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Just a fast question, since I have not been able to find any answer for it online.

Where are the zeros of Dedekind eta function $\eta(s)$ located?

Apart from the trivial one as $s \to i \infty$, are there any other zeros in the upper half of the complex plane? If so, where?

Thank you.

user3141592
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    "Apart from the trivial one at $s=0$", the function $\eta(s)$ is not defined at $s=0$. I suppose you mean $s=i\infty$. – pisco Nov 17 '18 at 06:50
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    Regarding your question, $\eta^{24}$ is a constant multiple of the modular discriminant, which is well-known to be non-zero at any finite value. – pisco Nov 17 '18 at 06:51

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If $\mathrm{Im}(\tau)>0$ then $q=e^{2\pi i\tau}$ has absolute value $<1$. Then $\eta(\tau)=q^{1/24}\prod_{n\ge1}(1-q^n)$, where $\prod_{n\ge1}(1-q^n)$ is an absolutely convergent infinite product since $\log|1-q^n|=O(|q|^n)$, so $\sum_{n\ge1}\log|1-q^n|<\infty$.

Thus, $\eta(\tau)\ne0$ for $\mathrm{Im}(\tau)>0$.

Kenta S
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Consider the 24th degree of the Dedekind eta-function; it is the modular discriminat $\Delta(\tau)$. Now, $\Delta$ is the disciminat of the cubic equation

$$ y^2 = x^3 - g_2(\tau) \; x -\; g_3 (\tau), $$

for the cubic curve which is isomorphic to $\mathbb{C} / L$, $L = Z + Z \tau$.

Since this 1-dimensional torus is smooth, this cubic curve is smooth; hence the cubic polynomial above has no multiple zeroes; and, hence, Delta is nowhere zero on the upper half-plane.

Maxim Leyenson
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