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I was going through this paper. Corollary 7.3.4 says the $L$-function $L(s,\pi, \rm{sym}^4)$ is holomorphic except possibly at $s=0,1$ and gives a necessary and sufficient condition for it to have a pole at $s=1$ where $\pi$ is a cuspidal representation of $\rm{GL}_2(\mathbb{A}_{\mathbb{Q}})$.

But it does not talk about the behaviour of the fucntion at $s=0$. Can we give a necessary and sufficient condition for the function to have a pole at $s=0$?

Also if $\pi$ is not monomial then, $L(s,\pi, \rm{sym}^4)$ does not have a pole at $s=1$. Can we say more? Can we say it is non-zero or zero at $s=1$?

Similarly, by corollary 5.1.7, $L(s,\rm{sym}^2(\pi)\otimes \chi)$ does not have a pole at s=1 iff $\pi$ is not monomial. Can it have a zero there or is it always non-zero at $s=1$ when $\pi$ is not monomial?

Thank you in advance.

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    $\begingroup$ Can you say what $\pi$ is to make your question self-contained? $\endgroup$
    – Kimball
    Commented Jun 7, 2021 at 3:47
  • $\begingroup$ @Kimball $\pi$ is a cuspidal representation of $\rm{GL}_2(\mathbb{A}_{\mathbb{Q}})$. $\endgroup$
    – user15243
    Commented Jun 7, 2021 at 14:19

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Let $n\geq 1$. The $L$-function of an automorphic representation of $\mathrm{GL}(n)$ is either (1) entire, or (2) holomorphic away from a pole of order $\leq n$ at $s=1+i\tau$ for some fixed $\tau\in\mathbb{R}$. Kim (Theorem B) proved that if $\pi$ is a cuspidal automorphic representation of $\mathrm{GL}(2)$, then $L(s,\pi,\mathrm{Sym}^4)$ is the $L$-function of an automorphic representation of $\mathrm{GL}(5)$. Therefore, there is no pole on the line $\mathrm{Re}(s)=0$.

The automorphy of $L(s,\pi,\mathrm{Sym}^4)$ follows from the automorphy of the exterior square lift from $\mathrm{GL}(4)$ to $\mathrm{GL}(6)$. The cuspidality criterion for this exterior square lift prove here directly determines the order of the pole at $s=1+i\tau$ of $L(s,\pi,\mathrm{Sym}^4)$.

If $\pi$ is an automorphic representation of $\mathrm{GL}(n)$, then $L(1+it,\pi)\neq 0$ for all $t\in\mathbb{R}$. See the appendix to this paper for a much stronger zero-free region.

With some modifications, all of these comments (holomorphic apart from possible pole at $s=1+i\tau$, classification of factorization that determines the order of the pole at $s=1+i\tau$, zero-free region that includes the line $\mathrm{Re}(s)=1$) apply to $L(s,\mathrm{Sym}^2(\pi)\otimes\chi)$. The cuspidality criterion follows from seminal work of Gelbart and Jacquet.

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    $\begingroup$ The pole can be anywhere on $\Re(s)=1$, even when the central character is unitary. This is because $L(s,\pi\otimes|\det|^{it})=L(s+it,\pi)$. $\endgroup$
    – GH from MO
    Commented Jun 7, 2021 at 15:09
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    $\begingroup$ @GHfromMO Good point. I edited the answer accordingly. $\endgroup$
    – 2734364041
    Commented Jun 7, 2021 at 16:06
  • $\begingroup$ @2734364041 Thank you so much for the reply. It has helped me a lot. I have few doubts which I want to clear. I did not understand how to find the order of the pole of $L(s,\pi, \rm{Sym}^4)$ at $s=1$ from the paper of Raghuram-Shahidi? Also what is the order of the pole of $L(s, \rm{Sym}^2(\pi)\otimes \chi)$ at $s=1$? $\endgroup$
    – user15243
    Commented Jun 7, 2021 at 23:25
  • $\begingroup$ @Kiddo For $L(s,\pi,\mathrm{Sym}^4)$, the order depends on $\pi$. Let $F=\otimes_v F_v$ be an automorphic representation of $\mathrm{GL}(4)$. If the $F_v$'s satisfy the hypotheses of Theorem A in Kim's paper, then the exterior square lift $\Lambda^2 F$ is an automorphic representation fo $\mathrm{GL}(6)$. If $\omega$ is the central character of $\pi$, then $L(s,\Lambda^2(\mathrm{Sym}^3\pi)\otimes\omega^{-1}) = L(s,\pi,\mathrm{Sym}^4)L(s,\omega)$. Raghuram-Shahidi tells you how the LHS factors. Similar game for $L(s,\mathrm{Sym}^2\pi\otimes\chi)$ using Gelbart-Jacquet. $\endgroup$
    – 2734364041
    Commented Jun 9, 2021 at 13:27
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    $\begingroup$ @Kiddo Whether or not $L(s,Sym^2\pi\otimes\chi)$ has a pole depends on BOTH $\pi$ and $\chi$, not just $\pi$. If you know the poles of $L(s,\Lambda^2(Sym^4\pi)\otimes\omega^{-1})$, then you can determine the poles of $L(s,\Lambda^2(Sym^4\pi)\otimes\omega^{-1}) / L(s,\omega)$, which equals $L(s,Sym^4\pi)$. $\endgroup$
    – 2734364041
    Commented Jun 10, 2021 at 15:06

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