Let $H$, $A$ be discrete abelian groups, and for simplicity suppose $A$ is given the trivial $H$-action.

When considering the second cohomology group $H^2(H,A)$, it is natural to talk about the symmetric cocycles $Z^2_{\operatorname{sym}}(H,A)$ and symmetric cohomology $H^2_{\operatorname{sym}}(H,A)$.

In the bar resolution, suitably coordinatised, this is as simple as saying a cocycle $\rho \colon H^2 \to A$ is symmetric if $\rho(x,y) = \rho(y,x)$. In terms of group extensions (which is where this usually arises) this corresponds exactly to the group operation on the extension being abelian, which is useful to know about.

More generally, it seems there is a natural action of $S_2$ on $H^2(H,A)$, and its interpretation in terms of extensions is replacing an extension $0 \to A \to G \to H \to 0$ of $H$ by $A$ with $G^{\operatorname{op}}$.

I am interested in a natural action of $S_n$ on $H^n(H,A)$ that generalizes this. Probably, that corresponds in the bar resolution to permuting the arguments of a cocycle $H^n \to A$, although if that is the "wrong thing to do" I would accept alternatives.

[I should mention that I have a convoluted reason for believing this action exists and/or is natural, but (i) it would take a very long time to write down, and (ii) I am interested in knowing more standard / less convoluted interpretations.]

Here are some specific questions.

**1. Is there a natural interpretation of this "cohomology operation" that does not depend on a particular resolution (the bar resolution), for larger $n$?**

I.e. I would like to know the good reason why a natural $S_n$ action on $H^n(H,A)$ should exist in general (if indeed it does). If so:—

**2. How do you compute the action in the context of standard tools?**

For example, suppose I want to understand $H_{\operatorname{sym}}^{n}(\mathbb{Z}^{k}, \mathbb{Z})$, the fixed subgroup of the putative $S_n$-action. I know by Kunneth that the full $H^{n}(\mathbb{Z}^{k}, \mathbb{Z})$ looks like $\mathbb{Z}^{\binom{k}{n}}$, and I know that when $n=2$ the whole thing is anti-symmetric (I guess, as there are no non-trivial abelian extensions), but I have no idea how to compute the $S_n$-action in general.

bothto get the signs right. If so, that might leave the $S_n$ thing looking pretty tenuous unless I can come up with another reason for it to be there. $\endgroup$ – Freddie Manners Nov 20 '19 at 17:18