How do I show that for a fixed $\omega$ in the upper half plane, the theta function

$\theta(z) = \sum_{n=-\infty}^{\infty} e^{\pi i(n^2\omega + 2nz)}$

is not identically zero? Is there an obvious choice of z that would work or would I have to resort to some fourier coefficient argument?



1 Answer 1


Letting $\omega=iy$ and $y\to +\infty$ gives limit $1$ (from the $n=0$ term), for arbitrary $\omega$. So it's not identically $0$. To prove it's not constant, subtract the $n=0$ term, divide by $e^{2\pi i n\omega}$, and let $\omega=iy$ with $y\to+\infty$ again, getting the $n=1$ coefficient.

But, yes, the Fourier series argument is instantaneous, so is better in the long run, even if one does want to see the guts of things in the short run.

(Also, you might consider interchanging the roles of $\omega$ and $z$, or letting $\omega$ by $\tau$, but this is inessential.)

EDIT: Sorry, due to my expectations about notational choices, I read it as though the roles of $\omega$ and $z$ were reversed from the question. To fix the questioner's $\omega$ and ask whether the resulting function of $z$ can be identically $0$, surely a Fourier series argument is optimal: a Fourier series with not-all-$0$ Fourier coefficients is not identically $0$ as a function (assuming the coefficients decay a bit, so that it has a good point-wise sense).

For generic $\omega$ in the upper half-plane, I don't think there's any clever choice of $z$ to see pointwise-non-zero-ness directly.

  • $\begingroup$ I'm not sure I understand your reasoning in the first paragraph. The OP wants to show $\theta (z)$ is never identically $0$ for each fixed $\omega$ in the upper half plane. $\endgroup$
    – zhw.
    Sep 25, 2015 at 23:54
  • $\begingroup$ @zhw., Well, there are many ways to certify this, and so many that perhaps none is "canonical". But/and it is not the case that one could show that various automorphic forms never vanish... but one can show that they do not identically vanish, that is, that they do not vanish for all inputs... Perhaps I do not understand your comment... $\endgroup$ Sep 25, 2015 at 23:59
  • $\begingroup$ Thanks for your answer. But I don't really understand how making z go to infinity along the imaginary axis makes the function go to 1. Don't the negative terms in the sum mess things up? Could you spell it out for me please? $\endgroup$
    – Liron Atia
    Sep 26, 2015 at 15:39
  • $\begingroup$ The parts $e^{i\pi n^2\omega}$, with $\omega=iy$ and $y\to+\infty$ all go to $0$ for $n\not=0$. (You are right that the $e^{2\pi inz}$'s would not go to $0$, because $n$, rather than $n^2$, can be negative.) $\endgroup$ Sep 26, 2015 at 16:33
  • $\begingroup$ But I want to show the function doesn't vanish identically for each $\omega$ in the upper half plane. $\endgroup$
    – Liron Atia
    Sep 26, 2015 at 16:42

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