type I” according to the value of A after an appropriate SL Z-transformation. 3

13.1.1 Type A1 To understand pairs (A ,B), we capitalize on the fact that A A over Q. Namely, the transformation 1 1 ∼ 0 1 T = 1  2   satisﬁes T A0T ⊤ = A1. Let a b c′ (A ,B)= A , b c d . 1  1   c′ d e Then the pair    1 1 1 1 T − A1 T − ⊤ ,T − B T − ⊤ = (A0,B′) is determined up to SO(Q, A0) by (A1,B), and the form

4 3 2 2 3 1 4 f(x, y)= ax +2bx y + (c + c′)x y + dxy + ey 4

104 1 is determined up to PSL2(Q). This is a form of a peculiar shape, the (1, 2, 1, 1, 4 )-forms. Since the Wood 1 embedding is resolvent-preserving, we see that the resolvent of a (1, 2, 1, 1, 4 )-form is actually integral, which can also be deduced directly from the formula for the resolvent of a binary quartic. 1 The (1, 2, 1, 1, 4 )-forms do not naturally have an action by PGL2Z, but rather by a group that we can reveal as SO(Z, A1):

Lemma 13.9. We have SO(Q, A1) ∼= PGL2(Q) via the isomorphism 2 1 2 a ab 2 b a b 1 1 ε : M = Tε (M)T − = 2ac ad + bc bd A1 c d 7→ A0 ad bc   − 2c2 2cd d2.   Under this map, the subgroup corresponding to SO(Z, A ) is G = GΓ0(2) τGΓ0(2), where 1 ⊔ a b GΓ0(2) = PGL (Z): b 0 mod2 c d ∈ 2 ≡ is a congruence subgroup, and 2 τ = , 1 Proof. The ﬁrst statement follows easily from considering the action of an element of SO(Q, A1) on the locus 2 1 of isotropic points for A1, a conic in P that is rationally isomorphic to P . Note that εA1 is compatible with the map εA0 found earlier: 1 εA1 = TεA0 T − . (114) As for the second statement, if a b M = PGL (Q) c d ∈ 2 is given such that εA1 (M) is integral, we ﬁrst multiply by τ if needed to make v2(det M) even, and then scale M so that a,b,c,d are coprime integers. Then we argue that if a prime p were to divide ad bc, it − must divide each of a,b,c,d by the integrality of εA1 (M), which is a contradiction.

We have now mapped each SL3-orbit of pairs of integral symmetric matrices of A1-type to a G-orbit of 1 binary quartic (1, 2, 1, 1, 4 )-forms; indeed, it is not hard to show that G is in fact the subgroup of PGL2(Q) that preserves the lattice of forms of this shape, and we have, by an argument similar to Lemma 13.7, 1 1 hA1-type C, (2) = = . AutC Q StabG f quartic rings Q 1 | | (1,2,1,1, 4 )-forms f(x,y) | | with resolventX C with resolventX , of type, g A1 up to G up to C-isom

13.1.2 Type I Following the same method, we can write 1 1 hI-type C, (2) = = . AutC Q Stab B quartic rings Q | | (I, B) with resolvent g SO(Z,I) with resolventX C up toXSO(Z, I) of A type, up to C-isom

At this point we make two striking observations:

• The resolvent condition det(xI yB)= g is equivalent to B having characteristic polynomial g(x, 1), so we have connected counting quartic− rings to another classical problem, namely counting symmetric matrices of given characteristic polynomial;

105 • Since SO(Z, I) is a finite group, isomorphic to 4 (in its representation as the group of rotations of a cube), there is no need to count orbits of symmetricS matrices; the matrices themselves will be finite in number. Thus I-type 1 3 3 h C, (2) = B Mat × Z : char poly(B)= g(x, 1) . 24 { ∈ } 13.1.3 Conditions at ∞ The interpretations of class numbers of quartic rings that we have here developed can be modified to take into account local conditions or weightings at a prime. Here we only consider the prime at infinity. Over R there are only two nondegenerate cubic algebras, R C and R R R. If C R = R C (that is, × × × g ⊗ ∼ × g has only one real root), then there is only one quartic algebra with resolvent Cg up to Cg-isomorphism, so it does not make sense to impose local conditions at the infinite place. If, on the other hand, C R = R R R, g ⊗ ∼ × × then the three factors of R are non-interchangeable, being labeled by the three real roots of g, and there are four non-C -isomorphic quartic algebras with resolvent C (one isomorphic to R R R R and three to g g × × × C C), parametrized by the four Kummer elements δ CN=1/(CN=1)2. The following is not hard to verify: × ∈ g g Lemma 13.10. Let g be a monic binary cubic form over R whose dehomogenization has three real roots ξ < ξ < ξ . Identify the corresponding R-algebra C with R R R with the coordinates ordered so that 1 2 3 g × × ξ (ξ1, ξ2, ξ3). Let L/R be a quartic algebra with resolvent Cg. The corresponding pair (A, B) of real 7→ N=1 N=1 2 symmetric matrices is related to the sign of the corresponding Kummer element δ Cg /(Cg ) in the following way: ∈

• If sgn δ = (+, +, +), then (A, B) is of type A1 and yields an indefinite binary quartic form with four real roots. • If sgn δ = (+, , ), then (A, B) is of type A and yields a positive definite binary quartic form. − − 1 • If sgn δ = ( , , +), then (A, B) is of type A and yields a negative definite binary quartic form. − − 1 • If sgn δ = ( , +, ), then (A, B) is of type I. − − 13.1.4 Statements of results We leave it to the reader to furnish the modifications of the condition at in the proof of Theorem* 13.3 ∞ to yield the following identities. Because Z is unramified at 2, we can prove them unconditionally, but only for tamely ramified resolvent at present. Theorem 13.11 (Quartic O-N for binary quartic forms). Let g be a monic integral binary cubic form whose splitting field is unramified at 2. Denote by h(g) the number of integral binary quartic forms of resolvent g, up to PGL2(Z)-equivalence and weighted by inverse of PGL2(Z)-stabilizer. Denote by h4(g) the 1 number of binary quartic (1, 2, 1, 1, 4 )-forms of resolvent g, up to G-equivalence and weighted by inverse of G-stabilizer. Denote by s(g) the number of integral 3 3 symmetric matrices of characteristic polynomial g. Then: × • If disc g < 0, then 2h(g)= h4(g). • If disc g > 0, then indef h(g)=2h4 (g) indef or pos def indef or pos def h (g)= h4 (g) indef or neg def indef or neg def h (g)= h4 (g) 24 hindef(g) hdef(g) = s(g) − where the superscripts instruct one to count only forms satisfying the indicated condition at infinity, with the same weighting.

106 Corollary 13.12. Let g be a monic integral binary cubic form with three real roots whose splitting field is unramified at 2. Among integral binary quartics with resolvent g, at least half are indefinite when we weight by inverse size of PGL2(Z)-stabilizer, with equality exactly when g is not the characteristic polynomial of an integral 3 3 symmetric matrix. × We state these unconditionally because they apply only to the number field K = Q. We do not attempt to generalize to other number fields. While the quartic rings of types corresponding to soluble conics (A0 and A1 in our notation) continue to be connected to binary quartic forms, the number of insoluble types grows with the degree of K.

13.2 The conductor property of the resolvent ring We conclude this part of the paper with a family of results which at first do not look at all like reflection theorems. Let R be an étale algebra over a local field K, and let E/R be an abelian extension whose Artin map ψ : R× Gal(E/R) vanishes on the base K×. Call an order R an admissible ring for E if E/R → O ⊆ ψ( ×)=0. Such rings exist (e.g. = K + f where f R is the conductor ideal) and are stable under passageO to suborders. If there is aO uniqueO maximal admissibl⊆ e ring, we call it the conductor ring of the extension E/R. In like manner, we define admissible and conductor rings for an abelian extension E/R over a global field K, if the Artin map ψE/R : I(R, m) Gal(E/R) vanishes on the ideals I(K, m) of the base. By Lemma 10.1, if E/R/K are fields, is an admissible→ ring of E if and only if E is contained in the ring class field of . O O In general, an extension E/R can have multiple maximal admissible rings, and there is no reason for a conductor ring to exist.

4 Example 13.13. Let p 5 be a prime. Take K = Q , R = Q and let χ : K× F× be any multiplicative ≥ p p → p homomorphism extending the natural projection from Z×. Deﬁne ψ : R× F× by p → p ab ψ(a; b; c; d)= χ . cd

This is the Artin map of a certain Fp× ∼= Z/(p 1)Z-torsor E/R. By construction, ψ vanishes on Qp×, and the orders −

= (a; b; c; d) Z4 : a c,b d mod p O1 { ∈ p ≡ ≡ } = (a; b; c; d) Z4 : a d, b c mod p O2 { ∈ p ≡ ≡ } are admissible rings for ψ. However, no ring strictly containing either or (of which there are very O1 O2 few) is an admissible ring for ψ; in particular, 1 2 generates the whole R, which is certainly not an admissible ring for ψ. Thus ψ has no conductorO ring.∪ O O However, in two special cases the conductor ring not only exists but has a striking characterization: it is the resolvent ring of a certain maximal order. These cases are those of general cubic and quartic algebras. Proposition 13.14. Let L/K be a cubic étale algebra over a global or local ﬁeld. Let T be its quadratic resolvent torsor and E = LT its 3-closure. Then the quadratic resolvent ring T of L is the conductor ring of E/T . S O⊆

Proof. The global case reduces immediately to the local one. The Artin map φE/T vanishes on K× by the Galois symmetry of the situation (the same argument is carried out in a global context in Nakagawa [38, p. 110]), so L/K has admissible rings. Now the orders in T are totally ordered: they are simply of the form i = K + π T for i 0. It is evident that the conductor ring of E/T must be K + fE/T T , which has OdiscriminantO O ≥ O O disc T f2 = disc T disc(E/T ). · E/T · The proposition is now reduced to the identity

107 disc L = disc T disc(E/T ). · This is a form of the “Brauer relation” between the absolute discriminants of L, T , and E and follows quickly from an Artin-conductor argument: see [12], equation (2.7). The above proof is not very deep and does not use reflection theorems at all. Let it be noted that over K = Q, a very similar result was proved, if not stated, by Hasse ([26], table on p. 568) and forms a foundation to Nakagawa’s proof of Ohno-Nakagawa [38, Lemma 1.3]. However, the quartic analogue of this statement, which we state in an identical way, is much deeper. Note that, in addition to the cubic resolvent R, a quartic étale algebra L/K has a natural sextic resolvent S coming from the map 4 6 that sends a permutation of 1, 2, 3, 4 to the corresponding permutation of its 2-element subsets. SS →is S naturally a quadratic étale extension{ of }R with the same Kummer element 2 δ RN=1/ RN=1 that parametrized L in 5.4. ∈ Theorem* 13.15 (Conductor rings). Assume Theorem* 13.2. Let L/K be a quartic étale algebra over a global number field or a p-adic field. Let R and S be its cubic and sextic resolvent algebras, respectively. Then the cubic resolvent ring S R of is the conductor ring of S/R. 0 ⊆ OL Proof. The global case reduces immediately to the local one. To see the vanishing of φS/R on K×, let S = R(√δ) where S (δ)=1. Then for a K×, R/K ∈ φ (a)= δ, a = a,δ = φ (δ)= φ S (δ) =1. S/R h i h i R(√a)/R K(√a)/K R/K The conjecture now has two parts:

(a) S0 is an admissible ring for S/R;

(b) Any admissible ring for S/R is contained in S0. As mentioned, this result does not on the surface look like a reﬂection theorem. But we will prove both (a) and (b) using Theorem* 13.2. Let g(L, S, t) denote the number of t-traced orders in L with reduced resolvent S. Then Theorem* 13.2 states that gˆ(L, S, (2)) = c g(L, S, (1)) (115) · for the appropriate positive constant c = /2 . Now g(L, S, (1)) is the number of orders in L of resolvent |OL OL| S. In particular, it is 1 if S = S and 0 if S S . On the other hand, g(L′, S, (2)) can be interpreted as the 0 6⊆ 0 number of ideals I R such that (S,I,δ′) is balanced, and overall ⊆ 1 gˆ(L, S, (2)) = 0 δ, δ′ g(Lδ′ , S, (2)) H (MR) h i δ H1(M ) | | ′∈ X R 1 = 0 δ, δ′ H (MR) 1 h i | | δ′ H (MR) ∈ X (S,I,δ′) balanced 1 = 0 φS/R(δ′). (116) H (MR) 1 | | δ′ H (MR) ∈ X (S,I,δ′) balanced (117)

Assume for the sake of contradiction that S = S is not an admissible ring for S/R. Then there exists ε S× 0 ∈ such that φS/R(ε) = 1. The rearrangement of terms (S,I,δ′) (S,I,εδ′) ﬂips the sign of the sum, so gˆ(L, S, (2)) = 0, a contradiction,− since g(L, S, (1)) = 1. This proves7→ (a). Now assume for the sake of contradiction that there is an admissible ring S S0. Choose such an S maximal for this property. Then divide the summands of (116) into two cases: 6⊆

108 2 • If I is invertible in S, then I = αS for some α, and δ = α ε for some ε S×. These terms contribute ∈

φS/R(δ)= φS/R(ε)=1, since S is an admissible ring. There is at least one term of this type, namely δ =1, I = S.

2 2 2 • If I is not invertible in S, then End I = S′ ) S. (If we had End I = S, then by Lemma 14.4, I would be invertible in S and then I would also.) By maximality, S′ is not an admissible ring for S/R and there is an ε S′× such that φ (ε) = 1. The rearrangement (S,I,δ′) (S,I,εδ′) permutes ∈ S/R − 7→ the terms with the same S′ and ﬂips their signs. So the terms of this type contribute nil. Overall, we get gˆ(L, S, (2)) > 0, a contradiction, since g(L, S, (1)) = 0. This proves (b).

14 Tame quartic rings with non-split resolvent, by multijection

In this section, we will adapt the methods of Section 11.1 to prove Theorem* 13.2 in the case that K is tame (not 2-adic) and R = K K K. 6∼ × × 14.1 Invertibility of ideals in orders We begin with a technical inquiry that has interest in its own right. It is well known that every Z-lattice a in a quadratic field is invertible with respect to some order, namely its endomorphism ring End a. Ina cubic or higher-degree field this is not so. However, the following two lemmas will help us understand the structure of orders and ideals in such a setting. Lemma 14.1. Let K be a local field, and let be an order in a finite-rank étale algebra L over K. Then there is a decomposition O L = L L , 1 ×···× s each Li being the product of some field factors of L, with the following properties: = O O1 ×···×Os is the product of orders in the Li, and each i has only a single prime pi above the valuation ideal p, so that every element of not lying in p is a unit.O Oi i Proof. Let L = K K be the field factor decomposition of L. Each K is a local field; let p be the 1 ×···× r i i pullback to L of the valuation ideal of Ki. Define an equivalence relation on the pi by p p p = p . i ∼ j ⇐⇒ i ∩ O j ∩ O Thus if p ≁ p , then there is an α such that either i j ∈ O p ∤ α, p α or p α, p ∤ α. (118) i j | i| j Suppose for the moment that it is the first. We first claim that we can take α 1 mod pi. Note that L/pi is a finite field extension of /p, so α satisfies a polynomial congruence ≡ O OK m m 1 α + um 1α − + + u0 0 mod pi, ui K , u0 × . − ··· ≡ ∈ O ∈ OK Then 1 m m 1 α′ = u0− (α + um 1α − + + u1α) − − ··· is 1 mod pi and 0 mod pj . Also, note that we can switch α′ with 1 α′ to satisfy these congruences for any i and j, regardless of which condition in (118) held to begin with. − Fix i and multiply the resulting values of α′, which are 1 mod p (and hence 1 mod any p p ) but 0 i j ∼ i mod pj for any chosen pj ≁ pi. We get a single α′′ such that

1 mod pj, pj pi α′′ ∼ ≡ ( 0 mod pj, pj ≁ pi.

109 As a ﬁnal step, we can iterate the polynomial

f(x)= x2(3 2x), − m 2m m 2m which takes pi to pi and 1+pi to 1+pi , and take the limit to derive that the idempotent e = (ej )j L, deﬁned by ∈ O

1, pj pi ej = ∼ ( 0, pj ≁ pi, lies in ( is closed in the p-adic topology). These ej ’s form a set of orthogonal idempotents decomposing intoO a directO product of , one for each -equivalence class, that have the properties we seek. O Oi ∼ Lemma 14.2. Let K be a local field, and let be an order in a finite-rank étale algebra L over K. Then a fractional ideal of is invertible if and only ifO it is principal. O Remark 14.3. This implies that a if is an order in a finite-rank algebra over a Dedekind domain, then a fractional ideal of is invertible if andO only if it is locally principal, where here “locally” denotes localization at each prime of OK. Thus our statement and proof differ slightly from the corresponding statement in Neukirch [39] (Theorem I.12.4), which is built by localization at the primes of (and also assumes that is a domain). O O Proof. By the preceding lemma, an ideal of a product is just a product a a , which is O1 ×···×Os 1 ×···× s principal (resp. invertible) if and only if every ai is: hence we can assume that s =1. The reverse direction is trivial (principal fractional ideals are invertible), so let a be an ideal of with 1 1 O inverse a− , aa− = . Since all ideals of L are principal, we can assume that a L = L and hence 1 O O O O a− = as well. We can express OL OL 1 1= α β , α a, β a− . i i i ∈ i ∈ i X Let p1,..., pr be the valuation ideals coming from the field factors of L. Since 1 / p1, some term αiβi, say α β , is nonzero mod p . But because p = p is the unique maximal ideal∈ of for all i, we have 1 1 1 i ∩ O 1 ∩ O O α β / p for all i. This implies that α and β are units, whose product lies in ×. Now since 1 1 ∈ i 1 1 O a α , b β ⊃ 1O ⊃ 1O and ab = , equality must hold. O Lemma 14.4. If c R is a lattice in a cubic algebra over a local field K, then c2 is invertible in its endomorphism ring End(⊆ c2).

Proof. First, c R is an invertible R-ideal, which, since R is a product of PID’s, we can scale to be R. O O O 2 O We first claim that c contains a unit, or else has a special form for which c = R is clearly invertible. Let p be a uniformizer for and k = /p the residue field. The units of areO those elements whose OK OK OK OR projections to the cubic k-algebra C1 = C1/pC1 are non-units (that is, zero divisors). The non-units of C1 are the union of at most three proper subspaces (the projections of the valuation ideals of each field factor). The projection ¯c of c down to C1 cannot lie in any of these subspaces since c R = C1. An easy theorem in linear algebra is that a vector space over a field k cannot be the union of fewerO than k +1 proper subspaces. We conclude that c contains a unit except if k = 2 and R = K K K has three| | field factors. In this case, the only ¯c C instantiating this case is| | × × ⊆ 1 ¯c = (a; b; c) F3 : a + b + c =0 . { ∈ 2 } 2 3 2 It is evident that ¯c is the whole of F2, whence by Nakayama’s lemma, c is the whole of R. Now we can assume that c contains a unit, which we scale to equal 1. We claim thatOc3 = c2. By the i j theory of modules over a PID, we can find a basis 1,α,β for K such that 1,p α, p β is a basis for c for some integers i, j 0. Now by translation, we can{ assume} thatOαβ = t {. We then have} ≥ ∈ OK c3 = c2 + α3, α2β,αβ2,β3 = c2 + α3,tα,tβ,β3

110 The elements tα and tβ are certainly already in c c2. As for α3, since α is an integral element of R, its 3 ⊆ 2 2 characteristic polynomial expresses α as an K -linear combination of α , α, and 1, all of which lie in c . So c3 = c2. O We conclude that c4 = c3 = c2, so c2 is closed under multiplication and hence is an order. In particular, it coincides with its endomorphism ring and in particular is invertible.

Remark 14.5. Although Lemma 14.4 is simple to state, we have not found it anywhere in the literature. In n 1 general, we suspect that if c is a lattice in an algebra L of rank n, then c − and all higher powers of c are invertible in their common endomorphism ring. This is not hard to prove if char kK >n. That the exponent n 1 is sharp is seen from the cute example − c = Z + (0; 1; ... ; n 1)Z + pZn Zn. p − p p ⊆ p i n The power c consists of all sequences (a0; ... ; an 1) Zp that are congruent modulo p to the values − ∈ (f(0); f(1); ... ; f(n 1)) − of a polynomial f of degree at most i with coefficients in k. If p>n, then this power stabilizes to the whole of Zn only for i n 1. p ≥ − 14.2 Self-duality of the count of quartic orders Theorem 14.6 (Local quartic O-N in the tame, not totally split case). Assume K is a local field of residue characteristic not 2. Let C R be a cubic étale order that is not totally split. Then the assignment f to each L of the number of orders⊆ L with resolvent C is self-dual. C O⊆ Proof. As in the cubic case, the proof proceeds by reduction to the zero case (i.e. that f(0) = fˆ(0)). Note that M ∼= M ′ as Galois modules (one can even make this canonical, using the unique alternating bilinear form on M). The fixity of this Galois module is easy to compute:

1 if R is a field H0 = H0(K, R) = 2 if R = K K for some quadratic field K | | | |  ∼ × 2 2  4 if R = K K K. ∼ × × This can be written concisely as  R×[2] H0 = | | | | 2 (the latter formula will work especially well for our case). Moreover, since the unramified cohomology is self-orthogonal, we have

2 R×[2] H1 = H0 2 = | | . | | | | 4 Since we are excluding the case R = K K K, there are just two possibilities: ∼ × × • If R is a field, then H1 =1, and there is nothing to prove, as any function on H1 is self-dual. | | 1 • If R is the product of two fields, then H =4. Pick an F2-basis σ1, σ2 . The Tate pairing is given by the unique alternating pairing on H1.| The| space of functions on Hh 1 is four-dimensional,i and a basis is 1 1 1 1 σ , σ , σ +σ , 0 . h 1i h 2i h 1 2i { } Note that the first three basis elements are self-dual, while the fourth differs from its dual even at 0. This proves that if f is a function on H1 with f(0) = fˆ(0), then f = fˆ. So we have reduced local O-N to the following lemma:

111 Lemma 14.7. Assume K is a local ﬁeld of residue characteristic not 2. Let C R be an order in an ⊆ étale algebra. Then the assignment gC to each L of the number of orders L with resolvent C satisﬁes self-duality at 0: O ⊆ gˆC(0) = gC(0). Proof. As in the cubic case, the proof is by explicit multijection. On the one hand, H0 gˆ (0) = g (σ) | | · C C σ X counts all quartic orders with cubic resolvent C, and using Theorem 6.18, these can be parametrized by 2 self-balanced ideals (C, c,δ), where δ ranges over a set of representatives for RN=1/ RN=1 . On the other hand, g (0) is the number of orders with resolvent C in K R. Write such an order as = +0 a, C K where a R is a lattice. The condition that be a ring is× (by Theorem 6.13) subsumedO by theO resolvent× conditions,⊆ namely that O

(i) NC(a)=1,

(ii) Φ (0; α)= α′α′′ C for all α a. 4,3 ∈ ∈ Our aim is to associate H0 values of (c,δ) to each value of c. | | 2 The multijection is as follows. First, c may not be invertible. Let C1 = End c and c1 = cC1, an invertible and thus a principal C1-ideal. Let [C1 : C] 1 2 − a1 = c1 . (119) δ[c1 : c] Finally, since c1 and a1 are both principal and thus scalar multiples of each other, we can take a to be an ideal that sits inside a1 as c sits inside c1: that is, if c1 = γ0C1 and a1 = α0C1, then α a = 0 c. γ0 Before checking that this a yields a valid ring, we check how many-to-one our multijection is. First note that 2 a determines C1 = End a and a1 = aC1, and in particular the index [c1 : c] = [a1 : a]. Then, by (119), the 2 2 2 2 “shadow” b = δc = δc1 is determined. Note that b is an invertible C1-ideal of norm [C1 : C] /[a1 : a] , a square. The pairs (c1,δ) satisfying b = δc1, where c1 is an invertible C1-ideal and δ is one of the representatives 2 for RN=1/ RN=1 , are found to be H0 in number by an argument identical to the cubic case. Finally, locating c within c involves the same| choice| as locating a within a . So we have a string of many-to-one 1 1 correspondences n to 1 H0 to 1 n to 1 (c,δ) (c ,δ) | | a a, (120) −−−−→ 1 −−−−−−→ 1 ←−−−− and thus overall there are H0 times as many (c,δ) as a. It remains to prove that| the| correspondence (120) preserves the resolvent and balancing conditions. As for the ﬁrst condition, regarding the discriminant of the ring, we leave it to the reader to verify that 1 NC (a)= 2 . (121) N(δ)NC (c)

Now we may assume that both sides of (121) are 1. Since c1 is principal, we may assume that c1 = C1, adjusting δ by a square if necessary. Then by the conditions of Theorem 6.18, N(δ) is a square t2 and

1 [C1 : C] = NC(c)= ; t [c1 : c] thus t a = C . 1 δ 1 We ﬁrst prove that if c satisﬁes its resolvent condition, so does a. Any α a a has the form α = t β, ∈ ⊆ 1 δ β C1, and then ∈ t2 α′α′′ = β′β′′ = δβ′β′′; δ′δ′′

112 and we note that if β C , then β′β′′ C as well, by the relation ∈ 1 ∈ 1 β′β′′ = ββ′ + ββ′′ + β′β′′ β(β + β′ + β′′ β). (122) − − ∈OK ∈OK 2 Thus if c satisfies the resolvent condition| (δc {z C), then} a|satisfies{z the} resolvent condition (α′α′′ C for all α a). To prove the converse, it suffices to⊆ show that ∈ ∈ B = β′β′′ : β c { ∈ } 2 spans C1 over K , where c C1 is a sublattice with c = C1. We first claimO that c contains⊆ a unit. Let p be a uniformizer for and k = /p the residue field. OK OK OK In the cubic k-algebra C1 = C1/pC1, the non-units are the union of at most three subspaces. The projection 2 ¯c of c down to C1 cannot lie in any of these subspaces since c = C1, so, since k 3, ¯c must contain a unit, which lifts to a unit in c. There is no harm in rescaling c so that 1 c. | |≥ Now 1 B. Also, for each β c, ∈ ∈ ∈ B β′β′′ (β′ + 1)(β′′ +1)+tr β 1= β, h i ∋ − − and thus 2 B β′β′′ (ββ′ + ββ′′ + β′β′′)+ β(β + β′ + β′′)= β . h i ∋ − But since char k =2, the elements β2, for β c, generate c2 = C , completing the proof. 6 ∈ 1 For the totally split case, the method of proof of Theorem 14.6 fails, because dim H1 =4 and fˆ(0) = f(0) is no longer enough to imply fˆ = f. But when we count by discriminant instead of resolvent, it can be rescued, due to the following symmetry argument. 1 Theorem 14.8. Fix a cubic algebra R over a local field K, and let gD : H Z count the number of orders 1 → in a quartic algebra L H with discriminant D. That is, gD is the sum of all the gC’s in Theorem 14.6 over C R of discriminant∈ D. Then g is self-dual. ⊆ D Proof. All cases are covered by Theorem 14.6 except for the totally split case R ∼= K K K, where dim H1 =4. We can write H1 = σ , τ , σ , τ , where the σ’s and τ’s correspond to Kummer× elements× h 1 1 2 2i σ1 : (u, u, 1), τ1 : (p, p, 1), σ2 : (1,u,u), τ2 : (1,p,p), where p is a uniformizer and u is a non-square unit. The Tate pairing is given, by Theorem 5.7, by ∈ OK the µ2-valued pairing σ , τ = σ , τ = 1, σ , σ = τ , τ = σ , τ = σ , τ =1. h 1 2i h 1 2i − h 1 2i h 1 2i h 1 1i h 2 2i The group Aut R = acts on H1, permuting σ , σ , σ + σ and τ , τ , τ + τ in the permutation manner. ∼ S3 1 2 1 2 1 2 1 2 There are five orbits, represented by 0, σ1, τ1, σ1 + τ1, and σ1 + τ2. Note that gD must be constant on each orbit, because its definition is S3-invariant. The functions 1 1 1 1 1 σ ,σ , τ ,τ , σ +τ ,σ +τ , π( σ ,σ +τ ), 0 h 1 2i h 1 2i h 1 1 2 2i h 2 1 2i { } π Aut R ∈X form a basis for the 5-dimensional space of -invariant functions on H1. The first four are self-dual, while S3 the last differs from its dual even at 0; so, since gD(0) =g ˆD(0) by Lemma 14.7), the coefficient of the last basis element must be 0 and gD is self-dual.

Part VII Counting quartic rings with prescribed resolvent

15 Introduction

Here end the cases in which a conceptual, bijective argument has been found to suﬃce for proving local reﬂection for quartic rings. To win the remaining cases, we attack a problem that has interest in its own

113 right: counting orders in a quartic algebra L over a local field K whose cubic resolvent ring C R is fixed. O ⊆ The index [ L : ] being fixed by the condition Disc = Disc C, we must analyze the resolvent condition Φ ( / ) OC/ O , where Φ : L/K R/K is theO resolvent map. Recall that, with respect to bases of 4,3 O OK ⊆ OK 4,3 → and C, Φ4,3 is given by a pair O ( , ) = ([M ], [N ]) M N ij ij of symmetric 3 3 matrices, and the resolvent condition can be viewed as the K -integrality of the entries (properly scaled× to account for the tracedness condition). By suitably choosingO coordinates, we can ensure that only the integrality of the entries

M11, N11,M12, and M22 is in doubt. Of these, the condition on M11 is the most challenging. It amounts to a quadratic condition on 2 the ﬁrst basis vector ξ1 of C, that is, a conic on some pixel (determined by the N11-condition) in P ( K ). The solubility of this conic over K is governed by the Hilbert symbol, which we analyze. It is very hardO in general to tell if any K-points of the conic lie in the requisite pixel, but if there is even one such K-point, then, using the rational parametrization of a conic with a basepoint, the volume of points in the pixel is easy to determine. Accordingly, our approach to solving the M11-condition is a three-step one:

• Determine the sum of the solution volumes for ξ1 over all quartic algebras L. • Find restrictions on what L can yield a nonzero volume and what that volume can be, providing an upper bound (the bounding step). • If the sums of these upper bounds agree, deducing that the bound is attained everywhere (the summing step).

The M12 and M22 conditions are essentially linear. We use the computer program LattE to sum the ring totals over all possible values of the discrete data and verify the local reflection theorem. The case of wildly ramified resolvent (splitting type 121) is still in progress. Except for brief remarks, it has been omitted from this edition. Also omitted are the adaptations to be made when char kK > 2, where, in view of Theorem 14.6, only splitting type (111) need be considered. It involves only the black, brown, beige, and white zones; the conics are all very easy to solve and yield the same answers as the wild case upon substituting e =0. So in the sequel, R is a tamely ramified étale algebra over a 2-adic local field K. Corresponding to R, there is a Galois module MR whose underlying group is 2 2. We will work 1 1 C ×C extensively with H (K,MR), which we abbreviate to H .

16 The group H1 of quartic algebras with given resolvent ¯ ¯ We fix a local field K and an separable closure K. Let K¯ be the ring of integers in K. (If the reader is uncomfortable with non-Noetherian rings, he can take K¯Oto be instead the compositum of all extensions of K of degree at most 4; the Galois cohomology and all proofs will be unaffected.) By Theorem 5.4(c), we can identify H1 naturally with

2 RN=1/ RN=1 .

Now there is a natural isomorphism

2 2 N=1 N=1 2 R×/(R×) = K×/(K×) R / R . ∼ × 1 3 Thus for any α R×, we can talk about the class [α] of α in H , that is, the class of α /N(α). ∈ 1 2 Hence the structure of H can be uncovered by taking a suitable Shafarevich basis of R×/(R×) and 2 removing a basis of K×/(K×) , which, by Lemma 7.13, maps in isometrically: 1 0 Lemma 16.1. If R/K is a cubic étale extension, then H is an F2-vector space of dimension 2 dimF2 H + 0 0 2ef. It has a basis of dimF2 H nonunits, dimF2 H intimate units, and 2ef generic units; the generic units can be chosen as follows:

114 (a) If R is unramiﬁed, we take 2ef units of the form 1+ xπ2i+1, where 0 i

H1 i = 1 1 2i − i = [α] H : α 1 mod π 0 i e L  { ∈ ≡ } ≤ ≤  [1] i = e +1, { } 0 2(e i)  noting that i = H q − for 0 i e and that i⊥ = e i for all i by Lemma 7.19. Likewise,|L in| the| ramiﬁed| case, we≤ deﬁne≤ the levelL space L −

H1 i = 1 1 i − i = [α] H : α 1 mod π 0 i 2e L  { ∈ ≡ } ≤ ≤  [1] i = e +1, { } 0 2e i  noting that i = H q − for 0 i e and that i⊥ = 2e i for all i by Lemma 7.19. We will occasionally|L | | | let ≤ ≤ L L − e, R unramified e′ = ( 2e, R ramified to shorten lemma statements. We define the level ℓ(α) of an element [α] H1 as the largest i 0 for which [α] . We have ∈ ≥ ∈ Li ℓ(1) = e′ +1. By convention, if [α] / , we set ∈ L0 ℓ(α)= 1/2 − to shorten some future statements.

