In order to define the cohomology of a topological group G, we first have to introduce the concept of a classifying space. A classifying space BG is the base space of a principal G bundle EG. The EG is the universal bundle: Any principal G bundle E over a manifold M allows a bundle map into the universal bundle, and any two such morphisms are smoothly homotopic. Given $$ \gamma: M \to BG $$ the induced map of the base manifolds, it is the so-called classifying map.

The topology of the bundle E is completely determined by the homotopy class of the classifying map $\gamma$. That is, the different components of the space Map(M,BG) correspond to the different bundles E over M. It can be shown that up to homotopy BG is uniquely determined by requiring EG to be contractible. That is, any contractible space with a free action of G is a realization of EG. In general, the classifying space BG of a compact group is an infinite-dimensional space.

Attempt 1:

Naively, we can have a generalized statement by modifying the map to a product of Eilenberg–MacLane space: $$ \gamma: M \to K(G_1,1) \times K(G_2,2) $$ where $G_1$ and $G_2$ are two different groups. The $G_1$ can be non-abelian.

  • How can we be more precise to state the similar structure, defining a "2"-group cohomology, by generalizing the relation between the "group cohomology" and the "topological cohomology of classifying space"?

  • Is it necessary to have $G_2$ be abelian?

Attempt 2

  • A more general classifying map may be: $$ \gamma: M \to BG' $$ where $BG'$ is the possible fibration $$1 \to K(G_2,2)\to BG' \to K(G_1,1) \to 1,$$ where we may classify the fibration by Postinikov class $$[a] \in H^d(K(G_1,1), G_2).$$ Is this formulation precise?
  • How can we be sure that $d=3$ is the only solution and $[a] \in H^d(K(G_1,1), G_2)$ classifies all sensible "2" group cohomology?
  • $\begingroup$ What do you mean by "2-group cohomology"? Note that $K(G, n)$ only makes sense for $n > 1$ if $G$ is abelian, because higher homotopy groups are abelian. $\endgroup$ – user98602 Jul 26 '18 at 7:41
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    $\begingroup$ I'm afraid its not at all clear what you are looking for. If you are looking for a cohomology theory of 2-groups, then these already exist. For example a google search for "crossed module cohomology" will give you results. Even searching for 2-group cohomology will give you a paper on Arxiv by Ginot. $\endgroup$ – Tyrone Jul 26 '18 at 9:40
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    $\begingroup$ As for your first comment note that $K(G_1,1)\times K(G_2,2)\simeq BG_1^\delta\times BK(G_2,1)\simeq B(G_2^\delta\times K(G_2,1))$, where $G_1^\delta$ is the (possibly non-abelian) group $G_1$ given the discrete topology, and $G_2$ is assumed abelian. Eilenberg-Mac lane spaces only give information on the underlying discrete group, and ignore any topological information it may hold. $\endgroup$ – Tyrone Jul 26 '18 at 9:41
  • $\begingroup$ Tyrone - thanks +1 $\endgroup$ – wonderich Jul 29 '18 at 17:58

I am not sure if the following gives the answer you want, but there is a notion of crossed module $\mathcal M =(\mu: M \to P)$ (due to JHC Whitehead) where $M,P$ are groups and $P$ acts on $M$ satisfyng two rules, which I won't detail here. These crossed modules classify pointed homotopy $2$-types. A crossed module $\mathcal M$ has a classifying space $B \mathcal M$. with first and second homotopy groups Coker $\mu$, Ker $\mu $ respectively. Full details are in the book Nonabelian Algebraic Topology.

This book does not contain much on the topological or Lie case.

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  • $\begingroup$ thanks for the comment +1 $\endgroup$ – wonderich Jul 26 '18 at 14:38

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