Timeline for Status of (global) Langlands conjecture for $\mathrm{GL}_2$ over $\mathbb{Q}$
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16 events
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| S May 2 at 17:21 | history | suggested | coLaideronnette | CC BY-SA 4.0 |
Fixing typos and making the answer more readable.
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| May 2 at 14:57 | review | Suggested edits | |||
| S May 2 at 17:21 | |||||
| S Feb 16, 2023 at 16:15 | history | suggested | Kenta Suzuki | CC BY-SA 4.0 |
improved formatting
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| Feb 16, 2023 at 15:32 | review | Suggested edits | |||
| S Feb 16, 2023 at 16:15 | |||||
| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 15, 2013 at 21:16 | comment | added | Joël | But then your question becomes: how much bigger is the Langlands group than the algebraic Langlands group", and on this I don't know much. I guess it should be much bigger to account for all the maass forms already for $Gl_2$ and their weird systems of Satake parameters... | |
| Apr 15, 2013 at 21:16 | comment | added | Joël | bigger than $G_{\mathbb Q}$. How much bigger? I don't known and I think it is not clear, but at least the algebraic L anglands group should have as a quotient an infinite product of copies of $SU_2(\R)$, which of course are not anywhere in the group $G_{\mathbb Q}$ (the reason for this lies in the theory of the Weil-Deligne group, which is the local analog of the algebraic Langlands group, and contains a $SU(2)$ factor, al least in some formulation)... | |
| Apr 15, 2013 at 20:32 | comment | added | Joël | ... $G_{\mathbb Q}$ suffices. However, in the formalism of the Langlands group, one consider not $\ell$-adic, but complex continuous representations. Due to the totally discontinuous nature of the Galois group, it has muck less complex than $\ell$-adic representations ($\ell$ being, as I should have said earlier, a fixed auxiliary prime). To compensates for this, the quotient of the Langlands group (let's call it the algebraic Langlands group, and it should be the same, I think, as the Tannakian Galois group of the categories of pure motives over $\mathbb Q$) that "suffices" for algebraic.. | |
| Apr 15, 2013 at 20:27 | comment | added | Joël | Dear Masoud: There is a somewhat technical yet important point which makes that the the absolute Galois group $G_{\mathbb Q}$ does not suffice even for algebraic representations, at least in the Langland's sense of "suffice". It is true that conjecturally, there should be a bijection between algebraic at $\infty$, cuspidal, automorphic representations of $\Gl_n$ over $\mathbb Q$ and irreducible continuous $\ell$-adic representations of $G_{\mathbb \Q}$ of dimension $n$, satisfying the condition of Fontaine-Mazue (that is unramified almost everywhere, and de Rham at $\ell$). In this sense... | |
| Apr 15, 2013 at 9:50 | vote | accept | Masoud | ||
| Apr 15, 2013 at 3:48 | comment | added | Masoud | @Joel: Thank you for the extensive reply. From what I can gather from Marc's and your reply, the issue of being algebraic at infinity is a condition about the restriction of the automorphic representation to the Archimedean place. So, in other words, if we include all automorphic representations we need Langlands group $L(\mathbb{Q})$, but if we want only algebraic at infinite automorphic representations, then the absolute Galois group $G_{\mathbb{Q}}$ suffices. Is there anyway one can see how much bigger this Langlands group is? | |
| Apr 12, 2013 at 0:37 | comment | added | Dror Speiser | @Joel: Yeah, I think so, isn't this a part of the Langlands-Tunnel theorem? | |
| Apr 11, 2013 at 15:49 | comment | added | Joël | @Dror: it does ignore it, and so do I. You are saying that from an even solvable galois representation one can construct an maass form? | |
| Apr 11, 2013 at 14:36 | comment | added | Marc Palm | Technically the limit of dicrete series reps is not a quotient as in the case $SL(2)$, so is it still common to call it that way? It is a principal series representation of $\GL_2(\mathbb{R})$. | |
| Apr 11, 2013 at 14:05 | comment | added | Dror Speiser | I hope I don't sound too ignorant, but doesn't this answer ignore the construction from (2) to (1c) for even solvable galois groups? | |
| Apr 11, 2013 at 0:59 | history | answered | Joël | CC BY-SA 3.0 |