Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This question is closely related to this previous question.

Chern and Stiefel-Whitney classes can be defined on bundles over arbitrary base spaces. (In Hatcher's Vector Bundles notes, he uses the Leray-Hirsch Theorem, which appears to require paracompactness of the base space. The construction in Milnor-Stasheff works in general, as does the argument given by Charles Resk in answer to the above question. A posteriori, this actually shows that Hatcher's construction works in general too, since he really just needs $w_1$ and $c_1$ to be defined everywhere.)

The proof of uniqueness (as discussed in Milnor and Stasheff, or in Hatcher's Vector Bundles notes, or in the answers to the above question) relies on the splitting principle, and hence (it seems to me) requires the existence of a metric on the bundle in question. More precisely, if we have two sequences of characteristic classes satisfying the axioms for, say, Chern classes, and we want to check that they agree agree on some bundle $E\to B$, the method is to pull back $E$ along some map $f: B'\to B$ (with $f^*$ injective on cohomology) so that $f^*E$ splits as a sum of lines. Producing the splitting seems to require a metric on $E$ (or at least on $f^*E$).

If $B$ is not paracompact, bundles over $B$ may not admit a metric (and may admit a classifying map into the universal bundle over the Grassmannian), so my question is:

Are Chern and/or Stiefel-Whitney classes unique for arbitrary bundles? If not, do $w_1$ and $c_1$ at least determine the higher-dimensional classes?

share|improve this question

1 Answer 1

up vote 8 down vote accepted

I'm going to assume that your characteristic classes are supposed to live in the singular cohomology of the base space. Then to show your uniqueness result, it should be enough if you can produce, for any space $B$, a map $f:B'\to B$ such that $B'$ is paracompact, and $f$ induces an isomorphism in singular cohomology.

For any $B$, you can find a CW-complex $B'$ and a weak equivalence $f: B'\to B$, by one of Whitehead's many theorems. Weak equivalences always induce isomorphisms in singular cohomology, and CW-complexes are paracompact (I think!). Hatcher's topology textbook proves all of these, except possibly for the paracompactness claim (for which I can't find a reference yet).

If you're talking about Cech cohomology, then this proof won't work.

share|improve this answer
2  
CW-complexes are paracompact. Milnor and Stasheff, p. 74, references Miyazaki, H., Paracompactness of CW complexes , Tohoku Math. J. 4 (1952), 309-313 and Dugundji, J., "Topology," Allyn and Bacon, Boston, 1996, p. 419. –  Cotton Seed Mar 18 '10 at 20:07
    
Excellent! I think I've actually used an argument along these lines before... Hatcher's Vector Bundles notes include a proof that CW complexes are paracompact. It's in the Appendix to Chapter 1. –  Dan Ramras Mar 18 '10 at 21:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.