# Cohomology of symmetric groups and the integers mod 12

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

• 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 $C_6$). – YCor Nov 5 '18 at 21:19
• 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. – John Baez Nov 5 '18 at 22:55
• 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 $\mathbb{Z}_{24}$. – Qiaochu Yuan Nov 5 '18 at 23:46
• @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. – Dev Sinha Nov 6 '18 at 19:30
• 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 $\zeta'(-1)=-1/12$. – François Brunault Nov 7 '18 at 10:50