Timeline for Are there motives which do not, or should not, show up in the cohomology of any Shimura variety?
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10 events
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| Dec 28, 2010 at 19:00 | comment | added | David Hansen | @Kevin (comment of Sep 22, 9:33): A motive with Hodge-Tate weights 0,1,3 should be "degenerate" - if the associated Galois representation were irreducible, and you believe Fontaine-Mazur, then you'd get a cuspidal GL3 automorphic form whose infinity type is not fixed by the Cartan involution, hence doesn't exist by Borel-Wallach. Also, how did this very interesting question peter out so quietly and without arriving at any consensus? | |
| Sep 22, 2010 at 13:48 | comment | added | Kevin Buzzard | @Joel: see the link at the top to "meta"? That's the place to ask questions about mathoverflow itself, such as yours. Or just go directly to tea.mathoverflow.net . | |
| Sep 22, 2010 at 13:28 | comment | added | Joël | @Kevin: You were right to ask the question (which is great) in a vague way. Making the question precise is part of answering it. About my question "is a motive with coeff in Q self-dual-up-to-a-Tate twist?", I meant "a simple motive". But I will try to clarify my mind and post again about your question. Meanwhile, sorry to pollute this discussion by a question to be erased when answered but : where on mathoevrflow can I ask the question "I have created a openId, how can I get back on this account the reputation points I earned as a non-registred user?" ? | |
| Sep 22, 2010 at 9:33 | comment | added | Kevin Buzzard | @Joel: I agree that for regular motives there is a chance that there is a simple conjecture! But I don't know it. As for motives being dual to Tate twists of themselves---just take a motive with Hodge-Tate weights 0,1,3 and I think this gives a counterexample. I can build reducible examples of this form easily; maybe building an irreducible one would be harder? About making the question precise: I do agree with you, but see my comments for Laurent for why I decided not to do this initially. I wonder if it really makes too much difference though? (at least conjecturally...) | |
| Sep 22, 2010 at 9:27 | comment | added | Kevin Buzzard | The reason I didn't mention the "potential" issue was because of the following construction: if $L/K$ is a finite Galois extension then I can consider something like $G=Res_{L/K}(GL(1))$ (or even $GL(0)$ if Deligne's axioms allow it; I forget) and get Shimura varieties whose $H^0$ is abelian over $L$ but which still give the Artin Galois representation over $K$ that I want as a subquotient. In general you're abelian over the reflex field but you can control the reflex field! | |
| Sep 22, 2010 at 9:24 | comment | added | Kevin Buzzard | @Laurent: I agree I should make it precise. The reason I didn't make it precise initially was that I was "fishing" for a precise statement that I couldn't quite remember, so it was to my advantage to be as vague as possible! I already found the answer to that in Milne's comment, so then I had to change the question a bit and I just figured I would leave it to see if anyone could formulate a precise negative result (e.g. "the etale cohomology of a Shimura variety always has this property, hence this Galois representation can never be a subquotient"). | |
| Sep 22, 2010 at 5:32 | comment | added | Minhyong Kim | As I mentioned to Kevin, you two should also ask Clozel about my comment, and let us know here if my memory (or my understanding at the time) is (or was) completely stupid. I'd quite like to know myself. | |
| Sep 22, 2010 at 4:22 | comment | added | Joël | That said, I'm pretty confused. What motives are supposed to appear (directly or potentially) in the cohomology of Shimura varieties? If we assume the motive regular, that is with distinct Hodge numbers, shouldn't this has a simple answer? Take F=Q, for example. Shouln't any regular motive that it is a twist of its dual appear in a Shimura variety (orthogonal, or symlectic, or unitary after restriction to quadratic imaginary field)? What about the converse? Now, another queston (that might be stupid): isn't any motive over Q (and with coefficients in Q) dual to a Tate twist of itself? | |
| Sep 22, 2010 at 4:04 | comment | added | Joël | I agree with Laurent that the question should be precised before attempting any answer. One precision is, as he said, are we talking of the intersection cohomology (which is the same as the L^2 cohomology, and which will see the motives attached to the discrete automorphic representations of the group G defining the Shimura varieties), or of the ordinary cohomology (which sees all the cuspidal automorphic representations, some of the discrete non-cuspidal, and some others, no one knows exactly which)? | |
| Sep 22, 2010 at 0:03 | history | answered | Laurent F. | CC BY-SA 2.5 |