Theta Series, Multiplier System

Consider the theta series

$\displaystyle \theta(\tau)=\sum_{n\in\mathbb Z}e^{2\pi i n^2\tau},\qquad \tau\in\mathbb H.$

We want to understand the classical half-integral weight transformation formula under an arbitrary matrix on $\Gamma_0(4)$ . For $\gamma=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ , in $\mathrm{SL}_2(\mathbb Z)$ with $4\mid c, c>0$ the result is

$\displaystyle \theta(\gamma\tau)=\varepsilon_d^{-1}\left(\frac{c}{d}\right)(c\tau+d)^{1/2}\theta(\tau).$

Here $d$ is odd,

$\displaystyle \varepsilon_d=\begin{cases}1,&d\equiv1\pmod4,\\ i,&d\equiv3\pmod4,\end{cases}$

and $\left(\frac{c}{d}\right)$ is the Jacobi symbol, understood as the signed Kronecker symbol when signs are allowed.

The theta series is one of the first places where several apparently unrelated ideas meet: Gaussian integrals, Fourier transform, lattice sums, quadratic residues, modular transformations, and eventually the functional equation of the Riemann zeta function. At first sight, its transformation law can look unnatural. Why should the simple seriestransform under a fractional-linear change of variable by a factor involving both a square root and a Jacobi symbol? Why should a complex-analytic object such as $(c\tau+d)^{1/2}$ occur beside an arithmetic object such as $\left(\frac{c}{d}\right)$ ? And why does the natural symmetry group turn out not to be all of $\mathrm{SL}_2(\mathbb Z)$ acting on one function, but rather the congruence subgroup $\Gamma_0(4)$ ? The answer is that theta is a sum of Gaussians over a lattice, and the Fourier transform of a Gaussian is another Gaussian. Poisson summation then says that summing a function over a lattice is equivalent to summing its Fourier transform over the dual lattice. In one dimension the dual of $\mathbb Z$ is again $\mathbb Z$ , so the resulting identity closes up and produces a functional equation for the theta series. The basic real-variable form of this principle is

$\displaystyle t^{-1/2} \sum_{n\in\mathbb Z}e^{-\pi n^2/t},\quad t>0.$

The factor $t^{-1/2}$ is already the origin of the square roots that later appear in modular transformation laws. It is simply the scaling factor in the Fourier transform of a Gaussian. The modular variable $\tau$ is a holomorphic version of the positive parameter $t$ , and the transformation $\tau\mapsto-1/\tau$ is the complex form of the substitution $t\mapsto1/t$ . Fourier transform of a Gaussian produces a square root. Poisson summation exchanges the original integer sum with a dual sum. After this exchange, a finite quadratic exponential sum appears, and that finite sum is exactly where the Jacobi symbol comes from. Thus the factor is not an arbitrary repair added to make a square root behave; it is the precise record of what Poisson summation does to the lattice Gaussian.

The expression

$\displaystyle J_\theta(\gamma,\tau)=\epsilon_d^{-1}\left(\frac{c}{d}\right)(c\tau+d)^{1/2}$

is called the automorphy factor, or multiplier factor, for $\theta$ . Its role is to measure the precise failure of $\theta$ to be invariant under the fractional-linear action of $\gamma$ . In other words, the transformation formula is not $\displaystyle \theta(\gamma\tau)=\theta(\tau),$ but rather

$\displaystyle \theta(\gamma\tau)=J_\theta(\gamma,\tau)\theta(\tau).$

The factor depends on $\tau$ as well as on $\gamma$ , chiefly through the term $(c\tau+d)^{1/2}$ . Thus it cannot be an ordinary character or homomorphism of $\Gamma_0(4)$ . Instead it must satisfy the compatibility rule appropriate to a group acting on a space:

$\displaystyle J_\theta(\gamma_1\gamma_2,\tau)=J_\theta(\gamma_1,\gamma_2\tau)J_\theta(\gamma_2,\tau).$

