Why is the integral of the second chern class an integer? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T00:15:27Z http://mathoverflow.net/feeds/question/59861 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59861/why-is-the-integral-of-the-second-chern-class-an-integer Why is the integral of the second chern class an integer? Greg Graviton 2011-03-28T16:42:47Z 2011-03-30T23:24:18Z <p>I'm currently reading the paper <a href="http://adsabs.harvard.edu/abs/1983PhRvL..51.2167S" rel="nofollow">"Holonomy, the Quantum Adiabatic Theorem, and Berry's Phase"</a> by Barry Simon.</p> <p>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, $C = \partial S$,</p> <p>$$\gamma(C) = \int_{S} F^{\nabla} .$$ </p> <p>I am interested in the integral of the curvature form over the whole manifold, which turns out to be an integer multiple of $2\pi$,</p> <p>$$\int_{M} F^{\nabla} = 2\pi k, k\in\mathbb{Z}$$</p> <p>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.</p> <p>What I would like to understand is the generalization to higher Chern classes. For instance,</p> <blockquote> <p>Why is the integral of the second Chern form an integer multiple of $4\pi^2$?</p> </blockquote> <p>$$\int_{M} F^{\nabla}\wedge F^{\nabla} = 4\pi^2 k, k\in\mathbb{Z}$$</p> <p>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?</p> <p>Obviously, my knowledge of vector bundles and characteristic classes is rather limited. I can find my way around the book <a href="http://adsabs.harvard.edu/abs/1983PhRvL..51.2167S" rel="nofollow">"From Calculus to Cohomology"</a>, 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 $1/(2\pi)^k$.</p> http://mathoverflow.net/questions/59861/why-is-the-integral-of-the-second-chern-class-an-integer/59866#59866 Answer by Jessica L for Why is the integral of the second chern class an integer? Jessica L 2011-03-28T16:56:17Z 2011-03-28T16:56:17Z <p>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.</p> http://mathoverflow.net/questions/59861/why-is-the-integral-of-the-second-chern-class-an-integer/59874#59874 Answer by Faisal for Why is the integral of the second chern class an integer? Faisal 2011-03-28T18:18:23Z 2011-03-28T18:29:32Z <p>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.</p> <p>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 <code>$$\int_S F^\nabla = 2\pi i k$$</code> given in the OP (i.e. the Gauss--Bonnet formula!) does just that. It follows that $a=2\pi i$.</p> http://mathoverflow.net/questions/59861/why-is-the-integral-of-the-second-chern-class-an-integer/59881#59881 Answer by Donu Arapura for Why is the integral of the second chern class an integer? Donu Arapura 2011-03-28T19:00:01Z 2011-03-30T23:24:18Z <p>Greg,</p> <p>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.</p> <p><strong>Postscript:</strong> 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.</p>