Theta functions and weighted theta functions of Euclidean lattices, with some applications

Noam D. Elkies1

March, 2009

0. Introduction and overview [...]

1. Lattices in Rn: basic terminology, notations, and examples By “Euclidean space” of dimension n we mean a real vector space of dimension n, equipped with a positive-definite inner product , . We usually call such a space “Rn” even when there is no distinguished choiceh· of·i coordinates. A lattice in Rn is a discrete co-compact subgroup L Rn, that is, a discrete sub- ⊂ group such that the quotient Rn/L is compact (and thus necessarily homeomorphic with the n-torus (R/Z)n). As an abstract group L is thus isomorphic with the free abelian group Zn of rank n. Therefore L is determined by the images, call them v ,...,v , of the standard generators of Zn under a group isomorphism Zn ∼ L. 1 n → We say the vi generate, or are generators of, L: each vector in L can be written as n a v for some unique integers a ,...,a . Vectors v ,...,v Rn generate i=1 i i 1 n 1 n ∈ a lattice if and only if they constitute an R-linear basis for Rn, and then L is the PZ-span of this basis. For instance, the Z-span of the standard orthonormal basis n n e1,...,en of R is the lattice Z . This more concrete definition is better suited for explicit computation, but less canonical because most lattices have no canonical choice of generators even up to isometries of Rn. An equivalent approach defines L as a free abelian group of rank n with a positive- definite symmetric pairing; the Euclidean space is then recovered as L Z R, ⊗ which inherits a symmetric bilinear pairing from L. With this approach we must be careful about the definition of a “positive-definite pairing” so that the extension to L Z R remains positive-definite. For most of the lattices we consider, the pairing takes⊗ values in Q, and then the usual definition for inner products suffices:

v L, v = 0 = v, v > 0. (1) ∈ 6 ⇒ h i 1Department of Mathematics, Harvard University, Cambridge, MA 02138 USA; e-mail: ([email protected]). Supported in part by NSF grant DMS-0501029. DRAFT1 But for a general R-valued pairing, (1) guarantees only that the pairing on L Z R ⊗ is positive semidefinite, not necessarily positive definite: a standard counterexam- ple is L = Z2 with v, w =(v tv )(w tw ) h i 1 − 2 1 − 2 for some irrational t, when the nonzero vector x = (t, 1) L Z R satisfies ∈ ⊗ x,x = 0. For this general case we say the pairing , is “positive-definite” if h i h· ·i it has a positive-definite Gram matrix A. The Gram matrix of , with respect to h· ·i generators v ,...,v of L is the symmetric matrix with (i,j) entry A = v , v . 1 n ij h i ji The Gram matrix depends on the choice of generators, but if w1,...,wn are any n other generators then each wk = i=1 bikvi for some integers bik forming a matrix B of determinant 1, and the Gram matrix with respect to w ,...,w is BTAB, ± P 1 n which is positive-definite if and only if A is. If v = m v + + m v for some 1 1 ··· n n m =(m ,...,m ) Zn then v, v is the value at m of the quadratic form 1 n ∈ h i n n n n (m,AmT) = A m m = v , v m m , (2) ij i j h i ji i j Xi=1 Xj=1 Xi=1 Xj=1 a homogeneous polynomial of degree 2 in n variables. We next recall some further invariants2 of L and the corresponding properties of the Gram matrix. The discriminant of L is

disc L = (Vol(Rn/L))2 = det A; (3) in particular, det A is independent of the choice of generators, which we can also verify directly: if det B = 1 then det A = det BTAB. The volume √disc L of ± the torus Rn/L is known as the covolume of L. A lattice of discriminant 1 is said to be unimodular. The dual lattice L∗ is defined by

n L∗ = v∗ R v L, v, v′ Z . (4) { ∈ | ∀ ∈ h i ∈ }

If L is the Z-span of v1,...,vn with Gram matrix A, then L∗ is the Z-span of the T 1 dual basis v1∗,...,vn∗ with Gram matrix (A )− ; in particular 1 disc L∗ = (disc L)− . (5)

We say L is integral if v, v Z for all v, v L; equivalently, if L L . In this h ′i ∈ ′ ∈ ⊆ ∗ case L∗/L is a finite group with #(L∗/L) = disc(L). In particular, L = L∗ if and only if L is integral and unimodular; we naturally say such a lattice is self-dual.

2These are “invariant” in the sense that they do not depend on a choice of generators, nor on other extrinsic features such as an embedding in Rn. The Gram matrix depends on the choice of generators, and is thus not an invariant: different Gram matrices may give rise to the same lattice. DRAFT2 The basic example of a lattice is Z with the pairing x, y = xy; this lattice is self- h i dual, and is the unique unimodular lattice in the 1-dimensional Euclidean space R. More generally, for every real D > 0 there is a unique lattice of discriminant D in R, namely D1/2Z, or equivalently Z with the pairing x, y = Dxy instead; h i this lattice is integral if and only if D Z. ∈ We next give some constructions of new lattices from old. A subgroup L′ of finite n n index in a lattice L in R is itself a lattice in R whose dual contains L∗. Com- paring covolumes, we see that [L′∗ : L∗]=[L : L′]; in fact much more is true: the n inner product on R induces a perfect pairing (L′∗/L∗) (L/L′) Q/Z on the quotient subgroups, so in particular these subgroups are× isomorphic,→ albeit not in general canonically isomorphic. If v ,...,v generate L and w ,...,w generate L , then [L : L ] = det B , 1 n 1 n ′ ′ | | where B is the integer matrix formed by the coefficients bik of the expansions n wk = i=1 bikvi. If A is the Gram matrix of L with respect to the vi, then as before det BTAB is the Gram matrix of L with respect to the w . Therefore P ′ k 2 disc L′ =[L : L′] disc L, (6) an identity that can also be obtained from the first equality in (3) because

n n Vol(R /L′)=[L : L′] Vol(R /L).

If L is integral then so is L′. In the other direction, if L′ is integral, let G be any set of generators of L/L′, and G˜ an arbitrary lift of G to a subset of L; then L is integral if and only if L L and g,˜ g˜ Z for all g,˜ g˜ G˜. ⊂ ′∗ h ′i ∈ ′ ∈ n1 n2 If n = n1 + n2 and L1 and L2 are lattices in R and R respectively, then the direct sum L = L L := (v , v ) v L , v L (7) 1 ⊕ 2 { 1 2 | 1 ∈ 1 2 ∈ 2} n is a lattice in R of discriminant disc(L1) disc(L2); if Ai (i = 1, 2) is a Gram matrix for Li then L has a block-diagonal Gram matrix with blocks A1, A2. The dual of L L is L L . The direct sum L is integral if and only if L and 1 ⊕ 2 1∗ ⊕ 2∗ 1 L2 are integral, and self-dual if and only if L1 and L2 are self-dual. For example, Z2 = Z Z R2 is self-dual; iterating the construction yields the self-dual lattice ⊕ ⊂ Zn Rn for each n =1, 2, 3,.... ⊂ Of course once n > 1 this self-dual lattice is no longer unique, because we can obtain uncountably many others by applying an orthogonal linear transformation to Rn; but the resulting lattices are isomorphic. It is known that for n 7 every ≤ self-dual lattice is isomorphic with Zn (see Proposition 7 below for one approach to this result), and for every n there are only finitely many isomorphism classes. DRAFT3 2 But for large n the number of isomorphism classes grows rapidly, exceeding (cn)n for some positive c. We conclude this section by constructing the self-dual lattice E R8 and showing that E = Z8. 8 ⊂ 8 6∼ For n =1, 2, 3,..., define a lattice D Rn by n ∈ n D = (x ,...,x ) Zn : x 0 mod 2 , (8) n 1 n ∈ i ≡ i=1  X 3 n a sublattice of Z of index 2. An explicit Z-basis consists of ei + ei+1 for 0 < n i < n together with 2e1. The dual lattice Dn∗ is the union of Z and the translate of Zn by the half-lattice vector

n 1 h := (1, 1,..., 1)/2 = e . (9) 2 i Xi=1

The 4-element quotient group Dn∗ /Dn is cyclic if n is odd, and of exponent 2 when n is even. In the latter case, 2h D , so ∈ n D+ := D (D + h) (10) n n ∪ n + n is a lattice, which is unimodular because [Dn : Dn]=2=[Z : Dn]. An explicit basis consists of e + e for 0 0, a sublattice mL. How are the Gram matrix, discriminant, and dual lattice of mL related to those of L? ii) For any lattice L in Rn and any positive α R, define L α to be the lattice ∈ h i obtained from L by multiplying the pairing by α. For example, mL = L m2 . ∼ h i How are the Gram matrix, discriminant, and dual lattice of L α related to those h i of L?

1.2 i) Verify that the 4-element group Dn∗ /Dn is cyclic if and only if n is odd, and that Dn is irreducible (not the direct sum of two sublattices of positive dimension) except for D = Z 2 Z 2 (= Z2 2 ). [For the irreducibility of D for n 3 2 ∼ h i ⊕ h i ∼ h i n ≥ you might consider the vectors x D with x,x = 2, which is the smallest ∈ n h i value of x,x for nonzero x D .] h i ∈ n ii) For all n =1, 2, 3,..., each of the lattices Zn and D contains 2(n2 n) vectors n − x with x,x =2, all equivalent under automorphisms of the lattice. h i iii) The same is true of D+ for each n 0 mod 4, except that D+ = E has 240 n ≡ 8 8 such vectors, not 112. Can you find an automorphism of E8 mapping e1 + e2 to h? [From the existence of such an automorphism we readily deduce that Aut(E8) acts transitively on all 240 vectors x E such that x,x = 2; we later give several ∈ 8 h i other approaches to this result, each of which also lets us calculate the number of automorphisms.]

1.3 i) The lattice An in the n-dimensional Euclidean space n (x ,x ,...,x ) Rn+1 : x =0 0 1 n ∈ i i=0  X is defined by n A = (x ,x ,...,x ) Zn+1 : x =0 . (11) n 0 1 n ∈ i i=0  X Describe the lattice A R2 geometrically. Show that each A is the Z-span of 2 ⊂ n the independent vectors ei ei 1 (i = 1,...,n). Use the Gram matrix for these − − generators to find disc(An) = n +1. In particular disc(A3) = disc(D3); show that in fact A3 ∼= D3 by finding an explicit isometry. ii) Prove that the group An∗ /An of order n +1 is cyclic, generated by the coset of the vector n 1 e + x A∗ . − 0 n +1 i ∈ n Xi=0 +d Thus for each positive factor d of n +1 there exists a unique lattice An An∗ +d 2 ⊆ that contains An with index d. Thus disc(An )=(n +1)/d . What is the dual of DRAFT5 +d +d 2 An ? Show that An is integral if and only if d n +1, and self-dual if and only 2 +2| Z3 if d = n +1. In particular we should have A3 ∼= ; prove this by finding an explicit isomorphism. Show on the other hand that A+3 = Z8 because there is no 8 6∼ x Z+3 such that x,x = 1. Thus we should have A+3 = E ; can you find an ∈ 8 h i 8 ∼ 8 explicit isometry?

1.4 Recall that a square matrix A is called “tridiagonal” if Aij =0 for all i,j such that i j > 1. Thus vectors v ,...,v have a tridiagonal matrix if and only if | − | 1 n any two vectors whose indices differ by more than 1 are orthogonal. An example is our generators vi = ei ei 1 of An. Show more generally that a lattice L has a − − tridiagonal Gram matrix whose entries are all integers of absolute value at most 2 if and only if L is the direct sum of lattices each isomorphic with either Z or some An. Find a tridiagonal Gram matrix for E8 each of whose entries is one of 0, 1, 2, 4, with 4 occurring only once.4 +2 2 1.5 i) The integral lattice A7 of discriminant (7+1)/2 =2 has the special name E . Show that E contains no vectors x such that x,x = 1, and 126 vectors x 7 7 h i such that x,x =2. h i ii) Find a self-dual lattice L E E , and prove that it is the unique such lattice. ⊃ 7 ⊕ 7 (Hint: start by showing that L must be contained in the dual of E E .) Prove 7 ⊕ 7 that L contains no vector x such that x,x =1, and thus that L is not isomorphic h i with any of the self-dual lattices Z14, Z6 E , and Z2 D+ that we already know ⊕ 8 ⊕ 12 in 14-dimensional space. We can now account for all the self-dual lattices tabulated in [CS2, p.49] for n < n n 8 n 12 + 16: they are Z , Z − E8 (for n 8), Z − D12 (for n 12), the lattice ⊕ ≥ ⊕+4 ≥ L of exercise 1.5ii for n = 14, and Z L and A15 for n = 15. As before, we +4 ⊕ can check that A15 is distinct from the other four self-dual lattices we know in this dimension by verifying that it contains no vectors x with x,x =1. We later h i give one approach to proving the completeness of that list of self-dual lattices for n < 16, see Proposition 7 and Exercise 3.6 below.

4 Iwaniec [Iw, p.176] exhibits a tridiagonal Gram matrix for E8 with much larger entries, namely 2, 2, 4, 4, 20, 12, 4, 2 on the main diagonal and 1, 1, 3, 5, 3, 1, 1 next to it. Curiously this already appeared in [H2, §7], presented as a quadratic polynomial in x1,...,x8, though with the coefficients 2, 2, 4, 4 misprinted as 1, 1, 2, 2. Perhaps Iwaniec presented this form only to silently correct Hecke’s error, but that still leaves the mystery of how the tridiagonal form was obtained in the first place, since a considerably simpler one was available; even if we require each off-diagonal entry to be odd, we quickly find the example with diagonal 2, 2, 2, 2, 2, 6, 2, 4 and off-diagonal 1, 1, 1, 1, 1, 1, 3, 1 by computer search. The same form appears (in matrix notation, with the correct coefficients) in [Sch, p.520], where it is attributed to Minkowski. Indeed Minkowski [Min, p.77] gives this tridiagonal matrix, as well as the Gram matrix for what we now call the E8 root system (with some sign changes), with Aii =2, Aij =1 when |i − j| =1, and all other Aij zero except for A25 = A52 = −1. We shall meet tridiagonal Gram matrices at least once later (Exercise 5.2). DRAFT6 2. Theta functions of lattices By the norm of a vector x in a lattice or inner-product space we mean x,x (and h i not the length x,x 1/2). For any lattice L Rn and each k R, define h i ⊂ ∈ N (L)=# v L v, v = k . (12) k { ∈ |h i } The set is finite; indeed for any k R the sum 0 ∈ N (L)=# v L v, v k (13) k { ∈ |h i ≤ 0} k k0 X≤ is finite because L is discrete and the subset x,x k in Rn is compact (a {h i ≤ 0} closed ball if k0 0, empty if k0 < 0). The Nk(L) are important invariants of L. For example, in≥ the context of sphere packing one is interested in the minimal (nonzero) norm of L, which is the smallest k > 0 for which Nk(L) > 0, and the 5 kissing number of L, which is the value of Nk(L) for that k. The theta function or theta series ΘL is a generating function that encodes these invariants Nk(L): v,v /2 k ΘL(q) := qh i = 1 + N2k(L)q . (14) v L k>0 X∈ X Note that N0(L)=1; the factors of 2 are convenient as we soon see. The sum in (14) converges absolutely if 0 q < 1, because the sum (13) is O (kn/2) as ≤ L 0 k (indeed it is asymptotic to c kn/2, where c is disc(L) 1/2 times the 0 → ∞ L 0 L − volume πn/2/Γ((n/2) + 1) of a unit sphere in Rn. This makes it easy to justify the following derivation of the product formula for the theta function of a direct sum:

v1+v2,v1+v2 /2 ΘL1 L2 (q) = qh i ⊕ (v1,v2) L1 L2 X∈ ⊕ ( v1,v1 + v2,v2 )/2 = q h i h i

(v1,v2) L1 L2 X∈ ⊕ v1,v1 /2 v2,v2 /2 = qh i qh i

v1 L1 v2 L2 X∈ X∈

= ΘL1 (q) ΘL2 (q). (15) For example, n n ∞ m2/2 ΘZn (q) = ΘZ(q) = q . (16) m= ! X−∞ 5The minimal norm is (2r)2 where r is the radius of the largest ball in Rn whose translates by L have disjoint interiors. These translates constitute the sphere packing associated with L. Each of them is tangent to κ others, where κ is the kissing number. DRAFT7 A more remarkable identity relates the theta functions of a lattice and its dual:

Proposition 1 (functional equation for theta series) For any lattice L in Rn we have

2πt 1/2 n/2 2π/t ΘL∗ (e− ) = disc(L) t− ΘL(e− ). (17) for all t > 0.

