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According to book of Eisenbud-Harris Page 332 and the following summary http://pbelmans.ncag.info/blog/2014/10/09/what-are-chern-classes/ one can describe Chern classes in terms of degeneracy loci.

Quote from the link: "Recall that we were trying to measure in which sense a vector bundle is non-trivial, and for ease of statement we will assume that our vector bundle is globally generated. If $r$ is the rank, then for any $k=1,\ldots,r$ we can consider global sections $s_1,\ldots,s_{r−k+1}$. If we evaluate these in a point we get $r−k+1$ vectors of length $r$, hence we can ask whether these are linearly (in)dependent. The degeneracy locus of our set of global sections is then exactly the set of points in which the evaluations become linearly dependent, hence they degenerate. Of course one has to choose these sections sufficiently generally in order to make the codimension of the degeneracy locus correct, but one can prove that if the codimension is correct and the vector bundle is globally generated, then the degeneracy locus is independent of the choice! This can then be one way of defining the Chern classes of a vector bundle."

Question: In the case of smooth complex vector bundles, there are plenty of generic sections, so I was wondering if the argument has an analogue in the smooth case (i.e. if it is written with details somewhere).

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  • $\begingroup$ For the highest chern class (that is the Euler class), you can find a proof in Bott and Tu (pay 135 of maths.ed.ac.uk/~aar/papers/botttu.pdf) that is not detailed at all if not wrong. I believe the general case follows analogously using basic intersection theory as long as you can pick enough transversal sections (I believe this follows from transversality theorem). Of course, you will have to use Poincaré duals instead of irreducible subvarieties. $\endgroup$ – user40276 Aug 31 '16 at 22:14
  • $\begingroup$ The case of Euler class is basic. I am definitely talking about the other cases here. I am asking for a reference not believes, otherwise I was told such statement exists. Individually transversal is not enough, the question is what "generic" means in this case for a set of sections. $\endgroup$ – Mohammad Farajzadeh-Tehrani Sep 1 '16 at 1:26
  • $\begingroup$ Maybe I'm being too naive, but doesn't a transversal (to the zero section) section of the form $s_1 \wedge … \wedge s_{r-k + 1}$ would suffice ?Furthermore aren't all section of the exterior product of a smooth vector bundle of this form (exterior product commutes with global sections in the smooth case?)? $\endgroup$ – user40276 Sep 2 '16 at 2:28
  • $\begingroup$ the best reference is the lecture note of Laurent Manivel , Symmetric Functions, Schubert Polynomials, and Degeneracy Loci $\endgroup$ – user21574 Jan 24 '17 at 20:50
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The construction you want via degeneracy loci definitely goes through in the general setting of a complex vector bundles over a smooth compact oriented manifold.

It is often referred to as Gauss-Bonnet type formulae.

You can find a nice account about that in "Principle of Algebraic Geometry" by Griffiths and Harris, Chapter 3, Section 3.

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