Indeed, begin with $\theta(\tau)$ and first apply $\gamma_2$ . The transformation law gives $\displaystyle \theta(\gamma_2\tau)=J_\theta(\gamma_2,\tau)\theta(\tau).$ Now apply $\gamma_1$ , but notice that the new base point is $\gamma_2\tau$ , not $\tau$ . Therefore $\displaystyle \theta(\gamma_1\gamma_2\tau)=J_\theta(\gamma_1,\gamma_2\tau)\theta(\gamma_2\tau).$ Substituting the first identity into the second gives $\displaystyle \theta(\gamma_1\gamma_2\tau)=J_\theta(\gamma_1,\gamma_2\tau)J_\theta(\gamma_2,\tau)\theta(\tau).$ On the other hand, transforming once by the product matrix $\gamma_1\gamma_2$ gives $\displaystyle \theta(\gamma_1\gamma_2\tau)=J_\theta(\gamma_1\gamma_2,\tau)\theta(\tau).$ Comparing the two expressions yields the cocycle identity. It says that the correction factors accumulate exactly as they must when modular transformations are composed. This is why $J_\theta$ , rather than its square-root or Gauss-sum pieces separately, is the natural object.

Poisson-summation

Assume initially that $c>0$ , and write $\displaystyle w=c\tau+d.$ Because $\tau$ lies in the upper half-plane, so does $w$ . We therefore use the holomorphic square root on the upper half-plane: if $w=re^{i\alpha}$ with $0<\alpha<\pi$ , then $w^{1/2}=r^{1/2}e^{i\alpha/2}$ . This branch is not chosen for cosmetic reasons. It is exactly the branch selected by the convergent Gaussian integral that will appear below. The determinant condition $ad-bc=1$ implies

$\displaystyle \frac{a\tau+b}{c\tau+d}=\frac{a}{c}-\frac{1}{c(c\tau+d)}.$

Substituting this into the theta series gives

$\displaystyle \theta(\gamma\tau)=\sum_{n\in\mathbb Z}e^{2\pi i a n^2/c}e^{-2\pi i n^2/(cw)}.$

The first exponential is periodic in $n$ modulo $c$ , so we split $n$ into residue classes: write $n=cm+r$ , with $r$ ranging modulo $c$ . Since $(cm+r)^2\equiv r^2\pmod c$ , this gives

$\displaystyle \theta(\gamma\tau)=\sum_{r\bmod c}e^{2\pi i ar^2/c}\sum_{m\in\mathbb Z}\exp\left(-\frac{2\pi i(cm+r)^2}{cw}\right).$

The inner sum is a Gaussian sampled along one arithmetic progression. This is the point at which Poisson summation enters.

Define $f(x)=\exp\left(-\frac{2\pi i x^2}{cw}\right).$ Using the Fourier-transform convention $\widehat f(\xi)=\int_{\mathbb R}f(x)e^{-2\pi i x\xi},dx,$ completion of the square gives

$\displaystyle \widehat f(\xi)=\left(\frac{cw}{2i}\right)^{1/2}\exp\left(\frac{\pi i cw\xi^2}{2}\right).$

The important analytic fact is that ${\text{Im}}w>0$ , which makes the original Gaussian decay and makes the integral convergent. Poisson summation on the shifted lattice $r+c\mathbb Z$ says

$\displaystyle \sum_{m\in\mathbb Z}f(cm+r)=\frac{1}{c}\sum_{\ell\in\mathbb Z}e^{2\pi i\ell r/c}\widehat f\left(\frac{\ell}{c}\right).$

Substituting this into the preceding expression for $\theta(\gamma\tau)$ , we find

