# What are cohomology of Lie algebra with coefficients geometrically?

I want to find analog of following two statements.

1. Let $G$ be a discrete group, $M$ is representation of $G$. Local systems on $BG$ are the same as $G$ representations (because $\pi_1 (BG) =G$). Let $\mathscr{M}$ be a local system corresponding to $M$. $$H^{\bullet}_{Grp} (G, M) = H^{\bullet} (BG, \mathscr{M})$$

2. Let $G$ be compact connected Lie group. $\mathfrak{g}$ is corresponding Lie algebra. $\mathbb{R}$ - trivial representation of $\mathfrak{g}$. $$H^{\bullet}_{Lie} (\mathfrak{g}, \mathbb{R}) = H_{dR}^{\bullet} (G)$$

Question: How to express $H^{\bullet}_{Lie} (\mathfrak{g}, M)$ geometrically? Here $M$ is a finite dimensional representation of $\mathfrak{g}$ (if you wish you can assume that it is integrated to Lie group representation).

Comment 1 : I am even not sure which geometric object corresponds to representation of $\mathfrak{g}$. Is it bi-D-module (bimodule over differential operators)? Is it $G \times G$ equivariant bundle on $G$?

Comment 2: I want to say in other words what I want. I want a geometric structure on group $G$ considered as a manifold, which counts $H^{\bullet}_{Lie} (\mathfrak{g}, M)$ . It can be sheaf, D-module, whatever. The point is that I want to forget that $G$ is a group. I want just $G$ considered as a manifold with extra geometric structure. And the way to get my cohomology back.

• "Here $M$ is an arbitrary representation" is rather vague: Presumably the representation is finite dimensional and derived from a continuous representation of $G$, but is it required to be irreducible? Also, "geometric" is somewhat open-ended in this framework. One direction might be a connection with the flag variety of $G$, for instance. – Jim Humphreys Feb 1 '15 at 17:04
• Just to clarify: in Comment 2, are you restricting yourself to the cases where G is compact and connected? – Yemon Choi Feb 1 '15 at 19:10
• Actually I am doubt whether my question makes sense. You see, the approach from statement 2 works only if Lie group is compact. But all finite dimensional representations of compact groups are direct sum of irreducible. And cohomology of irreducible not trivial representation is just 0. But it is still would be great to interpreter this 0. Or may be you can generalize my question for more general representations or groups? – quinque Feb 1 '15 at 19:25
• @Yemon Choi, compactness is really important. But I think that assumption that group simply connected is not (correct me if I am mistaking). – quinque Feb 1 '15 at 19:28

We assume $G$ is a compact connected Lie group with Lie algebra $\mathfrak{g}$. Let $\rho:\mathfrak{g}\to \mathrm{End}(E)$ is a finite representation.

We denote by $\underline{E}=G\times E$ the trivial bundle over $G$. Take $U\in \mathfrak{g}$, then $U$ define a left-invariant vector field $X_U$ on $G$.

For $s\in C^\infty(G,\underline{E})$, we define the $G$ action by, $(g\cdot s)(x)=s(g^{-1}x)$. Take $e\in E$, then $e$ define a $G$-invariant section $s_e$ of $\underline{E}$, that is $s_e(x)=e.$

We define a connection on $\underline{E}$ by $$\nabla_{X_U}s_e=s_{\rho(U)e}.$$ This is a flat connection. The $G$-invariant part of the de Rham cohomology associated to this flat bundle $(\underline{E},\nabla)$ is what you are looking for in comment 2, i.e., $$\Big(H^\cdot_{dR}(G,\underline{E})\Big)^G=H^\cdot(\mathfrak{g},E).$$

To show this, we identify $\Omega^\cdot(G,\underline{E})^G$ the left-G-invariant differential form with coefficients in $\underline{E}$, with $\mathrm{Hom}(\Lambda^\cdot(\mathfrak{g}),E)$. Under this identification, The de Rham differential operator $d$ become the differential of the complex $\mathrm{Hom}(\Lambda^\cdot(\mathfrak{g}),E)$. This means $$H^\cdot(\Omega^\cdot(G,\underline{E})^G,d)=H^\cdot(\mathfrak{g},E).$$

We apply the Hodge theorem. We denote by $\Box$ the Hodge Laplacian, we get $$\Omega(G,\underline{E})=H_{dR}^\cdot(G,\underline{E})\oplus \mathrm{im}(\Box)$$ Since $\Box$ commut with $G$, we have $$\Omega(G,\underline{E})^G=H_{dR}^\cdot(G,\underline{E})^G\oplus \mathrm{im}(\Box)^G.$$ From last equation, we get $$H_{dR}^\cdot(G,\underline{E})^G=H^\cdot(\Omega^\cdot(G,\underline{E})^G,d).$$

Examining the complex used to calculate $H^*(\mathfrak{g};\mathbb{R})$, one sees that it is precisely the complex of $G$-invariant differential forms on $G$. (To see this, take the left-trivialization of the tangent bundle of $G$, and re-write all exterior power bundles in terms of this trivialization.) I suppose the geometric content is that the cohomology of this complex coincides with that of all differential forms on $G$, namely $H^*_{dR}(G;\mathbb{R})$.

Let me briefly address your point about geometric interpretations of Lie algebra representations. Let $\mathfrak{g}\rightarrow End(V)$ is a $\mathfrak{g}$-representation. You can integrate it to a Lie group representation $G\rightarrow Aut(V)$, where $G$ is the connected, simply-connected Lie group with Lie algebra $\mathfrak{g}$. In this sense, the Lie algebra represention is an "infinitesimal" version of the Lie group representation.

Also, whenever a Lie group $G$ acts smoothly on a manifold $M$, there is a representation of its Lie algebra $\mathfrak{g}$ on the Lie algebra of all vector fields on $M$. One associates to each element of $\mathfrak{g}$ a fundamental vector field on $M$.

• I am asking about geometric interpretation of cohomology of Lie algebra. Not about geometric interpretation of representation itself. Sorry, but you answer some other question. – quinque Feb 1 '15 at 18:35
• I was responding to what you had called "Comment 1". It had sounded like a request for a geometric interpretation of a Lie algebra representation. Also, the first part of my answer attempts to give a geometric interpretation of Lie algebra cohomology. Perhaps it is not exactly the sort of answer you are seeking, but I think it at least attempts to address your question. Fundamentally, the problem is that a "geometric interpretation" of something like this is inherently ambiguous. Perhaps you could give more details about what precisely you are seeking. – Peter Crooks Feb 1 '15 at 18:38
• I understand now. I even wrote comment 2 to make it less ambiguous. You see, I need an answer, analogous to statement 1 (about cohomology of a group). In that example "geometric interpretation" of representation of discrete group is local system. – quinque Feb 1 '15 at 18:57