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I have read that $H^i(K(\mathbb{R},1)$) has rank $2^\omega$ for any $i\in \mathbb{N}$ (see Thurston's comment here Nontrivial finite group with trivial group homologies?) therefore $K(\mathbb{R},1)$ is not a finite CW complex and even the $n$-skeleton is not finite for any $n$.

On the other hand there are infinite (discrete) Lie groups $\pi$, such that $K(\pi,1)$ is finite. For example consider $\pi = \pi_1 (E)$ where $E\subset \mathbb{S}^3$ is the knot exterior of a knot. This seems to be a rather lucky case as we know that $K(F,1)$ is non finite for any discrete finite group $F$.

This made me wonder if the following is true:

Is $K(G,1)$ an infinite CW complex for any $G$ Lie group of dimension greater than 1?

What are other examples of $G$ such that $K(G,1)$ is finite?

Note: infinite CW complex is the same as being non compact.

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Overflowian
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1 Answers1

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For any finite CW-complex $X$ and any basepoint $x \in X$, the fundamental group $\pi_1(X,x)$ is finitely presented. (This is a consequence of the Seifert-van Kampen theorem.) In particular, the group itself is a quotient of a finitely generated free group, and hence must be a countable set.

However, if $G$ is a Lie group of positive dimension, then the underlying set of $G$ is uncountable. Therefore, no $K(G,1)$ can have the homotopy type of a finite CW-complex.

Tyler Lawson
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  • Thank you Tyler, do you know of other interesting classes of groups (a part from knot exteriors) that have finite $K(\pi,1)$? – Overflowian Mar 14 '20 at 19:49
  • @WarlockofFiretopMountain I'm afraid that the examples in the page that Steve D linked to above (esp. torsion-free, finitely-generated nilpotent or hyperbolic groups) cover most of the examples that I know. – Tyler Lawson Mar 15 '20 at 04:42
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    I think you should add that you consider $K(G^\delta,1)$, that is, $G$ is equipped with the discrete topology (which it is typically not if called a Lie group of positive dimension). Otherwise, I would not even know what $K(G,1)$ was supposed to mean. – Sebastian Goette Mar 15 '20 at 11:28
  • @SebastianGoette sorry maybe I am missing something, doesn't the definition of $K(G,1)$ depend just on the group structure of $G$? – Overflowian Mar 16 '20 at 21:59
  • The definition says $\pi_1(K(G,1))\cong G$ and $\pi_k(K(G,1))=0$ otherwise. Now everything hinges on your understanding of "$\cong$". If you say "Lie group of positive dimension", I think that you want the group with its topology and differentiable structure. On the other, $\pi_1(X)$ is classically just a discrete group. But there might be a context where the functor $\pi_1$ can take values in groups with additional structure (in which case $K(G,1)$ should be related to the classifying space $BG$ in that category). I just wanted to avoid any misunderstandings. – Sebastian Goette Mar 17 '20 at 09:50