Does the symmetric square L-function vanish at one?

Question

Take a cuspidal automorphic representation $\pi$ of $GL(3)$ over a number field. My question is quite straightforward and can be related to this one :

Can $L(1, \pi, \mathrm{sym}^2)$ be zero? If yes, is there any extra assumption ensuring it cannot?

I know that we have the splitting $L(s, \pi \times \tilde{\pi}) = L(s, \pi, \mathrm{sym}^2)L(s, \pi, \wedge^2 \pi)$. If we assume $\pi$ self-contragredient for instance, then the Rankin-Selberg convolution has a simple pole at $s=1$. Either $L(s, \pi, \mathrm{sym}^2)$ has a pole at $s=1$ (Gelbart-Jacquet lift case) and in that case it does not vanish ; or it has none and in that case the question rephrases as : can $L(s, \pi, \wedge^2 \pi)$ have a pole of order 2 at $s=1$?

Any new insight or reference on this precise question is welcome.

Answer 1

For $GL(3)$, the exterior square $L$-function $L(s,\wedge^2\pi)$ is entire as it agrees with $L(s,\tilde\pi\otimes\omega)$, where $\omega$ is the central character of $\pi$. Therefore, $L(1,\mathrm{sym}^2\pi)=0$ would imply that $L(1,\pi\otimes\pi)=0$, contradicting a result of Shahidi (1980). The best known zero-free region for general Rankin-Selberg $L$-functions is due to Brumley (2012): see the Appendix here.