$\displaystyle \theta(\gamma\tau)=\frac{1}{c}\left(\frac{cw}{2i}\right)^{1/2}\sum_{\ell\in\mathbb Z}e^{\pi i w\ell^2/(2c)}G(a,\ell;c),$

where $\displaystyle G(a,\ell;c)=\sum_{r\bmod c}e^{2\pi i(ar^2+\ell r)/c}.$

At this point, the analysis is over. Everything left is finite arithmetic. The original infinite theta sum has been transformed into another infinite sum, multiplied by a finite quadratic Gauss sum. The role of the condition $4\mid c$ now becomes transparent. Replace $r$ by $r+c/2$ in $G(a,\ell;c)$ . The quadratic part changes by an integer and therefore contributes nothing, while the linear term contributes $(-1)^\ell$ . Hence

$\displaystyle G(a,\ell;c)=(-1)^\ell G(a,\ell;c),$

so $G(a,\ell;c)=0$ whenever $\ell$ is odd. Only even frequencies remain.

Write $\ell=2m$ . Since $d\equiv a^{-1}\pmod c$ , completing the square modulo $c$ gives $\displaystyle ar^2+2mr\equiv a(r+dm)^2-dm^2\pmod c.$ Therefore

$\displaystyle G(a,2m;c)=e^{-2\pi i d m^2/c}G(a;c),\qquad G(a;c)=\sum_{r\bmod c}e^{2\pi i ar^2/c}.$

Now the payoff arrives. The $m$ -dependent exponential in the Poisson formula becomes $\displaystyle e^{2\pi iwm^2/c}e^{-2\pi i d m^2/c}=e^{2\pi i\tau m^2}.$ Thus the transformed infinite sum is exactly the original theta series again. We obtain

$\displaystyle \theta(\gamma\tau)=e^{-\pi i/4}\frac{G(a;c)}{\sqrt{2c}}(c\tau+d)^{1/2}\theta(\tau).$

This is already a complete elementary transformation formula. If one defines

$\displaystyle \mu(\gamma)=e^{-\pi i/4}\frac{G(a;c)}{\sqrt{2c}},$

then Poisson summation has proved $\displaystyle \theta(\gamma\tau)=\mu(\gamma)(c\tau+d)^{1/2}\theta(\tau).$ No Jacobi symbol was assumed. No multiplier was guessed. The finite coefficient $\mu(\gamma)$ has simply emerged from the calculation.

The remaining question is arithmetic: what is $G(a;c)$ ? For $4\mid c$ and $(a,c)=1$ , the standard quadratic Gauss-sum evaluation is

$\displaystyle G(a;c)=(1+i)\epsilon_a^{-1}\left(\frac{c}{a}\right)\sqrt c.$

A useful way to understand this formula is that it is itself a finite analogue of Fourier duality. One proves a reciprocity relation for quadratic Gauss sums by applying Poisson summation to a damped Gaussian, grouping the original side by residue classes modulo $c$ , grouping the Fourier-transformed side by residue classes modulo $2a$ , and then letting the damping tend to zero. In this finite reciprocity identity, the roles of $a$ and $c$ are exchanged. The usual odd-modulus Gauss-sum evaluation then produces the Jacobi symbol. Thus even the arithmetic ingredient is closely related to the same Fourier-transform mechanism. Substituting the Gauss-sum formula into the raw factor gives

$\displaystyle \mu(\gamma)=\epsilon_a^{-1}\left(\frac{c}{a}\right).$

Since $ad\equiv1\pmod c$ , we have $a\equiv d\pmod4$ because $4\mid c$ ; therefore $\epsilon_a=\epsilon_d$ . Also, $d\equiv a^{-1}\pmod c$ , and the quadratic character $u\mapsto\left(\frac{c}{u}\right)$ takes the same value on a unit and its inverse. Hence

$\displaystyle \epsilon_a^{-1}\left(\frac{c}{a}\right)=\epsilon_d^{-1}\left(\frac{c}{d}\right).$

This proves the formula

$\displaystyle \theta(\gamma\tau)=\epsilon_d^{-1}\left(\frac{c}{d}\right)(c\tau+d)^{1/2}\theta(\tau).$

The cases $c=0$ and $c<0$ are routine extensions. If $c=0$ , then $\gamma$ acts by an integral translation and the identity reduces to $\theta(\tau+1)=\theta(\tau)$ . If $c<0$ , replacing $\gamma$ by $-\gamma$ does not change its fractional-linear action on $\tau$ , and the standard signed conventions for the Kronecker symbol and square root give the same formula.

