Chern-Weil: why do we divide by $2\pi$?

Question

So here's a somewhat incoherent question.

To define characteristic classes in the Chern–Weil way, one takes a curvature form $\Omega$ on a vector bundle $E \to M$ and an invariant polynomial $f$ on $\mathrm{GL}(\mathrm{rk }(E),\mathbb R)$, and then forms the cohomology class $c = \Big[f\big(\!\frac 1{2\pi}\Omega\big)\Big]$.

Why do we divide by $2\pi$? I understand why in the sense that "it works": if we want an integral class, so that $\langle c, [M] \rangle = \int_M f\big(\!\frac 1{2\pi}\Omega\big) \in \mathbb Z$, and agreeing with other standard definitions of these classes, dividing by $2\pi$ works, and not doing it doesn't.

But why does it work? "Morally," why is this the right thing to do?

I suppose this is analogous to asking why one always divides by $2\pi i$ in complex analysis, but there I feel I have some grasp on the answer: Cauchy's theorem holds, the only power of $z$ whose antiderivative isn't well-defined everywhere a power is $1/z$, and $t \mapsto z_0 + re^{it}$ describes one loop around a point $z_0$ as $t$ ranges from $0$ to $2\pi$.

I don't have even that clear an understanding what's going on in the case of the Chern–Weil homomorphism.

Answer 1

The point is to think of dual integer lattices. This can be seen already in the case of the unit circle. The class in cohomology dual to the fundamental homology class will be $d\theta$ divided by $2\pi$ since the circle has length $2\pi$. On the unit 2-sphere the curvature is $1$ but the total area is $4\pi$, which is why the fundamental cohomology class has to be normalized in a similar way to the circle to get an integer class. Hope this helps.

Answer 2

The inclusion $\mathbb{Z}\hookrightarrow\mathbb{R}$ induces a homomorphism $\check{H}^2(M;\mathbb{Z})\longrightarrow \check{H}^2(M;\mathbb{R})$ and under the identification $\check{H}^2(M;\mathbb{R})\cong H_{dR}^2(M;\mathbb{R})$ one has a homomorphism $\check{H}^2(M;\mathbb{Z})\longrightarrow H_{dR}^2(M;\mathbb{R})$.

I think that the problem lies in the use of THE for $\mathbb{Z}\hookrightarrow\mathbb{R}$. This is related to user72694's answer about the unit circle. If one is willing to define $S^1$ by $\mathbb{R}/2\pi\mathbb{Z}$, one can also define the inclusion $\mathbb{Z}\hookrightarrow\mathbb{R}$ by $\mathbb{Z}\ni k\mapsto 2\pi k\in\mathbb{R}$. This choice can wash $2\pi$'s from some formulae, and makes them appear elsewhere.