# Differential version of $G\mapsto H^3(G,\mathbb Z)$?

Let $$\mathit{cLieGrp}^{\mathrm{inj}}$$ be the category of compact connected Lie groups, and injective continuous group homomorphisms. Is there a reasonable functor (some kind of degree $$3$$ differential cohomology?) $$F:\mathit{cLieGrp}^{\mathrm{inj}} \to \mathit{Ab}$$ to the category of abelian groups with the following features:

(1) On the full subcategory of compact semisimple Lie groups, this functor agrees with $$G\mapsto H^3(G,\mathbb Z)$$ (integral 3rd cohomology of the underlying manifold of $$G$$).

(2) On the full subcategory of compact abelian Lie groups (tori), this functor agrees with $$T\mapsto H^4(BT,\mathbb R)$$.

Motivation for the question:
The functor $$G \mapsto \{ \text{full WZW models with gauge group } G\}$$ seems to behave like the functor $$F$$ above (except that the set of full WZW models with gauge group $$G$$ needs to be group-completed in order to make it into an abelian group).

• Does "injective maps" mean "injective continuous group homomorphisms"? or "injective continuous maps" (or anything else)? – YCor Mar 11 at 14:42
• Well, $H^3(T,Z)=Z$ and $H^4(BT,R)=R$, this looks impossible... – user43326 Mar 11 at 16:22
• By the way I meant $T=S^1$ in the above. – user43326 Mar 11 at 18:08
• @user43326. For $T:=S^1$, we have $H^3(T,\mathbb Z)=0$. It's just the cohomology of the underlying manifold (and as the manifold is 1-dimensional, there is no $H^3$). – André Henriques Mar 12 at 0:06
• OK, I thought you meant "group cohomology". But in that case it still doesn't work since $H^4(BT,R)$ is still different from $H^3(T,Z)$. Or here does $B$ mean something other than the classifying space? – user43326 Mar 12 at 7:55