1
$\begingroup$

Let $\phi$ be an irreducible cuspidal automorphic representation of $GL_n(\mathbb{A})$ of symplectic type, that is, the exterior square $L$-function $L(s,\phi,\Lambda^2)$ has a pole at $s=1$.

Then I am wondering whether it is possible that $L(\frac{1}{2},\phi' \times \phi)=0$ for all $m \ge n$ and irreducible cuspidal automorphic representation of $GL_m(\mathbb{A})$ of orthogonal types? (Here, orthogonal type means that the symmetric square $L$-function $L(s,\phi',Sym^2)$ has a pole at $s=1$.)

I think it would not be true but I can't prove the existence of $\phi'$ such that $L(\frac{1}{2},\phi' \times \phi)\ne 0$.

Do you think is there a possibility that $L(\frac{1}{2},\phi' \times \phi)=0$ for all irreducible cuspidal automorphic representation $\phi'$ of $GL_m(\mathbb{A})$ of orthogonal types and for all $m \ge n$?

Any comments are welcome!

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ I'm guessing that this isn't known but it's expected that $L(1/2,\pi'\times\pi)$ is usually nonzero. A standard analytic approach to this question would be to fix $\pi$ and look at the average of $|L(1/2,\pi'\times\pi)|^2$ over a family of cuspidal automorphic representations $\pi'$ of orthogonal type (e.g. ordered by analytic conductor). If one could show that this is positive, then nonvanishing would follow for infinitely many $\pi'$. However, aside from low rank examples (e.g. $n = m = 2$), this kind of average is hard to calculate. $\endgroup$
    – Peter Humphries
    Commented Dec 3, 2021 at 15:50
  • 1
    $\begingroup$ @Peter, Thank you for the comment! Is there a reference for $n=m=2$ case? If it is, I would appreciate if you let me know it. And my precise question is to know the existence of any $m$ such that $m \ge n$ satisfying the above property. $\endgroup$
    – Monty
    Commented Dec 3, 2021 at 16:36
  • 1
    $\begingroup$ @Peter, Oh, I found some papers related with this. Those are Feigon and PD. Nelson's papers. Thank you! $\endgroup$
    – Monty
    Commented Dec 3, 2021 at 17:14

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .