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Theta Functions Theta Functions Lukas Lewark Seminar on Modular Forms, 31. Januar 2007 Abstract Theta functions are introduced, associated to lattices or quadratic forms. Their transformation property is proven and the conditions discussed under which theta functions are modular forms. As an application it is shown how theta functions can be used to analyse lattice sphere packings. The relation to the represantation number of natural numbers as sums of squares is mentioned. 1 Definition of the theta functions 1.1 The theta function of a lattice Let V be a n-dimensional real vector space equipped with an inner product (a bilinear, symmetric, positive definite form) , . Let µ be an invariant measure on V , i.e.h· ·i for any measurable subset U V, v V : µ(v + U) = µ(U) and λ R : µ(λU) = λ nµ(U). Furthermore⊂ ∀ ∈ ∀ ∈ | | µ is to be scaled so that for a basis (ε1,...εn) of V which is orthonormal with respect to , : µ( λ ε 0 λ 1 ) = 1. h· ·i { j j j | ≤ j ≤ } P Definitions A lattice Γ is a subset of V which can be written in the form Γ = e1Z + . + enZ where (e1,...en) is a basis of V , which is then called a basis of Γ. Equivalently it can be defined as a discrete subgroup which does not lie in a hyperplane. The volume of Γ is vol(Γ) := µ(V/Γ) = µ( λ e 0 λ 1 ). { j j j | ≤ j ≤ } The dual lattice Γ0 of Γ is a subset of the dualP space V 0 defined by h Γ0 : g Γ : h(g) Z ∈ ⇔ ∀ ∈ ∈ To the lattice Γ one assigns the theta function θ : H C, Γ → τ eπiτhg,gi (1) 7→ gX∈Γ 0 0 0 0 Remarks To verify that Γ is itself a lattice, let (e1,...,en) V denote ⊂ 0 0 the dual basis of (e1,...en), i.e. ej (ek) = δjk (Kronecker delta). (e1,...,en) is a basis of V 0 and a subset of Γ0 and any h Γ0 can be decomposed to h = h(e ) e0 e0 Z + . + e0 Z. So (e0 ,...e0 )∈ is a basis of Γ0. j j · j ∈ 1 n 1 n P V can be identified with V 0 via the inner product: v (w v, w ). Γ0 will also denote the image of Γ under this isomorphism, which7→ is another7→ h latticei in V . Lemma (1) converges uniformly on every compact subset and it converges absolutely. Thus θΓ is holomorphic. Proof (Sketch) Since τ H Re iτ < 0 and g, g > 0 g = 0, the exponent’s real part is negative.∈ The⇒ Lemma can thush bei proven∀ by6 finding a bound for the number of lattice points with g, g x which is polynomial in x. h i ≤ 1 1.2 Theta function of a quadratic form Definitions A quadratic form is a map S : V 2 R so that (v, w) S[v + w] S[v] S[w] is an inner product. → 7→ −Let Pos(− n, R) be the set of symmetric, positive definite n n-matrices and let S Pos(n, R). Such an S defines a quadratic form by ×v S[v] where S[v] :=∈v>Sv. All quadratic forms are given in such a way. 7→ One associates the theta function θ : H C to S, which is given by S → τ eπiτS[g] 7→ gX∈Zn Remark Let A be the Gram matrix of (e1,...en) with respect to , , i.e. A = e , e . h· ·i ij h i j i It follows that θΓ = θA because: n λ1 g = λ e g, g = (λ λ )A . j j ⇒ h i 1 · · · n . Xj=1 λ n So defining a theta function via a lattice or a quadratic form is just a difference of perspective. Lemma µ(V/Γ) = det(A) p > Proof Let B := (e1 en) be a n n matrix the rows of which are the basis vectors written in· terms · · of a basis× orthonormal with respect to , . B is called a generator matrix of the lattice. h· ·i det(B) = µ( λ e 0 λ 1 ) and { j j j | ≤ j ≤ } P B>B = A det(A) = det(B>) det(B) = det(B)2 = µ(V/Γ)2 ⇒ because the absolute value of the determinant of a matrix is the volume of the parallelepiped spanned by its row vectors. ¤ Lemma The inverse matrix A−1 is a Gram matrix of the dual lattice Γ0. 