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In Chern-Weil theory, we choose an arbitrary connection $\nabla$ on a complex vector bundle $E\rightarrow X$, obtain its curvature $F_\nabla$, and then we get Chern classes of $E$ from the curvature form. A priori it looks like these live in $H^*(X;\mathbb{C})$, but by an argument that I don't feel like I really understand, they're in the image of $H^*(X;\mathbb{Z})$, which is where they're usually considered to actually live. I've also recently been learning about the Atiyah-Singer index theorem, and I get the impression that whenever I see a arbitrary constants in geometry that end up having to live in $\mathbb{Z}$ I should ask myself whether the index theorem is lurking in there somewhere. Is there anything to this wild guess?

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    $\begingroup$ I'm not sure how often anyone gets to make a comment like this, but... this question might actually be more suited for MO than math.SE. :-) $\endgroup$ – Jesse Madnick Apr 29 '11 at 21:00
  • $\begingroup$ I think Jesse is right. An expert's answer to your question could help other non-experts in understanding the Atiyah-Singer theorem, and such an answer would better serve and interest the MO readers than the M.SE ones. Atiyah-Singer is strong voodoo, after all. $\endgroup$ – Gunnar Þór Magnússon Apr 29 '11 at 22:24
  • $\begingroup$ I don't know, I think it would also be good to leave the question here for at least a while and see if anyone has a go at answering it. I would also be quite interested in the answer. If you do post it on MO, please link to it here. $\endgroup$ – Glen Wheeler Apr 30 '11 at 9:40
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    $\begingroup$ @Eric: Certainly there's the algebraic topology proof (or even definition, really) that these are $\mathbb{Z}$-classes, but I saw an analytic argument in one my of lectures the other day that had nothing to do with classifying spaces or normalization. I think it's really cool when the same story can be carried all the way through in two such distinct settings. $\endgroup$ – Aaron Mazel-Gee May 2 '11 at 17:14
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    $\begingroup$ @Eric: The case of line bundles is Proposition 4.4.12 in Huybrechts' book, which just chases through the Cech-de Rham complex. I think the proof in my class was different, but maybe it was essentially equivalent. I'll try to find someone who took notes that day... $\endgroup$ – Aaron Mazel-Gee May 5 '11 at 9:20

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