Continuous cohomology of semi-simple Lie group

Question

Let $G$ be a real connected semi-simple Lie group. Let $M$ be a finite dimensional representation of it. Are there general criteria when the continuous cohomology groups $H_\text{cont}^q(G,M)$ vanish?

A situation of particular interest for me is $G=SO^+(n-1,1)$, namely the connected Lorentz group, and $M$ is the standard representation of it. Is it true that the first continuous cohomology $H^1_\text{cont}(G,M)=0$ ?

Answer 1

As pointed out by Konrad, this follows from the generalisation of van Est's theorem from Group cohomology and Lie algebra cohomology in Lie groups to the continuous case (see Hochschild and Mostow - Cohomology of Lie groups); namely, that $$ H_c^m(G,M) \cong H^m(\mathfrak{g},\mathfrak{k};M) $$ where $\mathfrak{g}$ is the Lie algebra of $G$ and $\mathfrak{k}$ is the Lie algebra of the maximal compact subgroup of $G$.

For the case in question, $m=1$, $\mathfrak{g} = \mathfrak{so}(n-1,1)$ and $\mathfrak{k}=\mathfrak{so}(n-1)$. I will take $n>2$.

According to Chevalley and Eilenberg - Cohomology theory of Lie groups and Lie algebras, the cohomology $H^m(\mathfrak{g},\mathfrak{k};M)$ is computed from ‘horizontal’ ‘equivariant’ cochains in $C^m(\mathfrak{g},M)$, where ‘horizontal’ means that the cochain vanishes whenever any of its entries belongs to $\mathfrak{k}$ and ‘equivariant’ means with respect to the action of $\mathfrak{k}$.

Now for the algebras in question, $\mathfrak{g}$ breaks up as $\mathfrak{k} \oplus V$ under the action of $\mathfrak{k}$, where $V$ is the fundamental vector representation of $\mathfrak{k}$, whereas $M = V \oplus \mathbb{R}$, with $\mathbb{R}$ the trivial one-dimensional representation.

Since $$ C^0(\mathfrak{g},\mathfrak{k};M) = M^{\mathfrak{k}} $$ it follows that $$ \dim C^0(\mathfrak{g},\mathfrak{k};M) = 1~. $$ The differential $\delta: C^0 \to C^1$ is injective, since if $T \in M^{\mathfrak{k}}$ ($T$ is ‘timelike’ hence the notation) $$ \delta T(X) = X \cdot T $$ which does not vanish identically.

On the other hand, $$ C^1(\mathfrak{g},\mathfrak{k};M) = \text{Hom}(V,M)^{\mathfrak{k}} $$ is again one-dimensional, hence $C^1 = \delta C^0$ and hence $H^1 =0$.

Answer 2

I think that for $G$ a connected compact Lie group, we have $$ H^m_{cont}(G,M) = 0 $$ for $m>0$.

This follows from the van Est-isomorphism $$ H^m_{cont}(G,M) \cong H^m(g,k; M), $$ which holds for $G$ a connected Lie group, $K$ a maximal compact subgroup and $g$, $k$ the Lie algebras of $G$, $K$, and the thing on the right hand side is the relative Lie algebra cohomlogy. For $K=G$ in the compact case, this relative Lie algebra cohomology is identically zero.

A good place to look is Stasheff's "Continuous cohomology of groups and classifying spaces".

Answer 3

I think that it is true if you use "smooth" instead of continuous.