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?](https://mathoverflow.net/questions/52552/nontrivial-finite-group-with-trivial-group-homologies#comment130307_52552)) 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.



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


