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I know that $H^3(Z_2\times Z_2, U(1))=Z_2^3$, I'd like to know all the cocycles explicitly. Is there a systematical way to find the cocycles (I guess one can always try to solve the cocycle conditions on computer, although here with $U(1)$ I'm not sure that can work out). Also, $H^3(Z_2\times Z_2, U(1))$ is computed using Kunneth formula. Can one say anything about the cocycles of product group in terms of its factors, in the spirit of Kunneth formula?

Sorry if this is too elementary...I'm a physicist and just get to know group cohomology.

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Computing $H^3$ explicitly is a good question and you could try this on MathOverflow. I have done some calculations but never with $U(1)$ as coefficients, so need to think about this! – Ronnie Brown Jul 28 '12 at 17:18
up vote 0 down vote accepted

There are several ways of looking at explicit representatives of $H^3(G,A)$, and a lot depends on your purpose.

A standard $3$-cocycle is a function on $G \times G \times G$ with values in $A$, so this can be difficult to work with.

Otherwise we look for a small free resolution of $G$, and then use tensor products of chain complexes to get a free resolution of the product. A small resolution of $Z_2$ has in dimension $n$ the group ring $Z[Z_2]$ and boundaries alternatively multiplication by $1+t, 1-t$ where $t$ is the generator of $Z_2$.

With coefficients in $S^1$, which is an injective abelian group, the Universal Coefficients Theorem gives, since Ext$(-,S^1)=0$, that $$H^3(G,S^1) \cong Hom (H_3(G), S^1), $$ which could be useful.

There are other representatives of elements of $H^3(G,A)$, namely as equivalence classes of so-called "crossed sequences"

$$0 \to A \to M \to^\mu P \to G \to 1$$ where $\mu: M \to P$ is a crossed module. It would be nice to think that this could be useful in physics! However this result is used mainly when $A$ is $G$-module, and in particular a discrete abelian group; so I have not seen such constructions when $A=S^1$.

You may find the work of Graham Ellis on Homological Algebra Programming relevant:

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