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Let $n,m$ be two different positive integers.

I heard that for cuspidal tempered automorphic representations $\pi_{n}$ and $\pi_m$ of $GL_n$ and $GL_m$, the Rankin-Selberg L-function $L(s,\pi_n \times \pi_m)$ is entire on $\mathbb{C}$.

I am wondering if the tempered or cuspidal condition is essential to ensure $L(s,\pi_n \times \pi_m)$ being entire. That means if one of $\pi_n$ or $\pi_m$ is non-tempered or non-cuspidal, then their $L$-function may have a pole? If this is the case, can we know the candidate for the location of poles?

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    $\begingroup$ @garret, from the beginning, I assumed $m \ne n$. I also fixed a question little bit. Would you please read it again? $\endgroup$
    – Monty
    Oct 7, 2014 at 1:17
  • $\begingroup$ With cuspidal data and $m\not=n$ it is relatively straightforward to prove that there is no pole in the right half-plane from the critical line. For $m=n-1$ the form of the integral representation (the "Hecke" form) also makes clear that the $L$-function is entire, for cuspidal data. For $m=n$ but non-contragredient cuspidal data, again the form of the integral repn makes clear the entire-ness. All other cases are more complicated... $\endgroup$ Oct 7, 2014 at 13:26
  • $\begingroup$ @garrett, Thanks for your comment. Indeed, my question comes from the consideration of the holomorphicity of the Rankin Selberg L-function $L(s,BC(\pi_{n+1}) \times BC(\pi_{n}))$ at $s=\frac{1}{2}$ appearing in the Gross-Prasad conjecture for the unitary group. (here, $\pi_i$'s are the cuspidal tempered representation of $U(n)(A_F)$ and $BC(\pi)$ its base change to $GL_n(A_E)$. I thoght it should be holomorphic at $s=\frac{1}{2}$, but since base change functor may not preserve cuspidality, we cannot guarantee its holomorphicity there. $\endgroup$
    – Monty
    Oct 7, 2014 at 18:34
  • $\begingroup$ Yes, I'd be entirely willing to believe that there is some delicacy in that situation... $\endgroup$ Oct 7, 2014 at 20:49

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