# An action of the symmetric group $S_n$ on group cohomology $H^n(G, A)$ of abelian groups

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.

• Isn't your action of $S_2$ on $𝐻^2(𝐻,𝐴)$ just multiplication by $-1$? – Gregory Arone Nov 20 '19 at 14:53
• @GregoryArone Do you mean multiplication by $-1$ on $H^2(H,A)$ (equivalently, on $A$), or the induced map given my multiplication by $-1$ on $H$? Actually I reckon it could be the action given by both to 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. – Freddie Manners Nov 20 '19 at 17:18

For your action symmetric cocycles are not a subcomplex. Just look at action of 3-cycle for, say, trivial action, and $$\alpha \in C^2$$. $$d\alpha(g, h, k) = \alpha(h, k) - \alpha(gh, k) + \alpha(g, hk) - \alpha(g, h); \\ d\alpha(h, k, g) = \alpha(k, g) - \alpha(hk, g) + \alpha(h, kg) - \alpha(h, k)$$. There is no relation whatsoever between those two; they share 2 summands, but that's it.
But, lo and behold! There's an action of $$S^{n+1}$$ on $$C^n$$, because there's a natural action on bar resolution $$B^{\bullet}[G]$$ of trivial module, whos $$n$$th term is a free abelian group on $$G^{n+1}$$ (do not forget to include sign of permutation). Take invariants and call the result $$SB^{\bullet}[G]$$
Now we can define $$HS^{\bullet}(G, M)$$ as cohomology of $$\rm{Hom_G}(SB[G], M)$$. Sad news: $$HS^2$$ has nothing to do with abelian extensions, it classifies extensions admitting a section commuting with $$g \mapsto g^{-1}$$. Clearly, if a group has no order 2 elements every extension always has such a section.