Is there a clear pattern for the degree $2n$ cohomology group of the $n$'th Eilenberg-MacLane space?

Question

Let $G$ be a finite abelian group, and its higher classifying space is $B^nG=K(G,n)$. For $n=1$ it is well known that $H^2(B G, \mathbb{R}/\mathbb{Z}) \cong H^2(G,\mathbb{R}/\mathbb{Z})$ is isomorphic to the group of alternating bicharacters over $G$, where the isomorphism sends a class $[\nu]\in H^2(G,\mathbb{R}/\mathbb{Z})$ to its commutator $$ \chi_\nu(g,h)=\nu(g,h)-\nu(h,g) \ . $$ An heuristic and intuitive way to understand this is that if $X_2$ is a 2-manifold and $A\in H^1(X,G)$, you can construct the pull-back $A^*[\nu]\in H^2(X,\mathbb{R}/\mathbb{Z})$, which should be represented by a 2-cocycle constructed out of $A$. The natural candidate is the cup product $A\cup _{\chi} A$ associated with a bilinear map $\chi :G\times G\rightarrow \mathbb{R}/\mathbb{Z}$, and since $A$ has degree-1 this product must be antisymmetric. This is heuristic and indeed antisymmetric is not enough, you need alternating, but it is just a way to guess a possible result.

For $n=2$ there is something similar: $$ H^4(B^2G,\mathbb{R}/\mathbb{Z})\cong \Gamma(G)^\vee $$ where $\Gamma(G)$ is the universal quadratic group of $G$, and its Pontryagin dual $\Gamma(G)^\vee$ can be understood as the group of $\mathbb{R}/\mathbb{Z}$-valued quadratic function over $G$. Also here there is an heuristic argument. Given a 4-manifold $X_4$, $[\nu]\in H^4(B^2G,\mathbb{R}/\mathbb{Z})$ and $A\in H^2(X_4,G)$ you construct $A^*[\nu]\in H^4(X_4,\mathbb{R}/\mathbb{Z})$ which is represented by a 4-cocycle constructed out of $A$, and a natural candidate is $A\cup _{\chi} A$, where now the cup product is constructed from a symmetric bilinear map, since $A$ has degree-2. Again this is not complete since we actually need more than a symmetric map, we need a quadratic refinement of it.

1. Is there an analogous result for $n=3$?. By the same heuristic arguments, since $\cup$ is antisymmetric on $H^3(X_6,G)$, the natural guess is that $H^6(B^3G,\mathbb{R}/\mathbb{Z})$ is isomorphic to a group somehow related with that of antisymmetric bicharacters on $G$.

2. Is there a generalization to any $n$? The heuristic arguments are only based on the parity of the degree, hence they seem to suggest that in general $$ H^{2n}(B^nG,\mathbb{R}/\mathbb{Z}) $$ should be isomorphic to a group associated with $G$, which is a kind of refinement of the group of symmetric (antisymmetric respectively) bilinear forms over $G$ for $n$ even (odd respectively). Is this true in some form? In case, what is the kind of refinement involved?

Answer 1

Let me turn my comment into an answer, since I think it qualifies as an answer to the literal question you ask at the end of 2. (although you might be disappointed about the non-explicitness)

Your heuristic guess is absolutely correct, for a fixed abelian group $A$, $H^{n+k}(B^n G;A)$ as a functor in $G$ behaves in a polynomial way. For $k<n$, it is additive in $G$, and in fact independent of $n$, agreeing with $H^k(HG;A)$ where $HG$ is the Eilenberg-MacLane spectrum of $G$, a kind of limiting case of the $B^n G$. For $k=n$, quadratic effects begin to contribute, for $k=2n$, cubic, and so on. This is explicit in something like $H^*(B^2G;\mathbb{Z})$, which, for finitely generated free abelian groups $G$ looks like a polynomial algebra.

Goodwillie calculus provides an explicit formalism for dealing with such polynomial functors. Concretely, it assigns to a functor $F$ a tower of approximations $P_d F$ where you should think of $P_d F$ as "best polynomial-of-degree-$d$ approximation to $F$". In the case of the cohomology of Eilenberg-MacLane spaces, we are interested in the functor $$ \mathrm{Sp}\to \mathrm{Sp}, X\mapsto \Sigma^\infty\Omega^{\infty - n}X, $$ since this takes $HG\mapsto \Sigma^\infty B^nG$, a spectrum whose cohomology is the cohomology of $B^n G$. Formally, one computes that $P_1 F(X) = \Sigma^n X$, and that the fiber of $P_2F(X)\to P_1F(X)$ is $(\Sigma^n X)^{\otimes 2}_{h\Sigma_2}$. The higher fibers will have even higher connectivity, and the tower will converge to $F$ for connective $X$ (both nontrivial statements!), so we find an exact sequence $$ 0 \to H^{2n}(\Sigma^n HG;A) \to H^{2n}(B^nG;A) \to H^{2n}((\Sigma^n HG)^{\otimes 2}_{h\Sigma_2};A) \to H^{2n+1}(\Sigma^n HG; A) $$ and since $(\Sigma^n HG)^{\otimes 2}_{h\Sigma_2}$ is $2n$-connective with $\pi_{2n} \cong (G\otimes G)_{\Sigma_2}$ with $\Sigma_2$-action given by the flip if $n$ even, the signed flip if $n$ odd, this simplifies to $$ 0 \to H^n(HG;A) \to H^{2n}(B^n G;A) \to \mathrm{Sym}'(G\times G; A) \to H^{n+1}(HG;A) $$ where $\mathrm{Sym}'(G\times G; A)$ denotes symmetric or antisymmetric bilinear forms depending on the parity of $n$. This identifies $H^{2n}(B^n G;A)$ with an extension of a subgroup of $\mathrm{Sym}'$ (those forms for which a certain invariant in $H^{n+1}(HG;A)$ vanishes), by the group $H^n(HG;A)$. Cohomology of Eilenberg-MacLane spaces has been computed explicitly (for general $G$ and $A$, I believe, but I don't know references), but the answer is necessarily complicated since the "linear part" $H^*(HG;A)$ is already quite complicated.