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I'm looking for explicit representatives of $H^3_{Borel}(G, R/Z)$, i.e. a measurable function $G^3\to R/Z$ representing a generator of the cohomology group. (Here $G$ is a compact (perhaps simple) Lie group and $R/Z$ is the circle group.) Surely someone has worked this out, but my literature search came up empty.

Alternatively, an explicit representative of the corresponding class in $H^4(BG, Z)$ would be useful, since Cor. 1.14 of Baker 1977 shows how to convert this to an $H^3_{Borel}(G, R/Z)$ representative.

I suspect this is not difficult, but I seem to be stuck, so I'm availing myself to MO.

[added later] (1) An explicit description of the corresponding Cheeger-Simons differential character would be another welcome alternative answer. (2) In lieu of a general answer, answers for particular groups, e.g. $SU(2)$, would be welcome.

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  • $\begingroup$ I'm not familiar with this definition of Borel cohomology, so I'm not certain if this helps any, or if it will come as news, but an explicit generator for the de Rham cohomology space $H^3(G)$, for $G$ compact simple Lie, is given by the Cartan 3-form (see, e.g <mathoverflow.net/questions/62998/…). $\endgroup$
    – jdc
    Commented Feb 28 at 13:46
  • $\begingroup$ Borel cohomology is (roughly) the $G$-equivariant cohomology of a point, or the ordinary cohomology of $BG$. Perhaps the Cartan 3-form is related to the explicit cocycle I'm looking for, but I don't see how to turn it (the Cartan 3-form) into a function $G^3\to R/Z$. Again, I might be missing something obvious. $\endgroup$
    – Kevin Walker
    Commented Feb 28 at 17:47
  • $\begingroup$ I think I understand what you mean by "Borel" now. I was brought up to interpret it as something different ("Borel equivariant: the (singular) cohomology of the Borel construction associated to an action). Did you wind up working this out? $\endgroup$
    – jdc
    Commented Jul 30 at 5:35

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