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.

  • $\begingroup$ Thank you Tyler, do you know of other interesting classes of groups (a part from knot exteriors) that have finite $K(\pi,1)$? $\endgroup$ – Warlock of Firetop Mountain Mar 14 at 19:49
  • $\begingroup$ @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. $\endgroup$ – Tyler Lawson Mar 15 at 4:42
  • 2
    $\begingroup$ 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. $\endgroup$ – Sebastian Goette Mar 15 at 11:28
  • $\begingroup$ @SebastianGoette sorry maybe I am missing something, doesn't the definition of $K(G,1)$ depend just on the group structure of $G$? $\endgroup$ – Warlock of Firetop Mountain Mar 16 at 21:59
  • $\begingroup$ 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. $\endgroup$ – Sebastian Goette Mar 17 at 9:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.