In particular, I am wondering if $H^1\_{DR}(G)\neq 0$ implies that the group can written as a semidirect product of $\mathbb{S^1}$ and something else, with the $\mathbb{S^1}$ factor being responsible for the first cohomology group. I have no idea if this is true. In case it is not clear by $H^1\_{DR}(G)$, I mean the De Rham cohomology of the underlying manifold associated to a group $G$. (I understand this is isomorphic to some kind of Lie algebra cohomology, but as of now know nothing about Lie algebra cohomology).


Note that I originally asked this on mathstackexchange and received no answers so I hope this was because perhaps it was more appropriate for here... I have deleted the question on stackexchange.



Which compact (finite dimensional) Lie groups have $H^1_{DR}(G)\neq 0$
The de Rham cohomology is non-trivial if and only if the connected component of $G$ contains a non-trivial central torus.


We can assume that $G$ is connected. Then $H^1\_{dR}(G)$ is isomorphic with $H\_1(G,{\mathbb R})^\*$ by Poincare duality. 
By Hurwitz's Theorem, $H\_1(G)$ is the maximal abelian quotient of $\Gamma$, so $H^1\_{dR}(G)\cong Hom(\Gamma,{\mathbb R})$, where $\Gamma$ is the fundamental group.
Now by, say, Theorem 7.1 of the book by Broecker and tom Dieck, the fundamental group is a quotient of two lattices, so is of the form $F\times {\mathbb Z}^r$, where $F$ is a finite abelian group and $r$ is the dimension of a maximal central torus in $G$. It follows that $H\_{dR}^1(G)\cong{\mathbb R}^r$.