Already the first example, with n = 1 and L = L∗ = Z, is surprising and impor- tant: ∞ πm2t 1/2 ∞ πm2/t e− = t− e− . (18) m= m= X−∞ X−∞ A famous application is Riemann’s proof of the analytic continuation and func- s 2πt tional equation of the zeta function ζ(s) = ∞ m : multiply ΘZ(e ) 1 m=1 − − − by ts/2 dt/t and integrate termwise over 0 < t < to find P ∞

∞ 2πt s/2 dt ∞ ∞ πm2t s/2 dt (ΘZ(e− ) 1) t = 2 e− t − t t 0 m=1 0 Z X Z ∞ 2 s/2 = 2 (πm )− Γ(s/2) m=1 X s/2 = 2π− Γ(s/2)ζ(s) = ξ(s) (19)

1 1 for Re(s) > 0; then split the integral as 0∞ = 0 + 1∞ and apply (18) to 0 to obtain R R R R 1 1 1 ∞ 2πt s/2 (1 s)/2 dt ξ(s) + + = (ΘZ(e− ) 1)(t + t − ) (20) s 1 s 2 − t − Z1 for 0 < Re(s) < 1. To recover the analytic continuation of ξ(s) to all of C (with simple poles at s = 0 and s = 1), observe that the integral in (20) is an analytic function of s on all of C, because ΘZ(e 2πt) 1 decays exponentially as t ; − − →∞ the functional equation ξ(s) = ξ(1 s) then follows from the symmetry of the − integral under s 1 s. ↔ − n Applying the same definite integral to ΘL for a general lattice L in R , but using ts instead of ts/2, we obtain

∞ 2πt s dt s (Θ (e− ) 1) t = π− Γ(s)ζ (s) = ξ (s), (21) L − t L L Z0 DRAFT8 where ζL is the zeta function of L, defined by s ζL(s) := v, v − , (22) v∈L h i Xv6=0 and ξL is defined by the last equality in (21). Transforming the integral as before, and using the functional equation (17) relating ΘL with ΘL∗ , we obtain the identity

n 1/2 ξ ∗ s = disc(L) ξ (s). (23) L 2 − L See Exercises 2.3 and 2.5ii below.  For a rather more frivolous application of (18) (and one admittedly unrelated to our main topic), differentiate both sides of (18) with respect to t and set t =1 to find6

d 2πt 1 2π ΘZ(e− ) = ΘZ(e− ), dt −4 t=1 whence

∞ πm2t 2π ∞ 2 πm2t 1+2 e− = ΘZ(e− )=8π m e− . m=1 m=1 X X Therefore 8πsummation formula in Rn, which relates the sums of a Schwartz function over L and of its Fourier transform over L∗. Here a Schwartz function is a C∞ function f : Rn C such that f and all its partial derivatives decay as o( x,x k) → h i for all k as x,x . The Fourier transform fˆ : Rn C is defined by h i→∞ → 2πi x,y fˆ(y) = f(x) e h i dµ(x), (25) x Rn Z ∈ and is a Schwartz function if f is. 6The next identity was certainly known to Riemann, if not to Poisson himself: an equivalent form appears in the last displayed equation before formula (2) in [Ed, p.17], where Edwards recites 1 Riemann’s derivation of a series representation for ξ( 2 + it). DRAFT9 Theorem 2 (Poisson summation in Rn) Let L be any lattice in Rn. Then 1/2 f(x) = (disc L)− fˆ(y) (26) x L y L∗ X∈ X∈ for all Schwartz functions f : Rn C. → Note that the y =0 term in (26) is 1 1/2 ˆ (27) (disc L)− f(0) = n f(x) dµ(x). Vol(R /L) x Rn Z ∈ After multiplying (26) by disc(L)1/2, we can thus interpret the left-hand side as a Riemann sum approximating the integral fˆ(0). The terms fˆ(y) for nonzero y L ∈ ∗ then measure the discrepancy between this Riemann sum and the integral. Proof of Poisson summation: Define F : Rn C by → F (z) = f(x + z). (28) x L X∈ Because f is Schwartz, the sum converges absolutely to a C∞ function, whose value at z =0 is the left-hand side of (26) Since F (z) = F (x + z) for all z Rn ∈ and x L, the function descends to a C function on Rn/L, and thus has a ∈ ∞ Fourier expansion 2πi y,z F (z) = Fˆ( y) e h i, (29) − y L∗ X∈ where ˆ 1 2πi z,y F (y) = n F (z) e h i dµ(z). (30) Vol(R /L) z Rn/L Z ∈ Note that the vectors y L are exactly those for which e2πi x,y is well-defined ∈ ∗ h i on Rn/L. Now choose a fundamental domain R for Rn/L; for instance, let v1,...,vn be generators of L and set R = a1v1 + + anvn : 0 ai < 1 . Then we have { ··· ≤ }

n 2πi y,z Vol(R /L)Fˆ(y) = F (z) e h i dµ(z) z R Z ∈ 2πi y,z = f(x + z) e h i dµ(z) Zz R x L ∈ X∈ 2πi y,z = f(z) e h i dµ(z) x L z R x X∈ Z ∈ − 2πi y,z = f(z) e h i dµ(z) = fˆ(y), (31) z Rn Z ∈ DRAFT10 where we used in the last step the fact that Rn is the disjoint union of the translates R x of R by lattice vectors. Thus (29) becomes − 1 2πi y,z F (z) = fˆ( y) e h i. (32) Vol(Rn/L) − y L∗ X∈ Taking z =0 we obtain (26), Q.E.D. The functional equation (17) is then the special case f(x) = exp( π x,x /t) − h i of (26). Proof of the functional equation (17) for theta series: Let f(x) = exp( π x,x /t) − h i in (26). We claim that fˆ(y) = tn/2 exp( π y, y t). Choosing any orthogonal n − h i coordinates (x1,...,xn) for R , we see that the integral (25) defining fˆ(y) factors as n 2 ∞ πx /t 2πixj yj e− j e dxj, jY=1 Z−∞ which reduces our claim to the case n =1, which is the familiar definite integral

∞ πx2/t 2πixy 1/2 πty2 e− e dx = t e− . Z−∞ Using these f and fˆ in the Poisson summation formula (26) we deduce the func- tional equation (17), Q.E.D.

Exercises 2.1 How are the theta series and zeta function of L α related with Θ and ζ ? h i L L 2.2 Having given in (16) a formula for ΘZn in terms of ΘZ, we find formulas for + 7 the theta functions of Dn, Dn∗, and Dn in terms of ΘZ and two further functions ΦZ, ΨZ, defined for 0 q < 1 by ≤

∞ 2 ΦZ(q) = ( 1)mqm /2 = 1 2q1/2 +2q4/2 2q9/2 + , (33) − − − −··· m= X−∞ ∞ (m+ 1 )2/2 1/8 9/8 25/8 49/8 ΨZ(q) = q 2 = 2 q + q + q + q . (34) ··· m= X−∞
Thus

4 1/4 Φ Z(q) = ΘZ(q) 2ΘZ(q ), ΨZ(q) = ΘZ(q ) ΘZ(q). (35) − − 7 The use of ΦZ, ΨZ for these functions is an ad hoc notation. Usually these would be written as the value at u =1 of Jacobi’s theta functions ϑ4, ϑ2; likewise ΘZ is Jacobi’s ϑ3 evaluated at u =1. DRAFT11 Prove that for n =1, 2, 3,...

1 n n n n Θ (q) = ΘZ(q) + ΦZ(q) , Θ ∗ (q) = ΘZ(q) + ΨZ(q) , (36) Dn 2 Dn   and for n even

1 n n n Θ + (q) = ΘZ(q) + ΨZ(q) + ΦZ(q) , (37) Dn 2   + 4 In particular, since Dn ∼= Z we have Jacobi’s identity 4 4 4 ΘZ(q) = ΨZ(q) + ΦZ(q) . (38)

2.3 Complete the derivation of the functional equation (23) relating ξL and ξL∗ , showing along the way that ξL is holomorphic except for simple poles at s =0 and s = n/2. What are the residues at these poles? 2.4 Show that A = A 1/3 . (The geometric description of A from Exercise 1.3i 2∗ ∼ 2h i 2 should help.) Use this and the functional equation for ΘA2 to prove

e2π/√3 > 8√3 π 6. − 2π/√3 The two sides are not nearly as close as in (24), despite e− being even smaller π than e− ; why? 2.5 [Sphere packing bounds from Poisson summation] 2.6 We can get further use of the formula (32) by evaluating both sides at z / L. ∈ Here are two examples for the case n =1: i) Taking z = 1/2, L = L = Z, and f(x) = exp( πx2/t), obtain a functional ∗ − equation relating ΨZ and ΦZ. Check that this equation is consistent with the result of applying the ΘZ functional equation (18) to the formulas (35) for ΨZ and ΦZ in terms of ΘZ. Verify also that your equation, and the formulas (36,37), are consis- + tent with (17) for L = Dn and L = Dn . ii) Let χ be the Dirichlet character mod 12 defined by χ (1) = χ ( 1) = 1, 12 12 12 − χ12(5) = χ12( 5) = 1 (and χ12(m)=0 if 2 m or 3 m). Show that if f is a Schwartz function− on R−then | |

∞ 1/2 ∞ ˆ χ12(m)f(m)=12− χ12(n)f(n/12). m= n= X−∞ X−∞ πx2t Letting f(x) = e− , obtain an identity analogous to (18), and deduce a func- s tional equation for the Dirichlet L-series L(s, χ12) = m∞=1 χ12(m)m− . P DRAFT12 2.7 The Schwartz condition is much more restrictive than is needed to justify Pois- son summation (though Schwartz functions suffice for all our applications to lat- tices and theta series).8 Show for instance that our derivation of (26) is valid also x 2 for n = 1 and f(x) = e−| |, fˆ(y)=2/((2πy) + 1), and use this to evaluate 2 2 in closed form n∞=1 1/(n + c ) for c > 0. Verify that your answer approaches ζ(2) = π2/6 as c 0. P→

3. Theta functions of self-dual lattices as modular forms for Γ+

We next consider ΘL as a function of a complex variable. For general lattices L we cannot make sense of ΘL(q) as a function of q in a neighborhood of q = 0 in C, because the exponents v, v /2 in (14) need not be integers. However, the change 2πt h i of variables q = e− suggested by the functional equation (17) yields a function of t that extends to a holomorphic function on the half-plane Re(t) > 0. That functional equation then extends to this half-plane, either by analytic continuation or by using the same proof.

Suppose now that L is self-dual. Then ΘL = ΘL∗ , so the functional equation (14) relates the values at t and 1/t of the same function. Also, each exponent v, v /2 is 1 2πt h i in 2 Z because L is integral. Thus ΘL(e− ) is also invariant under the imaginary translation t t +2i. Combining this invariance with the functional equation we 7→ then obtain further identities, one for each fractional linear transformation gener- ated by t 1/t and t t +2i. 7→ 7→ We make the coefficients of these transformations integral using the further change of variable t = iτ. Then τ is in the Poincare´ upper half-plane

= τ C Im(τ) > 0 , H { ∈ | } and q = e2πiτ . (39) Our transformations t 1/t and t t +2i then become 7→ 7→ S : τ 1/τ, T 2 : τ τ +2 7→ − 7→ acting on . We use the notation T 2 because we later need also the translation H T : τ τ +1. 7→ Recall that the group of orientation-preserving isometries of with respect to H the hyperbolic metric dτ / Im(τ) consists of the fractional linear transformations | | 8For example, Poisson summation holds if there exists δ > 0 for which both f(x) and fˆ(x) are O(hx, xi−(n/2)−δ) as hx, xi→∞ [SW, Ch.VII, Cor. 2.6]; this includes the example f(x) = e−|x| in Exercise 2.6 with n =1. DRAFT13 γ : τ (aτ + b)/(cτ + d) with a,b,c,d R and ad bc > 0, uniquely 7→ ∈ − determined up to scaling (a,b,c,d) (λa, λb, λc, λd). We may regard γ as the 7→ τ a b τ projective linear transformation of the (τ : 1)-line that takes 1 to cd 1 . Thus we compose γ’s by multiplying the corresponding 2 2 matrices a b . Requiring ×
cd

ad bc =1 determines λ up to sign; we thus identify our group of fractional linear −
transformation of with PSL (R)=SL (R)/ 1 . H 2 2 {± } Now the maps S : τ 1/τ and T : τ τ +2 become the integer matrices S 0 1 T2 7→12 − 7→ 2 = 1− 0 and = 01 of determinant 1. These generate a subgroup S, T of the full modular group h i

Γ := SL (Z)/ 1 = PSL (Z) 2 {± } 2 Because T2 is congruent mod 2 to the identity matrix I, every γ S, T 2 is 01 a b ∈ h i congruent to either I or 10 mod 2. (The reduction of cd PSL2(Z) mod 2 a b a b ∈ is well-defined because cd cd mod 2.) It turns out that this necessary −
≡2
condition is also sufficient: S, T is the group, call it Γ+, ofall PSL2(Z) matrices h
01i
congruent mod 2 to either I or 10 . See Exercise 3.1 below for one approach to the identification of S, T 2 with Γ . h i +
We are thus led to define for any lattice L in Rn the function

2πiτ πikτ θL(τ):=ΘL(e )=1+ Nk(L)e , (40) k>0 NkX(L)6=0 on . If L is self-dual then θ satisfies the functional equations H L θ (τ +2) = θ (τ), θ ( 1/τ)=(τ/i)n/2θ (τ), (41) L L L − L where “(τ/i)n/2” is the nth power of the principal square root of τ/i (that is, the square root with positive real part). Iterating these functional equations yields for each γ S, T 2 =Γ an identity ∈h i + n n/2 θL(γ(τ)) = ǫc,d(cτ + d) θL(τ) (42) for some ǫ C with ǫ8 =1.9 A holomorphic function φ : C satisfying c,d ∈ ∗ c,d H→ φ(γ(τ)) = ǫn (cτ + d)n/2φ(τ) for all γ = a b Γ is a “weakly modular c,d ± cd ∈ + form of weight n/2 for Γ ” (more fully, for Γ and the ǫn ; these factors are said + +
c,d to form a “multiplier system of weight n/2” [Iw, 2.6] — a multiplier system beinga

