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Let $G$ be a connected Lie group, without any further assumptions. Is it true, that its rational cohomology ring $$H^\bullet(G,\mathbb Q)$$ is finite dimensional? Is $G$ homotopy equivalent to a compact Lie group?

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Both questions have the answer of "yes". Of course, an answer of "yes" to the second must imply and answer of "yes" to the first because every compact manifold has a finitely generated cohomology ring.

So, why is $G$ homotopy equivalent to a compact Lie group? In fact, more is true. We have the following theorem (see wiki for more):

Suppose $G$ is a connected subgroup. Then, there exists a maximal compact subgroup $K\subseteq G$. Further, all such maximal compact subgroups are conjugate and $G$ is diffeomorphic to $K\times \mathbb{R}^n$ for some $n$.

(Note that while $G$ is diffeomorphic to $K\times\mathbb{R}^n$, $G$ is only rarely isomorphic (as a group) to a product of $K$ and $\mathbb{R}^n$.)

Finally, simply note that $K\times\mathbb{R}^n$ obviously deformation retracts onto $K$.

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  • $\begingroup$ Do you have a reference for this theorem other then Wikipedia? In fact my question was inspired by reading wikipedia and I wanted to make sure, that it is correct what is written there :) $\endgroup$ – Jan Jun 10 '13 at 14:29
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    $\begingroup$ mathoverflow.net/questions/53080/… seems to have a good bit of information, as well as more references in it. $\endgroup$ – Jason DeVito Jun 10 '13 at 15:38
  • $\begingroup$ Thank you very much for your answer. Can you say something about the idea of proof that $G=K\times \mathbb R^n$? $\endgroup$ – Jan Jun 11 '13 at 9:00

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