17 Reduced bases

Deﬁne a valuation on K¯ n by v(x ; ... ; x ) = min v(x ),...,v(x ) . 1 n { 1 n } Let R be a rank-n étale algebra over a local ﬁeld K. We can Minkowski-embed R into K¯ n.

Deﬁnition 17.1. Let I be an -lattice in R, and let ω K¯ n be a multiplier with the following property: OK ∈ (ι) (ι′) ( ) If ι, ι′ are two coordinates of the same ﬁeld factor of R, then ω and ω have the same valuation. ∗ A basis (ρ1,...,ρn) for ωI is called reduced if (a) v(ρ ) v(ρ ); 1 ≤···≤ n (b) If ρ ωR is decomposed as ∈ ρ = c ρ , c K, i i i ∈ i X then for each i, v(c ρ ) v(ρ). i i ≥ This notion has the following properties:

Proposition 17.2. Let ωI be as above.

115 (a) There exists a reduced basis (ρ1,...,ρn) for ωI.

(b) If (ρ′ ,...,ρ′ ) is any other basis for ωI, sorted so that v(ρ′ ) v(ρ′ ), then for each k, 1 n 1 ≤···≤ n

v(ρ′ ) v(ρ ). k ≤ k In particular, if both bases are reduced, equality holds.

(c) If (ρ1′ ,...,ρn′ ) is another reduced basis for ωI, then

ρi′ = cij ρj j X for some change-of-basis matrix

[c ] GL , v (c ) v(ρ ) v(ρ ). (123) ij ∈ nOK ij ≥ i − j

Conversely, any matrix [cij ] satisfying (123) yields a new reduced basis (ρi′ )i.

(d) As K¯ -modules, O v(ρ ) v(ρ) π− i ρ :1 i n = π− ρ : ρ ωR× . i ≤ ≤ ∈ D E D E Proof. (a) Choose a basis (ρ1,...,ρn) such that the sum of the valuations v(ρ1)+ + v(ρn) is maximal. This can be done because there are only ﬁnitely many possible valuations of primitive··· vectors in ωI. Sort the ρi in increasing order of valuation. We claim (ρi)i is reduced.

Let ρ = i ciρi be given. Let a be the minimal valuation v(ciρi) of a term, and suppose that a < v(ρ). Then we have a linear dependency P a c ρ 0 mod π m ¯ . i i ≡ K v(cXiρi)=a

Choose j such that v(cj ρj )= a and v(cj ) is minimal. Then

ci ρj′ = ρj cj v(cXiρi)=a

is an element of ωI whose valuation exceeds v(ρj ). Since the coeﬃcient of ρj in ρj′ is 1, replacing ρj by ρj′ does not change the span ωI but increases the valuation sum i ciρi, contradicting the choice of basis (ρi)i. P

(b) Since (ρi)i and (ρi′ )i are bases for the same module ωI, such a [cij ] GLn( K ) certainly exists. Applying the reducedness property to each decomposition ∈ O

ρi′ = cij ρj j X yields a bound v(c ) v(ρ′ ) v(ρ ). ij ≥ i − j Suppose that v(ρ ) < v(ρ′ ) for some k. Then for j k i, k k ≤ ≤

v(ρ ) v(ρ ) < v(ρ′ ) v(ρ′ ), j ≤ k k ≤ i

so cij has positive valuation. Thus, when the matrix [cij ] is reduced modulo π, it has an (n k + 1) k block of 0’s, large enough to make the determinant vanish, which is a contradiction. − ×

116 (c) By the preceding part, v(ρk′ )= v(ρk). So the associated matrix [cij ] must satisfy

v(c ) v(ρ′ ) v(ρ )= v(ρ ) v(ρ ). (124) ij ≥ i − j i − j

Conversely, if [c ] is an invertible matrix satisfying this inequality, we get a new basis ρ′ with v(ρ′ ) ij i i ≥ v(ρi). Equality must hold, and now since k v(ρk′ ) achieves the maximal value, (ρk′ )k is reduced by the proof of part (a). P (d) The direction is obvious. For the direction, let ρ ωR× be given. Since (ρ ) is reduced, ⊆ ⊇ ∈ i i ρ = c ρ , v(c ρ ) v(ρ), i i i i ≥ i X so v(ρ) v(ρi) v(ρ) v(ρi) π− ρ = π − ci π− ρi, i X and the parenthesized coeﬃcients belong to ¯ , as desired. OK We can ﬁnd reduced bases with added structure.

Definition 17.3. Fix an ordering R = R1 Rr of the field factors of R. For ρ ωR, let k be the minimal index such that v(ρ(k)) = v(ρ). We×···× say that ρ is R(k)-led, and we define the leader∈ of ρ to be the normalization ρ (i) ldr(ρ)= , πuω where u is the unique integer for which ldr(ρ) is a primitive vector in . We say that a reduced basis (ρ ) OR i i is well-led if the leaders ldr(ρ ) consist of a reduced basis for (k) for each k. i OR Proposition 17.4. Every ωI, as above, admits a well-led basis. Proof. Consider the element ε 2ε rε ηε = (π ; π ; ... ; π ) where ε is a positive rational number, small enough that if a1 < a2 are two valuations of elements in ωI, then rε < a a . By Proposition 17.2(a), there is a reduced basis ρ η ,...,ρ η for ωη I. Each basis 2 − 1 1 ε n ε ε element ρ η has some valuation u + kε, u Q, k 1,...,r , indicating that ρ is R(k)-led. Since replacing i ε ∈ ∈{ } i ε by 0 preserves non-strict inequalities among valuations in ηεωI, the ρi form a reduced basis for ωI, which we claim is well-led.

Given α (k) primitive, decompose ωα = c ρ as an element of ωR. We have ∈ OR j j j v(c ρ η ) Pv(αη )= v(α)+ iε, j j ε ≥ ε (k) and for a nonempty subset of j, equality must hold and, in particular, ρj must be R -led. Let Lk be the set of indices j for which ρ is R(k)-led, and let B = ldr(ρ ): j L . Now we have j k { j ∈ k} (k) ωα = cj ρj + ωα′ j L X∈ k with each term of valuation at least v(α), and where α′ with v(α′) > v(α). We can rewrite this as ∈ OR

α = cj′ ldr(ρj )+ α′ j L X∈ k u where cj′ = cj π K . We can iteratively decompose α′ the same way, and as its valuation goes to inﬁnity, we get a decomposition∈ O

α = cj′′ ldr(ρj ). (125) j L X∈ k

117 (k) So B generates (k) , and in particular, L [R : R ]. However, k OR | k|≥ [R : R(k)]= n = L . | k| Xk Xk So equality holds and each B is a basis for . Since the decomposition (125) has every term of valuation k OR at least v(α), and α was any primitive vector, Bk is in fact reduced. It is evident that reduced indices take a limited number of values modulo 1. Indeed, we have the following: Corollary 17.5. If I R is a lattice, then the multiset of valuations a = v(ρ ) mod 1 of reduced ⊆ { i}i { i }i basis elements for ωI depends only on R and ω, not on I. It consists of fRi/K copies of

(Ri) j v(ω )+ , j =0, 1,...,eRi/K 1, eRi/K − where R ranges over the field factors of R = R R . i 1 ×···× r Proof. Taking a well-led basis and passing to the leaders, we reduce to the case that R = Ri is a field. Then eR/K ai ω has equal valuations in all coordinates, and we may assume that ω = 1. Let ρi = πR ξi, where the ξ × differ only by a unit from the normalizations used before. For each congruence class of a modulo 1, i ∈ OR i note that if the set of corresponding ξi is linearly dependent modulo πR, then one of the ρi could be increased by an -linear combination of the others to increase its valuation, contradicting the hypothesis that our OK basis is reduced. So the ξi corresponding to each congruence class of ai modulo 1 are linearly independent, and in fact must form a basis for kR over kK in order for there to be the full number eR/K fR/K of ξi. This establishes the claimed multiset.

A reduced basis for ωI does not always remain reduced when we extend the ground ﬁeld K. To study this, we make the following deﬁnition.

Deﬁnition 17.6. An extender basis for ωI is a basis (ρ1,...,ρn) for ωI K K¯ such that the vectors ⊗O O

v(ρi) ξi = π− ρi

n form an K¯ -basis for K¯ . The valuations ai = v(ρi) are called the extender indices of the basis, and the ξi are calledO the extenderO vectors.

If it consists of ρi ωI, an extender basis is easily seen to be reduced. Fortunately, in the cases of tamely ramiﬁed resolvent, this∈ always holds:

Proposition 17.7. If R/K is tamely ramiﬁed, then any reduced basis of ωI is an extender basis. Proof. In view of Proposition 17.2(d), it is enough to show that

v(ρ) n ¯ π− ρ : ρ ωR× = . OK ∈ OK¯ D E We immediately reduce to the case that R is a ﬁeld, and then we may assume ω = 1. It suﬃces to prove that, for some reduced basis ρ1,...,ρn of R,

v(ρi) (j) det π− ρi 1. i,j ∼ h i But that follows from the familiar formula for the discriminant of a tamely ramiﬁed extension.

118 17.1 The extender basis of a cubic resolvent ring

Let C be a candidate resolvent for t-traced quartic rings, and let Ct be the corresponding reduced resolvent, 2 that is, the unique ring such that C = K + t Ct. First, look at the reduced basis of Ct R as a lattice; and ¯O 3 ⊂ look at its extender basis, a basis of Ct = Ct K K¯ . Because 1 Ct is an element of minimal valuation, there are not so many cases: ⊗O O ∈ • If R is tamely ramiﬁed, then the reduced basis

C = 1, πb1 θ , πb2 θ , b b t 1 2 1 ≤ 2 is also an extender basis. If R is unramified, the bi are of course integers; if R is tamely ramified, then by Proposition 17.4, we have 1 2 b ,b , mod Z. (126) { 1 2}≡ 3 3 Because θ1 is coprimitive to 1, at most one pair of its three coordinates can be congruent modulo mK¯ . We let s¯ =s ¯C , the idempotency index of C. Note that s¯ is infinite only when two coordinates of θ1 are exactly equal. Since the θi are determined only up to finite precision, we can, and will, assume that s¯ is finite. Lemma 17.8. If finite, the value of s¯ is constrained as follows: • If R is unramified, then s¯ is an integer. For simplicity we let s =s ¯. • If R has splitting type 13 (residue characteristic 2), then s¯ =0.

Proof. In the tame splitting types this is immediate, knowing that b0 = 1 is an extender vector for C of minimal valuation.

Notation 17.9. If s¯ > 0, then there is a unique coordinate of K¯ 3 = R K¯ at which ω has positive ∼ ⊗ C valuation. This defines a splitting R ∼= K Q into a linear and a quadratic (possibly split) factor. We denote the three coordinates of R by (K), (Q×1), (Q2), where (K) is the distinguished one; thus we can write an element ξ K¯ 3 as ∈ ξ = ξ(K); ξ(Q1); ξ(Q2) = ξ(K); ξ(Q) where ξ(Q) = ξ(Q1); ξ(Q2) K¯ 2 = Q K¯ . ∈ ⊗K A common tool in understanding nonmaximal orders is their duals under the trace pairing. Hence it is fitting that we should understand the element ω 3 , unique up to scaling, that satisfies the relations C ∈ OK¯ tr ωC = tr(θ1ωC) = 0; that is, K¯ ω = K¯ 1,θ1 ⊥ under the trace pairing. h i h i b1 (1) (2) (3) One explicit choice of ωC is as follows: If θ˜ = π θ1 = (θ˜ , θ˜ , θ˜ ) C is the second reduced basis vector, then ∈

ωˆ = θ˜(2) θ˜(3) θ˜(3) θ˜(1) θ˜(1) θ˜(2) θ˜(2) θ˜(3); θ˜(3) θ˜(1); θ˜(1) θ˜(2) . C − − − − − − 2 The symmetry ensures that ωˆ R. Note that N (ˆω ) = (disc θ˜) is a square in K×. Note also that C ∈ C tr (ωC ) = tr (θ1ωC)=0. Then ~v(ˆωC)=(4b1 +2¯s, 4b1 +¯s, 4b1 +¯s).

Two other rescalings of ωˆC , chosen for primitivity rather than the property of lying in R, will also be used:

4b1 s¯ ωC = π− − ωC , ~v(ωC)=(¯s, 0, 0) 4b 2¯s ω¯ = π− 1− ω , ~v(¯ω )=(0, s,¯ s¯). C C C − − 1 3 They have the properties that ω and ω¯− are primitive (that is, have valuation 0) in ⊕ . C C OK¯

119 18 Resolvent conditions

Let C, Ct R be a resolvent cubic ring and its corresponding reduced resolvent, whose reduced bases are related by⊆ C = 1, πb1+2tθ , πb2+2tθ and C¯ = 1, πb1 θ , πb2 θ , 0 b b , 1 2 t 1 2 ≤ 1 ≤ 2 where t = vK (t). Let L be a quartic algebra with resolvent R. As we noted in the proof of Theorem 6.13, an order L is completely determined by the lattice I R such that / = κ(I), where O⊆ ⊆ O OK κ : K K¯ 4 → ξ trK¯ 3/K ξω√δ 7→ ω is the map in Theorem 5.4(c). For reasons that will become clear below, we take the reduced and extender bases, not of I itself, but of s¯ π δωCI. Let the reduced basis be p πs¯δω I = πa1 ξ , πa2 ξ , πa3 ξ . C · h 1 2 3i and let the extender basis be p πs¯δω I¯ = πa¯1 ξ¯ , πa¯2 ξ¯ , πa¯3 ξ¯ . C · 1 2 3 Both δ and ωC satisfy property ( ) inp Deﬁnition 17.1, so the foregoing theory applies. (In this edition, since R will always be tamely ramiﬁed,∗ the overbars can be ignored.) We will take

ai′ = ai +2b1 and a¯i′ =¯ai +2b1, the reduced and extender indices of √δωˆCI. Then we can write the resolvent conditions as follows: Lemma 18.1. With respect to the above setup, a lattice I yields a ring with a t-traced resolvent to if and only if the following conditions hold: O C • Discriminant (Θ) condition:

a¯ +¯a +¯a = ¯b + ¯b +2¯s +4t 4e (127) 1 2 3 1 2 −

• Resolvent (Φ) conditions for θ¯2-coeﬃcients:

: tr(ξ¯ ξ¯ ) 0 mod πm¯ ij Mij i j ≡ where ¯b2 +2t 2e 2¯ai +¯s, i = j m¯ ij = − − ¯b +3t 3e a¯ a¯ +¯s, i = j. ( 2 − − i − j 6

• Resolvent (Φ) conditions for θ¯1-coeﬃcients:

1 n¯ : All coordinates of ω¯− ξ¯ ξ¯ are congruent mod π ij , Nij C · i j where ¯b1 +2t 2e 2¯ai +2¯s, i = j n¯ij = − − ¯b +3t 3e a¯ a¯ +2¯s, i = j. ( 1 − − i − j 6 Proof. The conditions that C is a t-traced resolvent for are that all coeﬃcients in the coordinate rep- 1 O resentations of Θ, Θ− , and Φ have nonnegative valuation. In particular, it is equivalent to study when ¯ ¯ C = C K K¯ is a resolvent of = K K¯ . We⊗ haveO O O O⊗O O 2t+¯b1 2t+¯b2 C¯ = 1, π θ¯1, π θ¯2 , D E 120 so 2 4t+¯b +¯b 4t+¯b +¯b 2 3 Λ (C/¯ ¯ )= π 1 2 θ¯ θ¯ = π 1 2 Λ ( / ¯ ) OK 1 ∧ 2 OK¯ OK and, by the formula for Θ in PropositionD 6.14, E

1 ¯ ¯ 2 ¯ 4t+b1+b2 3 Θ(Λ (C/ K¯ )) = π κ( ¯ ) O 16 N(δ) · OK

Meanwhile, p 1 I¯ = πa¯1 ξ¯ , πa¯2 ξ¯ , πa¯3 ξ¯ (128) s¯ 1 2 3 √π δωC so

3 1 a¯1+¯a2+¯a3 Λ I¯ = π ξ¯1 ξ¯2 ξ¯3 3¯s π N(δωC) ∧ ∧

1 a¯1+¯a2+¯a3 3 4 = p π Λ ( ¯ / K). 3¯s K π N(δωC) O O

2 3 So the condition for Θ to define an isomorphismp between Λ (C/¯ ¯ ) and Λ ( ¯/ ¯ ) is that OK O OK 1 3 1 4t + ¯b + ¯b 4e v(N(δ))=¯a +¯a +¯a s¯ v(N(δω )), 1 2 − − 2 1 2 3 − 2 − 2 C or, since v(N(ωC ))=s ¯, 4t 4e + ¯b + ¯b +2¯s =¯a +¯a +¯a , − 1 2 1 2 3 as desired. Likewise, we use the formula Φ(κ(α)) =4δα2 from Proposition 6.14 to transform the Φ-condition to the following: (i) For every ξ I¯, we have 4δξ2 C¯ + K¯ ∈ ∈ t (ii) For every ξ, η I¯, we have 8π− δξη C¯ + K¯ . ∈ ∈ In terms of the basis (128) for I¯, this is to say that the diagonal entries of the matrix representing Φ belong to C¯ + K¯ and the off-diagonal entries to tC¯ + K¯ . Hence it suffices to consider (i) for ξ a basis element and (ii) for ξ, η distinct basis elements. To test whether an α K¯ 3 lies in ∈

2t+¯b1 2t+¯b2 C¯ = 1, π θ¯1, π θ¯2 , D E we can pair it with a basis of the dual lattice C¯∨ with respect to the trace pairing. Let (λ0, λ1, λ2) be the dual basis to (θ¯ , θ¯ , θ¯ ) (that is, tr(θ¯ λ )= 1 ). Then (λ , λ ) is a basis for ( 3 )tr=0, and we have already 0 1 2 i j i=j 1 2 OK met λ2: it is ωC , up to a unit. Hence

2t ¯b1 2t ¯b2 C¯∨ = λ0, π− − λ1, π− − λ2

D 2t ¯b 3 tr=0 E 2t ¯b = λ + π− − 1 ( ) + π− − 2 ω h 0i OK¯ h Ci We actually wish to test not whether α C¯, but the weaker condition α C¯ + K¯ , so (due to the natural tr=0 ∈ ∈ duality between C/¯ ¯ and (C¯∨) ) we pair only with elements of OK tr=0 2t ¯b 3 tr=0 2t ¯b (C¯∨) = π− − 1 ( ) + π− − 2 ωˆ . OK¯ h Ci Hence α C¯ + K¯ if and only if ∈ 2t ¯b2 • π− − tr(ωC α) is integral, and

121 2t ¯b • π− − 1 tr(ακ) is integral for κ (1; 1;0), (0; 1; 1) ; that is, all coordinates of α are congruent ¯ ∈ { − − } modulo π2t+b1 .

2 t This is the origin of the - and -conditions respectively. Applying this to the values α = 4δξ , 8π− δξη derived from the basis aboveM yieldsN the desired form of all the Φ-conditions.

We say that the condition ij or ij is active if its corresponding modulus m¯ ij or n¯ij is positive. An M N 1 inactive condition is automatically satisﬁed (noting that ω¯C− has nonnegative valuations). Because the a’s and b’s have been sorted in increasing order, and because 0 t e, we have the following implications among the activity of the and : ≤ ≤ Mij Nij

/ / 33❄ 22❄ 11 N ❄❄ NK ❄ NO 33 / 22 / 11 M MK MO / / 23❄ 13❄ 12❄ N ❄❄ N ❄❄ N ❄ / / M23 M13 M12 The next lemma limits our concern to the four boxed conditions:

Lemma 18.2. (a) Suppose that the a¯i and ξ¯i come from the extender decomposition of a quartic ring and a resolvent thereof. Then: • No conditions are active except , , , , and . M11 M12 M22 N11 N12 • and are not both active. M12 M22 • is very weakly active, that is, n¯ s/¯ 2. N12 12 ≤ (b) Suppose that the a¯i and ξ¯i come from the extender decomposition of some lattice I R. Suppose that the inactivity restrictions from part (a) hold and that conditions , , ⊆, and are M11 M12 M22 N11 satisﬁed. Then 12 is satisﬁed, and the a¯i and ξ¯i actually come from a quartic ring. That is, “ 12 is automatic if it isN very weakly active.” N

Proof. (a) Suppose that the a¯i and ξ¯i come from a quartic ring. 3 If is active, so are and . Since the ξ¯ are supposed to form an ¯ -basis of , we M13 M12 M11 i OK OK¯ obtain for all ξ¯ 3 , ∈ OK¯ tr(ξ¯ ξ¯) 0 mod m ¯ . 1 ≡ K Since ξ¯1 is primitive, this is a contradiction.

If 12 and 22 are active, then so is 11. We have a 2-dimensional subspace V = ξ¯1, ξ¯2 of the 3-dimensionalM M space k3 that is isotropicM for the trace pairing. But the trace pairing is nondegenerate, K¯ so this is a contradiction. If is active, so are and . If char k = 2, then t = e =0 so is also active, and we M33 M22 M11 K 6 M12 have a contradiction as above. If char kK = 2, we use that squaring is a linear operation mod 2 to ¯ 3 obtain that for all ξ K¯ , ∈ O 2 tr(ξ¯ ) 0 mod m ¯ , ≡ K which is a contradiction. 1 If 22 is active, note that char kK =2 since 22 is active. There are two cases. If s¯ =0, then ω¯C− is a unit,N so the condition M

1 2 n¯ : All coordinates of ω¯− ξ¯ are congruent mod π ii Nii C · i ¯2 determines ξi mod mK¯ up to scaling. But since we are in characteristic 2, square roots are unique, 1 and ξ¯ and ξ¯ are scalar multiples mod m ¯ , a contradiction. Now assume s>¯ 0, so ω¯− (u;0;0) for 1 2 K C ≡

122 (K) some unit u × . Now gives ξ¯ 0 mod m ¯ . But now gives that ξ¯ is a unit multiple of ∈ OK¯ Nii i ≡ K Mii i (0;1;1) modulo mK¯ , a contradiction.

Finally, assume that 12 is active and not very weakly active: that is, n¯12 > s/¯ 2. Note that n¯11 > s¯ since otherwise wouldN be active. If s¯ =0, then implies that ξ¯2 ω mod m , up to scaling, N22 N11 1 ≡ C C and then 12 implies that ξ¯1ξ¯2 ωC mod mC , up to scaling. Since ωC is a unit, this is a contradiction. N ≡ s¯ 1 So assume s>¯ 0. Note that ωC is a unit multiple of (0; 1; 1) modulo π and that ω¯C− is a unit multiple of (1;0;0) modulo πs¯. Now implies that − N11 2 s¯ ξ¯ ω mod m ¯ ω = m ¯ (π ;1;1). 1 ≡ C K C K ¯(K) (K) ¯(K) So vK (ξ1 )=¯s/2 exactly (recalling the notion of ξ from Notation 17.9). If vK (ξ2 ) > 0 also, we ¯ ¯ ¯(K) get ξ1 ξ2 mod mK¯ by the same argument as when N22 is active. So ξ2 is a unit, and the 12 condition≡ N 1 n¯ : All coordinates of ω¯− ξ¯ ξ¯ are congruent mod π 12 N12 C · 1 2 is unsatisﬁed, because the K-coordinate has valuation s/¯ 2 and the others have higher valuation.

1 1 (b) Note that n¯12 2 n¯11 because otherwise 22 would be active. We have ω¯C− a unit multiple of (1;0;0) ≤ N (K) 2¯n12 2¯n12 ¯2 (K) n¯12 ¯ n¯12 1 ¯ modulo π , so 11 implies that π (ξ1 ) , that is, π ξ1 . Now π ω¯C− ξ1, so 12 is satisfied. N | | | N Based on this, we will count quartic rings with fixed resolvent. Since a lot will happen with various things being fixed and others varying, it is worthwhile to lay down the following: Conventions 18.3. We fix variables in the following order: • First, we fix the resolvent data, which comprise – a resolvent C R; ⊆ ¯b1 ¯b2 – an extender decomposition C = 1, π θ¯1, π θ¯2 , which fixes ωˆC and s¯. We can, and do, assume that s¯ is finite; D E – a tracedness parameter t, 0 t e, which defines a reduced resolvent C . ≤ ≤ t • Then we fix the discrete data of a quartic ring, which comprises

1 0 – a choice of coarse coset [δ] H / 0. There are H cosets δ0 0. Then δ = δ0τ, where τ R× can vary; ∈ L | | L ∈ O

– its extender indices a¯i, which are constrained by the integrality needed for a sublattice of √δωˆCR and the inactivity inequalities of Lemma 18.2. • Then we choose δ and ξ¯ , which are constrained by the and conditions. 1 M11 N11 • Then we choose ξ¯2, which is constrained by its coprimitivity with ξ¯1 and by the 12 and 22 conditions. M M

• Lastly, we choose ξ¯3, which is constrained by its coprimitivity with ξ¯1 and ξ¯2. Whenever we speak about possibilities for any of the items on this list, it will be implicitly assumed (if not stated) that all the previous items have been ﬁxed in conformity with their respective restrictions.

a¯′ Since ξ¯1 π− 1 √δωˆC R, conditions 11 and 11 can be viewed in another way, which will be simpler for some purposes:∈ M N Lemma 18.4. A ξ¯ √δωˆ R 3 satisﬁes the and resolvent conditions if and only if the quotient 1 ∈ C ∩OK¯ M11 N11 ¯2 ξ1 2¯a′ +¯s β = π− 1 R ωC ∈ is a linear combination of the reduced basis vectors of C of the form

n s¯ m 2¯a +¯s β = x + yπ 11− θ¯ + zπ 11 θ¯ , x π− 1 K, y,z . 1 2 ∈ ∈ OK

123 a¯′ π 1 ξ¯1 Proof. Since is the ﬁrst basis element for I¯, the 11 and 11 conditions are equivalent to √δωˆC M N 2 a¯1′ ¯ 2¯a1′ ¯2 π ξ1 4π ξ1 2¯a +4¯b 4¯b s¯ 2¯a s¯ K¯ + C¯ 4δ = =4π 1 1− 1− β =4π 1− β, ∋ √δωˆC ! ωC that is, s¯ 2¯a 2e n s¯ m β K¯ + π − 1− C¯ = K¯ + π 11− ¯ θ¯ + π 11 ¯ θ¯ . ∈ OK 1 OK 2 ¯b ¯b s¯ 2¯a Since (1, π 1 θ¯ , π 2 θ¯ ) form an -basis for C, we can always ﬁnd x π − 1 K and y,z K such that 1 2 OK ∈ ∈ n11 s¯ m11 β = x + yπ − θ¯1 + zπ θ¯2. Then the resolvent conditions simplify to y,z . ∈ OK 18.1 Transformation, and ring volumes in the white zone

We will proceed to compute the volumes of the solution sets in which the reduced vectors ξi lie. (We use reduced vectors ξ , not extender vectors ξ¯ , because the latter do not lie in a controllable -lattice.) For i i OK simplicity, we will transform everything to R itself, which we normalize so that R has volume 1, and to its 2 O projectivization P( R), to which we give a volume of 1+1/q +1/q , so that a distinguished aﬃne open has volume 1. O The simplest way to do this is as follows.

3 Lemma 18.5. Fix the discrete data. In particular, δ = τδ0 lies in a ﬁxed coarse coset.There is a γ = γi K¯ with the properties that, letting ∈ γ γ = i,0 , i √τ 1 we have that ξ′ = γ− ξ is a primitive vector in . i i i OR Proof. Note that ξi must lie in a′ 3 J = π− i δωˆ R = √τJ , (129) C ∩ OK¯ 0 where p a′ 3 J = π− i δ ωˆ R 0 0 C ∩ OK¯ is an R-lattice of dimension 1. As R is a principalp ideal ring (it’s a product of DVR’s), we obtain that O O 1 J0 = γi R for some γi, clearly not a zero-divisor. Since ξi J πJ, we get ξi′ = γi− ξi R π R, as desired.O ∈ \ ∈ O \ O The valuations of γ may be computed by observing the smallest nonnegative valuation of an element of J at each place. The ξi form a set of reduced vectors for a sublattice of √δωˆC I if and only if the ξi′ lie in certain explicit subsets of , computed below. OR Unramiﬁed.

• If [δωˆ ] , then all a are integers, and γ = γ is a unit. The three ξ′ must form a basis of . If C ∈ L0 i i OR they are found successively, their ring volumes are respectively 1+1/q +1/q2 , 1+1/q , and 1 .

• If [δωˆ ] (1; π; π) , then there is one a Z; there γ (1; √π; √π), and C ∈ L0 j0 ∈ j0 ∼ ξ′ × Q, j0 ∈ OK × O

a subset whose projectivization has volume 1 . Meanwhile, two aj1 ,aj2 lie in Z +1/2; there γjk (√π;1;1), and ∼ ξ′ K Q jk ∈ O × O

with their Q-coordinates forming a basis of Q; thus ξ′ has volume 1+1/q and ξ′ has volume O O j1 j2 1 .

124 Splitting type 13.

1 • Here [δωˆ ] . The a fill out the classes in Z/Z, but all γ are units and all ξ′ lie in ×, a subset C ∈ L0 i 3 i i OR whose projectivization has volume 1 . Lemma 18.6. In the white (i.e. free) zone where no or is active, we take all γ as in Lemma 18.5. Mij Nij i The ring volume of triples (ξ1′ , ξ2′ , ξ3′ ) is given in terms of the discrete data as follows: 1. If R is unramified and [δωˆ ] , the ring volume is (1+1/q +1/q2)(1 + 1/q). C ∈ L0 2. If R is unramified and [δωˆ ] / , the ring volume is 1+1/q. C ∈ L0 3. If R is totally ramified, the ring volume is 1.

18.2 From ring volumes to ring counts 2 3 Lemma 18.7. Let be a quartic ring. The set V of triples (ξ1′ , ξ2′ , ξ3′ ) in P ( R) whose associated ring is has a volume determinedO by the discrete data of alone. It is given by O O O tame 3d0 a2 a1 a3 a2 a3 a1 + + vK N(γ1γ2γ3) µ(V )= q−⌈ − ⌉−⌈ − ⌉−⌈ − ⌉ 2 c, · where dtame = e 1 0 Ri/K − R Xi is the standard lower bound for the discriminant valuation, attained for tame extensions, and

1 if a1 = a2 = a3 1 1 c =  1+ q + q2 if a1 = a2

πai Ξ πs¯δω I = πa1 ξ , πa2 ξ , πa3 ξ ; (130) i ∈ C h 1 2 3i

p s¯ ai and secondly, they generate the whole of √π δωCI. Since the π ξi form a K-basis for √δωˆC R, we can write

3 aj ai Ξi = π − cij ξi (131) j=1 X for some coeﬃcients c K. Condition (130) is then equivalent to ij ∈ v (c ) max a a , 0 , ij ≥ { i − j }

ai s while the condition that the π Ξi generate the whole of π ′ δωC I is equivalent to the change of basis being invertible: p v (det [cij ]) = 0. Thus we have parametrized V by the group

Γ= [c ] GL ( ): v (c ) a a . { ij ∈ 3 OK ij ≥ i − j }

125 3 More precisely, V is in continuous bijection with the cosets T Γ, where T = × is the subgroup of \ OK diagonal matrices, because the Ξ P( ) are deﬁned only up to scaling. i ∈ OR Without the invertibility condition, the volume of matrices in Mat3( K ) satisfying the valuation restrictions deﬁning Γ is O

max 0, a a − { ⌈ i − j ⌉} 1 i,j 3 a2 a1 a3 a2 a3 a1 q ≤X≤ = q−⌈ − ⌉−⌈ − ⌉−⌈ − ⌉.

