Timeline for Formal geometry
Current License: CC BY-SA 2.5
9 events
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| Jun 21, 2010 at 1:53 | history | edited | David Jordan | CC BY-SA 2.5 |
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| Jun 21, 2010 at 1:51 | comment | added | David Jordan | Yes, that was way wrong - serves me right for posting an answer secondhand. It was worth it, though, because you clarified exactly the point I was confused about - how the $\partial_i$'s relate to $X_{dR}$ to give the crystal structure. I tried to incorporate your corrections as best I could. Let me know if there are more - It's community wiki in case you wish to make any edits yourself. | |
| Jun 21, 2010 at 1:44 | history | edited | David Jordan | CC BY-SA 2.5 |
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| Jun 21, 2010 at 1:12 | history | edited | David Jordan | CC BY-SA 2.5 |
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| Jun 21, 2010 at 0:53 | history | made wiki | Post Made Community Wiki by David Jordan | ||
| Jun 21, 2010 at 0:34 | comment | added | David Ben-Zvi | The fiber product of X with itself over X_dR (which you calculate above) is the formal neighborhood of the diagonal in X x X -- in particular it's finite dimensional, unlike the group G.. | |
| Jun 21, 2010 at 0:29 | comment | added | David Ben-Zvi | Something's wrong here. The total space of the canonical G-torsor over X_dR is not X - G doesn't act on X. The way it's explained eg in my book with Frenkel is there's a G-torsor over X, which descends to X_dR, given by all formal coords on X (all nondegenerate maps from a disc). A nice formulation: the G action extends to the action of an ind-group G^ exponentiating all of W_n (including the del_i), and the quotient is X_dR (a G^ torsor over X_dR). Given an M as in your question, it carries a G^ action, and so we can take an associated bundle over X_dR, ie a sheaf on X with flat connection.. | |
| Jun 20, 2010 at 23:50 | history | edited | David Jordan | CC BY-SA 2.5 |
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| Jun 20, 2010 at 23:06 | history | answered | David Jordan | CC BY-SA 2.5 |