It is worth pausing here to explain why the proof was carried out on $\Gamma_0(4)$ , rather than on all of $\mathrm{SL}_2(\mathbb Z)$ . The restriction $4\mid c,$ is not an incidental hypothesis inserted only to make the Gauss sum easier to evaluate. It is precisely the condition under which this particular theta series returns to a scalar multiple of itself. Outside this subgroup, theta still transforms in a perfectly orderly way, but it generally transforms into a different theta function, or into a linear combination of several closely related theta functions. Thus $\Gamma_0(4)$ is the natural group for a scalar-valued transformation formula for the single function $\displaystyle \theta(\tau)$

One can see this directly in the Poisson-summation proof. Recall that, after splitting the original sum into residue classes and applying Poisson summation, the finite sum $\displaystyle G(a,\ell;c)=\sum_{r\bmod c}e^{2\pi i(ar^2+\ell r)/c}$ appeared. When $4\mid c$ , we may replace $r$ by $r+c/2$ . The summand changes by $\displaystyle \exp\left(2\pi i\left(\frac{ac}{4}+\frac{\ell}{2}\right)\right)=(-1)^\ell,$ because $ac/4$ is an integer. Hence $\displaystyle G(a,\ell;c)=(-1)^\ell G(a,\ell;c),$ and all odd frequencies vanish. The surviving frequencies have the form $\ell=2m$ , and this is exactly what allows the transformed sum to reassemble into $\displaystyle \sum_{m\in\mathbb Z}e^{2\pi i\tau m^2}=\theta(\tau).$ In other words, the divisibility condition $4\mid c$ is what makes the Fourier-transformed lattice sum land back in the same lattice sector. This also explains what changes when the condition fails. Suppose, for instance, that $c\equiv2\pmod4$ . Then $a$ is odd, and the same substitution $r\mapsto r+c/2$ changes the summand by $\displaystyle (-1)^{\ell+1}.$ Now the even frequencies vanish and the odd frequencies survive. The resulting infinite sum is no longer the original $\theta(\tau)$ ; it is a theta sum with a half-integral shift, or equivalently a theta function with a different characteristic. If $c$ is odd, there is not even an integral half-period $c/2$ available for this argument, and the transformed expression likewise does not close on the original one-dimensional space spanned by $\theta$ .

A clean way to organize the phenomenon is to introduce the companion series

$\displaystyle \theta_0(\tau)=\sum_{n\in\mathbb Z}e^{2\pi i n^2\tau},\quad \theta_1(\tau)=\sum_{n\in\mathbb Z}e^{2\pi i(n+1/2)^2\tau}.$

The first is our original theta function; the second is the theta series of the shifted lattice $\mathbb Z+\tfrac12$ . Under translation by $1$ , these functions transform diagonally: $\displaystyle \theta_0(\tau+1)=\theta_0(\tau),\qquad \theta_1(\tau+1)=i\theta_1(\tau).$ But under inversion they mix:

$\displaystyle \begin{pmatrix}\theta_0(-1/\tau)\\ \theta_1(-1/\tau)\end{pmatrix}=\left(\frac{-i\tau}{2}\right)^{1/2}\begin{pmatrix}1&1\\ 1&-1\end{pmatrix}\begin{pmatrix}\theta_0(\tau)\\ \theta_1(\tau)\end{pmatrix}.$

Thus the full modular group does act on theta functions, but it acts naturally on the two-dimensional vector $(\theta_0,\theta_1)$ , not on $\theta_0$ alone. The subgroup $\Gamma_0(4)$ is the familiar congruence subgroup for which the line spanned by $\theta_0$ is preserved, so that the vector-valued transformation law collapses to the scalar formula already proved.