The invertibility depends only on the cij modulo π, and the fraction of matrices over kK of the shapes

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ , 0 , and 0 ∗ ∗ ∗ 0 ∗ ∗ 0 0∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗       that are nondegenerate is seen to be 1 3 1 c, − q · accounting for the three cases in the deﬁnition of c. Projectivizing, T Γ is a subset of (P3 )3 of volume \ OK q a2 a1 a3 a2 a3 a1 c. −⌈ − ⌉−⌈ − ⌉−⌈ − ⌉ · It remains to compute how the volume transforms under the bijection Ψ: T Γ ∼= V that we have constructed. This map is K-linear and is a product of the three maps \

Ψ : K K K R i × × → 3 a a γj (c ,c ,c ) c π j − i ξ′ . i1 i2 i3 7→ ij γ j j=1 i X

ni On the domain where it sends primitive vectors to primitive vectors, Ψi scales volumes by q− , where ni is the determinant valuation, i.e. Ψ Λ3 3 = πni Λ3 . (132) i OK OR Extending scalars to K¯ , the left side of (132) becomes O

3 3 a a γj ξj′ Ψ Λ = π j − i i OK¯ γ j j i Y ^ a 3a π j j − i = ξ PN(γ ) j i j ^ 3 When R is tamely ramiﬁed, the wedge product of the ξj generates the whole of Λ K¯ , because the ξj are an extender basis. Indeed, in all cases, if we let O

dwild = d dtame, 0 0 − 0 then dwild/2 3 3 ξ ξ ξ = π 0 Λ . h 1 ∧ 2 ∧ 3i OK¯ Accordingly, we get

wild d0 j aj 3ai+ 2 3 π − 3 Ψi Λ K¯ = P Λ K¯ . O N(γi) O Meanwhile, the right side of (132) is

d0 ni 3 ni+ 2 3 π Λ ( R K K¯ )= π Λ K¯ . O ⊗O O O

126 Hence dtame n = a 3a v (N(γ )) 0 i j − i − K i − 2 j X so 3dtame n = v (N(γ γ γ )) 0 i − K 1 2 3 − 2 i X and tame 3d0 n a2 a1 a3 a2 a3 a1 + + vK (N(γ1γ2γ3)) µ(V )= µ (T Γ) q− i i = q−⌈ − ⌉−⌈ − ⌉−⌈ − ⌉ 2 c, \ · P · as desired. Consequently, we can compute the number of rings with any given discrete data by ﬁnding the volume of permissible (ξ1′ , ξ2′ , ξ3′ ), and dividing by µ(V ). We carry out the computation of this volume in the 2a1 2a3 succeeding sections. Observe that µ(V ) equals q − times a correction that depends only on the ai mod 1 and whether any ai are equal. This will simplify the entry of the ring volumes into Sage at the end of the proof.

19 The conic over OK For each α R×, the equation ∈ tr(αξ2)=0 deﬁnes a conic on the projectivization of R. Its determinant is D0N(α), up to squares of units, with respect to any K -basis of R, where D0 is the discriminant of R. As we will ﬁnd, it is preferable to transform the conicO so that itsO discriminant has as low valuation asO possible:

Definition 19.1. Let K be a local field, char K =2. Bya conic over K we mean a lattice V of dimension 3 over equipped with an integral bilinear form6 , or equivalentlyO an integer-matrix quadratic form OK C : V , up to scaling by × . We say that is C → OK OK C • unimodular if det 1 (note that det is uniquely defined up to squares of units); C ∼ C • tiny if det π and there exists a v V such that (v) is a unit; C ∼ ∈ C • relevant if it is either unimodular or tiny. Remark 19.2. Between changing basis and rescaling the whole form , we can scale det by any unit: hence we will sometimes assume that det is exactly 1 or π. C C C Let be a generator of the different ideal d . For instance, we can take ♦ R/K 1 R unramified = 2 (133) ♦ ( πR R totally tamely ramified Then by the definition of the different, the formula ξ λ♦(ξ)=tr ♦ defines a linear functional λ♦ : R K that is perfect, that is, λ♦ generates the dual R∨ as an R-module, and hence the pairing O → O O O (x, y) λ♦(xy) 7→ is a perfect -linear pairing on . If α , then the conic OK OR ∈ OR 2 λ♦(αξ )

127 is -integral on (because the corresponding bilinear form λ♦(αξη) is integral) with determinant N(α). OK OR We will put the conic defined by the 11- and 22-conditions in this form. The entities involved in transformationM will beM marked by the symbol (“odot”). This is the symbol for a circle in Euclidean geometry, and it is chosen to reflect a particular simplifying⊙ fact: after the transformation, 2 the conic is self-congruent, that is, any two points on it can be taken to one another by an isometry of P ( K ) preserving the conic. This will follow from the independence of basepoint in Lemmas 19.9 and 19.10.O The general conic over a p-adic field is not self-congruent. Lemma 19.3. Let i 1, 2 . Fix the discrete data such a way that is active. Recall that δ = δ τ is in ∈{ } Mii 0 a fixed coarse coset. Then there is a multiplier γ⊙ R× with the following property: i ∈

ξi′ ξi ξi⊙ = = γi⊙ γiγi⊙√τ

2 m11 is a primitive vector in R, and the 11 condition tr(ξ1 ) 0 mod π is equivalent to a condition of the form O M ≡ 2 m⊙ λ♦ δ⊙ξ⊙ 0 mod π 11 , ≡ where m11⊙ = m11 p⊙ is an integer and δ⊙ = δ0⊙τ R, where δ0⊙ depends on the discrete data alone and satisﬁes − ∈ O 1 [δ⊙] = [δ ωˆ ] H and v (N(δ⊙)) 0, 1 , 0 0 C ♦ ∈ K 0 ∈{ } 2 so that the conic ⊙(ξ⊙)= λ♦ δ⊙ξ⊙ is relevant. Moreover, ⊙ is unimodular exactly when M M [δ ωˆ ] . 0 C♦ ∈ L0 a 2b Proof. We have ξ π− 1− 1 √δωˆ R. 1 ∈ C · Note that whatever γi⊙ we pick, the conic takes the form

2 2 2 m¯ λ♦ γ⊙ γ τξ⊙ 0 mod π ii , ♦ i i i ≡ or, for any p⊙, 2 2 γi⊙ γi 2 m¯ p⊙ ⊙ ii λ♦ ♦ p τξi 0 mod π − . π ⊙ ! ≡

So we seek to pick γi⊙ and p⊙ so that 2 2 γi⊙ γi δ0 δ0⊙ = ♦ πp⊙ lies in with norm of valuation 0 or 1, and OR

m⊙ =m ¯ p⊙ ii ii − is an integer. The second condition is easily seen to follow from the ﬁrst.

Unramiﬁed. If [δ ωˆ ] , then all extender indices are in Z and γ is a unit, so δ⊙ is a unit as well, 0 C ∈ L0 i 0 choosing γi⊙ =1 and p⊙ =0. Thus we get If [δ0ωˆC ] (1; π; π) 0 for some ordering of the coordinates, then [δ] = [(1; π; π)δ′] for some unit δ′. By Corollary 17.5,∈ either L

• a′ Z and ξ √δ ( × √π ), or i ∈ i ∈ ′ OK × OQ 1 • a′ Z + and ξ √δ (√π ). i ∈ 2 i ∈ ′ OK × OQ

128 In the first case, is unsatisfiable if active, because αξ2 has exactly one coordinate of zero valuation. M11 i So we have the second case. Observe that ~v(γi) = (1/2, 0, 0), so choosing γi⊙ = 1 and p⊙ = 0, we get ~v(δ⊙) = (1, 0, 0): the conic has determinant π. Note that ⊙ 0 mod π as a quadratic form: after ∼ M 6≡ passing to an unramified extension we may assume that L = K K K, and then ⊙ is diagonal with × × M two of the three coefficients units. So ⊙ is tiny. For compatibility with the other splittingM types, we let

0, [δ0ωˆC] 0 hi = ∈ L ( 1, otherwise.

Splitting type (13). Here is generated by a uniformizer π with π3 = π (for a suitably chosen OR R R uniformizer π). Under the Minkowski embedding, π = √3 π ζ¯ , where R · 3 ¯ 2 ζ3 = (1; ζ3; ζ3 ). 2 We have = πR. ♦ 1 Here [δ0ωˆC] 0 always. The extender indices are in 3 Z, and γi is always a unit. Let h = hi 0, 1, 1 be the integer such∈ L that ∈{ − } h a′ Z . i ∈ − 3 Then h/3 h ξ π √δ R = ζ¯− √δ R; 1 ∈ ⊙ 3 ⊙ 3 indeed, since ξ1 is primitive in K¯ , O h ξ ζ¯− √δ ×. 1 ∈ 3 ⊙OR If h =0, then is unsatisﬁable because the trace of a unit in is always a unit. M11 OR If h =1, the choice γi⊙ =1, p⊙ =2/3 works, making δ⊙ a unit. If h = 1, we can no longer take γi⊙ = 1, because the maximal possible value for p⊙ is 1/3 and the − 2 2 corresponding conic has determinant π . Instead, take γ⊙ = π− . Then the corresponding values of ξ⊙, ∼ i R instead of being units, have valuation 2/3 and thus are still primitive in . Take p⊙ = 2/3 and observe OR − that δ⊙ is again a unit.

To summarize, the salient data of the transformation is shown here:

(K) (K) Q Q spl.t. a′ mod Z h [δωˆ ] v γ γ⊙ v γ γ⊙ p⊙ δ⊙ ξ′ 1 1 C♦ ∈ K K ∼ ∼ ur 0 0 0 0 0 0 1 ? ur 1/2 1 (1; πL; π) 1/2 0 0 (π;1;1) (?;1;1) L0 3 1 1/3 1 0 0 02/3 1 1 13 −1/3 1 L 2/3 2/3 2/3 1 π2 − L0 − − − R (134) The advantage of making the conic’s determinant associate to either 1 or π is that we have to solve very few isomorphism types of conics. Although we do not prove the following classiﬁcation, it animates the choice of what invariants we compute: Conjecture 19.4. Let be a relevant conic over the ring of integers of a local ﬁeld K. C OK (a) If is tiny, it is determined up to isomorphism by its Brauer class ε( ) 1 , the single bit telling whetherC (~x)=0 has a nonzero solution over K. C ∈ {± } C (b) If is unimodular, it is determined up to isomorphism by its Brauer class ε( ) and its squareness level C C ℓ( ), the largest ℓ Z, 0 ℓ e/2, such that C ∈ ≤ ≤ 2 min 2ℓ+1,e cλ mod π { } C ≡ as a quadratic form, for some constant c K× and linear form λ. Moreover, all combinations of values (ε,ℓ) occur, except that for e even, ℓ(∈)= O e/2 implies ε( )=1 by Proposition 19.7 below. Thus there are exactly e +1 isomorphism classesC of conics of determinantC 1.

129 Note that if is unimodular and 2ℓ+1 e, then is congruent to a cλ2 modulo π2ℓ if and only if modulo π2ℓ+1, as the x2C, y2, z2 coeﬃcients have square≤ ratiosC modulo π2ℓ+1 and the cross-terms are multiples of 2 anyway. So the squareness level carries the same amount of information as the squareness e 1 ( ) = max i : cλ2 mod πi as a quadratic form 0, 2, 4,..., 2 − e . C C ≡ ∈ 2 ∪{ } (We cannot have ( ) >e, or the determinant would vanish modulo π.) When we use coordinates,C we will generally use one of two explicit types of conics: the diagonal conic aX2 + bY 2 + cZ2 =0 and the basepoint conic 2XZ Y 2 + aZ2 =0, − so called because it passes through the basepoint [1 : 0 : 0] and is tangent to the line Z =0 there. We begin with results concerning the diagonal conic.

19.1 Diagonal conics

Lemma 19.5. Any relevant conic is diagonalizable, that is, there exists a basis (v1,...,v3) for the given lattice V such that C (x v + x v + x v )= a x2 + a x2 + a x2. C 1 1 2 2 3 3 1 1 2 2 3 3 Proof. When char k =2, we have that any conic is diagonalizable by an easy Gram-Schmidt procedure (in K 6 fact more is true: see O’Meara [46], 92:1). So we assume that char kK = 2. Here a quadratic space is not diagonalizable in general, and we must use the restrictions given on . Write the matrix of , with respect to any basis (ξ , ξ , ξ ), as C C 1 2 3 a f e = f b d . C e d c   In the case that is unimodular, we see from C det = abc +2def ad2 be2 cf 2 1 C − − − ∼ that at least one of the diagonal entries—say a—is a unit. Then we can use a to eliminate f and e (that, is, add multiples of ξ1 to ξ2 and ξ3). Now if b (or, symmetrically, c), is nonzero modulo π, we use it to eliminate d, and we are done, as we have found the requisite diagonal form. However, it is possible that a d mod π C ≡  d    for units a and d. Rescaling ξ3, we can assume that 1 a 1 mod π. C ≡ ·  1    At ﬁrst we are doubtful, because the unimodular form 0 1 = C2 1 0 on 2 is not diagonalizable. However, we can use the identity OK 1 1 1 1 1 1 1 1 1 1 1 1 1 1 mod 2 1 1  1  1 1 ≡  1         130 to change to a basis in which 1 a 1 mod π. C ≡ ·  1 Then the diagonalization proceeds without a hitch.  If is tiny, we proceed similarly. Taking a a unit (since we are given (ξ1) 1 for some ξ1), we can eliminateC f and e. Then since C ∼ det = a(bc d2) π, C − ∼ we must have at least one of b and c a unit, as otherwise det would be either a unit (if d is a unit) or a multiple of π2 (if π d). So we can eliminate d and again get theC desired diagonalization. | The following lifting lemma for solutions modulo 4π will be essential for us.

Lemma 19.6. Let be a diagonalized conic on an K -lattice V , and assume that v(det ) 1. Let v V be a primitive vectorC with O C ≤ ∈ (v) 0 mod πm, m> 2e. C ≡ Then there exists a v′ V such that ∈ m e v′ v mod π − and (v′)=0. ≡ C Proof. We may write the conic in diagonal form

(x v + + x v )= a x2 + + a x2 . C 1 1 ··· n n 1 1 ··· n n

Let v = x1v1 + + xnvn. Since v is primitive, not all the xi are zero modulo π. We claim that there is an i with ··· π ∤ ai and π ∤ xi. (135)

If not, then det π, and without loss of generality, a1,...,an 1 are units while an π; and xn is a unit C ∼ − ∼ while x1,...,xn 1 are multiples of π. Summing, we ﬁnd that − (v) a x2 0 mod π2, C ≡ n n 6≡ a contradiction. Choose i satisfying (135). We will construct v′ by changing only the xi coordinate of v to a diﬀerent value x′ . The desired condition (v′)=0 takes the form i C 2 xi′ = y

2 m for some y xi modulo π . Since m > 2e, we have that y is also a square and, indeed, has a (unique) ≡ m e square root x′ satisfying x′ x mod π − . This constructs the desired v′. i i ≡ i Here are two easy corollaries.

Proposition 19.7. If e is even and is a unimodular conic of squareness e (the maximal possible value), then ε( )=1, that is, has a rationalC point. C C Proof. We may assume that is diagonal: C (x ξ + x ξ + x ξ )= a x2 + a x2 + a x2. C 1 1 2 2 3 3 1 1 2 2 3 3 Then ε( ) is a Hilbert symbol, C a a ε( )= − 2 , − 3 . C a a 1 1 Both arguments are squares of units modulo 2. But since e is even, they are actually squares modulo 2π, so the Hilbert symbol is 1 by Lemma 7.19.

131 Proposition 19.8. Let and ′ be conics of determinant 1 and squareness level ℓ. Suppose that the C C 2e 2ℓ associated bilinear forms of and ′ are congruent modulo π − . Then ε(C)= ε(C′). C C Proof. We may assume that is diagonal: C (x ξ + x ξ + x ξ )= a x2 + a x2 + a x2. C 1 1 2 2 3 3 1 1 2 2 3 3

Although ′ need not be diagonal with respect to the same basis, the orthogonalization procedure furnished C 2e 2ℓ by the proof of Lemma 19.5 yields a basis (ξ1′ , ξ2′ , ξ3′ ) with ξj′ ξj mod π − such that ′ is diagonal with respect to it, ≡ C 2 2 2 ′(x ξ′ + x ξ′ + x ξ′ )= a′ x + a′ x + a′ x C 1 1 2 2 3 3 1 1 2 2 3 3 2e 2ℓ with a a′ mod π − . New compare j ≡ j

a2 a3 a2′ a3′ ε( )= − , − and ε( ′)= − , − C a a C a a 1 1 1′ 1′ The arguments to the Hilbert symbols are squares modulo π2ℓ+1, so the value of the Hilbert symbol is max 2ℓ+1,2e 2ℓ unchanged under multiplying by units that are squares modulo π { − } by Lemma 7.19, and that is exactly what we have done.

19.2 The solution volume of the conic We now use the basepoint form to determine volumes of conics.

Lemma 19.9 (Igusa zeta function of a conic). Let be a conic of determinant 1 on the projectivization C P(V ) of a 3-dimensional vector space V . Suppose that has Brauer class ε( )=1, that is, it admits a C C basepoint v0 such that (v0)=0. C 1 2 Let U be the volume of v P(V ) (counting the whole P(V ) to have volume 1+ q− + q− ) such that m⊙,n⊙ ∈ v v mod πn⊙ (136) ≡ 0 (v) 0 mod πm⊙ . (137) C ≡

(Note that these m⊙ and n⊙ correspond to the m11⊙ and n⊙ of Lemmas 19.3 and 19.18.) Then for m⊙ and n⊙ integers with m⊙ > 2e and m⊙ 2n⊙, the volume U depends only on m⊙, ≥ m⊙,n⊙ n⊙, and the squareness level ℓ = ℓ( ). It is given by C

2e m⊙ n⊙ U = q − − , n⊙ e m⊙,n⊙ ≥ and the recurrence

2 n⊙ = e 2ℓ 1 0 − − ≥ U q n⊙ e mod 2,n⊙ >e 2ℓ 1,n⊙ > 0 m⊙,n⊙ =  ≡ − −  e Um⊙,n⊙+1  q +1 n⊙ =0,ℓ =  2 1 otherwise.  Explicitly, 

e m⊙ n⊙ q − − n⊙ e (a black conic) e n⊙ ≥ m⊙+ −2  q− e n⊙ >e 2ℓ 1,n⊙ > 0 (a blue conic) j k U =  m⊙+ℓ ≥ − − m⊙,n⊙  2q 0 n e 2ℓ 1 (a green conic)  − ⊙  1 ≤ ≤ −e − 1+ q m⊙+e/2, n =0,ℓ = ,e even (a large conic) q − ⊙ 2   

132 Proof. For the black-conic case, we diagonalize the conic to

(X,Y,Z)= aX2 + bY 2 + cZ2. C

Let the basepoint be v0 = [X0 : Y0 : Z0]. Note that two coordinates of v0, say X0 and Y0, are nonzero modulo π. We may scale so that Y0 =1 and so that all solutions we seek have Y = 1. This eliminates the issue of scaling ambiguity. 2n⊙ n⊙ The n⊙-pixel of [X : 1 : Z] satisfying (136) has volume q− . For ﬁxed Z, with Z Z0 mod π 2 m⊙ 2 ≡n⊙+e the condition (137) simpliﬁes to X u mod π , where u is a unit with u X0 mod π . Hence its ≡ m⊙ e ≡ n⊙ e m⊙ e m⊙ n⊙ solutions form a congruence class mod π − , and overall, the solution volume is q q − = q − − . We use this as the base case to prove the recursive formula (and hence also the· explicit formula) for

Um⊙,n⊙ by downward induction on n⊙. Our aim is to determine the number r of (n⊙ + 1)-pixels within m⊙ the n⊙-pixel of v that contain a solution to (v)=0, or equivalently, to (v) 0 mod π . Then, by 0 C C ≡ induction, there is a volume Um⊙,n⊙+1 of solutions in each of those, so Um⊙,n⊙ = rUm⊙,n⊙+1 as desired. It remains to compute U r = m⊙,n⊙ . Um⊙,n⊙+1

We now abandon the diagonalized form and choose coordinates such that the basepoint is v0 = [1 : 0 : 0] and the tangent line there is Z =0. Then the conic has the form

(X,Y,Z)=2gXY +2dXZ cY 2 +2bYZ + aZ2. C − Note that π does not divide both d and g, for then the conic’s determinant

0 g d g c b =2bdg + cd2 ag2 − − d b a

2 would be divisible by π . So, by symmetry, we may assume d 1. We scale the conic so that d = 1, and then the transformation Z Z gY makes g = 0. Also, the∼ transformation X X bY makes b = 0. Now c =1 to make the determinant7→ − 1. Thus the conic takes the basepoint form 7→ −

(X,Y,Z)=2XZ Y 2 + aZ2, C − where a K is the only undetermined coefficient. By definition of squareness level, we know that a is ∈ O min 2ℓ+1,e 2ℓ+3 a square modulo π { }, but not modulo π if ℓ < e/2 . For any a′ K , the transformation 2 ⌊ ⌋ ∈ O Y Y + a′Z can be used to increment a by the square a′ , followed by another X X b′Y to remove 7→ 7→ − the YZ term. Picking a′ appropriately, we can arrange so that v(a) reveals the squareness: either • ℓ< e/2 and v(a)=2ℓ +1, or ⌊ ⌋ • ℓ = e/2 and 2 a. In this case, indeed, the transformation X X a/2 makes a = 0. (So we have proved⌊ one⌋ case| of Conjecture 19.4: for ℓ = e/2 and ε =1, the7→ conic− takes the fixed form 2XZ Y 2.) ⌊ ⌋ − To parametrize ( K ), we use the age-old trick of stereographic projection, that is, drawing lines of varying slope throughC O the known basepoint [1 : 0 : 0]. An easy calculation shows that the second intersection of the line sZ tY =0 with the conic is [s2 at2 :2st :2t2], yielding an isomorphism − C − P1(K) = (K) ∼ C [s : t] [s2 at2 :2st :2t2]. 7→ − 2 2 2 If [s : t] is in lowest terms over K , then [s at :2st :2t ] need not be in lowest terms over K , but will j O −2 2 2 O have cancellation by π , where j = min vK (s at ),e + vK (st),e + vK (t ) . Note that j e because s and t are coprime, so { − } ≤ j = min v (s2 at2),e . { K − }

133 Note also that the resulting point s2 at2 2st 2t2 [X : Y : Z]= − : : πj πj πj lies in the (e j)-pixel of [1 : 0 : 0], but not in the (e j + 1)-pixel if j > 0. Hence the points we are − − interested in, namely in the n⊙-pixel of the basepoint but outside the (n⊙ + 1)-pixel, correspond exactly to 2 2 values of [s : t] for which j = e n⊙. That is, the valuation v(s at ) must be exactly e n⊙ (if n⊙ > 0) or at least e (if j =0). − − − Observe that when n⊙ < e 2ℓ 1, there are no solutions. Also, when n⊙ > e 2ℓ 1 is of the same parity as e, there are no solutions,− because− − − 2 2 2 min 2ℓ+1,e s at s mod π { } − ≡ min 2ℓ+1,e has even valuation if nonzero mod π { }. The ratio Um⊙,n⊙ /Um⊙,n⊙+1 is thus 1 in these cases, as claimed. 2 2 • Suppose that n⊙ = e 2ℓ 1. If n⊙ > 0, we seek the [s : t] such that s at attains its maximal − − ℓ+1 e 2 − valuation 2ℓ +1: this happens when π s. If n⊙ = 0, we seek π s , which is still equivalent to ℓ+1 | ℓ+1 | π s. Hence we are looking at the [s : t] with t =1, s = π s′, where s′ K . Then the resulting point| on ( ) is ∈ O C OK s2 at2 2st 2t2 [X : Y : Z]= − : : π2ℓ+1 π2ℓ+1 π2ℓ+1 2ℓ+2 2 ℓ+1 2 π s′ a 2π s′ 2t = − : : π2ℓ+1 π2ℓ+1 π2ℓ+1 2 a 2s′ 2 = πs′ : : − π2ℓ+1 πℓ π2ℓ+1

2 a 2 n⊙+1 e 2ℓ πs′ : 0 : mod π = π − . ≡ − π2ℓ+1 π2ℓ+1 n⊙+1 2 We claim that this is actually the same point modulo π regardless of s′, that is, the πs′ term 2ℓ+1 contributes nothing. If 2ℓ +1= e, this is clear because n⊙ =0. Otherwise, a′ = a/π is a unit, and if we multiply all three coordinates by a ′ 1 mod π, a′ + πs′ ≡ the last two coordinates do not change mod πn⊙+1 because they are 0 mod πn⊙ . Thus all points

[X : Y : Z] obtained lie in a single (n⊙ + 1)-pixel, and hence the ratio Um⊙,n⊙ /Um⊙,n⊙+1 is 2.

• Suppose that n⊙ e mod 2, n⊙ > e 2ℓ 1, and n⊙ > 0. Then e n⊙ = j = 2j′ is even, with ≡ 2 −2 −j − j′ j′ j′ ℓ. The pairs [s : t] yielding s at π are exactly those with s π . Write t =1, s = π s′, ≤ − ∼ ∼ where s′ × . Then the resulting point on ( ) is ∈ OK C OK s2 at2 2st 2t2 [X : Y : Z]= − : : πj πj πj a 2 2s′ 2 = + s′ : : −π2j′ πj′ π2j′

2 2 n⊙+1 s′ : 0 : mod π , ≡ π2j′ where at the last step we multiplied all three coordinates by the unit 2 s′ 2 a 1 mod π. s′ 2j ≡ − π ′

We get q 1 diﬀerent (n⊙ +1)-pixels, one for each value of s′ mod π. Hence the ratio Um⊙,n⊙ /Um⊙,n⊙+1 is q. −

134 e e/2 • Finally, suppose that n⊙ =0 and ℓ = . Write t =1, s = π s′, where s′ . We get 2 ∈ OK s2 2st 2t2 [X : Y : Z]= : : πe πe πe 2s′ 2 2s′ 2 = + s′ : : πe/2 πe/2 πe

2 2 n⊙+1 s′ : 0 : mod π = π . ≡ πe 

We get q diﬀerent 1-pixels, one for each value of s′ mod π. Hence the ratio Um⊙,n⊙ /Um⊙,n⊙+1 is q +1.

For determinant π, we use the same method. Fortunately, everything comes out much simpler.

Lemma 19.10 (Igusa zeta function of a conic). Let be an integer-matrix conic over K of determinant π. Suppose that has Brauer class ε( )=1, that is, itC admits a basepoint v such that O(v )=0. C C 0 C 0 Let U be the volume of v P(V ) such that m⊙,n⊙ ∈ v v mod πn⊙ (138) ≡ 0 (v) 0 mod πm⊙ . (139) C ≡

Then for m⊙ > 2e and m⊙ 2n⊙, the volume U depends only on m⊙ and n⊙: ≥ m⊙,n⊙ qe m⊙ n⊙ n >e (a black conic) U = − − ⊙ m⊙,n⊙ m⊙ 2q− 0 n⊙ e (a green conic) ( ≤ ≤ Proof. Diagonalize the conic to the form

(X,Y,Z)= aX2 + bY 2 + cπZ2 =0, C where a,b,c K× . Observe that X0 and Y0 are units, and scale so that Y = Y0 =1. ∈ O n⊙ m⊙ 2 If n⊙ >e, then for each Z Z0 mod π , the condition (X,Y,Z) 0 mod π simpliﬁes to X u m⊙ 2 ≡ n⊙+e+1 C ≡ n⊙ ≡ mod π , where u X0 mod π , and hence the square roots X with X X0 mod π form a single ≡ m⊙ e e m⊙ n⊙ ≡ congruence class mod π − . So the volume is q − − . 2 m⊙ 2 If n⊙ = e, then we use the same method, but now the equation X u mod π , where u X0 mod 2e+1 m⊙ e ≡ ≡ π , has as solution set two classes mod π − , each the negative of the other. We claim that these are all the solutions mod πm⊙ ; that is, that the whole conic lies within an e-pixel. Suppose there is such an [X : 1 : Z] [X : 1 : Z ] mod2, and let v (X X )= i, v (Z Z )= j. Then 6≡ 0 0 K − 0 K − 0 a(X2 X2)= πc(Z2 Z2). − 0 − 0 If i

19.3 The Brauer class In this section we understand the Brauer class of conics of the form

: tr(αx2)=0. A This is a conic in the P2(K) of possible values of x. Over K, there are just two types of conic, one with points and one without. Our ﬁrst task will be to understand which case occurs for each α. Our main result will be the following.

135 Lemma 19.11. If α RN=1, deﬁne ∈ 1 if tr(αx2)=0 for some nonzero x L ε(α)= ∈ 1 otherwise. ( −

Then the map of F2-vector spaces

H1 µ → 2 α ε(α)/ε(1) 7→ is a nondegenerate quadratic form whose associated bilinear form is none other than the Hilbert pairing on H1. That is, ε(αβ)= ε(1) ε(α) ε(β) α, β . · · · h iε The proof is not especially difficult, but it uses different tools than the rest of the paper and so will be deferred. See Appendix A. Remark 19.12. ε comes up, in a related context, in the work of Bhargava and Gross ([7], §7.2), where it is stated to be a quadratic form, at least in the tamely ramified case. Remark 19.13. Over fields of characteristic not 2, a quadratic form is uniquely determined by its associated 1 1 1 bilinear form. However, over F2, the local Hilbert pairing , ε on H lifts to (H )∗ = H quadratic forms, thanks to the ambiguity by adding a linear functionalh•.•i It is not hard to show| that| these| | quadratic forms are exactly ε(ωα) α , 7→ ε(ω) for each ω H1. ∈ 19.4 The squareness For the cases in Lemma 19.3 in which the transformed conic is unimodular, we need also to compute its squareness. M If is a unimodular conic, the maximal C Lemma 19.14. Let [ ] = [1] H1. (The reason for this strange definition is that, in splitting type 121, we will need a [ ] = [1] in♥ general.)∈ The conic♥ 6 2 (ξ⊙)= λ♦ δ⊙ξ⊙ , δ⊙ × M ∈ OR has squareness

ℓ ( ) = max ℓ :[δ⊙ ] 1 mod π ; ℓ = e, or ℓ

Remark 19.15. As stated, the lemma only requires to be defined modulo e/2 (unramified types) resp. ♥ L⌊ ⌋ 2 e/2 (ramified types). We will mostly use in this way, but when we do the brown zone, we will need a finerL ⌊ definition⌋ and will mention this. ♥

Proof. In unramified splitting type, we first claim that going up to an unramified extension K′/K does not change either the left or the right side of (140). The right-hand side is less than e/2 only if [δωˆC ] 2i+1 ⌊ ⌋ ♥ is represented by a generic unit δ⊙ = 1+ απ , 2i +1 < e, and this generic unit remains generic in R′ = K′ R. As to the left side, we can diagonalize the conic to have the form ⊗K M (X,Y,Z)= aX2 + bY 2 + cZ2, abc =1. M

136 Then ( ) = min 2ℓ(b/a)+1, 2ℓ(c/b)+1,e , M { } and this remains invariant over K′. Therefore, we may assume that R ∼= K K K is totally split. Let δ⊙ = (a; b; c). Then is diagonal and × × M

( ) = min 2ℓ(b/a)+1, 2ℓ(c/b)+1,e M { } = min 2ℓ(a)+1, 2ℓ(b)+1, 2ℓ(c)+1,e { } = min 2ℓ(δ⊙)+1,e , as desired. 3 2 In splitting type 1 , we can scale δ⊙ by ( ×) so that its level is manifest: OR 2j+1 δ⊙ =1+ α π , · R where α R× and where ∈ O j 1, 3, 4, 6,..., 3e 2, 3e, ∈{ − ∞} 2 2 controls ℓ(δ⊙). Let δ⊙ = a + bπR + cπR and note that, in the basis (1, πR, πR), the conic

2 2 (ξ′) = tr(π− δ⊙ξ′ ) M R has matrix c b a b a πc . a πc πb Since a is a unit, we get  

c πb ℓ( ) = max i e : and are squares mod πi M ≤ a a j 1 = min 2 − ,e 3 ℓ(δ ) = min ⊙ ,e , 2 as desired.

