Why is the integral of the second chern class an integer? - MathOverflow

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

2011-03-28T16:42:47Z

I'm currently reading the paper "Holonomy, the Quantum Adiabatic Theorem, and Berry's Phase" 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", 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 $1/(2\pi)^k$.

Answer by Jessica L for Why is the integral of the second chern class an integer?

2011-03-28T16:56:17Z

Let $V$ be a complex vector bundle on a manifold $M$. Chern classes can be defined by topological means (see Milnor's book on characteristic classes), which yields elements $c_k(V) \in H^{2k}(M;\mathbb{Z})$. The normalization in the Chern-Weil theory is chosen so that the associated elements of de Rham cohomology groups $H^{2k}(M;\mathbb{R})$ agree with the integral elements, and thus integrate to give integers.

Answer by Faisal for Why is the integral of the second chern class an integer?

2011-03-28T18:18:23Z

Take a look at Appendix C in Milnor's book on characteristic classes. Essentially what is going on is that if you have a complex line bundle $L$ with connection $\nabla$ and curvature form $K_\nabla$, then the cohomology class of $\sigma_r(K_\nabla)$ is equal to $(2\pi i)^r c_r(L)$. Here $\sigma_r$ is the $r$th elementary symmetric function on the eigenvalues of the (matrix of the) connection.

The equality $\sigma_1(K_\nabla) = 2\pi i c_1(L)$ is rather transparent in case $L$ is a line bundle over a surface $S$ (as in the OP). Indeed, $\sigma_1 = \text{trace}$, and so what's being said is that $K_\nabla = 2\pi i c_1(L)$. And why is this true? Well, $K_\nabla$ is a closed $2$-form on $S$ that represents a characteristic cohomology class in $H^2(S;\mathbb{C})$, and therefore must be some multiple $a c_1(L)$ of the first Chern class. This constant $a$ is independent of $L$. So to compute it, all you need to do is work out some specific example. The formula $$ \int_S F^\nabla = 2\pi i k $$ given in the OP (i.e. the Gauss--Bonnet formula!) does just that. It follows that $a=2\pi i$.

Answer by Donu Arapura for Why is the integral of the second chern class an integer?

2011-03-28T19:00:01Z

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.