Quaternion Algebras Companion

Quaternion algebras companion

v1.0.2 August 13, 2021

Preface

This booklet is a supplement to the book Quaternion algebras, containing some additional material (statements and proofs, some remarks) and comments on exercises. Thanks to Joe Quinn for his comments on some of the exercises. Please contact me at [email protected] if you ﬁnd mistakes or have other suggestions.

i Contents

Contents ii

I Algebra1

3 Involutions3 3.6 Algorithmic aspects...... 3

4 Quadratic forms5 4.6 Algorithmic aspects...... 5

5 Ternary quadratic forms and quaternion algebras7 5.7 Algorithmic aspects...... 7

8 Simple algebras and involutions9 8.6 Algorithmic aspects...... 9

II Arithmetic 11

9 Lattices and integral quadratic forms 13 9.7 Normalized form...... 13

12 Ternary quadratic forms over local ﬁelds 15 12.5 Algorithmic aspects...... 15

14 Quaternion algebras over global ﬁelds 19 14.8 Algorithmic aspects...... 19

15 Discriminants 21 15.7 Algorithmic aspects...... 21

17 Classes of quaternion ideals 25 17.9 Algorithmic aspects...... 25

20 Integral representation theory 29

ii CONTENTS iii

20.5 Local Jacobson radical...... 29 20.6 Local integral representation theory...... 30

21 Hereditary and extremal orders 33 21.5 Classiﬁcation of local hereditary orders...... 33

22 Ternary quadratic forms 35 22.6 ∗ A functorial inverse to the even Cliﬀord map...... 35 22.7 ∗ Twisting and ﬁnal bijection...... 38 22.8 ∗ Rigidifying with isometries and class sets...... 41 Exercises...... 45

III Analysis 47

26 Classical zeta functions 49 26.2 Analytic class number formula...... 49

30 Optimal embeddings 51 30.10 Algorithmic aspects...... 51

IV Geometry and topology 53

33 Hyperbolic plane 55 33.3 Upper half-plane...... 55 33.6 Hyperbolic area and the Gauss–Bonnet formula...... 55

V Arithmetic geometry 59

40 Classical modular curves and modular forms 61 40.3 Classical modular forms...... 61

Exercises 63

A Hints and comments on exercises 67

Bibliography 83

Part I

Algebra

Chapter 3

Involutions

3.6 Algorithmic aspects

In this section, we exhibit an algorithm to determine if an algebra has a standard involution (and, if so, to give it explicitly as a linear map) [Voi2013, §2]; in the next section we will use this to recognize quaternion algebras. We begin with some basic deﬁnitions.

Definition 3.6.1. A field 퐹 is computable if 퐹 comes equipped with a way of encod- ing elements of 퐹 in bits (i.e. the elements of 퐹 are recursively enumerable, allowing repetitions) along with deterministic algorithms to perform field operations in 퐹 (addition, subtraction, multiplication, and division by a nonzero element) and to test if 푥 = 0 ∈ 퐹; a field is polynomial-time computable if these algorithms run in polynomial time (in the bit size of the input).

For precise deﬁnitions and a thorough survey of the subject of computable rings we refer to Stoltenberg-Hansen–Tucker [SHT99] and the references contained therein.

Example 3.6.2. A field that is finitely generated over its prime ring is computable by the theory of Gröbner bases [vzGG03]. Any uncountable field is not computable.

Let 퐵 be an 퐹-algebra with dim퐹 퐵 = 푛 and basis 푒1, 푒2, . . . , 푒푛 as an 퐹-vector 3 space. Suppose 푒1 = 1.A multiplication table for 퐵 is a system of 푛 elements (푐푖 푗푘 )푖, 푗,푘=1,...,푛 of 퐹, called structure constants, such that multiplication in 퐵 is given by 푛 ∑︁ 푒푖푒 푗 = 푐푖 푗푘 푒푘 푘=1 for 푖, 푗 ∈ {1, . . . , 푛}. An 퐹-algebra 퐵 is represented in bits by a multiplication table and elements of 퐵 are represented in the basis 푒푖. Basis elements in 퐵 can be multiplied directly by the multiplication table but multiplication of arbitrary elements in 퐵 requires 푂(푛3) arithmetic operations (additions and multiplications) in 퐹; in either case, note the output is of polynomial size in the input for ﬁxed 퐵.

3 4 CHAPTER 3. INVOLUTIONS

We now exhibit an algorithm to test if an 퐹-algebra 퐵 (of dimension 푛) has a standard involution. First, we note that if 퐵 has a standard involution : 퐵 → 퐵, then this involution and hence also the reduced trace and norm can be computed. Let {푒푖 }푖 be a basis for 2 퐵; then trd(푒푖) ∈ 퐹 is simply the coeﬃcient of 푒푖 in 푒푖 , and so 푒푖 = trd(푒푖) − 푒푖 for each 푖 can be precomputed for 퐵; one recovers the involution on 퐵 for an arbitrary element of 퐵 by 퐹-linearity. Therefore the involution and the reduced trace can be computed using 푂(푛) arithmetic operations in 퐹 and the reduced norm using 푂(푛2) operations in 퐹.

Algorithm 3.6.3. This algorithm takes as input 퐵, an 퐹-algebra given by a multiplication table in the basis 푒1, . . . , 푒푛 with 푒1 = 1. It returns as output true if and only if 퐵 has a standard involution, and if so returns the standard involution as a linear map.

2 2 1. For 푖 = 2, . . . , 푛, let 푡푖 ∈ 퐹 be the coeﬃcient of 푒푖 in 푒푖 , and let 푛푖 = 푒푖 − 푡푖푒푖. If some 푛푖 ∉ 퐹, return false. 2 2. For 푖 = 2, . . . , 푛 and 푗 = 푖 + 1, . . . , 푛, let 푛푖 푗 = (푒푖 + 푒 푗 ) − (푡푖 + 푡 푗 )(푒푖 + 푒 푗 ). If some 푛푖 푗 ∉ 퐹, return false. Otherwise, return true, and the linear map deﬁned by 푒푖 ↦→ 푡푖 − 푒푖.

Proof of correctness. Let 퐹[푥] = 퐹[푥1, . . . , 푥푛] be the polynomial ring over 퐹 in 푛 variables, and let 퐵퐹 [푥] = 퐵 ⊗퐹 퐹[푥]. Let 휉 = 푥1 + 푥2푒2 + · · · + 푥푛푒푛 ∈ 퐵퐹 [푥], and deﬁne 푛 ∑︁ 푡 휉 = 푡푖푥푖 푖=1 and 푛 ∑︁ 2 ∑︁ 푛 휉 = 푛푖푥푖 + (푛푖 푗 − 푛푖 − 푛 푗 )푥푖푥 푗 . 푖=1 1≤푖< 푗 ≤푛 Let 푛 2 ∑︁ 휉 − 푡 휉 휉 + 푛 휉 = 푐푖 (푥1, . . . , 푥푛)푒푖 푖=1 with 푐푖 (푥) ∈ 퐹[푥]. Each 푐푖 (푥) is a homogeneous polynomial of degree 2. The algorithm then veriﬁes that 푐푖 (푥) = 0 for 푥 ∈ {푒푖 }푖 ∪ {푒푖 + 푒 푗 }푖, 푗 , and this implies that each 푐푖 (푥) vanishes identically. Therefore, the specialization of the map 휉 ↦→ 휉 = 푡 휉 −휉 is the unique standard involution on 퐵.

2 3.6.4. Algorithm 3.6.3 requires 푂(푛) arithmetic operations in 퐹, since 푒푖 can be computed directly from the multiplication table and hence

2 2 2 (푒푖 + 푒 푗 ) = 푒푖 + 푒푖푒 푗 + 푒 푗 푒푖 + 푒 푗 can be computed using 푂(4푛) = 푂(푛) operations. Chapter 4

Quadratic forms

4.6 Algorithmic aspects

We conclude with two comments on algorithms arising naturally from the above; for an overview, see Voight [Voi2013]. First, the proof that a quadratic form can be diagonalized (Lemma 4.3.1) is algorithmic, requiring 푂(푛3) ﬁeld operations in 퐹. Second, Main Theorem 4.4.1 yields the following algorithm for algorithmic recognition of a quaternion algebra.

Algorithm 4.6.1. This algorithm takes as input 퐵, an 퐹-algebra with dim퐹 퐵 = 4, speciﬁed by a multiplication table. It returns as output true if and only if 퐵 is a quaternion algebra, and if so returns an isomorphism 퐵 '(푎, 푏 | 퐹).

1. Verify that 퐵 has a standard involution by calling Algorithm 3.6.3. If not, return false. 2. Compute a diagonalized basis 1, 푖, 푗, 푘 for the quadratic form nrd: 퐵 → 퐹. 3. Compute 푑 := disc(nrd) ∈ 퐹/퐹×2. If 푑 ≠ 0, return true and the quaternion algebra (푎, 푏 | 퐹) given by the standard generators 푖, 푗. Otherwise, return false.

Chapter 5

Ternary quadratic forms and quaternion algebras

5.7 Algorithmic aspects

In this section, we show how the equivalences of Main Theorem 5.4.4 involved in identifying the matrix ring (splitting of a quaternion algebra) can be made algorithmic. Algorithm 5.7.1. This algorithm takes as input 훼 ∈ 퐵 a zerodivisor and returns as output a nonzero element 휖 ∈ 퐵 such that 휖2 = 0.

1. If trd(훼) = 0, return 훼. 2. Compute 0 ≠ 훽 ∈ 퐵 orthogonal to 1, 훼 with respect to the quadratic form nrd. If 훼훽 = 0, return 훽; otherwise, return 훼훽.

Proof of correctness. The element 훼 ≠ 0 is a zerodivisor if and only if nrd(훼) = 훼훼 = 0. Since 훽 is orthogonal to 1, 훼 we have 훽 = −훽 and trd(훼훽) = − trd(훼훽) = 0. If 훼훽 = 0 then 훽 is as desired. If 훼훽 ≠ 0 then nrd(훼훽) = nrd(훼) nrd(훽) = 0, as desired. Algorithm 5.7.2. This algorithm takes as input 휖 ∈ 퐵 satisfying 휖2 = 0 and returns as output a standard representation 퐵 '(1, 1 | 퐹)' M2 (퐹).

1. Find 푘 ∈ {푖, 푗, 푖 푗} such that trd(휖 푘) = 푠 ≠ 0. Let 푡 := trd(푘) and 푛 := nrd(푘), and let 휖 0 := (1/푠)휖. 2. Let 푗 0 := 푘 + (−푡푘 + 푛 + 1)휖 0 and let

푖0 := 휖 0푘 − (푘 + 푡)휖 0

Return 푖0, 푗 0.

Proof of correctness. In Step 1, if trd(휖 푘) = 0 for all such 푘 then 휖 ∈ rad(nrd), contradicting Main Theorem 4.4.1. We have trd(휖 0푘) = trd(푘휖 0) = 1 so trd(휖 0푘) = −1. Consider 퐼 = 퐹휖 0 + 퐹푘휖 0. Note trd(푘휖 0) ≠ 0 implies that 휖 0, 푘휖 0 are linearly independent. Let 퐴 be the subalgebra of 퐵 generated by 휖 0 and 푘. We have 휖 0푘 + 푘휖 0 =

7 8 CHAPTER 5. TERNARY QUADRATIC FORMS AND QUATERNION ALGEBRAS

푡휖 0 + 1 from (4.2.14) and 푘2 = 푡푘 − 푛, and thus we compute that left multiplication yields a map

퐴 → End퐹 (퐼)' M2 (퐹) 0 1 0 −푛 휖 0, 푘 ↦→ , . 0 0 1 푡

0 1 1 0 A direct calculation then reveals that 푗 0 ↦→ and 푖0 ↦→ . It follows at 1 0 0 −1 0 once that 퐴 = 퐵, that 퐼 = 퐵휖 , and that the map 퐵 → M2 (퐹) is an isomorphism. In this way, we have seen that in deterministic polynomial time we can convert an isotropic vector (i.e., an 퐹-rational point on the associated conic) or a zerodivisor to ∼ an explicit splitting 퐵 −→ M2 (퐹) for all computable fields 퐹 with char 퐹 ≠ 2. On the other hand, the problem of finding such an isotropic vector depends in a serious way on the arithmetic of the field 퐹 [Voi2013, §4]. Chapter 8

Simple algebras and involutions

8.6 Algorithmic aspects

In section 4.6, we showed how to recover an isomorphism 퐵 ' M2 (퐹) from a nonzero nilpotent element 휖 ∈ 퐵 (or more generally, a zerodivisor 휖 ∈ 퐵). In a similar way, the proof of Proposition 8.2.3 can be made algorithmic: if 퐵1, 퐵2 are quaternion algebras over 퐹, then given a nonzero nilpotent element 휖 ∈ 퐵1 ⊗ 퐵2, we can exhibit explicitly a common embedded quadratic subfield. By the proof of Proposition 8.2.8, such a nilpotent element is given by a zero of the Albert form 푄(퐵1, 퐵2) when char 퐹 ≠ 2. From Exercise 8.2, we then find an explicit isomorphism 퐵1 ⊗ 퐵2 ' M2 (퐵3). We then have 퐵1 ' 퐵2 if and only if 퐵3 ' M2 (퐹), so from this method we have reduced the problem of testing for an isomorphism between quaternion algebras to the problem of splitting a quaternion algebra. For a more general point of view on the algorithmic problem of testing if two central simple algebras over a number field are isomorphic using norm equations, see work by Hanke [Hanke2007].

Part II

Arithmetic

Chapter 9

Lattices and integral quadratic forms

9.7 Normalized form

We give an algorithmic proof of Proposition 9.8.4. Algorithm 9.7.5. Let 푅 be a computable ring which is a local PID with (computable) valuation 푣 : 푅 → Z≥0 ∪ {∞}. Let 푄 : 푀 → 푅 be a quadratic form over 푅 and let 푒1, . . . , 푒푛 be a basis for 푀. This algorithm returns a basis of 푀 in which 푄 is normalized.

1. If 푇 (푒푖, 푒 푗 ) = 0 for all 푖, 푗, return 푓푖 := 푒푖. Otherwise, let (푖, 푗) with 1 ≤ 푖 ≤ 푗 ≤ 푛 be such that 푣푇 (푒푖, 푒 푗 ) is minimal, taking 푖 = 푗 if possible and if not taking 푖 minimal. × 2. If 푖 = 푗, let 푓1 := 푒푖 and proceed to Step 3. If 푖 ≠ 푗 and 2 ∈ 푅 , let 푓1 := 푒푖 + 푒 푗 and proceed to Step 3. Otherwise, proceed to Step 4. 3. Let 푒푖 := 푒1. For 푘 = 2, . . . , 푛 let

푇 ( 푓1, 푒푘 ) 푓푘 := 푒푘 − 푓1. 푇 ( 푓1, 푓1) Let 푚 = 2 and proceed to Step 5. 4. (We have 2 ∉ 푅× and 푖 ≠ 푗.) Let

휋푣 (푇 (푒푖 ,푒 푗 )) 푓1 := 푒푖, 푇 (푒푖, 푒 푗 )

2 푓2 := 푒 푗 , 푒푖 := 푒1 and 푒 푗 := 푒2. Let 푑 := 푇 ( 푓1, 푓1)푇 ( 푓2, 푓2) − 푇 ( 푓1, 푓2) . For 푘 = 3, . . . , 푛, let

푡푘 := 푇 ( 푓1, 푓2)푇 ( 푓2, 푒푘 ) − 푇 ( 푓2, 푓2)푇 ( 푓1, 푒푘 )

푢푘 := 푇 ( 푓1, 푓2)푇 ( 푓1, 푒푘 ) − 푇 ( 푓1, 푓1)푇 ( 푓2, 푒푘 ) and let 푡 푢 푓 := 푒 + 푘 푓 + 푘 푓 . 푘 푘 푑 1 푑 2 Let 푚 = 3.

13 14 CHAPTER 9. LATTICES AND INTEGRAL QUADRATIC FORMS

5. Recursively call the algorithm with 푀 = 푅 푓푚⊕· · ·⊕푅 푓푛, and return 푓1, . . . , 푓푚−1 concatenated with the output basis.

Given such a basis, one recovers the normalized quadratic form by factoring out in each atomic form the minimal valuation achieved. (One can also keep track of this valuation along the way in the above algorithm, if desired.) Remark 9.7.6. If 2 ∈ 푅×, then a quadratic form 푄 is atomic if and only if 푄(푥) = 푎푥2 for 푎 ∈ 푅×, so Algorithm 9.7.5 computes a diagonalization of the form 푄, ordering the coeﬃcients by their valuation.

Proof of correctness of Algorithm 9.7.5. In Step 3, we verify that

푣(푇 ( 푓1, 푓1)) ≤ 푣(푇 ( 푓1, 푒푘 )).

Indeed, 푇 ( 푓1, 푓1) = 푇 (푒푖, 푒푖) + 2푇 (푒푖, 푒 푗 ) + 푇 (푒 푗 , 푒 푗 ) and so 푣(푇 ( 푓1, 푓1)) = 푣(푇 (푒푖, 푒 푗 )) by the ultrametric inequality and the hypotheses that 푣(푇 (푒푖, 푒 푗 )) < 푣(푇 (푒푖, 푒푖)), 푣(푇 (푒 푗 , 푒 푗 )) and 푣(2) = 0. So Steps 2 and 3 give correct output. We have left to check Step 4. This is proven by letting 푓푘 = 푒푘 + 푡푘 푓1 + 푢푘 푓2 and solving the linear equations 푇 ( 푓1, 푓푘 ) = 푇 ( 푓2, 푓푘 ) = 0 for 푡푘 , 푢푘 . The result then follows from a direct calculation, coupled with the fact that 푣(푑) = 2푣(푇 ( 푓1, 푓2)) ≤ 푣(푡푘 ) (and similarly with 푢푘 ). This case only arises if (and only if)

푣(푇 ( 푓1, 푓2)) < 푣(푇 ( 푓1, 푓1)) = 푣(2푄( 푓1)) ≤ 푣(2푄( 푓2)) so the corresponding block is atomic. Algorithm 9.7.5 requires 푂(푛2) arithmetic operations in 푅, and can be modiﬁed suitably to operate directly on the Gram matrix (푇 (푒푖, 푒 푗 ))푖, 푗 of the quadratic form 푄. Chapter 12

Ternary quadratic forms over local ﬁelds

12.5 Algorithmic aspects

In this section, we discuss algorithms for computing the Hilbert symbol. For more details, see Voight [Voi2013, §5]. Let 퐹 be a local field with char 퐹 = 0 and 퐹 ; C. If 퐹 is nonarchimedean, let 푅 be the valuation ring, 픭 the maximal ideal, 휋 a uniformizer, 푘 := 푅/픭 the residue field, and ord: 퐹 → Z ∪ {∞} the valuation with ord픭 (휋) = 1. If 퐹 is archimedean, then 퐹 ' R and we let 푅 = 퐹 = 픭 and 휋 = −1 and let 푎 = (−1)ord(푎) |푎| for 푎 ∈ 퐹, so ord(푎) = 0, 1 according as 푎 > 0 or 푎 < 0. Remark 12.5.1. To be completely precise, a local field 퐹 is uncountable, so it is not computable. When we talk about computing in a local field, this can be interpreted either to mean we work in finite precision (and there are several ways to interpret this, giving different models) or we work in the algebraic closure of a global field (see section 14.4) inside its completion at a prime 픭. For 푎 ∈ 퐹×, we define the square symbol

×2 1, if 푎 ∈ 퐹 ; 푎  := −1, if 푎 ∉ 퐹×2 and ord(푎) is even; 퐹  × 0, if 푎 ∉ 퐹 2 and ord(푎) is odd.  푎 √ We have = −1 if and only if 퐹( 푎) is an unramified field extension of 퐹 and 퐹 푎 √ = 0 if and only if 퐹( 푎) is ramified; when 퐹 ' R is real, we follow the convention 퐹 that C is considered to be ramified over R. Accordingly, if 푣 is nonarchimedean, then 푎 = 0 if and only if ord(푎) is odd. The square symbol is not multiplicative. 퐹

15 16 CHAPTER 12. TERNARY QUADRATIC FORMS OVER LOCAL FIELDS

× Proposition 12.5.2. Let 푎, 푏 ∈ 퐹 . Then (푎, 푏)퐹 = 1 if and only if 푎 푏 −푎푏 푎 푏 −푎푏 = 1 or = 1 or = 1 or = = = −1. 퐹 퐹 퐹 퐹 퐹 퐹 Proof. The result is immediately veriﬁed if 퐹 is archimedean; if 퐹 is nonarchimedean, the result follows from (12.4.9). To conclude, we discuss the computability of the Hilbert symbol when char 푘 ≠ 2 using Proposition 12.5.2. We may suppose 퐹 is nonarchimedean. Then we can 푎 푎 evaluate by simply computing ord(푎) = 푒; if 푒 is odd then = 0, whereas 퐹 퐹 푎 푎 푎 푎 if 푒 is even then = 0 where 푎 = 푎휋−푒 ∈ 푅 and 0 = 0 is the usual 퐹 퐹 0 푣 픭 Legendre symbol, deﬁned by

0, if 푎0 ≡ 0 (mod 픭); 푎0  := 1, if 푎0 . 0 (mod 픭) and 푎0 is a square modulo 픭; . (12.5.3) 픭  −1, otherwise.  The Legendre symbol can be computed in deterministic polynomial time by Euler’s formula 푎 0 ≡ 푎 (푞−1)/2 (mod 픭) 픭 0 using repeated squaring, where 푞 = #푘. We ﬁnd that there exists a deterministic polynomial-time algorithm to evaluate the Hilbert symbol when char 푘 ≠ 2. The Hilbert symbol when char 푘 = 2 is somewhat more complicated; we follow Voight [Voi2013, §6].

