How do you obtain all the zeros in $z$ of the Jacobi Theta function $$\vartheta(z) = \sum_{n} e^{\pi i n^2 \tau + 2\pi i n z} \, ?$$

Probably the easiest way is to just read them of the Jacobi-Triple product, but I'm pretty sure they can also be derived from the series representation. The zeros are $$z=a\tau + b + \frac{\tau + 1}{2} \, ,$$ where $a,b \in {\mathbb Z}$, of which I lack to find the term $a\tau$. Since $\vartheta(z+1)=\vartheta(z)$, it is periodic with perdiod $1$ in $z$. So any zero $z_0$ will lead to a zero $b+z_0$ for any integer $b$. It can be seen that $z_0=\frac{\tau+1}{2}$ is a zero since $$\vartheta(z_0) = \sum_{n} e^{\pi i n^2 \tau + \pi i n (\tau+1) } \stackrel{n\rightarrow -n-1}{=} \sum_{n} e^{\pi i n^2 \tau + 2\pi i n \tau + \pi i\tau - \pi i (n+1)(\tau+1)} \\ = -\sum_{n} e^{\pi i n^2 \tau + \pi i n \tau - \pi i n} = -\sum_{n} e^{\pi i n^2 \tau + \pi i n (\tau + 1)} = - \vartheta(z_0) \, .$$

$\vartheta(z)$ has a period of $2$ in $\tau$, but that doesn't help to obtain the term $a\tau$. Any idea?

  • 2
    $\begingroup$ It is "quasi-periodic" with period $\tau$: $\theta(z+\tau)=G(z)\theta(z)$ with $G(z)$ some simple non-zero function. $\endgroup$ Apr 16, 2020 at 12:19
  • $\begingroup$ Of course... How could I forget :-( $\endgroup$
    – Diger
    Apr 16, 2020 at 16:01

1 Answer 1

  • You don't need the full Jacobi triple product, only that $$\prod_{m\ge 1}\left( 1 + x^{2m-1} y^2\right) \left( 1 +\frac{x^{2m-1}}{y^2}\right) = c_0(x)\sum_{n=-\infty}^\infty x^{n^2} y^{2n}, \qquad c_0(x)\ne 0, |x|< 1 $$ Which is elementary.

Finding $c_0(x)$ is harder.

  • Otherwise just from the transformation law of $\theta$ you can locate its zeros and prove there are no more with the residue theorem (integrating over the boundary of a parallelogram, finding there is one more zero than pole, and since $\theta$ has no poles..).
  • $\begingroup$ Yes I know that, but still I was wondering, because it should be possible. $\endgroup$
    – Diger
    Apr 16, 2020 at 15:43

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