9 ′ ′ 2k We write ǫc,d rather than ǫγ because if some γ has the same (c, d) as γ then γ = T γ for Z ′ ′ some k ∈ , so θL(γ(τ)) = θL(γ (τ)) so γ and Γ have the same ǫ; also, γ determines (c, d) only up to sign, and changing (c, d) to (−c, −d) multiplies ǫc,d by ±i. DRAFT14 system of factors in an identity such as (42) that is consistent with θL(γ1(γ2(τ)) = θL((γ1γ2)(τ)) for all choices of γ1, γ2.) We shall soon see that θL also satisfies the additional condition that make it modular, not just weakly modular; this will let us cite results from the theory of modular forms that for each n confine θL to an affine vector space of dimension n/8 . To motivate this additional condition we must ⌊ ⌋ first review a few further facts concerning the action of Γ and Γ on . + H The relevant results for Γ are very well known:

Theorem 3 i) Γ = PSL2(Z) is generated by S and T . ii) The action of Γ on has a fundamental domain H = τ : Re(τ) 1/2, τ 1 . (43) F { ∈H | | ≤ | | ≥ } The second assertion means that every Γ-orbit has a representative in , which is F unique if it is in the interior of . Thus the images of under Γ cover and do not overlap except on the boundaries;F see Figure 1 forF a picture of partH of this tiling of . See [Se, VII, Theorems 1 and 2] for an exposition that nicely proves H both parts of Theorem 3 together. While is not compact, it is closed in and a F H sequence z with no accumulation point in must approach the “cusp” i in { j} F ∞ the sense that Im z . j →∞

T 1 T − F F F

i t T 1S S T S − F F F

R 2 1 0 1 2 − − Figure 1: The fundamental domain for Γ and some of its nearby images F

The corresponding facts for Γ+ are:

2 Theorem 4 i) Γ+ is generated by S and T . ii) The action of Γ on has a fundamental domain + H = τ : Re(τ) 1, τ 1 . (44) F+ { ∈H | | ≤ | | ≥ } DRAFT15 This can be proved by adapting the argument of [Se, VII] for Theorem 3. Alterna- tively it can be derived from Theorem 3 as follows. For part (i), see Exercise 3.1 be- low. For part (ii), we first show that [Γ:Γ+]=3. (This, as well as part (i), is noted by Serre in the concluding “Complements” section of [Se, VII].) We saw already that reduction mod 2 gives a well-defined homomorphism Γ SL2(Z/2Z). The homomorphism is readily seen to be surjective; for example, check→ that the images of S and T generate SL2(Z/2Z). Since SL2(Z/2Z) has order 6, the preimage Γ+ of the 2-element subgroup 1, 01 has index 6/2=3 as claimed. We then obtain { 10 } a fundamental domain for Γ by forming the union of the images of under the +
F coset representatives 1,T,TS. The fundamental domain is then obtained by F+ applying T 2 Γ to the right half of T T S . See Figure 2. − ∈ + F ∪ F

F

i t

R 1 0 1 − Figure 2: The fundamental domain dissected into three images of F+ F We see that has two cusps, at and 1. For a general finite-index subgroup F+ ∞ ± Γ Γ, we can similarly get a fundamental domain by combining [Γ:Γ ] images ′ ⊆ ′ of , and the cusps may be identified with the orbits of the action of Γ on Q F ′ ∪ = P1(Q). {∞} Now our theta function θL not only satisfies the identity (42) that makes it weakly modular, but also remains bounded as Im τ — that is, as τ approaches the → ∞ cusp i — because q 0 as τ i . For each γ Γ it follows via (42) that ∞ → → ∞ ∈ + (Im τ) n/2θ (γ(τ)) remains bounded as τ i( ), as long as τ remains in . − L → ∞ F Changing variables we see that (Im τ)n/2θ (τ) remains bounded as τ a/c = L → γ(i ), as long as τ remains in γ . This does not constrain the growth of θ (τ) ∞ F L as τ approaches the cusp 1 or its images under Γ+. But we can prove directly:

Lemma 5 Let L be a lattice in Rn. Then for any t > 0 the function τ 0 7→ (Im τ)n/2θ (τ) is bounded on the strip τ : Im τ t . L { ∈H ≤ 0} DRAFT16 Proof : If Im τ = t then

n/2 n/2 2πiτ n/2 2πt t θ (τ) = t Θ (e ) t Θ (e− ) , | L | | L | ≤ | L | πi v,v τ 2πt because each of the terms e h i in the sum defining ΘL(e− ) has absolute π v,v t 2πt value equal to the corresponding term e− h i in ΘL(e− ). Hence by the func- tional equation (17) we have

n/2 1/2 2π/t t θ (τ) disc(L) Θ (e− ), | L | ≤ L and Θ (e 2π/t) Θ (e 2π/t0 ), again because the inequality holds termwise. L − ≤ L − This gives the upper bound disc(L)1/2Θ (e 2π/t0 ) on (Im τ)n/2θ (τ) , Q.E.D. L − | L | This suggests the following definitions: fix a finite-index subgroup Γ′ of Γ and some multiplier system ε (for all (c,d) occurring as the bottom row of some { c,d} γ Γ ). A holomorphic function φ : C satisfying ∈ ′ H→ n/2 φ(γ(τ)) = εc,d(cτ + d) φ(τ) for all γ = a b Γ is a weakly modular form of weight n/2 for Γ and the ε ; ± cd ∈ ′ ′ c,d if moreover (Im τ)n/2φ(τ) is bounded in strips Im τ t then φ is a modular |
| ≤ 0 form of weight n/2 for Γ′ and the εc,d. The boundedness condition is equivalent to one growth condition for each cusp of /Γ . Thus Lemma 5, combined with H ′ the preceding analysis, states that if L is self-dual then θL is a modular form of weight n/2 for Γ and the multiplier system ǫn . + { c,d} Given Γ′, m, and the εc,d, these modular forms constitute a vector space over C, denoted by M (Γ , ε ), or simply M (Γ ) if the ε are known. This is n/2 ′ { c,d} n/2 ′ c,d a finite-dimensional vector space, and much is known about the coefficients of its elements. We shall see that for any lattice L Rn with rational inner products ⊂ the theta function θ is in M (Γ , ε ) for some Γ and ε . This can be used L n/2 ′ { c,d} ′ c,d to obtain very precise statements on the Nk(L), that is, on the radial distribution of lattice vectors. We shall also study the angular distribution by generalizing θL to “weighted theta functions” and showing and show that they, too, are modular forms. For the rest of this section and the next we illustrate this by using the structure of Mn/2(Γ′) for Γ′ = Γ+ and Γ′ = Γ to study self-dual lattices via their vectors’ radial distribution. If ε(1) and ε(2) are multiplier systems of weights n /2 and n /2 for Γ , then { c,d} { c,d} 1 2 ′ ε(1)ε(2) is a multiplier system of weight (n +n )/2, and the product of modular { c,d c,d} 1 2 forms in M (Γ , ε(1) ) and M (Γ , ε(2) ) is in M (Γ , ε(1)ε(2) ). n1/2 ′ { c,d} n2/2 ′ { c,d} (n1+n2)/2 ′ { c,d c,d} This happens in our setting where Γ′ = Γ+ and we use for each weight n/2 the DRAFT17 n multipliers ǫc,d of (42). (Note that this is consistent with the identity (15): the change of variable q = e2πiτ transforms that identity to θ θ = θ , equat- L1 L2 L1 L2 ⊕ ing a modular form of weight (n1 + n2)/2 with the product of forms of weights n1/2 and n2/2.) This gives the direct sum

∞ M+ := Mn/2(Γ+) (45) n=0 M the structure of a graded algebra. For general Γ′ such algebras can be quite com- plicated, but our M+ is known to have the following simple description:

Theorem 6 The algebra M+ is freely generated over C by the modular forms θZ of weight 1/2 and θ of weight 4; equivalently, by θ and10 E8 Z

1 8 1/2 3/2 2 ∆ := (θZ θ ) = q 8q + 28q 64q + . (46) + 16 − E8 − − −··· Thus each M (Γ )(n 0) has dimension 1 + n/8 and basis n/2 + ≥ ⌊ ⌋ n 8m m θ − ∆ : m =0, 1, 2,..., n/8 . (47) { Z + ⌊ ⌋}

This theorem can be proved starting from the description of in much the same F+ way that Serre obtains the generators of the algebra of modular forms for Γ ([Se, VII, Theorem 4]; Theorem 11 below). Corollary. Let L be a self-dual lattice in Rn. Then

n/8 ⌊ ⌋ n n 8m m θL = θZ + cmθZ− ∆+ (48) m=1 X for some constants c (m =1, 2,..., n/8 ). m ⌊ ⌋ n/8 n 8m m Proof : By Theorem 6 we have θL = m⌊ =0⌋ cmθZ− ∆+ for some cm (m = 0, 1, 2,..., n/8 ); so we need only show c0 =1. But c0 is the value of the sum at ⌊ ⌋ P q =0, which is N0(L)=1 as desired. In particular, the coefficients N (L) of θ for k = 1, 2,..., n/8 determine θ , k L ⌊ ⌋ L because we can use them to calculate the cm iteratively. We can already use this to prove our earlier claim:

10 The notation “∆+” for this form is not standard; we use it here in analogy to the weight-12 form ∆ which plays a similar role in the next section. DRAFT18 Proposition 7 If n < 8 then every self-dual lattice in Rn is isomorphic with Zn.

Proof : Here n/8 = 0 so our Corollary determines the theta series completely: ⌊ ⌋ n θL = θZn . Comparing coefficients, we deduce Nk(L) = Nk(Z ) for all k. In n particular, N1(L)=2n, because N1(Z )=2n, as may be seen either directly or by expanding (16) in powers of q1/2. Thus N (L)=2n, so L contains n pairs v 1 ± i (1 i n) of vectors with v , v =1. For i = j we then have v , v < 1 by ≤ ≤ h i ii 6 |h i ji| Cauchy-Schwarz; since L is integral, it follows that v , v =0. That is, the v are h i ji i orthonormal. Therefore L contains their Z-span, call it L0, which is isomorphic with Zn. But then L = L L = L , so L = L = Zn, Q.E.D. ∗ ⊆ 0∗ 0 0 ∼ See Exercises 3.5 and 3.6 for the classification of self-dual L Rn with n =8 and ⊂ 9 n 15; Exercise 3.7 for an application of the case n = 2 to prove Fermat’s ≤ ≤ 2 two-squares theorem, which determines Nk(Z ) when k is an odd prime; and Ex- 2 ercise 3.8 for a different construction of θZ2 that yields a formula for Nk(Z ) for all k.

We conclude this section with a warning: we used the fact that θL = θZn to prove 2 L ∼= Z , but in general it is possible for lattices L and L′ to have the same theta se- ries (equivalently: to satisfy Nk(L) = Nk(L′) for all k) without being isomorphic. We say more about this later, and give an example with self-dual lattices in R16.

Exercises S 0 1 T 11 S2 ST 3 3.1 Verify that the matrices = 1− 0 and = 01 satisfy = ( ) = I, 2 3 − and thus that their images S, T in PSL2(Z)=Γ satisfy S = (ST ) = 1. Use

2 this, and the fact that Γ = S, T , to prove that Γ+ = S, T . [Write any γ Γ+ h i 1 h i ∈ as g1g2g3 ...gr where each gj S, T, T − , and eliminate any occurrences of 1 1 ∈ { } SS, T − T , or T T − in the product. Then, working from one side (say from the 1 left), combine pairs with gj = gj+1 = T ± to obtain a product of factors in 2 2 3 1 1 S, T , T − . If an impasse arises, use (ST ) = I to replace T ST or T − ST − { 1 } 1 2 1 by ST − S or STS respectively; note that T − ST = T − T ST and TST − = T 2T 1ST 1. This process either terminates or proves that γ / Γ by writing γ as − − ∈ + the product of an element of S, T 2 with either T or T S.] h i 3.2 Define a q-series

∞ n2/24 η = χ12(n)q n=1 X = q1/24(1 q q2 + q5 + q7 q12 q15 + q22 + + ), (49) − − − − − − ··· where χ12 is the even Dirichlet character mod 12 introduced in Exercise 2.5(ii). Use the identity proved in that Exercise to show that η satisfies, for each γ = DRAFT19 a b Γ, a functional equation ± cd ∈ 1/2
η(γ(τ)) = εc,d(cτ + d) η(τ) (50) for some ε C with ε24 =1. We thus say that η is a modular form of weight c,d ∈ ∗ c,d 1/2 for Γ and the εc,d (not just weakly modular, because η is clearly bounded at the one cusp q 0). → 3.3 i) Let L be an integral lattice in Rn that contains a vector v of norm 1. Show that every x L can then be written uniquely as x + nv for some n Z and ∈ ′ ∈ x L with x , v =0, and thus that such vectors x constitute a lattice L in the ′ ∈ h ′ i ′ ′ orthogonal complement of Rv, whence L = L Z. ∼ ′ ⊕ ii) More generally, let L be an integral lattice in Rn, and L L a lattice in some 1 ⊂ subspace V of dimension n . If L is self-dual, show that L = L L where 1 1 1 ⊕ ′ L = L V . (Hint: given x L, use the homomorphism L R, v x, v . ′ ∩ ⊥ ∈ 1 → 7→ h i Part (i) is the special case L1 = Zv.) 3.4 Either by iterating the construction in part (i) of the previous exercise, or using part (ii) directly, prove that every integral lattice L in Rn can be uniquely written n1 as L′ L1 where L1 ∼= Z for some integer n1 [0, n] and L′ is a lattice in n n⊕1 ∈ R − that contains no vectors of norm 1. (This gives an alternative conclusion of the proof that for n < 8 every self-dual lattice in Rn is isomorphic with Zn; here n1 = n and L′ is the zero lattice.) 3.5 Show that if L is a self-dual lattice in R8 then either L = Z8 or θ = θ . ∼ L E8 [In particular, in the latter case N2(L) = N2(E8) = 240. If you know about root systems then this is all you need to deduce L ∼= E8, because E8 is the only root lattice of rank at most 8 with as many as 240 vectors of norm 2. Thus L has a sublattice L1 of finite index isomorphic with E8, and then [L : L1]=1 because disc(E8)=1, so L = L1.] 3.6 More generally, show that if 0 < n < 16 and L is a self-dual lattice in R8 n n 8 without vectors of norm 1 then n 8 and θ = θZ 2nθ − ∆ . In particular, ≥ L − Z + N (L)=2n(23 n). [Using this plus the classification of root systems we can 2 − show that E is the only such lattice with n < 12. For 9 n 15, it follows from 8 ≤ ≤ Proposition 7 and Exercise 3.3ii that the root system cannot contain E8; this leaves only the root systems D for n = 12, E2 for n = 14, and A or A D for 12 7 15 4 ⊕ 11 n = 15. In each case we then know a sublattice of finite index in L. Because this sublattice must have square discriminant, A D cannot occur. In each of the 4 ⊕ 11 remaining cases the root lattice determines L uniquely as we saw in Exercises 1.2, 1.3, and 1.5 above.] 3.7 (Fermat’s two-squares theorem) Let p be an odd prime. Fermat’s two-squares 2 2 theorem asserts that p can be representedas a sum x1+x2 of squares of two integers DRAFT20 if and only if p 1 mod 4, in which case the representation is unique up to the ≡ obvious transformations x x , x x , and x x . Equivalently: if 1 ↔ − 1 2 ↔ − 2 1 ↔ 2 p 1 mod 4 then N (Z2)=0; and if p +1 mod 4 then N (Z2)=8 and the ≡ − p ≡ p dihedral group of symmetries of Z2 acts simply transitively on the 8 vectors with x,x = p. We give a version of a standard proof of this theorem that concludes h i by invoking the case n =2 of Proposition 5. If p = x2 + x2 then x = 0 so we may let r = x /x Z/pZ, so that r2 = 1. 1 2 2 6 1 2 ∈ − By Legendre’s criterion, 1 is a square in Z/pZ if and only if ( 1)(p 1)/2 − − − ≡ +1 mod p. This already proves the case p 1 mod 4. If p +1 mod 4 then ≡ − ≡ there are two square roots of 1 in Z/pZ; let r be one of them. The vectors − (x ,x ) Z2 such that x rx mod p constitute a lattice, call it L . We 1 2 ∈ 1 ≡ 2 p want x Lp such that x,x = p. Prove that Lp 1/p is integral. Check that 2 ∈ h i 2 h i [Z : Lp] = p, and thus that disc(Lp) = p . Conclude that Lp 1/p is self-dual. 2 h i Thus Lp 1/p ∼= Z by Proposition 5. In particular Lp 1/p contains four unit vectors. Useh thisi to complete the proof of the two-squaresh theorem.i 2 11 3.8 (The coefficients Nk(Z ) of θZ2 ) Recall that the hyperbolic cosine is de- z z fined by cosh z = cos(iz)=(e + e− )/2, and the hyperbolic secant is sech z = z z 1/ cosh(z) = sec(iz)=2/(e + e− ). It is a known application of contour inte- gration that for t > 0 (or even Re(t) > 0) the Fourier transform of sech(πtx) is 1 1 t− sech(πt− y). Use this to prove that

