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Consider the extension $1 \to C_3 \to S_3 \to C_2 \to 1$. I am trying to see intuitively why the action induced by the conjugation action $C_2 \curvearrowright C_3$ on $H^{\bullet}(C_3, \mathbb{Z})$ is multiplication by $(-1)^m$ for $n = 2m$.

It suffices to consider $n=2$. I was able to verify the statement by considering the group action $C_2 \curvearrowright H^2(C_3, \mathbb{Z})$, where the action is given by:

$$(g\phi)(h_1,...,h_n) = g \phi(gh_1g^{-1}, ..., gh_ng^{-1}) = \phi(gh_1g^{-1}, ..., gh_ng^{-1})$$

as the action on $\mathbb{Z}$ is trivial. If $\phi$ is a ($2$-)cocycle, then by direct computation, we can show that $\phi(1, h) = \phi(h',1)$ for any $h, h'$. Then, letting $\psi = \phi(-, h_2)$, it follows that $\phi + g\phi = d^1(\psi)$ i.e. that $[\phi] = -[g\phi]$ in $H^n$.

I suppose that, in this instance, there are not too many possibilities for the action, so it would have been feasible to determine the map by exhaustion. But is there a way to see that this is the case without exhaustion -- should I find it obvious that the map is multiplication by $-1$?

J. S.
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  • @FShrike It's not too well-motivated in the notes which I'm reading where the map comes from either. It's just quoted that there is an action $S_3 \curvearrowright H^2(C_3, \mathbb{Z})$ as above, for which the action of $C_3 \le S_3$ is invariant. Therefore, this induces an action $C_2 \cong S_3 /C_3 \curvearrowright H^2$. The proof that $C_3$ acts trivially is not entirely immediate. This is in the context of LHS spectral sequences, if that helps, and in the notation above, I've used $g$ for an element of $C_2$, and variants of $h$ for elements of $C_3$. – J. S. May 20 '24 at 18:14
  • Right. It feels like this is not quite an LHS situation but maybe I’m just confusing myself. I wrote a long post about the actions in LHS (and why they have to be the way they are) here – FShrike May 20 '24 at 18:21
  • Ok never mind I see what you mean. This is an LHS situation – FShrike May 20 '24 at 18:25
  • @FShrike Ah, with the bar resolution, I used the fact that we can consider the set of $\mathbb{Z}G$-maps from the free group on the bar symbols with $n$ entries to $M$ as isomorphic to the set of set-theoretic functions $G^n \to M$. Sorry if that was the issue. – J. S. May 20 '24 at 18:29
  • So, is the question now to see whether or not the action is by $(-1)$ in all degrees? – FShrike May 20 '24 at 18:37
  • @FShrike Right, no: by $(-1)^m$ for $\bullet = 2m$ even. By the structure of $H^{\bullet}(C_3, \mathbb{Z})$, it suffices to consider $\bullet = 2$, but I realise this is not written very clearly in the post. I'll edit it now. – J. S. May 20 '24 at 18:39

1 Answers1

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So I have an indirect argument. Consider the exact sequence

$$0\to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0.$$

Viewing all of them as a trivial $S_3$ module, we can take cohomology to obtain

$$H^1(S_3,\mathbb{Q}) \to H^1(S_3,\mathbb{Q}/\mathbb{Z}) \to H^2(S_3,\mathbb{Z}) \to H^2(S_3,\mathbb{Q}).$$

Since $\mathbb{Q}$ is uniquely divisible, it’s cohomology is trivial, so this induces isomorphisms

$$H^2(S_3,\mathbb{Z}) \simeq H^1(S_3,\mathbb{Q}/\mathbb{Z}) \simeq \text{Hom}(S_3,\mathbb{Q}/\mathbb{Z})\simeq \mathbb{Z}/2\mathbb{Z}.$$

Now we consider the inflation restriction exact sequence on the group extension

$$0\to C_3 \to S_3 \to C_2 \to 0$$

Applying inflation-restriction to this exact sequence (and noting that $H^1(C_3,\mathbb{Z})$ vanishes) we have

$$0\to H^2(C_2, \mathbb{Z}) \to H^2(S_3,\mathbb{Z}) \to H^2(C_3, \mathbb{Z})^{C_2} \to H^3(C_2, \mathbb{Z})$$

Substituting in well-known values of $H^i(C_2,\mathbb{Z})$, and the previous calculated value of $H^2(S_3,\mathbb{Z})$, we get

$$0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to H^2(C_3, \mathbb{Z})^{C_2} \to 0$$

Which implies $H^2(C_3, \mathbb{Z})^{C_2}\simeq 0$. Since $H^2(C_3, \mathbb{Z})\simeq \mathbb{Z}/3\mathbb{Z}$, the only action that makes this possible is multiplication by $-1$.

We remark that this generalizes to the extension

$$0 \to C_p \to D_{p} \to C_2 \to 0.$$

Snacc
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  • Thanks for the answer! The lecture notes I've been reading haven't mentioned coefficients in $\mathbb{Q} / \mathbb{Z}$, but in fairness, they are quite incoherent. – J. S. May 27 '24 at 01:55