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 $H^1\_\text{cont}(G,M)=0$ ?



Continuous cohomology of semi-simple Lie group
As [pointed out](https://mathoverflow.net/a/79001/2383) by Konrad, this follows from the generalisation of van Est's theorem from [Group cohomology and Lie algebra cohomology in Lie groups](https://mathscinet.ams.org/mathscinet-getitem?mr=59285) to the continuous case (see [Hochschild and Mostow - Cohomology of Lie groups](https://doi.org/10.1215/ijm/1255632500)); 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](https://www.ams.org/journals/tran/1948-063-01/S0002-9947-1948-0024908-8/home.html), 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 $H^1 =0$.


