# smooth cohomology of Lie groups

Let $G$ be a Lie group and $A$ a smooth $G$-module. Define $C^n(G,A)=\{ f: G^n \to A|~f~\text{is smooth}\}$ and $\partial^n: C^n \to C^{n+1}$ by the standard formula as used in the cohomology of abstract groups. I think this cohomology must be well studied. Can somebody provide me some references for this cohomology. A similar cohomology for topological groups has been studied by Hu and Heller using continuous cochains.

Also, can somebody tell me what kind of Lie group extensions $H^2(G,A)$ correspoind to.

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Edited the LaTeX so that the braces would show up. –  Zev Chonoles Jan 24 '11 at 7:21
I believe that at least for compact groups, but may be more generally, this complex is quasi-isomorphic to the corresponding Chevalley- Eilenberg complex for the Lie algebra. I have some notes at work and I will check later today. –  José Figueroa-O'Farrill Jan 24 '11 at 8:16
@JFF: For compact groups, one should expect most reasonably cohomology theories to agree, because you can average cochains over the group to get left-invariant cochains, and hope that the averaging doesn't change the cohomology (note that left-invariance implies smoothness). –  Theo Johnson-Freyd Jan 24 '11 at 8:23
Many thanks. But, what kind of Lie group extensions does $H^2(G,A)$ characterize. –  gurs Jan 24 '11 at 9:28

This is indeed very well studied. The standard references are Borel-Wallach, Continuous cohomology, discrete subgroups and representations of reductive groups, Annals of Math. Studies 94, Princeton University Press (1980) and the more gentle book by A. Guichardet, Cohomologie des groupes topologiques et des algèbres de Lie Textes Mathématiques 2 Fernand Nathan, Paris (1980). But there has been a lot of progress since. For instance, there is some recent work by Crainic extending the theory to Lie groupoids and algebroids.

The continuous cohomology is usually attributed to Hochschild-Mostow and coincides with the smooth cohomology under reasonably weak hypotheses.

The interesting fact is that there is an intimate interplay between the continuous cohomology of, say a semi-simple Lie group and the cohomology of its Lie algebra and the de Rham cohomology of the associated symmetric space. One of the most important results is the van Est-isomorphism:

Let $G$ be a semi-simple Lie group with finite center and no compact factors. Let $M = G/K$ be the associated symmetric space. Then the de Rham complex $\mathbb{R} \to \Omega^{0}(M) \to \Omega^{1}(M) \to \cdots$ is an injective resolution of $\mathbb{R}$. In particular, since a $G$-invariant differential form on $M$ is automatically closed, there is a natural isomorphism $H_{c}^{\ast}(G,\mathbb{R}) \cong \Omega^{*}(M)^{G}$ where the right hand side are the $G$-invariant differential forms on $M$. If $A$ is a sufficiently nice smooth $G$-module then the $A$-valued differential forms need not be closed, but one still has $H_{c}^{\ast}(G,A) \cong H^{\ast}(\Omega^{\ast}(M,A)^{G})$ Moreover, one may identify this with relative Lie algebra cohomology $H^{\ast}(\mathfrak{g},\mathfrak{k};A)$.

This immediately gives us an interpretation of $H_{c}^{2}(G,\mathbb{R}) = \Omega^{2}(M)^{G}$. Namely, if $G$ is simple, then $\Omega^{2}(M)^{G}$ is one-dimensional if and only if $G$ is Hermitian (when it is generated by the Kähler form on $M$), and otherwise it is zero. So the dimension of $H_{c}^{2}(G,\mathbb{R})$ corresponds to the number of of Hermitian simple factors of $G$.

Of course, one may also bring discrete subgroups into play, together with all their rich and beautiful connections to geometry and number theory. There are far too many things to mention here, so I'd better stop now.

Before I forget: I don't know of a direct interpretation of $H_{c}^{2}(G,A)$ as equivalence classes of suitable extensions of $G$. One problem is that in the topological category, it is not clear at all that there should be a smooth (or continuous) section of non-trivial extension.

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Thanks for your comments and providing the references. I would like to add that, in the continuous cohomology for topological groups due to Hu and Heller, they showed that the second cohomology group classifies topologically split group extensions. More precisely, if $G$ is a topological group and $A$ is a topological $G$-module, then $H_{cont}^2(G,A)$ classies all extensions $1 \to A \to E \to G \to 1$ such that as a space $E$ is $A \times G$. I am guessing that the similar result is true for Lie groups when we work with smooth cohains as mentioned in my question. Is is true? –  gurs Jan 24 '11 at 10:56
I really don't know for sure but have you looked at the proof? I mean, if this is true for topological groups then you can bet that it is also true for Lie groups. I mean, it all boils down to having a smooth section $\sigma: G \to E$ to get a smooth $2$-cocycle $c: G \times G \to A$ by writing $c(g,h) = c(gh)c(h^{-1})c(g^{-1})$. Conversely, given a cocycle you should be able to write down the extension. If two cocycles are cohomologous then this should give equivalent extensions and vice versa. So from the bar-resolution viewpoint, I'd be very surprised, if it didn't work. –  Theo Buehler Jan 24 '11 at 11:16
Thanks for your comments. If $E$ is $A \times G$ as a smooth manifold, then the extension has an obvious smooth section and the proof works as per your suggestion. I am also feeeling that it should be true. Thanks again. –  gurs Jan 24 '11 at 12:48