Heisenberg interpretation

There is a concise conceptual explanation for the square root and for the central role of Fourier transform. Translation and modulation operators on functions of one real variable, $\displaystyle (T_xf)(u)=f(u+x),\qquad (M_\xi f)(u)=e^{2\pi i\xi u}f(u),$ do not commute: exchanging their order introduces a scalar phase. The Heisenberg group is the group that keeps track of these translations, modulations, and their central phase. The group $\mathrm{SL}_2(\mathbb R)$ , viewed as the symplectic group of the plane, acts on the underlying position-frequency phase space; it therefore acts on the Heisenberg group by automorphisms. On functions, the standard generators of this action are familiar. A shear acts by multiplying a function by a quadratic Gaussian phase. A scaling acts by rescaling the variable and inserting a square-root Jacobian. The inversion matrix acts, up to a constant phase, by Fourier transform. These operators satisfy the relations of $\mathrm{SL}_2(\mathbb R)$ only up to signs, which is why the natural representation belongs to its double cover, the metaplectic group. The factor $(c\tau+d)^{1/2}$ in theta transformation is the visible remnant of this double cover. The theta series may be regarded as the Gaussian evaluated on the integer lattice. Poisson summation says that the integer lattice is fixed by Fourier transform after passing to the dual lattice, which in one dimension is again $\mathbb Z$ . The finite Gauss sum records the residual phase created by the integral arithmetic. This is why the analytic square root and the arithmetic Jacobi symbol appear together in one multiplier.

The Riemann zeta functional equation

The same Poisson identity gives the functional equation of the Riemann zeta function. Define the real theta function

$\displaystyle \vartheta(t)=\theta\left(\frac{it}{2}\right)=\sum_{n\in\mathbb Z}e^{-\pi n^2t},\qquad t>0.$

Poisson summation for the Gaussian $e^{-\pi tx^2}$ gives $\displaystyle \vartheta(t)=t^{-1/2}\vartheta(1/t).$ Remove the zero term by defining

$\displaystyle \psi(t)=\sum_{n\ge1}e^{-\pi n^2t}=\frac{\vartheta(t)-1}{2}.$

For ${\text{Re}}s>1$ , term-by-term Mellin integration is justified by absolute convergence and gives

$\displaystyle \int_0^\infty\psi(t)t^{s/2-1}dt=\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s).$

The Mellin transform converts the Gaussian $e^{-\pi n^2t}$ into $\Gamma(s/2)(\pi n^2)^{-s/2}$ , and summing over $n$ is exactly what creates $\zeta(s)$ . The theta relation gives

$\displaystyle \psi(t)=t^{-1/2}\psi(1/t)+\frac12(t^{-1/2}-1).$

Split the Mellin integral at $t=1$ , use this relation on the interval $(0,1)$ , and substitute $t\mapsto1/t$ . One gets

$\displaystyle \Lambda(s)=I(s)+I(1-s)+\frac{1}{s-1}-\frac{1}{s},$

where $\displaystyle \Lambda(s)=\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s)$ and $\displaystyle I(s)=\int_1^\infty\psi(t)t^{s/2-1},dt.$ Because $\psi(t)$ decays exponentially as $t\to\infty$ , the function $I(s)$ is entire. This identity therefore continues $\Lambda(s)$ meromorphically to the whole complex plane, and its symmetry immediately yields

$\displaystyle \pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{-(1-s)/2}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s).$

The conceptual message is that the zeta functional equation is not separate from theta modularity. It is the Mellin-transform version of the same Poisson-summation symmetry. Theta converts Fourier symmetry into modular symmetry; Mellin transform converts the same symmetry into a relation between $s$ and $1-s$ .

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