Subject: Re: Borel, continuous and smooth group cohomology From: David J Green Date: Wed, 1 Sep 2004 21:46:34 +0200 (MEST) To: John Baez , Donald M Davis Dear John Baez, dear fellow topologists, My knowledge is limited to "algebraic" cohomology for finite groups, and I'm away from my desk, but I suspect that the point is that R -> U(1) is a covering map with fibre Z. As you will know, the key to getting a Bockstein exact sequence is translating the short exact sequence of coefficient modules into a short exact sequence of cochain modules 0 -> C^n_X(G,Z) -> C^n_X(G,R) -> C^n_X(G,U(1)) -> 0. For X=algebraic it's enough that R -> U(1) splits as a map of sets. X=continuous: if the path component of the identity in the Lie group G is simply connected [or more generally, admits no non-trivial maps to Z = \pi_1(U(1))], then you can lift continous maps h:G -> U(1) to continuous maps H:G -> R as follows: for one element g of each path component of G one just picks a lift H(g) of h(g), and then extends H to the whole path component by lifting paths; this is well-defined by the assumption on pi_1(G). The same principle works for lifting cocycles. And I guess it covers the case X=smooth too. As for X=Borel, I've forgotten the exact definition of "Borel measurable" (shame!), but perhaps it's enough here that R -> U(1) is split by a map which is continous almost everywhere. Of course, the argument I propose for X=continuous fails spectacularly for G=U(1). Hope this is correct and of some use. Best wishes, David Green