# Direct proof that Chern-Weil theory yields integral classes

Suppose $$E$$ is a complex vector bundle of rank $$n$$ on a compact oriented manifold (both assumed smooth). Let $$h$$ be a Hermitian metric on $$E$$, and let $$A$$ be a Hermitian connection on $$E$$ and $$F_A$$ its curvature form, an element of $$\Omega^2(M,\text{End }E)$$. Chern-Weil theory produces a closed even differential form $$c(A) = \det(1 + \frac i{2\pi}F_A) = c_0(A)+c_1(A) + \cdots + c_n(A)$$.

These classes have the property that for all compact oriented submanifolds $$\Sigma\subset M$$ of dimension $$2k$$, the expression $$\int_\Sigma c(A)$$ is an integer. The usual way to prove this seems to be to first show that the de Rham cohomology class of $$c(A)$$ is independent of the connection $$A$$ and functorial under pullbacks. We then check that it has the desired integrality property for the tautological bundles on Grassmannians by direct computation (which says this class lies in singular cohomology with $$\mathbb Z$$ coefficients) and then use the fact that every bundle on $$M$$ is a pullback of the tautological bundle from a sufficiently large Grassmannian. Alternatively, we show that the above definition satisfies the axiomatic characterization of Chern classes and then we identify it with the image in $$H^\bullet(-,\mathbb C)$$ of Chern classes constructed in $$H^\bullet(-,\mathbb Z)$$ by using a computation of the cohomology ring of infinite Grassmannians.

Is there a direct proof that $$\int_\Sigma$$ c(A) is an integer for all compact oriented submanifolds $$\Sigma\subset M$$?

By a direct proof, I mean something which only uses only local reasoning about curvature and connections and doesn't have to pass through much algebraic topology. For instance for the case $$n=1$$ (i.e., a line bundle) and taking $$\Sigma = M$$ to be a compact oriented surface, we note by elementary local computations (using Stokes' theorem) that for any disc $$D$$ inside $$\Sigma$$ with oriented boundary circle $$\gamma$$, the number $$e^{\pm\int_D F_A}\in U(1)$$ measures the holonomy of the parallel transport along the loop $$\gamma$$. From this, it easily follows that we can extend the above statement to the case when $$D$$ is a more general region with smooth oriented boundary $$\gamma$$. As $$\Sigma$$ has no boundary, applying this to $$D = \Sigma$$ gives $$\int_\Sigma F_A\in \frac{2\pi}i\mathbb Z$$ as desired. (This proof is similar to the moral proof of Stokes' theorem given by checking it at the infinitesimal level -- essentially the definition of the exterior derivative -- and then triangulating the domain of integration and noting that boundary terms with opposite orientations cancel.) The $$c_1$$ case seems to be simple because of $$U(1)$$ being abelian.

• Is the splitting principle too indirect for you? It should reduce the general case to the abelian case. – Mike Miller Jan 24 at 20:28
• This seems to be related to finding a "direct proof" of the following assertion: given $n$ line bundles $L_i$ ($1\le i\le n$) with connections $A_i$, how do we show that $\int_\Sigma\prod_i c(A_i)\in\mathbb Z$ for each compact oriented $\Sigma\subset M$? I'd like to avoid using singular (co)homology with $\mathbb Z$ coefficients in the proof. – Mohan Swaminathan Jan 24 at 20:49
• This seems surprisingly non-trivial to me, essentially requiring the Kunneth formula mod torsion for integral homology... – Mike Miller Jan 26 at 4:16