Given a $G$-principal bundle $ P \to M $, and an invariant polynomial $\mathcal{P}$ on the Lie algebra, we can define an element of the cohomology of the base space by plugging the curvature (of an arbitrary connection) in the polynomial and looking for the form on the base which pulls back to the answer on total space. Thus there is a homomorphism $I_G \to H^{*}(M)$.

So far so good. However may times I have seen the 'universal' Chern-Weil homomorphism $I_G \to H^{*}(B_G)$ being referred to (e.g. at the beginning of 'Characteristic forms and geometric invariants' by Chern & Simons). I do not understand what this means. One could not have started with the classifying bundle and done the same construction because the base space $B_G$ need not be a manifold. I am assuming that $B_G$ denotes the classifying space obtained by Milnor's construction. Are there any other constructions of classifying spaces which yield spaces with manifold structure ? If not, how does one obtain the 'universal' Chern-Weil homomorphism ? I will appreciate if someone gives any references that discuss this. Thanks.

  • $\begingroup$ $BG$ has (homotopy equivalent) manifold models; they'll just be infinite dimensional manifolds. $\endgroup$ – user98602 Nov 29 '16 at 23:21
  • $\begingroup$ @MikeMiller Can you please suggest a reference where I can leanrn and understand it from ? Thanks $\endgroup$ – user90041 Nov 30 '16 at 7:31
  • $\begingroup$ Did you figure out what is this $H^*(BG)$?? I do not know clearly.. $\endgroup$ – Praphulla Koushik Dec 26 '18 at 17:39
  • $\begingroup$ Is $H^*(BG)$ the singular cohomology ring (coefficients ignored) something like $H^*(BG,A)$ for some abelian group $A$? $\endgroup$ – Praphulla Koushik Dec 27 '18 at 4:10

This is an old question but one reference worth looking into is Dupont's "Curvature and Characteristic Classes" book. He introduces a Simplicial De Rham theory with a version of the Chern-Weil homomorphism, and then constructs classifying spaces as simplicial manifolds.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.