0 −1 0 0 Proof Let ej = A ej . Then (e1,...,en) is the dual basis (which is a basis of Γ0) because e0 , e = (A−1e )>Ae = e>e = δ . h j ki j k j k jk So indeed e0 , e0 = (A−1e )>A(A−1e ) = e A−1e . ¤ h j ki j k j k 2 2 The Transformation Formula Theorem Let Γ be a lattice. Then n 1 τ 2 τ H : θΓ0 = µ(V/Γ)θΓ(τ) (2) ∀ ∈ µ−τ ¶ ³ i ´ where that branch of √ is to be chosen that maps positive reals to positive reals. To prove the theorem one requires the following Lemma (Poisson formula) A bounded function f : V R is called rapidly decreasing if it can be partially differentiated arbitrarily→ often and all these derivatives are bounded. The Fourier transform f of such a function is defined as b f : V 0 R, h e−2πih(g)f(g) dµ(g) → 7→ Z b V For any such f and lattice Ω the Poisson formula states that µ(V/Ω) f(g) = f(h) gX∈Ω hX∈Ω0 b Proof Use a basis of Ω to identify V with the Rn; Ω is mapped to Zn, a lattice with volume 1. Identify µ with the product measure dx1 . dxn. Then use Rn0 = Rn and Zn0 = Zn and notice that the fourier transform as defined above is identified with the known fourier transform (using the right scaling of µ with respect to , ). So this more generalh· ·i form of the Poisson formula follows from the classical f(g) = f(h) gX∈Zn hX∈Zn b ¤ Proof of the Transformation Formula By the identity theorem, for two holomorphic functions defined on a open connected set – such as H – to be equal it is sufficient that they agree on a subset which has an accumulation point contained in that subset. So it is sufficient to prove the transformation formula for τ = iy with y R+, that is to prove ∈ −1 yn/2µ(V/Γ) e−πyhg,gi = e−πy hh,hi gX∈Γ hX∈Γ0 Let f : V R, g e−πhg,gi. f is rapidly decreasing and f = f if you identify V with V 0→using ,7→. This can be proven by doing the same identifications as in h· ·i b 3 \2 2 the proof of the Poisson formula, thus reducing the claim to e−πkxk = e−πkxk for x Rn, a known analytic result. Let∈ Ω = y1/2Γ. Then Ω0 = y−1/2Γ0 and µ(V/Ω) = yn/2µ(V/Γ). The Poisson formula, applied to f and Ω yields yn/2µ(V/Γ) e−πhg,gi = e−πhh,hi g∈Xy1/2Γ h∈yX−1/2Γ0 −1 yn/2µ(V/Γ) e−πyhg,gi = e−πy hh,hi ⇔ gX∈Γ hX∈Γ0 And that was to be proven. ¤ 3 The theta functions as modular forms The transformation property of θΓ is similar, but not equal to that of a modular form. What conditions has Γ to satisfy for θ to be a modular form? If g, g is an even integer g Γ, say, for a given g, g, g = 2c, then h i ∀ ∈ h i eπi(τ+1)hg,gi = e2πi(τ+1)c = e2πiτc = eπiτhg,gi i.e. θ would have period 1: θΓ(τ + 1) = θΓ(τ) (3) From the matrix point of view this means that the diagonal entries are even; such matrices are called even. If Γ is selfdual, i.e. Γ = Γ0, its theta function satisfies the modified transfor- mation property n 1 τ 2 θΓ = µ(V/Γ)θΓ(τ) (4) µ−τ ¶ ³ i ´ Lemma If Γ is selfdual and even then 8 n. | 1 1 Proof The modular group is generated by the matrices T = 0 1 and S = 0 −1 1 0 . ¡ ¢ ¡ To¢ prove the lemma, one makes use of (ST )3i = i: 4 Choose τ = i and apply alternatingly (3) and (4): (3) θΓ(i) = θΓ(i + 1) n/2 (4) i + 1 1 = θ µ i ¶ Γ µ−i + 1¶ i 1 = (1 i)n/2θ − − Γ µ 2 ¶ (3) i + 1 = (1 i)n/2θ − Γ µ 2 ¶ n/2 (4) 1 i + 1 2 = (1 i) θ µ i 2 − ¶ Γ µ−i + 1¶ = ( i)n/2θ (i 1) − Γ − (3) = ( i)n/2θ (i) − Γ n/2 Because θΓ(i) is a sum of positive terms it is non-zero. So ( i) = 1 8 n follows. − ⇒ |¤ Lemma Γ=Γ0 det(A) = 1 and j, k : A Z. ⇔ ∀ jk ∈ Proof ” ”: Γ=Γ0 det(A) = det(A−1) det(A) = 1. And A = ⇒ ⇒ ⇔ jk ej , ek Z. p p h ” i”: ∈ Let g,h Γ, then g,h = g>Ah Z Γ Γ0. A−1 is itself a matrix with⇐ determinant∈ 1 and integralh i entries, so∈ switching⇒ ⊂ the roles of Γ = Γ00 and Γ0 one deduces Γ0 Γ Γ=Γ0. ¤ ⊂ ⇒ Corollary The theta function of a selfdual, even lattice Γ is a modular form n of weight 2 . Proof θΓ is 1-periodic and using the two prior lemmas the transformation property (4) of a selfdual lattice becomes 1 n θ = τ 2 θ (τ) Γ µ−τ ¶ Γ ¤ 4 Sphere packings A sphere packing is a set of non-overlapping balls of same radius in the Rn.
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