푒 Algorithm 12.5.4. Suppose char(푘) = 2 and let 2푅 = 픭 so 푒 := ord픭 (2). Let 푎, 푏 ∈ 퐹 be such that ord픭 (푎) = 0 and ord픭 (푏) = 1. This algorithm outputs a solution to the congruence 1 − 푎푦2 − 푏푧2 ≡ 0 (mod 픭2푒) with 푦, 푧 ∈ 푅/픭2푒 and 푦 ∈ (푅/픭)×.

푓 1. Let 푓 ∈ Z≥1 be the residue class degree of 픭 (so that #(푅/픭) = 2 ) and let 푓 푞 = 2 . Let 휋 be a uniformizer√ at 픭. 2. Initialize (푦, 푧) := (1/ 푎, 0). 2 2 3. Let 푁 := 1 − 푎푦 − 푏푧 ∈ 푅/4푅 and let 푡 := ord픭 (푁). If 푡 ≥ 2푒, return 푦, 푧. Otherwise, if 푡 is even, let √︂ 푁 푦 := 푦 + 휋푡/2 푎휋푡 and if 푡 is odd, let √︂ 푁 푧 := 푧 + 휋 b푡/2c . 푏휋푡−1 Return to Step 3. 12.5. ALGORITHMIC ASPECTS 17

√ 2푒 × √ In this algorithm, when we write 푢 for 푢 ∈ (푅/픭 ) we mean a choice of a lift of 푢 ∈ (푅/픭)× to 푅/픭2푒. We reduce to the above Hensel lift by the following algorithm.

Algorithm 12.5.5. Let 픭 an even prime with ramiﬁcation index 푒 = ord픭 2 and let 푎, 푏 ∈ 퐹× be such that 푣(푎) = 0 and 푣(푏) ∈ {0, 1}. This algorithm outputs 푦, 푧, 푤 ∈ 푅/픭2푒 such that

1 − 푎푦2 − 푏푧2 + 푎푏푤2 ≡ 0 (mod 픭2푒) and 푦 ∈ (푅/픭)×. Let 휋 be a uniformizer for 픭.

1. If 푣(푏) = 1, return the output (푦, 푧, 0) of Algorithm 12.5.4 with input 푎, 푏. 푒 ×2 푒 ×2 2 푒 2. Suppose 푎 ∈ (푅/픭 푅) and 푏 ∈ (푅/픭 푅) . Let (푎0) 푎 ≡ 1 (mod 픭 ) and 2 푒 (푏0) 푏 ≡ 1 (mod 픭 ). Return

푦 := 푎0, 푧 := 푏0, 푤 := 푎0푏0.

3. Swap 푎, 푏 if necessary so that 푎 ∈ (푅/픭푒 푅)× \(푅/픭푒 푅)×2. Let 푡 be the largest integer such that 푎 ∈ (푅/픭푡 )×2 but 푎 ∉ (푅/픭푒)×2. Then 푡 is odd; write 2 푡 푎 = 푎0 + 휋 푎푡 with 푎0, 푎푡 ∈ 푅. Let 푦, 푧 be the output of Algorithm 12.5.4 with 0 0 input 푎 := 푎, 푏 := −휋푎푡 /푏. Return

1 휋 b푡/2c 푦휋 b푡/2c 푦0 := , 푧0 := , 푤0 := 푎0 푎0푧 푎0푧 (reswapping if necessary).

For a proof of correctness of these two algorithms, see Voight [Voi2013, Algo- rithms 6.2, 6.5]. Deﬁnition 12.5.6. We say that 휋−1 ∈ 퐹× is an inverse uniformizer for the prime −1 −1 픭 ⊆ 푅 if ord픭 (휋 ) = −1 and ord픮 (휋 ) ≥ 0 for all 픮 ≠ 픭. We are now prepared to evaluate the even Hilbert symbol. 푎, 푏 Algorithm 12.5.7. Let 퐵 = be a quaternion algebra with 푎, 푏 ∈ 퐹×, and let 픭 퐹 be an even prime of 퐹. This algorithm returns the value of the Hilbert symbol (푎, 푏)픭.

1. Scale 푎, 푏 if necessary by an element of Q×2 so that 푎, 푏 ∈ 푅. 2. Let 휋−1 be an inverse uniformizer for 픭. Let

푎 := (휋−1)2bord픭 (푎)/2c 푎 and 푏 := (휋−1)2bord픭 (푏)/2c 푏.

−1 2 If ord픭 푎 = ord픭 푏 = 1, let 푎 := (휋 ) (−푎푏). Swap if necessary so that ord픭 푎 = 0. 3. Call Algorithm 12.5.5, and let 푖0 := (1 + 푦푖 + 푧 푗 + 푤푖 푗)/2. Let 푓 (푇) = 푇 2 − 푇 + nrd(푖0) be the minimal polynomial of 푖0. If 푓 has a root modulo 픭, return 1. 0 0 0 2 0 4. Let 푗 := (푧푏)푖 − (푦푎) 푗 and let 푏 := ( 푗 ) . If ord푣 푏 is even, return 1, otherwise return −1. 18 CHAPTER 12. TERNARY QUADRATIC FORMS OVER LOCAL FIELDS

Proof of correctness. If in Step 2 we have a root modulo 픭, then by Hensel’s lemma, 0 푓 has a root 푡 ∈ 퐹픭, hence 푡 − 푖 is a zerodivisor and we return 1 correctly. Otherwise, 0 0 퐾픭 = 퐹픭 [푖 ] is the unramiﬁed ﬁeld extension of 퐹픭. We compute that trd( 푗 ) = 0 0 0 0 trd(푖 푗 ) = 0, so 퐵픭 '(퐾픭, 푏 | 퐹픭) and 퐵픭 is split if and only if ord픭 푏 is even. Chapter 14

Quaternion algebras over global ﬁelds

14.8 Algorithmic aspects

In this seciton, we show how to make the classification of quaternion algebras (Main Theorem 14.1.3, and more generally Main Theorem 14.6.1) algorithmic, giving a computable bijection between quaternion algebras and ramification sets. First, we showed in Proposition 14.2.7 how to exhibit explicitly a quaternion algebra 퐵 over Q with a given ramification set Ram 퐵 = Σ. In general, we need to be able to find a prime 푞 satisfying certain congruence conditions (14.2.11)–(14.2.12), and this may be done with a probabilistic algorithm. The generalization of this algorithm to number fields is as follows [GV2011, Algorithm 4.1]. We state Exercise 14.16 as an algorithm.

Algorithm 14.8.1. This algorithm takes as input a ﬁnite set Σ ⊂ Pl(퐹) of noncomplex places of a number ﬁeld 퐹 of even cardinality, and returns as output 푎, 푏 ∈ Z퐹 such 푎, 푏 that the quaternion algebra 퐵 = has Ram 퐵 = Σ. 퐹

Î 1. Let 픇 := 픭∈Σ 픭 be the product of the (ﬁnite) primes in Σ. Find 푎 ∈ 픇 such that for all real places 푣 we have 푣(푎) < 0 for all 푣 ∈ Ram 퐵 and such that 푎Z퐹 = 픇픟 with 픇 + 픟 = Z퐹 and 픟 odd. 2. Factor the ideal 픟 into primes. Find 푡 ∈ Z퐹 /8푎Z퐹 such that the following hold: 푡 a) For all primes 픭 | 픇, we have = −1; 픭 푡 b) For all primes 픮 | 픟, we have = 1; and 픮 푒 푒 c) For all prime powers 픯 k 8Z퐹 with 픯 - 픇, we have 푡 ≡ 1 (mod 픯 ).

3. Find 푚 ∈ Z퐹 such that 푏 := 푡 + 8푎푚 ∈ Z퐹 satisﬁes the following conditions: a) 푏 is prime (i.e., (푏) is a prime ideal); b) 푣(푏) < 0 for all real places 푣 ∈ Ram 퐵; and c) 푣(푏) > 0 for all real places 푣 ∉ Ram 퐵 such that 푣(푎) < 0.

19 20 CHAPTER 14. QUATERNION ALGEBRAS OVER GLOBAL FIELDS

푎, 푏 It can be veriﬁed in a manner similar to the proof over Q that the algebra 퐵 = 퐹 output by Algorithm 14.8.1 has the correct set of ramiﬁed places. The steps in Algorithm 14.8.1 for working with elements in 퐹 and ideals in the ring of integers Z퐹 are standard, and they are described in the books on computational algebraic number theory by Cohen [Coh93, Coh2000].

Remark 14.8.2. When possible, it is often helpful in practice to take 푎Z퐹 = 픇 in × Algorithm 14.8.1: for example, if 픇 = Z퐹 and there exists a unit 푢 ∈ Z퐹 with the right real signs as in Step 3 and such that 푢 ≡ 1 (mod 8), then we may simply take −1, 푢 퐵 = . In any event, in Step 2, one may find the element 푡 by deterministic or 퐹 probabilistic means; moreover, one may wish to be alternate between Steps 2 and 3 in searching for 푏. 14.8.3. Algorithm 14.8.1 may be generalized to the case where 퐹 is a global function field is analogous, but at the present time the literature is much less complete in describing a suite of algorithms for computing with integral structures in such fields analogous to those mentioned in section 9.8—particularly in the situation where one works in a relative extension of such fields. (See Hess [Hess2002] for a start.) There- fore, in this book we will often consider just the case of number fields and content ourselves to notice that the algorithms we provide will generalize with appropriate modifications to the global function field setting. 14.8.4. Next, the converse: given a quaternion algebra 퐵 = (푎, 푏 | 퐹) over a number field 퐹, we compute the ramification set Ram 퐵. For this, we simply factor 2 and (the numerator and denominator of) 푎 and 푏, and for each prime 픭 occuring in these factorizations and each real place 푣 of 퐹, we compute the corresponding Hilbert symbol as described in section 12.5. In the special case where this computation reveals that Ram 퐵 = ∅, we may ask ∼ further for an explicit isomorphism 퐵 −→ M2 (퐹). (See Voight [Voi2013, Section 7] for discussion of the algorithmic problem of “recognizing the matrix ring”.) There are several points of view on this problem, relating it to important algorithmic problems in algorithmic number theory. First, as discussed in 4.6, it is equivalent to compute a zerodivisor in 퐵. One method to find such a zerodivisor is to seek appropriate “small” integral elements. Second, as explained in section 5.5, one can equivalently find a rational point on a conic over 퐹; in the case 퐹 = Q, algorithms for this problem are due to Cremona–Rusin [CR2003] and Simon [Sim2005], and they run in probabilistic polynomial time given the factorization of 푎, 푏 ∈ Z (probabilism only occurs in the need to compute square roots modulo 푝). Aspects of these approaches have been extended to number fields, but there is no as yet definitive reference. Third and finally, by Main Theorem 5.4.4, one can also equivalently solve a norm equation in a relative quadratic extension 퐾 ⊇ 퐹; there are a number of approaches to this problem (see Simon [Sim2002] and the references therein, and Bartels [Bar80] for more on the theory). Unfortunately, the running time for these algorithms is often poor (as they involve the computation of an S-unit and S-class group). Nevertheless, this point of view generalizes cleanly to splitting central simple algebras over number fields (see work of Hanke [Hanke2007]). Chapter 15

Discriminants

15.7 Algorithmic aspects

Let 퐹 be a number ﬁeld and Z퐹 its ring of integers. Let 퐵 = (푎, 푏 | 퐹) be a quaternion algebra over 퐹, and let O ⊂ 퐵 be a Z퐹 -order. Recall we represent O by a pseudobasis (9.3.7) O = Z퐹 ⊕ 픞푖 ⊕ 픟 푗 ⊕ 픠푘 as in section 9.8.

15.7.1. The reduced discriminant discrd(O) can be computed using 15.2.8 and Lemma 15.4.7, without taking a square root:

discrd(O) = 픞픟픠 푚(푖, 푗, 푘).

We now discuss some algorithmic aspects of computing maximal orders in this setting, following Voight [Voi2013, §7]. We say an order O ⊂ 퐵 is 픭-maximal for a prime 픭 of Z퐹 if O픭 = O ⊗Z퐹 Z퐹,픭 is maximal. We begin with an order to start with: rescaling we may suppose 푎, 푏 ∈ Z퐹 , and then we may take O = Z퐹 h푖, 푗i where 푖, 푗 are standard generators. Given an order O, to compute a maximal order O0 ⊇ O we compute the reduced discriminant discrd(O), factor this ideal, and recursively compute a 픭-maximal order for every prime 픭 | discrd(O), proceeding in two steps.

Deﬁnition 15.7.2. An order O is 픭-saturated if nrd |O픭 has a normalized basis 1, 푖, 푗, 푘 (see Proposition 9.8.4) such that each atomic block has valuation at most 1; we then say that 1, 푖, 푗, 푘 is a 픭-saturated basis for O. We compute a 픭-saturated order in the following straightforward way.

Algorithm 15.7.3. Let O = Z퐹 ⊕ 픞푖 ⊕ 픟 푗 ⊕ 픠푘 ⊂ 퐵 be an order and let 픭 be prime. This algorithm computes a 픭-saturated order O0 ⊇ O and a 픭-saturated basis for O0.

1. Choose 푑 ∈ 픞 such that ord픭 (푑) = ord픭 (픞) and let 푖 := 푑푖; compute similarly with 푗, 푘. Let O0 := O.

21 22 CHAPTER 15. DISCRIMINANTS

2. Run Algorithm 9.7.5 over the localization Z퐹,(픭) with the quadratic form nrd ∗ ∗ ∗ on the basis 1, 푖, 푗, 푘; let 1, 푖 , 푗 , 푘 be the output. Let 푐 ∈ Z퐹 be such that ∗ ∗ ord픭 푐 = 0 and such that 푐푖 ∈ O, and let 푖 := 푐푖 ; compute similarly with 푗, 푘. −1 3. Let 휋 be an inverse uniformizer for 픭. For each atomic form 푄 in nrdO, let 푒 be the valuation of 푄, and multiply each basis element in 푄 by (휋−1) b푒/2c. 0 Return O := O + (Z퐹 푖 ⊕ Z퐹 푗 ⊕ Z퐹 푘) and the basis 1, 푖, 푗, 푘.

Proof of correctness. In Step 3, we verify that the output of Algorithm 9.7.5 leaves 1 as the first basis element. We note that ord픭 trd( 푗) ≤ ord픭 trd(푖(푖 푗)) since trd(푖(푖 푗)) = 2 trd(푖) − trd( 푗) nrd(푖) and similarly ord픭 trd(푖) ≤ ord픭 trd((푖 푗) 푗). Let 1, 푖, 푗, 푘 be the basis computed in Step 3. By definition, this basis is 픭- saturated; we need to show that O is an order. But O is an order if and only if O픮 is an order for all primes 픮, and O픮 = Λ픮 for all primes 픮 ≠ 픭. For all 훼, 훽 ∈ 퐵 we have 훼훽 + 훽훼 = trd(훽)훼 + trd(훼)훽 − 푇 (훼, 훽), so if O is an order then O + Z퐹 훼 is multiplicatively closed if and only if 푇 (훼, 훽) ∈ Z퐹 for all 훽 ∈ O. We have 푇 (훼, 훽) = 0 if 훼, 훽 are orthogonal, and if 훼, 훽 are a basis for an atomic block 푄 then by definition the valuation of 푇 (훼, 훽) is at least the valuation of 푄 and we can multiply each by (휋−1) b푒/2c, preserving integrality.

One can compute a 픭-maximal order as follows.

Algorithm 15.7.4. Let O be an order and let 픭 be prime. This algorithm computes a 픭-maximal order O0 ⊇ O.

1. Compute a 픭-saturated order O0 ⊇ O and let 1, 푖, 푗, 푘 be a 픭-saturated basis for O0. Let 휋−1 be an inverse uniformizer for 픭. 2 2. Suppose 픭 is odd. Swap 푖 for 푗 or 푘 if necessary so that 푎 := 푖 has ord픭 (푎) = 0. 푎 Let 푏 := 푗2. If ord 푏 = 0, return O0. Otherwise, if ord 푏 = 1 and = 1, 픭 픭 픭 solve 푥2 ≡ 푎 (mod 픭) −1 0 0 for 푥 ∈ Z퐹 /픭. Adjoin the element 휋 (푥 − 푖) 푗 to O , and return O . 3. Otherwise, 픭 is even. Let 푡 := trd(푖), let 푎 := − nrd(푖), and let 푏 := 푗2. 0 a. Suppose ord픭 푡 = 0. If ord픭 푏 = 0, return O . If ord픭 푏 = 1 and there is a 2 −1 solution 푥 ∈ Z퐹 to 푥 −푡푥 + 푎 ≡ 0 (mod 픭), and return O+Z퐹 휋 (푥 −푖) 푗. b. Suppose ord픭 trd(푖) > 0. Let 푦, 푧, 푤 be the output of Algorithm 12.5.5 with input 푎, 푏. Let

푖0 := (휋−1)푒 (1 + 푦푖 + 푧 푗 + 푤푖 푗).

Adjoin 푖0 to O, and return to Step 1.

Proof of correctness. At every step in the algorithm, for each prime 픮 ≠ 픭 the order O픮 does not change, so we need only verify that O픭 is a maximal order. In Step 2, 푏 is a uniformizer for 픭 and discrd(O픭) = 4푎푏Z퐹,픭. If ord픭 (푏) = 0 then ord픭 discrd(O픭) = 0 so O is maximal. Otherwise, discrd(O픭) = 픭 and 퐵픭 '(퐾픭, 푏 | 15.7. ALGORITHMIC ASPECTS 23

퐹픭) where 퐾픭 = 퐹픭 [푖]. We conclude that 퐵픭 is a division ring (and hence O픭 is maximal) if and only if (푎/픭) = −1. If (푎/픭) = 1 and 푗 0 = 휋−1 (푥 −푖) 푗, then 1, 푖, 푗 0, 푖 푗 0 0 2 −1 2 2 form the Z퐹,픭-basis for a maximal order, since ( 푗 ) = (휋 ) (푥 − 푎)푏 ∈ Z퐹,픭 and 푗 0푖 = −푖 푗 0. In Step 3, first note that 푖 푗 is also orthogonal to 1, 푖: since 푖 is orthogonal to 푗 we get trd(푖 푗) = 0, so 푖 푗 is orthogonal to 1, and similarly trd(푖 푗푖) = trd(nrd(푖) 푗) = 0. In particular, 퐵픭 = (퐾픭, 푏 | 퐹픭) where 퐾픭 = 퐹픭 [푖]. By a comparison of discriminants, using the fact that the basis is normalized, we see that 1, 푖, 푗, 푖 푗 is a 픭-saturated basis for O as well, so without loss of generality we may take 푘 = 푖 푗. Suppose first that ord픭 trd(푖) = 0, so we are in Step 3a. If ord픭 푏 = 0, then ord픭 discrd(O픭) = 0 so O픭 is maximal. If ord픭 푏 > 0, then since the basis is 픭- saturated, ord픭 푏 = 1. Thus as in the case for 픭 odd, 퐵픭 is a division ring if and only if −1 퐾픭 is not a field, and as above the adjoining the element 휋 (푥 − 푖) 푗 yields a maximal order. So suppose we are in Step 3b, so ord픭 trd(푖) > 0. Since 1, 푖, 푗, 푘 is normalized, 0 ord픭 trd(푖) = ord픭 푇 (1, 푖) ≤ ord픭 푇 ( 푗, 푘). Adjoining 푖 to O gives a Z퐹,픭-module 0 0 × 0 with basis 1, 푖 , 푗, 푖 푗 since 푦 ∈ (Z퐹 /픭) ; adjoining 푗 gives a module with basis 0 0 0 0 1, 푖 , 푗 , 푖 푗 for the same reason. We verify that O픭 after these steps is an order: 0 −1 푒 0 −1 2푒 2 2 2 trd(푖 ) = 2(휋 ) ∈ Z퐹,픭 and nrd(푖 ) = (휋 ) (1 − 푎푦 − 푏푧 + 푎푏푤 ) ∈ Z퐹,픭 by 0 2 0 construction, so at least Z퐹,픭 [푖] = Z퐹,픭 ⊕ Z퐹,픭푖 is a ring. Similarly ( 푗 ) = 푏 ∈ Z퐹,픭. Finally, trd(푖0푖) = 2(휋−1)푒 푦푎 and trd(푖0 푗) = 2(휋−1)푒푧푏; it follows that trd(푖0 푗 0) = 0, and hence 푗 0푖0 = −푖0 푗 0 = −푖0 푗 0 − trd(푖0) 푗 0, so we have an order.

