# $G$ uncountable implies $K(G,1)$ is not a finite CW complex

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.

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.
• Thank you Tyler, do you know of other interesting classes of groups (a part from knot exteriors) that have finite $K(\pi,1)$? – Warlock of Firetop Mountain Mar 14 at 19:49
• 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 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 $G$? – Warlock of Firetop Mountain Mar 16 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 at 9:50