Hello, this is a request for literature/a reference. I'm looking to do some calculations with the symmetric group ($S_6$ and higher) and would be interested in explicit expressions for 3-cocycles, i.e. elements of $H^3(S_6, U(1))$. Does anyone know whether these have already been calculated somewhere?
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$\begingroup$ Do you know what $H^3(S_6,U(1))$ is? $\endgroup$– André HenriquesCommented Mar 30, 2012 at 14:53
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$\begingroup$ I would use $H^3(S_6,U(1)) \cong H^4(S_6,\mathbb Z)$. $H^4(S_6,\mathbb Z/2)$ is known (Adem-Milgram) and the Sylow p-subgroups for p=3,5 are ablian or even cyclic. Thus the calculation of the mod-p cohomology of $S_6$ shouldn't be to hard. Then I would use universal coefficients to obtain (at least parts of) $H^4(S_6,\mathbb Z)$. $\endgroup$– RalphCommented Mar 30, 2012 at 20:08
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$\begingroup$ When you say you want a 3-cocycle on $G$ I presume you mean you want a function on the 3-fold product of $G$ with itself; or do you want any convenient, whatever that may mean, representation of elements of the cohomology group? So the further purpose of the question could be important. $\endgroup$– Ronnie BrownCommented Mar 30, 2012 at 21:33
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1$\begingroup$ $H^3(S_6,U(1))=Z_2\times Z_2 \times Z_{12}$, this I computed using GAP. I am looking for an explicit representative, i.e. a function $f:G\times G \times G \rightarrow U(1)$ satisfying the 3-cocycle condition (that is not a coboundary, obviously). $\endgroup$– henrikCommented Mar 31, 2012 at 6:22
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1$\begingroup$ So you don't care which element of $\mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z\times\mathbb Z/12\mathbb Z$ your cocycle represents??? You just want it to be non-trivial? That seems strange. Why do you want such a cocycle? $\endgroup$– André HenriquesCommented Apr 1, 2012 at 14:41
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2 Answers
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I hope it's ok to advertize some GAP code on this site.
Let G be a finite group and let
$$R_*: \cdots \rightarrow R_4 \rightarrow R_3 \rightarrow R_2 \rightarrow R_1 \rightarrow R_0$$
be a free $\mathbb ZG$-resolution of $\mathbb Z$.
A 3-cocycle with coefficients in $U(1)$ is a $\mathbb ZG$-linear homomorphism $c\colon R_3 \rightarrow U(1)$ such that the composite $R_4 \rightarrow R_3 \rightarrow U(1)$ is trivial. Using the Universal Coefficient Theorem and the divisibility of $U(1)$ we get isomorphisms $H^3(G,U(1)) \cong Hom(H_3(G,\mathbb Z),U(1)) \cong Hom(H_3(G,\mathbb Z),A)$ where $A$ is the cyclic group generated by an $m$-th root of unity with $m$ the exponent of $H_3(G,\mathbb Z)$. We thus get a surjection $H^3(G,A) \rightarrow H^3(G,U(1))$ with kernel of order $|Ext(H_2(G,\mathbb Z),A)|$. In short, we can represent the $\mathbb ZG$-linear homomorphism $c$ by a $\mathbb ZG$-linear homomorphism $ f\colon R_3 \rightarrow A$. Here $f$ is a 3-cocycle of $G$ with coefficients in the cyclic group $A$.
Let's call the 3-cocycle $f\colon R_3 \rightarrow A$ a "standard 3-cocycle" in the case where the resolution $R_*$ is the standard bar resolution. A standard 3-cocycle can then be thought of as a function $F\colon G\times G\times G\rightarrow A$ .
If we have any small resolution $R_*$, endowed with a contracting homotopy, then we can (in principle) construct a standard cocycle for each cohomology class. This can be done in GAP for many groups and provides examples of explicit cocycles which can be analyzed.