We have the following corollary:

Lemma 19.16. The Brauer class ε(δ) takes the same value for all δ in the coset ωˆC e /2 . ♥♦L⌈ ′ ⌉ Proof. Let [δ] = [κωˆ ], where C♥♦ 2 e/2 +1 π ⌈ ⌉ , R unramiﬁed κ 1 mod 2 e/2 ≡ ( π ⌈ ⌉, R ramiﬁed.

Then [δ⊙ ] e′/2 , so the conic = δ has maximal squareness e. If e is even, we know that ε(δ)=1 by Proposition♥ ∈ L 19.7.⌈ ⌉ So we may assumeM thatM e is odd. Now κ 1 mod2π, so the associated bilinear forms ≡

(ξ, η)= λ♦(δ⊙ξη) and ′(ξ, η)= λ♦(κδ⊙ξη) M M are congruent modulo 2π. So by Proposition 19.8, the two conics have the same Brauer class.

137 19.5 11 N 1 2 In this section, we will transform the -condition, which says that all coordinates of ω¯− ξ are congruent N11 C · 1 modulo πn11 , into a more manageable form. We will sometimes need to make some subtle reductions, and thus we make the following definition: Definition 19.17. A first vector problem P consists of a choice of resolvent algebra R and as much of the discrete data as is needed to make and meaningful: the coarse coset δ , the resolvent extender M11 N11 0L0 vector θ1 (which determines ωˆC and s¯), and the moduli m11 and n11. These are required to satisfy the requisite integrality properties, which essentially say that

m n s¯ B (m ,n s¯)= π 11 θ + π 11− θ θ1 11 11 − OK 1 OK 2 is a subset of R, but are otherwise untethered from a cubic or a quartic ring. The answer to a ﬁrst vector problem is the weighting WP = Wθ ,m ,n : δ0 0 Q 0 1 11 11 L → ≥ that attaches to each quartic algebra δ δ the volume of ξ′ P( ) such that the corresponding ∈ 0L0 1 ∈ OR ξ1 = ξ1′ γ1 satisﬁes the resolvent conditions

: tr(ξ2) 0 mod πm11 M11 1 ≡ 1 2 n : All coordinates of ω¯− ξ are congruent mod π 11 . N11 C · 1 We normalize volumes so that 1 1 µ(R)=1 and µ(P(R))=1+ + . q q2

We write W ⊙ instead of W when we wish to normalize instead by the vector ξ1⊙ = ξ1′ /γ⊙ in Lemma 19.3. Thus v(NR/K (γ⊙)) WP⊙ = q WP.

First vector problems will be sorted into zones, given by linear inequalities on m11 and n11, and having the properties that within each zone, the answer has a uniform description. Zones will be named by colors in such a way that a brightening of the color correlates with a lowering of m11 and/or n11 and an increase in the answer. Brightening is governed by the following poset:

gray s9 ●● sss ●● ss ●● sss ●● ss ●# black / plum / purple / blue / green / red (spl.t. 121 only)

 brown / yellow ❥❥❥ ❥❥❥❥ ❥❥❥❥ ❥❥❥❥ u❥❥❥ lemon / beige / white (spl.t. 121 only)

Lemma 19.18. Fix the data of a ﬁrst vector problem P in such a way that is active with N11 0

138 • In unramiﬁed splitting types, n11 n⊙ = . 2 l m • In splitting type (13), n11 h1 n⊙ = . 2 − 3 

Remark 19.19. The condition n11 2e (which, as we will see, restricts us to the blue, green, red, yellow, and lemon zones) can be removed,≤ but then our conclusion must be that there is a family ξ ,...,ξ { 0(1) 0(r)} of basic solutions, r 1, 2, 4 . The formula for n⊙ becomes more complicated, and we will be able to solve these zones by other∈{ means. }

Proof of Lemma 19.18. In view of Lemma 19.6, we may assume m = , replacing ξ⊙ by a value in the 11 ∞ 1 same e-pixel that satisﬁes ⊙(ξ1⊙)=0 exactly. M 2 2 Let ξ0 be a ﬁxed solution to P, and let ξ1 be any solution to 11. Observe that ξ1 and ξ0 are both 2 2 M 3 2 3 traceless, so their wedge product ξ1 ξ0 is a scalar multiple of (1;1;1) K¯ . (Here we identify Λ K¯ with 3 ∧ ∈ O O K¯ via the trace pairing and standard orientation, so that the wedge product is given by the same formula O 3 2 3 as the cross product on R .) Let ξ , α be an ¯ -basis for the traceless plane in . Write { 0 } OK OK¯ 2 2 ξ1 = c0ξ0 + c1α.

The coeﬃcient c controls how far ξ deviates from ξ and thus the satisfaction of : 1 1 0 N11 1 2 n All coordinates of ω¯− ξ are congruent mod π 11 N11 ⇐⇒ C · 1 1 n All coordinates of c ω¯− α are congruent mod π 11 ⇐⇒ 1 C 1 3 We claim that the element ω¯− α does not have all coordinates congruent mod m ¯ : C ∈ OK¯ K 2 • If s¯ =0, then ω¯C is a unit so this is equivalent to α and ξ0 being linearly independent modulo mK¯ ;

s¯ s¯ (K) (K) (K) • If s¯ > 0, then ω¯C (1; π ; π ), so ξ0 has positive valuation. Hence α and (¯ωC α) are units, (Q) ∼ while (¯ωC α) is not. Consequently

c 0 mod πn11 (143) N11 ⇐⇒ 1 ≡ ξ2 ξ2 0 mod πn11 . (144) ⇐⇒ 1 ∧ 0 ≡ 2 2 Now (144) is advantageous, because the wedge product ξ1 ξ0 has all its coordinates equal, so we can test by looking at any one of them. We have (coordinate indices∧ mod 3) N11 2 2 2 2 2 2 (i) (i+1) (i 1) (i 1) (i+1) ξ ξ = ξ ξ − ξ − ξ 1 ∧ 0 1 0 − 1 0 (i+1) (i 1) (i 1) (i+1) (i+1) (i 1) (i 1) (i+1) = ξ ξ − ξ − ξ ξ ξ − + ξ − ξ 1 0 − 1 0 1 0 1 0 and the two factors are congruent modulo 2, so, since n 2e, 11 ≤ (ξ ξ )(i) 0 mod πn11/2. (145) N11 ⇐⇒ 1 ∧ 0 ≡ We now examine this for each coordinate i in turn, and for each splitting type.

139 Unramified. We first dispose of the case h1 =1. Here the conic is tiny, and by Lemma 19.10, all solutions ξ1 satisfy e ξ⊙ ξ⊙ mod π 1 ≡ 0 ξ ξ mod πe+1/2; πe; πe 1 ≡ 0 ξ2 ξ2 mod π2e+1; π2e; π2e . 1 ≡ 0 Since we are assuming n11 2e, we find that 11 is automatic, andn⊙ e may be chosen at will. ≤ N √ ≤ Now assume that h1 =0. Here δ⊙ R× and ξ0, ξ1 δ⊙ R are K¯ -primitive. We scale ξ1 by K× to be as close to ξ as possible. Then k =∈v( Oξ ξ ) is an∈ integer,O and O O 0 1 − 0 ξ ξ ξ and 1 − 0 0 πk are linearly independent elements of √δ . In particular, their wedge product is primitive, so ⊙OR v(ξ ξ )= k. 0 ∧ 1 Hence n k 11 N11 ⇐⇒ ≥ 2 n k 11 ⇐⇒ ≥ 2 l m n11/2 ξ ξ mod π⌈ ⌉ ⇐⇒ 1 ≡ 0 n /2 ξ⊙ ξ⊙ mod π⌈ 11 ⌉, ⇐⇒ 1 ≡ 0 as desired.

3 ¯ h1 √ Splitting type (1 ). Scale ξ1 ζ3− δ⊙ R× to be as close to ξ0 as possible, and consider the valuation 1 ∈ O k = v(ξ1 ξ0) 3 Z. If k e, then both 11 and its claimed transformation are easily seen to hold, so assume that− k

140 20 Boxgroups

If 11 is strongly active, then N ξ2 β = 1 ωC is a unit. By Lemma 18.4, the solutions to P arise from the β that lie in the box n s m 2a +s x + yπ 11− θ + zπ 11 θ , x π− 1 K, y,z . 1 2 ∈ ∈ OK 1 Necessarily 2a1 + s Z and x K× . Also, [δ] = [β] H . So the support of δ is bound up with the H1-classes of− units in∈ various boxes.∈ O Certain boxes have∈ pride of place: those for which the corresponding subset of H1 is a group, which we will call a boxgroup. In this section, our aim is to define certain subgroups of H1. We fix the resolvent data. We do not fix the discrete data, but we will reference the transformation γ1,0 of the conic that occurs in Lemma 19.3 when L = K R is the algebra for δ and m is large enough. ∼ × ∈ L0 11 20.1 Signatures Recall that in Lemma 16.1, we filtered H1 by level spaces , where L0 ⊃ L1 ⊃···⊃Le′ e, R unramified e′ = ( 2e, R ramified. We would like to define some additional subgroups of H1. We use the following notion. 1 Definition 20.1. If S H is a subgroup, define the signature of S to be the sequence of e′ +2 subgroups ⊆ S S = ∩ Li / , 1 i e. i S ⊆ Li Li+1 − ≤ ≤ ∩ Li+1 The following subgroups S / will occur frequently and will be given names: i ⊆ Li Li+1 • 0 denotes the zero subgroup / ; Li+1 Li+1 • ⋆ denotes the entire group / ; Li Li+1 • , in unramified resolvent for 0 i

ℓ0 ℓ1 ℓ2 ℓ0 ℓ1 ℓ2 0.(00) (0⋆) (⋆⋆) .⋆ and 0.(00) (⋆0) (⋆⋆) .⋆, ℓi = e i X and will be denoted by T 1(ℓ0,ℓ1,ℓ2) and T1(ℓ0,ℓ1,ℓ2) respectively. −

141 20.2 Boxgroups in unramiﬁed splitting type Lemma 20.2. Let m,n N+ , m n> 0. Let B (m,n) be the box ∈ ∪ {∞} ≥ θ1 B (m,n)= πc + πnc θ + πmc θ : c . θ1 { 0 1 1 2 2 i ∈ OK } (a) For every ξ 1+ B , ∈ θ1(m,n)

1 Bθ1 (m,n)= ξBξ− θ1 (m,n) (146)

1 1+ Bθ1 (m,n)= ξ(1 + Bξ− θ1 (m,n)). (147)

(b) If m 2n + s, ≤

then Bθ1 (m,n) is closed under multiplication and the translate 1+ Bθ1 (m,n) is a group under multiplication.

1 1 Proof. (a) Since ξ 1 mod π, we can take ξ− θ2 in the role of θ2 for deﬁning ξBξ− θ1 (m,n), which is a ≡ n n m lattice with basis [π ξ, π θ1, π θ2], all of which are contained in Bθ1 (m,n). This proves the reverse inclusion of (146), and equality follows by comparing volumes. To get (147), we add ξ to both sides and use ξ 1 B (m,n) to simplify the left-hand side. − ∈ θ1 n n m (b) Observe that Bθ1 (m,n) is a lattice with basis [π , π θ1, π θ2]. Since m n, the only product that n 2 ≥ 2n does not clearly lie in the lattice is (π θ1) , whose 1- and θ1-components are divisible by π , and whose θ -component is divisible by π2n+s. Since m 2n + s, this product lies in the lattice. 2 ≤ Thus Bθ1 (m,n) is closed under multiplication and so is 1+Bθ1 (m,n). To show the existence of inverses, simply note that 1 =1 ξ + ξ2 1+ ξ − − · · ·

converges to an element of 1+ Bθ1 (m,n) for every ξ Bθ1 (m,n). ∈

Lemma 20.3. Let = min 2ℓ(ˆω )+1,e C { C } be the squareness of the conic for δ =1; put =0 if [ˆω ] / (i.e. s is odd). C C ∈ L0 If θ1 is translated by a suitable element of K and scaled by a suitable element of K× (neither of which change the associated resolvent C), then thereO is an η such that O ∈ OR η2 θ mod πs+ C (148) ≡ 1 and s η 1,θ , π 2 θ , 2θ . (149) ∈ 1 ⌈ ⌉ 2 2 D E Proof. If s is odd, then we can scale and translate θ1 so that

θ (1;0;0) mod πs. 1 ≡

Then η = θ1 satisﬁes the desired conditions. If s is even, then by deﬁnition of C , there is a linear form λ such that

2 2 tr(1⊙ξ⊙ ) c λ(ξ⊙) mod π C ≡ · 1 as functions of ξ⊙ . Here 1⊙, the transform of δ =1 under Lemma 19.3, is a unit whose class in H is ∈ OR [ˆωC ]; for concreteness, we may take ωˆC 1⊙ = . (πs;1;1)

142 s/2 Since ωˆC is traceless, the conic has a distinguished basepoint, namely ξ0′ = [π ; 1; 1]. Pick a ξ′ = ξ1′ in the kernel of λ that does not lie in the same 1-pixel as the basepoint. We claim that the choice

s/2 s/2 η = (1; π ; π )ξ1′ fulﬁlls the conditions. 2 The θ2-coeﬃcient of η′ is given by

2 s 2 s s+ tr(ω η )= π tr(1⊙ξ′ ) π λ(ξ′ ) = 0 mod π C . C 1 ≡ 1 Hence there are a,b such that ∈ OK η2 a + bθ mod πs+ C . ≡ 1

We claim that π ∤ b, which makes it possible to replace θ1 by a + bθ1. Suppose not. If s =0, we get ξ1′ ξ0′ (K) (K) ≡ modulo π, contrary to hypothesis. If s> 0, we get π a so π (ξ1′ ) ; we also know that π (ξ0′ ) . But 1⊙ is | |2 (|K) a unit, so λ′ is a perfect linear functional, and its kernel in P (kK ) intersects the line (ξ′) 0 in only one 1-pixel, a contradiction. ≡ As for (149), it can be rewritten as s v (η(2) η(3)) min ,e . K − ≥ 2 n o To prove this, observe that

η(2) η(3) η(2) + η(3) = (η(2))2 (η(3))2 θ(2) θ(3) 0 mod πs, − − ≡ 1 − 1 ≡ with the two factors on the left-hand side congruent modulo 2 πe. ∼ Lemma 20.4. Let 0

1 Then the projection [1 + Bθ1 (m,n)] of 1+ Bθ1 (m,n) onto H is a subgroup of signature

0.0ℓ0 ℓ1 ⋆ℓ2 .⋆ ⊤ where n ℓ = 0 2 jmk n ℓ = 1 2 − 2 j k m j k ℓ = e . 2 − 2 j k

Proof. The gray-red inequality (150) ensures that the projection T = [1 + Bθ1 (m,n)] is a subgroup. It is clear that

n T m , L⌊ 2 ⌋ ⊆ ⊆ L⌊ 2 ⌋ ℓ0 ℓ1 ℓ2 so the signature of T has the shape 0.0 ? ⋆ .⋆. Those middle ℓ1 components of the signature are at least , because for i n/2 and for all a K, we have ⊤ ≥⌊ ⌋ ∈ [1 + aπ2i+1θ ] T. 1 ∈ Thus the signature is at least the one claimed.

143 To prove that equality occurs, we ﬁx m 2e and proceed by downward induction on n. The base case ≤ n = m is clear since T = m . When moving from n +1 to n, note that T can grow by at most a factor of L⌊ 2 ⌋ | | [1 + B(m,n):(1+ B(m,n + 1))(1 + πn )] = q. OK If n is odd, there is nothing to prove, as we claim that T actually grows by a factor of q. If n is even, we claim that T does not change. It suﬃces to prove that each| | of the q cosets in (1 + B(m,n)) (1 + B(m,n + 1))(1 + πn ) OK contains a square. For c , consider ∈ OK n 2 n n 2 2 (1 + π 2 cη) =1+2π 2 cη + π c η .

n 2 n+s+C m The last term is π c θ1 up to an error in π R, which is in π R by the gray-green inequality. We n/2 O O claim that the cross term 2π cη lies in B(m,n + 1) also. If s is odd, this is trivial since we took η = θ1. Otherwise, we have n s n 2πn/2η 2πn/2 1,θ , πs/2θ , 2θ = B e + + min ,e ,e + ∈ 1 2 2 θ1 2 2 2 D E n o We get the needed inequality n s e + + m 2 2 ≥ from the gray-blue inequality, the diﬀerence of whose sides lies in Z +1/2 by parity considerations. So we n 2 n 2 have found a square in the coset 1+ π c θ1 + B(m,n +1)=(1+ π c θ1)(1 + B(m,n + 1)), as desired. As a corollary, we have:

Lemma 20.5. For every triple (ℓ0,ℓ1,ℓ2) of nonnegative integers satisfying

ℓ0 + ℓ1 + ℓ2 = e (153) s ℓ ℓ + + 1 (the gray-red inequality) (154) 1 ≤ 0 2 s + +1 ℓ C (the gray-green inequality) (155) 1 ≤ 2 s ℓ ℓ + +1, (the gray-blue inequality) (156) 1 ≤ 2 2

1 ℓ0 ℓ1 ℓ2 there is a boxgroup T (ℓ0,ℓ1,ℓ2) H of signature 0.0 ⋆ .⋆. such that, if m, n are integers satisfying the conditions of Lemma 20.4, then⊆ ⊤ n m n m [1 + B (m,n)] = T , ,e . θ1 2 2 − 2 − 2 j k j k j k j k Proof. If ℓ =0, take T (ℓ , 0,ℓ )= , the unique subgroup with the correct signature. 1 0 2 Lℓ0 Otherwise, let m =2ℓ0 +2ℓ1, n =2ℓ0 +1 in the preceding lemma. The transformation of the gray-red, gray-green, and gray-blue conditions is routine. For the last claim, note that decreasing m or increasing n can only make the conditions of Lemma 20.4 truer, with the exception of the condition m n. If m/2 = n/2 , then clearly [1 + Bθ1 (m,n)] = m/2 , so we can assume that ≥ ⌊ ⌋ ⌊ ⌋ L⌊ ⌋ n m n 2 +1 < 2 m. ≤ 2 2 ≤ Clearly j k j k m [1 + B (m,n)] 1+ B 2 ,n , θ1 ⊇ θ1 2 but both sides have the same signature, so equalityh holds. Likewise,j k i m m n n m n m 1+ B 2 ,n 1+ B 2 , 2 +1 = T , ,e , θ1 2 ⊆ θ1 2 2 2 2 − 2 − 2 h j k i h j k j k i j k j k j k j k but both sides have the same signature, so equality holds.

144 20.2.1 Supplementary boxgroups

As thus deﬁned, all boxgroups T satisfy e T 0. Groups not satisfying these inclusions occur will be denoted as follows. L ⊆ ⊆ L If s> 0, so that a distinguished splitting R = K Q exists, consider the image ι(K×) of the map ∼ × 2 1 ι: K×/(K×) H → a [(a; 1)] = [(1; a)]. 7→

We find that ι(K×) has signature . e. ⊤ ⊤ ⊤ where the middle e-many ’s denote the usual subgroups ⊤ = 1+ π2i+1aθ = 1+ π2i+1a(1; 0) / ⊤i { 1} { } ⊆ Li Li+1 1 and where the initial 1 = [1], [(1; π)] H / 0 and the final e = 1+4a : a K e have size 2 and H0 /2 respectively.⊤− In particular,{ }⊆ L ⊤ { ∈ O } ⊆ L | | 0 e 1 ι(K×) = H q = H . | | | | | | p From the explicit description in terms of the Hilbert pairing, we find that ι(K×) is isotropic and hence maximally isotropic for , . h• •i The group ι(K×) is always important, but it does not behave well with respect to boxgroups unless s> 2e, in which case we give it the name T ( , e, ). ∅ ∅ If s> 2ℓ1, we let

1 2e 2ℓ1+1 T (e ℓ1,ℓ1, )= ι(K×) ℓe ℓ = [(a; 1)] H : a 1 mod π − − ∅ ∩ L − 1 { ∈ ≡ } 1 2ℓ +1 T ( ,ℓ ,e ℓ )= ι(K×) = [(a; α)] H : α 1 mod π 1 . ∅ 1 − 1 · Lℓ1 { ∈ ≡ } ℓ e ℓ ℓ e ℓ Their signatures are, respectively, 0.0 0 − 0 . and . 1 x − 1 .x. The restrictions on s ensure that these⊤ boxgroups⊤ ⊤ satisfy⊤ such natural relations as

T (e ℓ ,ℓ , ) = T (e ℓ ,ℓ , 0) − 1 1 ∅ · Le − 1 1 T ( ,ℓ ,e ℓ ) = T (0,ℓ ,e ℓ ), ∅ 1 − 1 ∩ L0 1 − 1 which we will often use without comment. Finally, in all cases, we let

T (e, , )= 1 ∅ ∅ { } T ( , ,e)= H1. ∅ ∅

It will turn out that T (ℓ0,ℓ1,ℓ2) and T (ℓ2,ℓ1,ℓ0) are orthogonal complements whenever both are deﬁned (Lemma 21.13). Actually, this is simple to prove in the case that one of ℓ0,ℓ1,ℓ2 is the symbol . The sizes of these groups follow immediately from their signatures: ∅ Lemma 20.6. If T (ℓ ,ℓ ,ℓ ) is deﬁned and ℓ = , then 0 1 2 1 6 ∅ 0 e+ℓ ℓ T (ℓ ,ℓ ,ℓ ) = H q 2− 0 , | 0 1 2 | | | where if occurs as either ℓ or ℓ , it must be replaced by 1/[k : F ] = log (1/2). ∅ 0 2 − K 2 q 20.3 Boxgroups in splitting type 13 Let h 1, 1 be the integer such that ∈{ − }

h h h b Z + , θ ζ¯ ×, θ ζ¯− ×. 1 ∈ 3 1 ∈ 3 OR 2 ∈ 3 OR

145 Note the tight connection with the h of Lemma 19.3. Namely, if is strongly active, then β is a unit in i N11 Lemma 18.4, from which we get m11 Z, a1 Z, and h1 = h. Let ∈ ∈ ℓ(ˆω ) = min 2 C +1,e C 2 be the squareness of the conic 2 (ξ⊙)= λ♦(1⊙ξ⊙ ) M that occurs for δ =1. Note that there is just one conic, with [1⊙]=[ˆωC], although we turn our attention to 2 the part where ξ⊙ 1 resp. ξ⊙ π according as h =1 resp. 1. Then ∼ ∼ R i − ¯ h Lemma 20.7. Fix a resolvent C. If θ1 is translated and scaled appropriately, there is an η ζ3− R× such that ∈ O 2 2h η θ mod π C − 3 . ≡ 1 Proof. By Lemma 19.14, has squareness , which means that there is a linear form λ and a scalar M C c K× such that ∈ O 2 (ξ⊙) c λ(ξ⊙) mod π C M ≡ · The zero locus of modulo π, or equivalently of λ modulo π, consists of (q + 1)-many 1-pixels, q of which M 2 consist of units and the remaining one of elements ξ⊙ πR. If we were searching for a ξ⊙ = ξ1⊙ with h1 = h, 1 h ∼ 1+h we would have v(ξ⊙)= −3 . We pick a ξ⊙ of the other valuation v(ξ⊙)= 3 which lies in the kernel of λ, ensuring that (ξ⊙) 0 mod π C . M ≡ Then take

h 1⊙ 2b1 3 ¯ h η = π − ξ⊙ ζ3− R. sωˆC ∈ ♦ (See pp. XI.304–05 for motivation and details.) Note that

2 4b s+2(2b h ) 2 coef (η )= π− 1− 1− 3 ξ⊙ θ2 M 2h 0 mod π C − 3 ≡ Hence there are √3 π2 , h =1 a OK ∈ √3 π , h = 1 ( OK − and b K such that ∈ O 2 2h η a + bθ mod π C − 3 . ≡ 1 Since η is a unit, we must have π ∤ b, so θ1 can be replaced by a + bθ1. The following lemma is proved just like Lemma 20.2. Lemma 20.8. If m and n satisfy h h 0

Let Bθ1 (m,n) be the box

B (m,n)= πc + πnc θ + πmc θ : c ,c ,c . θ1 { 0 1 1 2 2 0 1 2 ∈ OK } ⊆ OR (a) For every ξ 1+ B , ∈ θ1(m,n)

1 Bθ1 (m,n)= ξBξ− θ1 (m,n) (157)

1 1+ Bθ1 (m,n)= ξ(1 + Bξ− θ1 (m,n)). (158)

146 (b) If m 2n, ≤

then Bθ1 (m,n) is closed under multiplication and the translate 1+ Bθ1 (m,n) is a group under multiplication.

1 Our goal is to study the projection of 1+ Bθ1 (m,n) onto H . The following yields the conditions under which a useful group is formed thereby: Lemma 20.9. Let 0

.(00)ℓ0 (⋆0)ℓ1 (⋆⋆)ℓ2 . or .(00)ℓ0 (0⋆)ℓ1 (⋆⋆)ℓ2 . for h =1 and h = 1 respectively, where − n ℓ = △ 0 2 m n ℓ = △ △ 1 2 − 2 m ℓ = e △ 2 − 2 except for the case h =1, m△ =2i +1, n△ =2i +2 (i Z), where T = . ∈ L2i+1 Proof. Suppose h =1. For a × , there are elements in T of the form ∈ OK

2i+ 1 3n 1 n△ [1 + aπ 3 θ ], i − = 1 ≥ 6 2 2i+ 5 3m 5 m△ [1 + aπ 3 θ ], i − = . 2 ≥ 6 2 Likewise, in the case h = 1, there are elements in T of the form −

2i+ 5 3n 5 n△ [1 + aπ 3 θ ], i − = 1 ≥ 6 2 2i+ 1 3m 1 m△ [1 + aπ 3 θ ], i − = . 2 ≥ 6 2 This shows that the signature of T is at least as large as claimed. To show equality, we ﬁx m< 2e and proceed by downward induction on n. The base case n = m 1/3 − (for h =1) or n = m 2/3 (for h = 1) is clear since T = m△ 1 or 2 m△/2 +1 respectively. When moving from n +1 to n, note− that T can grow− by at most a factorL of− L ⌊ ⌋ | | n [1 + B(m,n):(1+ B(m,n + 1))(1 + π⌈ ⌉ )] = q. OK

147 If n△ is odd, there is nothing to prove, as we claim that T actually grows by a factor of q. If n△ is even, we are claiming that T does not change. It suﬃces to prove| | that each of the q cosets in

(1 + B(m,n)) (1 + B(m,n + 1))(1 + π n ) ⌈ ⌉OK h contains a square. Recall the approximate square root η ζ¯− × from Lemma 20.7, which satisﬁes ∈ 3 OR 2 2h η θ mod π C − 3 . ≡ 1

For c K , consider ∈ O n 2 n n 2 2 (1 + π 2 cη) =1+2π 2 cη + π c η . (162)

n 2 n+ 2h m The last term is π c θ up to an error in π C − 3 . To say that this is in π , we need the inequalities 1 OR OR 2h 2h m n + and m n + e . ≤ C − 3 ≤ − 3 The ﬁrst of these is (160), and the second follows easily from (159) and (161). We claim that the middle m term of (162) lies in π R also, that is, O n m + e. ≤ 2 n 2 This follows from (161) and the fact that m n/2 Z. So we have found a square in the coset 1+ π c θ1 + n 2 − ∈ B(m,n +1)=(1+ π c θ1)(1 + B(m,n + 1)), as desired. As a corollary, just like Lemma 20.5, we get the following:

Lemma 20.10. For every triple (ℓ0,ℓ1,ℓ2) of nonnegative integers satisfying

ℓ0 + ℓ1 + ℓ2 = e (163) ℓ ℓ (the gray-red inequality) (164) 1 ≤ 0 +1 ℓ C h (the gray-green inequality) (165) 1 ≤ 2 − ℓ ℓ , (the gray-blue inequality) (166) 1 ≤ 2 there is a boxgroup T (ℓ ,ℓ ,ℓ ) H1 of signature h 0 1 2 ⊆ .(00)ℓ0 (⋆0)ℓ1 (⋆⋆)ℓ2 . or .(00)ℓ0 (0⋆)ℓ1 (⋆⋆)ℓ2 . for h = 1 and h = 1 respectively, such that, if m, n are rational numbers satisfying the conditions of Lemma 20.9, then −

[1 + Bθ1 (m,n)] = T (ℓ0′ ,ℓ1′ ,ℓ2′ ) where ℓ0′ ,ℓ1′ ,ℓ2′ are the numbers ℓ0,ℓ1,ℓ2 deﬁned in Lemma 20.9. Proof. If ℓ =0, take T (ℓ , 0,ℓ )= . Otherwise, take 1 0 2 Lℓ0

n△ =2ℓ +1 2Z +1, m△ =2e 2ℓ 2Z, 0 ∈ − 2 ∈ In other words, 2h 2h n =2ℓ +1 , m =2e 2ℓ + . 0 − 3 − 2 3 Conditions (164)–(166) immediately imply (159)–(161). Just as in Lemma 20.5, we then argue that increasing m by 1 (resp. decreasing n by 1), if it does not violate (159)–(161), yields a boxgroup of the same signature that is contained in (resp. contains) Th(ℓ0,ℓ1,ℓ2) and thus must equal Th(ℓ0,ℓ1,ℓ2).