Remark 15.7.5. In the proof of correctness for Algorithm 15.7.4, in each case where 픭 is ramiﬁed in 퐵 we have in fact written 퐵픭 '(퐾픭, 휋 | 퐹픭) where 퐾픭 is the unramiﬁed extension of 퐹픭. The reader will note the similarity between this algorithm and the algorithm to compute the Hilbert symbol: the former extends the latter by taking a witness for the fact that the algebra is split, namely a zerodivisor modulo 픭, and uses this to compute a larger order (giving rise therefore to the matrix ring).

15.7.6. Combining Algorithm 15.7.3 and 15.7.4, we have the following immediate consequence: if O ⊂ 퐵 is an order in a quaternion algebra 퐵 over a number ﬁeld 퐹 and 픭 is prime of Z퐹 which is unramiﬁed in 퐵, then there exists an algorithm to compute an explicit embedding O ↩→ M2 (O픭). Such an algorithm is sometimes called an algorithm to recognize the 푝-matrix ring.

The algorithmic complexity in factoring cannot be avoided in this context, according to the following theorem.

Theorem 15.7.7. For a ﬁxed number ﬁeld 퐹, the problem of computing maximal orders in quaternion algebras over 퐹 is probabilistic polynomial-time equivalent to the problem of factoring integers.

For a proof, see Voight [Voi2013, Theorem 7.15] (following a hint of Ronyai [Ron92, §6]). Remark 15.7.8. More generally, there are algorithms to compute maximal orders in semisimple algebras over number fields that run in deterministic polynomial time 24 CHAPTER 15. DISCRIMINANTS given oracles for the problems of factoring integers and factoring polynomials over finite fields: see Ivanyos–Rónyai [IR93, Theorem 5.3], Nebe–Steel [NS2009], and Friedrichs [Fri2000]. Chapter 17

Classes of quaternion ideals

17.9 Algorithmic aspects

In this section, we exhibit an algorithm to compute the size of the class set of an order in a totally definite quaternion algebra. A more sophisticated algorithm, inspired by the notion of Hecke operators acting on modular forms, will be discussed in Chapter 41; our goal in this section is just to try to convince the reader that computations can be carried out easily in practice. Let 퐹 be a number field with ring of integers 푅, let 퐵 be a division quaternion algebra over 퐹, and let O ⊂ 퐵 be an 푅-order. Then by Main Theorem 17.7.1, there is an effective constant 퐶 > 0 such that

ClsR O = {[퐼]R : 퐼 ⊆ O invertible and N(퐼) ≤ 퐶}. We compute Cls O in two steps: first, we compute the set of invertible integral O-ideals 퐼 ⊆ O with bounded absolute norm, and second we sort them according to their right class. For the first step, we first note that N(퐼) = N(nrd(퐼))2; we can loop over those ideals 픞 ⊆ 푅 with bounded N(픞) by factoring in 푅, and so it suffices to enumerate all invertible 퐼 ⊆ O with nrd(퐼) = 픞. In general, we appeal to a slight modification of Exercise 16.6: every such ideal is represented as 퐼 = 픞O + 훽O with 훽 ∈ O, and conversely the 푅-lattice 픞O + 훽O is locally principal if nrd(훽)푅 + 픞 = nrd(훽). Since 훽 is well-defined as an element of O/픞O, we can simply enumerate representatives of the finite quotient. We can stay more organized in our task by factoring. If 퐼 ⊆ O and 퐼 0 ⊆ O are invertible integral O-ideals with reduced norms nrd(퐼) = 픞 and nrd(퐼 0) = 픞0 such that 픞 + 픞0 = 푅, then 퐼 ∩ 퐼 0 is an invertible integral O-ideal with reduced norm 픞픞0: this follows by looking locally. Conversely, if 퐽 has reduced norm 픞픞0 with 픞 + 픞0 = 푅, then 퐼 = 픞O + 퐽 has reduced norm 픞. So it suffices to compute the set of ideals whose reduced norm is a power of a prime 픭. When 픭 is unramified in 퐵 and O is 픭-maximal for all primes 픭 | 픞, we have more direct control over the set of right ideals as follows. To compute the set of right O-ideals of reduced norm 픭푒 with 픭 unramified, using 15.7.6 we compute an embedding 휄픭 : O ↩→ M2 (푅픭), and then we take the set of ideals 푒 푒 퐼 = 픭 O + 훼O where 휄픭 (훼) is congruent to an element in the set (17.6.2) modulo 픭 .

25 26 CHAPTER 17. CLASSES OF QUATERNION IDEALS

Now we turn to the second step: given invertible right O-ideals 퐼, 퐽, we need to check if [퐼] = [퐽] ∈ Cls O. We ﬁrst appeal to (17.3.3): we see it is algorithmically equivalent to check if (퐽 : 퐼)L is principal. The colon ideal itself can be computed using standard methods for pseudobases. To check for principality, we follow Kirschmer– Voight [KV2010, Algorithm 4.10] and employ tactics from lattice reduction. Algorithm 17.9.1. Let 퐼 be an integral 푅-lattice and suppose that 퐼 is principal. Then this algorithm exhibits a generator for 퐼. 1. Compute nrd(퐼) ⊂ 푅 and let 푐 ∈ 푅 be such that nrd(퐼) = 푐푅. Initialize 훼 := 1. If 푐 ∈ 푅×, return 훼. 2. View 퐼 as a lattice equipped with the absolute reduced norm 푄 (17.7.11). Reduce 퐼 using the LLL algorithm [LLL82]. By exhaustively enumerating short elements in 퐼, ﬁnd 훾 ∈ 퐼 such that nrd(훾) = 푐푑 with N(푑) < N(푐). Let 훼 := 훾훼/푑, let 퐼 := 푑훾−1퐼, and let 푐 := 푑, and return to Step 2.

Proof. In Step 2 we have nrd(푑훾−1퐼) = 푑2/(푐푑) nrd(퐼) = 푑푅, and so if the algorithm terminates then it gives correct output by Lemma 16.3.8 since 푑훾−1퐼 = 훼O if and only if 퐼 = (훾훼/푑)O. The algorithm terminates because at each stage in Step 2, N(푑) < N(푐) a decreasing sequence of positive integers so this is executed only ﬁnitely many times, and if 퐼 is principal a generator will be found eventually by exhaustive enumeration. In practice, Algorithm 17.9.1 runs better than naive enumeration; however, we are unable to prove a rigorous runtime bound. With that proviso, given the generator 푐 as in Step 1, we can measure the value of the LLL-step as follows [KV2010, Lemma 4.11].

Lemma 17.9.2. There exists 퐶 ∈ R>0 such that for every invertible 푅-lattice 퐼, the ﬁrst basis element 훾 in the 퐿퐿퐿-reduced basis in Step 2 of Algorithm 17.9.1 satisﬁes

2 |Nm퐵/Q (nrd(훾))| ≤ 퐶N(discrd(OR (퐼)))· N(퐼).

Proof. The output of the LLL algorithm [LLL82, Proposition 1.9] is an element 훾 ∈ 퐼 which satisﬁes 푄(훾) ≤ 2(4푛−1)/2 covol(퐼)1/(2푛) . We argue as in Proposition 17.7.19:

covol(퐼) = N(퐼) covol(O) so by (17.7.18)

(4푛−1)푛 2 1 2푛 2 |Nm / (nrd(훾))| ≤ 푄(훾) ≤ N(퐼) covol(O) 퐹 Q 푛2푛 푛2푛 so the result follows taking

2(4푛−1)푛 covol(O) 퐶 = . 푛2푛 N(discrd(O)) 17.9. ALGORITHMIC ASPECTS 27

Lemma 17.9.2 indicates that, up to a constant depending on the quaternion algebra (choice of 푎, 푏), Algorithm 17.9.1 examines elements that are close to being generators. Now suppose that 퐵 is totally definite, so in particular 퐹 is totally real. Then we can improve on Algorithm 17.9.1 to provide a rigorous algorithm with an estimate on the running time, as follows [KV2010, Algorithm 6.3]. In this case, we defined the absolute reduced norm 푄 (17.7.11) independently of choices. For 푐 ∈ 퐹×, we define −1 푄푐 (훼) := 푄(푐 훼) for 훼 ∈ 퐵.

Algorithm 17.9.3. Let 퐵 be a totally deﬁnite quaternion algebra. Let 퐼 ⊂ O be an invertible 푅-lattice. This algorithm determines if 퐼 is principal and, if so, returns a generator for 퐼.

1. Compute nrd(퐼) ⊆ 푅 and test if nrd(퐼) is principal; if not, then return false. Otherwise, let 푐 ∈ 푅 be such that nrd(퐼) = 푐푅. Initialize 훼 := 1. × 2. Determine if there exists a unit 푢 ∈ Z퐹 such that 푣(푢푐) > 0 for all real places 푣. If so, let 푐 := 푢푐; if not, return false. × ×2 3. For each totally positive unit 푧 ∈ 푅>0/푅 : a. Let 훼 be a shortest vector of the lattice 퐼 with respect to the rational quadratic form 푄푢푐푧. b. If 휑푢푐푧 (훼) = 푛 then return true and the element 훼. 4. Return false.

Remark 17.9.4. Note that if 퐹 = Q then in Steps 3, 4 we have 푧 = 푢 = 1. Hence the algorithm simply looks for a shortest vector in the lattice 퐼 (with respect to the reduced norm form).

Proof of correctness of Algorithm 17.9.3. In Step 1, if 퐼 is principal, then nrd(퐼) is generated by a totally positive element 푢푐 where 푢 ∈ 푅×. Then Lemma (16.3.8) × implies that 훼 ∈ 퐼 generates 퐼 if and only if nrd 훼 = 푢푐푧 for some 푧 ∈ 푅>0. To ﬁnd such an element 훼, we only need to search for elements of norm 푢푐푧 where 푧 runs × ×2 through some arbitrary transversal of 푅>0/푅 . × −1 Let 푧 ∈ 푅>0 and 훼 ∈ 퐼. Then nrd 훼 ∈ nrd 퐼 = (푢푐푧)푅, so 푚 = (푢푐푧) (nrd 훼) ∈ 푅 is totally positive. The arithmetic-geometric mean inequality implies

1/푛 푛 ≤ 푛 Nm퐹/Q (푚) ≤ Tr퐹/Q 푚 = 푄푢푐푧 (훼).

0 Moreover, equality holds throughout if and only if 1 = Nm퐹/푄 푚 and 푣(푚) = 푣 (푚) for all real places 푣, 푣0, so equality holds if and only if 푚 = 1. Hence nrd(훼) = 푢푐푧 if and only if 훼 ∈ 퐼 satisﬁes 푄푢푐푧 (훼) = 푛 is a shortest vector.

Algorithm 17.9.3 runs in deterministic polynomial time in the size of the input for a fixed totally real field 퐹 [KV2010, Proposition 6.9]: Steps 1,2 involve some precomputation which can be done in constant time for fixed 퐹, and the shortest vector computation can be performed in constant time for a fixed dimensional lattice by employing the LLL-algorithm [LLL82] (see e.g., Kannan [Kan87, Section 3]). 28 CHAPTER 17. CLASSES OF QUATERNION IDEALS

17.9.5. By the surjective map ClsR O → Typ O in (17.4.14), these algorithms also give representatives for Typ O: they are the set of orders {OL (퐼) : [퐼] ∈ ClsR O}, since 0 0 −1 0 OL (퐼)' OL (퐼 ) for 퐼, 퐼 invertible right O-ideals if and only if OL (퐼) = 훼 OL (퐼 )훼 = −1 0 0 OL (훼 퐼 ) if and only if [퐼] = [퐼 ]. In other words, we need only check equality of the left orders. Chapter 20

Integral representation theory

20.5 Local Jacobson radical

We ﬁll in some of the proofs in this section.

Theorem 20.5.1. Let 휙 : O → O/픭O be reduction modulo 픭. Then

rad O = 휙−1 (rad O/픭O) ⊇ 픭O, and (rad O)푟 ⊆ 픭O for some 푟 > 0.

Proof. We follow Reiner [Rei2003, Theorem 6.15]. We ﬁrst show rad O ⊇ 픭O. Let 푀 = O푥 be a simple left O-module 20.4.2; then either 픭푀 = 푀 or 픭푀 = {0}. The former implies the latter by Nakayama’s lemma. Thus by deﬁnition, we conclude

픭O ⊆ rad O. (20.5.2)

Next, we have a surjective homomorphism 휙 : O → O/픭O. By Corollary 20.4.10, we have 휙(rad O) ⊆ rad(O/픭O) and an induced map

휙 : O/rad O → (O/픭O)/(rad O/픭O).

At the same time, we have an surjective map 휓 : O/픭O → O/rad O by (20.5.2). By Lemma 20.4.7, O/rad O is Jacobson semisimple, so by Corollary 20.4.10 again, 휓(rad(O/픭O)) = {0}. Thus 휓 factors through a surjective map

휓 : (O/픭O)/rad(O/픭O) → O/rad O.

Putting 휙 and 휓 together as surjective homomorphisms between ﬁnite-dimensional 푘- vector spaces, we conclude that both are isomorphisms, and rad O = 휙−1 (rad O/픭O). Finally, rad O/픭O is a nilpotent ideal by Lemma 7.4.8, so (rad O/픭O)푟 = {0} for some 푟 > 0. Thus 휙((rad O)푟 ) = {0}, and (rad O)푟 ⊆ 픭O.

Corollary 20.5.5. Let 퐼 ⊆ O be a two-sided ideal. Then the following are equivalent: (a) 퐼 ⊆ rad O;

29 30 CHAPTER 20. INTEGRAL REPRESENTATION THEORY

(b) 퐼푟 ⊆ rad O for some 푟 > 0; and (c) 퐼 is topologically nilpotent.

Proof. We follow Reiner [Rei2003, Exercise 39.1, Exercise 6.3]. The implication (i) ⇒ (ii) is immediate. For (ii) ⇒ (iii), by Theorem 20.5.1, we have (rad O)푠 ⊆ 픭O for some 푠 > 0. Therefore if 퐼푟 ⊆ rad O then 퐼푟푠 ⊆ (rad O)푠 ⊆ 픭O. Finally (iii) ⇒ (i), suppose 퐼푟 ⊆ 픭O for some 푟 > 0. Let 휙 : O → O/픭O be the reduction map. Then 휙(퐼) is a nilpotent ideal in O/픭O, so 휙(퐼) ⊆ rad(O/픭O); by Theorem 20.5.1, 퐼 ⊆ 휙−1 (rad O/픭O) = rad O.

20.6 Local integral representation theory

Let 퐽 = rad O; then O/퐽 is a semisimple 푘-algebra, by Corollary 20.5.2. We now follow Hijikata–Nishida [HN94, §1].

Lemma 20.6.13. Let 퐼 ⊆ 퐵 be a left O-submodule.

(a) 퐼 has a unique maximal O-submodule 퐼 0 ⊆ 퐼 if and only if 퐼/퐽퐼 is simple as a O/퐽-module. In this situation, 퐼 is indecomposable and 퐼 0 = 퐽퐼. (b) If 퐼 is projective, then it has a unique maximal O-submodule if and only if it is indecomposable.

Proof. Statement (a) follows since O/퐽 is semisimple. Statement (b) follows from Lemma 20.6.8.

Corollary 20.6.14. Let 푀 be a left O-module with a unique maximal O-submodule, and let 푉 := 푀 ⊗푅 퐹. Suppose that ℓ(푉) ≥ ℓ(퐵푒) for all primitive idempotents 푒 of O. Then 푀 is projective and indecomposable.

Proof. By Lemma 20.6.13(a), 푀/퐽푀 is simple and therefore 푀/퐽푀 ' 퐼/퐽퐼 for some projective indecomposable 퐼 by Lemma 20.6.8. Write 퐼 = O푒 where 푒 is a primitive idempotent, so we have a O-module homomorphism O푒 → 퐼/퐽퐼 ' 푀/퐽푀. Choosing a lift of the image of 푒, we get a map 휙 : O푒 → 푀. By Nakayama’s lemma, 휙 is surjective. Therefore induced map 휙퐹 : 퐵푒 → 푉 is a surjective 퐹- algebra homomorphism. But by hypothesis ℓ(퐵푒) ≥ ℓ(푉), so 휙퐹 is injective. Thus 휙 is injective and thus gives an isomorphism of 푀 with a projective indecomposable module.

Corollary 20.6.15. Let 푀 be a ﬁnitely generated left O-module. The following are equivalent:

(a) 푀 has a unique composition series; (b) 푀 ⊇ 퐽푀 ⊇ 퐽2 푀 ⊇ ... is a composition series; and (c) 퐽푖 푀/퐽푖+1 푀 is simple for all 푖 ≥ 0.

If these equivalent conditions hold, then 푉 := 푀 ⊗푅 퐹 is simple as a 퐵-module. 20.6. LOCAL INTEGRAL REPRESENTATION THEORY 31

Proof. The equivalences are immediate from Lemma 20.6.13. For the second statement, suppose 푉 is not simple, with 푉 ) 푊 ) {0}. We consider the O-submodule Ñ∞ 푖 푀 ∩ 푊 ⊆ 푀; since 푖=1 퐽 푀 = {0} and nontriviality, there exists 푟 ≥ 0 such that 푀 ∩ 푊 ⊆ 퐽푟 푀 but 푀 ∩ 푊 * 퐽푟+1 푀. Now consider the inclusion

퐽푟+1 푀 ( 퐽푟+1 푀 + (푀 ∩ 푊) ( 퐽푟 푀, (20.6.16) the strictness of inclusions following the hypotheses. This contradicts uniqueness of the composition series. 20.6.17. Suppose that 푀 has a unique composition series. Then 픭푀 ⊆ 푀 is a submodule, so by uniqueness 픭푀 = 퐽푡 푀 for some 푡 ≥ 1. We call 푡 the period of the composition series.

20.6.18. Let 퐼 ⊆ 퐵 be a left O-module. Suppose that 퐼/퐽퐼 is simple. Let 퐼 0 ⊇ 퐼 be a minimal O-supermodule. We claim that if 퐼 is the minimal O-supermodule of 퐽퐼, then 퐼 is the maximal O-submodule of 퐼 0. Indeed, 퐽퐼 0 ( 퐼 0 so by minimality, 퐽퐼 0 ⊆ 퐼; since 퐼/퐽퐼 is simple, we have either 퐽퐼 0 = 퐼 or 퐽퐼 0 = 퐽퐼. We rule out the former because then 퐼 0/퐽퐼 0 = 퐼 0/퐽퐼 ' 퐼 0/퐼 ⊕ 퐼/퐽퐼 is decomposable, so 퐼 is not the minimal O-supermodule.

