Timeline for Why is the integral of the second chern class an integer?
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
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| Mar 30, 2011 at 15:12 | vote | accept | Greg Graviton | ||
| Mar 29, 2011 at 12:56 | comment | added | Charlie Frohman | 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$. | |
| Mar 29, 2011 at 12:48 | comment | added | Charlie Frohman | 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? | |
| Mar 28, 2011 at 19:00 | answer | added | Donu Arapura | timeline score: 11 | |
| Mar 28, 2011 at 18:59 | comment | added | Donu Arapura | Sorry, I guess I misunderstood what you meant by "second Chern form". I realize your expression $\int F\wedge F$ is what I would write as $c_1^2$, and this certainly doesn't have to be zero. Integrality is also easier to see in this case. See below. | |
| Mar 28, 2011 at 18:18 | answer | added | Faisal | timeline score: 6 | |
| Mar 28, 2011 at 17:57 | comment | added | Greg Graviton | Donu, just to be sure: you are saying that the differential 4-form $F\wedge F = (\sum_{ij} F_{ij}dx^i\wedge dx^j) \wedge F$ coming from a connection of a complex line bundle on, say, a four-dimensional torus must be exact? I did not know that. Could you give a quick pointer as to why? | |
| Mar 28, 2011 at 17:39 | comment | added | Donu Arapura | I mean that $p$ is a symmetric function of the eigenvalues. | |
| Mar 28, 2011 at 17:37 | comment | added | Donu Arapura | 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. | |
| Mar 28, 2011 at 17:25 | comment | added | Greg Graviton | 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? | |
| Mar 28, 2011 at 16:58 | comment | added | Donu Arapura | 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. | |
| Mar 28, 2011 at 16:56 | answer | added | Jessica L | timeline score: 17 | |
| Mar 28, 2011 at 16:42 | history | asked | Greg Graviton | CC BY-SA 2.5 |