Skip to main content
9 events
when toggle format what by license comment
Sep 18, 2019 at 11:48 vote accept CommunityBot
Sep 17, 2019 at 17:43 answer added skupers timeline score: 4
Sep 17, 2019 at 17:29 comment added YCor For the continuous one, it's a standard fact about central extensions of semisimple Lie groups. For the non-continuous one, it's related to (symplectic) K-theory and more complicated (oh, well I'm sure that $H_2$ is huge; that $H^2$ valued in $\mathbf{C}$ is still huge, should be checked); see mathoverflow.net/questions/63529 for a related discussion.
Sep 17, 2019 at 16:32 comment added user39380 @YCor Thanks for pointing out, let's focus on continuous cohomology. And would you explain the calculation of the two versions of $H^2(\mathrm{Sp}_{2n}(\mathbb{C}),\mathbb{C})$? Thanks!
Sep 17, 2019 at 16:00 comment added YCor As regards (2), do you mean group cohomology or continuous group cohomology? For instance $H^2(\mathrm{Sp}_{2n}(\mathbf{C}),\mathbf{C})$ is huge but the continuous cohomology is trivial. Maybe the same phenomenon occurs with the module $\mathbf{C}^{2n}$, I'm not sure. In general guess that the continuous cohomology in degree $\ge 1$ of any finite-dimensional $G$-module is zero, for $G$ semisimple.
Sep 17, 2019 at 15:54 comment added YCor @Bombyxmori If it's isomorphic to a Lie algebra cohomology group, this is a theorem, not a definition. In the $\mathbf{Z}$ case this sounds quite doubtful, by the way.
Sep 17, 2019 at 15:53 comment added Bombyx mori Isn't that defined by the cohomology of the Lie algebras? I think Chevally studied it during 1950s...
Sep 17, 2019 at 15:53 history edited YCor
edited tags
Sep 17, 2019 at 15:49 history asked user39380 CC BY-SA 4.0