Chapter 21

Hereditary and extremal orders

21.5 Classiﬁcation of local hereditary orders

We provide a proof of the following result. Theorem 21.5.1. Let 푅 be a complete DVR with 퐹 = Frac 푅. Let 퐵 be a ﬁnite- dimensional 퐹-algebra, and let O ⊆ 퐵 be an 푅-order. Let 퐽 = rad O. Then the following are equivalent, along with the conditions 0 where ‘left’ is replaced by ‘right’: (i) O is extremal; (ii) Every projective indecomposable left O-submodule 푃 ⊆ 퐵 is the minimum O-supermodule of 퐽푃; (iii) Every projective indecomposable left O-module 푃 has a unique composition series; (iv) Every projective indecomposable left O-module 푃 has a unique composition series consisting of projectives; (v) O is hereditary; (vi) 퐽 is projective as a left O-module; (vii) If 푃 is a projective indecomposable left O-module, then 퐽푃 is also projective indecomposable; and (viii) 퐽 is invertible as a (sated) two-sided O-ideal. Proof. We follow Hijikata–Nishida [HN94, §1]. (i) ⇒ (ii). Let 푃 = O푒 ⊆ O ⊆ 퐵 (see 20.6.4) and suppose that (ii) does not hold for 푃. Then there is a minimal O-supermodule 푀 ) 퐽푃 = 퐽푒 such that 푀 ≠ 푃. We cannot have 푀 ⊆ 푃 = O푒 by minimality, so 푀 * O by projection onto 퐵푒. Now 푀 + 퐽 ) 퐽 is a minimal O-supermodule because (푀 + 퐽)/퐽 ' 푀/퐽푒 and 푀 ⊇ 퐽푒 is minimal. Therefore by 20.6.12, we have 퐽(푀 + 퐽) ⊆ 퐽, i.e., 푀 + 퐽 ⊆ OR (퐽). But 푀 + 퐽 * O, so OR (퐽) ≠ O, contradicting Proposition 21.2.3. (ii) ⇒ (iii). Among the projective indecomposables 푃 = O푒, we choose 푃 so that ℓ(푉) is maximal, where 푉 = 퐹푃. By (ii), there exists 푃 ⊇ 푃1 a minimal O- supermodule. Then 푃 = 퐽푃1 is the maximum O-submodule of 푃1. By Corollary 20.6.14, 푃1 is projective indecomposable since ℓ(푉) is maximal, and repeating the process we get a period 푃 ⊆ 푃1 ⊆ · · · ⊆ 푃푟 where 푃푟 ⊆ 푃 as left O-modules,

33 34 CHAPTER 21. HEREDITARY AND EXTREMAL ORDERS

and 푃푟−1 = 퐽푃푟 , and 푃 has a unique composition series. Therefore 푉 is a simple 퐵-module, and ℓ(푉) = 1. Since ℓ(푉) ≥ ℓ(퐵푒) for all 푒, we conclude ℓ(퐵푒) = 1 for all idempotents 푒. Now let 푃 be a projective indecomposable. But we just showed that ℓ(퐹푃) ≥ ℓ(퐵푒) = 1 for all idempotents 푒, so the same argument applying Corollary 20.6.14 in the previous paragraph works for 푃. (iii) ⇒ (iv). The same argument as in the previous implication applies. (iv) ⇒ (v). Let 퐼 be an indecomposable left O-ideal; writing 퐼 as a direct sum of indecomposables, we may suppose 퐼 is indecomposable. The 퐵-module 퐹퐼 has a simple quotient. By (iv), 퐼 has unique composition series, so 퐹퐼 = 퐵푒 is simple. We therefore have an exact sequence of 퐵-modules

휙 0 → ker 휙 → 퐹퐼 −→ 퐵푒 → 0 which intersectiong with O gives an exact sequence of O-modules

휙 | 0 → 퐼 ∩ ker 휙 → 퐼 −−−→퐼 휙(퐼) → 0.

We have 휙(퐼) ⊆ 퐼; by (iv), iterating, we conclude 휙(퐼) is projective and so is a direct summand. But 퐼 is indecomposable, so 퐼 ' 휙(퐼) and 퐼 is projective. (v) ⇒ (vi). Immediate. (vi) ⇒ (iv). Let 푃 ' O푒 be a projective indecomposable; then 퐽푃 = 퐽푒 is the maximum O-submodule. By (vi), 퐽푃 is also projective; so we can iterate to a unique composition series. (vi) ⇒ (vii). We have 푃 = O푒 for an idempotent 푒, so 퐽푃 = 퐽푒 and 퐽 = 퐽푒⊕퐽(1−푒). É ⊕푛푖 (vii) ⇒ (vi). Write O ' 푖 푃푖 as a ﬁnite direct sum of indecomposable left É ⊕푛 O-ideals 푃푖. Then 퐽 ' (퐽푃푖) 푖 . By (vi), each 퐽푃푖 is projective indecomposable, so isomorphic to 푃휏 (푖) for some 휏(푖). Thus 퐽 is projective. (vii) ⇒ (i). We continue with the previous paragraph, so we know (vi) and hence (iv). Therefore 퐽푃푖 ⊆ 푃푖 is the unique projective indecomposable, and so 휏 must be a permutation of the indices. Therefore by 20.6.6, OL (퐽) = O. It follows that O is extremal, by Proposition 21.2.3. For the left-right symmetry in (i)–(vii), we note that extremal (i) is left-right symmetric, and we can repeat all of the above arguments on the right instead. 0 (i) + (vi) + (vi ) ⇔ (viii): by Theorem 20.3.3, we have OL (퐽) = OR (퐽) = O and O projective as a two-sided O-ideal if and only if 퐽 is invertible as a (sated) two-sided O-ideal. Chapter 22

Quaternion orders and ternary quadratic forms

22.6 ∗ A functorial inverse to the even Cliﬀord map

We have seen that the even Cliﬀord functor associates to a nondegenerate ternary quadratic module a quaternion 푅-order. In this section, we show how to do the converse, furnishing an inverse to the Cliﬀord functor. The construction is due to Voight [Voi2011a, §2], following Bhargava [Bha2004b] (who considered the case of commutative rings of rank 4) and a footnote of Gross–Lucianovic [GrLu2009, Footnote 2]. Let 푅 be a (noetherian) domain with 퐹 = Frac 푅. Let O ⊂ 퐵 be a projective 푅-order in a quaternion algebra 퐵 over 퐹. Lemma 22.6.1. O/푅 is projective of rank 3 as an 푅-module.

Proof. For every prime ideal 픭 of 푅, there exists a basis for the algebra O픭/픭O픭 over the ﬁeld 푅픭/픭푅픭 which includes 1, and by Nakayama’s lemma this lifts to a basis for O픭. In particular, the quotient O/푅 is locally free hence projective of rank 3. Proposition 22.6.2. There exists a unique quadratic map

Ó2 Ó4 휓 = 휓O : (O/푅) → O with the property that 휓(푥 ∧ 푦) = 1 ∧ 푥 ∧ 푦 ∧ 푥푦 (22.6.3) for all 푥, 푦 ∈ O. Proof. We first prove the proposition for 퐵 over 퐹; then the restriction of the map to O/푅 then has image in Ó4 O and satisfies (22.6.3). We first define the map on sets:

휑 : 퐵 × 퐵 → Ó4퐵 (푥, 푦) ↦→ 1 ∧ 푥 ∧ 푦 ∧ 푥푦.

35 36 CHAPTER 22. TERNARY QUADRATIC FORMS

The map 휑 descends to a map 휑 : 퐵/퐹 × 퐵/퐹 → Ó4 퐵. We have 휑(푎푥, 푦) = 휑(푥, 푎푦) for all 푥, 푦 ∈ 퐵 and 푎 ∈ 퐹. Furthermore, we visibly have 휑(푥, 푥) = 0 for all 푥 ∈ 퐵, so the map is well-defined on the source. Finally, the map 휑 when restricted to each variable 푥, 푦 separately yields a quadratic map 퐵/퐹 → Ó4 퐵. Now let 1, 푖, 푗, 푘 be an 퐹-basis for 퐵. Then 푖 ∧ 푗, 푗 ∧ 푘, 푘 ∧푖 is a basis for Ó2 (퐵/퐹). We define 휓 : Ó2 (퐵/퐹) → Ó4퐵 휓(푖 ∧ 푗) = 휑(푖, 푗) (22.6.4) 휓(푖 ∧ 푗 + 푗 ∧ 푘) = 휑(푖 − 푘, 푗) = 휑( 푗, 푘 − 푖) together with the cyclic permutations of (22.6.4); we obtain a uniquely defined quadratic map by scaling and 푅-bilinearity (see Remark 9.7.2). We claim that 휓(푥 ∧ 푦) = 휑(푥, 푦) for all 푥, 푦 ∈ 퐵. By definition and the skew commutativity relation (4.2.16), we have that this is true if 푥, 푦 ∈ {푖, 푗, 푘}. For every 푦 ∈ {푖, 푗, 푘}, consider the maps

Ó4 휑푦, 휓푦 : 퐵/퐹 → 퐵 푥 ↦→ 휑(푥 ∧ 푦), 휓(푥 ∧ 푦) restricted to the ﬁrst variable. Note that each of these maps are quadratic and they agree on the values 푖, 푗, 푘, 푖 − 푘, 푗 − 푖, 푘 − 푗, so they are equal. The same argument on the other variable, where now we may restrict 휑, 휓 with 푥 ∈ 퐵, proves the claim.

Ó2 Ó4 Deﬁnition 22.6.5. The quadratic module 휓O : (O/푅) → O in Proposition 22.6.2 is called the canonical exterior form of O.

Proposition 22.6.6. The association O ↦→ 휓O yields a functor from the category of projective quaternion orders over 푅, under isomorphisms to the category of nondegenerate ternary quadratic modules, under similarity.

Proof. An isomorphism 휙 : O → O0 of quaternion 푅-orders induces a similarity

휓 Ó2 (O/푅) / Ó4 O

o ∧2 휙 o ∧4 휙 휓0 Ó2 (O0/푅) / Ó4 O0 because for all 푥, 푦 ∈ O we have

(∧4휙)(휓(푥 ∧ 푦)) = 1 ∧ 휙(푥) ∧ 휙(푦) ∧ 휙(푥푦) = 1 ∧ 휙(푥) ∧ 휙(푦) ∧ 휙(푥)휙(푦) (22.6.7) = 휓0(휙(푥) ∧ 휙(푦)) = 휓0((∧2휙)(푥 ∧ 푦)) as desired. 22.6. ∗ A FUNCTORIAL INVERSE TO THE EVEN CLIFFORD MAP 37

22.6.8. Suppose that O is free with a good basis 1, 푖, 푗, 푘 (see 22.4.7) and multiplication laws (22.3.7). We now compute the canonical exterior form

Ó2 Ó4 휓 = 휓O : (O/푅) → O.

We choose bases, with

Ó2 ∼ (O/푅) −→ 푅( 푗 ∧ 푘) ⊕ 푅(푘 ∧ 푖) ⊕ 푅(푖 ∧ 푗) = 푅푒1 ⊕ 푅푒2 ⊕ 푅푒3 and the generator −1 ∧ 푖 ∧ 푗 ∧ 푘 for Ó4 O. With these identiﬁcations, the canonical exterior form 휓 : 푅3 → 푅 has

휓(푒1) = 휓( 푗 ∧ 푘) = 1 ∧ 푗 ∧ 푘 ∧ 푗 푘 = 1 ∧ 푗 ∧ 푘 ∧ (−푎푖) = 푎(−1 ∧ 푖 ∧ 푗 ∧ 푘) and

휓(푒1 + 푒2) − 휓(푒1) − 휓(푒2) = 휓(푘 ∧ (푖 − 푗)) − 휓( 푗 ∧ 푘) − 휓(푘 ∧ 푖) = −1 ∧ 푘 ∧ 푗 ∧ 푘푖 − 1 ∧ 푘 ∧ 푖 ∧ 푘 푗 = −푤(1 ∧ 푘 ∧ 푖 ∧ 푗) = 푤(−1 ∧ 푖 ∧ 푗 ∧ 푘).

Continuing in this way, we see that

휓(푥( 푗 ∧ 푘) + 푦(푘 ∧ 푖) + 푧(푖 ∧ 푗)) = 푄(푥푒1 + 푦푒2 + 푧푒3) = 푄(푥, 푦, 푧) with 푄(푥, 푦, 푧) = 푎푥2 + 푏푦2 + 푐푧2 + 푢푦푧 + 푣푥푧 + 푤푥푦, (22.6.9) so that 0 Clf (휓O)' O. (22.6.10) Therefore 휓 furnishes an inverse to the map in Main Theorem 22.4.1 when 푅 is a PID, and combined with Proposition 22.4.4 provides another proof of this theorem. In particular, we have obtained the same quadratic form as constructed in 22.4.8, so

♯ 푁 nrd(O ) ∼ 휓O (22.6.11)

In particular, from Proposition 22.4.12 we have for every nondegenerate ternary 3 quadratic form 푄 : 푅 → 푅 we have 휓Clf0 (푄) is similar to 푄. (We reprove this for general 푅 in the next section.)

22.6.12. Suppose that 푅 is a Dedekind domain with ﬁeld of fractions 퐹. By Exercise 22.5, there is a good pseudobasis for O

O = 푅 ⊕ 픞푖 ⊕ 픟 푗 ⊕ 픠푘. (22.6.13)

The canonical exterior form of O, by the same argument as in 22.6.8 but keeping track of scalars, is given by

휓O : 픟픠푒1 ⊕ 픞픠푒2 ⊕ 픞픟푒3 → 픞픟픠 38 CHAPTER 22. TERNARY QUADRATIC FORMS under the identiﬁcation Ó4 O −→∼ 픞픟픠 (22.6.14) 1 ∧ 푖 ∧ 푗 ∧ 푘 ↦→ −1; we again have

2 2 2 휓O (푥푒1 + 푦푒2 + 푧푒3) = 푎푥 + 푏푦 + 푐푧 + 푢푦푧 + 푣푥푧 + 푤푥푦 as in (22.6.9), but now with 푥, 푦, 푧 are restricted to their respective coeﬃcient ideals. Repeating the same argument as in 22.6.8, we obtain again a similarity

♯ nrd(O ) ∼ 휓O (22.6.15) as in (22.6.11).

22.7 ∗ Twisting and ﬁnal bijection

We are now ready for the final bijection, one that uses the canonical exterior form from the previous section. The even Clifford algebra and the canonical exterior form carry with them one global property (Steinitz class) that must be taken into account before we obtain an equivalence of categories. Briefly, in addition to similarities one must also take into account twisted similarities, obtained not by a global map but by twisting by an invertible module.

Deﬁnition 22.7.1. A quadratic module 푑 : 푃 → 퐼 with 푃, 퐼 projective of rank 1 is called a twisting quadratic module if the associated bilinear map 푃 ⊗ 푃 → 퐼 is an 푅-module isomorphism.

Example 22.7.2. The quadratic module 푑 : 푅 → 푅 by 푧 ↦→ 푧2 is a (trivial) twisting. If 푃 is an invertible 푅-module, then the quadratic module

푃 → 푃⊗2 (22.7.3) 푧 ↦→ 푧 ⊗ 푧 is twisting.

Deﬁnition 22.7.4. Let 푄 : 푀 → 퐿 be a quadratic module and let 푑 : 푃 → 퐼 be a twisting quadratic module. The twist of 푄 by 푑 is the quadratic module

푄 ⊗ 푑 : 푀 ⊗ 푃 → 퐿 ⊗ 퐼 푥 ⊗ 푧 ↦→ 푄(푥) ⊗ 푑(푧).

A twisted similarity between quadratic modules 푄 : 푀 → 퐿 and 푄0 : 푀 0 → 퐿0 is tuple ( 푓 , ℎ, 푑) where 푑 : 푃 → 퐼 is a twisting quadratic module and ( 푓 , ℎ) is a similarity between 푄 ⊗ 푑 and 푄0. 22.7. ∗ TWISTING AND FINAL BIJECTION 39

Example 22.7.5. Let 푄 : 푀 → 퐿 be a quadratic module and let 픞 ⊆ 푅 be an invertible fractional ideal of 푅. Then 푑 : 픞 → 픞2 is twisting, and the twist of 푄 by 픞 can be identiﬁed with

푄 ⊗ 푑 : 픞푀 → 픞2 퐿 푧푥 ↦→ 푧2푄(푥).

If 픞 = 푎푅 is principal, then 푄 is similar to 푄 ⊗ 푑 via the similarity ( 푓 , ℎ) obtained by scaling by 푎. However, if 픞 is not principal, then 푄 may not be similar to 푄 ⊗ 푑.

Lemma 22.7.6. Let 푄 : 푀 → 퐿 be a quadratic module and let 푑 : 푃 → 퐼 be twisting. Then there is a canonical isomorphism of 푅-algebras

Clf0 (푄) −→∼ Clf0 (푄 ⊗ 푑).

Proof. First, we have a canonical isomorphism

(푀 ⊗ 푃) ⊗ (푀 ⊗ 푃) ⊗ (퐿 ⊗ 퐼)∨ −→∼ 푀 ⊗ 푀 ⊗ 퐿∨ (22.7.7) coming from rearranging, the canonical map 푑 : 푃 ⊗ 푃 → 퐼 followed by the evaluation map 퐼 ⊗ 퐼∨ −→∼ 푅. Now recall the deﬁnition of the even Cliﬀord algebra Clf0 (푄) and (22.3.3): Clf0 (푄) = Ten0 (푀; 퐿)/퐼0 (푄). The canonical isomorphism (22.7.7) induces an isomorphism Ten0 (푀 ⊗ 푃; 퐿 ⊗ 퐼) that ∼ maps 퐼0 (푄 ⊗ 푑) −→ 퐼0 (푄), and the result follows.

We are now ready to state the ﬁnal result of this chapter.

Main Theorem 22.7.8. Let 푅 be a noetherian domain. Then the associations ( Nondegenerate ternary quadratic ) ( Projective quaternion ) modules over 푅 ↔ orders over 푅 up to up to twisted similarity isomorphism (22.7.9) 푄 ↦→ Clf0 (푄)

휓O ← 휓O are mutually inverse, discriminant-preserving bijections that are also functorial with respect to 푅.

Before we begin the proof of the theorem, we need one preliminary lemma.

Lemma 22.7.10. Let 푀 be a projective 푅-module of rank 3. Then there are canonical isomorphisms

⊗ Ó3 Ó2 푀 −→∼ Ó3 푀 2 (22.7.11) Ó2 Ó2 푀 −→∼ 푀 ⊗ Ó3 푀. (22.7.12) 40 CHAPTER 22. TERNARY QUADRATIC FORMS

Proof. The proof is a bit of fun with multilinear algebra; the details are requested in Exercise 22.14. To illustrate, we give the proof in the special case where 푀 is completely decomposable 푀 = 픞푒1 ⊕ 픟푒2 ⊕ 픠푒3—so in particular the result holds when 푅 is a Dedekind domain (and, more generally, see Remark 22.7.13). In this case, we have Ó2 푀 ' 픞픟(푒1 ∧ 푒2) ⊕ 픞픠(푒1 ∧ 푒3) ⊕ 픟픠(푒2 ∧ 푒3) and so Ó3 Ó2 2 푀 '(픞픟픠) (푒1 ∧ 푒2) ∧ (푒1 ∧ 푒3) ∧ (푒2 ∧ 푒3) agreeing with Ó3 ⊗2 2 ⊗2 푀 '(픞픟픠) (푒1 ∧ 푒2 ∧ 푒3) by rearranging the tensor in a canonical way. The second isomorphism follows similarly.

Remark 22.7.13. There is a general method, called the splitting principle (see e.g. Elman–Karpenko–Merkurjev [EKM2008, Proposition 53.13] or Fulton [Ful96, §2]), that allows one to reduce questions about modules (or vector bundles) to the case of a sum of invertible modules (line bundles). For more on the connection to symmetric powers of wedge powers of modules and the relationship to problem of inner plethysm (related to Proposition 22.6.2), see Weyman [Wey2003, p. 63]. For a 퐾-theory version of the splitting principle, see Atiyah [Ati67, Corollary 2.7.11].

22.7.14. The canonical map

Ó3 O → Ó4 O 푥 ∧ 푦 ∧ 푧 ↦→ 1 ∧ 푥 ∧ 푦 ∧ 푧 by Lemma 22.6.1 induces a canonical 푅-module isomorphism

Ó3 (O/푅) −→∼ Ó4 O of invertible 푅-modules.

