Editing Chern class (section)

=== Via the Chern–Weil theory ===
{{main|Chern–Weil theory}}

Given a complex [[Hermitian metric|hermitian]] [[vector bundle]] ''V'' of [[vector bundle|complex rank]] ''n'' over a [[smooth manifold]] ''M'', representatives of each Chern class (also called a '''Chern form''') <math>c_k(V)</math> of ''V'' are given as the coefficients of the [[characteristic polynomial]] of the [[curvature form]] <math>\Omega</math> of ''V''.

<math display="block">\det \left(\frac {it\Omega}{2\pi} +I\right) = \sum_k c_k(V) t^k</math>

The determinant is over the ring of <math>n \times n</math> matrices whose entries are polynomials in ''t'' with coefficients in the commutative algebra of even complex differential forms on ''M''.  The [[curvature form]] <math>\Omega</math> of ''V'' is defined as
<math display="block">\Omega = d\omega+\frac{1}{2}[\omega,\omega]</math>
with ω the [[connection form]]  and ''d'' the [[exterior derivative]], or via the same expression in which ω is a [[gauge field]] for the [[gauge group]] of ''V''.  The scalar ''t'' is used here only as an [[indeterminate (variable)|indeterminate]] to [[generating function|generate]] the sum from the determinant, and ''I'' denotes the ''n'' × ''n'' [[identity matrix]].

To say that the expression given is a ''representative'' of the Chern class indicates that 'class' here means [[up to]] addition of an [[exact differential form]]. That is, Chern classes are [[cohomology class]]es in the sense of [[de Rham cohomology]].  It can be shown that the cohomology classes of the Chern forms do not depend on the choice of connection in ''V''.

If follows from the  matrix identity <math>\mathrm{tr}(\ln(X))=\ln(\det(X))</math> that <math> \det(X) =\exp(\mathrm{tr}(\ln(X)))</math>. Now applying the  [[Taylor series|Maclaurin series]] for <math>\ln(X+I)</math>, we get the following expression for the Chern forms:

<math display="block">\sum_k c_k(V) t^k = \left[ 1 
       + i \frac{\mathrm{tr}(\Omega)}{2\pi} t 
       +   \frac{\mathrm{tr}(\Omega^2)-\mathrm{tr}(\Omega)^2}{8\pi^2} t^2
       + i \frac{-2\mathrm{tr}(\Omega^3)+3\mathrm{tr}(\Omega^2)\mathrm{tr}(\Omega)-\mathrm{tr}(\Omega)^3}{48\pi^3} t^3
       + \cdots
       \right].</math>