Recall that the theta function with character $(a,b)\in \mathbb{R}^2$ is defined by $$ \vartheta_{a,b}(z, \tau) :=\sum^\infty_{n=-\infty} e^{\pi i (n + a)^{2} \tau + 2 \pi i(n + a)(z + b)},\quad a,b \in \mathbb{R}. $$ Where are the zeros of $\vartheta_{a,b}(z, \tau)$?


1 Answer 1


The zeros are at $z=\!(k-\!\tfrac 12\!-\!a)\tau+\!\tfrac 12\!-b-n$, with $k$ and $n$ being integers. First complete the square to rewrite the exponential as $e^{i\pi\tau(n+a+(z+b)/\tau)^2-i\pi(z+b)^2/\tau}$ , then pull out $e^{-i\pi(z+b)^2/\tau}$ and reconsider the rest of the sum as $\vartheta(a+\!\frac{z+b}\tau,-i\tau)$, with $\vartheta(u,-i\tau)=\sum_{k\in\mathbb{Z}}e^{i\pi\tau(k+u)^2}$ (Note $\vartheta(-u,-i\tau)=\vartheta(u,-i\tau)=\vartheta(u+1,-i\tau)$). Now using the transformation $\vartheta(u,-i\tau)=\frac 1{\sqrt{-i\tau}}e^{i\pi\tau u^2}\vartheta(\tau u,\frac i\tau)$, show that $\vartheta(u+\frac 1\tau,-i\tau)=e^{\pi/q+2\pi\tau}\vartheta(u,-i\tau)$ and then plug in $u=\frac 12-\!\frac 1{2\tau}$ (which leads to $$\vartheta\left(\frac 12+\frac 1{2\tau},-i\tau\right)=-\vartheta\left(\frac 12-\frac 1{2\tau},-i\tau\right)=-\vartheta\left(-\frac 12-\frac 1{2\tau},-i\tau\right)=-\vartheta\left(\frac 12+\frac 1{2\tau},-i\tau\right)=0$$ due to $\vartheta(u,-i\tau)$ having no singularities). Finally substitute $u=a+\frac{z+b}\tau$ and use $\vartheta(u,-i\tau)=\vartheta(u+1,-i\tau)$ and $\vartheta(u+\frac 1\tau,-i\tau)=e^{\pi/q+2\pi\tau}\vartheta(u,-i\tau)$ to fetch the other zeros (then solve for $z$).

The identity $\vartheta(u,-i\tau)=\frac 1{\sqrt{-i\tau}}e^{i\pi\tau u^2}\vartheta(\tau u,\frac i\tau)$ follows from the Poisson summation formula for $f(n+x)=e^{i\pi\tau(n+x)^2}$

  • $\begingroup$ Why the zeroes of theta functions are only of this form? Are there any other zeroes? $\endgroup$
    – wzstrong
    Feb 20 at 7:07

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