Timeline for Invariant differential forms on commutative group schemes are closed!?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Feb 22, 2022 at 18:32 | answer | added | Nils Matthes | timeline score: 1 | |
| Sep 11, 2011 at 14:03 | answer | added | Torsten Ekedahl | timeline score: 6 | |
| Sep 11, 2011 at 8:37 | answer | added | inkspot | timeline score: 2 | |
| May 30, 2011 at 4:13 | comment | added | user2490 | Philipp, it seems the usual Maurer-Cartan equations are all you need. See, for example, Proposition 51 in Chapter 3 Section 14 of Bourbaki's Lie Groups and Lie Algebras, which is stated in terms of Lie groups and Lie algebras, of course---possibly over a field of characteristic > 0---but which also applies to smooth algebraic groups. (Bourbaki's formulation avoids the "1/2" that may have worried you in other treatments.) The starting point is the familiar formula d\omega(X,Y) = X\omega(Y) - Y\omega(X) - \omega([X,Y]): if X,Y, and \omega are left invariant, then the first two terms vanish... | |
| May 27, 2011 at 11:27 | comment | added | naf | I think that some sort of lifting argument might be unavoidable. A more elementary method might be to use Grothendieck's theorem that there exist formal lifts over the Witt vectors or just $W_2(k)$ (see for example, Oort: Finite group schemes, local moduli,...) and Corollaire 2.5 of Deligne, Illusie: Relèvements modulo $p^2$ et décomposition du complexe de de Rham. Invent. Math. 89 (1987), no. 2, 247–270. (If one forms are closed, all p-forms are closed in the case of abelian varieties.) | |
| May 27, 2011 at 9:35 | history | edited | Philipp Hartwig | CC BY-SA 3.0 |
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| May 27, 2011 at 9:34 | comment | added | Philipp Hartwig | I'm afraid I wouldn't be comfortable using that. My goal is to really understand some of the proofs in BBM and this would not be achieved by quoting far more difficult results. | |
| May 24, 2011 at 9:22 | comment | added | naf | For abelian varieties, the claim (in any characteristic) follows from the fact that they can be lifted to characteristic 0, a result of Norman and Oort (and Grothendieck and Mumford). However, I believe that there should be a more elementary proof that also works in characteristic 2. | |
| May 23, 2011 at 15:44 | comment | added | Philipp Hartwig | Thanks, I've simply added the assumption that the group scheme should be commutative. | |
| May 23, 2011 at 15:38 | history | edited | Philipp Hartwig | CC BY-SA 3.0 |
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| May 21, 2011 at 17:21 | answer | added | naf | timeline score: 12 | |
| May 21, 2011 at 16:10 | history | edited | Philipp Hartwig | CC BY-SA 3.0 |
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| May 21, 2011 at 16:06 | comment | added | José Figueroa-O'Farrill | I suppose it depends on what it means to be "invariant". Neither the left- or right-invariant differential forms on a Lie group are necessarily closed, but the bi-invariant forms are. | |
| May 21, 2011 at 15:31 | history | asked | Philipp Hartwig | CC BY-SA 3.0 |