∞ ∞ s2(τ) := sech(mπiτ) = sec(mπτ) (52) m= m= X−∞ X−∞ is a modular form of weight 1 for Γ . Since s (τ) 1 as τ i we must have + 2 → → ∞ 11 2 The formula s2(τ) = θZ(τ) = θZ2 (τ) proved in this Exercise is over a century old; the proof via modular forms is also not new but its origins are harder to track down. Dickson [Di, p.235] attributes the identity

q q3 q5 (1+2q +2q4 +2q9 + ··· )2 =1+4 − 4 +4 −··· (51) 1 − q 1 − q3 1 − q5 to Jacobi, in a “letter to Legendre, Sept. 9, 1828”; the right-hand side is obtained from the formula (51) for s2(τ) by interchanging the sums over m and d. For Jacobi and others, including Dickson himself [KD], the identity arises in the theory of elliptic functions and the proof does not explicitly 2 use the fact that s2 and θZ are modular. Much more recently, the development in [SS, Ch.10, §3.1, pages 297–304] closely follows the modular approach of our Exercise; this approach is also noted 2 in [Coh, p.106, Remark (6)] though without deducing the formula for Nk(Z ). But this too was probably known much earlier.

DRAFT21 2 s2(τ) = θZ(τ) = θZ2 (τ). Then expand s2(τ) in powers of q:

∞ qm/2 ∞ ∞ s (τ)=1+4 =1+4 χ (d)qdm/2, (53) 2 1 + qm 4 m=1 m=1 X X Xd=1 where χ4 is the Dirichlet character mod 4 defined by χ4( 1) = 1 and χ4(0) = ±2 ± χ4(2) = 0. Deduce the formula for the number Nk(Z ) of representations of 2 an integer k > 0 as the sum of two squares: Nk(Z )=0 if there is a prime p 3 mod 4 such that the p-valuation v (k) is odd; otherwise, ≡ p 2 Nk(Z )=4 (1 + vp(k)). (54) p 1 mod 4 ≡ Y Recover Fermat’s two-squares theorem as the special case that k is an odd prime.

4. The characteristic coset and shadow of a self-dual lattice For any integral lattice L the identity

v + w, v + w = v, v +2 v, w + w, w (55) h i h i h i h i yields a homomorphism

L Z/2Z, v v, v mod 2. (56) → 7→ h i Suppose further that disc L is odd. Then , mod 2 gives an isomorphism from h· ·i L/2L to L Hom(L, Z/2Z); in particular there is a unique coset C of 2L in L, called the haracteristic coset [Se, Ch. V], that maps to the homomorphism (56). Thus w C if and only if v, w v, v mod 2 for all v L. Scaling by ∈ 1 h i≡h i ∈ 1/2 yields a coset of L in 2 L called the shadow of L; we denote the shadow by s(L). For example, the lattice Z has characteristic coset C =2Z +1 and shadow s(Z) = 1 C = Z + 1 , and likewise for Z α for any odd integer α > 0. The 2 2 h i characteristic coset and shadow are additive: if L1 and L2 are integral lattices of odd discriminant with characteristic cosets C1 and C2, then the integral lattice L L has odd discriminant and its characteristic coset and shadow are C C 1 ⊕ 2 1 ⊕ 2 and s(L ) s(L ) respectively. For example, the characteristic vectors of Zn are 1 ⊕ 2 the vectors all of whose coordinates are odd. Now the vectors in any coset of 2L in L have the same norm mod 4, but for the characteristic coset we actually get congruence mod 8. Thus if w, w C we may ′ ∈ write w = w +2v for some v L, so ′ ∈ w′, w′ = w +2v, w +2v = w, w + 4( v, w + v, v ) w, w mod 8. h i h i h i h i h i ≡h i DRAFT22 Thus we have a lattice invariant σ(L) mod 8 which is the common residue mod 8 of the norms of all characteristic vectors. (Equivalently, all shadow vectors have quarter-integral norm congruent to 1 σ(L) mod 2Z.) For example, σ(Z α ) = α 4 h i because every odd square is congruent to 1 mod 8. Again we have additivity: if lattices L ,L have odd discriminant then σ(L L ) = σ(L ) σ(L ). For 1 2 1 ⊕ 2 1 ⊕ 2 example, σ(Zn) = n. The self-dual lattices D+ (4 n) also have characteristic n | vectors of norm n mod 8 (see Exercise 4.1). In fact it turns out that σ(L) = n ≡ for any self-dual lattice L Rn. This can be proved algebraically, for example ⊂ by extending σ to indefinite quadratic forms of discriminant 1 and studying the arithmetic of such forms as in [Se, Ch.I–V]. In our positive-±definite setting the characteristic coset and shadow arise naturally when we study the transformation of θ under arbitrary γ Γ (not just γ Γ ). In the course of this study we shall L ∈ ∈ + give a proof of σ(L) = n for self-dual L Rn using theta functions. ⊂ We already know θ (γ(τ)) for all γ Γ . We shall determine θ (T (τ)) and L ∈ + L θ (T S(τ)); this will suffice to obtain θ (γ(τ)) for all γ Γ because T and T S L L ∈ are representatives of the two nontrivial cosets of Γ+ in Γ. The action of T : τ τ +1 is easy: this map takes eπiτ to eπiτ , so for any 7→ − integral lattice L we have simply

∞ θ (T (τ)) = θ (τ +1)=1+ ( 1)kN (L)eπikτ . L L − k Xk=1

The formula for θL(T S(τ)) can be expressed in terms of the theta series for the shadow of L; as suggested by our notation ΨZ in Exercise 2.2, we define

v,v /2 ∞ k ΨL(q) := qh i = 1 + N2k(s(L))q , (57) v s(L) k=1 ∈X X 2πiτ πikτ ψL(τ) := ΨL(e ) = Nk(s(L))e , (58) k>0 Nk(Xs(L))6=0 where N (s(L))=# v s(L) v, v = k . (59) k { ∈ |h i } Then we have:

Proposition 8 For any self-dual lattice L in Rn we have

n/2 θL(T S(τ))=(τ/i) ψL(τ) (60) for all τ . ∈H DRAFT23 Proof : Let w be a characteristic vector. Then

v,v πi v,v τ πi( v,v τ+ v,w ) θ (T (τ)) = θ (τ +1) = ( 1)h ie h i = e h i h i (61) L L − v L v L X∈ X∈ because ( 1)c = eπic for any integer c. We now apply Poisson summation to the − sum. The Fourier transform of exp πi( x,x τ + x, w ) is the integral over x Rn h i h i ∈ of exp(πi( x,x τ + x, w +2y )), which is to say the value at y+ w of the Fourier h i h i 2 transform of exp πi x,x τ, which we alreadyknow is (τ/i) n/2 exp( πi x,x /τ). h i − − h i Poisson summation then gives

n/2 πi v+ w ,v+ w /τ n/2 πi c,c /τ θL(T (τ))=(τ/i)− e− h 2 2 i =(τ/i)− e− h i v L v s(L) X∈ ∈X (62) n/2 which is (τ/i)− ψL(S(τ)); replacing τ by Sτ we recover (60), Q.E.D. Corollary. We have σ(L) = n for any self-dual lattice L in Rn. πiσ(L)/4 Proof : Because ψL(t +1) = e ψL(t), the claim σ(L) = n is equivalent to πin/4 ψL(t +1) = e ψL(t). (63) 2 3 2 Using Proposition 8, together with S = (ST ) = 1 and θL(T τ) = θL(τ), we calculate n/2 t +1 1 ψ (t +1) = θ (T ST (t)) = θ (ST − S(t)) i L L L   n/2 1 n/2 1 i(t + 1) =(T − S(t)/i) θ (T − S(t)) = θ (T S(t)) L t L   2 3 2 (in which we used S =(ST ) = 1 and the invariance of θL under T , and again write n/2 power to mean nth power of principal square root). Comparing with (60) yields the desired identity (63), Q.E.D. See Exercise 4.4 for an alternative approach, and Exercise 4.3 for an application to the determination of σ(L) for lattices L Rn of (odd) prime discriminant. ⊂ The identity (60) can be used to study the norm distribution of unimodular lattices and their shadows. For instance, we have seen that the characteristic coset of Zn consists of the vectors all of whose coordinates are odd, whence it contains 2n vectors of norm n but no vectors of norm less than n. We showed in [E1] that this latter property characterizes the Zn lattice:

Theorem 9 [E1] Let L Rn be a self-dual lattice with no characteristic vector ⊂ w such that w, w < n. Then L = Zn. h i ∼ DRAFT24 Proof :12 By Theorem 6 we can write

n/8 ⌊ ⌋ n 8m m θL = cmθZ− θE8 (64) m=0 X for some constants c . We shall see that ψ = θ (Exercise 4.4 below, and at m E8 E8 greater length in the next section). Hence by (60) we have

n/8 ⌊ ⌋ n 8m m ψL = cmψZ− θE8 . (65) m=0 X We have ψ =2q1/4 + O(q9/4) as q 0, while θ = 1 + O(q2). By hypothesis Z E8 n/4 → n ψL = O(q ). Hence cm = 0 for m > 0, and θL = c0θZ. Since N0(L)=1 n it follows that c0 = 1, so L has the same theta function as Z . In particular n N1(L)=2n. It follows as in Exercises 3.3 and 3.4 that L ∼= Z , Q.E.D. This characterization of Zn answered a question that arose in the geometry of 4-manifolds (where self-dual lattices can arise via the intersection pairing on the second homology group). See also [E2] for the determination of all self-dual L Rn whose characteristic coset has minimal norm n 8, and [CS1, RS] for ⊂ − the use of (60) to obtain upper bounds on the minimal nonzero norm of L.

Exercises + 4.1 Prove directly that σ(Dn ) = n for all n 0 mod 4 by showing that (n/2)ei is + ≡ a characteristic vector of Dn for any i = 1, 2,...,n. Find a characteristic vector +4 of the self-dual lattice A15 (see Exercises 1.3 and 1.5 above), and check that its norm is 7 mod 8. ≡ 4.2 i) If L is an integral lattice of odd discriminant, and L L is a sublattice of 1 ⊂ odd index, then σ(L1) = σ(L). ii) Let L Rn be an integral lattice of odd discriminant, and L L a lattice of ⊂ 1 ⊂ odd discriminant in some subspace V Rn. Then L := L V also has odd ⊂ ′ ∩ ⊥ discriminant and σ(L ) = σ(L) σ(L ). [Show that L L has (finite and) odd ′ − 1 1 ⊕ ′ index in L.] 4.3 Let L Rn be an integral lattice whose discriminant is an odd prime p. ⊂ Suppose L contains a vector v such that p v , v 1 mod p. Prove that ∗ ∗ h ∗ ∗i ≡ − 12The proof in [E1] uses an alternative, and probably preferable, approach to the key fact that such a lattice must have θL = θZn , which avoids the explicit determination of Mn/2(Γ+) in Theorem 6 or indeed any explicit mention of modular forms. Instead we use that fact that θZ vanishes at the cusp τ = ±1 but nowhere in H, deducing in effect that θL/θZn is in M0(Γ+) and thus constant. DRAFT25 σ(L) = n +1 p. (Apply 4.2i to a suitable lattice containing L Z p .) What − ⊕ h i happens when L∗ does not contain such v∗? [In general if L is an integral lattice with disc L = p, and v L is any dual vector not in L, then p v , v is an inte- ∗ ∈ ∗ h ∗ ∗i ger not divisible by p, and all choices of v∗ yield integers with the same quadratic character mod p; thus p v , v 1 mod p is one of only two possibilities.] h ∗ ∗i ≡ − 4.4 Show that 0 is a characteristic vector of E8, again consistent with σ(L) = n. Use Theorem 6 to derive (63), and thus the fact that σ(L) = n, for all self-dual lattices L from the special cases L = Z and L = E8. 4.5 Generalize formula (60) to integral lattices of odd discriminant. 4.6 Let L Rn be an integral lattice whose characteristic coset has minimal ⊂ n 8m m norm n 8m0. Show that the coefficient cm of θZ− ∆+ in the expansion (48) − m0 12m0 n of θL vanishes for each m>m0, while cm0 is ( 1) 2 − Nn 8m0 (C). In − − particular ( 1)mc > 0. − m0 4.7 i) [E2] Use this to prove without using root systems that if L Rn is a uni- ⊂ modular lattice for some n < 12 such that N (L)=0 then n = 8 and θ = θ . 1 L E8 (Use the formula for θL obtained in Exercise 3.6 and the fact that Nn 8m0 (C) is − an integer which is even if n 8m =0.) − 0 6 ii) Deduce further that a nonzero integral lattice with no vectors of norm 1 or 2 must have rank at least 23, and if L R23 is an integral lattice with no vectors of ⊂ norm 1 or 2 then

23 15 3/2 2 5/2 θ = θZ 46θZ ∆ = 1+4600q + 93150q + 953856q + . L − + ··· [It is known that there is a unique such lattice L R23, the “odd Leech lattice” ⊂ or “shorter Leech lattice”; it can be obtained from the Leech lattice Λ R24 24 ⊂ by choosing v Λ of norm 4 (all such v are equivalent under Aut(Λ )), and 0 ∈ 24 0 24 projecting all v Λ such that 2 v , v to the 23-dimensional space (Rv ) .] ∈ 24 |h 0 i 0 ⊥

5. Even self-dual lattices and their theta functions An important special case is an integral lattice L (with no restriction on disc(L)) for which the homomorphism (56) is zero; that is, lattices such that v, v 2Z for h i ∈ all v L. Such a lattice is said to be even an integral lattice that is not even is said ∈ to be odd (it contains vectors of odd norm). By (55), L is automatically integral if v, v 2Z for all v L. More commonly we check that L is even by verifying h i ∈ ∈ that it is integral and has generators of even norm; equivalently, that it has a Gram matrix with integer entries and even diagonal entries. As with integral lattices, the direct sum of even lattices is even, as is any sublattice of an even lattice. A nonzero vector in an even lattice must have norm at least 2. A norm-2 vector in DRAFT26 an integral lattice L is called a root. A lattice generated by its roots is known as a root lattice; a root lattice is even because it is integral and generated by vectors of norm 2. Such lattices are surprisingly ubiquitous, playing crucial roles in the theory of Lie groups and algebras, in the study of singularities in algebraic geometry, and elsewhere; they are also important building blocks of even lattices and other integral lattices. The direct sum of root lattices is again a root lattice, and the irreducible root lattices are precisely the lattices A (n 1), D (n 4), and n ≥ n ≥ En (n =6, 7, 8). We have already encountered all of these except E6, which may be defined as the sublattice of E8 orthogonal to any isometric copy of A2 in E8 + (all are equivalent under Aut(E8)). For instance, having constructed E8 as D8 we may choose the copy spanned by e1 e2 and e2 e3, and then E6 is the sublattice consisting of vectors whose first three− coordinates− are equal. See Exercise 5.1 for the discriminants and roots of the lattices An, Dn, En. For any integral lattice L the “root (sub)lattice of L” is the sublattice generated by the roots of L. If the integral lattice L has odd discriminant then L is even if and only if 0 is a characteristic vector, which is to say L is its own shadow. In this case σ(L)=0, whence we have:

Theorem 10 (Schoeneberg [Sch, p.520]) If Rn contains an even self-dual lattice then n 0 mod 8. ≡ Proof : We have seen that σ(L) = n for every self-dual L Rn. If L is also even ⊂ then σ(L)=0 so n vanishes in Z/8Z. Q.E.D. Theorem 10 can be used as the starting point for proving congruences mod 8 on the ranks of even lattices of other discriminants; see for example 5.4i–iii (discrim- n/2 inant 2) and 5.7i–ii (even lattices L such as D4 for which L∗/L ∼= (Z/2Z) ). Exercise 5.2 gives a more offbeat application of Theorem 10 to Diophantine equa- tions. A self-dual lattice is said to be of Type I or Type II depending on whether it is odd or even. The necessary condition 8 n on the existence of Type II lattice in Rn | is also sufficient: if 8 n then Rn contains the Type II lattice En/8 (direct sum of | 8 n/8 copies of E8). For n = 8 this is the only such lattice. For n = 16 there + 2 is another Type II lattice, namely D16, which is not isomorphic with E8 because 2 + E8 is generated by its roots and D16 is not. We shall soon see that these are the only Type II lattices in R16 up to isomorphism. In R24 there are 24, as first shown with great effort by Niemeier [Ni]. The classification was considerably simplified by Venkov [Ve], using the weighted theta functions we introduce in the next section. The Niemeier lattices are ubiquitous in many other classification problems for lattices of rank up to 24. See Exercise 5.4iv for an example of one DRAFT27 such use.

