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?


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 $K$.

  • $\begingroup$ 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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