Proof of Main Theorem 22.7.8. We have two functors, by Theorem 22.3.1 and Propo- sition 22.6.6, that are functorial with respect to the base ring 푅. We now compose them. Let 푄 : 푀 → 퐿 be a quadratic module with O = Clf0 (푄), and consider its canonical exterior form 휓 : Ó2 (O/푅) → Ó4 O. As 푅-modules, we have canonically O/푅 ' Ó2 푀 ⊗ 퐿∨. (22.7.15) By the isomorphism 22.7.12 of Lemma 22.7.10, we have as the domain of 휓 the 푅-module

Ó2 (O/푅)' Ó2 (Ó2 푀 ⊗ 퐿∨)' Ó2 (Ó2 푀) ⊗ (퐿∨) ⊗2 (22.7.16) ' 푀 ⊗ Ó3 푀 ⊗ (퐿∨) ⊗2 22.8. ∗ RIGIDIFYING WITH ISOMETRIES AND CLASS SETS 41 and as codomain we have by 22.7.14 and the isomorphism (22.7.11) of Lemma 22.7.10

Ó4 O ' Ó3 (O/푅)' Ó3 (Ó2 푀 ⊗ 퐿∨)' Ó3 (Ó2 푀) ⊗ (퐿∨) ⊗3 (22.7.17) '(Ó3 푀) ⊗2 ⊗ (퐿∨) ⊗3. Now we twist. Let 푃 := Ó3 푀 ⊗ (퐿∨) ⊗2 and let 푑 : 푃∨ → (푃∨) ⊗2 be the natural twisting quadratic module. Then the twist 휓 ⊗ 푑 has domain and codomain canonically isomorphic to

Ó2 (O/푅) ⊗ 푃∨ ' 푀 (22.7.18) Ó4 O ⊗ (푃∨) ⊗2 ' 퐿 by (22.7.16)–(22.7.17) so we have a quadratic form 휓Clf0 (푄) ⊗ 푑 : 푀 → 퐿. 0 We show that the composition 푄 ↦→ Clf (푄) = O ↦→ 휓O is naturally isomorphic to the identity, via the twist 푑. But to do this (and show the induced maps are similarities), since the above construction is canonical, we can base change to 퐹 and check within the quadratic space 푄퐹 : 푉 → 퐹 where 푉 := 푀 ⊗푅 퐹. Choosing a basis, we find that the composition is the identity by 22.6.8. We may then conclude that the map of sets in (22.7.9) is a well-defined bijection: functoriality shows that the map is well-defined and that both maps are injective, and the composition shows that it is surjective.

22.7.19. One can also prove Main Theorem 22.7.8 by extending the deﬁnition in 22.4.8 using the reduced norm to a domain 푅 as follows: we deﬁne

nrd♯ (O) : (O♯)0 → discrd(O)−1 훼 ↦→ nrd(훼).

The fact that (O♯)0 is projective of rank 3, that discrd(O) is invertible as an 푅-module, and that the quadratic module takes values in discrd(O) follow locally, the latter from (22.4.14). Locally, this form is similar to the canonical exterior form (22.6.11), so this should come as no surprise.

Remark 22.7.20. When 2 ∈ 푅× (and more generally for certain schemes with 2 invertible), Balmer–Calmès [BC2012] develop a categorical theory of lax-similitude that coincides with our notion of twisting.

22.8 ∗ Rigidifying with isometries and class sets

In this section, we shift to isometries and upgrade the bijection to an equivalence of categories; this has the further consequence of relating the class set of a ternary quadratic module (Deﬁnition 9.7.13) and the type set of a quaternion order. Throughout, 푅 is a noetherian domain with 퐹 = Frac 푅. 42 CHAPTER 22. TERNARY QUADRATIC FORMS

Remark 22.8.1. In section 5.6, we used orientations to get an equivalence of categories: the crux of the problem being that the isometry −1 : 푉 → 푉 maps to the identity map on Clf0 푉 under the Clifford functor. However, −1 acts nontrivially on the odd Clifford module Clf1 푉, and the point of an orientation is to keep track of this action on piece of Clf1 푉 coming from the center. In this section, we avoid orientations and add a different structure on quaternion orders to allow for these extra morphisms. Let 푄 : 푀 → 퐿 be a nondegenerate quadratic module with 푀 of finite odd rank 푛. When we restrict to morphisms as isometries, we require that the map act as the identity on the codomain 퐿; to this end, we will have to somehow remember this codomain, and its compatibility with the other structures of the even Clifford algebra. We make the following definition. Definition 22.8.2. Let 푁 be an invertible 푅-module. A parity factorization of 푁 is a pair of invertible 푅-modules 푃, 퐿 and an isomorphism 푝 : 푁 −→∼ 푃⊗2 ⊗ 퐿. If 푁 has a parity factorization, we call it paritized. An isomorphism (푁, 푝) to (푁 0, 푝0) of paritized invertible 푅-modules is a pair of isomorphisms 푁 ' 푁 0, 푃 ' 푃0 such that the diagram

푝 푁 / 푃⊗2 ⊗ 퐿

o o (22.8.3) 푝0 푁 0 / 푃0⊗2 ⊗ 퐿 commutes. We do want to ﬁx the module 퐿 in order to work with isometries; this serves as an anchor for our construction.

Definition 22.8.4. Let O be a quaternion 푅-order. Then O is paritized if O is equipped with a parity factorization of Ó4 O. An isomorphism of paritized quaternion orders is an isomorphism 휙 : O ' O0 and an isomorphism of parity factorizations with the isomorphism Ó4 O ' Ó4 O0 given by ∧4휙. 22.8.5. Every invertible 푅-module 푁 has the identity parity factorization, with 푃 = 푅 and 퐿 = 푁; up to isomorphism, every other differs by a choice of isomorphism class of 푃. So parity factorizations can be thought of as factorizations according to parity inside Pic 푅. 22.8.6. The main example of a parity factorization comes from the even Clifford construction. Let 푄 : 푀 → 퐿 be a nondegenerate ternary quadratic module. Let O = Clf0 (푄), and let 푃 = Ó3 푀 ⊗ (퐿∨) ⊗2. Then by (22.7.17), we have a canonical parity factorization

Ó4 Ó3 ⊗2 ∨ ⊗3 ⊗2 푝푄 : O '( 푀) ⊗ (퐿 ) ' 푃 ⊗ 퐿. (22.8.7)

Lemma 22.8.8. Let (O, 푝) be a paritized quaternion 푅-order. Then

Aut푅 (O, 푝)' Aut푅 (O) × {±1}. 22.8. ∗ RIGIDIFYING WITH ISOMETRIES AND CLASS SETS 43

Proof. By deﬁnition, an automorphism of (O, 푝) is a pair of automorphisms (휙, ℎ) with 휙 ∈ Aut푅 (O) and ℎ ∈ Aut푅 (푃) such that the diagram

푝 Ó4O / 푃⊗2 ⊗ 퐿

∧4 휙 o o 푝 Ó4O / 푃⊗2 ⊗ 퐿 commutes. The choice 휙 = idO and ℎ = −1 gives us an automorphism we denote by −1. Since 퐿 is fixed, it plays no role in this part. We claim that if 휙 ∈ Aut푅 (O) is an 푅-algebra isomorphism, then there is a unique ℎ ∈ Aut푅 (푃) up to −1 such that (휙, ℎ) ∈ Aut푅 (O, 푝). We will prove this over each localization 푅(픭) , including 푅(0) = 퐹: once we choose such an ℎ over 퐹, it follows that ±ℎ ∈ Aut푅(픭) (푃(픭) ) and hence by intersecting ℎ ∈ Aut푅 (푃). So we may suppose O is free with good basis and 푃 ' 푅, and Ó4휙 acts by det 휙; we want to show that det 휙 = ℎ2 is a square. This follows from 22.3.15: we have 휙 = adj(휌) where 3 휌 ∈ GL3 (푅) is the action on the ternary quadratic module 푀 ' 푅 associated to O, and det 휙 = det adj(휌) = (det 휌)2. 22.8.9. We recall the twist construction employed in (22.7.17) that we now define in the above terms. Let (O, 푝) be a paritized quaternion 푅-order, with the parity factorization 푝 : Ó4 O −→∼ 푃⊗2 ⊗ 퐿. Ó2 Ó4 Let 휓O, 푝 : (O/푅) → O be the canonical exterior form. We then define the quadratic module

∨ Ó2 ∨ 휓O, 푝 := 푝 ◦ 휓O ⊗ 푃 : (O/푅) ⊗ 푃 → 퐿 (22.8.10) where 푝 induces an isomorphism Ó4 O ⊗ (푃∨) ⊗2 ' 퐿. We now have the following theorem. Theorem 22.8.11. The associations

0 푄 ↦→ (Clf (푄), 푝푄)

휓O, 푝 ← (O, 푝) are functorial and provide a discriminant-preserving equivalence of categories between

nondegenerate ternary quadratic modules over 푅 under isometries and

paritized projective quaternion 푅-orders under isomorphisms that is functorial in 푅. 44 CHAPTER 22. TERNARY QUADRATIC FORMS

Proof. First, we show the associations are functorial. If 푄 : 푀 → 퐿 and 푄0 : 푀 0 → 퐿 are isometric (nondegenerate ternary) quadratic modules under 푓 : 푀 → 푀 0, then by functoriality of the even Cliﬀord algebra, this induces an isomorphism 푓 : O ' O0 and thereby an isomorphism

푃 = Ó3 푀 ⊗ (퐿∨) ⊗2 ' 푃0 = Ó3 푀 0 ⊗ (퐿∨) ⊗2 and then of parity factorizations

푝푄 Ó4 O / 푃⊗2 ⊗ 퐿 .

o o (22.8.12)

푝푄0 Ó4 O0 / 푃0⊗2 ⊗ 퐿

Conversely, if (O, 푝) and (O0, 푝0) are isomorphic paritized quaternion 푅-orders under 휙 : O → O0 and 푃 ' 푃0, then we get an isometry

휓O, 푝 Ó2O/푅 ⊗ 푃∨ / 퐿

o (22.8.13)

휓O0, 푝0 Ó2O0/푅 ⊗ (푃0)∨ / 퐿 by Proposition 22.6.6 (see (22.6.7)): the similitude factor is the identity by construction. We now tackle the two compositions and show they are each naturally isomorphic to the identity. Let (O, 푝) be a paritized quaternion 푅-order. We ﬁrst associate Ó2 ∨ (휓O, 푝, id), and let 푀 = O/푅 ⊗ 푃 be the domain of 휓O, 푝. We then associate its even Cliﬀord algebra. As 푅-modules, we have canonical isomorphisms

0 Ó2 ∨ Ó2 ∨ ∨ Clf (휓O, 푝) = 푅 ⊕ 푀 ⊗ 퐿 = 푅 ⊕ (O/푅 ⊗ 푃 ) ⊗ 퐿 ' 푅 ⊕ Ó2 (O/푅) ⊗ (푃⊗2 ⊗ 퐿)∨ (22.8.14) ' 푅 ⊕ O/푅 ⊗ Ó3 (O/푅) ⊗ (푃⊗2 ⊗ 퐿)∨ ' 푅 ⊕ O/푅 where we have used 22.7.12 and in the last step we used the parity factorization 푝 giving a natural isomorphism of the last piece to 푅. To check that the corresponding map is an 푅-algebra homomorphism, by functoriality we can do so over 퐹, and we suppose that 퐵 is given by a good basis, and then the verification is as in 22.6.8. To finish, we show that the parity factorization is also canonically identified: we have

Ó3 푀 ⊗ (퐿∨) ⊗2 = Ó3 (Ó2 (O/푅) ⊗ 푃∨) ⊗ (퐿∨) ⊗2 ' Ó3 (Ó3O/푅) ⊗2 ⊗ (푃∨) ⊗3 ⊗ (퐿∨) ⊗2 ' 푃 22.8. ∗ RIGIDIFYING WITH ISOMETRIES AND CLASS SETS 45 where now we use (22.7.11) and then again (twice) the parity factorization. Therefore we have a natural isomorphism of parity factorizations

푝휓 Ó4 0 O, 푝 Ó3 ∨ ⊗2 Clf (휓O, 푝) / ( 푀 ⊗ 퐿 ) ⊗ 퐿

o o (22.8.15) 푝 Ó4 O / 푃⊗2 ⊗ 퐿

This completes the veriﬁcation that the composition in this order is naturally isomorphic to the identity. Now for the second composition. Let 푄 : 푀 → 퐿 be a (nondegenerate ternary) 0 quadratic module over 푅. We associate O = Clf 푄 and 푝푄 its parity factorization, and then 휓O, 푝푄 . In (22.7.18), using (22.7.16)–(22.7.17), we showed that we had an natural isometry

휓O, 푝푄 Ó2O/푅 ⊗ 푃∨ / 퐿

푓 o (22.8.16) 푄 푀 / 퐿 and this completes the proof.

Exercises

I 14. Prove Lemma 22.7.10, as follows. Let 푅 be a noetherian domain and let 푀 be a projective 푅-module of rank 3. We ﬁrst show that Ó3 (Ó2 푀) −→(∼ Ó3 푀) ⊗2. Deﬁne the map

⊗ 푠 : 푀∧6 → Ó3 푀 2 푥 ∧ 푥0 ∧ 푦 ∧ 푦0 ∧ 푧 ∧ 푧0 ↦→ (푥 ∧ 푥0 ∧ 푦0) ∧ (푦 ∧ 푧 ∧ 푧0) − (푥 ∧ 푥0 ∧ 푦) ∧ (푦0 ∧ 푧 ∧ 푧0)

with 푥, 푥0, 푦, 푦0, 푧, 푧0 ∈ 푀. (a) Show that 푠 descends to a map from (Ó2 푀) ⊗3. (b) Show that 푠 descends to a map from Ó3 (Ó2 푀). [Hint: Localize so 푀 is free with basis 푒1, 푒2, 푒3, and argue by linearity.] (c) Conclude that the induced map is an isomorphism. (d) Repeat to show that Ó2 (Ó2 푀) −→∼ 푀 ⊗ Ó3 푀 via the map

푀 ⊗4 → 푀 ⊗ Ó3 푀 (22.8.17) 푥 ⊗ 푥0 ⊗ 푦 ⊗ 푦0 ↦→ 푥0 ⊗ (푥 ∧ 푦 ∧ 푦0) − 푥 ⊗ (푥0 ∧ 푦 ∧ 푦0).

15. Let 푄 : 푀 → 퐿 be a quadratic module. 46 CHAPTER 22. TERNARY QUADRATIC FORMS

(a) Suppose that 푄0 : 푀 0 → 퐿0 is similar to 푄. Show that 푄 is primitive if and only if 푄0 is primitive. (b) Let 푑 : 푃 → 퐼 be a twisting quadratic module. Show that 푄 is primitive if and only if 푄 ⊗ 푑 is primitive. 0 16. Let 푄 be a nondegenerate ternary quadratic module over 푅 and let (Clf (푄), 푝푄) be the associated paritized quaternion 푅-order by Theorem 22.8.11. Show that 0 0 O(푄)(푅)' Aut푅 (Clf (푄), 푝푄) and SO(푄)(푅)' Aut푅 (Clf (푄)). 17. Let O be a quaternion 푅-order. Then the standard involution deﬁnes an isomorphism : O −→∼ Oop. This isomorphism induces a self-similarity on the exterior form 푄 : Ó2 (O/푅) → Ó4 O, since O = Oop as 푅-modules. Compute the similarity factor of this self-similarity. Part III

Analysis

Chapter 26

Classical zeta functions

26.2 Analytic class number formula

We provide a proof of the following technical result.

푛 Theorem 26.2.19. Let 푋 ⊆ R be a cone. Let 푁 : 푋 → R>0 be a function satisfying

푛 푁 (푡푥) = 푡 푁 (푥) for all 푥 ∈ 푋, 푡 ∈ R>0.

Suppose that 푛 푋≤1 := {푥 ∈ 푋 : 푁 (푥) ≤ 1} ⊆ R (26.2.20) 푛 푛 is a bounded subset with volume vol(푋≤1). Let Λ ⊆ R be a (full) Z-lattice in R , and let ∑︁ 1 휁 (푠) := . Λ,푋 푁 (휆)푠 휆∈푋∩Λ

Then 휁Λ,푋 (푠) converges for Re 푠 > 1 and has a simple pole at 푠 = 1 with residue

∗ vol(푋≤1) 휁Λ,푋 (1) = lim(푠 − 1)휁Λ,푋 (푠) = . 푠&1 covol(Λ)

Proof. We have

( 1 Λ ∩ ) vol(Λ) 1 # 푡 푋≤1 vol(푋≤1) = lim #( Λ ∩ 푋≤1) = vol(Λ) lim . 푡→∞ 푡푛 푡 푡→∞ 푡푛 By the homogeneity condition on 푁,

1 #( 푡 퐿 ∩ 푋≤1) = #(퐿 ∩ 푋≤푡 푛 ).

Label the points of Λ ∩ 푋 = {휆1, 휆2,...} so that 푁 (휆1) ≤ 푁 (휆2) ≤ ...; we claim that 푘 vol(푋 ) lim = ≤1 = 푣. 푘→∞ 푁 (휆푘 ) covol(Λ)

49 50 CHAPTER 26. CLASSICAL ZETA FUNCTIONS

To prove this claim, write 푏(푥) = #(Λ ∩ 푋≤푥푛 ) for 푥 > 0. From the previous 푛 푛 paragraph, 푏(푥)/푥 → 푣. Let 푥푘 = 푁 (휆푘 ) for 푘 ≥ 1. Then for all 휖 > 0, we have 푏(푥푘 − 휖) < 푘 ≤ 푏(푥푘 ). So 푏(푥 − 휖) 휖 푘 푏(푥 ) 푘 1 − < ≤ 푘 . ( − )푛 푛 ( ) 푛 푥푘 휖 푥푘 푁 휆푘 푥푘 푛 Taking the limit as 푘 → ∞, since 푥푘 → ∞ we have lim푘→∞ 푘/푁 (휆푘 ) = 푣 by the sandwich theorem, proving the claim. Now, for all 휖 > 0, there exists 퐾 such that for 푘 ≥ 퐾 we have 1 1 1 ( − )푠 ( + )푠 푣 휖 푠 < 푠 < 푣 휖 푠 ; 푘 푁 (휆푘 ) 푘 summing over 푘 ≥ 퐾, we multiply by (푠 − 1) and let 푠 → 1+ to get

∗ ∗ ∗ (푣 − 휖)휁Q (1) ≤ 휁Λ,푋 (1) ≤ (푣 + 휖)휁Q (1)

∗ where 휁Q (1) = 1 is the residue of the Riemann zeta function (25.2.4). Now letting 휖 → 0, vol(푋≤1) 휁 ∗ (1) = 푣 = . Λ,푋 covol(Λ) Chapter 30

Optimal embeddings

30.10 Algorithmic aspects

In Lemma 30.6.17, we computed local embedding numbers for Eichler orders when the residue field has odd cardinality. Suppose now that 푅 is local, with residue field of even cardinality. Then the calculation of the local embedding number is quite painful and as such is not conducive to a formula that is both compact and intelligible; nevertheless, the representation in Proposition 30.6.12(c) shows that the local embedding number is effectively computable.

30.10.1. We can improve upon the brute force method of calculating 푚(푆, O) by calculating 푀(푠) more directly as follows. The map 푆/2푆 → 푆/2푆 given by 푥 ↦→ 푥2−푡푥 is F2-linear, and therefore by linear algebra over F2 one can compute all solutions to 2 푓 푓 (푥) = 푥 − 푡푥 + 푛 ≡ 0 (mod 픭 ) if 푓 ≤ ord픭 (2). For each of these solutions, one can then use Hensel lifting to test which among them give rise to solutions modulo 픭 푓 for 푓 ≤ ord픭 (4); and then Hensel’s lemma implies that each such solution lifts to a 푓 unique solution modulo 픭 whenever 푓 > ord픭 (4).

Part IV

Geometry and topology

Chapter 33

Hyperbolic plane

33.3 Upper half-plane

We provide a further elaboration of the proof of the following theorem.

2 Theorem 33.3.14. The group PSL2 (R) acts on H via orientation-preserving isome- + 2 tries, i.e., PSL2 (R) ↩→ Isom (H ).