148 The subscript “h” in Th(ℓ0,ℓ1,ℓ2) is logically superfluous, because θ1 is fixed. But it allows the following manipulation. Define T 1(ℓ0,ℓ1,ℓ2)= T1(ℓ0 + ℓ1, ℓ1,ℓ1 + ℓ2) − − for all (ℓ0,ℓ1,ℓ2) for which either side has been defined. Note that T1(ℓ0, 0,e ℓ0)= T 1(ℓ0, 0,e ℓ0)= 2ℓ0 already fulfills this relation, while allowing − − − L

T1(ℓ0, 1,e +1 ℓ0)= T 1(ℓ0 1, 1,e ℓ0)= 2ℓ +1 − − − − − L 0 saves us the trouble of excluding the case h = 1, m△ = 2i +1, n△ = 2i +2 from Lemma 20.9. We do not use any other boxgroups with negative ℓ1 within this paper, but in the code we do, converting everything to a T 1. − 20.4 The recentering lemma

When θ1 is nearly a square, we will sometimes be able to assume that it is a square, thanks to the following lemma, which we state separately for each splitting type. Lemma 20.11. If C is unramiﬁed and s is even, then there is an element ψC 1+Bθ1 ( , C ) and a K , 1 2 ∈ ∞ ∈ O b × such that θ′ = a + bψ− θ = η is a square in . ∈ OK 1 C 1 OR Proof. It is easy to see that tweaking θ1 by addends in K or multipliers in K× , which do not change the underlying resolvent ring C, do not aﬀect the truth ofO the lemma either. WeO may therefore assume, by Lemma 20.3, that there is an η such that η2 θ mod πs+ C . ≡ 1 2 2ℓ+1 In particular, η θ mod π [θ ]. Let k = . Notice that θ′ can take all values in the orbit ≡ 1 OK 1 C 1 a11θ1 + a12 a12 (a11a22 a12a21)θ1 θ1′ = = + − a21θ1 + a22 a22 a22(a21θ1 + a22) of θ1 under the congruence subgroup

a11 a12 1 0 k Γ(k)= γ = GL2( K ): γ mod π . a21 a22 ∈ O ≡ 0 1 k We claim that this orbit is precisely the pixel θ1 + π K [θ1]. The orbit is clearly contained in this pixel and contains all elements of the form O k k (1 + a11′ π )θ1 + a12′ π k 2 2k γ(θ1)= k k θ1 + π (a12′ + (a11′ a22′ )θ1 + a21′ θ1) mod π K [θ1]. a21′ π θ1 +(1+ a22′ π ) ≡ − O

2k So at least the orbit contains a point in each congruence class mod π K [θ1] in the claimed pixel. But O 4k applying general elements of Γ(2k) to each of those, we get a point in each congruence class mod π K [θ1], and so on. Hence, the orbit is dense in the pixel, and being compact, it coincides with the pixel, establishingO the desired result. 3 Lemma 20.12. If R is of splitting type 1 , there is an element ψC 1+ Bθ1 ( , C 2h/3) and a h/3 1 2 ∈ h ∞ − ∈ π− K ¯ , b × such that θ′ = a + bψ− θ = η is a square, η ζ¯− . ∩ OK ∈ OK 1 C 1 ∈ 3 OR Proof. It is easy to see that tweaking θ1 by addends in K or multipliers in K× , which do not change the underlying resolvent ring C, do not aﬀect the truth ofO the lemma either. WeO may therefore assume, by Lemma 20.7, that there is an η such that

η2 θ mod πk ≡ 1 where k = 2h/3. Notice that θ′ can take all values in the orbit C − 1 a11θ1 + a12 a12 (a11a22 a12a21)θ1 θ1′ = = + − a21θ1 + a22 a22 a22(a21θ1 + a22)

149 of θ1 under the congruence subgroup

a11 a12 h 1 0 3 (i j) k Γ(k)= γ = GL2( K¯ ): aij π − K,γ mod π . a21 a22 ∈ O ∈ ≡ 0 1 We claim that this orbit is precisely the pixel

h k θ′ ζ¯ × : θ′ θ mod π . { 1 ∈ 3 OR 1 ≡ 1 } The orbit is clearly contained in this pixel and contains all elements of the form

k k (1 + a11′ π )θ1 + a12′ π k 2 2k γ(θ1)= k k θ1 + π (a12′ + (a11′ a22′ )θ1 + a21′ θ1) mod π a21′ π θ1 +(1+ a22′ π ) ≡ −

h (i j) k where the aij′ are integral in the appropriate groups π 3 − − K. 2k So at least the orbit contains a point in each congruence class mod π K [θ1] in the claimed pixel. But O 4k applying general elements of Γ(2k) to each of those, we get a point in each congruence class mod π K [θ1], and so on. Hence, the orbit is dense in the pixel, and being compact, it coincides with the pixel, establishingO the desired result.

20.5 Charmed cosets Lemma 20.13. Let H be a ﬁnite 2-torsion group, and let ε : H 1 be a nondegenerate quadratic form →{± } over F2. Let V H be a subspace that is coisotropic; that is, V ⊥ is isotropic, or equivalently, V contains a maximal isotropic⊆ subspace. Then: (a) There is exactly one coset αV H such that ⊆ ε(α) =0. 6 δ αV X∈ Indeed, the sum is H , ±ε | | where ε 1 is an invariant of the quadraticp space (V,ε). We call αV the charmed coset of V . ± ∈ {± } (b) On any coset βV ⊥ inside the charmed coset αV , ε is constant. By contrast, on any coset βV ⊥ outside the charmed coset, ε is equidistributed.

Proof. (a) Assume ﬁrst that V is maximal isotropic. Then V = H . (If an F2-space admits a nondegenerate quadratic form, its dimension is even if ﬁnite.)| On| each| coset| αV , ε looks like a linear form, p that is, there is a λ V ∗ such that α ∈ ε(αβ)= ε(α) λ (β). · α Note that the linear form λα is independent of coset representative, so we have a mapping

λ : H/V V ∗. → If λ takes the same value on two diﬀerent cosets α1V, α2V , then we see that ε is linear on the union α1V α2V . Then the associated bilinear form , ε is isotropic on the space V 0, α2/α1 , which is too big⊔ to be isotropic. So λ is injective. Comparingh i sizes, we see that λ is surjective⊕{ also.} So there is one coset αV on which ε is identically 1 or 1. This is the charmed coset. On the remaining cosets, the values of ε are those of a nontrivial linear− functional on V and hence are equidistributed between 1 and 1. − For a general V , take N V maximal isotropic. Every V -coset decomposes into N-cosets, and only the one containing the charmed⊆ coset of N will yield a nonzero sum for ε, namely V = H . ±| | ± | | A priori the sign of the sum on the charmed coset depends on both ε and V . But if Np V are coisotropic, then it± is easy to see that N and V yield the same sign. Then, taking V = H, we⊆ obtain that one sign holds for all coisotropic subspaces.

150 (b) Take N V maximal isotropic. Then the charmed coset αV of V is the one containing a charmed coset αN⊆of N on which ε = 1 is constant. ±ε If β/α V , then for all γ V ⊥, we have αβ,γ =1 so ∈ ∈ h i ε(βγ)= ε(β)ε(α)ε(αγ) αβ,γ h i = ε(β),

since α, αγ αN. By contrast, if β/α / V , there exists γ V ⊥ such that αβ,γ = 1. Then ∈ ∈ ∈ h i − ε(βγ) = ε(β), so ε is nonconstant on βV ⊥. But V ⊥ is isotropic, so ε is a linear form (plus a − constant) on βV ⊥ and is therefore equidistributed.

As you might expect, we apply this lemma to the space H = H1 with its quadratic form ε. We will eventually ﬁnd that ε =1 (it is “positively charmed,” one might say) though this is not obvious. Let ε be the following± translation of ε: for [α] H1, C ∈ 2 1, coefθ¯2 (αξ )=0 for some ξ R× εC(α)= ε (ˆωCα)= ∈ 1, otherwise. ( − Note that ε is still a quadratic form on H1 whose associated bilinear form is the Hilbert symbol , . C h• •iR Note that εC (1) = 1. Lemma 19.16 can be interpreted as saying that

e /2 ♥♦L⌊ ′ ⌋ is charmed for εC . If V H1 is a subspace, we let ⊆ FV = 1V be its characteristic function. If V is coisotropic, we denote by Gεa,V , resp. GεC ,V the characteristic function of its charmed coset with respect to one of the quadratic forms εa,εC whose associated bilinear form is the Hilbert pairing. The εa will be omitted if clear. The following results will power the computation of Fourier transforms of ring totals, a necessary step in our desired reﬂection theorems. Lemma 20.14. Let V be a subspace of H1. (a) V FV = | | F . H0 V ⊥ | | (b) If V is coisotropic, then c V G\ε ,V = | | εa(1)εaF . a H0 V ⊥ | | (c) If V is coisotropic, then ε\G = qeε (1)ε F . a V ±εa a a V Proof. Part (a) is a standard property of the Fourier transform. For parts (b) and (c), let N V be a maximal isotropic subspace, so ⊆ V N = N ⊥ V ⊥. ⊇ ⊇ Let βN be the charmed coset of N, so βV is the charmed coset of V . For (b), we compute 1 G (δ)= αβ,δ V H0 h i α V | | X∈ d 1 = β,δ α, δ H0 h i h i α V | | X∈ V = | | β,δ F (δ). H0 h i V ⊥ | |

151 Hence it remains to prove that, for δ V ⊥, ∈ β,δ = ε (δ). h i a By the deﬁnition of the associated bilinear form,

β,δ = ε (1)ε (δ)ε (β)ε (δβ). h i a a a a

But εa(β)= εa(δβ) since both arguments lie in the charmed isotropic coset βN. This establishes (b). For (c), we compute 1 ε\G = ε (αβ) αβ,δ a V H0 a h i α V | | X∈ 1 = ε (1)ε (δ)ε (αβδ) H0 a a a α V | | X∈ ε (1)ε (δ) = a a ε (αβδ). H0 a α V | | X∈ 1 The last sum equals εa H if βδV is charmed, 0 otherwise. But βV is charmed, so the relevant condition is that δ V . So ± | | ∈ p ε (1)ε (δ) ε\G = a a H1 F (δ) a V H0 ·±εa | | V | | = qeε (1)ε (δ)Fp (δ), ±εa a a V as desired.

20.6 The projectors On the space of complex- (or even rational-) valued functions on H1, we can deﬁne certain projectors that divide up the work to be done. First look at the cosets of . Let L0 (1; π; π) , s>¯ 0 I = L0 ∪ L0 , s¯ =0 ( L0 be the union 0 α 0 of up to two cosets, using the distinguished splitting R ∼= K Q if s>¯ 0. Let I0, I1, I be the restrictionL ∪ L operators that restrict the support of a function to , I , and× H1 I, respectively. 2 L0 \ L0 \ They are orthogonal idempotents (I1 and/or I2 may vanish). Let Ji be the conjugate of Ii under the Fourier transform. Each J is convolution by a certain function supported on ; J is none other than the smear i Le′ 0 operator Se′ which will occur below. Since e′ 0, each Ii commutes with each Jj , so the IiJj form a system of nine orthogonal idempotents. ForL orderliness⊆ L of presentation, we transform all ring answers to a sum of terms each in the image of one idempotent. The Fourier transform interchanges the images of IiJj and Ij Ji. In splitting type (111), all nine idempotents are nonzero, although I1J2 and I2J1 will be found to annihilate every ring total. In the remaining splitting types, some of the idempotents vanish, and correspondingly some terms of our answers can be ignored.

Definition 20.15. In unramified splitting types, we define the use of a symbol x as follows, where the ℓi are such that the relevant boxgroups are well defined: • xF (0,ℓ ,ℓ )= I (F ( ,ℓ ,ℓ )), so that 1 2 1 ∅ 1 2 F ( ,ℓ ,ℓ )= F (0,ℓ ,ℓ )+ xF (0,ℓ ,ℓ ). ∅ 1 2 1 2 1 2 • xxF (0, 0,e)= I (F ( , ,e)), so that 2 ∅ ∅ F ( , ,e)= F (0, 0,e)+ xF (0, 0,e)+ xxF (0, 0,e). ∅ ∅

152 • F x(ℓ ,ℓ , 0)=2 J (F (ℓ ,ℓ , )), so that, if there is a distinguished coarse coset, 0 1 · 1 0 1 ∅ 2F (ℓ ,ℓ , )= F (ℓ ,ℓ , 0) + F x(ℓ ,ℓ , 0). 0 1 ∅ 0 1 0 1 • F xx(e, 0, 0) = H0 J (F (e, , )), so that | | · 2 ∅ ∅ H0 F (e, , )= F (e, 0, 0)+ F x(e, 0, 0)+ F xx(e, 0, 0). | | ∅ ∅

• xF x(0, e, 0)=2 I1J1(F ( , e, )) (note there must be a distinguished coarse coset for this to be meaningful), so that · ∅ ∅ 2F ( , e, ))=2F (0, e, )+2xF (0, e, ) ∅ ∅ ∅ ∅ = F ( , e, 0)+ F x( , e, 0) ∅ ∅ = F (0, e, 0)+ xF (0, e, 0)+ F x(0, e, 0)+ xF x(0, e, 0).

• The same definitions with G replacing F , as appropriate. We find that xG, xxG, and xGx are applicable. Observe that these definitions are crafted so that the following cute rule applies:

Lemma 20.16. Let N be one of the symbols F x, xF , F xx, xxF , xF x, and let N ′ be the symbol made by spelling N backward. If the ℓi are integers such that N(ℓ0,ℓ1,ℓ2) is meaningful, then

ℓ2 ℓ0 N(ℓ\0,ℓ1,ℓ2)= q − N ′(ℓ2,ℓ1,ℓ0). (This lemma also holds for N = F , though this will be proved later: see Lemma 21.13.) This will allow us to write the ring totals for all three unramified splitting types in a uniform way and verify reflection for them simultaneously. For the unshifted quadratic form ε, i is charmed for i e′/2, an easy consequence of for i e′/2, from which we derive: L ≥ ≥ Lemma 20.17. The function ε = ε F ( , ,e) has at most the following projections nonzero: C C ∅ ∅ (a) If [ˆω ] , then I J , I J , and I J . C ∈ L0 0 0 1 1 2 2 (b) If [ˆω ] / , then I J , I J , and I J . C ∈ L0 0 1 1 0 2 2 Proof. It suffices to prove that I (ε F ( , ,e)) = (xiF )(0, 0,e) i C ∅ ∅ is in the image of J , for each pair (i, j) mentioned in the lemma. Now is charmed for the unshifted j L0 quadratic form ε, an easy consequence of Proposition 19.8. Hence ωˆC 0 is charmed for εC , from which we get L (xiF )(0, 0,e) = (xj G)(0, 0,e). Taking the Fourier transform by Lemma 20.14(c),

(xiF\)(0, 0,e)= (xj F )(0, 0,e), ±εC i which lies in the image of Ij . Hence the original (x F )(0, 0,e) lies in the image of Jj , as desired.

20.7 Notation

If T (ℓ0,ℓ1,ℓ2) is a boxgroup, we let F (T ) = F (ℓ0,ℓ1,ℓ2) be its characteristic function, and G(T ) = G(ℓ0,ℓ1,ℓ2) be the characteristic function of its charmed coset, if applicable. Let T ×(ℓ0,ℓ1,ℓ2) denote the subset of elements of T (ℓ0,ℓ1,ℓ2) having minimal level, assuming this level is less than e′: T (ℓ ,ℓ ,ℓ ) T (ℓ +1,ℓ 1,ℓ ), ℓ 0,ℓ 1 0 1 2 \ 0 1 − 2 0 ≥ 1 ≥ T ×(ℓ0,ℓ1,ℓ2)= T (ℓ ,ℓ ,ℓ ) T (ℓ +1, 0,ℓ 1), ℓ 0,ℓ =0  0 1 2 \ 0 2 − 0 ≥ 1  T ( ,ℓ1,ℓ2) T (0,ℓ1,ℓ2), ℓ0 = ∅ \ ∅ Let F ×(ℓ0,ℓ1,ℓ2) and G×(ℓ0,ℓ1,ℓ2)(δ) be the characteristic functions of T ×(ℓ0,ℓ1,ℓ2) and ψC T ×(ℓ0,ℓ1,ℓ2), respectively. This will provide enough notation to write the ring totals in the succeeding sections.

153 21 Ring volumes for ξ1′

In this section, we will compute the volume of vectors ξ1′ R satisfying the 11 and 11 conditions. A sample of our answers are tabulated in Appendix B. ∈ O M N Because ξ¯1 = ξ1, we freely omit the bar on m11 = m11 and n11 = n11.

21.1 The smearing lemma

We now prove a simple lemma that allows us to reduce to the case m11 large.

Lemma 21.1. For a ﬁrst vector problem P, deﬁne u1 to be the unique value such that, when the conic is 2 u1 u1+1 transformed to minimal discriminant in accordance with 19.3, β⊙ = δ⊙ξ1⊙ must lie in π R π R, to wit: O \ O 3 1 if (σ, h1) (1 , 1) u1 = ∈{ − } ( 0 otherwise.

Assume that the m11 and n11 of P satisfy m11⊙ >u1 ♯ (so 11 is active even after transformation) and m11⊙ > n11. Let m11 > m11 lie in the same class mod Z, M ♯ ♯ ♯ and let m11⊙ be the corresponding value of m11′ . Let P be the ﬁrst vector problem with m11 = m11 and the rest of the data the same. Then the answer to P can be computed from that of P♯ by the following formula:

♯ m m11 WP = q 11− SrWP♯

Here the smear operator Sr is deﬁned by the following convolution: 1 SrW (δ)= W (αδ), r α |L | X∈Lr and r is the level for which m⊙ u = [η]: η 1 mod π 11− 1 . Lr ≡ n o We call this the smearing lemma because it states that the function WP can be obtained from WP♯ by averaging over the cosets of r, like reducing the resolution of a picture by averaging over larger pixels. Here the symbol ♯ is used to markL the “sharper” image given by the solutions of P♯ and should not be confused with the use of the same symbol in the context of tilting. The level r is given explicitly as follows: • In unramiﬁed type, m r = 11 , 0

m⊙ u η 1 mod π 11− 1 , ≡ ♯ 2 m⊙ = (η, ξ′): λ♦(ηδ⊙ξ′ ) 0 mod π 11 ,  R× P( R). (167) S  ≡  ⊆ O × O  1 2 2 n   ω¯− γ ξ′ a mod π 11 for some a  C ≡ ∈ OK   m⊙ u1  First, ﬁx η 1+ π 11− R. The conditions on ξ′ are seen to be the 11 and 11 conditions for δ replaced ∈ O M N u1 by ηδ; the omission of the η factor in 11 makes no diﬀerence, since the left-hand side is a multiple of π and the addition would have valuationN at least

(m⊙ u )+ u = m⊙ n . 11 − 1 1 11 ≥ 11

154 So the volume of ξ′ for ﬁxed η is W ♯ (δη), and since [η] takes all classes in equally often while ranging P Lr 3(m⊙ u ) in a pixel of volume q− 11− 1 , we get

3(m⊙ u ) q− 11− 1 µ( )= WP♯ (δη). S r η |L | X∈Lr On the other hand, a fixed ξ′ has a chance of being the second coordinate of a pair in only if ∈ OR S 1 2 2 n • it satisfies the condition ω¯− γ ξ′ a mod π 11 for some a , and N11 C ≡ ∈ OK m⊙ u • it satisfies the condition for some η =1+ π 11− 1 η′, η′ ; in particular, M11 ∈ OR 2 0 λ♦(ηδ⊙ξ′ ) ≡ 2 m⊙ u 2 = λ♦(δξ′ )+ π 11− 1 λ♦(η′δξ′ )

2 m⊙ u 2 = λ♦(δξ′ )+ π 11 λ♦ π− 1 δξ′

2 m⊙ λ♦(δξ′ ) mod π 11 . ≡

The volume of ξ′ satisfying these conditions is, by deﬁnition, none other than WP. For ﬁxed ξ′, the value of η′ is constrained by alone: M11 ♯ m⊙ u 2 m⊙ π 11− 1 λ♦(η′δ⊙ξ′ ) 0 mod π 11 ≡ ♯ ♯ u 2 m⊙ m⊙ m m λ♦ η′ π− 1 δ⊙ξ′ 0 mod π 11 − 11 = π 11− 11 . · ≡ u1 2 Since λ♦ is a perfect linear functional and π− δ⊙ξ′ is a primitive vector in R, the volume of η′ satisfying ♯ ♯ m m 3(m⊙ u )+m m O this congruence is q 11− 11 , which makes a volume of q− 11− 1 11− 11 for η. So

♯ 3(m⊙ u )+m m µ( )= π− 11− 1 11− 11 W (δ). S P Comparing the two expressions for µ( ), the result follows. S

21.2 The zones when 11 is strongly active (black, plum, purple, blue, green, and red) N In this section, we solve ﬁrst vector problems in which is strongly active, that is, n > s¯. In view of N11 11 the smearing lemma, we assume that m11 > 2e. Let nc (“n for the colorful zones”) be n11 s¯. The following little symmetry will be occasionally useful: −

n11 Lemma 21.2. Let ψ = a + bθ1 R with a K× , b π K× , and let P′ be the ﬁrst vector problem ∈ O ∈ O ∈ O 1 derived from P by replacing the pertinent extender vector θ1 by θ1′ = ψ− θ1. Then: 1 (a)ω ˆP′ = ψ− ωˆP. 1 (b) εP′ (δ)= εP(ψ− δ). (c) W (δ)= W (ψ 1δ). θ1′ ,m11,n11 θ1,m11,n11 −

Proof. The left-hand side is the volume of ξ′ for which 2 ξ′ δ × + B 1 (m ,n ) ξ ∈ OK ψ− θ1 11 11 0′ 1 = ψ− × + B (m ,n ) · OK θ1 11 11 by Lemma 20.2(a), since ψ × + B (m ,n ). So the condition on ξ′ can be written as ∈ OK θ1 11 11 2 ξ′ δψ × + B (m ,n ), ξ ∈ OK θ1 11 11 0′ of which the solution volume is seen to be Wθ1,m11,n11 (ψδ).

155 The following formula for the sum of the values of Wm11,n11 will be essential:

Lemma 21.3. If the values of m11 and n11 make 11 strongly active, then Wm11,n11 =0 unless the chosen coarse coset is . In this case N L0 0 2e m n + s¯ + d0 +v(N(γ)) W (δ)= H q − 11 − 11 2 2 m11,n11 | | δ X∈L0 where d0 = vK (DiscK R); equivalently,

s¯ d0 0 2e m n + + +v(N(γγ⊙)) W ⊙ (δ)= H q − 11− 11 2 2 . m11,n11 | | δ X∈L0 1 Proof. By Lemma 18.4, the support of Wm11,n11 consists of the classes [δ] = [β] in H of elements β of the box n m × + B = x + yπ c θ + zπ 11 θ , x × , y,z . OK m11,nc { 1 2 ∈ OK ∈ OK }

Our method is to show that Wm11,n11 (δ) can be interpreted as the volume of β of class [δ] in the box, up to a scalar. Then since each β belongs to just one square-class, the sum is known.

Since m11,nc > 0, β must be a unit, explaining why [δ] 0. For ﬁxed δ R×, as ξ′ ranges over the solution set of its transformed conditions, β ranges over the∈ elements L of its box∈ Oof class [δ], up to scaling. The correspondence is given by a relation of the form

ξ 2 β = δ ′ , (168) ξ 0′ where 1 ξ′ = δω 0 γ C is an element of constructed from the γ of Lemmap 18.5 whose valuations represent a lower bound on the OR valuations of ξ′. Now we compare the projective volumes of β and ξ′ satisfying the 11 and 11 conditions. In the sequence M N 2 2 ξ′ ξ′ ξ′ ξ′ δ = β, 7−→ ξ 7−→ ξ 7−→ ξ 0′ 0′ 0′ each member is a primitive vector in , so we can speak of projective volumes. OR Dividing by ξ0′ is a one-to-one operation that scales both aﬃne and projective volumes by

v (N (ξ′ )) s/¯ 2 v(N(γ)) q K R/K 0 = q − .

s/¯ 2 v(N(γ)) So there is a volume q − Wm11,n11 (δ) of ξ/ξ0′ . 2e Squaring, on units, multiplies small projective volumes by q− (since it takes the i-pixel about 1 to the (i + e)-pixel for i>e). But it is H0 -to-one since there are H0 -many square roots of 1 in R, up to scaling | | | | by 1. Since the resolvent conditions are invariant under multiplying ξ′ by a square root of 1, the volume ± of (ξ/ξ0′ ) is q 2e+¯s/2 v(N(γ)) − − W (δ). H0 m11,n11 | | Lastly, δ is a unit, so multiplying by it does not change volumes. Hence

q 2e+¯s/2 v(N(γ)) − − W (δ) H0 m11,n11 δ | | X∈L0 is the volume of the box P( × + B ). Thus it suﬃces to prove that OK m11,nc m n +¯s+ d0 µ(P( × + B )) = q− 11− 11 2 . OK m11,nc

156 Converting to aﬃne volumes, with µ(R)=1, 1 µ(P( × + Bm ,n )) = µ( × + Bm ,n ) OK 11 c 1 1 OK 11 c − q = µ( + B ) OK m11,nc = qi where i is the integer such that

Λ3( + B )= πiΛ3 OK m11,nc OR 3 n m i 3 Λ ( ¯ 1, π c θ , π θ )= π Λ OK h 1 11 2i OR d n + m = i + 0 . c 11 2 So d0 0 2e s/¯ 2+v(N(γ)) m n + 0 2e m n +¯s/2+d′ +v(N(γ)) W (δ)= H q − q− 11− c 2 = H q − 11− 11 0 m11,nc | | · · | | · δ X∈L0 as desired. We tabulate: spl.t. h ξ′ v(N(γγ⊙)) 0 ∼ ur 0 (πs/2;1;1), s even 0 ur 1 (π(s 1)/2;1;1), s odd 1/2 − (169) 13 1 1 0 13 1 π2 2 − R −

As shown, we obtain as a by-product that s determines h in unramiﬁed splitting types.

We now come to our main lemma, which computes Wm11,n11 for large m11.

21.2.1 Unramiﬁed Lemma 21.4. Suppose R is unramiﬁed over K Q . Let m and n be integers, m > 2e n > 0. Let ⊇ 2 11 c 11 ≥ c C = min 2ℓ(ˆωC)+1,e be the squareness of the δ = 1 conic if s is even; let C = 0 if s is odd. If s is even, let { } 2e s n +2 e s n11 n˜ = − − c = − 2 − 2 . 4 2 & '

Then Wm11,n11 is given as follows:

(a) If nc > 2e (black zone), then

0 2e m n s W = H q − 11− c− 2 F (e, , ) m11,n11 | | ⌊ ⌋ ∅ ∅ s 2e m11 nc = q − − −⌊ 2 ⌋[F (e, 0, 0)+ F x(e, 0, 0)+ F xx(e, 0, 0)]

(b) If 2e s

2e s nc m + − − nc nc W = q− 11 4 F ,e n˜ , n˜ m11,n11 ⌊ ⌋ 2 − − 2 j k j k 157 (d) If

m +ℓ nc s nc s W = q− 11 (1 + ε )G× ℓ, + ,e ℓ + m11,n11 C 2 2 − 2 − 2 − nc 2 ℓ

n m 1+ B = 1+ yπ c θ + zπ 11 θ , x × , y,z . m11,nc { 1 2 ∈ OK ∈ OK }

Since m11 > 2e, the z term has no eﬀect on [β], and we ignore it. In particular, the conic

2 2 coefθ2 (βξ )=0, that is, tr(ˆωC βξ )=0 has a solution ξ =1, so εC (β)= ε(ˆωC β)=1 for all β in the box.

Black zone. In the black zone, we have β 1 mod4π, so [β]=1. Hence only δ = 1 yields a nonzero volume, which is, by Lemma 21.3, ≡

d0 s 0 2e m n + (s/2 v(N(γ))) 0 2e m11 nc H q − 11− c 2 − − = H q − − − 2 . | | | | ⌊ ⌋

For the remaining zones, let Wm′ 11,n11 denote the claimed value of Wm11,n11 in each case. Our proof method will consist of two steps:

• We prove that Wm ,n (δ) W ′ (δ) for every δ (the bounding step). 11 11 ≤ m11,n11 • We check that 0 2e m n W ′ (δ)= H q − 11− c = W (δ) m11,n11 | | m11,n11 δ H1 δ H1 X∈ X∈ (the summing step), implying that equality must hold for every δ.

Purple zone. In the purple zone, s> 0 deﬁnes a splitting R = K Q. If we translate θ so that θQ 0 × 1 1 ≡ mod πs, then we get βQ 1 mod4π. Also, β 1 mod πnc , and indeed, β(K) can achieve any value ≡ ≡ 1 mod πnc , each congruence class modulo 4π achieved equally often. So [δ] = [β] ranges uniformly over ≡ n n T c ,e c , , and for each class [δ] that is attained, 2 − 2 ∅

s s 0 2e m11 nc 0 2e m11 nc H q − − − 2 H q − − − 2 nc s ⌊ ⌋ ⌊ ⌋ e m11 Wm ,n (δ)= | | = | | =2q − − 2 − 2 , 11 11 nc nc H0 e nc ⌈ ⌉ ⌊ ⌋ T ,e , | | q − 2 | 2 − 2 ∅ | 2 ⌊ ⌋ as claimed.

158 Blue zone. Note that s < 2e and that s is even (as the bounds imply C > 0). Our strategy is to note that β 1+ B 1+ B ∈ m11,nc ⊆ m′,nc for some m′ m11 for which [1 + Bm′,nc ] is a boxgroup. Here, we ﬁnd that the gray-blue condition (152) in Lemma 20.4≤ is the most stringent one, so we take nc s m′ = + e + 2 2 and get l m n s β 1+ B c + e + ,n ∈ θ1 2 2 c h l m i n nc + e + s n nc + e + s = T c , 2 2 c ,e 2 2 2 $ 2 % − 2 − $ 2 %! j k j k n n = T c ,e n˜ c , n˜ . 2 − − 2 For each such δ, the value ofj k j k

Wm11,n(δ) is controlled by the conic via Lemma 19.9, once we know the level e ℓ = min ℓ(δωˆ ), . C ♦♥ 2 We claim that all these conics are blue in the sensen of Lemmaj 19ko.9; this requires

? n11 nc + s 2ℓ +1 e n′ = e = e , ≥ − − 2 − 2 l m that is, ? n +s n +s+1 2e nc s 1 δωˆC e c = e c = − − − . ♦♥ ∈ L − 2 L − 2 L 4 ⌈ 2 ⌉ ⌊ 2 ⌋ ⌈ ⌉ & ' & ' When δ =1, the required relation

2e nc s 1 ωˆC − − − ♦♥∈L⌈ 4 ⌉ follows from the given inequality n > 2e s 2 . So it suffices to show that c − − C 2e nc s 1 β − − − . ∈ L⌈ 4 ⌉ Since β 1 mod πnc , it suffices to show that ≡ n ? 2e n s 1 c − c − − . 2 ≥ 4 j k 2e s But the given red-blue inequality n − gives c ≥ 3 n 1 2e n s 1 c − − c − − , 2 ≥ 4 from which the desired inequality follows by taking ceilings. So all conics are blue, and for every δ, 2e s nc m11+ − 4− Wθ1,m11,n11 (δ)= q− ⌊ ⌋ if nonzero. The summing step is now straightforward: 2e s nc m11+ − − nc nc W ′ (δ)= q− 4 T ,e n˜ , n˜ m11,n11 ⌊ ⌋ 2 − − 2 δ H1 X∈ j k j k 2e s nc nc 0 m + − − +e n˜ +2ñ = H q− 11 4 − − 2 | | · ⌊ ⌋ ⌊ ⌋ nc 2e s nc 2e s nc+2 0 m +e + − − + − − = H q− 11 − 2 4 4 | | · ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ nc 2e s nc 0 m +e + − − = H q− 11 − 2 2 | | · ⌊ ⌋ ⌊ ⌋ 0 m +e nc +e s n + nc = H q− 11 − 2 − 2 − c 2 | | · ⌊ ⌋ ⌊ ⌋ 0 2e m n s = H q − 11− c− 2 . | | ·

159 nc nc This completes the proof, and in particular shows that ε = 1 identically on ωˆC T 2 ,e n˜ 2 , n˜ . This result will be important in proving the remaining zones. · − − Green zone. Again, we write β 1+ B 1+ B ∈ m11,nc ⊆ m′,nc where m′ is as large as possible to make a boxgroup. This time, we ﬁnd that the gray-green inequality (151) nc is the most stringent of the conditions in Lemma 20.4, so we take m′ =2 2 + C + s (noting that m′ is an odd integer) and ﬁnd that the support of W is contained in m11,n11 n n s s n 1+ B 2 c + + s,n = T c ,ℓ + + 1 ,e ℓ c . θ1 2 C c 2 C 2 2∤nc − C − 2 − 2 h l m i j k l m We claim all conics are green of the same squareness . The zone boundaries easily imply ℓ < e/2 , so C C ⌊ ⌋ [ˆωC ] ℓ ℓ +1. ♦♥ ∈ L⌊ C ⌋ \ L⌊ C ⌋ nc We then note that [β] ℓ +1, because β 1 mod π and we have the inequality nc > C. So [βωˆC ] ∈ L⌊ C ⌋ ≡ ♦♥ is also of exact level ℓC . Thus all conics are green of the same squareness, and by Lemmas 19.9 and 19.10, we have the bound m + ℓ nc s s nc W 2q− 11 ⌈ C ⌉T ,ℓ + + 1 ,e ℓ , m11,nc ≤ 2 C 2 2∤nc − C − 2 − 2 indeed j k l m m + ℓ nc s s nc W q− 11 ⌈ C ⌉(1 + ε )T ,ℓ + + 1 ,e ℓ . m11,nc ≤ C 2 C 2 2∤nc − C − 2 − 2 j k l m where εC (δ)= ε(ˆωCδ). This completes the bounding step. To perform the summing step, we need to compute the sum

m11+ ℓC q− ⌈ ⌉ (1 + εC (δ)). nc s 1 s nc δ T ( ,ℓC + + 2∤n ,e ℓC ) ∈ ⌊ 2 ⌋ 2 Xc − − 2 −⌈ 2 ⌉ s 0 2e m11 nc The term 1 is found to sum to the desired total H q − − − 2 . We claim that | | · ⌊ ⌋ εC(δ)=0, nc s 1 s nc δ T ( ,ℓC + + 2∤n ,e ℓC ) ∈ ⌊ 2 ⌋ 2 Xc − − 2 −⌈ 2 ⌉ in other words that ε is equidistributed between 1 and 1 in this boxgroup. This follows from Lemma − 20.13(b): because ℓ +1 =ω ˆC ℓ +1 is an uncharmed coset, we have ε equidistributed on cosets of L⌊ C ⌋ 6 L⌊ C ⌋ e ℓ 1. L −⌊ C ⌋−

1 Red zone. We first recenter. Changing θ1 to the element a + bψC− θ1 from Lemma 20.11, keeping the rest of the resolvent data fixed, gives us a new first vector problem P′ whose associated cubic ring C′ has first 2 extender vector θ1′ = η is a square. By Lemma 21.2, the ring-count function WP simply shifts by ψC . Note that all boxgroups T (ℓ0,ℓ1,ℓ2) in claimed totals satisfy the gray-green inequality s + +1 ℓ C 1 ≤ 2 s+ Since θ′ θ mod π C , the replacement does not change any of the boxgroups, and their charmed cosets 1 ≡ 1 merely translate by [ψ] along with the quadratic form εC′ (δ)= εC (ψδ). So it suffices to prove the result in the case that θ1 is a square. Note that [ˆωC ] e/2 since ωC θ1 mod 2. ∈ L⌊ ⌋ ≡ We will actually prove something stronger:

2e+s nc m +ℓ nc s nc s m11+ − W = q− 11 (1+ε )F × ℓ, + ,e ℓ +q− 4 F (˜n,e 2˜n, n˜), m11,n11 C 2 2 − 2 − 2 − ⌊ ⌋ − nc ℓ

160 • εC is identically 1 on T , and hence • T is maximal isotropic for the Hilbert pairing, and also

• the identity coset T is charmed, so FT = GT , and

• the quadratic form εC is positively charmed.

