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

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.

• I am planning to study some Chern-Weil theory. Are you interested in one-one communication on this topic..
– user537667
Dec 25, 2018 at 13:09

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.

• Thank you, but I feel like I knew that much. This is largely my problem, in that I am having such difficulty making clear what my target is.
– jdc
May 17, 2014 at 1:02

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.

• This is sort of (and I apologize for the imprecision in the question itself) not what I was asking. I see that it works in the sense that one must normalize to get an integer class, and the total curvatures of the circle and the 2-sphere are $2\pi$ and $4\pi$, but it's not clear to me why it's always a power of $2\pi$ one needs to divide by. Right now it has the status of "mysterious coincidence" so far as I'm concerned, and that's something less than real understanding.
– jdc
May 5, 2014 at 19:51
• The "powers" are not really relevant because integer cohomology is a ring, and you expect that the powers of an integer class will also be integer classes. So what you seem to be asking is, why is the surface of a 2-sphere divisible by $2\pi$? I am not sure, but perhaps the explanation is that the moment map to the interval $[-1,1]$ has the property that the inverse image of each subinterval has area $2\pi$ times the length of the subinterval. May 6, 2014 at 11:46
• I know how to inductively find the volume of $S^n$, so I guess what I was asking was more why it's always right to divide $f(\Omega)$ by $(2\pi)^n$ for $f$ homogeneous of degree $n$. I agree that it's clear that $2\pi$ is right for $f$ of degree one on the circle, and that once that is known, it makes sense, to get integral classes, that $f(\Omega)^2$ should be divided by $4\pi^2$, etc. What I don't see is why the $2\pi$ applying to the circle should apply to all other manifolds. Perhaps the right argument occurs in $BU(n)$, $BO(n)$, and $BSO(n)$...
– jdc
May 7, 2014 at 12:50
• It may be helpful to stick with the ratio $\frac{\Omega}{2\pi}$ and its powers $(\frac{\Omega}{2\pi})^n$, rather than viewing the latter as $\frac{\Omega^n}{(2\pi)^n}$. Then your question is reduced to understanding why cohomology with integer coefficients is a ring, with a cup product that's compatible with that in de Rham cohomology. This is actually somewhat nontrivial and involves analysis of graded associative differential algebras. This happened to have been analyzed briefly in my book here. May 7, 2014 at 13:23
• I'm obviously missing something, because I'm happy with singular cohomology with integer coefficients being a ring, with the natural transformation $H^*(-,\mathbb Z) \to H^*(-, \mathbb R)$, and with the isomorphism $H^*(-, \mathbb R) \cong H^*_{\mathrm{de\ Rham}}(-)$ for manifolds (as presented, for instance, in Warner), but it's not helping me.
– jdc
May 7, 2014 at 14:09