Proof elaboration. Let 휐 be a (piecewise continuously diﬀerentiable) path in H2 given by 푧(푡) for 푡 ∈ [0, 1]; then by deﬁnition

∫ 1 d푧 d푡 ℓ(휐) = . 0 d푡 Im 푧(푡) The path 푔휐 is given by 푧0(푡) = 푔(푧(푡)), and by the chain rule

∫ 1 0 d푧 (푡) d푡 ℓ(푔(휐)) = 0 0 d푡 Im 푧 (푡) ∫ 1 d푔(푧(푡)) d푧(푡) d푡 = 0 d푧 d푡 Im 푔(푧(푡)) ∫ 1 d푧(푡) d푡 = , 0 d푡 Im 푧(푡) the latter equality from (33.3.15). The fact that lengths are preserved immediately implies the invariance of the hyperbolic metric.

33.6 Hyperbolic area and the Gauss–Bonnet formula

We prove the Gauss–Bonnet formula.

Theorem 33.6.1 (Gauss–Bonnet formula). Let 푇 be a hyperbolic triangle with angles 훼, 훽, 훾. Then 휇(푇) = 휋 − (훼 + 훽 + 훾).

55 56 CHAPTER 33. HYPERBOLIC PLANE

Proof. We ﬁrst consider the case where 푇 has at least one vertex in bd H2. Since the 2 group PSL2 (R) acts transitively on the boundary bd H , by applying an element of PSL2 (R), we may suppose this vertex is ∞ (without changing the area). Then there is a diagram as follows:

The fact that the angles are duplicated along the real axis is explained by the following diagram:

The semicircular segment lies along a circle with some radius 푐; applying the isometry √ 1/ 푐 0 푔 = √ ∈ PSL (R) 0 푐 2 with the eﬀect 푔(푧) = 푧/푐, and then translating, we may suppose that this segment lies along the unit circle. Then

∫ ∫ d푥d푦 ∫ 푏 ∫ ∞ d푦 ∫ 푏 −1 ∞ = d푥 = d푥 2 √ 2 √ 푇 푦 푎 1−푥2 푦 푎 푦 1−푥2 ∫ 푏 d푥 ∫ 훽 = √ = −d휃 = 휋 − (훼 + 훽) 푎 1 − 푥2 휋−훼 where we make the substitution 푥 = cos 휃. If 푇 has two or three vertices in bd H2, the same argument applies, but with possibly 훼 = 0 (and 푎 = −1) or 훽 = 0 (and 푏 = 1). So we are left with the case where 푇 has all vertices in H2. We then consider the following diagram: 33.6. HYPERBOLIC AREA AND THE GAUSS–BONNET FORMULA 57

The triangle with vertices 퐵, 퐶, ∞ has area 휋 − (훿 + (휋 − 훾)) = 훾 − 훿, and the triangle with vertices 퐴, 퐵, ∞ has area 휋 − 훼 − 훽 − 훿, so our triangle with vertices 퐴, 퐵, 퐶 has area equal the diﬀerence,

휋 − (훼 + 훽 + 훿) − (훾 − 훿) = 휋 − (훼 + 훽 + 훾).

Part V

Arithmetic geometry

Chapter 40

Classical modular curves and modular forms

40.3 Classical modular forms

We prove (most of) the following proposition. Proposition 40.3.4. Let 푓 : H2 → C be a meromorphic modular form of weight 푘, not identically zero. Then ∑︁ 1 푘 ord∞ ( 푓 ) + ord푧 ( 푓 ) = (40.3.5) 푒푧 12 Γ푧 ∈Γ\H2 where 푒푧 = # StabΓ (푧). The sum (40.3.5) has only ﬁnitely many terms, by 40.3.1, and the stabilizers are given in 40.3.2. Proof. See Serre [Ser73, §3, Theorem 3]. To prove this theorem, we perform a contour 1 d 푓 integration on the boundary of . More precisely, ﬁrst suppose that 푓 has 2휋푖 푓 neither zero nor pole on the boundary of◊ except possibly at 푖, 휔, −휔2. We consider the contour 퐶 containing all zeros or poles◊ of 푓 in int( ). ◊

61 62 CHAPTER 40. CLASSICAL MODULAR CURVES AND MODULAR FORMS

By the argument principle, 1 ∫ d 푓 ∑︁ = ord ( 푓 ). (40.3.6) 2휋푖 푓 푧 퐶 푧 ∈int( ) We write 퐶 as the sum of several contours as◊ indicated. In the change of variable 2휋푖푧 푧 ↦→ 푞 = 푒 , the contour 퐶1 transforms into a circle centered at 푞 = 0 with negative orientation whose only enclosed zero or pole is ∞. Thus 1 ∫ d 푓 = − ord∞ ( 푓 ). (40.3.7) 2휋푖 퐶1 푓 −1 We have 푇 (퐶8) = 퐶2 with opposite orientation; since 푓 (푧 + 1) = 푓 (푧), these contributions cancel. On 퐶3, as the radius of this arc of a circle tends to 0, 1 ∫ d 푓 1 −휋푖 1 → ord휔 ( 푓 ) = − ord휔 ( 푓 ) (40.3.8) 2휋푖 퐶3 푓 2휋푖 3 6 as the angle formed with 휔 by the endpoints of 퐶2 is 휋/3 (Exercise 40.1). Similarly, 1 ∫ d 푓 1 1 ∫ d 푓 1 → − ord푖 ( 푓 ) and → − ord−휔2 ( 푓 ). (40.3.9) 2휋푖 퐶5 푓 2 2휋푖 퐶7 푓 6 푘 Finally, 푆(퐶6) = 퐶4 with opposite orientation; but now 푓 (푆푧) = 푧 푓 (푧) so d 푓 (푆푧) d 푓 (푧) = 푘푧푘−1 푓 (푧) + 푧푘 d푧 d푧 and hence d 푓 (푆푧) d푧 d 푓 (푧) = 푘 + , 푓 (푆푧) 푧 푓 (푧) so 1 ∫ d 푓 1 ∫ d 푓 1 ∫ d푧 d 푓 = − 푘 + 2휋푖 퐶 ∪퐶 푓 2휋푖 퐶 푓 2휋푖 퐶 푧 푓 4 6 4 4 (40.3.10) 1 ∫ d푧 −푘 −휋푖 푘 = −푘 → = 2휋푖 퐶4 푧 2휋푖 6 12 as the angle formed with 0 is 휋/6. Summing, we obtain the result. If 푓 has a zero or pole on the boundary of , we repeat the same argument with a contour modiﬁed as follows: ◊ 40.3. CLASSICAL MODULAR FORMS 63

The details are omitted.

Hints and comments on exercises

Appendix A

Hints and comments on exercises

1.2. For part (a), see May [May66, p. 290]. 1.3(b). The minimal polynomial is irreducible, because 퐷 is a division algebra. The minimal polynomial divides the characteristic polynomial of degree 3 which factors over R, so the minimal polynomial has degree at most 2. If the minimal polynomial has degree 2 (irreducible), then since every irreducible factor of the characteristic polynomial is a factor of the minimal polynomial, we conclude that the characteristic polynomial has even degree, a contradiction. So the minimal polynomial has degree 1, and this implies that 훼 ∈ R for all 훼 ∈ 퐷, contradicting 퐷 ≠ R. 0 1 2.3. For such a map, we must have 푖 푗 ↦→ . Check that the four matrices −1 0 are linearly independent, so the map is an 퐹-linear isomorphism. Then, using the universal property of algebras given by generators and relations, show that the given matrices satisfy the relations in 퐵, so the map is an 퐹-algebra homomorphism. 2.5. Try 푗 0 = 푖0 푗 − 푗푖0, and show that 푗 0푖0 + 푖0 푗 0 = 0. If ( 푗 0)2 = 0, consider instead 푗 0 = 푖0푘 − 푘푖0. 2.7. Use Exercise 2.4(c) and show that the center over 퐹 has dimension 1 or compute directly with 훼푖 − 푖훼 = 훼 푗 − 푗훼 = 0 for 훼 = 푡 + 푥푖 + 푦 푗 + 푧푘 ∈ 퐵. 2.11. 퐵 acts on itself by left multiplication, which in the standard basis gives a map

퐵 ↩→ M4 (퐹) 푡 푎푥 푏푦 −푎푏푧 ©푥 푡 −푏푧 푏푦 ª 푡 + 푥푖 + 푦 푗 + 푧푘 ↦→ ® . − ® 푦 푎푧 푡 푎푥 ® 푧 푦 −푥 푡 « ¬ 2.20. Use the left regular representation either to 퐹 or a subfield 퐾, and use the block matrix determinant. See also Aslaksen [Asl96]. 3.2. Define 푥 + 푦훼 = 푥 + 푦(푡 − 훼) = (푥 + 푡푦) − 푦훼 for 푥, 푦 ∈ 퐹. 3.4. The standard involution on 퐹 × 퐹 is given by (푥, 푦) ↦→ (푦, 푥). (Note that 퐹 embeds diagonally in 퐹 × 퐹, so 푎 ↦→ (푎, 푎), and so 퐹 is indeed fixed under this map.)

67 68 APPENDIX A. HINTS AND COMMENTS ON EXERCISES

3.6. (The first part holds even if 퐵 = M2 (퐹).) 3.8. 푔 ↦→ 푔−1 is a standard involution if and only if 퐺 has exponent 2 and char 퐹 = 2 (so the standard involution is the identity and 퐹[퐺] is commutative). This remains true even if 퐺 is an infinite group (but then dim퐹 F[퐺] = ∞). 3.10. Let 푖, 푗 ∈ 퐾 \ 퐹. Then 푖 + 푗 satisfies a quadratic polynomial, but 푗푖 = 푖 푗, so we have (푖 + 푗)2 = 푖2 +2푖 푗 + 푗2 ∈ 퐹(푖 + 푗) + 퐹 hence 2푖 푗 = 푐(푖 + 푗) + 푑 with 푐, 푑 ∈ 퐹: but then since 2 ≠ 0, we have 2푖 ≠ 푐 ∈ 퐹 so 푗 = (푐푖 + 푑)/(2푖 − 푐) ∈ 퐾. 3.12. For part (a), Suppose 퐵 has degree 2. Choose a basis 1, 푥2, . . . , 푥푚. For each 푖, the quadratic 퐹-algebras 퐹[푥푖] have a standard involution, and so extending by 퐹-linearity we obtain a map : 퐵 → 퐵. For 푥 ∈ 퐵, let 푡(푥) = 푥 + 푥 and 푛(푥) = 푥푥. By induction and 퐹-linearity, we may suppose 1, 푥, 푦 are 퐹-linearly independent. Suppose (푥 + 푦)2 − 푠(푥 + 푦) + 푚 = 0 with 푠, 푚 ∈ 퐹. We show that 푠 = 푡(푥) +푡(푦). We have (푥 − 푦)2 = 푥2 − (푥푦 + 푦푥) + 푦2 = 2(푥2 + 푦2) − 푠(푥 + 푦) + 푚 = (2푡(푥) − 푠)푥 + (2푡(푦) − 푠)푦 + (푚 − 2푛(푥) − 2푛(푦))

But (푥 − 푦)2 ∈ 퐹(푥 − 푦) + 퐹 so 2푡(푥) − 푠 = 푠 − 2푡(푦), i.e. 2푠 = 2푡(푥) + 2푡(푦). Since char 퐹 ≠ 2, we have 푠 = 푡(푥) + 푡(푦) as desired. To conclude, we show 푥푦 = 푦 푥. We may suppose 푥푦 ∉ 퐹. We verify that both (푥푦)2 − (푥푦 + 푦 푥)푥푦 + (푦 푥)(푥푦) = 0 and (푥푦)2 − (푥푦 + 푥푦)푥푦 + 푥푦(푥푦) = 0, so the result follows by uniqueness of the minimal polynomial. For part (b), by the uniqueness of the standard involution, we have 푥 = 푥 + 1 if 푥 ∉ 퐹. But then if 1, 푥, 푦 are 퐹-linearly independent we have 푥 + 푦 + 1 = 푥 + 푦 =

푥 + 푦 = (푥 + 1) + (푦 + 1) = 푥 + 푦, a contradiction. So dimF2 퐵 ≤ 2. Since a 2 Boolean ring consists of idempotents, we have 퐵 = F2 or 퐵 F2. 3.13. Under right multiplication by 퐵 = M푛 (퐹), a matrix is nothing other than the direct sum of its rows, so in particular, the characteristic polynomial of right multiplication by 퐴 ∈ M푛 (퐹) acting on M푛 (퐹) will be the 푛th power of the usual characteristic polynomial of 퐴 acting on row vectors 푉 퐹푛. (In the language 푛 푛 of Chapter 7, 퐵 = M푛 (퐹) as a right 퐵-module is 퐵 푉 where 푉 퐹 is the unique simple right 퐵-module.) 3.15. By 퐹-linearity, it suﬃces to verify these statements on a basis for 퐵. 3.16. The class equation reads 푞4 − 1 = (푞 − 1) + 푚(푞2 + 1)

2 for some 푚 ∈ Z≥0. Thus (푞 + 1) | (푞 − 1), a contradiction. This argument can be generalized in a natural way to prove Wedderburn’s theorem in full: see Schue [Schu88], for example. 3.18. See van Praag [vPr68, vPr02]. 4.9. For part (c), by the transitivity of trace, we may assume 퐾/퐹 is purely inseparable and [퐾 : 퐹] is a multiple of 푝. But then all roots of the minimal polynomial of 푥 ∈ 퐾 over 퐹 are equal, so the characteristic polynomial of multiplication by 푥 ∈ 퐾 has all roots equal and there are a multiple of 푝 of them and thus the trace is zero. 69

√ For part (d), tr((푎 + 푏 5)2) = 2(푎2 + 5푏2) and tr((푎 + 푏훼 + 푐훼2)2) = 2(푎2 − 2푎푏 + 10푎푐 + 5푏2 − 8푏푐 + 13푐2). 0 4.10. Choosing bases for 푉, 푉 and writing 푓 as a matrix 퐴 ∈ GL푛 (퐹) in these bases, we find that 푢푥푡 [푇]푥 = 푥푡 (퐴푡 [푇 0] 퐴)푥 for all 푥 ∈ 푉 ' 퐹푛, so 푢[푇] = 퐴푡 [푇 0] 퐴. Taking determinants, we find det 푇 0 = 푢푛 det 푇 ∈ 퐹/퐹×2. 4.11. See Lam [Lam2005, Chapter X] for more on Pfister forms; in particular, for (c) see Lam [Lam2005, Theorem X.1.7]. 5.3. The implication (vi) ⇒ (v) follows similarly: if 푎 ∈ 퐹×2 then already h푎i represents 1; if 푎 ∉ 퐹×2 and we have 푥2 −푎푦2 = 푏 then either 푥 = 0, in which case h푎, 푏i is isotropic and thus represents 1, or 푥 ≠ 0 and then 푎(푦/푥)2 +푏(1/푥)2 = 1 as desired. 푎, 푏 5.4. If 퐵 = then 퐾 = F푞 (푖) F푞2 and Nm : F푞2 → F푞 is surjective so F푞 × 푏 ∈ Nm퐾 /퐹 (퐾 ). 5.9. See Lam [Lam2005, Examples III.2.12–13]. For the first, exhibit an explicit isometry h1, 1, 1i h2, 3, 6i. For the second, note that h2, 5, 10i represents 7 but h1, 1, 1i does not (by showing 푥2 +푦2 +푧2 +푤2 . 0 (mod 8) for 푥, 푦, 푧, 푤 ∈ Z with gcd(푥, 푦, 푧, 푤) = 1); or note that h1, 1, 1i represents 1 but h2, 5, 10i (looking modulo 5, and arguing similarly). 5.12. Indeed, for any 푥, 푦, 푧 ∈ 푉, by Clifford multiplication we have (푥 + 푦푧)(푥 + 푦푧) = (푥 + 푦푧)(푥 + 푧푦) = 푞(푥) + 푦푧푥 + 푥푧푦 + 푞(푦)푞(푧) = 푞(푥) + 푞(푦)푞(푧) − 푇 (푥, 푦)푧 + 푇 (푥, 푧)푦 + 푇 (푦, 푧)푥. Suppose that : Clf(푄) → Clf(푄) is a standard involution and rk(푉) ≥ 3. If 푥, 푦, 푧 are 퐹-linearly independent, then we must have 푇 (푥, 푦) = 푇 (푥, 푧) = 푇 (푦, 푧) = 0. Then the fact that (푥 + 1)(푥 + 1) = 푄(푥) + 1 + 2푥 for all 푥 ∈ 푉 implies that 2 = 0 ∈ 퐹, a contradiction. A similar argument works for Clf0 (푄). 5.21. See Auel [Auel2015, Theorem 1.8] for a proof of a more general result. 6.2. Since 퐾 is separable, the restriction of the reduced norm to 퐾 is nondegenerate. Let 푗 ∈ 퐾⊥ \{0} be a nonzero element in the orthogonal complement of 퐾. Then 퐵 = 퐾 +퐾 푗 since dim퐹 (퐾 +퐾 푗) = 4. Since 1 ∈ 퐾, we have 푇 (1, 푗) = trd( 푗) = 0 (recall (4.2.14)) so 푗 = − 푗 and 푏 = 푗2 ∈ 퐹×. By (4.2.16) we have trd( 푗훼) = 0 so 푗훼 + 훼 푗 = 훼 푗 − 푗훼 = 0 so 푗훼 = 훼 푗. 6.12. Let 푒1, . . . , 푒푛 be a normalized basis for 푉, and let 푛 = 2푚 + 1. By Example 6.3.10, since 푄 is nondegenerate we may take 푎1 ··· 푎푚 = 1 and 푐1 = 푑, i.e., 푄 '[1, 푏1] ⊥ · · · ⊥ [1, 푏푚] ⊥ h1i. Then

푒푖푒 푗 − 푒 푗 푒푖 = 푒푖푒 푗 + 푒푖푒 푗 = 푇 (푒푖, 푒 푗 )

for all 푖, 푗. For 퐼 = {푖1, . . . , 푖푟 } ⊆ {1, . . . , 푛}, let 푒퐼 = 푒푖1 ··· 푒푖푟 . By the orthogonal decomposition, 푒푖 centralizes 푒퐼 if and only if 푖 ∉ 퐼 for each 푖 = 1,..., 2푛, 2 and 푒푛 centralizes Clf(푄). Therefore 푍 (Clf(푄))' 퐹[푒푛]' 퐹[푥]/(푥 − 푑). If 2 1 푑 = 1, the unique solution to 휁 = 1 in 푍 (Clf(푄)) ∩ Clf (푄) is 푒푛. 70 APPENDIX A. HINTS AND COMMENTS ON EXERCISES