All the remaining terms use FV where V T , so F is interchangeable with G there too. We now prove (170). ⊇ Begin with an arbitrary nc β =1+ π bθ1.

n e s nc +1 n Assume ﬁrst that π c b is not a square modulo π − 2 −⌈ 2 ⌉ , and let k be the largest integer such that π c b is a square modulo π2k+1. Note that nc k< n˜. We may write 2 ≤

nc 2 nc 2k+1 π b = π⌈ 2 ⌉a + π c, π ∤ c. nc Let ζ =1+ π⌈ 2 ⌉aη. We claim that β n 1+ B 2 c + s +2k +1, 2k +1 , ζ2 ∈ θ1 2 l m implying that [β] T k, nc + s ,e nc s k (compare the kth term of the sum). Write ∈ 2 2 − 2 − 2 − 2 nc 2k+1 β 1+ π⌈ 2 ⌉a θ1 + π cθ1 = ζ2 ζ2 2 nc 2 nc +2k+1 1+ π 2 a θ 1+ π2k+1cθ π 2 a2cθ2 ⌈ ⌉ 1 1 − ⌈ ⌉ 1 = ζ2 2 nc nc 2 nc +2k+1 2 1+ π 2 aη 2π 2 aη 1+ π2k+1cθ π 2 a2cθ ⌈ ⌉ − ⌈ ⌉ 1 − ⌈ ⌉ 1 = ζ2 nc nc 2 2 2k+1 2 2 +2k+1 2 2 ζ 2π⌈ ⌉aη 1+ π cθ1 π ⌈ ⌉ a cθ1 = − − . ζ2 We ﬁrst claim that the denominator ζ2 belongs to [θ ]= B (s, 0). Since OK 1 θ1 2 2 nc 2 nc 2 nc ζ = 1+ π⌈ 2 ⌉aη =1+ π ⌈ 2 ⌉a θ1 +2π⌈ 2 ⌉aη, only the last term is in question, and since vK (coefθ2 η) s/2 (as we saw in Lemma 20.3), the inequality needed is ≥ n s e + c + s, 2 2 ≥ a consequence of s< 2e. l m Therefore the last term of β/ζ2 is

nc 2 2 +2k+1 2 2 π a cθ1 2 nc +2k+1 − ⌈ ⌉ π 2 [θ ] ζ2 ∈ ⌈ ⌉ OK 1 n = B 2 c + s +2k +1,n +2k +1 θ1 2 c ln m B 2 c + s +2k +1, 2k +1 . ⊆ θ1 2 l m 

161 Thus it is enough to prove that

2 nc 2k+1 ζ 2π 2 aη 1+ π cθ1 − ⌈ ⌉ nc 1+ Bθ1 (nc + s +2k +1, 2k + 1) 1+ Bθ1 2 + s +2k +1, 2k +1 . ζ2 ∈ ⊆ 2 l m 2k+1 Since the right-hand side is a group and contains 1+ π cθ1, it is enough to show that

nc ζ2 2π 2 aη − ⌈ ⌉ 1+ B (n + s +2k +1, 2k + 1), ζ2 ∈ θ1 c that is, nc 2π 2 aη ⌈ ⌉ B (n + s +2k +1, 2k + 1). ζ2 ∈ θ1 c 2 But, since η belongs to the ring Bθ1 (s/2, 0) and ζ is a unit in that ring,

nc 2 2π aη e+ nc s ⌈ ⌉ π 2 B , 0 ζ2 ∈ ⌈ ⌉ θ1 2 n s n = B e + c + ,e + c θ1 2 2 2 B (n +ls +2mk +1, 2k +l 1)m, ⊆ θ1 c where the last step uses k< n˜. This establishes the claim that n s n s [β] T k, c + ,e c k . ∈ 2 2 − 2 − 2 − l m l m To replace the T by T ×, note that β 1+ π2k+1cθ mod π2k+2, ζ2 ≡ 1 a generic unit of exact level k. Hence

nc s nc s β T × k, + ,e k . ∈ 2 2 − 2 − 2 − l m l m Also ℓ(βωˆC )= k, so at δ = β there is a green conic of squareness k and

m11+k Wm11,n11 (β)=2q− .

s nc nc 2 nc e +1 nc 2ñ If it so happens that bπ is a square modulo π − 2 −⌈ 2 ⌉ , then writing π b = π⌈ 2 ⌉a + π c and carrying out the above computations, mutatis mutandis, shows that [β] T (ñ, e 2ñ, n˜) ∈ − and ℓ(βωˆ ) n,˜ C ≥ so there is a blue conic at β and 2e s nc m11+ − 4− Wm11,n11 = q− ⌊ ⌋. Overall,

2e+s nc m +ℓ nc s nc s m11+ − W q− 11 (1 + ε )F × ℓ, + ,e ℓ + q− 4 F (˜n,e 2˜n, n˜) m11,n11 ≤ C 2 2 − 2 − 2 − ⌊ ⌋ − nc 2 ℓ

162 1 Further remarks on the red zone. For general C, we end up proving that the coset ψC− T is charmed for each T = T (ℓ0,ℓ1,ℓ2) appearing (either positively or negatively) in the sum. Now ψC 1+ Bθ1 ( , C ), which, if ∈ ∞ ℓ ℓ , (171) 0 ≤ C is contained in the box deﬁning T . Thus the identity coset is charmed and we can replace G by F .

21.2.2 Splitting type 13

3 Lemma 21.5. Suppose R has splitting type 1 . Let m11 and n11 be rational numbers with h h m Z , n Z + , m > 2e n > 0. 11 ∈ − 3 11 ∈ 3 11 ≥ 11 Write 2h 2h m△ = m Z, n△ = n + Z. 11 − 3 ∈ 11 3 ∈ Let = min 2 ℓ(ˆω )/2 +1,e be the squareness of the δ =1 conic. Also let C { ⌊ C ⌋ }

2e n△ +2+2h n˜ = − . 4 

Then Wm11,n11 is given as follows:

(a) If n11 > 2e (black zone), then

2e m11 n11+1 Wm11,n11 = q − − F (e, 0, 0) .

4h 2e (b) If 2e 2 +2+ n 2e and n (blue zone), then − C 3 ≤ 11 ≤ 11 ≥ 3

2e n△+2h m△+1 h+ − 4 n△ n△ Wm ,n = q− − Fh ,e n,˜ n˜ 11 11 j k 2 − 2 − 2h 4h (c) If +1 n 2e 2 + (green zone), then C − 3 ≤ 11 ≤ − C 3

m△+ℓC +1 h n△ n△ Wm ,n = q− − (1 + εC)Fh ,ℓC h + 1 ,e ℓC + h 11 11 2 − 2∤n△ − 2 − 

C 1 where ℓC = − Z 0. 2 ∈ ≥ 2e 2h (d) If n < and n (red zone), then 11 3 11 ≤ C − 3

m△+1 h+ℓ n△ n△ W = q− − (1 + ε )G× ℓ, h,e ℓ + h + m11,n11 C h 2 − − 2 − n△ ℓ

1 Proof. By Lemma 18.4, the support of Wm11,n11 consists of the classes [δ] = [β] in H of elements β of the box n m 1+ B = 1+ yπ 11 θ + zπ 11 θ , x × , y,z . m11,n11 { 1 2 ∈ OK ∈ OK } Since m11 > 2e, the z term has no eﬀect on [β], and we ignore it. In particular, the conic

2 2 coefθ2 (βξ )=0, that is, tr(ˆωC βξ )=0

163 has a solution ξ =1, so ε(ˆωC β)=1 for all β in the box. In this case, Lemma 21.3 yields

2e m11 n11+1 Wm11,n11 (δ)= q − − , δ X∈L0 in particular verifying the total in the black zone.

Again, let Wm′ 11,n11 denote the claimed value of Wm11,n11 in each case. Our proof method will consist of bounding and summing, as in the preceding splitting types.

Blue zone. Our strategy is to note that

β 1+ B 1+ B ∈ m11,n11 ⊆ m′,n11 for some m′ m11 for which [1 + Bm′,n11 ] is a boxgroup. Here, we ﬁnd that the gray-blue condition (161) in Lemma 20.9≤ is the most stringent one, so we take

n△ +1 2h m′△ = + e h and m′ = m′△ + 2 − 3 We easily verify that m e △ =n, ˜ − 2 so [1 + Bθ1 (m′,n11)] is exactly the support of Wm′ 11,n11 . We claim all conics are blue. This requires that the squareness ℓ(βωˆ ) = min 2 C +1,e 2 satisfy ? n△ e n′ = e + h, ≥ − − 2 which simpliﬁes to

βωˆC 2e n△+2h . ∈ L − 4 j k When β = 1, we have ℓ = ℓC, and the required relation follows from the given blue-green inequality 4h n 2e 2 +1+ . So it suﬃces to show that 11 ≥ − C 3

β 2e n△+2h . ∈ L − 4 j k Using the known relation ℓ(β) 2 n△/2 and the blue-red inequality n11 2e/3, this is not hard to prove. So all conics are blue, and the≥ only possible nonzero value of W (δ) is≥ m11,n11

e n11′ m⊙ + − 2e n△+2h − 11 2 m△+ − 4 q = q− . j k This completes the bounding step. The summing step is routine. This completes the proof, and in particular shows that ε = 1 identically on boxgroups of the shape in the lemma. This result will be important in proving the remaining zones.

Green zone. Again, we write β 1+ B 1+ B ∈ m11,n11 ⊆ m′,n11 where m′ is as large as possible to make a boxgroup. This time, we ﬁnd that the gray-green inequality (160) 2h is the most stringent of the conditions in Lemma 20.9, so we take m′ = n + , that is, 11 C − 3 2h m′△ = m = n△ + 2h Z. 11 − 3 C − ∈

164 and ﬁnd that the support of Wm11,n11 is contained in

n△ n△ [1 + Bθ (m′,n11)] = Th ,ℓC h + 1 ,e ℓC + h 1 2 − 2∤n△ − 2 − We claim all conics are green of the same squareness . It is easy to prove that

We then note that [β] 2ℓC +2, because [β] 2 n /2 and we have the inequality n△ 2ℓC +2. So ∈ L ∈ L ⌊ △ ⌋ ≥ [βωˆC ] is also of exact level 2ℓC or 2ℓC +1. Thus all conics are green of the same squareness, and by Lemmas♦♥ 19.9 and 19.10, we have the bound

m△+1 h+ℓC n△ n△ Wm ,n 2q− − Fh ,ℓC h + 1 ,e ℓC + h , 11 11 ≤ 2 − 2∤n△ − 2 − indeed

m△+1 h+ℓC n△ n△ Wm ,n q− − (1 + εC )Fh ,ℓC h + 1 ,e ℓC + h . 11 11 ≤ 2 − 2∤n△ − 2 − This completes the bounding step. m△+ℓC To perform the summing step, we need to compute the sum of q− (1+εC) over the stated boxgroup. 0 2e m n +h The term 1 is found to sum to the desired total H q − 11− 11 . We claim that | | ·

εC(δ)=0,

n 1 h n 3h+1 δ T 2 △ + − ,2ℓC 1 h+1 ,2e 2ℓC 2 △ + ∈ 2 2 − − X2∤n△ − − 2 2 j k l m in other words that ε is equidistributed between 1 and 1 in this boxgroup. This follows from Lemma 20.13(b): because =ω ˆ is an uncharmed− coset, we have ε equidistributed on cosets of L2ℓC +2 6 C L2ℓC +2 2e 2 ℓ 2, of which the boxgroup in question is a union by the green-blue inequality. L − ⌊ C ⌋− Red zone. Considerations of space prevent us from writing out the proof, which is like that in the unram- iﬁed splitting types with the following changes:

2 • We reduce to the case that θ1 = η is a square using Lemma 20.12, and there we will prove the result with the G’s replaced by F ’s. We begin with an arbitrary

n n△ 2h/3 β =1+ π 11 bθ =1+ π b π− θ , b . 1 · 1 ∈ OK

e n△ +1+h • We assume ﬁrst that πn△ b is not a square modulo π − 2 , and let k be the largest integer such l m that πn△ b is a square modulo π2k+1. We ﬁnd that

n△ n△ [β] T × ℓ, h,e ℓ + h ∈ 2 − − 2 − is in the support of the kth term of the claimed answer and that there is a green conic of level k there.

n11 n△ e +1+h • If π b in fact is a square modulo π −⌈ 2 ⌉ , then we ﬁnd that [β] is in the support of the last term of the claimed answer and that there is a blue conic there. The rest of the proof, including the summing step, is completely like the unramiﬁed splitting types.

The G’s can be replaced by F ’s when the index k is at most ℓC, for reasons just like those named above. Another corollary is the following.

165 21.2.3 The level parity lemma

Note that in ramiﬁed splitting type, conics of given squareness k< e/2 occur for [δ⊙] of exactly two levels: 2k and 2k +1. The following lemma tells when each occurs, at least⌊ when⌋ δ =1.

Lemma 21.6. Take δ =1. If [δ⊙] / , then either ∈ Le • δ⊙ has even level and b Z +1/3, or 1 ∈ • δ⊙ has odd level and b Z 1/3. 1 ∈ − Proof. Consider the ﬁrst vector problem P with the given θ1, with m11 > 2e and with 11 as minimally active as can be: N 1/3, b1 Z +1/3 n11 = ∈ 2/3, b Z 1/3. ( 1 ∈ − This lies in the red zone, and we get an answer of the form (using ellipses to mark unimportant portions) e e W = ( )G× (ℓ, . . .) + ( )G , 1 , . m11,n11 ··· h ··· h 2 − 2∤e 2 0 ℓ< e/2 ≤ X⌈ ⌉ l m l m

Since ℓ(δ⊙) < e, the black-red comparison shows that δ = 1 lies in the support of the kth summand, k = ℓC = ℓ(δ⊙)/2 . Now simply note that the kth summand consists entirely of elements of exact level 2k (for h =1⌊) or 2k +1⌋ (for h = 1). −

21.3 The zones when 11 is weakly active (brown and yellow) N We now turn our attention to first vector problems such that N11 is weakly active and m11 > 2e. Here we use a significantly different framework. Note that s¯ > 0 induces a distinguished splitting R = K Q. The resolvent conditions and simplify to × M11 N11 : tr(ξ2) 0 mod πm11 M11 1 ≡ (K) : ξ 0 mod πn11/2 N11 1 ≡

For ξ1 to satisfy this, its two Q-components must be units. Under the transformation of Lemma 19.3, the conditions can also be written as

2 m⊙ : tr(δ⊙ξ⊙ ) 0 mod π 11 M11 1 ≡ n11 (K) (K) v (γγ⊙) : ξ⊙ 0 mod π 2 − N11 1 ≡ n11 (K) (K) v (γγ⊙) ξ⊙ 0 mod π 2 − . ⇐⇒ 1 ≡ ⌈ ⌉ For simplicity we let n11 (K) n⊙ = v (γγ⊙) , 11 2 − l m which sometimes diﬀers slightly from the n⊙ of Lemma 19.18. (K) In each of the three applicable splitting types, we will ﬁnd a brown zone where n11 is so high that ξ1⊙ (Q) can be taken to be 0, so the only δ⊙ for which there is a solution are those where δ is the class of an element on the traceless line of Q. Let be such an element of the form ♣ = (α +α ¯; α α¯;¯α α) ♣ − − where α is a generator for as an -module. Remarks are in order: OQ OK • In splitting type (111), we can take = (1; 1; 1). ♣ − • In splitting type (12), we can take α = (1+ √D )/2, giving = (1; ζ¯ √D ). 0 ♣ 2 0

166 Denote by Y (δ♣) the volume of ξ′ satisfying the and conditions when P 1 M11 N11

[δ♣]=[ˆω δ], C♣ that is, Y (δ♣)= W ωˆ δ♣ . P P C♣ Observe that the dependence on θ1 has been nulliﬁed and, if m11⊙ > 2e, that YP is supported on the vanishing locus of the quadratic form ε♣(δ♣) := ε( δ♣). ♣ Let YP⊙ be the corresponding volume of ξ1⊙. As in the strong zones, we need a summation lemma.

Lemma 21.7. We have (Q) 0 2e m⊙ n⊙ +2vK (δ⊙ ) Y (δ♣)= H q − 11− 11 , P | | δ δ♣ ♣∈X0 L0 1 where m⊙ , γ, and δ⊙ are given by Lemma 19.3 and depend only on the class δ♣ H / . 11 L0 ∈ L0 (Q) (Q) Proof. Since ξ1 must be a unit, ξ1⊙ has ﬁxed valuation

(Q)) (Q) v ξ⊙ = v γγ⊙ =: r. 1 − (K) (i) Meanwhile, v ξ⊙ n⊙ can vary. For i n⊙ , let Y (δ♣) denote the volume of ξ⊙ satisfying the ﬁrst 1 ≥ 11 ≥ 11 P 1 (K) vector problem P and having vK (ξ⊙ )= i. We have

∞ (i) YP = YP , i=i X0 (K) since only the measure-zero set where ξ⊙ =0 has been dropped. Let

2 2r β = δ⊙ξ⊙ π (172) ·

Observe that 2r is an integer (by reference to (134)) so β R. Moreover, as we vary [δ♣] δ♣ 0 and ξ1⊙ in (i) ∈ ∈ L the set whose volume is Ym ,n , we get that β varies in the region of primitive members of such that 11 11 OR (K) (K) v (β )= v δ⊙ +2i 2r K K − Q vK (β )=0

m⊙ +2r λ♦(β) 0 mod π 11 . ≡ (K) n m m n Conditions of the form β 0 mod π 11 and λ♦(β) 0 mod π 11 cut out a box of volume q− 11− 11 ; so the volume of β is ≡ ≡

1 (K) (vK (δ⊙ )+2i+2r) (m⊙ 2r) 1 q− − 11− − q 1 (K) 2i m⊙ vK (δ⊙ )+4r = 1 q− − 11− . − q Consider the sequence 2 2 ξ′ ξ′ δ⊙ξ′ ξ′ = β. 7−→ i 2r 7−→ 2i 4r 7−→ π2r π ; πQ π ; πQ

167 Every term is a primitive vector in R, so we can consider projective volumes. The linear map of dividing by πi; π2r scales volumes by qi+2r. SquaringO by units scales volumes by 1/ H0 q2e on regions symmetric Q | | · under multiplication by R [2], as we noted above in the proof of Lemma 21.3. Finally, multiplication by × 2i 4r δ⊙ π ; π · Q 2r π scales volumes by (K) (Q) vK (δ⊙ ) 2vK (δ⊙ ) 2i+2r q− − − . (i) So overall, a volume YP (δ) of ξ′ transforms to a volume of

1 (K) (Q) 2e i vK (δ⊙ ) 2vK (δ⊙ )+4r (i) q− − − − Y (δ). H0 P | | Summing over [δ] δ , ∈ 0L0 1 (K) (Q) 1 (K) 2e i vK (δ⊙ ) 2vK (δ⊙ )+4r (i) 2i m⊙ vK (δ⊙ )+4r q− − − − Y (δ)= 1 q− − 11− , H0 P − q δ δ0 0 | | ∈XL that is, (Q) (i) 1 2e i m⊙ +2v δ⊙ Y (δ)= 1 q − − 11 K ( ). P − q δ δ0 0 ∈XL Summing over i n⊙ , the right-hand side becomes a geometric series and we get ≥ 11 (Q) 2e m⊙ n⊙ +2vK (δ⊙ ) YP(δ)= q − 11− 11 , δ δ ∈X0L0 as desired. We now use this to power the summing step in each splitting type.

21.3.1 Unramiﬁed Here = (1; ζ¯ √D ), where D is scaled so that D 1 mod4. Observe that Q is traceless. ♣ 2 0 0 0 ≡ ♣ Lemma 21.8. There is an α such that 0 ∈ OQ 2 ker tr R/ K = (0; 1), (1; α0) O O ♣ ♣ as -modules. OK Proof. This is a notable example of a lemma of simple form that can be proved using the machinery we have got. (Incidentally, if R is totally split, the choice α0 = (0; 1) works, so we’re really only concerned about splitting type (12): but we have no need to separate the splitting types here.) The conic (ξ) = tr( ξ2) A ♣ has determinant 1. Since 1 mod2, has maximal squareness e/2 . It has Brauer class ε( )=1, ♣ ≡ A ⌊ ⌋ A since ξ = (0; 1) is a solution. Hence, by Lemma 19.9, not all the K -points of lie in a single 1-pixel. The reduction of modulo π consists of (q + 1)-many 1-pixels, onlyO one of whichA has vanishing K-coordinate. A Hence there is a solution ξ = (a; α) with π ∤ a. Rescaling, we can take a =1.

Lemma 21.9. In splitting types 111 and 12. If h1 =0, let

2e n +2 e n⊙ n˜ = − 11 = − 11 . 4 2 Then YP is given in terms of n11 = n11 and as follows:

168 (a) If n11 > 2e (brown zone), then

e m11 n11⊙ 2q − − F (0, e, )[δ0♣] 0 YP = ∅ ∈ L e m n⊙ 2q − 11− 11 xF (0, e, )[δ♣] (1; π; π) ( ∅ 0 ∈ L0 (b) If 0

2e n11 m +ℓ m11+ − q− 11 (1 + ε♣)F × ℓ,n⊙ ,e n⊙ ℓ + q− 4 F (˜n,e 2˜n, n˜), h =0 11 − 11 − ⌊ ⌋ − 1 YP =  0 ℓ

2 Q 2e+1 (δ⊙ξ⊙ ) a mod π ≡ ♣ for some a K , necessarily in K× . So [δ⊙] ι(K) = T ( , e, ). It’s easy to see that all values occur, and the∈ O solution volume is constantO within∈ the ♣ appropriate♣ ∅ coarse∅ coset, because the K-coordinate 2 h1 n (Q) (Q) of β = δ⊙ξ⊙ /π can range over all of π K× (for n 2n11⊙ h1 of the correct parity) while β mod π2e+1 remains of constant class. So O ≥ − ≡ ♣

e m11 n⊙ YP(δ♣)=2q − − 11 for each δ♣ in the support, as desired. In particular, T ( , e, ) is charmed for ε♣. In the yellow zone,∅ the∅ conditions imply that

Q 2n⊙ +h β a mod π 11 1 ≡ ♣ for some a in , necessarily in × . Hence OK OK 1 ε♣− (1) T × ,n11⊙ ,e n11⊙ h1 =1 [δ♣] 1 ∩ ∅ − ∈ ε♣− (1) T 0,n⊙ ,e n⊙ h =0. ( ∩ 11 − 11 1 Note that the boxgroups have ℓ = n⊙ s/2 , so they are well deﬁned. Also, F = G in the answer, 1 11 ≤ ⌈ ⌉ because everything contains ι(K×) , which is charmed for ε♣. · Le In the case h1 =1, the conics are all tiny and green, and we get the bound

m Y q− 11 (1 + ε♣)xF 0,n⊙ ,e n⊙ , P ≤ 11 − 11 which is exactly as desired. The summing step precedes routinely, noting that ε♣ is equidistributed because everything is contained in a non-charmed coarse coset. In the case h = 0, that is, [δ⊙] , some further analysis must be done to narrow the support. By 1 ∈ L0 Lemma 19.6, we can assume that 11 is an equality on the nose and also that the K-component of ξ′ is not M 2 2 exactly 0 (to allow recovery of [δ♣] = [δ⊙ ξ′ ]). Then tr(δ⊙ξ′ )=0, so there are u K× , b K such that ♦♣ ∈ O ∈ O 2 2n′ 2 δ♣ξ′ = u(0;1) + bπ (1; α0). 1 1 Hence we are curious about the H -class of the right-hand side. Since scaling by K× preserves H -class, we may assume that u =1. We have b =0. 6 Assume ﬁrst that b is not a square modulo πvK (b)+2˜n. Write

b = a2(1 + π2k+1c), π ∤ c, 0 k< n.˜ ≤

We will show that [δ♣] lies in the kth term

T × k,n⊙ ,e n⊙ k . 11 − 11 − 169 We compute:

2n11⊙ 2 [δ♣] = [(0; 1) + bπ (1; α0)]

2n11⊙ 2 = [(b;1+ bπ α0)]

2k+1 2 2n11⊙ 2 2 2n11⊙ +2k+1 2 = [1 + cπ ;1+ a π α0 + a cπ α0]

2k+1 2 2n11⊙ 2 [1 + cπ ;1+ a π α0] mod k+n⊙ ≡ L 11 2k+1 n11⊙ 2 [1 + cπ ;(1+ aπ α0) ] mod (e+n⊙ )/2 k+n⊙ ≡ L⌈ 11 ⌉ ⊆ L 11 = [1 + cπ2k+1; 1]

T × k,n⊙ ,e n⊙ k . ∈ 11 − 11 − Also, the conics here are green of squareness k. Likewise, if b is a square modulo πvK (b)+2ñ, the same computation shows that [δ♣] S(ñ,e 2ñ, n˜). Here the conics are blue, and we have the bounding step. ∈ − For the summing step, we note that ε is equidistributed on each support T ×(ℓ,n⊙ ,e n⊙ ℓ), since it 11 − 11 − is a union of cosets of e ℓ 1 inside a non-charmed coset of ℓ+1. The sum is then easy to compute and compare against the totalL − of− Lemma 21.7. L When we translate back to W ’s counting the volumes by the value of δ, the answers change but slightly. In the brown zone, the support is contained in ωˆ ι(K), but using s> 2e (or the black-brown comparison, C♣ when the chosen coarse coset is 0), this coset is the identity. Intersecting this with the two possible coarse cosets yields the answers L

e m n /2 2q − 11−⌈ 11 ⌉F (0, e, )[δ] , s even ∅ ∈ L0 e m11 n11/2 2q − −⌈ ⌉xF (0, e, )[δ] (1; π; π) 0, s odd WP = ∅ ∈ L  e m11 n11/2 2q − −⌊ ⌋F (0, e, )[δ] 0, s odd  ∅ ∈ L e m11 n11/2 2q − −⌊ ⌋xF (0, e, )[δ] (1; π; π) 0, s even  ∅ ∈ L   Further remarks on the yellow zone. In the “long” yellow zone answer (h1 =0), the F ’s can be changed to G’s, because the answer implies that T (ñ,e 2ñ, n˜) is charmed. When translating from Y back to W , − we keep these G’s, switching quadratic forms from ε♣ to εC. However, some of these G’s and xG’s admit simplifications that are of importance to us. • If h =1 and s is odd, the fact that δ =1 yields a nonzero ring volume implies that the xG simplifies to F : m n11 n11 W = q− 11 (1 + ε )F 0, ,e m11,n11 C 2 − 2 j k j k • Still assuming h =1 and s is odd, in the very special case that n11 =2e (on the yellow-brown border), the term εC F (0, e, 0)

admits a curious simpliﬁcation. Since the charmed coset of T (0, e, 0) is T ×( , e, 0), we have ∅ G(0, e, 0) = xF (0, e, 0). Taking Fourier transforms of both sides,

εCF (0, e, 0) = F x(0, e, 0). This accounts for the “Fx-yellow-special” zone in the code.

• If h = 1 and s is even, we claim that the xG simpliﬁes to xF . Note that ωˆC is constrained by s: we have θ (1;0;0) mod πs, so ω (0; Q) mod πs and 1 ≡ ≡ ♣ s n n11 n11 [ˆω ] ι(K×) (1 + π ) (1 + π 11 )= T , ,e . C ♣ ∈ · OR ⊆ · OR ∅ 2 − 2 j k j k So the regions in which δ♣ lay are also good for δ.

170 • If h = 0 and s is even, then δ = 1 must also lie in some term of the sum. Which term it is can be determined using the levels of the conics. If

ℓ n,˜ C ≥ then δ =1 lies in the last term and all G’s can be made F ’s. Otherwise, the terms for

ℓ ℓ ≤ C so simplify.

21.4 The beige zone 21.4.1 Unramiﬁed The following is immediate from Lemmas 19.9, 19.10, and 19.14.

Lemma 21.10. In unramiﬁed splitting type, if m > 2e but is inactive (beige zone), then the ring 11 N11 volume for ξ1′ is given by

m11 ℓ  q− (1 + εC ) q G×× (ℓ, 0,e ℓ) + (core) [δ⊙] 0 Wm11,0 = − ∈ L  0 ℓ e 1   ≤ ≤X2 −   m11 ⌊ ⌋ q− (1 + ε ) xG(0, 0,e)+ xxG(0, 0,e)  [δ⊙] / C ∈ L0  where  e 1 e e q 2 1+ G , 0, e even q 2 2 (core)=  e 1 e 1 e +1  q −2 (1 + ε )G − , 0, e odd  C 2 2  Further remarks on the beige zone. In the unramiﬁed h =1 case, the answer would more strictly be written as a restriction to the particular coarse coset speciﬁed by the discrete data, but we do not do so, as all ( H0 1)-many non-charmed coarse cosets admit the same extender indices and will be immediately summed.| We|− know which coset of is charmed, and hence: L0 xF (0, 0,e)+ xxF (0, 0,e) s even xG(0, 0,e)+ xxG(0, 0,e)= ( F (0, 0,e)+ xxF (0, 0,e) s odd.

21.4.2 Splitting type 13

Here, a little more care is required to deal with the restrictions on the valuation of ξ⊙ that remain active in the beige zone. We first use a summation lemma. (We could have proved a summation lemma in the unramified splitting type, but it was unnecessary for finding the answers.)

3 Lemma 21.11. In splitting type 1 , let m11 > 0. Then

h 1 2e+ −2 m11△ Wm11,0 = q − . δ H1 X∈ 2 Proof. Analogous to Lemma 21.3, map each ξ′ R× in the solution set to a corresponding β = δ⊙ξ′ . The ∈h O1 − m△ 11-condition restricts β to a space of volume q 2 − 11 , and the squaring multiplies projective volumes by M2e q− , establishing the result.

