1 Theta Functions

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1 Theta Functions - Massachusetts Institute of Technology

April 22, 1:00 pm

We've previously seen connections between modular forms and Ramanujan's work by verifying that Eisenstein series are indeed modular forms, and showing that the Discriminant function is a weight 12 cusp form with a product expansion inq. As we said then, the extent to which we can express modular forms in terms of products and quotients of thefunction begin to explain many of theq-series iden- tities given in Berndt. Using facts like the additivity of weights of modular forms under multiplication, we can even begin to predict where to look for such impressive identities. Many of the identities we used for solving the problem of representations by sums of squares could be boiled down to identities about the theta function, aq- series supported on powersn2. In this section, we'll begin a study of theta functions and their connection to quadratic forms.

2 Poisson Summation for Lattices

The theory of the Fourier transform is often stated for functions of a real variable, but is really no dierent for a real vector space. Given a functionfonRthat is suciently well-behaved (For examplefis piece-wise continuous with only nitely many discontinuities, of bounded total variation, satisfying f(a) =12 lim x!af(x) + lim x!a+f(x) for alla, and jf(x)j< c1min(1;xc2) for somec1>0;c2>1. One can fuss with these conditions some, but these will be more than sucient for us.) Now for such a function, dene the Fourier transform f(x) =Z 1 1 f(y)e2ixydy: 1 Proposition 1 (Poisson Summation forR)Given a functionfas above with

Fourier transform^f, then

1 X n=1f(n) =1X n=1^ f(n) Remember that we're working with functions onRhere, which is non-compact. The theory is somewhat dierent from, sayR=Z, a compact domain for which we can express suciently nice functions as a Fourier series. We don't have such a series here, but we can regard the above formula as a partial analogue of this property.

ProofConsider the function

F(x) =1X

n=1f(x+n); which is suciently nice (in the above sense) iffis suciently nice, and is periodic of period 1. That is, we can considerFas a function on the compact domainR=Z, so it has a Fourier expansion such that

F(x) =1X

n=1a me2imx; where, as usual, a m=Z 1 0

F(x)e2imxdx=Z

1 01 X n=1f(x+n)e2imxdx Sincee2imx=e2im(x+n), we may interchange the order of summation and integra- tion, giving a m=1X n=1Z 1 0 f(x+n)e2im(x+n)dx=Z 1 1 f(x)e2imxdx=^f(m):

Thus takingx= 0, we have

1 X n=1f(n) =F(0) =1X n=1a n=1X n=1^ f(n); as required. 2 More generally, to a real vector spaceVof dimensionnwith a translation invari- ant measure, we may dene (again for a rapidly decreasing smooth functionf) the

Fourier transform

^f(x) =Z V e2ihy;xif(y)(y): (Note that comparing withV=Rindeed reduces to our earlier denition.) Then^f is a suciently nice function onV, the dual vector space ofVconsisting of linear functionsV!R. If is a lattice inV, then the dual lattice 0inVis dened as the set ofx2Vsuch thathy;xi 2Zfor ally2. Proposition 2 (Poisson Summation for Lattices)The volume of the lattice inVis(V=). Forfsuciently nice as above, we have X y2f(y) =1(V=)X x20^f(x) ProofWe just want to reduce to the usual Poisson summation overRn, whose proof is essentially identical to the one we gave above forR. We can rescale the measure by the volume, so we may assume(V=) = 1. Lete1;:::;enbe a basis of . Then we may identifyVwithRnaccording to the coecients of these basis vectors, may be identied withZn, andwithdx1:::dxn. The space of linear maps may also be identied withRnand 0withZn, so we reduce to Poisson summation forRn.