7.3. The map 푎 ⊗ 푏 ↦→ (푥 ↦→ 푎푥푏) gives an 퐹-algebra homomorphism 퐵 ⊗퐹 퐵 → End퐹 (퐵) M4 (퐹), which is injective since 퐵 ⊗퐹 퐵 is simple and therefore an isomorphism by a dimension count. 7.13. See e.g. Drozd–Kirichenko [DK94, Theorem 6.1.2]. Í Í 7.14. The augmentation ideal is the kernel of the surjective map 푔 푎푔푔 ↦→ 푔 푎푔, so is nontrivial. 7.15. See Reiner [Rei2003, Exercise 7.8], Lam [Lam2001, Theorem 6.1], etc. 7.20. Lemma 7.4.8 characterizes rad 퐵 as the largest two-sided ideal in which every element is nilpotent, so we show that rad nrd has this property. First, rad nrd is a two-sided ideal: if 휖 ∈ rad nrd and 훼 ∈ 퐵, then trd(훽훼휖) = 0 for all 훽 ∈ 퐵 so 훼휖 ∈ rad nrd and similarly trd(훽휖훼) = trd(훼훽휖) = 0 so 휖훼 ∈ rad nrd. Next, every 휖 ∈ rad nrd is nilpotent: we have trd(휖) = 0, so 휖2 = − nrd(휖) and trd(휖2) = −2 nrd(휖) = 0 so nrd(휖) = 0 and 휖2 = 0. 7.23. This exercise was given in a course by Bjorn Poonen in Spring 2000 at the University of California, Berkeley. First, parts (a) and (b). Choose 푥 ∈ 퐷 \ 퐹. Then 퐾 = 퐹(푥) is a purely inseparable extension of 퐹 so the minimal polynomial of 푥 in 퐷 (or in 퐹) is of 푛 the form 푇 푝 − 푎. In particular, 푝 | [퐾 : 퐹], but 퐷 is a left 퐾-vector space and [퐷 : 퐹] = [퐷 : 퐾][퐾 : 퐹] so 푝 | [퐷 : 퐹]. For part (c), all roots of the minimal polynomial of 푥 are equal, hence all eigenvalues of 푥 ⊗ 1 ∈ M푛 (퐹) are equal, and the number of them is divisible by 푝 by (a), so the trace is zero. For part (d), by (c), all elements of M푛 (퐹) have trace zero, which is a contradiction. 7.26. Apply the Skolem–Noether theorem to a nontrivial automorphism of 퐾; verify that the conjugating element has trace zero. Let 푗 ∈ 퐵× satisfy 푗훼 푗−1 = 훼. Then 퐵 = 퐾 ⊕ 퐾 푗, but 푗2훼 푗−2 = 훼 so 푗2 ∈ 푍 (퐵) so 푗2 = 푏 ∈ 퐹×. 7.29. By Corollary 7.7.11, every maximal subfield 퐾 of 퐵 has the same dimension, so since 퐹 is a finite field they are isomorphic (as abstract fields). But then by the Skolem–Noether theorem, since every element lies in a maximal subfield, × Ð −1 × we have 퐵 = 훼∈퐵× 훼 퐾 훼, which is a contradiction. One can also proceed without using the maximal subfield dimension theorem. Suppose 퐵 is a minimal counterexample (by cardinality); then 퐵 is a division ring, but every subalgebra of 퐵 is a field. Let 퐹 = 푍 (퐵). Let 푖 ∈ 퐵 \ 퐹; then by minimality, the centralizer of 푖 is a maximal subfield 퐾. We may assume 퐾 = 퐹(푖). If 퐵 = 퐾, we are done. Otherwise, let 푖 have multiplicative order 푚. Consider 퐿 : 퐵 → 퐵 by 퐿(훼) = 푖훼푖−1. Then 퐿 is a 퐾-linear map with 퐿푚 equal to the identity. We may therefore decompose 퐵 into eigenspaces for 퐿. Arguing as in the case of quaternion division rings, we show that each such nonzero eigenspace has dimension 1 as 퐾-vector space. Now consider the normalizer × 푁 = 푁퐵 (퐾). Then there is a bijection between the set of cosets of 푁/퐾 and the eigenspaces of 퐿. But 푁 acts on 퐾 as 퐹-linear automorphisms with kernel 퐾×, so 푁/퐾× is a subgroup of the Galois group Gal(퐾/퐹). It must be the full Galois group, otherwise 푁/퐾× fixes some subfield and its centralizer is a noncommutative 퐹-subalgebra, contradicting minimality. Therefore dim퐾 퐵 = dim퐹 퐾. We now proceed as above. 71

There are a large number of proofs of Wedderburn’s little theorem: see for example Kaczynski [Kacz64]. 8.2. If 푖, 푗 and 푖0, 푗 0 are standard generators of 퐵 and 퐵0, respectively, then consider the subalgebras generated by the pair 푖 ⊗ 1 and 푗 ⊗ 푗 0 and then pair 푖 ⊗ 푖0 and 1 ⊗ 푗 0. 8.11. See Weil [Weil60, §7, Propositions 2–3]. 10.6. Using the matrix units, show that if 푀 = (푚푖 푗 )푖, 푗 ∈ O then 푚푖 푗 · 1 ∈ O, but then 푚푖 푗 is integral over 푅 so in fact 푚푖 푗 ∈ 푅 and hence 푀 ∈ M푛 (푅). 10.9. The converse is true if char 퐹 ≠ 2 and 푅 is integrally closed. It is immediate if 1/2 ∈ 푅 since trd(훼2) = trd(훼)2 − 2 nrd(훼), so 2 nrd(훼) ∈ 푅. But for the same reason more generally we have 2 nrd(훼푛) = 2 nrd(훼)푛 ∈ 푅 so 푅[nrd(훼)] ⊆ (1/2)푅; so if 푅 is integrally closed we have in fact nrd(훼) ∈ 푅. The statement is false if char 퐹 = 2: take 퐵 = 퐹 × 퐹 (with char 퐹 = 2) and 훼 = (푎, 푎) with 푎 ∈ 퐹 not integral over 푅. Then trd(훼푛) = 2푎푛 = 0 for all 푛 but 훼 is not integral. 12.6. Let 푘 = F푞. If 푄 is not nondegenerate, then it is isotropic already. Otherwise, choosing a normalized basis for 푉 we may assume the quadratic form 푄 is of the form 푧2 = 푓 (푥, 푦) where 푓 (푥, 푦) is a quadratic form in two variables. If 푞 is even we are now done since every element of 푘 is a square. So suppose 푞 is odd. Then function 푓 (푥, 1) takes exactly (푞 + 1)/2 values in 푘, but there are (푞 − 1)/2 nonsquares in 푘×, so at least one of the values must be a square. 12.7. The quadratic form h−1, 푒, −1i is isotropic by a previous exercise, so diagonal- izing we have h−1, 푒i h1, 푠i for some 푠 ∈ 푘×. But disc(h−1, 푒i) = −푒 = 푠 = disc(h1, 푠i) ∈ 푘×/푘×2, so h1, 푠i h1, −푒i. More generally, this argument shows that two nonsingular binary quadratic forms over a finite field are isometric if and only if they have the same discriminant. 12.4. For (a), under the multiplication map 푚 : 퐺×퐺 → 퐺, we have 푚(1, 1) = 1; since multiplication is continuous, there exists an open neighborhood 푉1 ×푉2 3 (1, 1) −1 2 with 푉1 × 푉2 ⊆ 푚 (푈), i.e., 푉1푉2 ⊆ 푈. Letting 푉 = 푉1 ∩ 푉2 3 1, then 푉 ⊆ 푈. Statement (b) follows similarly, using that inversion is continuous. 12.5. Let 푥퐻, 푦퐻 ∈ 퐺/퐻 be distinct. Then 푦퐻푥−1 ⊆ 퐺 is closed and 1 ∉ 푦퐻푥−1. By Exercise 12.4, there is an open neighborhood 푉 3 1 in 퐺 such that 푉−1푉 ⊆ 퐺 r 푦퐻푥−1. So 푉푥퐻 3 푥퐻 and 푉 푦퐻 3 푦퐻 are disjoint open neighborhoods as desired. 12.10. Exercise suggested by Grant Molnar. 퐾, 2 13.1. Write 퐵 in the form 퐵 = with 퐾 ⊇ Q2 the unique unramified extension Q2 of Q2. 13.5. The standard involution on 퐵 has O = O and 푃 = 푃 since 푗 = − 푗, therefore it induces a standard involution on the quotient O/푃. Recall the classification of these algebras (Theorem 3.5.1, extended to Theorem 6.2.8 in all characteristics). 13.16 First proof: the symbol is trivial if and only if 퐹 is a splitting field for (−1, −1 | Q2) if and only if [퐹 : Q2] is even, by Proposition 13.4.4. Second, more direct proof: Let 푒 be the ramification degree of 퐹 over Q2 and 푓 the inertial degree, so [퐹 : Q2] = 푒 푓 . We need to show (−1, −1)퐹 = 1 if and only if 푒 is even or 푓 is even. By a change of basis, we have (−1, −1 | 퐹) = 72 APPENDIX A. HINTS AND COMMENTS ON EXERCISES

(−3, −2 | 퐹), so we need to show (−3, −2)퐹 = 1 if and only if 푒 is even or 푓 is even. The 퐹-algebra 퐾 = 퐹[푥]/(푥2 + 3) is not a field if and only if 푓 is even, and then the quaternion algebra splits. So suppose 푓 is odd. Then 퐾 is a field, the unramified quadratic extension of 퐹, so the algebra is split if and only if −2 is a norm from 퐾 to 퐹 if and only if 푒 = ord푣 (2) is even. 13.18. The proof that addition and multiplication are continuous with respect to the absolute value | | induced by 푤 is identical to the commutative case. We have a filtration O ⊃ 푃 ⊃ 푃2 ⊃ ... where 푃 is generated by 푗 and thus to show that 퐵 is 2 complete it suffices to note that the limit of the partial sums 푥0 +푥1 푗 +푥2 푗 +· · · = (푥0 + 푥2휋 + ... ) + (푥1 + 푥3휋 + ... ) 푗 ∈ 퐾 + 퐾 푗 exists since 퐾 is complete. The set O is compact since it is complete and totally bounded. By translating, since O is open we have that 퐵 is locally compact. Finally, if 푥 ∉ O then 푤(푥) < 0 so the ring generated by O and 푥 is equal to 퐵; but 퐵 is not compact, since the Ð −푖 open cover 푖 휋 O has no subcover. See Vignéras [Vig80a, Lemme II.1.6]. Î ord푝 (푡푝) 14.10. Take 푡 = ±푞 푝∈Σ\{∞} 푝 . Select the prime 푞 to satisfy congruences to ensure that the conditions hold. [See also Cassels [Cas78, Corollary to Theorem 6.5.1].] 14.13. There are infinitely many separable quadratic splitting fields of 퐵 (by the Hasse– Minkowski theorem), and only finitely many of them can be contained in 퐿. Check that a separable quadratic field 퐾 ⊇ 퐹 that is not contained in 퐿 is linearly disjoint with 퐿 over 퐹. 15.18. See Reiner [Rei2003, Theorem 41.3]; the result generalizes to the statement that 0 0 0 0 if O ⊇ O is a maximal order containing O, then O ' O1 × · · · × O푟 , and 푟 Ê 푛 (O0 : O) = codiff(O0) L 푛 푖 푖=1 푖 a result due to Jacobinski [Jaci66]. See also Reiner [Rei2003, Exercises 41.1– 41.3]. 16.16. See Shimura [Shi71, Proposition 4.11, (5.4.2)]. 16.18. This exercise is due to Kaplansky [Kap69]. We compute that 푅 푅 (푎) 푅 푅 푅 ( ) ©( ) ( 2)ª ( ) © ª OL 퐼 = 푎 푅 푎 ® and OR 퐼 = 푅 푅 푅® 푅 푅 푅 (푎2)(푎2) 푅 « ¬ « ¬ and 푅 푅 푅 −1 © ª 퐼 = 푅 푅 푅 ® (푎) 푅 (푎2) « ¬ −1 has 퐼 퐼 = OR (퐼) but (푎) 푅 (푎) −1 ©( ) ( 2)ª ( ) 퐼퐼 = 푎 푅 푎 ® ≠ OL 퐼 . 푅 푅 푅 « ¬ 17.7(a). The norm is Euclidean because 퐵∞ ' H and H has the standard Euclidean norm. The order Zh푖, 푗, 푘i is discrete in 퐵∞ (taking coordinate neighborhoods); 73

it follows that O is discrete in 퐵∞ as well, since O is commensurable with Zh푖, 푗, 푘i (a coordinate neighborhood contains only finitely many points). 17.7(b). See also Goren–Lauter [GL2007, Lemma 2.1.1]. 17.7(c). See also Goren–Lauter [GL2007, Corollary 2.1.2]. 18.2. Suppose that 퐼퐽 ⊆ {0} with 퐼, 퐽 two-sided O-ideals. Then (퐼퐹)(퐽퐹) = 퐹(퐼퐽) = {0}. If 퐼, 퐽 are both nonzero, then by Paragraph 18.2.1, 퐼퐹 = 퐽퐹 = 퐵, so 퐵 = {0}, impossible. 18.3. 퐽 has the structure of an 푅-module, since 푅 ⊆ 푍 (O). Since 푅 is noetherian and O is finitely generated as an 푅-module, 퐽 is finitely generated as an 푅-module; and 퐽퐹 ⊆ 퐵 is a nonzero two-sided ideal of 퐵, so since 퐵 is simple, we have 퐽퐹 = 퐵. 18.6. See Fröhlich [Frö73], with thanks to Ardakov [Ard-MO]. 20.6. We first show the inclusion (⊆) following (a)–(c). Since rad 퐴 is a two-sided ideal, it suffices to show that 1 − 훽 ∈ 퐴×. Suppose 퐴(1 − 훽) ( 퐴, then there is a maximal left ideal 퐼 ⊆ 퐴 such that 퐴(1 − 훽) ⊆ 퐼 and so 1 − 훽 ∈ 퐼; but since rad 퐴 ⊆ 퐼, we conclude 1 ∈ 퐼, a contraduction. Therefore 퐴(1 − 훽) = 퐴, so there exists 훼 ∈ 퐴 such that 훼(1 − 훽) = 1 and 1 − 훽 has the left inverse 훼. Further, 1 − 훼 = −훼훽 ∈ rad 퐴. Repeating the argument again, we conclude that 퐴(1 − (1 − 훼)) = 퐴훼 = 퐴, so there exists 훾 ∈ 퐴 such that 훾훼 = 1, and 훾 = 훾(훼(1− 훽)) = 1− 훽; thus 훼 is also a right inverse of 1− 훽. Thus 1− 훽 ∈ 퐴×. Conversely, we show (⊇) to show (d). Let 훽 ∈ 퐴 be such that 1 − 훼훽훾 ∈ 퐴× for all 훼, 훾 ∈ 퐴. Let 푀 be a simple left 퐴-module; we show that 훼푀 = {0}. Let 푥 ∈ 푀, 푥 ≠ 0; if 훽푥 ≠ 0, then by simplicity 푀 = 퐴훽푥 so 훽 = 훼훽푥 for some 훼 ∈ 퐴 and (1 − 훼훽)푥 = 0; since 1 − 훼훽 ∈ 퐴×, we have 푥 = 0, a contradiction. 21.4. Let 퐼 be a two-sided integral O-ideal. Then since O is a finitely generated 푅- module and 푅 is noetherian, we conclude that 퐼 is contained in a proper maximal (integral) O-ideal 푀. From 퐼 ⊆ 푀 we conclude that 퐼푀−1 ⊆ O, so 퐼푀−1 is integral. But nrd(퐼푀−1) = nrd(퐼)/nrd(푀) | nrd(퐼). It follows that 퐼 can be written as the product of maximal ideals 푀 by induction on the reduced norm. We will now show that in fact a maximal O-ideal is prime. For suppose that 퐼퐽 ⊆ 푀 and that 퐼 * 푀. Then 퐼 + 푀 is a two-sided O-ideal strictly containing 푀 so 퐼 + 푀 = O. But then 퐽 = 퐼퐽 + 푀퐽 ⊆ 푀. Conversely, if 푃 is prime and 푃 ⊆ 퐼 where 퐼 is a proper two-sided integral O-ideal. Since O is hereditary, 퐼 is invertible and 푃 = 퐼(퐼−1푃); but 퐼−1푃 ⊆ 푃 implies 퐼−1 ⊆ O which is impossible so 퐼 ⊆ 푃 hence 푃 = 퐼. To conclude, we show that this group is abelian. Let 푃, 푄 be prime ideals. Then 푃푄 ⊆ 푃, so as above 푃푄푃−1 is integral, say 푃푄푃−1 = 푄0. If 푄0 = O then 푄 = O, a contradiction. But then choosing 0 ≠ 푝 ∈ 푃∩푅 then 푄 = 푝푄푝−1 ⊆ 푄0, but 푄 is maximal so 푄 = 푄0. Thus 푃푄 = 푄0푃, so the group is abelian. 21.6(a). We follow Reiner [Rei2003, Exercise 39.2]. Let 퐽 = rad O. Certainly O + 퐽0 is an 푅-order, since O퐽0O ⊆ O0퐽0O0 ⊆ 퐽0. 21.6(b). We want to show that 퐽 ⊆ rad(O + 퐽0), and for that we can show 퐽 + 퐽0 ⊆ rad(O + 퐽0). By Corollary 20.5.5, for 푟 large we have 퐽푟 ⊆ 픭O. Thus

(퐽 + 퐽0)푟 ⊆ 퐽푟 + 퐽0 ⊆ 픭O + 퐽0 ⊆ 픭O0 + 퐽0; 74 APPENDIX A. HINTS AND COMMENTS ON EXERCISES

and for 푟 possible larger by Theorem 20.5.1, we have (퐽0)푟 ⊆ 픭O0 so

(픭O0 + 퐽0)푟 ⊆ 픭O0 ⊆ (퐽0)푟 ⊆ 픭O0.

Combining these, and making 푟 even larger,

3 (퐽 + 퐽0)푟 ⊆ (픭O0)푟 ⊆ 픭O ⊆ 픭(O + 퐽0).

Now by Corollary 20.5.5, we have 퐽 + 퐽0 ⊆ rad(O + 퐽0). 21.6(c). We have shown (퐽0)푟 ⊆ 픭O; if 퐽0 ⊆ O, then by Corollary 20.5.5 we conclude 퐽0 ⊆ 퐽. 21.9. See Small [Sma66]. 21.10. See Reiner [Rei2003, Theorem 40.7]. 21.13. See [AG60, Theorem 2.3]. 22.14(a). See Voight [Voi2011a, Lemma 3.7]. The map 푠 descends ﬁrst to (Ó2 푀) ⊗3 since

푠(푥 ⊗ 푥0 ⊗ 푦 ⊗ 푦0 ⊗ 푧 ⊗ 푧0) = 0

whenever 푥 = 푥0 and

푠(푥 ⊗ 푥0 ⊗ 푦 ⊗ 푦0 ⊗ 푧 ⊗ 푧0) = −푠(푥0 ⊗ 푥 ⊗ 푦 ⊗ 푦0 ⊗ 푧 ⊗ 푧0),

with similar statements for 푦, 푧. 22.14(b). To show that 푠 in fact descends to Ó3 (Ó2 푀), we observe that

푠(푥 ∧ 푥0 ⊗ 푦 ∧ 푦0 ⊗ 푧 ∧ 푧0) = 0

whenever 푥 = 푦 and 푥0 = 푦0 (with similar statements for 푥, 푧 and 푦, 푧). To ﬁnish, we show that

푠((푥 ∧ 푥0) ⊗ (푦 ∧ 푦0) ⊗ (푧 ∧ 푧0)) = −푠((푦 ∧ 푦0) ⊗ (푥 ∧ 푥0) ⊗ (푧 ∧ 푧0)).

