Timeline for Universal deformations of modular Galois representations
Current License: CC BY-SA 3.0
9 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| May 14, 2014 at 14:20 | comment | added | Olivier | Look, all I'm pointing out is that a positive answer to your question implies a positive answer for the "different" question, which is already hard (read: false in general) but at least admits a positive answer if you invert $p$. That's why I said you have better chances in that case. But these are just comments: my official answer remains that I think your problem is a hard one. | |
| May 14, 2014 at 9:27 | comment | added | David Loeffler | No, I am trying to do mathematics here, not philosophy. I am fully aware of the existence of sheaves of Gal.reps. on the eigencurve coming from completed cohomology, but that is the answer to a different question: this is not the right rigid space -- it is 2-dimensional (if you include twists by characters) and parametrizes deformations of $\bar\rho$ with a choice of $p$-refinement, but I want the 3-dimensional space parametrizing arbitrary deformations which are not necessarily finite-slope at $p$. | |
| May 14, 2014 at 9:23 | comment | added | Olivier | If your "why" was more of a philosophical nature, to me it all boils down to the fact that the Hecke algebra is "very nice" after inverting $p$ (étale over $\mathbb Q_{p}$, formally smooth in a classical neighborhood...). | |
| May 14, 2014 at 9:19 | comment | added | Olivier | Because in that case the results of Coleman-Mazur tell you that you have a locally free sheaf on the eigencurve. Emerton's construction through completed cohomology tells you furthermore that this sheaf comes from the (image through the Jacquet functor of the locally algebraic vectors of the) direct limit of the cohomology of modular curves, so satisfies your requirement. In an affinoid neighborhood of a classical non-critical point, you have an infinitesimal R=T theorem by Kisin identifying this sheaf with the universal deformation, as you wished for. | |
| May 13, 2014 at 16:08 | comment | added | David Loeffler | Why is one then in better shape? | |
| May 13, 2014 at 15:25 | comment | added | Olivier | I thought you might say this. This is why I added the final paragraph. You are in much better shape if you are fine with inverting $p$ and with free sheaves on the corresponding rigid analytic space. | |
| May 13, 2014 at 14:55 | comment | added | David Loeffler | OK, so the non-$p$-distinguished case is bad; but I'm happy to rule this out -- I'd be content if there were any non-empty set of $\bar\rho$ for which such a construction were possible. | |
| May 13, 2014 at 14:28 | history | answered | Olivier | CC BY-SA 3.0 |