TITLE: Chern-Weil: why do we divide by $2\pi$?
QUESTION [11 upvotes]: 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.

REPLY [4 votes]: 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.