171 Lemma 21.12. In splitting type 13, a ﬁrst vector problem with m > 2e and n 0 (beige zone) has the 11 11 ≤ answer, for h1 =1, e/2 1 ⌈ ⌉− m△ k W = q− 11 (1 + ε ) q G (k, 0,e k) G (k +1, 1,e k) + m11,0 C 1 − − 1 − − k=0 X m△ + e/2 e e + q− 11 ⌈ ⌉G , 1 , 1 2 − 2∤e 2 and, for h = 1, l m l m 1 − e/2 1 ⌊ ⌋− m△ +2 k Wm ,0 = q− 11 (1 + εC ) q G 1(k, 1,e k 1) G 1(k +1, 0,e k 1) + 11 − − − − − − − k=0 X e e m11△ +2+ e/2 1 + q− ⌊ ⌋G 1 , 2∤e, . − 2 2 Proof. Since is charmed for ε, all ofj the kG’s canj bek viewed as selectors for the coset of the indicated Le boxgroup containing [ˆωC ]. 2 For each [δ⊙] = [δωˆ ] with ε(δ⊙)=1, the conic : λ♦(δ⊙ξ⊙ ) has some rational point ξ⊙. Let h′ C Mδ⊙ be the value of h1 to which ξ⊙ contributes to the beige-zone answer:

1 ξ⊙ 1 (here ξ⊙ lies in a generic 1-pixel) h′ = ∼ 2 1 ξ⊙ π (here ξ⊙ lies in the special 1-pixel) ( − ∼ R ¯h′ (A potential third case ξ⊙ πR does not satisfy δ⊙ even mod π.) Then let θ1′ ζ3 R whose associated traceless ω is the known ∼ M ∈ θ1′ 2 δ⊙ξ⊙ ω′ = ♦ π4+2h

When [δ⊙] = [δωˆC ] 2 e/2 , the corresponding conic δ⊙ has squareness k < e/2 . The solutions to the conic lie in a single 6∈1-pixel, L ⌊ ⌋ which must be either genericM or special. By the level⌊parity⌋ result in Lemma 21.6 applied to θ1′ , values with ℓ(δ⊙) even and odd give the generic and special pixels respectively. This establishes the claimed result when [δ⊙] / 2 e/2 , which corresponds to the terms k e/2 1 of each sum. ∈ L ⌊ ⌋ ≤ ⌊ ⌋− We now turn to the case that [δ⊙] 2 e/2 , so the conic has squareness e/2 . Refer to Lemma 19.9 for the analysis of conics. If e is even, the∈ L conic⌊ ⌋ has solutions in (q + 1)-many 1⌊-pixels.⌋ These form the line in P( R/π R) given by reducing the conic mod π; they are q generic and 1 special. Because the conic has equalO volumeO in each 1-pixel, the volume can be computed explicitly and contributes

m△ +e/2 e e q− 11 G , 0, 1 2 2 to the h = 1 case and q times as much to the h = 1 case (the special pixel gets inﬂated by q in the 1 1 − ξ⊙ ξ′ transition), as desired. 7→If e is odd, the conic has solutions in two 1-pixels and we need to know whether one of them is special. For δ⊙ of exact level e 1, we know, again by Lemma 21.6, that both pixels are generic. For δ⊙ e, we get at least one generic− pixel and at most one special pixel, leading to the inequalities ∈ L e 1 e +1 1 e 1 e 1 h1=1 m11△ +(e 1)/2 Wm ,0 q− − (1 + εC ) G 1 − , 0, G 1 − , 1, − 11 ωˆC e 1 ≥ − 2 2 − 2 − 2 2 L −

m△ +(e 1)/2 e 1 e +1 e 1 e 1 = q− 11 − (1 + εC ) G 1 − , 0, G 1 − , 1, − + − 2 2 − − 2 2 m△ +(e 1)/2 e 1 e 1 + q− 11 − G 1 − , 1, − − 2 2 and e 1 e 1 h1= 1 m11△ +2+(e 1)/2 Wm ,−0 q− − G 1 − , 1, − . 11 ωˆC e 1 ≤ − 2 2 L −

Then, summing and comparing against Lemma 21.11, we ﬁnd that equality must hold.

172 21.5 Orthogonality

Before proceeding to ﬁrst-vector problems with m11 2e, we prove the following result, which will enable us to compute Fourier transforms of ring totals. ≤

Lemma 21.13. For any boxgroup Tθ1 (ℓ0,ℓ1,ℓ2), its orthogonal complement is given by Tθ1 (ℓ2,ℓ1,ℓ0). Remark 21.14. This is an example of an explicit reciprocity law, that is, a formula for the Hilbert symbol in a certain region. There is a wide literature on explicit reciprocity laws, but we suspect that this one is new. In our proof, the only fact we use about the Hilbert pairing is that it is the associated bilinear form to ε (and εC ). This enables us to use various facts about ε gleaned in the preceding sections. We will be concocting various values of the resolvent datum θ1 and of the discrete datum n11 to plug into the lemmas regarding the ring volumes.

Proof of Lemma 21.13. We carry out the proof in the unramified splitting type only, the proof in the other types being very similar. We first note that if any of ℓ ,ℓ ,ℓ is the symbol , the result follows easily from the self-orthogonality 0 1 2 ∅ of ι(K×) (if applicable), as mentioned above. So we can assume that the ℓi are integers. By definition, they must satisfy

ℓ0 + ℓ1 + ℓ2 = e (173) s ℓ ℓ + + 1 (the gray-red inequality) (174) 1 ≤ 0 2 s + +1 ℓ C (the gray-green inequality) (175) 1 ≤ 2 s ℓ ℓ + +1, (the gray-blue inequality) (176) 1 ≤ 2 2

We ﬁrst reduce to the case C = e. If s is even, this is accomplished, as in the proof of the red zone, by replacing θ by θ = η2 and noting that, by the gray-green inequality, the boxgroups T ( )= T ( ) 1 1′ θ1 θ1′ max 2e+1,s ··· ··· are unchanged. If s is odd, we simply replace θ1 by a θ1′ (1;0;0) mod π { } whose corresponding 2s ¯ ≡ s ωˆC′ is in e/2 . (For example, θ1′ = (1;8π ζ2√D0)) is found to work.) Then since θ1′ θ1 mod π and all L⌊ ⌋ ≡ boxgroups satisfy the gray-green inequality ℓ1 (s + 1)/2, the boxgroups are unchanged. Incidentally, we can also assume as a result of this reduction that≤ s is even. Now the gray-green inequality is subsumed by the gray-blue and gray-red ones. Using i⊥ = e i, the truth of the lemma for a triple (ℓ ,ℓ ,ℓ ), ℓ 1, implies its truth for the triples (ℓ +1L,ℓ 1L,ℓ−) and 0 1 2 1 ≥ 0 1 − 2 (ℓ0,ℓ1 1,ℓ2 + 1). Hence we can run these reductions backward, increasing ℓ1 until we reach an obstruction. This usually− happens if either the gray-blue or the gray-red inequality becomes an equality, but it can also happen in two special cases, which we dispatch now:

• ℓ = ℓ = 0. Here s 2e 2, and the self-orthogonality of T (0, e, 0) follows from that of T ( , e, )= 0 2 ≥ − ∅ ∅ ι(K×), unless e =1 and s =0, in which case the gray-blue and gray-red inequalities are also equalities. • Both the gray-blue and gray-red inequalities are 1 away from equality, that is,

2e s e + s 2e s (ℓ ,ℓ ,ℓ )= − , , − . 0 1 2 6 3 6 As we see, this is only possible if 2e s mod 3. This space shows up as the support in the blue zone for the ﬁrst vector problem ≡ 2e s m =2e +1, n s = − , 11 11 − 3 right on the blue-red boundary. It is therefore isotropic and, by virtue of its size, maximal isotropic. So we are left with the case that, without loss of generality, the gray-blue inequality is an equality s ℓ = ℓ + +1. 1 2 2

173 We thus have s s (ℓ ,ℓ ,ℓ )= e 2ℓ 1,ℓ + +1,ℓ . 0 1 2 − 2 − 2 − 2 2 2 Let V = T (ℓ0,ℓ1,ℓ2) and W = T (ℓ2,ℓ1,ℓ0) be the claimed orthogonals. By the gray-red inequality, ℓ0 ℓ2 and V W . ≥ ⊆ If ℓ0 = ℓ2, we again have a unique group

2e s 2 e + s +2 2e s 2 (ℓ ,ℓ ,ℓ )= − − , , − − 0 1 2 6 3 6 which shows up as the support in the blue zone for the first vector problem 2e s 2 m =2e +1, n = − − , 11 c 3 right on the blue-red boundary. It is therefore isotropic and, by virtue of its size, maximal isotropic. So we may assume that ℓ0 > ℓ2 and V ( W . The space V shows up as the blue-zone support for the first vector problem m =2e +1, n =2e 4ℓ 1. 11 c − 2 − Hence V is isotropic. Also, the first vector problem

m11 =2e +1, nc =2ℓ2 +1 lies in the red zone. The ﬁrst term of its answer is a positive multiple of

(1 + εC)G×(W ), but since C = e, the G can be replaced by F . Our strategy is as follows. Since W × = T ×(ℓ2,ℓ1,ℓ0) generates W as a group, it’s enough to show that any α W × and β V are orthogonal. We may write ∈ ∈ α, β = ε (1)ε (β)ε (α)ε (αβ) h i C C C C = εC(α)εC (αβ).

So if α and β are not orthogonal, then, applying the transformation α αβ if need be, we may assume 7→ ε (α)=1 and ε (αβ)= 1. C C −

Since εC (α)=1, α is in the support of the red-zone answer Wm11=2e+1,nc=2ℓ2+1. So there is a ψ in the box 1 1+ Bθ1 ( , 2ℓ2 + 1) representing the class [α] H . Recenter, using Lemma 21.2, and consider the P′ with 1∞ ∈ θ1′ = ψ− θ1, m11′ =2e +1, and n′ =2e 4ℓ 1. c − 2 − Since

ℓP′ = ℓ(α)= ℓ2, this problem is still in the blue zone (right on the blue-green boundary), and we get that εC′ = 1 on the boxgroup 1+ B (ℓ ,ℓ ,ℓ ). θ1′ 0 1 2

2ℓ2+1 Since θ1′ θ1 mod π K [θ1], the subscript can be changed from θ1′ to θ1 without changing the boxgroup. So ≡ O

1= εP′ (β)= εP(αβ), a contradiction. This completes the proof in the unramiﬁed splitting types.

174 21.5.1 E-forms of the long answers In the red, yellow, and beige zones when h =0, the answer, as announced in 21.4, 21.9, and 21.10, is a sum of G and εCG terms. It is capable of a simpliﬁcation.

Definition 21.15. If T (ℓ0,ℓ1,ℓ2) is defined, define

E(ℓ0,ℓ1,ℓ2)= εCF (ℓ0,ℓ1,ℓ2) Note that if the ℓ are integers, we have i db

G(ℓ0,ℓ1,ℓ2), ℓ0 ℓ2 E(ℓ0,ℓ1,ℓ2)= ℓ ℓ ≤ q 2− 0 ε G(ℓ ,ℓ ,ℓ ), ℓ ℓ . ( C 2 1 0 2 ≤ 0 Lemma 21.16. (a) A sum of the form

b/2 1 ⌈ ⌉− ℓ b/2 b b b q (1 + ε )G×(ℓ,e b,b ℓ)+ q−⌊ ⌋G× ,e 2 , , C − − 2 − 2 2 Xℓ=a where b 2a, can be rewritten as ≥ b a b a 1 − − − qℓE(ℓ,e b,b ℓ) qℓE(ℓ +1,e b 1,b ℓ). − − − − − − Xℓ=a Xℓ=a (b) In unramiﬁed splitting types, a sum of the form

ℓ (1 + ε ) q G×× (ℓ, 0,e ℓ) + (core), C − a ℓ e 1 ≤ ≤X⌊ 2 ⌋− where (core) denotes the beige-zone core (see Lemma 21.10) and a< e/2 , can be rewritten as ⌊ ⌋ e a e a 1 − ℓ − − ℓ 1 q E(ℓ, 0,e ℓ) q − E(ℓ, 0,e ℓ) − − − Xℓ=a ℓ=Xa+1 where the second sum may be empty or may have to be interpreted according to the natural convention

1 − x = x . i − 0 Xℓ=1 Proof. The proof is straightforward, converting each G and εCG into an E and merging the ranges of summation. Corollary 21.17. For unramiﬁed splitting type, we get in the red zone:

e n + s − 11 2 m +ℓ n11 n11 W = q− 11 E ℓ, ,e ℓ m11,n11 2 − 2 − nc ℓ= 2 l m l m X⌊ ⌋ e n + s 1 − 11 2 − m +ℓ n11 n11 q− 11 E ℓ +1, 1,e ℓ , − 2 − − 2 − nc ℓ= 2 l m l m X⌊ ⌋ and in the yellow zone:

e n /2 −⌈ 11 ⌉ m +ℓ n11 n11 W = q− 11 E ℓ, ,e ℓ m11,n11 2 − 2 − Xℓ=0 l m l m e n /2 1 −⌈ 11 ⌉− m +ℓ n11 n11 q− 11 E ℓ +1, 1,e ℓ , − 2 − − 2 − Xℓ=0 l m l m

175 and in the beige zone:

e e 1 − m +ℓ m +k 1 W = q− 11 E (k, 0,e k) q− 11 − E(k, 0,e k). m11 − − − Xℓ=0 Xℓ=1 For splitting type 13, we get in the red zone:

e n△+h − m△+1 h+ℓ n△ n△ W = q− − E ℓ, h,e ℓ + h m11,n11 h 2 − − 2 − ℓ= n△ X2 j k e n△+h − m△+1 h+ℓ n△ n△ q− − E ℓ +1, h 1,e ℓ + h , − h 2 − − − 2 − ℓ= n△ X2 j k and in the beige zone (h1 = 1):

e e 1 − m△+ℓ m△+ℓ W = q− E (k, 0,e k) q− E (k +1, 1,e k) m11 1 − − 1 − − Xℓ=0 Xℓ=0 and in the beige zone (h = 1): 1 − e 1 e 2 − − m△+2+ℓ m△+2+ℓ Wm = q− E 1(k, 1,e k 1) q− E 1(k +1, 0,e k 1). 11 − − − − − − − Xℓ=0 Xℓ=0

The following are to be kept in mind when manipulating terms E(ℓ0,ℓ1,ℓ2):

• When ℓ0 ℓC and ℓ0 ℓ2, the E came from a G(ℓ0,ℓ1,ℓ2) with ℓ0 ℓC. As we observed in the “Further remarks”≤ sections≤ following the red, yellow, and beige zones, such≤ a G is interconvertible with an F . Hence such an E will be changed to F if it appears in the ﬁnal answer (after smearing and applying the ξ2 restrictions: see below).

• When ℓ2 ℓC and ℓ2 < ℓ0, the E came from an εCG(ℓ2,ℓ1,ℓ0) with ℓ2 ℓC . Such an εCG is ≤ ≤ e interconvertible with εC F and hence is its own Fourier transform, up to the inevitable factor of q . We annotate it as Ebal (“bal” for “balanced”). The same can happen to εC F x and εC F xx terms, which we accordingly notate as Ebalx and Ebalxx.

• When ℓC <ℓ0 and ℓC <ℓ2, the above transformations do not apply. We annotate the E as Eside and note that, for reﬂection to hold, either

– The Eside pairs with its Fourier transform, an εC F from the green zone, or

– The Eside cancels with a like term for a diﬀerent value of the discrete data. Indeed, we notice that increasing n11 by 2 in the red or yellow zone causes most of the positive terms to reappear with a negative sign.

21.6 Smeared answers

It is now necessary to compute Sr(W ) to solve ﬁrst vector problems with small m11. It is convenient to express as much as possible in terms of sparks that vanish suddenly as the smear index r, or equivalently m11, is decreased.

1 Deﬁnition 21.18. A function W : H C is a spark of level r if for all r, 0 r e′ +1, → 0 ≤ ≤

W r r0 Sr(W )= ≥ ( 0 r < r0.

176 Lemma 21.19. Let 0 r e′ +1. A function W is a spark of level r if and only if its Fourier transform ≤ 0 ≤ 0 W is supported on the set e r e r +1 of elements of exact level r0 (or 1/2, in the case r0 = e′ +1). L ′− 0 \L ′− 0 − Proof. Using the familiar Fourier duality between multiplication and convolution, we have the relation c 1 Sr(W )= e r W . (177) L ′− · So W is a spark of level r if and only if 0 c b 1 W ℓ e′ r0 ℓ W = ≤ − L · ( 0 ℓ>e′ r0. c − c This evidently happens exactly when W is supported on e r e r +1, as desired. L ′− 0 \L ′− 0 Drawing on the repertory of Fourier transforms we computed in Lemma 20.14, as well as the deﬁnition c of E-functions, we get the following.

Lemma 21.20. Let r(ℓ) be the minimal level of elements in a boxgroup denoted T (ℓ,ℓ1,ℓ2); to wit, • r(ℓ)= ℓ in unramiﬁed splitting type • r(ℓ)=2ℓ in splitting type 13 if h =1 • r(ℓ)=2ℓ +1 in splitting type 13 if h = 1 − The following functions are sparks of the indicated levels. (a) If T (ℓ ,ℓ ,ℓ ) is deﬁned and ℓ ,ℓ ℓ +1, then 0 1 2 0 2 ≥ C W = εC F (ℓ0,ℓ1,ℓ2)

is a spark of level e′ r(ℓ ). This will be used in the green zone. − C (b) If both terms are defined and ℓ 1, then 1 ≥ W = q E(ℓ ,ℓ ,ℓ ) E(ℓ ,ℓ 1,ℓ + 1) · 0 1 2 − 0 1 − 2 is a spark of level r(e ℓ2). This will be used in the red and yellow zones, as well as the beige zone in ramified splitting types.− (c) In unramified splitting type, if both terms are defined, then W = q2 E(ℓ , 0,ℓ ) E(ℓ +1, 0,ℓ 1) · 0 2 − 0 2 − is a spark of level r(ℓ0). This will be used in the beige zone. (d) Expressions of the form W = Gx(), xGx(), and Gxx()

are sparks of level e′ +1.

Although a single F is not generally a spark, we do have the relation e r(ℓ ) T (ℓ0,ℓ1,ℓ2), from which L ′− 2 ⊆ F (ℓ ,ℓ ,ℓ ) and G(ℓ ,ℓ ,ℓ ) are stable under smears of levels r e′ r(ℓ ). 0 1 2 0 1 2 ≥ − 2 We are now ready to compute explicit answers for the smear. Note that we do not try to write Wm11,n11 for each value of m11 and n11. Instead, we express Wm11,n11 for m11 large as a sum of sparks and stable terms whose appearance and disappearance can be coded simply. One region that we do not have to work out is the gray zone where m11 is so low as to satisfy all the conditions of Lemma 20.4, resp. 20.9, for the deﬁning of boxgroups. There the zone total is simply

0 2e m n + d0 (s/2 v(N(γ))) H q − 11− c 2 − − W = | | F (ℓ ,ℓ ,ℓ )), (178) m11,n11 T (ℓ ,ℓ ,ℓ ) · 0 1 2 | 0 1 2 | where T (ℓ0,ℓ1,ℓ2) is the corresponding boxgroup. We now consider each zone in turn:

177 • The black and brown zones need no smear, as m n > 2e automatically in them. 11 ≥ 11 • The purple and blue zones have as answer a single F . It is stable as long as m11 is above the gray zone, since we computed the support by relaxing until we hit the gray zone. M11 • The green-zone answer is a sum of type F + ε F . The F is stable, as we got it by relaxing until C M11 we hit the gray zone. The εC F is a spark by Lemma 21.20(a). • In the red, yellow, and beige zones in the charmed coarse coset, the answer is a diﬀerence of two series of E’s. The E’s pair up to form sparks of the types in 21.20, leaving one singleton (two in the beige zone), a mostly stable F . In the yellow and beige zones, if the positive sum gets cut down to 1 or 0 terms respectively, the negative sum has 1 term and must be coded up in a special zone. This happens in a few cases, as shown in the code.−

21.7 The average value of a quadratic character on a box The results in this subsection, coupled with the strong-zone answers in Section 21.2, yield a quick solution to a problem that, at ﬁrst glance, is unrelated to the topic of this paper.

Theorem 21.21. Let K Q2 be a finite extension, and let R K be a tamely ramified étale extension of degree 3. In other words, ⊇R is one of the following: ⊃ • K K K × × • K Q, where Q is the unramified quadratic extension field × • the unramified cubic extension field • a totally ramified cubic extension field. Let χ : R× 1 →{± } be a character, that is, a group homomorphism, such that χ(a)=1 for all a K×. (All such characters can be put in the form ∈ χ (ξ)= α N (α), ξ , α · R/K where α R× and , is the Hilbert symbol.) Let B be an -sublattice contained in the Jacobson radical ∈ h• •i OK of R. (That is, B is a subgroup of R of finite index closed under multiplication by K , and all elements of BO have positive valuation at everyO field factor of R.) Then the average value O 1 χ(1 + ξ) dξ 1 B ZB takes on one of the following values: R • 0 • 1

i • q− for some i, 1 i e. (Here q = k , and e = v (2) is the absolute ramiﬁcation index.) ≤ ≤ | K | K Proof. We may assume that B contains π , as enlarging 1+ B to (1 + π )(1 + B) does not change the OK OK average of a character that vanishes on K×. Now we can take a reduced basis B = π + πnθ + πmθ , OK 1 2 and observe that B is one of the boxes that came up in Lemma 18.4. The strong-zone total Wm,n(δ) that we have computed in Section 21.2 can also be interpreted (up to scaling) as the volume of β B of class δ. Hence the average in question is ∈

χ(δ)W (δ) W (χ) I(χ)= δ m,n = m,n . W (δ) P δ m,n Wm,n(1) c P 178 c We wish to understand the possible values of this as χ ranges over H1. In view of the smearing lemma (Lemma 21.1), increasing m only makes the theorem stronger, so we can assume that we are in the case of Lemma 21.4 or 21.5. In the black, purple, and blue zones, Wm,n(δ)= c F (T ) for some subgroup T H1, so I takes the value 1 or 0 according as χ(T )=1 or not. · ⊆ In the green zone, Wm,n(δ)= c(F (T )+ εC F (T )) for some boxgroup T = T (ℓ0,ℓ1,ℓ2) on which εC is equidistributed, so

F (T ⊥)+ G(T ⊥),T ⊥ T ⊇ I =  V F (T ⊥)+ εC | |G(T ),T ⊥ ( T.  T p| | In either case, the identity coset T is uncharmed, so the two terms have disjoint supports. In the ﬁrst case, I is either 1 or 0. In the second case, we can additionally get a value of

V ℓ ℓ ε | | = q 0− 2 . C T ± p| |

The value of i = ℓ2 ℓ0 evidently satisﬁes 1 i e by our setup of boxgroups. Finally, in the red− zone, the Fourier transform≤ ≤ is easier to compute using the E-form (Corollary 21.17), which is of the form

b a b a 1 − c+ℓ − − c ℓ W = q E(ℓ,e b,b ℓ) q − E(ℓ +1,e b 1,b ℓ). m,n − − − − − − Xℓ=a Xℓ=a The Fourier transform, by deﬁnition of E, is

b a b a 1 − − − b ℓ b ℓ 1 Wm,n = cεC q − F (b ℓ,e b,ℓ) q − − F (b ℓ,e b 1,ℓ + 1) . · − − − − − − ! Xℓ=a Xℓ=a c b a Scaling by Wm,n(1) = cq − , we get

b a b a 1 c − a ℓ − − a ℓ 1 I = εC q − F (b ℓ,e b,ℓ) q − − F (b ℓ,e b 1,ℓ + 1) . − − − − − − ! Xℓ=a Xℓ=a The boxgroups on which the terms are supported form a nested chain

T (b a,e b,a) T (b a,e b 1,a + 1) T (b a 1,e b,a + 1) T (a,e b,a) − − ⊂ − − − ⊂ − − − ⊂···⊂ − that appear alternately with positive and negative coeﬃcients. So there are two types of behavior upon plugging in any individual χ: • χ lies in an even number of boxgroups in the chain, and they cancel in pairs to yield I =0. • χ lies in an odd number of boxgroups in the chain, and only the smallest one yields a contribution a ℓ I = q − . The negative of the exponent satisﬁes ± 0 ℓ a b a e. ≤ − ≤ − ≤ We must exclude the possibility that I = 1. This can be done by noting that I is the average value of a character that takes only the values 1 and −1, and that in a small neighborhood of 1 1+B, χ is identically 1 (continuity of χ is automatic, because all− values near 1 are squares). ∈

179 22 Ring volumes for ξ2′

Fix ξ1 and all the data leading up to it. By Lemma 18.2, there are only three possibilities for ξ¯2: either it is unrestricted, in which case the volume is given by the white-zone answer in Lemma 18.6, or it is restricted by or . M12 M22

22.1 12 M Lemma 22.1. Assume that R is tamely ramiﬁed and 12 is active. Fix ξ1 satisfying the 11, 11 conditions, and normalize γ as in Lemma 18.5. Then M is equivalent to a relation of the formM N 2 M12 m λ♦(α ξ′ ) 0 mod π⌊ 12⌋ 1 2 ≡ where α is primitive. 1 ∈ OR Proof. The condition says that M12 tr(ξ ξ ) 0 mod πm12 . 1 2 ≡ We have tr(ξ ξ ) = tr(ξ γ ξ′ )= λ♦( ξ γ ξ′ ). 1 2 1 2 2 ♦ 1 2 2 Observe that a′ a′ ξ γ π− 1− 2 R. ♦ 1 2 ∈ Let r a′ + a′ + Z be the unique value such that 12 ∈ 1 2 α = πr12 ξ γ 0 ♦ 1 2 is a primitive vector in . Then OR m +r : λ♦(α ξ′ ) 0 mod π 12 12 , M12 1 2 ≡ and the exponent is seen to be an integer. To show that it is m , it’s enough to prove that ⌊ 12⌋ 1 < r 0. − 12 ≤ We examine the cases.

• If R is unramified and [δωˆ ] , then , γ are units and ξ is primitive, so r =0. C ∈ L0 ♦ 2 1 12 • If R is unramified and [δωˆ ] (1; π; π) , then is a unit. Since is satisfied, we have j =1 and C ∈ L0 ♦ M11 0 6 (with respect to the naïve choice of γ1 from Lemma 18.5, which differs from how we actually found Q the ring volume for ξ ), γ (√π;1;1), (ξ′ ) is primitive. There are then two subcases, j = 2 and 1 1 ∼ 1 0 j0 =3. In both cases we find that the scaling of α0 is controlled by the Q-components and r12 = 1/2 or 0 respectively. − 2 ¯ • If R is totally ramified, then πR, ξ1 1, γ2 1. According as the product ξ1γ2 lies in R, ζ3R, or ζ¯2R, we must take r =0,♦∼2/3, or ∼1/3 respectively.∼ ♦ 3 12 − − Thus in all cases 1 < r 0, as desired. − 12 ≤ This allows us to compute the ring volume for ξ2:

Lemma 22.2. If is active, the solution volume for ξ′ is M12 2 m q−⌊ 12⌋, except when m Z in splitting type 13, in which case has no solutions. 12 ∈ M12

180 Proof. By Lemma 22.1, the -condition is given by one of the form M12 m λ♦(α ξ′ ) 0 mod π⌊ 12⌋. 1 2 ≡

m12 Here α1 is primitive, so we have a linear relation modulo π⌊ ⌋ which yields a solution volume of (1 + m12 1/q)q−⌊ ⌋. The solutions must be further whittled down using the restrictions on ξ2′ in Lemma 18.5 as well as the condition that ξ1 and ξ2 be linearly independent mod mK¯ . Here it’s important to note that the solutions to are distributed equally among (q + 1)-many 1-pixels. M12

• If R is unramiﬁed and [δωˆC ] 0, then a1′ Z. One of the 1-pixels is that of ξ1, which violates the linear independence, so we eliminate∈ L it. ∈

• If R is unramiﬁed, [δωˆ ] (1; π; π) , and C ∈ L0 1 1 (a ,a ,a ) , 0, mod 1, 1 2 3 ≡ 2 2 (K) Q then m12 Z +1/2, and the condition that π ∤ (ξ2′ ) eliminates one 1-pixel (no more, because (ξ1′ ) is primitive).∈

• If R is unramiﬁed, [δωˆ ] (1; π; π) , and C ∈ L0 1 1 (a ,a ,a ) , , 0 mod 1, 1 2 3 ≡ 2 2 then m Z and γ = γ (√π;1;1). When the -condition 12 ∈ 1 2 ∼ M12 2 m tr(γ ξ′ ξ′ ) 0 mod π 12 1 1 2 ≡ Q Q is looked at mod π, it uniquely determines ξ2′ ξ1′ mod π (by 11, the value ξ2′ = ξ1′ is a solution). But then ξ ξ mod √π, violating linear independence.≡ So Mis unsatisﬁable in this case. 2 ≡ 1 M12

• If R is totally ramiﬁed, then the condition that ξ2′ be a unit eliminates one 1-pixel, unless all its solutions are non-units. This happens exactly when α π2 , which is seen to be equivalent to m Z. 1 ∼ R 12 ∈ m Thus, in all but the stated exceptional case, we eliminate one 1-pixel, leaving a ring volume of q−⌊ 12⌋.

22.2 22 M When 22 is active, of course 11 is also, and we normalize both γ1 and γ2 (that is, ξ1′ and ξ2′ ) according to LemmaM 19.3. In tame splittingM types, we ﬁnd that the conic

2 : λ♦(δ⊙ξ′ ) 0 mod m′ M i ≡ ii is actually the same conic, but that the ξi′ cannot even lie in the same 1-pixel. The following two lemmas detail when this can happen.

Lemma 22.3. Let be a conic of determinant 1 on a lattice Λ over , and let m be an integer, 1 m e. A OK ≤ ≤ Then there are coprimitive ~x , ~x P(Λ) satisfying 1 2 ∈ (~x ) (~x ) 0 mod πm (179) A 1 ≡ A 2 ≡ if and only if the squareness ( ) satisﬁes A ( ) m. A ≥ Moreover, if ( )+1 m, then for ﬁxed ~x1, the volume of ~x2 satisfying (179) and coprimitive to ~x1 is m/2 A ≥ q−⌈ ⌉, split evenly among q-many 1-pixels.

181 Proof. If coprimitive ~x1, ~x2 satisfy (179), we can complete them to a basis (~x1, ~x2, ~x3) of Λ. Note that π ∤ (~x3), or else the determinant could not be 1. So we may scale so that (~x3)=1, and now we see that isA a square modulo πm. A A Conversely, suppose that is a square λ2 of a linear form modulo πm. Then for ~x P(Λ), A ∈ (~x) 0 mod πm A ≡ λ(~x)2 0 mod πm ⇐⇒ ≡ m/2 λ(~x) 0 mod π⌈ ⌉. ⇐⇒ ≡ m/2 Since λ 0 mod π, this has solution volume (1+1/q)q−⌈ ⌉, split evenly among (q + 1)-many 1-pixels. If 6≡ ~x1 is given, then ~x2 can occupy any 1-pixel except the one containing ~x1.

The following lemma limits m22:

Lemma 22.4. The conditions on ξ′ can be satisfied only if m e and m⊙ e. 2 22 ≤ 22 ≤ Proof. For the first part, note that if m22 >e, then m + m e + e m = 11 22 e + t> e =0, 12 2 − 2 − so 12 is active, contradicting Lemma 18.2. In unramified splitting type, this is the entire content of the M 3 lemma, since m22⊙ = m22. In splitting type 1 , note that

2h1 2h2 m⊙ = m and m⊙ = m 11 11 − 3 22 22 − 3 satisfy m⊙ +m⊙ = m +m , since h ,h = 1, 1 . Also, m⊙ >m⊙ since m m = 2(a a ) 2/3. 11 22 11 22 { 1 2} { − } 11 22 11− 22 2− 1 ≥ So if m⊙ >e, then m⊙ >e and is active as above. 22 11 M12

This enables us to compute the ring volume for ξ2′ .

Lemma 22.5. Let R be tamely ramiﬁed. Fix the discrete data of a quartic ring and a ξ1′ satisfying its conditions. The -condition is solvable for ξ′ if and only if the following conditions are satisﬁed: M22 2 • [δωˆ ] ; C ∈ L0

• the value of a2 mod Z allows for a γ2 and m22⊙ according to Lemma 19.3, and

• m⊙ min 2ℓ( )+1,e . 22 ≤ { M }

In such cases, the volume of ξ2′ is

1 m⊙ /2 • q − 22 if R is totally ramiﬁed, h =1 and h = 1, ⌈ ⌉ 1 2 − m⊙ /2 • q−⌈ 22 ⌉ otherwise.

