I'm currently reading the paper ["Holonomy, the Quantum Adiabatic Theorem, and Berry's Phase"](http://adsabs.harvard.edu/abs/1983PhRvL..51.2167S) by Barry Simon.


Imagine a vector bundle with a connection $\nabla$. For simplicity, we assume that this is a $U(1)$ vector bundle. Parallel transport gives rise to the holonomy group, which assigns to each curve $C$ a number $e^{i\gamma(C)}$ that indicates how a vector is "rotated" when transporting it along the curve. In turns out that the phase change $\gamma(C)$ can be expressed as an integral of the curvature form over any surface $S$ that delimits the curve, $C = \partial S$,


$$ \gamma(C) = \int\_{S} F^{\nabla} .$$ 


I am interested in the integral of the curvature form over the whole manifold, which turns out to be an integer multiple of $2\pi$,


$$ \int\_{M} F^{\nabla} = 2\pi k, k\in\mathbb{Z}$$


Simon notes that this "standard fact" is a consistency condition on the holonomy group. I can understand that: integrating over the whole manifold is like taking the holonomy of the constant path, which must be the identity.


What I would like to understand is the generalization to higher Chern classes. For instance,



> 
> Why is the integral of the second Chern form an integer multiple of $4\pi^2$?
> 
> 
> 


$$ \int\_{M} F^{\nabla}\wedge F^{\nabla} = 4\pi^2 k, k\in\mathbb{Z}$$


I have a pedestrian proof for special cases, but I would like to understand a general reason behind this phenomenon. Is there a "higher holonomy" at work here?


Obviously, my knowledge of vector bundles and characteristic classes is rather limited. I can find my way around the book ["From Calculus to Cohomology"](http://adsabs.harvard.edu/abs/1983PhRvL..51.2167S), but have by no means absorbed all the material. Basically, my question is why the Chern classes defined via connections are normalized with a factor of $1/(2\pi)^k$.



Why is the integral of the second chern class an integer?
Greg,


I realize that I may as well write an answer rather than a series of comments.
Although Jessica has given a good answer, I'll try to say this as concretely as possible,
since I now think I understand the question more clearly. The question was actually 
about the integrality of 
$$\frac{1}{4\pi^2}\int\_M F\wedge F$$
where $F$ is the curvature of line bundle $V$ on a $4$-manifold $M$. This is what mathematicians
(I'm assuming you're a physicist) would call $c\_1(V)^2$. 
The first thing is observe that
$c\_1(V)\in H^2(M,\mathbb{Z})$, and that it's image in de Rham cohomology is
 given by $1/(2\pi i)[F]$. To see this in explicit terms, note that the classifying space
for line bundles in $\mathbb{C}\mathbb{P}^\infty$. This implies that the $V$ is the pull
back of the tautological bundle under a $C^\infty$ map $f:M\to \mathbb{C}\mathbb{P}^N$,
for $N\gg 0$. Working on projective space, we can check integrality of the class
$1/(2\pi i)[F]$ by doing a direct calculation to see that this integrates to $1$ over a complex line (aka $2$-sphere). This suffices because the line generates $H\_2(\mathbb{C}\mathbb{P}^N)$. After this, $c\_1(V)^2=-1/4\pi^2[F]^2$ is automatically integral. That's it.


**Postscript:** If you are unhappy with the last part, you can replace $f$ with its composition with a generic projection to obtain $f:M\to \mathbb{C}\mathbb{P}^2$. Then your integral
becomes the degree of $f$ which is certainly an integer. Hopefully, you can take it from
here.


