Timeline for Is an automorphic form of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ determined by its L-function?
Current License: CC BY-SA 4.0
15 events
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| Mar 26, 2021 at 8:36 | comment | added | GH from MO | @PeterScholze: Thank you, I agree of course. I delete my comment regarding number fields. | |
| Mar 26, 2021 at 8:26 | comment | added | Sylvain JULIEN | But aren't they the same as functions from $C$ to itself? | |
| Mar 26, 2021 at 8:15 | comment | added | Peter Scholze | 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 $\mathbb Q$. | |
| Mar 26, 2021 at 8:11 | comment | added | Sylvain JULIEN | @Peter Scholze: are these permutations of local L-factors the analogue of permutations of roots of the minimal polynomial of an algebraic number? If yes, can one deduce from it that any "symmetry" of an L-function is an element of some Galois group? | |
| Mar 26, 2021 at 8:02 | comment | added | Peter Scholze | @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. | |
| Mar 26, 2021 at 7:18 | history | edited | Sylvain JULIEN | CC BY-SA 4.0 |
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| Mar 20, 2020 at 7:11 | comment | added | Watson | @GHfromMO : just out of curiosity, what happens for automorphic form over $\mathrm{GL}_n(\Bbb A_K)$, when $K$ is a number field? | |
| Mar 20, 2020 at 2:53 | comment | added | GH from MO | @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 $L(s,\pi_p)$. | |
| Mar 19, 2020 at 18:42 | comment | added | D_S | @GHfromMO can you conclude the local L-functions are equal if the global L-functions are? | |
| Mar 19, 2020 at 15:17 | comment | added | GH from MO | Certainly true for irreducible cuspidal representations by the multiplicity one theorem: en.wikipedia.org/wiki/Multiplicity-one_theorem | |
| Mar 19, 2020 at 14:19 | comment | added | Sylvain JULIEN | Thank you. This should be also true for primitive L-functions, from Lemma 4.2 in M. Ram Murty, Selberg conjectures and Artin L-functions (1994). | |
| Mar 19, 2020 at 14:07 | comment | added | D_S | I think it's true at least for cuspidal automorphic reps. If $L(s,\pi) = L(s,\pi')$, there should be some way to conclude that the local components $L(s,\pi_p) = L(s,\pi'_p)$ for almost all primes $p$. At most of the primes $\pi_p$ is unramified and determined by its local L-function. Then strong multiplicity one gives you $\pi = \pi'$. | |
| Mar 19, 2020 at 12:19 | comment | added | reuns | We are dealing with irreducible representations so it is generated by the $GL_n(A_Q)$-translates of one "vector" (which concretely is a left-$GL_n(Q)$-invariant function $GL_n(A_Q)\to C$), an automorphic form, that you can choose with some concrete desirable properties, translate, or make abstract by decomposing the representation into local parts. | |
| Mar 19, 2020 at 10:41 | history | edited | Sylvain JULIEN | CC BY-SA 4.0 |
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| Mar 19, 2020 at 10:31 | history | asked | Sylvain JULIEN | CC BY-SA 4.0 |