Explicit 3-cocycles for the symmetric group $S_6$ 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?
 A: 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! 
A: 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
