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I am a physics student trying to self-learn Chern numbers and Chern class. The book I am learning (Nakahara) introduces the total Chern class as an invariant polynomial of local curvature form $F$ $P(F) = \det (I + t\frac{{iF}}{{2\pi }}) = \sum\limits_{r = 0}^k {{t^r}{P_r}(F)} $
and each ${P_j}(F)$ defines the j-th Chern class ${c_j}(F) \in {H^{2j}}(M)$

The book didn't mention anything about the Chern number. According to some other material I found (may be wrong), the Chern number is defined as an integral over 2$r$-cycle,

$\int_\sigma {{c_{{j_1}}}(F)} \wedge {c_{{j_2}}}(F) \cdots {c_{{j_l}}}(F) $

where ${j_1} + {j_2} + \cdots {j_l} = r$

The material also said that this integral is always an integer. Due to my limited knowlege, I cannot see how is this proved and I cannot find some reference that is easy enough to me. So can anybody help ?

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  • $\begingroup$ Chern classes can be refined to integral cohomology classes. A good reference is the book "Characteristic classes" Milnor-Stasheff. $\endgroup$
    – David C
    Oct 29, 2016 at 9:34
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    $\begingroup$ The Chern classes are Poincare duals of certain integral homology classes described by so called degeneracy loci. The Chern numbers can then be described as intersection numbers of integral homology classes. The book of Griffiths and Harris discuses this in detail. $\endgroup$
    – Liviu Nicolaescu
    Oct 29, 2016 at 9:40
  • $\begingroup$ Thanks very much, it looks there is no simple way to show it, I will take a look of those book first . $\endgroup$
    – Chen Li
    Oct 29, 2016 at 9:50
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    $\begingroup$ Something relatively elementary that you shoud try to do: prove that the Chern classes (as defined in the question) do not depend on the choice of connection but only depend on the underlying complex vector bundle. This will show you the "topological nature" of the Chern classes and so make plausible some integrality property. $\endgroup$
    – user25309
    Oct 29, 2016 at 11:14
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    $\begingroup$ almost by definition, Chern classes of a vector bundle are expressed as integral linear combinations of products of Chern classes of line bundles; the latter are patently integral : these are the images of $H^1(X,\mathcal{O}^*)\rightarrow H^2(X,\mathbb{Z})$. $\endgroup$
    – Venkataramana
    Oct 29, 2016 at 13:33

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Although the question may be a bit borderline for this site, I thought I'd contribute an answer since I think that it is surprising (it certainly surprised me as a student). My understanding is that Chern arrived at his classes while trying to generalize the Gauss-Bonnet theorem. This theorem says that the integral of the Gaussian curvature of any metric on a compact surface equal $2\pi$ times the Euler characteristic. So in particular, the curvature normalized by $1/2\pi$ always integrates to an integer. But of course, the real explanation is the previous statement that this integral is topological in nature.

Jumping ahead to the present, the integrality of Chern numbers is an artifact of the fact that Chern classes can be defined purely topologically, in several ways, as classes in $H^{2*}(X,\mathbb{Z})$. I'm not really saying anything that hasn't already been said in the comments.

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