Analytic Chern classes

Question

I have two questions on Chern classes, following Huybrechts' Complex Geometry.

1. Are the analytic Chern forms just the elementary symmetric polynomials of the eigenvalues of the curvature?
2. I googled some stuff with regards to the Chern–Weil homomorphism, and I'm confused about one thing. I keep seeing an extra coefficient of $i/2\pi$ when finding the characteristic polynomial of an element of a Lie algebra. For instance, $P(t)= \text{det}(I - t * ig/2\pi)$, where $g$ is an element of the Lie algebra. What is the reason for this coefficient?

Answer 1

1 Sort of. It's not totally obvious what "the eigenvalues of the curvature" are because the curvature is not a matrix, but a matrix valued in 2-forms. Also, there's the $1/2\pi i$ thing you mention. Once you understand those caveats I think it's correct.

2 The point is just to make sure that the Chern classes are actually integral cohomology classes. To figure out that $i/2 \pi$ is the right ratio, you probably want to consider the tautological bundle on the projective line. You want the integral of the Chern class over the projective line to be the degree of the bundle, which is $-1$. Instead you will find that it is $2 \pi i$, with the $2 \pi$ coming because you integrate over the projective line, which is a sphere.

You have to multiply the $k$th Chern class by the $k$th power of this to keep their multiplicativity properties.