30

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

John Baez
  • 21,373
  • 1
    Would you accept the following rephrasal? The generator of your summand in $H^3(S_n,\mathbb{Z}/12)$ corresponds to a 4-term exact sequence (a crossed module extension) $0\to \mathbb{Z}/12\to N\to E\to S_n\to 0$, from which you can write down an explicit 3-cocycle using a choice of cross-section $S_n\to E$. I am not sure if this is due to MacLane (could check later). – Chris Gerig Nov 05 '18 at 20:58
  • 3
    The homology is isomorphic to $C_{12}\oplus C_2(\oplus C_2)$ for $n\ge 4$ (without the second $\oplus C_2$ for $n=4,5$ https://groupprops.subwiki.org/wiki/Group_cohomology_of_symmetric_groups) 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 05 '18 at 21:19
  • 2
    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 05 '18 at 22:55
  • @ChrisGerig - an explicit 4-term exact sequence would be a big step in the right direction and in some ways even better than the cocycle. – John Baez Nov 05 '18 at 22:57
  • 1
    OK. Note that this is equivalent to independently asking for (a) an explicit 3-cocycle $c_4$ valued in $Z/4Z$, such that $2c_4$ is not a coboundary; (b) an explicit 3-cocycle $c_3$ valued in $Z/3Z$, which is not a coboundary. [The solutions $c_3$ and $2c_4$ are unique modulo coboundaries, but not $c_4$.] – YCor Nov 05 '18 at 23:43
  • 3
    Here's a thing I wrote which is maybe relevant to your question: https://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 05 '18 at 23:46
  • @QiaochuYuan - that's a great explanation of what's secretly motivating my question! However, I'm looking for something maximally concrete: a 'formula', as explicit as possible, that computes an element of $\mathbb{Z}_{12}$ from 3 permutations, obeying the conditions I listed. If you have to use conformal nets to get that element, okay... but I'm hoping there's a formula that combinatorists or finite group theorists could more easily enjoy. Sort of like the sign of a permutation, only a lot subtler. – John Baez Nov 06 '18 at 00:28
  • Should the domain of $f$ be two copies of $S_{n}$? – Minseon Shin Nov 06 '18 at 03:43
  • @MinseonShin - yes, I'll fix that. Thanks! – John Baez Nov 06 '18 at 04:26
  • I believe the last (non-coboundary) condition needs one more summand, $-f(g,g')$ – მამუკა ჯიბლაძე Nov 06 '18 at 05:20
  • Yes, it needs that summand. Fixed, thanks! – John Baez Nov 06 '18 at 05:26
  • 1
    And I think that answer linked by @QiaochuYuan contains also a recipe for an explicit 3-cycle representing a generator in $H_3$. Take the Cayley graph for some presentation of $S_5$ by generators and relations - e. g. the Coxeter one. It gives a polyhedral subdivision of the 3-sphere (for the Coxeter presentation it consists of all (hyper)faces of a permutahedron). Triangulate it, and take the sum of all obtained 3-simplices (with signs according to the orientation). For larger $n$, just choose any $S_5$ subgroup on a 5-element subset of ${1,...,n}$ – მამუკა ჯიბლაძე Nov 06 '18 at 05:30
  • Hmmm I don't actually see clearly a proof that for larger $n$ the cycle I describe will continue to be a generator - in fact that it will not even bound. In detail one should do this: realise $S_5$ as a subgroup in $Pin(3)$; for larger $n$, find an $m$ such that there is an embedding of $Pin(3)$ into $Pin(m)$ that (1) induces isomorphism on $H_3$ and (2) carries $S_5$ into a subgroup of a $S_n$ subgroup. While this will give what is needed, what I don't see is whether you can choose such a thing in such a way that $S_5$ will land in a subgroup as indicated in my previous comment... – მამუკა ჯიბლაძე Nov 06 '18 at 06:00
  • I didn't see that recipe in there. I like it but I don't quite understand it. It seems there should be a direct way to see how 3-simplices in the triangulation you describe give 3-chains in some chain complex that computes the homology of $S_5$. But I'm not seeing it. – John Baez Nov 06 '18 at 06:11
  • 1
    Our comments crossed paths. Okay, so your idea (in the case $S_5$) is based on embedding $S_5$ as a subgroup of $\mathrm{Pin}(3)$. This is closely related to Epa and Gantner's method embedding of the double cover of $A_n$ in $\mathrm{Spin}(n)$. – John Baez Nov 06 '18 at 06:17
  • @JohnBaez Yes I think one might also obtain a generator of $H_3(A_n)$ this way. I said a little bit about it in an answer here – მამუკა ჯიბლაძე Nov 06 '18 at 07:23
  • On the afterthought - I think I had (at least) one thing imprecise: $S_5$ embeds in $O(3)$, and its pullback to $Pin(3)$ is actually a double cover of $S_5$, so you get a 3-cycle for that double cover, which must then be descended back to $S_5$ via transfer/induction/Gysin/whateveritscalled... – მამუკა ჯიბლაძე Nov 06 '18 at 07:29
  • 1
    @მამუკაჯიბლაძე $H_3$ is not cyclic in this case, so I'm confused what you want to call a generator in $H^3(S_n,Z/12Z)$. You maybe mean an element of order 12. – YCor Nov 06 '18 at 07:49
  • @YCor Yes thank you, I guess I constructed an element of $H_3(S_n)$ coming from the generator of $H_3(A_n)$ but maybe I am confused here too... At any rate I did not think at all about how to move to $H^3$ – მამუკა ჯიბლაძე Nov 06 '18 at 07:58
  • 1
    This doesn’t really address your question, but I once saw Fernando Rodriguez-Villegas give a talk called “All 12s are the same.” The theme was exactly your motivation here - that of describing deep connections that underlie various occurrences of the number 12 throughout mathematics. I don’t recall if this occurrence was discussed, but he might be a good guy to hit up as a curator of 12s. – Ramsey Nov 06 '18 at 12:35
  • 2
    It would be equivalent (and also interesting) to demonstrate a $\Bbb Z/12$-valued invariant for bordism classes of oriented 3-manifolds equipped with $n$-fold covering space, $n \geq 4$. It's not so clear to me that this is any easier, though. – mme Nov 06 '18 at 15:38
  • @JohnBaez -- I guess you've seen the formulae for 3-cocycles on dihedral groups in the math-phys literature? It doesn't directly help, since you assume $n \geq 4$... but it gives some idea of what might be possible. – Marty Nov 06 '18 at 17:26
  • 4
    @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 06 '18 at 19:30
  • 3
    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 07 '18 at 10:50
  • @JohnBaez Why do the second and the third summand in the cocycle condition have the same sign? Should the sign not alternate? – Philipp Lampe Nov 07 '18 at 16:31
  • @PhillippLampe - another typo! I'll fix it, thanks. – John Baez Nov 08 '18 at 15:44
  • @DevSinha - can you explain that class and how it works in more detail, or point me toward an explanation? I wanted a "formula", but an explicit geometrical description like this would also be nice. – John Baez Nov 09 '18 at 22:54

0 Answers0