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When $n \ge 4$, the third homology group $H_3(S_n,\mathbb{Z})$ of the symmetric group $S_n$ contains $\mathbb{Z}_{12}$ as a summand. Using the universal coefficient theorem we get $\mathbb{Z}_{12}$ as a summand of the cohomology group $H^3(S_n,\mathbb{Z}_{12})$.

What's a nice 'formula' for a $\mathbb{Z}_{12}$-valued 3-cocycle on $S_n$ that generates this summand of the third cohomology?

In more concrete terms, I'm looking for a nontrivial recipe to get an element $c(g,g',g'') \in \mathbb{Z}_{12}$ from three elements of $S_n$, such that

$$ c(g',g'',g''') - c(gg',g'',g''') + c(g,g'g'',g''') - c(g,g',g''g''') + c(g,g',g'') = 0 $$

for all $g,g',g'',g''' \in S_n$. Here "nontrivial" means that for all nonzero $\alpha \in \mathbb{Z}_{12}$, we do not have

$$ \alpha c(g,g',g'') = f(g',g'') - f(gg',g'') + f(g,g'g'') - f(g,g') $$

for some $f \colon S_n \times S_n \to \mathbb{Z}_{12}$.

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    $\begingroup$ The homology is isomorphic to $C_{12}\oplus C_2(\oplus C_2)$ for $n\ge 4$ (without the second $\oplus C_2$ for $n=4,5$ groupprops.subwiki.org/wiki/…) but there is no reason that the $C_{12}$ summand to be canonical a priori. If you're looking for a nice formula it's likely that you don't want non-canonical choices and hence have to accept these few $C_2$ terms (on the other hand, modding out the 2-torsion should give something canonical, in $C_6$). $\endgroup$ – YCor Nov 5 '18 at 21:19
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    $\begingroup$ Since I'm trying to get to the bottom of why the numbers 12 and 24 show up in many contexts in mathematics, I greatly prefer a $\mathbb{Z}_{12}$-valued cocycle and suspect there will be a good answer to my question, even if technically speaking it involves some noncanonical choices. However, beggars can't be choosers! So, a $\mathbb{Z}_6$-valued cocycle would also be okay. $\endgroup$ – John Baez Nov 5 '18 at 22:55
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    $\begingroup$ Here's a thing I wrote which is maybe relevant to your question: mathoverflow.net/a/196130/290 In the frame of that answer you are looking for something like a suitably nontrivial action of $S_n$ on an object in a suitably nice 3-category, and you can try to get such a thing by taking powers of an invertible object in a symmetric monoidal 3-category (for example, as suggested in that answer, conformal nets). The action of $S_n$ on such powers factors through the $3$-truncation of the sphere spectrum which naturally introduces a $\mathbb{Z}_{24}$. $\endgroup$ – Qiaochu Yuan Nov 5 '18 at 23:46
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    $\begingroup$ @JohnBaez, I can tell you about the geometry of this cocycle, using Fox-Neuwirth cochains in the unordered configuration space model for $BS_n$: this class is represented by having four points which share a coordinate, and three points which all share a coordinate, with two sharing another coordinate. There is probably a prettier subvariety which represents this, but I don't have any techniques/ideas to get at those. $\endgroup$ – Dev Sinha Nov 6 '18 at 19:30
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    $\begingroup$ From a number theoretic perspective, the third homology group of the general linear group is related to algebraic $K$-theory, namely $K_3$ (this seems relevant because $S_n$ embeds into $\mathrm{GL}_n$). In general the definition of $K$-group is highly non-constructive but $K_3$ is related to the Bloch group which is very explicit and some people have studied its torsion. The group $K_3(\mathbb{Z})=K_3(\mathbb{Q})$ is cyclic of order 48. By conjectures of Quillen-Lichtenbaum and Bloch-Kato the size of $(K_3)_\mathrm{tors}$ is linked with $\zeta'(-1)=-1/12$. $\endgroup$ – François Brunault Nov 7 '18 at 10:50

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