there are two definitions of the first Chern class that I don't know how to relate - hints and references are both welcome.

So, first approach: say that I have a complex vector bundle $E\to M$; I can pass to the frame bundle $F(E)\to M$, where I have a $U(n)-$action, for some $n$. Then I consider the Chern-Weil map \begin{equation} S({\frak u}(n)^*)^{U(n)}\to H^*_{U(n)}(F(E))\simeq H^*(M) \end{equation} and pick the image of the invariant polynomial $tr$, the trace. (up to some constant)

Second approach: I pick a connection $D:\Gamma(E)\to \Gamma(E\otimes TM)$, with $\Gamma(\cdot)$ representing the space of sections, and consider its curvature. Work locally on some $U$ and pick a section $s$ which generates the bundle: we get $D(s)=\theta(s)\cdot s$, with $\theta(s)$ a $1-$form, and apparently the first Chern class is proportional to $d\theta(s)$.

I am a bit confused: the very idea of connection seems to be absent from the first approach, so either there's a natural choice of connection, or I get the same result no matter which one I pick. Moreover, it would seem to me that also the choice of generator $s$ matters.

Honestly, I don't have much of a clue about where to start tackling the problem - the two things seem quite unrelated.

  • $\begingroup$ What is your definition of the Chern-Weil map in the first approach? The approach I know also involves a connection (although the classes do end up being independent of this choice). $\endgroup$ – Qiaochu Yuan May 14 '16 at 17:58
  • $\begingroup$ @QiaochuYuan The induced map on (equivariant) cohomology from f: F(E) \to {pt}. you then use H^*_U(n)(pt.) = the polynomials on the Lie algebra of U(n) invariant under the adjoint action, and get the Chern-Weil map $\endgroup$ – nelv May 14 '16 at 20:42
  • $\begingroup$ Sure. Then yes, the point is that the Chern classes end up being independent of the choice of connection in the second approach. $\endgroup$ – Qiaochu Yuan May 14 '16 at 21:32
  • $\begingroup$ that's good to know. but how does one show, from the first approach, that the image of the trace is a the curvature of a connection? btw, does this mean that in this case all the connections have the same curvature? nice! As I said, I'm a bit confused because the very idea of connection seems to be absent from the first approach $\endgroup$ – nelv May 15 '16 at 6:50
  • $\begingroup$ @nelv: It does not mean that all the connections have the same curvature, but rather that the cohomology class of the curvature is the same for all connections. $\endgroup$ – Jesse Madnick May 16 '16 at 11:15

Your Answer

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

Browse other questions tagged or ask your own question.