Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a positive integer $n\geq 2$ and an abelian group $G$, is it possible to find a finite dimensional $K(G,n)$? In case it does, which are some examples?


share|cite|improve this question
up vote 8 down vote accepted

There is never a $K(G, n)$ for $n \geq 2$ which is a finite complex (and $G$ nontrivial).

In fact, a finite simply connected complex has finitely generated homotopy groups in all dimensions (by Serre's mod $\mathcal{C}$ theory applied to the universal cover). So one reduces to seeing that a $K(\mathbb{Z}, n)$ or a $K(\mathbb{Z}/p^k, n)$ cannot be a finite complex. In both cases this can be established by computing the cohomology ring and showing that it is infinite-dimensional.

In the first case, one can use $\mathbb{Z}/2$-cohomology to get a polynomial ring: namely, Serre established that the $\mathbb{Z}/2$-cohomology ring of $K(\mathbb{Z}, n)$ is a polynomial ring generated by the following elements: take the "universal" class $\iota_n \in H^n$, and take all the Steenrod squares $\mathrm{Sq}^I \iota_n$ as $I$ ranges over appropriate sequences (I'll refer you e.g. to Mosher-Tangora for the precise statement). This is not too hard to prove, once you invoke the Serre spectral sequence inductively on $n$ and use the fact that the Steenrod squares commute with the transgression. You can do something similar for $\mathbb{Z}/2$. I don't know offhand what the answer is for the other primes (it'll involve the more complicated Steenrod powers), but it should look similar.

One reason you should not expect these spaces to be simple or finite-dimensional is that they represent cohomology. There is a "universal" $G$-cohomoloy class in $H^n(K(G, n), G)$ such that any $G$-cohomology class is a pull-back of this. If $K(G, n)$ were simple, then that would mean there were not many things you could do with $G$-cohomology classes, when there are in general lots of things you can do with them, such as the Steenrod operations. There are only a few Eilenberg-MacLane spaces with simple geometric constructions (the only ones I know are the circle, infinite real and complex projective space, compact surfaces of positive genus, $K(S_n, 1)$ as the spaces in the little cubes operad, and examples such as the one below); this seems to express the fact that cohomology operations are complicated. Note that all but one of the examples I cited were in dimension one.

If you only want the complex to be finite-dimensional, though, you can get a $K(\mathbb{Q}, 1)$ by the following procedure. Take the sphere $S^1$, consider the sequence of maps $f_n: S^1 \to S^1$ of degree $n$, and form an "infinite mapping telescope" of the $f_n$. (This is a special case of a homotopy colimit.) It's not hard to check, because $\mathbb{Q}$ is the filtered colimit $\mathbb{Z} \to \mathbb{Z} \to \dots$ where successive maps are multiplication by $1, 2, \dots$, that this is a simple example of a $K(\mathbb{Q}, 1)$. I don't know how to rule out getting $K(G, n)$ for $n\geq 2$ for some group $G$ (though a $K(\mathbb{Q}, 2)$ is out, again by looking at cohomology).

Here is a (harder) result of Serre which also proves the claim I made above: a simply-connected, finite CW complex which is not contractible has infinitely many nonzero homotopy groups (in fact, infinitely many homotopy groups containing either a copy of $\mathbb{Z}$ or $\mathbb{Z}/2$).

share|cite|improve this answer
Thanks a lot for your detailed answer! – MBL Nov 22 '11 at 21:55
"A finite complex has finitely generated homotopy groups in all dimensions": this is not true, as $\pi_2(S^1 \vee S^2) \cong \Bbb Z[t,t^{-1}]$ shows. To apply Mod C theory to the universal cover, it would also need to be (homotopy equivalent) to a finite complex; unless $\pi_1(X)$ is finite, there's no reason we should be so lucky. – Mike Miller Apr 10 '15 at 18:10
@MikeMiller: thanks for the correction. – Akhil Mathew Apr 11 '15 at 0:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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