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?


2 Answers 2


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.

  • 5
    $\begingroup$ 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. $\endgroup$ Mar 18, 2010 at 20:07
  • 1
    $\begingroup$ 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. $\endgroup$
    – Dan Ramras
    Mar 18, 2010 at 21:47

You do not need to have metric to have splitting principle. Let $p: E \to B$ be a $n$-dimensional vector bundle and as usual let $E_0$ consist of nonzero vectors. Then consider the $\bar{E}= E_0/ \sim$ where $\sim$ identifies any two vectors in the fiber if they lie on the same line passing through origin(which just requires vector space structure)(Milnor and Stasheff describe this as defining fiber of new bundle as the quotient vector space $F/(\mathbb{C}v), v \neq 0$ pg157). Then $\bar{p}: \bar{E} \to B$ defines the associated projective bundle. We obtain splitting map by repeating this $n$-times.

But the problem with non-paracompact space $B$ is that it may not admit a classifying map $f : B \to Gr_n$ such that $f^{\ast}\gamma^n \cong E$(bundle isomorphism) where $p : E \to B$ is $n$-plane bundle. Since for a given cover there is no locally finite refinement, in the local trivializations of vector bundle $p: E \to B$ we cannot use the Urysohn lemma properly to get a bundle map $f' :E \to E(\gamma^n)$ which restricts to linear map in each fiber as Milnor-Stasheff do in the proof of lemma 5.3. So we cannot prove theorem 5.6(Milnor-Stasheff) for non-paracompact spaces and cannot define $w_1,c_1$(characteristic classes) as they define in page 69.

In the case of long line there is a paper showing that its tangent bundle is nontrivial without using any characteristic classes.

  • $\begingroup$ When you pull back E over $\bar{E}$, the resulting bundle contains the tautological line bundle as a subbundle, but how do you know that this subbundle has a complement? If there's a metric, you can take the orthogonal complement; otherwise it's unclear. I'll note that a metric on E itself is enough; you can then pull this metric back. $\endgroup$
    – Dan Ramras
    Mar 6, 2017 at 20:06
  • $\begingroup$ Metric is definitely enough. You are right when we pull back $E$ over $\bar{E}$ the resulting bundle contains the tautological line bundle as a subbundle $\gamma^1$, in other words, it injects in pullback $\bar{p}^{\ast}(E)$. So we can consider the exact sequence of vector bundles $0 \to \gamma^1 \to \bar{p}^{\ast}(E) \to \bar{p}^{\ast}(E)/\gamma^1$. $\endgroup$
    – 512122
    Mar 7, 2017 at 3:22

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