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To an automorphic representation $\pi$ of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ one can associate its L-function $s\mapsto L_{\pi}(s)$.

Is the map $\pi\mapsto L_{\pi}$ bijective?

Edit March 26th, 2021: can such a bijection be seen as a fully faithful fonctor between Tannakian categories $\mathcal{C}$ whose objects are automorphic representations and $\mathcal{D}$ whose objects are L-functions suggesting every automorphism of a given object of the latter arises from an intertwinning operator? If so, can such an automorphism $\varphi$ be written as $\varphi_{\sigma}:L_{\pi}\mapsto\sigma\circ L_{\pi}\circ\sigma^{-1}=L_{\pi}\mapsto{L_{\sigma'\circ\pi\circ\sigma'^{-1}}}$ where $\sigma$ is a field automorphism of $\mathbb{C}$ and $\sigma'$ the corresponding intertwinning operator?

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    $\begingroup$ Certainly true for irreducible cuspidal representations by the multiplicity one theorem: en.wikipedia.org/wiki/Multiplicity-one_theorem $\endgroup$
    – GH from MO
    Commented Mar 19, 2020 at 15:17
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    $\begingroup$ @GHfromMO can you conclude the local L-functions are equal if the global L-functions are? $\endgroup$
    – D_S
    Commented Mar 19, 2020 at 18:42
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    $\begingroup$ @D_S: Yes, since the local $L$-functions are just the Euler factors of the global $L$-function. To put another way, if you restrict the Dirichlet coefficients of $L(s,\pi)$ to powers of $p$, you get $L(s,\pi_p)$. $\endgroup$
    – GH from MO
    Commented Mar 20, 2020 at 2:53
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    $\begingroup$ @GHfromMO Not quite: For number fields, you can't necessarily recover the local $L$-factors from the global $L$-function -- for example if $K$ is Galois over $\mathbb Q$, a Galois twist of $\pi$ will have the same $L$-function as $\pi$, but the local $L$-factors are permuted. This has been studied, for example by Harry Smit. $\endgroup$
    – Peter Scholze
    Commented Mar 26, 2021 at 8:02
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    $\begingroup$ It's not a symmetry of any given $L$-function -- rather, two distinct $L$-functions happen to agree (by virtue of the local $L$-factors being permuted around). But I think the whole phenomenon is even more subtle than this. One way to think about it is that the $L$-function only knows about the automorphic induction from $K$ to $\mathbb Q$. $\endgroup$
    – Peter Scholze
    Commented Mar 26, 2021 at 8:15

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