Here is the definition of $\xi(s,\chi)$:

$\xi(s,\chi)= \left(\frac{s(s-1)}{2} \right)^{1_{\chi=1}} (q/\pi)^{\frac{s+a}{2}} \Gamma \left( \frac{s+a}{2} \right) L(s,\chi)$

Here is the definition of the Hadamard product applied to $\xi(s,\chi)$:

Since it is an entire function of order one, there exists constants $A_\chi, B_\chi \in \mathbb{C}$ such that $\xi(s,\chi)=e^{A_\chi+B_\chi s} \prod_{p} \left(1-\frac{s}{p} \right) e^{s/\rho} $

with the product running over every non-trivial zero of $L(s,\chi)$.

Why does this imply there are infinitely many non-trivial zeros for $L(s,\chi)?$

FYI: I know there non trivial zeros are in one to one correspondence with the zeros of $\xi(s,\chi)$


1 Answer 1


The existence of the Hadamard product by itself doesn't show the function has any zeroes (it might just be an exponential!) -- you need a little more.

The idea is to study the logarithmic derivative, which by your two expressions above satisfies ($\delta_\chi = 1$ for the principal character):

$$ -\frac{\xi'}{\xi}(s;\chi) = \delta_\chi\left(\frac{1}{s}-\frac{1}{1-s}\right)-\frac{1}{2}\log\frac{q}{\pi} -\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{s+a}{2}\right) -\frac{L'}{L}(s;\chi) $$ $$ = -B_\chi - \sum_\rho \left(\frac{1}{s-\rho}+\frac{1}{\rho}\right)\,.$$

Now plug in $s=2+iT$ and examine the terms as $T\to\infty$. In the first row, all terms are bounded except for the logarithmic derivative of the Gamma function, which grows like $\log T$. On the second row, individual terms are bounded in $T$ so there must be infinitely many.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.