If L is a Type II lattice then its theta function θL is invariant under T , not just T 2, because q never appears raised to half-integral powers, only integral ones. Be- cause θL still transforms under S by Proposition 1, we conclude that it transforms under the full modular group Γ generated by S and T . Moreover, since 8 n the n/2 | S transformation simplifies to θL(S(τ)) = τ θL(τ), whence we have for all γ = a b Γ the identity ± cd ∈
n/2 θL(γ(τ))=(cτ + d) θL(τ). (66) In other words, the theta function of a Type II lattice L Rn is a modular form of ∈ weight n/2 for Γ with trivial multiplier system. k While the weight of θL is a multiple of 4, the transformation (cτ + d) φ(τ) = φ(γ(τ)) is well-defined for all even k (because τ determines (cτ + d)); as before ± we say φ is a modular form of weight k for Γ if it is a holomorphic function on H that satisfies φ(γ(τ)) = (cτ + d)kφ(τ) for all γ Γ and φ remains bounded as ∈ τ i . We shall later need such forms also for k 2 mod 4, so we consider → ∞ ≡ them together. Again the modular forms constitute a graded algebra, which we call

M := Mk(Γ); (67) k≥0 kMeven and again the algebra turns out to be freely generated. Here these generators are the normalized Eisentein series13

∞ m3qm E = 1+240 = 1+240q + 2160q2 + 6720q3 + , (68) 4 1 qm ··· m=1 X − ∞ m5qm E = 1 504 =1 504q 16632q2 122976q3 , (69) 6 − 1 qm − − − −··· m=1 X − of weights 4 and 6 respectively. The analogue of Theorem 6 thus reads:

Theorem 11 The algebra M is freely generated over C by the modular forms E4 and E6. In other words, each Mn(Γ) (n =0, 2, 4,...) has basis EaEb : a, b 0, 4a +6b = n . (70) { 4 6 ≥ } 13Warning: there is an unfortunate but unavoidable notational collision here between the lattice E E8 and the Eisenstein series k. Even worse, the usual indexing of Eisenstein series makes θE8 16 equal E4, not E8 which is the theta function of each of the Type II lattices in R . We always use a sanserif E for the Eisenstein series — which also suggests the mnemonic that this E, being thin, has half the weight. DRAFT28 Hence dim M (Γ) is the number of possible (a, b) in (70), which is n/24 if n ⌊ ⌋ n 2 mod 12 and 1 + n/24 for other even n 0: ≡ ⌊ ⌋ ≥ n dim M n dim M n dim M n n n ··· 0 1 12 2 24 3 ··· 2 0 14 1 26 2 ··· 4 1 16 2 28 3 (71) ··· 6 1 18 2 30 3 ··· 8 1 20 2 32 3 ··· 10 1 22 2 34 3 ··· See [Se, VII, Theorem 4] for a classic exposition of the proof. Corollary. Let L be an even self-dual lattice in Rn. Then

n/24 ⌊ ⌋ En/8 En 24m m θL = 4 + cm 4− ∆ (72) m=1 X for some constants c (m =1, 2,..., n/8 ), where m ⌊ ⌋ E3 E2 ∆ := 4 − 6 =1 24q+252q2 1472q3+4830q4 6048q5 16744q6 (73) 123 − − − − ···

Proof : Because 8 n any element of M (Γ) is a polynomial in E and E2. We may | n/2 4 6 n/24 n 8m m thus write θ = ⌊ ⌋ c θZ− ∆ for some c (m = 0, 1, 2,..., n/24 ), L m=0 m m ⌊ ⌋ and as with (48) we use evaluate at q =0 to obtain c = N (L)=1 as desired. P 0 0 Again it follows that the coefficients N (L) of θ for k =1, 2,..., n/24 deter- 2k L ⌊ ⌋ mine θL, because we can use them to calculate the cm iteratively.

In particular, when n < 24 there are no undetermined coefficients, so θL = θn/8 = 1+30nq + O(q2), whence L has 30n roots. For n = 8 we already E8 know that N2(L) = 240, and observed that this forces L ∼= E8. The power series expansion (68) then determines N2d(E8) for all integers d > 0, namely 3 8 N2d(E8) = 240 d n d [Sch, p.521]. This lattice yields a sphere packing of R that is densest in its| dimension among lattice packings (this was proved in 1935 P by Blichfeldt [Bl], who also obtained the corresponding results for E6 and E7; in dimensions n =1, 2, 3, 4, 5 the densest lattice packings come from the root lattices A1, A2, A3,D4,D5). The E8 packing is also expected to be the densest among all 8 14 packings of identical spheres in R , and known to be within a factor 1+10− of the maximal density ([CK], extending the computations of [CE] in the course of giving a new proof of its optimality among lattice packings). DRAFT29 For n = 16 we have N2(L) = 480, and the only root systems of rank at most 16 2 that have 480 roots are E8 and D16. In the former case the root lattice already has discriminant 1, so L = E2; in the latter case, D L D with each inclusion ∼ 8 16 ⊂ ⊂ 16∗ having index 2; there are three such lattices, of which one is the odd lattice Z16 + + and the other two are isomorphic with D16, so L ∼= D16. Since dim M4(Γ) = 1 it E also follows that θL is the normalized Eisenstein series 8, which yields the closed form 2 +2 7 N2d(E8 ) = N2d(D16 ) = 480 d (74) d n X| for the theta function coefficients. 2 + The existence of two inequivalent lattices E8 and D16 that cannot be distinguished by their theta functions has a nice consequence in differential geometry: as Mil- 16 2 16 + nor [Mil] observed, the tori R /E8 and R /D16 are non-isometric but “isospec- tral”, in that their Laplacians have the same eigenvalues with the same multiplic- ities. In the language of [Ka], we cannot hear the difference between these two 16-dimensional drums. In general, for any lattice L Rn the Laplacian on the n ⊂ 2πi v∗,x torus R /L has an eigenbasis consisting of functions x e h i for v∗ L∗; 2 7→ n n ∈ the corresponding eigenvalue is (2π) v∗, v∗ , so the tori R /L1 and R /L2 are 14 − h i isospectral if and only if θ ∗ = θ ∗ , or equivalently (by (17)) θ = θ . Thus L1 L2 L1 L2 if L = L we obtain non-isomorphic isospectral tori. Subsequently examples 1 6∼ 2 with n < 16 were discovered; it is now known that θL uniquely determines L for lattices L Rn if n 3, but already for n = 4 (and thus for each n 4, using ⊂ ≤ ≥ direct sums with Z α n 4) there is a two-dimensional family of non-isomorphic h i − pairs L ,L with the same theta functions. { 1 2} Once n > 16 we need at least one coefficient in (72). The first example is n = 24, 3 when (72) gives θ = θ (720 N2(L))∆. For example, if L has no roots then L E8 − − θ = E3 720∆ = 1 + 196560q2 + 16773120q3 + 398034000q4 + . (75) L 4 − ··· In particular, the minimal norm of a Type II lattice in R24 is at most 4, and if it is that large then the kissing number is 196560. Leech [Le] costructed such a lattice Λ24, and Conway [Con], while studying his sporadic simple group Co1 = Aut(Λ )/ 1 , showed that Λ is the unique Type II lattice of minimal norm 4. 24 {± } 24 As with E8, the Leech lattice yields a sphere packing expected to be the densest in its dimension, now known to be the densest among lattice packing (proved in

14 ∗ ∗ ∗ ∗ Clearly if Nk(L1) = Nk(L2) for all k then θL1 = θL2 . To show the converse, note that if ∗ ∗ Nk(L1) = Nk(L2) does not hold for all k then there is a least counterexample k0 because the set πk0t ∗ ∗ ∗ ∗ of lattice norms is discrete. But then e (θL1 (it) − θL2 (it)) → (Nk0 (L1) − Nk0 (L2)) 6= 0 as ∗ ∗ t →∞, so θL1 6= θL2 . DRAFT30 2004 [CK], and the only example of a lattice packing of Rn that has been proved optimal for any n > 8); it is known to be within a factor 1 + ǫ of the maximal 30 density for any 24-dimensional sphere packing, here with ǫ =1.65 10− ([CK], again extending the computations of [CE]). ·

For n = 32 and n = 40 there is again one coefficient c1 to choose. If N2(L)=0 then c = 30n and we find N (L)=90n(211 5n), so again the minimal norm 1 − 4 − is 4 and the kissing number is 4. Already for n = 32 it is known [Ki] that there are literally millions of such lattices, and thus millions of isospectral 32-dimensional tori. Each of these lattices has kissing number 146880, and we shall see that there are enumerative properties beyond the theta coefficients Nk(L) shared by all these lattices’ configurations of short vectors. Similarly for n = 40, with an even larger number of lattices, each of kissing number 39600. 2 For n = 48 there are two unknown coefficients c1,c2; the q and q terms in θL vanish if and only if (c ,c )=( 1440, 125280), giving 1 2 − θ = E6 1440E3∆ + 125280∆2 L 4 − 4 = 1+52416000q3 + 39007332000q4 + . (76) ··· Thus any such lattice has minimal norm 6 and kissing number 52416000. Several such lattices have been constructed (see [CS2] for more information), and the as- sociated sphere packings are the densest known in R48. We shall see that their configurations of short vectors must satisfy many combinatorial constraints. One might hope to use these constraints to fully describe such lattices, but so far nei- ther a full classification nor a large lower bound on the number of lattices has been achieved. For any n 0 mod 8, we can likewise use (72) to determine the theta series ≡ of any Type II lattice in Rn with no nonzero vectors of norm 2m or less where m = n/24 . The resulting modular form is sometimes called the “extremal theta ⌊ ⌋ m+1 function” in Mn/2(Γ). One might hope that the q coefficient could vanish as well, so that the minimal norm might exceed 2m +2; but Siegel [Si] showed that this cannot happen. See Exercise 5.8 for one approach to this result, which Siegel proved using the Lagrange-Burmann¨ theorem. It follows that a Type II lattice L Rn has minimal norm at most 2 n/24 + 2, and for any such lattice θ ⊂ ⌊ ⌋ L is the extremal theta function in Mn/2(Γ). Such L is called extremal. Past n = 8, extremal lattices yield particularly good sphere packings when n = 24m, when (2 n/24 +2)/n has a local maximum. Unfortunately none are known in this case ⌊ ⌋ past m =2; the existence of any such lattice, and in particular the m =3 case (an extremal Type II lattice in R72), is a long-standing open question. They are known for all other n 88 (subject to 8 n as always) and for a few other values. It is ≤ | DRAFT31 known that for very large n (above roughly 41000) there are no extremal Type II lattices because the extremal theta series has negative the qm+2 coefficient [MOS].

Exercises 5.1 i) We have already seen that D has 2(n2 n) roots (Exercise 1.2ii), and that n − E7 and E8 have respectively 126 and 240 roots. Find the 72 roots of E6, and the 2 n + n roots of An. ii) We already know that disc(An) = n +1 (Exercise 1.3i) and disc(Dn)=4 n (because [Z : Dn]=2). Show that E6 has generators ei ei 1 (i = 4, 5, 6, 7), 1 n − − e7 + e8, and h = 2 i=1 ei, and use this to compute that disc(E6)=3. iii) More generally we can define for each n = 2, 3,..., 8 a sublattice En of E8 P by requiring the first 9 n coordinates to be equal. This agrees with the usual − En for n = 6, 7, 8. Identify the lattices E5, E4, E3 with the root lattices D5, A4, A A respectively, and show that E is a lattice with Gram matrix 21 (not a 1 ⊕ 2 2 14 root lattice); in each case disc(E )=9 n. n −
5.2 [Another application of Theorem 10]15 For positive integers a,b,c,d, let M be the tridiagonal matrix 2a 1 0 0 1 2b 1 0 M =   . 0 1 2c 1  0 0 12d   Show that M is positive definite (hint: the special case a = b = c = d = 2 should be familiar), and is thus the Gram matrix of an even lattice L. Note that M mod p has rank at least 3 for any prime p, and deduce that L∗/L is cyclic. Use Theorem 10 to conclude that det(M) cannot be a square. [This could also have been done using Proposition 7 (Z4 is the only self-dual lattice in R4), but the approach here generalizes to n n matrices for all even n 0 mod × 6≡ 8. Only the case n 4 mod 8 is of interest: if n 2 mod 4 then det(M) ≡ ≡ ≡ 3 mod 4 so det(M) can never be a square. Square values can be attained when 8 n; for example, M could be a tridiagonal Gram matrix for the lattice A of | 8 discriminant 9=32. The case n =4 can also be solved more laboriously by rewriting det(M) = x2 as

15I do not know the original source of this problem. Henri Cohen posted it May 25, 1990 to the NMBRTHRY mailing list, writing that it is “linked to some problems of algebraic topology” and was posed to him about 15 years earlier. Cohen no longer recalled the poser’s name, but was sure that he or she knew how to solve the problem. (In later communication Cohen dated the event to a graduate course taught in 1971, but still did not know who originated the problem.) I posted the solution outlined here to the mailing list, and Atkin posted a solution using quadratic reciprocity. Henri also reported that H. Lenstra solved it, but did not specify the method. DRAFT32 (4ab 1)(4cd 1) = x2 +4ad. We always have det(M) 1 mod 4; numerical − − ≡ computation suggests that all non-square positive integers congruent to 1 mod 4 occur as det(M) for some choice of a,b,c,d, with the exception of 105 and 2961. If a,b,c,d are not required to be positive then det(M)=1 can be attained by taking a = b = c =0 to obtain a Gram matrix of II2,2 (the indefinite self-dual even lattice of signature (2, 2)).]