To prove this, we may do so locally and hence assume that 푀 is free with basis 푒1, 푒2, 푒3; by linearity, it is enough to note that

푠((푒1 ∧ 푒2) ⊗ (푒2 ∧ 푒3) ⊗ (푒3 ∧ 푒1)) = (푒1 ∧ 푒2 ∧ 푒3) ⊗ (푒2 ∧ 푒3 ∧ 푒1)

= (푒2 ∧ 푒3 ∧ 푒1) ⊗ (푒2 ∧ 푒3 ∧ 푒1)

= −푠((푒2 ∧ 푒3) ⊗ (푒1 ∧ 푒2) ⊗ (푒3 ∧ 푒1)). 22.14(c). It follows then also that 푠 is an isomorphism, since it maps the generator

Ó3 Ó2 (푒1 ∧ 푒2) ∧ (푒2 ∧ 푒3) ∧ (푒3 ∧ 푒1) ∈ ( 푀)

Ó3 ⊗2 to the generator (푒1 ∧ 푒2 ∧ 푒3) ⊗ (푒2 ∧ 푒3 ∧ 푒1) ∈ ( 푀) . 22.8. Exercise suggested by Asher Auel. 22.17. We have

휓(푥 ∧ 푦) = 1 ∧ 푥 ∧ 푦 ∧ 푥푦 = 1 ∧ (−푥) ∧ (−푦) ∧ (−푥푦) = −(1 ∧ 푥 ∧ 푦 ∧ 푥푦)

for all 푥, 푦 ∈ O, with similitude factor ℎ = −1 : Ó4 O −→∼ Ó4 O. 75

23.6. By the computation

푎 푏 픭 푅 푎픭 + 푏픭푒 푎푅 + 푏픭 = 푐 푑 픭푒 픭 푐픭 + 푑픭푒 푐푅 + 푑픭

푎 푏 we see that ∈ O (퐽) if and only if 푎 ∈ 푅, 푏 ∈ 픭−1, 푐 ∈ 픭푒−1, and 푑 ∈ 푅. 푐 푑 L A similar calculation holds on the right. 23.7. Start with any 푅-basis 푥1, 푥2 of 퐿; writing a basis of 푀 in terms of 푥1, 푥2 yields the columns of a matrix in GL2 (퐹). If we change the basis of 퐿 or of 푀, we are applying the group GL2 (푅) on the left or right to this matrix, i.e., we can perform integral row and column operations on this matrix. Now by direct manipulation with 2 × 2-matrices (more abstractly, the theory of elementary divisors), we can transform this matrix into a diagonal matrix of the desired form. 24.2. See Faddeev [Fad65, Proposition 24.2]. 25.1(b). The inverse is given by √︄ √︄ 1 − 푥2 1 − 푦2 (푥, 푦) ↦→ ©cos−1 , cos−1 ª . 1 − 푥2푦2 1 − 푥2푦2 ® « ¬ To prove that this inverse map has the correct codomain, recall that if 0 < 휃, 휙 < 휋/2, then 휃 + 휙 < 휋/2 if and only if cos(휃 + 휙) > 0. 25.3. See Weston [Wes, Lemma 1.19]. 25.4. We follow Weston [Wes, Proposition 4.5], working through each part. Write

∞ 1 ∑︁ 푏 휁 (푠) = 푛 퐾 , [픟] 푤(Nm(픟)푠) 푛푠 푏=1 where −1 푏푛 := #{푎 ∈ 픟 : Nm(푎) = 푛}. Since Nm(푎) = |푎|2, for all 푥 > 1

∑︁ −1 √ 푏푛 = #{푎 ∈ 픟 : 0 < |푎| ≤ 푥}; 푛≤푥 from Lemma 25.2.11, we conclude

∑︁ 휋푥 √ 푏 − ≤ 퐶 푥 푛 퐴 푛≤푥

where 퐴 is the coarea of 픟−1 and 퐶 is a constant that does not depend on 푥. We compute that √︁|푑| 퐴 = Nm(픟−1) . 2 This proves (a). 76 APPENDIX A. HINTS AND COMMENTS ON EXERCISES

Now consider the Dirichlet series

∞ 1 ∑︁ 휋 1 푓 (푠) := 푏 − . 푤푁 (픟)푠 푛 퐴 푛푠 푛=1

Then the estimate

∑︁ 휋 ∑︁ 휋푥 √ 푏 − = 푏 − ≤ 퐶 푥 푛 퐴 푛 퐴 푛≤푥 푛≤푥

by the comparison test implies that 푓 (푠) converges for all Re 푠 > 1/2 and in particular 푓 (푠) converges at 푠 = 1, proving (b). For 푠 > 1, 휋 푓 (푠) = 휁 (푠) − 휁 (푠) 퐾 , [픟] 퐴푤 Nm(픟)푠 so 2휋 1−푠 휁퐾 , [픟] (푠) = 푓 (푠) + Nm(픟) 휁 (푠). 푤√︁|푑| hence

res푠=1 휁퐾 (푠) = lim(푠 − 1)휁퐾 , [픟] (푠) 푠&1 2휋 = lim(푠 − 1) 푓 (푠) + lim(푠 − 1) Nm(픟)1−푠 휁 (푠) 푠&1 푤√︁|푑| 푠&1 2휋 2휋 = 0 + · 1 = . 푤√︁|푑| 푤√︁|푑|

This proves (c). Finally, (d): 휁퐾 , [픟] (푠) has a simple pole at 푠 = 1 with residue independent of [픟]. Summing the residues over [픟] ∈ Cl(퐾), from (25.2.9) we conclude the result. 25.6. For further reading, see Fitzgerald [Fit2011] and Clark–Jagy [CJ2014]. 25.7(b). We follow Lenstra [Len79, Lemma 1.5]. Let 퐽 be an invertible right O-ideal; without loss of generality, we may suppose 퐽 ⊆ O is integral. We argue by induction on nrd(퐽) > 0. The base case nrd(퐽) = 1 implies that 퐽 = O, which is true. Since 퐼, 퐽 are invertible, we have 퐽−1퐼 ≠ 퐼, so there exists 훾 ∈ 퐽−1퐼 \ 퐼. Since 퐼 is Euclidean, there exists 휇 ∈ 퐼 such that nrd(훾 − 휇) < nrd(퐼). Let 휈 = 훾 − 휇. Then 휈 ≠ 0, else 훾 = 휇 ∈ 퐼. Since 훾 ∈ 퐽−1퐼 and 휇 ∈ 퐼 we have 휈 ∈ 퐽−1퐼, so 퐽0 = 휈퐼−1퐽 ⊆ O is an integral, invertible right O-ideal. We have

nrd(퐽0) = nrd(휈) nrd(퐼)−1 nrd(퐽) < nrd(퐽).

So by induction, for [퐽0] ∈ Cl(O) we have either [퐽0] = 1 or [퐽0] = [퐼], and thus [퐽] = [퐼−1] = [퐼] or [퐽] = 1, the latter using the fact that Pic(O) has exponent dividing 2. 77

× 26.3 First (a). Let 푉 = 퐹R be the ambient space. The group 푅 acts by preserving the × × norm, so we can write (푉/푅 )≤1 = 푉≤1/푅 . Choose a system of fundamental units for 푅× and let Z푟+푐−1 ' 퐸 ≤ 푅× be the group generated by them; then × × 푅 = 퐸푅tors, and so 푐 ∫ × 1 2 vol(푉≤1/푅 ) = vol(푉≤1/퐸) = d푥 d푧 푤 푤 푉≤1/퐸 푟 푐 with 푥푖, 푧 푗 standard coordinates on R × C —and we use multi-index notation to simplify. 푐 Now (b). Let 휌 푗 , 휃 푗 be polar coordinates on C , and for symmetry restrict the + domain 푉 to the domain 푉 with 푥푖 > 0 for all 푖. Then ∫ ∫ ∫ d푥 d푧 = 2푟 d푥 (휌 d휌 d휃) = 2푟 (2휋)푐 휌 d푥 d휌 / + / +,≤1/ 푉≤1 퐸 푉≤1 퐸 푊 퐸 where 푊+ is the projection of 푉+ onto the 푥, 휌-coordinate plane. Now let 2 푥푟+ 푗 = 휌 푗 to get ∫ ∫ 2푟 (2휋푐) 휌 d푥 d휌 = 2푟 휋푐 d푥 + / +,≤1/ 푊≤1 퐸 푊 퐸 and the norm is now simply the product of 푟 + 푐 (positive) coordinates. Î Next, (c): we apply the change of variables 푢푖 = log 푥푖; the condition 푖 푥푖 = Í 푡 ≤ 1 becomes 푖 푢푖 = log 푡 ≤ 0, and we obtain ∫ ∫ ∫ 0 ∫ ∫ d푥 = 푒푢 d푢 = 푒푡 d푡 d푢 = d푢 푊 +,≤1/퐸 log(푊 +,≤1/퐸) −∞ 푃 푃 where 푃 is the fundamental parallelogram for the additive (logarithmic) action of 푅×. Finally (d): by definition, 푃 has covolume Reg퐹 , and putting all of these together, we conclude that 푐 푟 푐 × 2 푟 푐 2 (2휋) Reg퐹 vol(푉≤1/푅 ) = 2 휋 Reg퐹 = 푤 푤퐹 as claimed. 28.4(a). See Newman [New72, Theorem II.7]. × × × 28.7. We show that b훼 ∉ 퐵 Ob . Indeed, suppose that b훼 = 훽휇b with 훽 ∈ 퐵 and ∈ × ( ×) ∩ Q× = { } ∈ 1 휇b Ob . Since nrd Ob >0 1 , we must have 훽 퐵 . But then −1 2 ℓb훼휇b = ℓ훽 = 훾 ∈ 퐵 ∩ Ob = O; thus nrd(훾) = ℓ . But nrd |O = h1, −푝, −푞, 푝푞i only represents ℓ2 by ±ℓ, a contradiction. 29.3(a). It suffices to show that 퐻 has a compact neighborhood in 퐺/퐻. Since 퐺 is locally compact, there is a compact neighborhood 푈 3 1 in 퐺. By Exercise 12.4(b), there exists an open neighborhood 푉 3 1 such that 푉−1푉 ⊆ 푈. The projection map 휋 : 퐺 → 퐺/퐻 is open by definition of the quotient topology, so 휋(푉) ⊆ 퐺/퐻 is an open neighborhood of 퐻. The closure cl(휋(푉)) ⊆ 휋(푈) is therefore compact. 78 APPENDIX A. HINTS AND COMMENTS ON EXERCISES

29.11. We know that 퐵\퐵 is compact, so 휇(퐵\퐵) < ∞. Let 퐸 ⊆ 퐵× be a compact set with measure 휇(퐸) > 휇(퐵\퐵). Then (generalizing Minkowski), we claim that the map 퐵 → 퐵\퐵 is not injective on 퐸: otherwise, integrating the characteristic function Φ of 퐸 we ﬁnd 휇(퐸) ≤ 휇(퐵\퐵), a contradiction:

∫ ∫ ∑︁ 휇(퐸) = Φ(훼) d휇(훼) = Φ(훼 + 훽) d휇(훼) 퐵 퐵\퐵 훽∈퐵 ∫ ≤ d휇(훼) = 휇(퐵\퐵). 퐵\퐵

30.7. Referring to Lemma 30.6.16, we need to consider the case where 푆 is not maximal; then without loss of generality (translating), we may assume 훾/휋 is 2 × integral so 푓훾 (푥) ≡ 푥 (mod 픭). Then by Proposition 30.6.12, 푚(푆, O; O ) ≥ 1. 32.2. First compute all elements in O of norm 2, then show the product of any two of these elements belongs to 2O. 32.4. Write 푗−1훾 푗 = 푎훾/nrd(훾) with 푎 ∈ 퐹×, and taking reduced norms we get nrd(훾) = 푎2/nrd(훾), so nrd(훾) = ±푎 and 푗−1훾 푗 = ±훾; then taking reduced trace to get trd(훾) = ± trd(훾); if trd(훾) ≠ 0 then we are done, otherwise trd(훾) = 0 so 훾 = −훾 and the result is true anyway. 32.5. See Chinburg–Friedman [CF2000, Lemma 2.8]. 32.7. See Hallouin–Maire [HM2006, Proposition 5]. 33.1(a). Estimate the integral defining the length above and below. 33.5. We refer to the method of proof in the Iwasawa decomposition (Proposition 33.4.2). First, we translate by − Re 푧 to assume that 푧 = 푦푖 and then stretch to obtain 푧 = 푖. To conclude, we rotate (fixing 푧) to obtain 푧0 purely imaginary; that this is possible is easiest to see in D2, or it can be verified directly. Alternatively, if 푧, 푧0 are on a vertical line we can translate; otherwise there is a unique circle through 푧, 푧0 that is orthogonal to the real axis, having center 푐, 0 −1 and the matrix acting by 푧 ↦→ −1/(푧 − 푐) maps this circle to a vertical 1 −푐 line, so we reduce to the previous case. 푎 푏 33.6. If 푔 = , then the image of 푔−1 (R 푖) is equal to 푐 푑 >0

푎푧 + 푏 Re = 0 푐푧 + 푑

so by (33.3.8) is given by 푎푐|푧|2 + (1 + 2푏푐) Re 푧 + 푏푑 = 0, and this is a circle whose center is on the real axis if 푎푐 ≠ 0 and a vertical line if 푎푐 = 0. 33.8. By Exercise 33.5, we may assume that the points lie on the imaginary axis. We then move to the unit disc, taking the center to be the unique midpoint of the geodesic between these two points; then the points are −푡, 푡 ∈ D2 with 푡 ∈ R>0. We then compute using the formula (33.7.5) for distance that the line 퐿 is described by |푤 − 푡|2 = |푤 + 푡|2, and expanding this consists of the set of points Re 푤 = 0. This set is geodesic and is the perpendicular bisector of the 79

geodesic segment [−푡, 푡]. It follows that 퐻(훾; 푧0) is geodesic, because for two distinct points 푤, 푤0 in the right half-plane Re(푤), Re(푤0) ≥ 0, say, the geodesic between them is an arc of a circle also in the right half-plane. 33.10. Checking this on the generators in Lemma 33.4.4 make the result almost imme- 푎 푏 diate. Alternatively, note that if 푔 = and 푧, 푧0 ∈ H2 then 푐 푑 푧 − 푧0 푔푧 − 푔푧0 = (푐푧 + 푑)(푐푧0 + 푑) so plugging in recovers the result. 33.11. We have 2 cosh(log(푥)) = exp(log(푥)) + exp(− log(푥)) = 푥 + 1/푥 for 푥 ∈ R>0, and 푥 + 푦 푥 − 푦 2푦2 + = 2 1 + 푥 − 푦 푥 + 푦 푥2 − 푦2 for 푥, 푦 ∈ R with 푥 ≠ ±푦. Now simplify, and enjoy the magical cancellation. 33.12. We have 4 Im(푧) 1 − |휙(푧)|2 = |푧 + 푖|2 and 휙0(푧) = (2푖)/(푧 + 푖)2, so 2|휙0(푧)| 1 = . 1 − |휙(푧)|2 Im 푧 Part (b) follows from plugging in d푤 = 휙0(푧)d푧 into part (a). 33.15. Let 푤 = 푢 + 푖푣, so the map has (푢, 푣) = (푥/(1 + 푡), 푦/(1 + 푡)). By the chain rule, we have 푥 1 d푢 = − d푡 + d푥 (1 + 푡)2 1 + 푡 푦 1 d푣 = − d푡 + d푡 (1 + 푡)2 (1 + 푡)2 Now square; and then substitute −푡d푡 + 푥d푥 + 푦d푦 = 0 from diﬀerentiating −푡2 + 푥2 + 푦2 = −1, get 푥2 + 푦2 − 2푡(1 + 푡) d푥2 + d푦2 d푢2 + d푣2 = d푡2 + . (1 + 푡)4 (1 + 푡)2 Finally, from −푡2 + 푥2 + 푦2 = −1 show that 1 − 푢2 − 푣2 = 2/(1 + 푡), and substitute to get the result. 33.16. Or work directly, with 푝 = (푡, 푥, 푦) show we may assume 푣 = (1, 푎, 푏), in which case 푡 − 푎푥 − 푏푦 = 0, and given that 푡2 − 푥2 − 푦2 = 1, then show 푎2 + 푏2 − 1 > 0 by the Cauchy–Schwarz inequality. 34.4. We check that multiplication is continuous. Let 푈 = 푉 (ℎ; 푥, 휖) ⊆ Isom(푋) be an open ball. Suppose 푓 푔 = ℎ ∈ 푈. Let 푔(푥) = 푦 and let

푉 푓 = 푉 ( 푓 ; 푦, 휖/2) 3 푓 and 푉푔 = 푉 (푔; 푥, 휖/2) 3 푔.

We claim that 푉 푓 푉푔 ⊆ 푈, so that (푉 푓 , 푉푔) 3 ( 푓 , 푔) is an open neighborhood, 0 0 and thus the inverse image of 푈 is open as desired. So let 푓 ∈ 푉 푓 and 푔 ∈ 푉푔. 80 APPENDIX A. HINTS AND COMMENTS ON EXERCISES

Then by the triangle inequality, we have

휌( 푓 0푔0(푥), ℎ(푥)) ≤ 휌( 푓 0푔0(푥), 푓 0(푦)) + 휌( 푓 0(푦), ℎ(푥)) 휖 휖 = 휌(푔0(푥), 푔(푥)) + 휌( 푓 0(푦), 푓 (푦)) < + = 휖. 2 2

34.6. See Shimura [Shi71, Theorem 1.1, Lemma 1.2]. 34.7. To show that it is a local homeomorphism, let 푥 ∈ 푋. By definition, we find an open neighborhood 푈 3 푥 such that 푔푈 ∩ 푈 ≠ ∅ if and only if 푔 ∈ Stab퐺 (푥); since 퐺 acts freely, we have Stab퐺 (푥) = {1}, so in fact 푔푈 ∩ 푈 = ∅ for all 푔 ≠ 1. Therefore 푈 maps injectively into 퐺\푋: if 휋(푦) = 휋(푦0) for 푦, 푦0 ∈ 푈 then 푦0 = 푔푦 ∈ 푈 ∩ 푔푈 for some 푔 ∈ 퐺, so 푔 = 1 and 푦 = 푦0. To conclude, we need to show that this injection is continuous, and for that it suffices to show that if 푉 ⊆ 푈 then 휋(푉) is open; and indeed, 휋(푉) is open if and only if −1 Ð 휋 (휋(푉)) = 푔∈퐺 푔푉 is open by definition of the quotient topology, and the latter is open as each 푔푉 is open (퐺 acts continuously). (Indeed, one can show that the quotient topology is the unique topology on 퐺\푋 such that the quotient map is continuous and a local homeomorphism.) 34.8. Let 푉 3 푥 be an open neighborhood with cl(푉) compact; replacing 푈 by 푈 ∩ 푉, we may suppose 푈 has cl(푈) compact. The boundary bd(푈) ⊆ cl(푈) is closed so compact. Now there exists an open neighborhood 푉 3 푥 and an open set 푊 ⊇ bd(푈) such that 푉 ∩ 푊 = ∅. (Since 푋 is Hausdorff, the compact set bd(푈) is covered by finitely many separating open sets, so we can take an intersection.) In particular, cl(푉) ∩ bd(푈) = ∅. Let 푉 0 = 푈 ∩ 푉. Then

cl(푉 0) = cl(푈) ∩ cl(푉) = (푈 ∪ bd(푈)) ∩ cl(푉) = 푈 ∩ cl(푉) ⊆ 푈.

푛+1 34.14. Let 푥 = (1, 0,..., 0) ∈ R . We have SO(푛)' Stab푥 (SO(푛 + 1)) ≤ SO(푛 + 1). Gram–Schmidt orthogonalization implies that SO(푛 + 1) acts transitively on S푛. Next, SO(푛 + 1) is compact, because it is closed (deﬁned by polynomial equations) and bounded; thus SO(푛+1) acts properly on S푛 (Proposition 34.4.9). The result follows then from 34.4.11. 34.16. See Lee [Lee2011, Proposition 12.25]. 81

34.17. From the formula (33.5.3) for distance, we have:

2 푎푖 + 푏 푖 − |푖 − 푔푖|2 푐푖 + 푑 2 cosh 휌(푖, 푔푖) = 2 + = 2 + Im 푔푖 1 |푐푖 + 푑|2 = 2 + | (푐푖 + 푑)푖 − (푎푖 + 푏)|2 = 2 + (푏 + 푐)2 + (푑 − 푎)2 = 푎2 + 푏2 + 푐2 + 푑2 − 2푎푑 + 2푏푐 + 2 = k푔k2.

35.1. We have

∫ d푥d푦 ∫ 1/2 ∫ ∞ d푦d푥 ∫ 1/2 d푥 = = √ 2 √ 2 푦 −1/2 1−푥2 푦 −1/2 1 − 푥2 휋 ◊ = sin−1 (1/2) − sin−1 (−1/2) = . 3 36.3. Compute

(푎푤 + 푏)(푐푤 + 푑)−1 = (푎푤 + 푏)(푐푤 + 푑) nrd(푐푤 + 푑) = 푎푤푐푤 + 푏푐푤 + 푎푤푑 + 푏푑

and continue. 36.4. To show that 푔 is a hyperbolic isometry, compute the Jacobian. 푑 −푏 37.4(c). Under the map to the unit disc, 푔−1 = becomes −푐 푎

1 푎 + 푑 + 푖(−푏 + 푐) −푎 + 푑 + 푖(푏 + 푐) . 2 −푎 + 푑 − 푖(푏 + 푐) 푎 + 푑 + 푖(푏 − 푐)

Mapping back to the unit disc, the bisector is described by

| (−푎 + 푑 − 푖(푏 + 푐))(푧 − 푖) + (푎 + 푑 + 푖(푏 − 푐))(푧 + 푖)| = |푧 + 푖|

which simpliﬁes to

| (푑 − 푖푐)푧 + 푖(푎 + 푖푏)| = |푧 + 푖|.

Expanding with 푧 = 푥 + 푦푖, we get

(푐2 + 푑2 − 1)(푥2 + 푦2) − 2(푎푐 + 푏푑)푥 + (푎2 + 푏2 − 1) = 0

and we can read the result from this. 37.6. We have

푧0 = 푔푧 ∈ 퐼(푔−1) ⇔ 휌(푔−1푧0, 0) = 휌(푧0, 0) ⇔ 휌(푧, 0) = 휌(푔푧, 0) ⇔ 푧 ∈ 퐼(푔)

and the result follows.

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