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](https://mathoverflow.net/questions/178654/characterizing-the-newforms-s-t-the-associated-symmetric-square-l-function-ha) :



> 
> 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.



Does the symmetric square L-function vanish at one?
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)](https://projecteuclid.org/euclid.bams/1183546358). The best known zero-free region for general Rankin-Selberg $L$-functions is due to Brumley (2012): see the Appendix [here](https://www.esi.ac.at/static/esiprpr/esi2403.pdf).


