Timeline for Cohomology of a space with coefficients in a Lie group?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 4, 2012 at 20:21 | comment | added | John Klein | Justin: I didn't mean to normalize (unfortunately, one can't edit comments on MO), but I thought the Moore complex is always defined for a simplicial group, and it seemed to me therefore the same would be true for any cosimplicial group. Am I missing something? | |
| Jan 4, 2012 at 19:56 | comment | added | Justin Noel | If I try to form the analogue of the normalized cochain complex using a cosimplicial group I don't see why the iterated boundary map should be trivial. | |
| Jan 4, 2012 at 17:12 | comment | added | John Klein | Kevin: you are probably right. | |
| Jan 4, 2012 at 15:58 | comment | added | Kevin Walker | In your definition of $(g\cdot f)(\alpha)$, did you want to put the $f$ factor between the $g$ and $g^{-1}$ factors? | |
| Jan 4, 2012 at 12:21 | comment | added | John Klein | David: If $X$ is a simplicial set, then one can form the cosimplicial topological group $n \mapsto G^{X_n}$, where the target is the functions $X_n\mapsto G$ (here $G$ can be any topological group). Couldn't one take something like a normalized Moore complex of this to obtain a cochain complex of topological groups? (I.e., the intersection of the kernels of all the coface operators except the last in each degree, where the coboundary is given by the last coface operator.) | |
| Jan 4, 2012 at 2:41 | history | edited | John Klein | CC BY-SA 3.0 |
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| Jan 4, 2012 at 2:18 | comment | added | David Roberts♦ | If you want higher dimensional cohomology groups, you need $G$ abelian, see my answer mathoverflow.net/questions/36466/… | |
| Jan 4, 2012 at 2:13 | history | edited | John Klein | CC BY-SA 3.0 |
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| Jan 4, 2012 at 2:06 | history | asked | John Klein | CC BY-SA 3.0 |