3 Functional Equations for Theta Functions

Suppose thatVhas a symmetric, bilinear formB(x;y) =xywhich is positive and non-degenerate (xx >0 ifx6= 0). This simplies the situation above, asVcan be identied withVusing the form, and 0is now a lattice inV. To any lattice , dene (t) =X x2e txx;wheret >0,t2R. TakingV=R; =Z;q=e2iz;andz=iyfory >0 (or actually, somewhat annoyingly,z=iy=2) recovers the theta function'dened in Berndt's book. Theorem 1The functionsatises the functional equation (t) =tn=21(V=)0(t1): 3

In particular, if = 0=ZR, we have

(t) =1pt (1t ProofWe will apply Poisson summation to the functionf(x) =exx, a rapidly decreasing smooth function onV. To determine the Fourier transform off, choose an orthonormal basis forVto identify the vector space withRnso that the measure becomesdx=dx1dxnand the inner product simplies to givef=e(x21++x2n).

Hence the Fourier coecient

f(x) =Z R ne2i(x1y1++xnyn)e(y21++y2n)dy can be realized as an iterated integral which is identical in each coordinate. Choose one such integral, complete the square in the exponent and evaluate. We nd the Fourier transform ofex2is againex2, sofis equal to^f. Our theta function has summandsetxx. Again, use the functionfdened above, now for the latticet1=2, all translates of elements of byt1=2. It's volume in Vistn=2(V=) wherenis the dimension ofV, and its dual ist1=20according to the denition of the dual. Applying Poisson summation for lattices gives the desired result.

4 Matrix Description of Theta Functions

Note that in our theorem, we need not takee1;:::;ento be an orthonormal basis of . More generally, to obtain a symmetric bilinear form which is positive and non-degenerate, then settingaij=eiej, the matrixA= (aij) must be positive, non-degenerate, and symmetric. With this choice of basis and bilinear form, xx=Xa ijxixjforx=x1e1++xnen and the corresponding theta function is (t) =X x2ZnetPaijxixj: By comparing the basise1;:::;enchosen above to an orthonormal basis, we see that the volume of our lattice is det(A)1=2(see Serre, p. 108 for a slick proof using wedge products). Further ifB= (bij) is the inverse matrix toA, thenBplays the

same role asAbut now for the dual basis. That is, settinge0i=Pbijei, then thee0iare a dual basis whose inner products recoverB, and whose volume is 1=det(A)1=2.

5 Theta Functions as Modular Forms

First we analyze convergence of the theta functions. Given any integerm0, let r (m) denote the number of elementsx2 such thatxx= 2m. Thenr(m) is bounded by a polynomial inm. This is clear since we're counting lattice points on the surface of a sphere of radius 2minndimensions. Serre quotes the boundO(mn=2). What's the best you can prove? For our purposes, we just need polynomial growth for convergence since the terms in our theta functions have exponential decay. This implies that theq-series1X m=0r (m)qm converges forjqj<1. So lettingq=e2izas usual, forz2 Hdenes a holomorphic function (z) =1X m=0r (m)qm=X x2e iz(xx)=X x2q (xx)=2: Because we want integral powers ofqonly in ourqexpansion, we see now that we need to require an additional condition on our lattice :xx0 (mod 2) for all x2. In matrix terms, this impliesA= (eiej) has even entries along the diagonal. Note further the intentional use of the lower case, as this function is slightly dierent from the functions dened in the previous sections. In those sections, we made careful study of theta functions with real parametert, which we now see recovered by settingz=it. In fact, we'd like to use the symmetry property of the real-valued theta function ast7!1=tto conclude a similar property forz7! 1=z for(z).

Using our new lower-case notation, we have

(t) =(it);(t1) =(1=it) We'd like to apply Poisson summation for lattices, to obtain a relation for. To do this, we require the dual lattice

0to be equal to { in terms of the denition

of the dual lattice, this means thatxy2Zfor allx;y2Zso thatxydenes an isomorphism between and its dual. In matrix terms,A= (aij) = (eiej) has integer coecients with determinant equal to 1. Given this assumption, we may apply Theorem 1 to give (1=it) =tn=2(it): Because both(1=z) and (iz)n=2(z) are analytic functions inz, and we have just proved they are equal forzon the positive imaginary axis, then by analytic continuation, it is true for allz2 H. We record this result as: 5 Proposition 3Givena self-dual lattice with notation as above, for anyz2 H,quotesdbs_dbs2.pdfusesText_2

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