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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 $1/(2\pi)^k$.

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A standard trick in the trade is the splitting principle, which says in effect that you can pretend that your vector bundle (with connection) is a sum of line bundles. There ways to justify it. Apply it, then you'll see that the factors of $2\pi (i)$ in the higher classes will have to multiply. –  Donu Arapura Mar 28 '11 at 16:58
    
The question is already restricted to complex line bundles. Or are you saying that there is a way to interpret the second form $F^{\nabla}\wedge F^{\nabla}$ as a form of lower degree in a higher-dimensional bundle? –  Greg Graviton Mar 28 '11 at 17:25
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Greg, you can't mean this! $c_2$ and higher for a line bundle are zero in cohomology. In fact, your integrand is zero. You want to have at least rank $2$, and take $\int p(F\wedge F)$ where $p$ is an elementary symmetric function. –  Donu Arapura Mar 28 '11 at 17:37
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The title of the question misrepresents the content. Its not about the second Chern class. In fact the question seems to be about the Chern-Weil formula for characteristic classes of a complex vector bundle. Here is a more succinct statement. Let $t$ be a formal variable, and consider $$det(\frac{it\Omega}{2\pi}+Id)$$ where $\Omega$ is the curvature form of a connection on a complex vector bundle over a smooth manifold. Why do the coefficients of $t$ lie in the image of the integral cohomology of the manifold inside the DeRham cohomology? –  Charlie Frohman Mar 29 '11 at 12:48
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Donu's remark at the beginning answers the question modulo checking that the integral over $i\Omega/2\pi$ over $\mathbb{C}P(1)$ for the standard connection on the canonical line bundle over $\mathbb{C}P(n)$ is $ 1$. –  Charlie Frohman Mar 29 '11 at 12:56

3 Answers 3

up vote 8 down vote accepted

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.

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D'oh! I mixed up the nomenclature. $F \wedge F$ would be the second Chern character class according to "From Calculus to Cohomology". I'm not familiar with classifying spaces (why is there a map $f : M \to \mathbb{CP}^N$ for any manifold $M$), but I don't care much about that. –  Greg Graviton Mar 29 '11 at 16:44
    
What seems more interesting is the integrality property. Assuming I only know de-Rham cohomology, could you elaborate on how I can detect whether a differential form from $H^k(M,\mathbb{C})$ is already a member of $H^k(M,\mathbb{Z})$? Is there a short reason why the wedge (cup) product of two integral forms is again integral? –  Greg Graviton Mar 29 '11 at 16:44
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If you only know deRham cohomology, then you don't know what $H^k(X, \mathbb{Z})$ means! OK, that's a bit unfair. It looks like in practice, you are interested in the subspace of $\omega$ in $H^k(X, \mathbb{R})$ such that, for every $k$-cycle $\sigma$, we have $\int_{\sigma} \omega \in \mathbb{Z}$. This is the image of $H^k(X, \mathbb{Z})$ in $H^k(X, \mathbb{R})$. So it sounds like what you want to know is why wedge product preserves this property of differential forms. –  David Speyer Mar 29 '11 at 19:33
    
I'd be curious to see a direct proof of this, which doesn't go through the construction of $H^k(X, \mathbb{Z})$. Why don't you ask this as a separate question over on math.stackexchange.com ? (It will get closed here, because people here will have no problem assuming that integral cohomology exists.) –  David Speyer Mar 29 '11 at 19:34
    
Thank you, David, I have done as you suggested. –  Greg Graviton Mar 29 '11 at 20:17

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.

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That is certainly true, but kind of dodges the question. One would have to show why the two agree up to a constant factor and then one would have to determine the constant factor to be precisely $(2\pi)^k$, and not, say $3\pi^{k/2}$. Both are currently beyond my understanding. –  Greg Graviton Mar 28 '11 at 17:08
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Let's say that you knew everything (functoriality, Whitney sum...) but the correct normalization. Then you'd have to calculate essentially one example: the universal bundle on a Grassmanian to get the correct constant. To do carry this out, pull it up to the flag manifold (you won't lose anything, since cohomology injects), split it has a sum of line bundles, and apply Whitney sum. That's what I meant in my previous comment. –  Donu Arapura Mar 28 '11 at 17:30
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Or, as the poser of the question is familiar with "From Calculus to Cohomology" you can quote Theroem 18.4 on page 184. –  Charlie Frohman Mar 29 '11 at 15:44

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 $a=2\pi i$.

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Thank you for pointing me to appendix C in Milnor's book! –  Greg Graviton Mar 29 '11 at 20:18

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