Cohomology of Lie groups

Remarks

Example: $G = S^1$, the circle group. Then $H^0(S^1; \R) = \R$, $H^1(S^1; \R) = \R x$, and all other groups vanish. So $H^*(S^1; \R)$ is the exterior algebra on one generator $x$ of degree $1$, with $\mu^*(x) = x \otimes 1 + 1 \otimes x$.

Non-example: the cohomology of $G$ as an abstract group, $H^*(BG^\delta; k)$, or the continuous cohomology of $G$, is a different object. For $G = S^1$ the continuous cohomology with coefficients in $\R$ vanishes in positive degrees, while $H^1(S^1; \R) \ne 0$.

For $k = \R$, the de Rham theorem identifies $H^*(G; \R)$ with the de Rham cohomology of $G$. If $G$ is compact and connected, averaging over $G$ shows that each class has a left- and right-invariant representative. This gives an isomorphism $H^*(G; \R) \cong H^*(\mathfrak{g}; \R)$ with the Lie algebra cohomology of the Lie algebra $\mathfrak{g}$.