Timeline for Get $H^1(S,sl(2,R)_{Ad\phi}$) dimension directly from differential forms
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 19, 2012 at 14:57 | comment | added | Qiongling Li | Dr.Misha, that is really a nice reference. Thank you. | |
| Apr 19, 2012 at 4:21 | comment | added | Misha | A good reference for differential forms with twisted coefficients is Raghunathan's book "Discrete subgroups of Lie groups." | |
| Apr 19, 2012 at 3:45 | comment | added | Qiongling Li | Dr.Misha, thank you, that gives me hint to find the right elements in $H^1$. | |
| Apr 19, 2012 at 0:18 | comment | added | Misha | Actually, Teichmuller space naturally identifies with a certain component of $Hom(π,SL(2,{\mathbb R}))/SL(2,{\mathbb R})$ consisting of discrete and faithful representations. Then the simplest computation of $H^1$ is combination of Poincare duality and Euler characteristic: $\chi(S, {\mathbb V})= dim(V) \chi(S)$, where ${\mathbb V}$ the the flat bundle over $S$ associated with the surface group action on $V=psl(2,{\mathbb R})$. You also have $0=H^0\cong H^2$ by Poincare duality and the fact that centralizer of $\phi(\pi)$ is trivial. Thus, $dim H^1= (2g-2) dim(V)= 6g-6$. | |
| Apr 18, 2012 at 23:34 | history | edited | Qiongling Li | CC BY-SA 3.0 |
deleted 76 characters in body
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| Apr 18, 2012 at 20:59 | history | asked | Qiongling Li | CC BY-SA 3.0 |