The following example constructs, for each of the 96 cohomology classes in $H^3(S_6,A)$, a representative standard 3-cocycle with coefficients in the cyclic group $A$ of order 12 (since we know that $H_3(S_6,\mathbb Z)$ has exponent 12). To run the example the HAP package (v 1.10.1 http://hamilton.nuigalway.ie/Hap/www) needs to be loaded into GAP.
EXAMPLE
gap> G:=SymmetricGroup(6);;
gap> A:=Group((11,12,13,14,15,16,17,18,19,20,21,22));;
gap> A:=TrivialGModuleAsGOuterGroup(G,A);; #This is the cyclic group of order 12 encoded as a trivial G-module
gap> R:=ResolutionFiniteGroup(G,4);;
gap> C:=HomToGModule(R,A);;
gap> CH:=CohomologyModule(C,3);;
gap> classes:=Elements(ActedGroup(CH));; #This is the list of cohomology classes
gap> Length(classes); #This gives the number of distinct cohomology classes
96
gap> c:=CH!.representativeCocycle(classes[2]); #This gives a 3-cocycle representing the second cohomology class
Standard 3-cocycle
gap> f:=Mapping(c);; #A cocycle f:GxGxG-->A corresponding to the second cohomology class
END OF EXAMPLE
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$\begingroup$ Thank you, I had a look at GAP/HAP already, but is there a way to find out which cohomology class the cocycle computed this way is in? It seems to me like you can never be sure to even get a nontrivial element. $\endgroup$– henrikCommented Apr 2, 2012 at 7:15
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$\begingroup$ I've changed the GAP example in my answer to better illustrate how one can construct a representative 3-cocycle for each of the classes in the cohomology group $H^3(G,A)$. The cocycle function f at the end of the example can be evaluated by typing $f(x,y,z)$ for any three permutations $x,y,z \in S_6$. $\endgroup$ Commented Apr 2, 2012 at 17:31
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$\begingroup$ This seems like a very nice way to obtain specific class representatives! I assume the fact that we get 96 classes instead of 48 comes from $|Ext(H_2(G,\mathbb{Z}),A)|=2$? $\endgroup$– henrikCommented Apr 2, 2012 at 19:49
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1$\begingroup$ @Graham, given two 3-cocycles with values in $A$, how do we know whether they represent the same 3-cocycle with values in $U(1)$? If we think of $A$ as a subgroup of $U(1)$ in the obvious way, how can we take the kernel (the Ext group) into account? $\endgroup$ Commented Jun 17, 2014 at 18:23
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I think Graham's answer is relevant to the question.
A 3-cocycle on $G$ with values in an abelian group $A$ my also be described as a morphism of chain complexes from the standard free resolution $F(G)$ of $G$ to the chain complex say $(A,3)$ which is $A$ in dimension 3 and zero elsewhere. But it may be more convenient to compute another free resolution $C$ of $G$ and describe a morphism $C \to (A,3)$. A standard 3-cocycle then comes from the description of a morphism of chain complexes $F(G) \to C$, which "in principle" is standard homological algebra.
But the issue remains of the form required for the answer, and for this one probably needs to know the place of the question in a research project. Why is the question asked?
There are other ways of representing elements of $H^3(G,A)$, for example by "crossed sequences"
$$0 \to A \to C_2 \to C_1 \to G \to 1$$
where $C_2 \to C_1$ is a crossed module. There are good examples where $G,A, C_2,C_1$ are finite. But intriguingly, for the current question $A$ is a topological abelian group!
answered Apr 1, 2012 at 10:52
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1$\begingroup$ Graham's paragraph just before the EXAMPLE is the key one; this role of contracting homotopies for computational purposes is I suspect not in the standard texts, as it seems to be fairly new, but is in the theory behind that paragraph. See Ellis, G. Computing group resolutions.J. Symbolic Comput.38 (3) (2004) 1077--1118. for description, discussion and references. $\endgroup$ Commented Apr 1, 2012 at 22:14