5.3 [Uniqueness of E8 (again) and # Aut(E8)] We use theta functions to prove 8 that E8 is the unique Type II lattice in R and count its automorphisms. Let L 8 E be a Type II lattice in R . Thus θL = 4. Consider the distribution of vectors of norm at most 4 among the 28 cosets of 2L in L. The zero coset contains only 0; if r, r are roots in the same coset then r = r. Thus zero and the roots account ′ ′ ± for 1+(240/2) = 121 cosets, leaving only 28 121 = 135. Show that each coset − contains at most 16 vectors of norm 4, and if there are as many as 16 vectors then they are 2e for some orthonormal basis e 8 of R8. But the q2 coefficient ± i { i}i=1 of E is exactly 2160 = 16 135. Hence each of our 135 leftover cosets contains 16 4 · vectors of that form. Since they are congruent mod 2L we deduce that L contains each of the norm 2 vectors ei ej, whence L D8. We have seen already that ± ⊃ 8 this implies that L is one of the two lattices other than Z contained between D8 + and D8∗, whence L ∼= D8 = E8 . Show that Aut D8 is the hyperoctahedral group of order 288!, and deduce that E has 135 278! = 696729600 automorphisms. 8 · 5.4 [Bimodular even lattices] Suppose L Rn is an even lattice of discriminant 2. ⊂ (Integral lattices of discriminant 2 are sometimes called “bimodular”.) Let v ∗ ∈ L L and ς =2 v , v . ∗ − h ∗ ∗i i) Show that ς is an integer whose residue mod 4 does not depend on the choice of v . We may thus write ς = ς(L) Z/4Z. ∗ ∈ ii) Show that ς is odd. [Else L∗ would be an integral lattice of discriminant 1/2.] Check that ς(A1)=1 and ς(E7)=3. iii) Let L′ = E7 or A1 according as ς(L)=1 or ς(L)=3. Construct an even unimodular lattice L˜ containing L L with index 2. Conclude that n 1 mod 8 ⊕ ′ ≡ if ς(L)=1 while n 7 mod 8 if ς(L)=3. ≡ iv) If n = 1 then clearly L = Z 2 = A . Show that if n = 7 then L = E ; if ∼ h i 1 ∼ 7 n = 9 then L = A E ; while if n = 15 then either L = E E or L is the ∼ 1 ⊕ 8 ∼ 7 ⊕ 8 unique even lattice containing A D with index 2. 1 ⊕ 14 [You’ll use several times the existence of a bijection between copies of A1 and E7 in E8, namely the orthogonal slice. For n = 17 such lattices correspond to Niemeier lattices L˜ containing E ; there are four, with root systems D E2, 7 10 ⊕ 7 A E , E3, and D E . The first of these yields L with root sublattice 17 ⊕ 7 8 16 ⊕ 8 D E , which has index 2 in L; the other three yield the lattices A+3, A E2, 10 ⊕ 7 17 1 ⊕ 8 and A D+2 respectively. For n = 23 we need a Niemeier lattice with a root, 1 ⊕ 16 DRAFT33 and there are 31 distinct possibilities. Borcherds [Bo] enumerated 121 lattices for n = 25 using his analysis of the even Lorentzian lattice II25,1. For n 31 an enumeration is likelye hopeless: by the mass formula there are millions≥ of them for each n = 32 1, and many more for even larger n.] ± 5.5 [The theta function of a bimodular even lattice, and the Kohnen space] Let n L R be an even lattice of discriminant 2, and this time set ψ = θ ∗ θ , the ⊂ L L − L theta function of the nontrivial coset of L in L∗. Thus by the functional equation ψ (τ)=21/2(τ/i)n/2θ (S(τ)) θ (τ). L L − L ς(L) Also θL(T (τ)) = θL(τ) as usual, and ψL(T (τ)) = i ψL(τ). i) Use these transformation rules and the identity (ST )3 = 1 to give an analytic proof that n 1 or 7 mod 8 according as ς(L)=1 or ς(L)=3. ≡ ii) Show that θ and ψ are modular forms of weight n/2 for Γ(2) = ker(Γ L L → SL2(Z/2Z)). 7 7 iii) [E7 via “Construction A”] Define a lattice L such that A1 L A1∗ as 7 7 7 ⊂ ⊂ follows: identify A1∗ /A1 with (Z/2Z) , and let L be the preimage of the subspace of (Z/2Z)7 consisting of zero and the cyclic permutations of (0, 0, 1, 0, 1, 1, 1). Check that this is indeed a subspace; it is known in coding theory as the dual Hamming code. Deduce that L is a lattice, and verify that it is even and bimodular. Thus by the previous Exercise L ∼= E7 (can you find an explicit isomorphism?). 7 3 4 Show that its theta function is θZ +7θZψZ, and use this to compute the first few coefficients of θ . E7

5.6 [Uniqueness of E7 (again) and # Aut(E7)] Now take n = 7. Then θL = 1+126q + 756q2 + O(q3). Again we consider the distribution of vectors of norm at most 4 among the cosets of 2L in L. There are 27 cosets. The two cosets that constitute 2L∗ contain the zero vector and no other vector of norm less than 6. Of the remaining 27 2 = 126 cosets, N (L)/2 = 63 contain only a pair of roots, − 2 leaving 63 others. As in Exercise 5.2, each of these remaining cosets is represented by vectors 2e for some orthonormal e ; but this time we cannot have a full ± i i orthonormal frame because then L would contain 2Z7 with finite index, which is impossible because 2Z7 has square discriminant and L does not. Thus there are at most 6 pairs, and since N (L)=756=2 6 63 each of the cosets must have 4 · · exactly 6. Thus L D . Construct a map L/D D /D by sending v L ⊃ 6 6 → 6∗ 6 ∈ to the class of the homomorphism D Z, w v, w . Since L/D is infinite 6 → 7→ h i 6 cyclic and D6∗/D6 has exponent 2, the kernel of the map has index 1 or 2 in L/D6. Compare discriminants to show that the index is 2 and L contains D A . Finally 6 ⊕ 1 check that there are only two even integral lattices containing D A with index 2, 6⊕ 1 which are related by Aut(D A ). This shows that all bimodular even L R7 6 ⊕ 1 ⊂ are isomorphic. Since we already know such a lattice E7 we conclude that L ∼= E7, DRAFT34 and moreover that # Aut(E ) = 63#Aut(D )=63 266! = 2903040. 7 6 · 5.7 [2-modular lattices and their theta functions] 5.8 [Extremal theta series] Let f(q) and g(q) be power series of the form

f(q)=1+ O(q), g(q) = q + O(q2).

For a nonnegative integer k and any a0, a1,...,ak there exists a unique homoge- neous polynomial P ( , ) of degree k such that P (f,g) = k a qi + O(qk+1). · · i=0 i [Construct this polynomial iteratively starting from the leading coefficient a0, as 8 E3 P we did for (f,g)=(θZ, ∆+) or ( 4, ∆).] In particular, taking a0 =1 and ai =0 for i =1, 2,...,k, we find P such that P (f,g)=1+Cqk+1 +O(qk+2). We shall ( show that C is k/(k + 1) times the 1/q coefficient of f ′/g k + 1). In particular, C > 0 for all k > 0 if f and 1/g have positive coefficients. This is the case for E3 E (f,g)=( 4, ∆), using the expansion (68) of 4 and the infinite product formula n 24 ∆ = q ∞ (1 q ) . n=1 − k+1 k+2 k DivideQ both sides of P (f,g)=1+ Cq + O(q ) by g to obtain p(f/g) = 1/gk+Cq+O(q2) where p is the degree-k univariate polynomial p(X) = P (X, 1). Since g and g/f are both of the form q + O(q2), we can find a (unique) power se- ries F (z) = z + b z2 + b z3 + such that g = F (g/f). (This power series is 2 3 ··· holomorphic in a neighborhood of z = 0 if f and g have positive radii of conver- gence, but we need F only as a formal power series.) Taking z = g/f, we find 1/F (z)k = p(1/z) Cz + O(z2), in which the right-hand side is the beginning of − the Laurent expansion of 1/F (z)k about z =0. Therefore C is the z coefficient − of that expansion. But then C is the residue of (1/F (z)k) dz/z2 at z =0. −

6. Weighted theta functions We next introduce weighted theta functions Θ , θ of a lattice L Rn. These L,P L,P ⊂ generalize the theta functions ΘL, θL: they are generating functions that encode not just the number N2k(L) of lattice vectors of each norm 2k but also their distribution on the sphere x,x =2k. h i Weighted theta functions are defined as follows. The weight P is a harmonic poly- nomial on Rn, that is, a homogeneous polynomial whose Laplacian vanishes (we shall give a fuller description of such polynomials soon). Then

v,v /2 k ΘL,P (q) := P (v) qh i = N2k(L,P ) q , (77) v L k 0 X∈ X≥ πi v,v τ πikτ 2πiτ θL,P (τ) := P (v) e h i = Nk(L,P ) e = ΘL,P (e ), (78) v L k 0 X∈ X≥ DRAFT35 where we define Nk(L,P ) := P (v). (79) v∈L hv,vXi=k

Let d = deg(P ). If d =0 then P is a constant, so ΘL,P and θL,P reduce to scalar multiples of Θ and θ . If d is odd then Θ = θ = 0 because the v and v L L L,P L,P − terms cancel. For even d > 0 we get a nontrivial generalization of ΘL and θL; in this case N0(L,P )=0 so the sums over k in (77, 78) may be taken over k > 0.

The definitions of ΘL,P and θL,P make sense for any polynomial P , harmonic or not. We require that P be harmonic so that we can generalize Proposition 1 (the functional equation (17)) to weighted theta functions. Again we shall prove the functional equation using Poisson summation; here the relevant functions on Rn are π x,x t f(x) = P (x) e− h i . (80)

Theorem 12 Suppose that t > 0 and P is a harmonic polynomial on Rn of de- gree d, and define a function f : Rn R by (80). Then the Fourier transform → of f is d ( n +d) π y,y /t fˆ(y) = i t− 2 P (y) e− h i . (81)

This will yield:

Proposition 13 (functional equation for weighted theta series) For any lattice L in Rn, and any harmonic polynomial f of degree d, we have

2πt d 1/2 (n/2) d 2π/t ΘL∗,P (e− ) = i disc(L) t− − ΘL(e− ) (82) for all t > 0.

Proof (modulo Theorem 12): Apply Poisson (26) to L and the function (80), and use the formula (81) for the Fourier transform of this function. Q.E.D.

Note that since ΘL,P and ΘL∗,P vanish identically for odd d we can write the factor id as ( 1)d/2. − To prove Theorem 12, and then to use Proposition 13 to study lattices, we next review some key properties of harmonic polynomials. Let be the C-vector space of polynomials on Rn, and (d = 0, 1, 2,...) its P Pd subspace of homogeneous polynomials of degree d, so that = ∞ . The P d=0 Pd L DRAFT36 Laplacian is the differential operator16

n 2 ∂ n n ∆ = 2 : ∞(R ) ∞(R ), , d d 2. (83) ∂xj C →C P → P P → P − Xj=1 Here x ,...,x are any orthonormal coordinates on Rn, and is taken to be 0 1 n Pd { } for d < 0. The space of harmonic polynomials of degree d is then

0 d := ker(∆ : d d 2); (84) P P → P − this is the degree-d homogeneous part of

0 := ker(∆ : ). (85) P P → P Examples. 0 and 0 are the spaces of constant and linear functions respectively, P0 P1 of dimensions 1 and n. If n = 1 then 0 = 0 for all d > 1. If n = 2 then 0 Pd { } Pd is 2-dimensional for each d > 0; see Exercise 6.3 below for an explicit basis. For any n, A quadratic polynomial P = 1 j k n ajkxjxk is harmonic if and only n ≤ ≤ ≤ n if ajj =0, because ∆P is the constant polynomial 2 ajj. j=1 P j=1 WeP shall see that ∆: d d 2 is surjective, whence P P → P −

0 n + d 1 n + d 3 dim d = dim( d) dim( d 2) = − − . (86) P P − P − d − d     Indeed we shall give a more precise result using two further operators on (Rn) C∞ and on its subspace . The first is17 P n ∂ E := x = xj . (87) ·∇ ∂xj Xj=1 Euler proved that if P (Rn) is homogeneous of degree d then EP = d P ; ∈ C∞ · in particular d is the d-eigenspace of E . The second operator is multiplication by the norm:P |P n F := x,x = x2, P x,x P. (88) h i j 7→ h i Xj=1 16 24 ∞ n 24 The use of ∆ for both this operator and the modular form η = q Qn=1(1 − q ) may be unfortunate, but should not cause confusion because the two ∆’s never appear together outside this footnote. The alternative notation L for the Laplacian would be much worse when we regularly use L for a lattice. 17 Fortunately this operator will never appear together with an Eisenstein series E2k. . . DRAFT37 F 0 F Clearly injects each d into d+2. Thus d = ker( ∆ : d d); that is, 0 P P F P P → P d is the zero eigenspace of the operator ∆ on d. We next show that the other P Fk 0 P eigenspaces are d 2k for k =1, 2,..., d/2 , and that d is the direct sum of P − ⌊ ⌋ P these eigenspaces, from which the surjectivity of ∆: d d 2 will follow as a P → P − corollary. We begin with by finding the commutators of ∆, E, F. Recall that the commutator of any two operators A, B on some vector space is

[A, B] = AB BA = [B, A]. (89) − − For example, [xj,xk]=[∂/∂xj,∂/∂xk]=0 for all j,k, while [∂/∂xj,xk] = δjk (Kronecker delta).

Lemma 14 (Commutation relations for ∆, E, F). We have

[∆, F]=4E +2n, [E, ∆] = 2∆, [E, F]=+2F. (90) − Proof : These are direct computations using the pairwise commutators of the oper- 2 2 2 ators xj and ∂/∂xk. For example, [∂ /∂xj ,xk]=0 unless j = k, and then we calculate

2 2 ∂ 2 ∂ ∂ ∂ ∂ ∂ 2 xj = xj + ,xj xj = xj + xj ∂xj ◦ ∂xj ◦ ∂xj ∂xj ◦ ∂xj ◦ ∂xj ◦     2  2 2 ∂ ∂ ∂ ∂ 2 ∂ ∂ = xj +1 +xj +1 = xj +3xj +2 = xj 2 +4xj +2, ∂xj ∂xj ∂xj ∂xj ∂xj ∂xj     whence 2 ∂ 2 ∂ 2 ,xk = δjk 4xj +2 . ∂xj ∂xj     Summing over j,k = 1,...,n yields [∆, F]=4E +2n as claimed. We complete the proof of Lemma 14 by verifying the remaining two identities in (90) via a similar but easier calculation (see Exercise 6.1 below), Q.E.D. For a further check on the formulas for [E, ∆] and [E, F], note that they are consis- tent with the action of ∆, E, F on the d: if P d then ∆P d 2 yields P ∈ P ∈ P − [E, ∆]P = E∆P ∆EP =(d 2)∆P ∆(d P ) = 2∆P, − − − · − consistent with [E, ∆] = 2∆, and likewise for the third part [E, F]=+2F of (90). − See the further Remarks at the end of this section for the interpretation of the commutation relations (90) (and Lemmas 17,18(ii) below) in terms of sl2 and other Lie algebras and groups. DRAFT38 Now suppose P is in the λ-eigenspace of F∆ for some λ. Then x,x P = ∈ Pd h i FP is in the (λ +4d +2n)-eigenspace of F∆ acting on , because Pd+2 F∆FP = F(F∆+[∆, F])P = F(F∆+4E +2n)P = F(λ +4d +2n)P.

