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June 18th, 2018, 09:59 AM   #1
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De Rham Cohomology of Lorentz Group


My ultimate goal is to compute the de Rham cohomology groups $H^k(G)$ of the Lorentz group $G=O(3,1)$.

I know that $H^0(G)$ is simply given by $\mathbb R^n$ where $n$ is the number of connected components of $G$.

To show $n$ is four is it enough to show that the map $f:G\to \mathbb R^2$ given by $X\mapsto (\det X, \text{sgn}X^0_0)$ has an image consisting of a discrete set of four points (namely $(1,1)$, $(1,-1)$, $(-1,1)$, $(-1,-1)$)?

(In particular, I am wondering what particular theorem or lemma one states if proving it in this manner.)

Also, how does one go about calculating the higher order cohomology groups?

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