When working with characteristic classes (more concretely Chern classes), one finds at least four essentially distinct approaches:

  1. Axiomatic Approach. See, for instance, Vector Bundles and K-Theory, page 78, th. 3.2.

  2. Characteristic Classes as Pullbacks of the Cohomology of Grassmannians. See, for instance, Vector Bundles and K-Theory, page 84, th. 3.9.

  3. Locus of Linear Dependency of General Sections (and translating into cohomology by intersection). See, for instance, 3264 & All That: Intersection Theory in Algebraic Geometry, page 59, prop-def. 1.36.

  4. Characteristic Polynomial of the Curvature Matrix (at least when working with smooth manifolds). See, for instance, Lecture Notes on Seiberg-Witten Invariants, page 19, sec. 1.5.

The approaches 2, 3 and 4 are proven to be equivalent by comparing the obtained objects with the axiomatic definition. But I wonder

How can be seen geometrically, in an intuitive way, that the approaches 2, 3 and 4, however different may seem, describe the same phenomena in the non-triviality of vector bundles?

Any intuitive explanation is welcome.

  • $\begingroup$ I am not certain what you are asking, but it is not true that vanishing of the Chern classes is equivalent to triviality of a complex vector bundle. For instance, consider complex vector bundles on an odd-dimensional sphere, where the associated even-degree cohomology groups are all zero. $\endgroup$ – Jason Starr Dec 28 '14 at 12:41
  • $\begingroup$ @JasonStarr I do not mean that the vanishing of the Chern classes is equivalent to the triviality of a complex vector bundle, but that Chern classes is a very useful tool for describing the non-triviality of vector bundles (although, as you say, it is not exhaustive), and that (this is the question) there should be a hard geometric background underlying all the concepts and approaches involved. $\endgroup$ – Jjm Dec 28 '14 at 13:21
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    $\begingroup$ I am not aware of any (short) intuitive explanation. As Euclid said: "There is no royal road to geometry." The best I could think of is that all of the above are cohomological invariants of U(n)-bundles, and polynomials in the Chern classes are the only such invariants (because the real cohomology algebra of $BU(n)$ is the polynomial algebra in the Chern classes). To conclude that these invariants are precisey the Chern classes (not merely polynomials in them) takes more work. $\endgroup$ – Igor Belegradek Dec 28 '14 at 15:54
  • $\begingroup$ Check that they all satisfy the axioms. Check that the axioms have at most one solution. This is all in Huybrecht's book or Demailly's book, I am pretty sure. $\endgroup$ – Ben McKay Dec 28 '14 at 19:48
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    $\begingroup$ (Re: unifying 2 and 4) I don't remember any details but there certainly is an algebraic version of $H(BG)\to H(M)$: the algebraic version of (de Rham complex of) $G\to EG\to BG$ is a contractible complex $\Lambda g[1]\otimes Sg[2]$ — and a connection gives a map from this guy to the de Rham complex of $M$ — something like that... $\endgroup$ – Grigory M Jan 10 '15 at 21:51

I would compare at least 2 & 3, if not 4, using maps into classifying spaces.

For a space $X$ homotopic to a finite CW-complex (at least), the pullback map $Map(X,Gr_n(\mathbb C^\infty)) \to \{$isomorphism classes of $n$-plane bundles$\}$ is bijective. So we should study the induced maps on cohomology, and get definition 2.

Over $Gr_n(\mathbb C^\infty)$ we have the universal bundle $\mathcal V$, and its $k$th power $\mathcal V^{\oplus k}$. Putting an $n$-plane bundle on $X$ is the same as giving a map $\phi$ into the Grassmannian, but putting on the $n$-plane bundle and choosing $k$ sections is the same as factoring $\phi$ through a map $X \to \mathcal V^{\oplus k}$.

Inside $\mathcal V^{\oplus k}$ is a universal degeneracy locus $\Omega$, where the $k$ sections are dependent. The genericity condition on the sections needed for definition 3 is the same as requiring $\phi$ to be transverse to $\Omega$. Then definition 3 amounts to pulling back $[\Omega]$ along $\phi$.

To compare to definition 2, one need only notice that $\mathcal V^{\oplus k}$ is homotopic to its base, the Grassmannian, and that $[\Omega]$ pulled back from $\mathcal V^{\oplus k}$ to the Grassmannian gives the expected Schubert class.

I guess another way to say this is that you only need to establish 2 = 3 = 4 for the tautological vector bundles on Grassmannians, and then you get it for all spaces.

  • $\begingroup$ If you could relate it somehow with curvature, perhaps user63660 wants to give the bounty +100 for you. Is it really difficult? $\endgroup$ – Jjm Jan 12 '15 at 12:37
  • $\begingroup$ Its easy to show that there is a norm in the Grassmannian case for which the curvature represents the right de Rham class, but I have always found it a bit mysterious that curvature is closed, and that the deRham class of curvature doesn't depend on the choice of norm. $\endgroup$ – David E Speyer Jan 12 '15 at 17:18

(A lot of people here can write a better comment about connection b/w 2 and 4 — but per Jjm's request here is a rough sketch.)

Algebraically the universal $G$-bundle is modelled by Weil (DG-)algebra $W(g)=\Lambda(g^\ast)\otimes S(g)$.

For a principle $G$-bundle $E\to M$ a connection on $E$ gives a map $g^\ast\to\Omega^1(E)$ which together with curvature define a map $W(g)\to C_{de Rham}(E)$. This map agrees with filtrations so it induces a map of $E_2$-terms of corresponding spectral sequences, $H^q(g;S^p(g^\ast))\to H^{2p}(B;H^q(G))$. For $q=0$ we get a map from $S(g^*)^G=H(BG)$ to $H(M)$ — namely, an invariant polynomial $P$ goes to $P(\Omega)$. For $gl_n$ this gives the usual definition of Chern class via curvature.


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