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## Continuous cohomology of semi-simple Lie group.

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

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As pointed out by Konrad, this follows from the generalisation of van Est's theorem to the continuous case (due to Hochschild and Mostow); 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, 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 $H^1 =0$.

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 Great! Thank you very much!! – semyon alesker Oct 25 2011 at 10:38 You are welcome! If I may ask, in what context do you need this calculation? – José Figueroa-O'Farrill Oct 25 2011 at 10:42 To be short, I want to understand when invariant vectors in a quotient representation of the Lorentz group can be lifted to invariant vectors in the given representation. – semyon alesker Oct 25 2011 at 18:58 This problem comes from my interest in the theory of valuations on convex sets. These are fintiely additive measures on convex sets, see e.g. these survey arxiv.com/PS_cache/arxiv/pdf/1008/1008.0287v3.pdf It is of interest, both intrinsically to the theory and sometimes for applications in integral geometry, to classify valuations invariant under various groups. Right now my student is trying to classify valuations invariant under the Lorentz group. Vanishing of the first cohomology is sufficient for this purpose. – semyon alesker Oct 25 2011 at 19:00 Many thanks for this. I'm always happy to see groups I like appear in seemingly unrelated work. – José Figueroa-O'Farrill Oct 25 2011 at 20:38

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

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 Van Est's original theorem was indeed for smooth cohomology, but I believe that this was extended to the continuous case by Hochschild and Mostow in ams.org/mathscinet-getitem?mr=147577 – José Figueroa-O'Farrill Oct 24 2011 at 17:53

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".

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 The group in the question is not compact, though. Still, the calculation of $H^1(\mathfrak{g},\mathfrak{k};M)$ is not hard in this case and one finds that indeed it vanishes. – José Figueroa-O'Farrill Oct 24 2011 at 17:50 Oh, oops, of course you're right, José. – Konrad Waldorf Oct 24 2011 at 17:54