By induction on k =0, 1, 2,... it follows that FkP is an eigenvector of F∆ |Pd+2k with eigenvalue

k 1 − λ + 4(d +2j)+2n = λ + k 4(d + k 1)+2n . − j=0 X
Replacing d by d 2k and taking λ =0, we see that if P 0 then FkP is an − ∈ Pd 2k eigenvector of F∆ with eigenvalue − |Pd λ (k) := k (4(d k 1)+2n) . (91) d − − We next prove that this accounts for all the eigenspaces of F∆ . |Pd

Lemma 15 Fix d 0 . For integers k,k such that 0 k

Proof : By induction it is enough to check this for k′ = k +1. We compute

λ (k + 1) λ (k)=2n + 4(d 2k′) 2n > 0, d − d − ≥ Q.E.D. Fk 0 Corollary. The sum of the subspaces d 2k of d over k = 0, 1,..., d/2 is direct. P − P ⌊ ⌋ Fk 0 F Proof : Since d 2k is a subspace of the λd(k) eigenspace of ∆, it is enough to P − prove that the λd(k) are pairwise distinct. By Lemma 15, they are strictly increas- ing, Q.E.D.

k Fk 0 Proposition 16 For k =0, 1,..., d/2 , let d = d 2k. Then: ⌊ ⌋ P P − i) The map ∆: d d 2 is surjective. P → P − d/2 k 0 F Fk 0 ii) d = k⌊=0 ⌋ d = d d 2, and = k∞=0 . P k P P ⊕ P − PF FP iii) d is the entire λd(k) eigenspace of ∆ d , and ∆ d has no eigenvalues P L |PL |P other than the λd(k) for k =0, 1,..., d/2 . 0 ⌊ ⌋ iv) dim d = dim( d) dim( d 2) as claimed in (86). P P − P − DRAFT39 Proof : The sum in (ii) is direct by the previous Corollary. We prove that it equals F k 0 d by comparing dimensions. Since is injective we have dim d = dim d 2k; Pmoreover P P − 0 dim d 2k dim d 2k dim d 2k 2, P − ≥ P − − P − − 0 with equality if and only if ∆ : d 2k d 2k 2 is surjectve, because d 2k = P − → P − − P d/2 k − ker(∆ : d 2k d 2k 2). Hence dim ⌊ ⌋ d is P − → P − − k=0 P d/2 d/2 d/L2 ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ k 0 dim d = dim d 2k dim d 2k dim d 2k 2, (92) P P − ≥ P − − P − − Xk=0 Xk=0 Xk=0 and the last sum telescopes to dim d. Thus equality holds in the last step of (92) d/2 P and dim ⌊ ⌋ k = dim . The first of these proves (i) (by taking k =0). The k=0 Pd Pd second yields L d/2 ⌊ ⌋ = k, (93) Pd Pd Mk=0 k 0 as claimed in (ii); taking the direct sum over d yields = ∞ F , also P k=0 P claimed in (ii). to complete the proof of (ii) we compare the decompositions (93) k F k 1 L of d and d 2 and note that d = − for each k > 0. Claim (iii) follows P P − P Pd 2 because the decomposition (93) diagonalizes− F∆ . Finally (iv) is the case k =0 |Pd of equality in (92), Q.E.D. 0 F 0 Remark: Part (ii) implies d d 2 = 0 , and thus that contains no nonzero P ∩ P − { } P multiple of x,x . Proving this was set as problem B-5 on the 2005 Putnam exam [KAL, p.736],h whichi was the hardest of the 12 problems that year, solved by only five of the top 200 scorers [KAL, p.741]. The solution printed in [KAL, p.742] uses some of the ingredients used here to prove Proposition 16.18 π x,x t We next characterize the functions f(x) = P (x) e− h i of (80) using the opera- tors ∆, E, F. For t C define an operator ∈ n n πt x,x G : ∞(R ) ∞(R ), g e− h ig (94) t C →C 7→ 18 Suppose hx, xi|P and ∆P =0. Write P = Pd≥0 Pd with each Pd ∈ Pd. Then hx, xi|Pd and ∆Pd = 0 for each d, and we may choose d so that Pd 6= 0. Let m be the largest integer such that m Pd = hx, xi Q for some polynomial Q; by assumption m > 0. In our notations, then, Q ∈ Pd−2m with ∆FmQ = 0 and Q∈ / FP. Using in effect the formula for [∆, F] and Euler’s description of E, compute that ∆FmQ = Fm−1[F∆+2m(n + 2(d − m − 1))]Q m for all Q ∈ Pd−2m. Thus ∆F Q =0 implies F∆Q = −2m(n + 2(d − m − 1))Q, and the factor n + 2(d − m − 1) is positive (because d ≥ 2m ≥ m +1), so Q ∈ FP, contradiction. DRAFT40 that multiplies every function by the Gaussian e πt x,x ; these operators con- C∞ − h i stitute a one-parameter group: GtGt′ = Gt+t′ for all t,t′. We are then interested in f = G P for P in the intersection of the kernel of ∆ with an eigenspace t ∈ P of E. If P then ∈ Pd

d f = Gt(d P ) = GtEP =(GtEG t)GtEP =(GtEG t)f, (95) · · − − so f is in the d-eigenspace of GtEG t; likewise f ker Gt∆G t. Since our one- − ∈ − parameter group G has infinitesimal generator πF, we expect that conjugation { t} − by Gt will take ∆, E to some linear combination of ∆, E, F. Indeed we find:

Lemma 17 (Conjugation of ∆, E, F by Gt) The operators Gt commute with F, and we have

2 GtEG t = E +2πtF, Gt∆G t = ∆ + πt(4E +2n)+(2πt) F. (96) − −

Proof : As with Lemma 14, this comes down to an exercise in differential calculus. Here we start from the fact that Gt commutes with each xj while Gt(∂/∂xj)G t = − 2πtxj, whence the first formula in (96) quickly follows, while GtF = FGt is immediate. A somewhat longer computation (see Exercise 6.2 below) establishes the second formula, Q.E.D. Corollary. The operators ∆, E, F act on G , and the subspace G 0 is the inter- tP tPd section of ker(∆ + πt(4E +2n)+(2πt)2F) with the d-eigenspace of E +2πtF in G . tP We next relate the Fourier transform of a Schwartz function f with the Fourier transforms of its images under ∆, E, F, and use this to prove Theorem 12.

Lemma 18 (Conjugation of ∆, E, F by the Fourier transform). Let f : Rn → C be any Schwartz function. Then: 1 i) For each j =1,...,n, the Fourier transform of xjf is (2πi)− ∂f/∂yˆ j, and the Fourier transform of ∂f/∂x is 2πiy fˆ. j − j ii) The Fourier transforms of ∆f, (2E + n)f, and Ff are respectively (2π)2Ffˆ, − (2E + n)fˆ, and (2π) 2∆fˆ. − − − Proof : Again a calculus exercise, this time with definite integrals. The formula for the Fourier transform of ∂f/∂xj is obtained by integrating by parts with re- spect to xj. The Fourier transform of xjf can be obtained from this using Fourier inversion, or directly by differentiation with respect to yj of the integral (25) that defines fˆ(y). We then obtain (ii) by iterating the formulas in (i) to find the Fourier DRAFT41 2 2 2 E transform of ∂ f/∂xj , xj∂f/∂xj, or xj f, and summing over j. The case of f E n can be explained by writing the operator 2 +n as j=1 xj(∂/∂xj)+(∂/∂xj) xj, Q.E.D. ◦ P ˆ ˆ We first use this to show that if f Gt then f G1/t , that is, that f is some π y,y /t∈ P ∈ P polynomial multiplied by e− h i ; more precisely:

Proposition 19 Let t C with Re(t) > 0. If f = GtP for some P d then ˆ ˆ ˆ∈ d ˆ ˆ ˆ d ∈( Pn +d) f = G1/tP for some P = d′=0 Pd′ with each Pd′ d′ and Pd = i t− 2 P . ( n +d) ∈ P As before t− 2 denotes the (n +2d) power of the principal square root of t. P −

Proof : We use inductionon d. Thebasecase d =0 is the fact that the Fourier trans- πt x,x n/2 π y,y /t form of e− h i is t− e− h i , which we showed already. Suppose we have established the claim for P . By linearity and the fact that is spanned by ∈ Pd Pd+1 its subspaces x , it is enough to prove the Proposition with P replaced by x P . jPd j By part (i) of Lemma 18, the Fourier transform of GtxjP = xjGtP is ˆ 1 ∂ ˆ 1 ∂P 2π ˆ (G1/tP ) = G1/t yjP . (97) 2πi ∂yj 2πi ∂yj − t   d ( n +d) By the inductive hypothesis Pˆ has degree d and leading part Pˆd = i t− 2 P . Therefore the right-hand side of (97) has degree d +1 and leading part

1 2πt− i d+1 ( n +d+1) − Pˆ = Pˆ = i t− 2 y P. 2πi d t d j This completes the induction step and the proof, Q.E.D. Proof of Theorem 12: Suppose P 0 and f(x) = P (x) e π x,x t = G P. By the ∈ Pd − h i t Corollary to Lemma 17,

(∆ + πt(4E +2n)+(2πt)2F)f =0, (E +2πtF)f = d f. (98) · Taking the Fourier transform and applying Lemma 18(ii), we deduce t ( (2π)2F πt(4E +2n) t2∆)fˆ =0, (E + n + ∆)fˆ = d f.ˆ (99) − − − − 2π ·

Eliminating ∆fˆ, we find d fˆ = (E + 2π F)fˆ; that is, fˆ is in the d-eigenspace · t of E +2πt 1F. By Proposition 19 we know fˆ = G Pˆ for some Pˆ . By − 1/t ∈ P Lemma 17, then, Pˆ is in the d-eigenspace of E; that is, Pˆ d. By Proposition 19 d ( n +d) ∈ P we conclude that Pˆ = i t− 2 P , Q.E.D. DRAFT42 2 Remark: Multiplying the first equation in (99) by t− , we recover the first equa- 1 − tion of (98) with t replaced by t− ; this lets us show without Euler’s theorem that 0 0 the Fourier transform takes Gt to G1/t . See the further Remarks at the end of this section for a sketch ofP an alternativeP approach to Theorem 12, using the connection with sl2 and SL2 and avoiding Proposition 19. A natural application of weighted theta functions is to the question of equidistribu- tion of lattice points in spherical shells. The N (L) lattice vectors v L on the k ∈ sphere x Rn x,x = k yield a configuration, call it { ∈ |h i } 1/2 S (L) := k− v L v, v = k , (100) k { ∈ |h i } of N (L) vectors on the unit sphere Σ Rn. As k through the lattice k ⊂ → ∞ norms, are the Sk asymptotically equidistributed on Σ? Recall that a sequence C of nonempty finite subsets of Σ are asymptotically equidistributed if, { m}m∞=1 for every continuous function ϕ : Σ C, the average of ϕ over C approaches → m the average of ϕ over Σ as m : →∞ 1 ϕ(x) dνx = lim ϕ(x). (101) x Σ m #Cm Z →∞ x Cm ∈ X∈

Here dν is the invariant measure on Σ such that x Σ dµx = 1; for instance we may define ∈ R x ϕ(x) dν = tn/2 e πt x,x ϕ dµ (102) x − h i 1/2 x x Σ x Rn x,x ! Z ∈ Z ∈ h i for any t > 0. The coefficients of weighted theta functions ΘL,P give us the sum in (101) when P is a harmonic polynomial. Using the decomposition (93) from Proposition 16(ii), we show that these are enough to test equidistribution:

Proposition 20 0 is dense in (Σ); that is, for every continuous ϕ : Σ C P |Σ C → and any ǫ > 0 there exists P 0 such that ∈ P x Σ: P (x) ϕ(x) < ǫ. (103) ∀ ∈ | − | Proof : By the Stone–Weierstrass theorem there exists P satisfying (103). It ∈ P is thus enough to prove that for every P there exists Q 0 such that P = Q ∈ P ∈ P on Σ. Applying Proposition 16(ii) to each homogeneous part of P we write

deg(P )/2 ⌊ ⌋ P = x,x kQ (104) h i k Xk=0 DRAFT43 for some Q 0. Since x,x = 1 on Σ, the polynomial Q = Q 0 k ∈ P h i k k ∈ P agrees with P on Σ, Q.E.D. P Theorem 21 A sequence C of nonempty finite subsets of Σ is asymptoti- { m}m∞=1 cally equidistributed if and only if 1 lim P (x)=0 (105) m #Cm →∞ x Cm X∈ for all harmonic polynomials P of positive degree.

Proof : For the “only if” direction, assume (101) holds for all ϕ (Σ), and take ∈ C ϕ = P . We claim (101) is then equivalent to (105), i.e., that P (x)dν =0. |Σ x Σ x Equivalently (see Exercise 6.5), we claim P (x) e πt x,x∈ dµ = 0. But x Rn − hR i x the integral is the value at y = 0 of the Fourier∈ transform of P (x) e πt x,x . We R − h i obtained this Fourier transform in Theorem 12; it vanishes at y = 0 as claimed, because P (0) = 0 for P of positive degree. Since both sides of (101) are linear, we have thus proved the converse implication for functions ϕ that are the restriction to Σ of any finite linear combination of harmonic polynomials of positive degree. We can drop the condition of positive degree, because (101) holds automatically for ϕ = 1: its left-hand side equals 1, and the right-hand side reduces to limm 1. Thus (105) implies (101) for all ϕ →∞ of the form P with P 0. By Proposition 20, every continuous ϕ : Σ C |Σ ∈ P → can be uniformly approximated by such P . Hence (101) holds for all ϕ (Σ) |Σ ∈C by the following standard argument. Changing ϕ to P moves both x Σ ϕ(x) dνx and each average (#C ) 1 ϕ(x) by at most ǫ. For large enough∈ m, the m − x Cm ∈ R average of P over Cm is within ǫ of P (x) dνx. Therefore the average and P x Σ integral of ϕ are within 3ǫ of each other.∈ Since ǫ is arbitrary, we are done. Q.E.D. R [...] Further Remarks:19 The commutation relations in Lemma 14 are tantamount to an isomorphism of Lie algebras from sl to the span of ∆, E + n , F that takes the 2 { 2 } standard basis (X,H,Y ) = 01 , 1 0 , 00 of sl to ( 1 ∆, (E+ n ), πF). 00 0 1 10 2 4π − 2 − d/2 k − The decomposition d = k⌊=0
⌋ d in
Proposition

16ii then says in effect that 0 P P = ∞ V n +d, where for any real m > 0 we write Vm for the infinite- P ⊕d=0Pd ⊗ 2 L dimensional irreducible representation of sl with basis Y kv where Xv =0 2 { }k∞=0 and Hv = mv. − 19The reader anxious to reach the application of weighted theta series to the study of extremal lattices etc. will likely want to skim or skip these remarks, and the last exercise for this section, at least on first reading(s). DRAFT44 Moreover, we noted already that G = exp( πtF), and it is known that the Fourier t − transform is e πin/4 exp πi (πF 1 ∆).20 Thus Lemmas 17 and 18, which give − 2 − 4π the action on ∆, E, F of conjugation by Gt and the Fourier transform, correspond tY 10 πi(X+Y )/2 to the action on sl2 of conjugation by the elements e = t 1 and e− = 0 i of SL . − i 0 2
In Proposition
19, we obtain Pˆ by applying to P the operators Gt, then the Fourier πin/4 transform, then G 1/t; up to the constant factor e− , this corresponds to the product −

