Let $X$ be some smooth manifold and let $p:E\to X$ be a complex vector bundle of finite rank. Our goal is to classify $E$ up to bundle-isomorphisms.

According to Atiyah's K-Theory book, there is a a bijection between the isomorphism classes $Vect_n(X)$ of rank $n$ vector bundles over $X$ and $$[X\to G_n(\mathbb{C}^\infty)]$$ the homotopy classes of maps $f:X\to G_n(\mathbb{C}^\infty)$ (the infinite Grassmannian).

Thus, for each $E$ there is a unique homotopy class $[f_E:X\to G_n(\mathbb{C}^\infty)]$ determined by $E$.

This in turn induces a map on any contravariant functor $\mathcal{F}$ from the homotopy category to an algebraic category $$ \mathcal{F}([f_E]): \mathcal{F}(G_n(\mathbb{C}^\infty))\to \mathcal{F}(X) $$

Then $\mathcal{F}([f_E]) \neq \mathcal{F}([f_\tilde{E}])$ implies $E \ncong\tilde{E}$, because $E\cong\tilde{E}$ implies $[f_E]=[f_\tilde{E}]$.

Question 1: Is there any functor $\mathcal{F}$ such that $\mathcal{F}([f_E]) = \mathcal{F}([f_\tilde{E}])$ implies $E\cong \tilde{E}$? (perhaps for sufficient conditions on $X$?)

Question 2: Is it the case that the $k$th Chern class of $E$ is given by $\mathcal{F}([f_E])(c_k)$ where $c_k\in\mathcal{F}(G_n(\mathbb{C}^\infty))$, and $\mathcal{F}$ is either the de Rham or the singular $k$th cohomology functor? If yes, is there a difference between using de Rham cohomology or singular cohomology? Also, how to determine $c_k$ concretely? I realize that $c_k$ is defined as the $k$th Chern class of the classifying bundle (the bundle quotient $(G_n(\mathbb{C}^m)\times \mathbb{C}^m)/F$ where $F$ is the tautological $n$-plane bundle), but is there a way to explain the definition of $c_k$ using this language without reverting to other constructions such as the Euler characteristic or invariant polynomials?

Question 3: What could one obtain by using other functors? For instance, the singular homology functor? The $n$th homotopy group functor? Why does one never hear about these? Also, why is it that one only uses the even-degree cohomology functor in this context?

Is there a textbook that is very similar to Chern's original 1945 paper, but which uses regular vector bundles instead of sphere bundles?

  • $\begingroup$ Yes you could take other functors. You have such things as $K$ theory characteristic classes, or bordism characteristic classes. The even cohomology functor arises because the complex grassmannian only has even cohomology, the rest vanishes. $\endgroup$
    – Thomas Rot
    Nov 4, 2016 at 9:43

1 Answer 1

  1. It depends on what you mean by an algebraic category. It's certainly not enough to take anything that looks like cohomology, since there are nontrivial spaces which have no nontrivial cohomology (with respect to any generalized cohomology theory).

  2. Yes for singular cohomology. For de Rham it's unclear what one might to mean by the de Rham cohomology of an infinite-dimensional thing.

  3. Homology gives you the same information as cohomology. Homotopy is hard to compute with. The infinite Grassmannian has no odd-dimensional cohomology. Look up K-theory.

  • $\begingroup$ Thanks for your comment. However, could you elaborate on question 2 in regards to $c_k$? $\endgroup$
    – PPR
    Dec 8, 2015 at 16:54
  • $\begingroup$ @PPR: there are a ton of ways of defining Chern classes so it really depends on your background. $\endgroup$ Dec 8, 2015 at 16:56
  • $\begingroup$ As in the question, is there a way to define it without reverting to Euler characteristics or invariant polynomials? Could you perhaps briefly explain what is the meaning of the Chern classes of the infinite Grassmannian in this context? Because so far, it seems like any non-zero element in $H^k(G_n(\mathbb{C}^\infty))$ could potentially give me something interesting, just by checking to see that its images under two different maps $\mathcal{F}([f_E])$ and $\mathcal{F}([f_\tilde{E}])$ are unequal. $\endgroup$
    – PPR
    Dec 8, 2015 at 17:00
  • $\begingroup$ @PPR: yes, but that won't give you any more information. The cohomology of $BGL_n(\mathbb{C})$ (which is a more invariant name for that Grassmannian) is a polynomial algebra on the Chern classes $c_1, \dots c_n$, so looking at the Chern classes already gives you the same information as looking at every integral cohomology class. $\endgroup$ Dec 8, 2015 at 17:20

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