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Let $\pi$ be an irreducible cuspidal automorphic representation of $GL(n)$.

Suppose the symmetric square $L$-function of $\pi$ $L(s,\pi,Sym^2)$ has a pole at $s=1$.

Then since $L(s,\pi \times \pi)=L(s,\pi,Sym^2)L(s,\pi,Ext^2)$, $\pi$ is self-dual and so the central character $\chi_{\pi}$ is quadratic. (i.e. $\chi_{\pi}^2=1$.)

Then I am wondering whether that $n$ should be greater than 1. Because when $n=1$ and $\pi$ is the trivial character of $GL(1)$, I guess $L(s,\pi \times \pi)=L(s,\pi,Sym^2)$ is just the completed Riemann zeta function and it is known to be holomorphic and non-zero at $s=1$.

So for having $L(s,\pi,Sym^2)$ has a pole at $s=1$, $n$ should be greater than 1?

Any comments are highly appreciated!

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    $\begingroup$ It depends on what you mean by completed Riemann zeta function. If you just mean $\pi^{-s/2} \Gamma(s/2) \zeta(s)$, then this has a pole at $s = 1$, whereas if you mean $\frac{1}{2} s(s - 1) \pi^{-s/2} \Gamma(s/2) \zeta(s)$, then this is holomorphic and nonzero at $s = 1$. Usually (especially in the context of automorphic representations) people mean the former. $\endgroup$
    – Peter Humphries
    Commented Jul 24, 2021 at 14:17
  • $\begingroup$ @Peter, Oh thank you very much. I confused the completed Riemann zeta function latter. It is possible that $L(s,\pi,Sym^2)$ has a pole at $s=1$ even when $n=1$. $\endgroup$
    – Andrew
    Commented Jul 24, 2021 at 18:17

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