1 0 0 i 1 0 t 1 − = i , 1/t 1 i 0 t 1 − 0 1/t  −   −     −  in SL2, which can be written as

t 1 1 1/t t/i 0 1 i/t i = 1 . − 0 1/t 0 1 0 (it)− 0 1  −      0 1 Now if P then P is fixed by the one-parameter subgroup ∗ of SL gener- ∈ Pd 01 2 ated by X. Thus Pˆ is a multiple of the image of P under a diagonal matrix in SL .
2 Such diagonal matrices constitute the one-parameter group generated by H. Since P is an eigenvector of H, we deduce that Pˆ is proportional to P . More precisely, n β β βH β( n +d) since HP = ( + d)P we have diag(e ,e )P = e P = e− 2 P ; taking − 2 − eβ = t/i (with Im(β) = π/2),21 and restoring the factor e πin/4, we recover − − πin/4 ( n +d) d ( n +d) Pˆ = e− (t/i)− 2 P = i t− 2 P, which is Theorem 12. 2 1 The differential operators ∂ /∂xj∂xk, xj∂/∂xk + 2 δjk, and xjxk generate a Lie E n F algebra isomorphic with sp2n, which contains the span of ∆, + 2 , as the sub- algebra invariant under the orthogonal group of Rn. See the final exercise for this

20This has a memorable physical interpretation: running the Schrodinger¨ equation on a quantum harmonic oscillator for 1/4 of its classical period applies a multiple of the Fourier transform to the wave function. The distribution of factors of π in this formula is the reason we chose the homomor- 0 1 0 0 1 F 1 1 F phism from sl2 that maps `0 0´ and `1 0´ to 4π ∆ and −π , rather than 2 ∆ and − 2 which seems more natural at first. Alternatively, we could use Lemma 18 to identify the Fourier transform with conjugation by 0 1 ±i`1 0´. 21 We must specify the path because in general a representation of sl2 lifts to a representation not of SL2 but of its universal cover. A more honest treatment would either carefully check that we are d working in a contractible patch of SL2(C), or compute the constant factor i of (81) in some other way. DRAFT45 2 section. Multiplying ∂ /∂xj∂xk and xjxk by i (to make all the generators skew- Hermitian) yields infinitesimal generators of the Lie algebra of Weil’s projective 2 n representation [Wei] of Sp2n(R) on L (R ).

Exercises 6.0 Verify directly that Theorem 21 holds for n =1. 6.1 Check the commutation relations [E, ∆] = 2∆ and [E, F]=+2F of (90). − 6.2 Verify the second formula in (96). 6.3 (Harmonic polynomials on R2) Let n =2, and identify R2 with C as usual by d d x1 + ix2 = z. Show that for each d 1 the polynomials z and z¯ constitute a basis for 0. [Hint: recall that ≥ Pd ∂ ∂ ∂ ∂ ∆ = + i i . ] ∂x1 ∂x2 ∂x1 − ∂x2    Deduce that the case n = 2 of Theorem 21 is tantamount to Weyl’s characteriza- tion [Wey] of asymptotic equidistribution in R/Z, or equivalently in the unit circle Σ = z C : z = 1 : a sequence of finite nonempty subsets C Σ is 1 ∼ { ∈ | | } m ⊂ 1 asymptotically equidistributed in Σ1 if and only if for each d =1, 2, 3,... we have (#C ) 1 zd 0 as k .22 m − z Cm ∈ → →∞ 6.4 (a weightedP theta function twisted by a Dirichlet character) Show that if f is a Schwartz function on R then

∞ 1 ∞ χ (m)f(m) = χ (n)fˆ(n/4), 4 2i 4 m= n= X−∞ X−∞ where χ4 is the Dirichlet character mod 4 introduced in Exercise 3.7. Since χ4 is odd, we would get only the trivial identity 0=0 using the even function f(x) = πtx2 πtx2 e− . However f(x) = xe− yields a new identity. Use this identity to prove that

∞ 2 nχ (n)qn /8 = q1/8(1 3q +5q3 7q6 +9q10 11q15 + ) (106) 4 − − − −··· n=1 X 3 is a modular form of weight 3/2 for Γ and the 8th roots of unity εc,d, with εc,d as in Exercise 2.5; deduce that this modular form is η3.

22 Weyl also shows that that this is equivalent to the condition that #(I ∩ Cm)/#Cm → |I|/|Σ| for all arcs I ⊆ Σ (where |·| denotes the length). See also [Ko,¨ Ch.3]. Analogous statements can be made on the unit sphere in Rn for any n, but we do not need those versions of equidistribution here. DRAFT46 6.5 Prove that there are positive constants c0,c2,c4,... (depending on n) such that if ϕ is the restriction to Σ of a homogeneous polynomial P of even degree d then

(n+d)/2 πt x,x ϕ(x) dνx = cdt e− h i P (x) dµx. (107) x Σ x Rn Z ∈ Z ∈ for all t > 0. We defined the integral over Σ so that c0 = 1; can you evaluate the other cd explicitly? (For odd d, (107) holds for all cd because both sides vanish...) 6.6 Define an inner product , on the space of Schwartz functions on Rn by h· ·i

f,g := f(x) g(x) dµx. (108) h i x Rn Z ∈ Recall that an operator A on this space is self-adjoint if Af,g = f,Ag for all h i h i Schwartz functions f,g. Prove that the operators ∆ and F are self-adjoint. [This is immediate for F, and uses integration by parts for ∆. Warning: E is not self- adjoint; indeed the fact that ∆ and F are self-adjoint implies that 4E +2n = [∆, F] 0 is anti-self-adjoint.] Use this and the Corollary to Lemma 17 to prove that d 0 P and ′ are orthogonal for d = d , that is, that P (x) Q(x) dν = 0 if P Pd 6 ′ x Σ x and Q are harmonic polynomials of different degrees.∈ (We showed the special case R d = 0 during the proof of Theorem 21.) Recover the orthogonal decomposition 2 0 L (Σ,dν ) = ∞ using this and Proposition 20. x d=0 Pd 6.7 (Failure ofL equidistribution for n = 2) Equidistribution of Skm (L) can fail for n = 2 even if N (L) , and already for the familiar lattice L = Z2. To km → ∞ study this question we must use the intimate connection between the Diophantine equation x,x = x2 + x2 = k and arithmetic in the Gaussian integers x + ix h i 1 2 { 1 2 | x ,x Z that was kept in the background of Exercises 3.7 and 3.8. 1 2 ∈ } (i) Let p be a prime such that p 1 mod 4, and k a sequence of integers with ≡ { m} pm k . Then if N =0 then N > 4m; in particular N . Show that in | m km 6 km km →∞ this case S is asymptotically equidistributed on the unit circle. { km }

(ii) Prove that there exist km such that Nkm but Skm is not asymptotically equidistributed on the unit circle. [Use the following→∞ spec{ ial} case of Hecke’s theo- rem [H1] on equidistribution of prime ideals in number fields: for all ǫ there exist 2 2 integers x1,x2 with 0

6.8 (Identification of a Lie algebra of differential operators with sp2n) Let V be n the 2n-dimensional R-vector space of differential operators on R with basis xj and ∂/∂x (1 j n); and let g be the space of dimension 2n2 + n with basis j ≤ ≤ ∂2/∂x ∂x , x ∂/∂x + 1 δ , and x x (1 j,k n). j k j k 2 jk j k ≤ ≤ i) Check that if A, B g then [A, B] g, while if A g and v V then ∈ ∈ ∈ ∈ [A, v] V . Thus g is a Lie algebra and V is a representation of g. ∈ DRAFT47 ii) The map [ , ] : V V R is a perfect alternating pairing on V . Use the · · × → Jacobi identity [[A, B],C]+[[B,C], A]+[[C, A],B]=0 to prove that [gv,v′] + [v,gv ]=0 for all g g and v, v V . Thus g is contained in the Lie algebra ′ ∈ ′ ∈ of the symplectic group for our pairing; since dim g = dim sp2n, this gives an isomomorphism g ∼ sp . → 2n 7. Extremal lattices and spherical designs

8. Further directions Lattices with arbitrary discriminants; periodic weights Higher θ functions; coding analogues Murphy's Law Acknowledgements

References

[Bl] Blcihfeldt, H.F.: The minimum values of positive quadratic forms in six, seven and eight variables. Math. Z. 39 (1935), 1–15.

[Bo] Borcherds, R.E.: Table 2: the norm 2 vectors in II25,1. Online at http://math.berkeley.edu/− reb/lattices/table2.html− . ∼ [Coh] Cohen, H.: Number theory, Vol. II: Analytic and Modern Tools, New York: Springer, 2007 (GTM 240).

[CE] Cohn, H., and Elkies, N.D.: New upper bounds on sphere packings I, An- nals of Math. 157 (2003), 689–714 arXiv:math/0110009 [math.MG]).

[CK] Cohn, H., and Kumar, A.: Optimality and uniqueness of the Leech lattice among lattices, to appear in the Annals of Math. (arXiv:math/0403263 [math.MG]).

[Con] Conway, J. H.: A characterisation of Leech’s lattice, Invent. Math. 7 (1969), 137–142.

[CS1] Conway, J.H., and Sloane, N.J.A.: A new upper bound for the minimum of an integral lattice of determinant one, Bull. Amer. Math. Soc. 23 (1990), 383–387; Erratum: 24 (1991), 479. DRAFT48 [CS2] Conway, J.H., and Sloane, N.J.A.: Sphere Packings, Lattices and Groups (3rd ed.). New York: Springer 1999.

[Di] Dickson, L.E.: History of the Theory of Numbers, Vol. II: Diophantine Analysis. New York: Stechert & Co., 1934.

[Ed] Edwards, H.M.: Riemann’s zeta function. New York: Academic Press, 1974.

[E1] Elkies, N.D.: A characterization of the Zn lattice, Math. Research Letters 2 (1995), 321–326 (arXiv:math/9906019v1 [math.NT]).

[E2] Elkies, N.D.: Lattices and codes with long shadows, Math. Research Let- ters 2 (1995), 643–651 (arXiv:math/9906086v1 [math.NT]).

[E3] Elkies, N.D.: Lattices, Linear Codes, and Invariants, Notices of the Ameri- can Math. Soc. 47 (2000), 1238–1245 and 1382–1391.

[EV] Ellenberg, J., and Venkatesh, A.: Local-global principles for representa- tions of quadratic forms, Invent. Math. 171 #2 (2008), 257–279.

[H1] Hecke, E.: Eine neue Art von Zetafunktionen und ihre Beziehungen zur Verteilung der Primzahlen II, Math. Z. 6 (1920), 11–51.

[H2] Hecke, E.: Analytische Arithmetik der positiven quadratischen Formen, Kgl. Danske Vid. Selsk. Math.-Fys. Medd. 17 (1940) #12 [= pages 789–918 in Mathematische Werke, Gottingen:¨ Vandenhoeck & Ruprecht, 1959].

[Iw] Iwaniec, H.: Topics in Classical Automorphic Forms. Providence, RI: American Math. Society, 1997.

[KD] Kaba, M., “in collaboration with” Dickson, L.E.: On the Representation of Numbers as the Sum of Two Squares, American Math. Monthly 16 #5 (May 1909), 85–87.

[Ka] Kac, M.: Can one hear the shape of a drum? American Math. Monthly 73 (1966), 1–23.

[KAL] Klosinski, L.F., Alexanderson, G.L., and Larson, L.C.: The Sixty- Sixth William Lowell Putnam Mathematical Competition, American Math. Monthly 113 #8 (October 2006), 733–743.

[Ki] King, O.D.: A mass formula for unimodular lattices with no roots, Math. Comp. 72 (2003), 839–863 (arXiv:math/0012231v1 [math.NT]). Online tables at http://math.berkeley.edu/ reb/lattices/table.txt . ∼ DRAFT49 [Ko]¨ Korner,¨ T.W.: Fourier Analysis. Cambridge, England: Cambridge Univer- sity Press, 1988.

[Le] Leech,J.: construction of Leech lattice, c. 1960

[Mu] Munroe,R.: E TO THE PI MINUS PI, xkcd 217 (2007) (http://www.xkcd.com/217/).

[MOS] Mallows, C.L., Odlyzko, A.M., and Sloane, N.J.A.: Upper Bounds for Modular Forms, Lattices, and Codes. J. Alg. 36, 68–766 (1975).

[Min] Minkowski, H.: Grundlagen fur¨ eine Theorie der quadratischen Formen mit ganzzahligen Koeffizienten, pages 3–145 in Gesammelte Abhandlun- gen (Leipzig, 1991, republished 1967 by Chelsea, New York); German translation of “Memoire´ sur la theorie´ des formes quadratiques a` coeffi- cients entiers”, Memoirs´ present´ es´ par divers savants a` l’Academie des Sciences de l’Institut de France 29 (1887), 1–180.

[Mil] Milnor, J.: Eigenvalues of the Laplace operator on certain manifolds, Proc. Nat. Acad. Sci. USA 51 (1964), 542.

[Ni] Niemeier, H.-V.: Definite quadratische Formen der Dimension 24 und Diskriminante 1, em J. Number Theory 5 (142–178), 1973.

[Po] Pommerenke, C: Uber¨ die Gleichverteilung von Gitterpunkten auf m- dimensionalen Ellipsoiden, Acta Arith. 5 (1959), 227–257.

[RS] Rains, E.M., and Sloane, N.J.A.: The Shadow Theory of Modular and Unimodular Lattices, J. Number Theory 73 (1998), 359–389 [= http://www.research.att.com/ njas/doc/mod.pdf]. ∼ [SP] Schulze-Pillot, R.: Representation by integral quadratic forms — a sur- vey. Pages 303–321 in Algebraic and arithmetic theory of quadratic forms (Contemp. Math.) 344, Providence, RI: Amer. Math. Soc., 2004.

[Sch] Schoeneberg, B.: Das Verhalten von mehrfachen Thetareihen bei Modul- substitutionen, Math. Annalen 116 (1939), 511–523.

[Se] Serre,J.-P.: A Course in Arithmetic. New York: Springer, 1973.

[Si] Siegel, C.L.: Berechnung von Zetafunktionen an ganzzahligen Stellen. Nachr. Akad. Wiss. Gottingen¨ Math.-Phys. Kl. II 1969, 87–102 (1969) [= pages 82–97 in Gesammelte Abhandlungen IV, Berlin: Springer 1979]. DRAFT50 [SS] Stein, E.M., and Shakarchi, R.: Complex Analysis. Princeton, NJ: Prince- ton University Press, 2003.

[SW] Stein, E.M., and Weiss, G.L.: Introduction to Fourier Analysis on Eu- clidean Spaces, Princeton Math. Series 32, Princeton, NJ: Princeton Uni- versity Press, 1971.

[Ve] Venkov, B.B.: On the classification of integral even unimodular 24-dimensional quadratic forms. Trudy Mat. Inst. Steklov. 148 65–76 (1978). [In Russian; trans. by the Amer. Math. Soc. as Proc. Steklov Inst. Math. 148 63–74 (1980); also the source of [CS2, Ch.18]]

[Wei] Weil, A.: Sur certaines groupes d’operateurs´ unitaires. Acta Math. 111, 143–211 (1964).

[Wey] Weyl, H.: Uber¨ ein Problem aus dem Gebiete der diophantischen Approxi- mationen, Nachrichten der Koniglichen¨ Gesellschaft der Wissenschaften zu Gottingen.¨ Mathematisch-physikalische Klasse, 1914, 234–244 [= Gesammelte Abhandlungen I (Springer: Berlin 1968), 487–497].

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