Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

For $G$ an abelian group, the Eilenberg-Maclane space $K(G, n)$ represents singular cohomology $H^n(-; G)$ with coefficients in $G$ on the homotopy category of CW-complexes. If $n > 1$, then $G$ must be abelian, but for $n = 1$ there are also Eilenberg-Maclane spaces for $G$ nonabelian, so one might imagine that they represent some kind of nonabelian cohomology $H^1(-; G)$. Is this functor known by a better name, and what's known about it? In particular,

  • Does it have an alternate definition along the lines of the usual definition of singular cohomology?
  • Is there a universal coefficient theorem for it?
share|cite|improve this question
What happened you deleted a Galois theory question of yours which you just now posted. – user9413 May 9 '11 at 8:21
I figured it out. – Qiaochu Yuan May 9 '11 at 8:58
up vote 13 down vote accepted

K(G,1) aka BG classifies G-bundles — i.e. G-coverings, if G is discrete. (Details can be found e.g. in May's Concise Course in Algebraic Topology.)

Usual definition of Cech cohomology works for $H^1(X;G)$ even in non-abelian case (but it's just the usual cocycle definition of G-bundle).

As for universal coefficient theorem, even if $H_1(X;\mathbb Z)$ is trivial, $H^1(X;G)$ needn't be; but (if G is discrete) $H^1(X;G)=\operatorname{Hom}(\pi_1(X);G)/\text{conjugation}$ (reference: Hatcher, 1B.9). (But if one wishes to consider BG for general G, things get worse — "$H^1$" is no longer defined by 2-skeleton of X. Perhaps, AHSS from cohomology to K-theory can be viewed as kind of "universal coefficient spectral sequence" for $G=U=\lim U(n)$.)

share|cite|improve this answer
Thanks. Do you have a reference? This material doesn't seem to be covered in Hatcher, for example. – Qiaochu Yuan Apr 30 '11 at 11:45
Also, I googled -- which looks good – Grigory M Apr 30 '11 at 12:07

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.