Proof. The necessity of the restrictions on m22⊙ is shown by the foregoing lemmas. If they are satisﬁed, then m⊙ /2 transforms to a linear condition with solution volume (1+1/q)q− 22 , distributed equally among M22 ⌈ ⌉ (q + 1)-many 1-pixels. We must check its solutions against the other restrictions on ξ2′ :

• If R is unramified and [δωˆC ] 0, then ξ1′ , ξ2′ are required to be coprimitive, eliminating one of the 1-pixels. ∈ L • If R is unramified and [δωˆ ] (1; π; π) then j must be 3 in order for both γ and γ to exist. Then C ∈ L0 0 1 2 since m⊙ = m 1, we must have two solutions ξ′ , ξ′ to the transformed conic modulo π whose 22 22 ≥ 1 2 M Q-components are coprimitive. But (ξ′) mod π only depends on the Q-component and has only a M unique solution in P( /π ), so is unsatisfiable if active. OQ OQ M22

182 • If R has splitting type 13, then for γ ,γ to exist, we must have h ,h = 1, 1 . Here, coprimitivity 1 2 { 1 2} { − } between the ξi is subsumed by the condition that each ξi⊙ lie in its correct domain

1 hi ξ⊙ π − . i ∼ R As we noted in the proof of Lemma 20.7, the solutions to mod π comprise q-many 1-pixels of ξ⊙ 1 2 M ∼ and one 1-pixel of ξ⊙ π . Hence if h = 1 and h = 1, we retain q of the (q + 1)-many 1-pixels, ∼ R 1 − 2 m22⊙ /2 getting a volume q−⌈ ⌉. But if h1 = 1 and h2 = 1, then ξ2⊙ is restricted to one 1-pixel. This 1 m⊙ /2 − 22 2 2 gives a volume of q− −⌈ ⌉, but we multiply back by q since ξ2′ = πR− ξ2⊙. Of the conditions, only m⊙ min 2ℓ( )+1,e (180) 22 ≤ { M } is not trivial to verify. The following solves it: Lemma 22.6. Suppose that the discrete data is ﬁxed in such a way that • is active, M22 • [δωˆ ] (so the conic has determinant 1), C ∈ L0 • the value of a2 mod Z allows for a γ2 and m22⊙ according to Lemma 19.3.

Also suppose that ξ1 is ﬁxed, satisfying the conditions 11, 11 governing it. Then the remaining condition (180) can be checked as follows: M N • In the black, purple, and blue zones, it is automatic. • In the green zone, it is equivalent to m⊙ 22 ≤ C • In the red, yellow, and beige zones, it restricts the sum to only use terms G×(ℓ0,ℓ1,ℓ2) with

m⊙ 2ℓ +1. 22 ≤ 0

Proof. When the conic is green, its level was computed as part of the finding of the zone total for ξ1′ . So it remains to prove thatM if the conic is black or blue, (180) is satisfied. That the conic is black or blue implies that n 2e 4ℓ( ) 1. 11 ≥ − M − We already know m e. Suppose that 22 ≤ m 2ℓ( )+2. 22 ≥ M Since m12 is inactive, m =2m m +2e 2t 2e 2ℓ( ) 2. 11 12 − 22 − ≤ − M − But then n = m m + n 3, 22 22 − 11 11 ≥ so is active and we have a contradiction. N22 Remark 22.7. In the code, o2 and h2 are defined by o = 3(b mod 1) 0, 1, 2 , h =2o 3 3, 1, 1 . 2 2 ∈{ } 2 2 − ∈ {− − } The condition m△ 2k +1 22 ≤ (where k = ℓC in the green zone, or k is the index of summation in the red zone) is coded as 1 2o 2k + m + 2 0. 3 − 22 3 ≥

One veriﬁes that this is the same thing when o1 1, 2 , while when o1 = 0, it reduces to m22 0, as desired, since is unsatisﬁable if active. ∈ { } ≤ M22

183 23 Further remarks on the code

In the attached code, we use the computer programs SAGE and LattE to compute the generating function of rings. First we count “zone tuples” consisting of integer values of the following variables:

• e,t,b1,b2,s. • a1f = a , a2f = a , o , o , o . Here we’ve decomposed ⌊ 1⌋ ⌊ 2⌋ 1 2 3 o a = a + i , i ⌊ i⌋ o where 1 in splitting types (111), (12), and (3) o = 3 . ( 3 in splitting type (1 )

The oi belong to one of a ﬁnite number of “ﬂavors” coding the classes of the ai and bi mod Z.

Note that a3 is missing from the variable list, as its value is uniquely determined by the discriminant identity (Lemma⌊ ⌋ 18.1) • 1, =0 − C lCf =  ℓC , 0 < C

Because each variable is bounded below in terms of the preceding variables, the power series is formally convergent. Because each zone is delimited by ﬁnitely many linear inequalities with Z-coeﬃcients, the generating function is a rational function, computed by Barvinok’s algorithm as implemented in LattE. We then encode ring totals as substitutions that land us in a common ring

RINGS_RING = Z[[E_, T_, LCF_, SFL_, B1_, B2_, q, L0, L2]].

The trailing underscores are to prevent the computer from confounding certain elements of RRR and RINGS_RING, although the reader can think of them as identiﬁed. Two new variables L0, L2, whose exponents are the ℓ0 and ℓ2 of a boxgroup, complete the description of a ring total, along with a string Ftype that tells the kind of weighting (F , F xx, Ebal, etc.).

184 Part VIII Unanswered questions

24 Doubly traced quartic rings

It is an open question to classify, analogously to Theorem 11.1, the lattices in the space VK of pairs of ternary quadratic forms over K invariant under GL ( ) GL ( ). This problem is not quite the most relevant 2 OK × 3 OK to us because the factor GL2( K ), which changes the coordinates of the resolvent, is not relevant when we count rings with a ﬁxed resolvent,O as we have done in this paper. We therefore look at lattices invariant under GL3( K ) alone. By inspection we ﬁnd the following examples, which we conjecture exhaust all of them: O

Conjecture 24.1. Let K be a number ﬁeld, and let ( K ) be the lattice of pairs of ternary quadratic forms over . A primitive, GL ( )-invariant latticeV LOin ( ) is GL ( )-equivalent to one whose OK 3 OK V OK 2 OK completions Lp are as follows: (a) For p 2, the lattices L of (pt, ps)-traced pairs of ternary quadratics | p,t,s

L = L = a x x , b x x : a 0 mod pt,b 0 mod ps (i < j) t,s p,t,s  ij i j ij i j  ij ≡ ij ≡  1 i j 3 1 i j 3  ≤X≤ ≤ ≤X≤ ≤    for 0 t s e =v (2);  ≤ ≤ ≤ p

(b) For all other p, the maximal lattice V p only. OK, In other words, they are of the form

L = L = a x x , b x x : a 0 mod t,b 0 mod s (i < j) t,s p,tp,sp  ij i j ij i j  ij ≡ ij ≡  p 2 1 i j 3 1 i j 3 \|  ≤X≤ ≤ ≤X≤ ≤    for ideals (1) t s (2).  ⊇ ⊇ ⊇ As in the cubic case, the same lattices L necessarily appear in the analogues of ( ) with any Steinitz t,s V OK class a, appropriately adjusting the ideals that the aij , bij must lie in. For now, we look at the (2-adic) local case K/Q2. An element of Lt,s can be visualized as a pair of symmetric matrices 1 1 1 1 a11 2 a12 2 a13 b11 2 b12 2 b13 1 1 1 1 ( , )= 2 a12 a22 2 a23 , 2 b12 b22 2 b23 A B  1 1   1 1  2 a13 2 a23 a33 2 b13 2 b23 b33    1 t  1 s with diagonal entries aii,bii K and oﬀ-diagonal entries aij 2− πK K , bij 2− πK K . It is easy to check that the cubic resolvent∈f O(x, y) = 4 det( x y) has the∈ form O ∈ O A −B f(x, y)= π2sax3 + π2sbx2y + π2tcxy2 + π2tdy3 with discriminant D π4t+4s. Now (A, B) parametrizes a quartic ring with a resolvent Φ: C, and f(x, y) is the index form∈ of C. The divisibility conditions on f(x, y) can be interpreted as a non-maximalityO → condition on C: since t s, ≤ t t f(π− x, π− y) 2s 2t 3 2s 2t 2 2 3 2t = π − ax + π − bx y + cxy + dy π− and t s f(π− x, π− y) 3 s t 2 2 s t 3 t s = ax + π − bx y + cxy + π − dy π− −

185 are integral and thus are the index forms of certain overrings C1, C2 which we call the reduced resolvent and the reduced coresolvent. Appropriately lifting the basis ξ,¯ η¯ of C/ K in which (A, B) and hence f are t t t O s written, we have C = 1,ξ,η , C1 = 1, π− ξ, π− η , and C2 = 1, π− ξ, π− η . Note that C1 is a subring of s t h i h i h i s t C2 of index π − and moreover is a unidirectional subring in the sense that C1/C2 ∼= K /π − is generated by one element. The integrality properties of A and B translate readily into relationsO between the resolvent and the rings C1 and C2. Deﬁnition 24.2. Let be a Dedekind domain, and let L/K be a quartic algebra over its ﬁeld of fractions OK with resolvent Φ: L R. Let C1 C2 R be a pair of subrings with C2/C1 = K /d for some divisor d of → ⊆1 ⊆ ∼ O 2 in K . Let t be a divisor of 2d− and let s = td. AnO order L is (t, s)-traced with reduced resolvent C and reduced coresolvent C if O⊆ 1 2 (a) C = + t2C is a resolvent for , that is, disc = disc C and Φ( ) C; OK 1 O O O ⊆ (b) The associated bilinear form

Φ(˜ x, y)=Φ(x + y) Φ(x) Φ(y)= xy′ + x′y + x′′y′′′ + x′′′y′′ − − maps into + t3sC . O × O OK 2 1 1 1 1 If C has a basis 1,ξ,η for which C1 = 1, t− ξ, t− η and C2 = 1, t− ξ, s− η , as always happens when is a PID, thenh this isi easily seen to be equivalent to the condition that Φ: C is a resolvent whose OK O → matrix under this basis is in Lt,s. The functional equation for the Shintani zeta functions on VZ was stated and proved by Sato and Shintani ([48]). It relates pairs of integer-coeﬃcient ternary quadratic forms (over Z) with pairs of integer-matrix forms, that is, L(1),(1) to L(2),(2) in our notation. In contrast to the lattice of cubic forms, there is no SL3(K)-invariant inner product in VK : as a representation of SL3(K), VK is not isomorphic to its dual. However, VK is isomorphic to its dual twisted by the automorphism of SL3(K) given by inverse transpose, 1 1 and under this duality, it is easy to see that Lt,s is interchanged with L2s− ,2t− . Therefore, it is a pretty conjecture that the corresponding composed varieties are naturally dual.

Conjecture 24.3. Let K be a 2-adic local ﬁeld and 0 t s e = vK (2) be integers. Let C1 C2 ≤ s ≤t ≤ ⊆ be orders in a cubic K-algebra R such that C /C = /π − is an -module with one generator. For 2 1 ∼ OK OK each quartic K-algebra L with resolvent R, denote by g(L, C1, C2,t,s) the number of (t,s)-traced quartic rings in L with reduced resolvent and coresolvent C1 and C2, respectively. Then the dual of g(L, C1, C2,t,s), considered as a function of L, is

gˆ(L, C , C ,t,s)= qt+sg(L, C , C ,e s,e t) (181) 1 2 1 2 − − where q = k is the order of the residue ﬁeld. | K | t Note that the quartic rings counted on either side of (181) actually have resolvents K + π C1 and e s 2t 2e 2s O K + π − C1 and discriminants π disc C1 and π − disc C1, respectively. If t + s = e, this conjecture Oasserts the self-duality (up to the correct scaling) of the indicated local weighting. If this conjecture is true, then by our reﬂection engine, we immediately get the following corresponding global result.

Conjecture 24.4. Let K be a number ﬁeld. Let C C be orders in a cubic K-algebra R such that 1 ⊆ 2 C /C = /d is an -module with one generator. Let t, s be ideals such that s = dt (2). Let 2 1 ∼ OK OK ⊆ OK | h(C1, C2, t, s) count the number of (t, s)-traced quartic rings with reduced resolvent and coresolvent C1 and C2, respectively, each weighted by the reciprocal of its number of resolvent-preserving automorphisms. Let ntc h (C1, C2, t, s) count the subset of the foregoing that are ntc, weighted in the same way. Then

N(ts) ntc 1 1 h(C1, C2, t, s)= r h (C1, C2, 2s− , 2t− ), 2 ∞ · where r is the number of real places of K over which R is not totally real plus twice the number of complex places of∞K.

186 25 Reflection for 2 n n boxes × × We close with a conjectural generalization to pairs of symmetric matrices ( , ) of any odd order n. These play an important role in understanding 2-torsion in n-ic rings, owing toA a parametrizationB of Wood [60] as well as the Selmer groups of hyperelliptic curves of genus (n 1)/2 (see [6]). The connection to quartic rings, on the other hand, has no known analogue. − Let K be a field, char K =2 and n be an odd integer. The group Γ=SLn acts on the space V of pairs of symmetric matrices ( , )6 over K, preserving the resolvent A B n 1 f(x, y)=2 − det( x y). A −B n 1 The point stabilizer of an orbit with resolvent f is isomorphic to − , with Galois action permuting a C2 hyperbasis as it permutes the roots of f, and the variety (Vf , Γ) of pairs with fixed resolvent is a composed variety (closely related to the “third representation” of SO in [7]). If K is a Dedekind subring with n OK ⊆ fraction field K, there are integral forms ( a,t,s, a) of V for each [a] Cl( K ) and all ideals (1) t s (2), defined by a straightforward extension ofV the 2G 3 3 case. For simplicity∈ O we look only at the⊇ case⊇ ⊇ when [a] = [1], t = (τ)= s are principal: × ×

1 1i=j ( )= [a ,b ]: a ,b (2− τ) 6 . Vτ OK ij ij ij ij ∈ OK n 1 Note that the resolvent f of such a box is divisible by τ − , since

n 1 1 1 f(x, y)= τ − 2− τ det 2τ − ( x y) , · A −B and the argument to the determinant is an K -integral matrix that is skew-symmetric, hence singular (being 1 O of odd order), modulo 2τ − . Conjecture 25.1 (Local O-N for 2 n n boxes). If K is a local ﬁeld, char K =2, then × × 6

n 1 and 1 1 n 1 Vτ,τ − f V2τ − ,(2τ − ) − f (n 1)v (τ) are naturally dual with duality constant q − K ; in order words, for every binary n-ic form f, the associated local orbit counters g : H1(K,M ) N τ,f f → satisfy the local reﬂection theorem

n 1 gˆ = /τ − g 1 . τ,f |OK OK | · 2τ − ,4f If this conjecture is true, our local-to-global reflection engine yields the following. Conjecture 25.2 (O-N for 2 n n boxes). Let K be a number field, and let τ be a divisor of 2 in . × × OK Denote by hτ (f) the number of Γ( K )=SLn( K )-orbits of pairs ( , ) of n n symmetric matrices whose O O 1 A B × on- and off-diagonal elements belong to K and 2− τ K respectively, each ( , ) weighted by the reciprocal of the order of its stabilizer in Γ( ). DenoteO by hntcO(f) the count (by the sameA B weighting) of the subset of OK τ orbits whose corresponding self-balanced ideal (Rf ,I,δ) has δ > 0 at every real place of K. Then we have a global reflection theorem 2 n 1 N(t) ntc 1 n 1 h (τ − f)= h 1 (2τ − ) − , τ r 2τ − 2 ∞ · where r Z depends only on the splitting type of f at each of the real places of K. ∞ ∈ While some low-discriminant cases of this conjecture can be verified using the known structure of self- balanced ideals, the general case—which involves the Igusa zeta function of an intersection of two quadrics in n 1 P − —is quite far from solvable using the techniques in this paper. If true, it furnishes a more satisfactory answer to the question addressed by Cohen–Rubinstein-Salzedo–Thorne [12], namely the production of a family of O-N-like reflection theorems for representations whose dimensions go to infinity.

187 Part IX Appendices

A The Grothendieck-Witt ring and the proof of Lemma 19.11

In this section we prove Lemma 19.11, which is of a different character than the other results in this paper. To prove this lemma, we must recall some facts about the Grothendieck-Witt ring of a local field. The nondegenerate quadratic forms over a field K (char K =2), up to isomorphism, form a semiring under the operations of orthogonal direct sum and tensor product6 ; it is cancellative (the so-called Witt cancellation theorem), and the ring obtained by⊥ adjoining formal additive⊗ inverses is called the Grothendieck-Witt ring GW (K) of K. In the case that K is a local field, a form is determined (see O’Meara [46], Theorem 63:20) by 2 three invariants: its dimension n Z 0, its determinant D K×/(K×) , and one other bit of information, ∈ ≥ 2 ∈ 2 the Hasse symbol ε 1 . For a diagonal form a1x1 + +anxn, the determinant is given by D = a1 an and the Hasse symbol∈ {± by } ··· ··· ε = a ,a . h i j iK i

dim(f g) = dim f + dim g (182) ⊥ det(f g) = det f det g (183) ⊥ ε(f g)= ε(f)ε(g) det f, det g (184) ⊥ h iK dim(f g) = dim f dim g (185) ⊗ det(f g) = (det f)dim g(det g)dim f (186) ⊗ dim g dim f dim g dim f ( 2 ) ( 2 ) dim f dim g 1 ε(f g)= ε(f) ε(g) det f, 1 det g, 1 det f, det g − . (187) ⊗ h − iK h − iK h iK

We denote by EllK the unique class in GW (K) of dimension 0, determinant 1, and Hasse symbol 1 (the “elliptic class”). Note that if f and g are nonisomorphic quadratic forms over K with the same dimension− and determinant, then [f] = [g] + EllK in GW (K). Suppose that L/K is a ﬁeld extension. If q : W L is a quadratic form over L, we can view W as a → K-vector space and postcompose with the trace trL/K to get a quadratic form trL/K q. Since trL/K respects orthogonal direct sums, it induces a group homomorphism (though not a ring homomorphism) from GW (L) to GW (K) (the opposite direction to the more familiar extension-of-scalars morphism). We easily compute that dim tr q = [L : K] dim q and det tr q = (disc(L/K))dim q N (det q). L/K · L/K · L/K We wish to understand how ε(trL/K q) behaves. The following is the most important result needed. Lemma A.1. For any extension L/K of local ﬁelds not of characteristic 2,

trL/K EllL = EllK .

In other words, trL/K : GW (L) GW (K) preserves the Hasse symbol on classes of dimension 0 and determinant 1. → Proof. We may assume that L/K is a primitive extension, that is, has no nontrivial intermediate extensions, since tr tr = tr for a tower L/E/K. E/K ◦ L/E L/K

188 Since trL/K EllL is of dimension 0 and determinant 1, the only other possibility is that trL/K EllL = 0. We prove that this cannot hold. Let a K× be an element that does not become a square in L. Such an a exists because L/K is primitive; if not, then∈ L/K would contain both an unramiﬁed and a ramiﬁed quadratic extension. Then choose θ L ∈ such that a,θ L = 1. We also have a,NL/K(θ) K = 1 by the standard relation a,θ L = a,NL/K(θ) K (a K, θ h L).i Consider− the following quadratic forms− over L: h i ∈ ∈ f(x, y)= x2 θy2, g(x, y)= ax2 aθy2. − − 2 In other words, f = q1 q θ and g = qa + q aθ where qa(x) = ax . Both f and g have dimension 2 and discriminant θ, but their⊥ Hasse− symbols are 1−and a, aθ = 1, respectively. Hence [f] = [g] + Ell . But − h − i − L g = af, so trL/K (g)= a trL/K (f)= qa trL/K (f). So by the formula (187) for the Hasse symbol of a tensor product, ⊗ ε(tr g)= ε(q tr f) L/K a ⊗ L/K 2[L:K] (1) 4[L:K] 1 = ε(q )2[L:K] ε(tr f)1 det q , 1 ( 2 ) det tr f, 1 2 det q , det tr f − a · L/K · h a − i · L/K − · a L/K [L:K] 4[L:K] 1 = ε(tr f) a, 1 a, (disc(L/K))2 N (det f) − L/K · h − i · · L/K [L:K] = ε(trL/K f) a, 1 a,NL/K( θ) · h − i · − = ε(tr f) a,N (θ) L/K · L/K = ε(tr f), − L/K So [trL/K f] = [trL/K g] + EllK , yielding the desired conclusion.

Proof of Lemma 19.11. We now relate ε(α) to Hasse symbols. Denote by qR,α the quadratic form 2 qR,α(ξ)= αξ over R. Then ε(α)=1 if and only if the form trR/K (qR,α) is isotropic, where trR/K is to be interpreted in the obvious way if R is not a ﬁeld. Given α, β R× of norm 1, the forms ∈ f = q q and g = q q R,1 ⊥ R,αβ R,α ⊥ R,β have dimension 2 and determinant αβ over R, and in GW (K), ε(1)ε(α)ε(β)ε(αβ)= ε([tr f] [tr g]). R/K − R/K r Decompose R = i=1 Ri into its ﬁeld factors (1 r 3), and let α = (α1; ... ; αr) and β = (β1; ... ; βr). Then ≤ ≤ Q r [tr f] [tr g]= tr ([q ] [q ] [q ] + [q ]). R/K − R/K Ri/K Ri,αiβi − Ri,αi − Ri,βi Ri,1 i=1 X

Since the invariants of each qRi,γ are known, the invariants of the class in parentheses can be computed by repeated application of (182)–(184). We ﬁnd that it has dimension 0, determinant 1 and Hasse symbol α ,β . By the preceding lemma, its trace has the same invariants. Hence the whole sum has Hasse i i Ri symbolh i

αi,βi = α, β , h iRi h iR i Y as desired.

B Examples of zone totals

The following tables serve to illustrate some of the totals computed in Section 21. The following are valid in the unramified splitting types (111, 12, and 3): more specifically, the first table (with s =0) applies to all three, the remaining ones to splitting types 111 and 12. For brevity, the following conventions have been observed:

189 • In the red zone, the answer (which is independent of ℓC ) is the sum of the red-colored entries of the corresponding row.

• In the other color zones, the answer is the single entry corresponding to the appropriate values of ℓC and n11.

m • The invariable factor of q− 11 has been omitted. For e =8,s =0 (similar results hold whenever s =0): 2) 2) 2) 2) 3) 4) 1) 1) 1) 1) 4) 0) 0) , , , , , , , , , , , , , 1 2 2 3 2 0 0 1 1 2 0 0 1 , , , , , , , , , , , , , — — — (5 (4 (4 (3 (3 (4 (7 (6 (6 (5 (4 (8 (7 F F F F F G F F F F G F F 1 1 2 2 2 3 0 0 1 1 3 0 0 q q q q q q q q q q q q q 4) 4) , , 1 1 , , 2) 2) 2) 2) 3) 3) 1) 1) 1) 1) 0) 0) 3) 3) (3 (3 , , , , , , , , , , , , , , × × 1 2 2 3 2 2 0 1 1 2 0 1 2 2 , , , , , , , , , , , , , , G G 3 3 (5 (4 (4 (3 (3 (3 (7 (6 (6 (5 (8 (7 (3 (3 q q ) ) F F F F F G F F F F F F G G 1 1 2 2 2 0 0 1 1 0 0 3 2 3 C C q q q q q q q q q q q q q q ε ε 4) 5) (1+ 5) (1+ 3) 4) 3) , , , , , , 2 1 1 3 2 2 , , , , , , 2) 2) 2) 2) 1) 1) 1) 1) 0) 0) (2 (2 (2 (2 (2 , , , , , , , , , , (3 × × × × × 1 2 2 3 0 1 1 2 0 1 F , , , , , , , , , , G G G G G 2 2 2 2 2 2 q (5 (4 (4 (3 (7 (6 (6 (5 (8 (7 q q q q q ) ) ) ) ) ) F F F F F F F F F F C 1 1 2 2 0 0 1 1 0 0 C C C C C ε q q q q q q q q q q ε ε ε ε ε (1 + (1 + 5) (1+6) (1+ 6) (1+ 2) 2) 3) 3) 4)4) (1+ 5) , , , , , , , , , , 2 1 1 1 2 1 2 1 2 1 , , , , , , , , , , 1) 1) 1) 1) 0) 0) (1 (1 (1 , , , , , , (5 (4 (4 (3 (3 (2 (2 × × × 0 1 1 2 0 1 F F F F F F F , , , , , , G G G 1 1 1 1 1 1 1 1 1 1 q q q q q q q (7 (6 (6 (5 (8 (7 q q q ) ) ) ) ) ) ) ) ) ) F F F F F F C C C C C C C 0 0 1 1 0 0 C C C ε ε ε ε ε ε ε q q q q q q ε ε ε (1 + (1 + 7) (1+ 1) 1) 2) 2) 3)3) (1+ 4) (1+ 4) (1+ 5) (1+ 5) (1+ 6) (1+ 6) (1+ 7) , , , , , , , , , , , , , , 1 0 1 0 1 0 1 0 1 0 1 0 1 0 , , , , , , , , , , , , , , 0) 0) (0 , , (7 (6 (6 (5 (5 (4 (4 (3 (3 (2 (2 (1 (1 × 0 1 F F F F F F F F F F F F F , , G 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 34 0 q q q q q q q q q q q q q (8 (7 q ) ) ) ) ) ) ) ) ) ) ) ) ) ) F F C C C C C C C C C C C C C 0 0 C ε ε ε ε ε ε ε ε ε ε ε ε ε q q ε (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + 9 8 7 6 5 4 3 2 1 → 15 14 13 12 11 10 C = 16 ℓ 11 n

190 For e =6,s = 14 (similar results hold whenever s 2e): ≥ ℓC any → 0 n11 = 12 2q F (6, 0, ) 11 2q0F (5, 1, ∅) 10 2q1F (5, 1, ∅) 9 2q1F (4, 2, ∅) 8 2q2F (4, 2, ∅) 7 2q2F (3, 3, ∅) 6 2q3F (3, 3, ∅) 5 2q3F (2, 4, ∅) 4 2q4F (2, 4, ∅) 3 2q4F (1, 5, ∅) 2 2q5F (1, 5, ∅) 1 2q5F (0, 6, ∅) ∅ For e = 11,s =7 (similar results hold whenever s< 2e is odd): ℓC 3 → 0 n11 = 22 2q F (11, 0, ) 21 2q0F (10, 1, ∅) 20 2q1F (10, 1, ∅) 19 2q1F (9, 2, ∅) 18 2q2F (9, 2, ∅) 17 2q2F (8, 3, ∅) 16 2q3F (8, 3, ∅) 3 ∅ 15 (1 + εC)q F (7, 4, 0) 3 14 (1 + εC)q F (7, 3, 1) 3 13 (1 + εC)q F (6, 4, 1) 3 12 (1 + εC)q F (6, 3, 2) 3 11 (1 + εC)q F (5, 4, 2) 3 10 (1 + εC)q F (5, 3, 3) 3 9 (1 + εC)q F (4, 4, 3) 3 8 (1 + εC)q F (4, 3, 4) 3 7 (1 + εC)q F (3, 4, 4) 3 6 (1 + εC)q F (3, 3, 5) 3 5 (1 + εC)q F (2, 4, 5) 3 4 (1 + εC)q F (2, 3, 6) 3 3 (1 + εC)q F (1, 4, 6) 3 2 (1 + εC)q F (1, 3, 7) 3 1 (1 + εC)q F (0, 4, 7)

191 For e = 10,s =4 (similar results hold whenever s< 2e is even): ) ) ) ) ∅ ∅ ∅ ∅ , 1) 1) 1) 1) 2) 2) 2) 2) 0) 0) 3) 4) 4) , , , , , , , , , , , , , 0 , , , 1 1 2 , 2 3 3 4 3 4 4 5 2 3 4 2 2 , , , 4 , , , , , , , , , , , , , (9 (9 (8 — — — (7 (6 (6 (5 (5 (4 (4 (3 (8 (7 (10 (3 (4 (4 ≥ F F F F F F F F F F F F F F F G G 0 1 1 0 5 2 2 3 3 3 3 4 4 5 2 2 4 q q q q q q q q q q q q q q q q q 4) 4) , , 3 3 ) 2 , , ))) 2 2 2 ∅ ∅ ∅ ∅ , 3) 1) 1) 1) 1) 2) 2) 2) 2) 3) 3) 0) 0) 3) (3 (3 , , , , , , , , , , , , , , , , 0 , 1 1 2 × × , 4 2 3 3 4 3 4 4 5 4 4 2 3 4 , , , , , , , , , , , , , , , , , G G (9 (9 (8 5 5 (3 (7 (6 (6 (5 (5 (4 (4 (3 (3 (3 (8 (7 (10 (3 q q F F F ) ) G F F F F F F F F G G F F F F 0 1 1 5 2 2 3 3 3 3 4 4 4 5 2 2 0 4 C C q q q q q q q q q q q q q q q q q q ε ε 4) 5) (1+ 3) 4) 5) (1+ 3) , , , , , , 4 3 5 4 3 ) 2 4 , , , , , ))) 2 2 2 ∅ , ∅ ∅ ∅ , 1) 1) 1) 1) 2) 2) 2) 2) 0) 0) (2 (2 (2 (2 (2 , , , , , , , , , , , , , 0 (3 1 1 2 × × × × × , 2 3 3 4 3 4 4 5 2 3 , , , F , , , , , , , , , , G G G G G 4 (9 (9 (8 4 4 4 4 4 q (7 (6 (6 (5 (5 (4 (4 (3 (8 (7 (10 q q q q q ) F F F ) ) ) ) ) F F F F F F F F F F F C 0 1 1 2 2 3 3 3 3 4 4 2 2 0 C C C C C ε q q q q q q q q q q q q q q ε ε ε ε ε (1 + (1 + 5) (1+6) (1+ 6) (1+ 2) 2) 3) 3) 4)4) (1+ 5) , , , , , , , , , , 4 3 3 ) 2 3 4 3 4 3 4 3 , , , ))) 2 2 2 ∅ , , , , , , , ∅ ∅ ∅ , 1) 1) 1) 1) 0) 0) (1 (1 (1 , , , , , , , , , 0 (5 (4 (4 (3 (3 (2 (2 1 1 2 × × × , 2 3 3 4 2 3 , , , F F F F F F F , , , , , , G G G 3 3 3 3 3 3 3 3 3 3 (9 (9 (8 q q q q q q q (7 (6 (6 (5 (8 (7 (10 q q q ) ) ) ) ) ) ) F F F ) ) ) F F F F F F F C C C C C C C 0 1 1 0 2 2 3 3 2 2 C C C ε ε ε ε ε ε ε q q q q q q q q q q ε ε ε (1 + (1 + 7) (1+ 1) 1) 2) 2) 3)3) (1+ 4) (1+ 4) (1+ 5) (1+ 5) (1+ 6) (1+ 6) (1+ 7) , , , , , , , , , , , , , , 3 ) 2 2 3 2 3 2 3 2 3 2 3 2 3 2 , ))) 2 2 2 ∅ , , , , , , , , , , , , , ∅ ∅ ∅ , 0) 0) (0 , , , , , 0 (7 (6 (6 (5 (5 (4 (4 (3 (3 (2 (2 (1 (1 1 1 2 × , 2 3 , , , F F F F F F F F F F F F F , , G 2 2 2 2 2 2 2 2 2 2 2 2 2 0 1 2 3 2 (9 (9 (8 q q q q q q q q q q q q q (8 (7 (10 q ) ) ) ) ) ) ) ) ) ) ) ) ) F F F ) F F F C C C C C C C C C C C C C 0 1 1 0 2 2 C ε ε ε ε ε ε ε ε ε ε ε ε ε q q q q q q ε 2 2 2 2 (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + (1 + 9 8 7 6 5 4 3 2 1 → 19 18 17 16 15 14 13 12 11 10 C = 20 ℓ 11 n

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