64
$\begingroup$

Chern-Weil theory gives characteristic classes (e.g. Chern class, Euler class, Pontryagin) of a vector bundle in terms of polynomials in the curvature form of an arbitrary connection. There seems to be no hope in getting Stiefel-Whitney classes from this method since Chern-Weil gives cohomology classes with real coefficients while Stiefel-Whitney classes have $\mathbb Z/2$ coefficients. Further, since any vector bundle over a curve has vanishing curvature, classes obtained by Chern-Weil can't distinguish, for example, the Mobius bundle from the trivial bundle over the circle (while Stiefel-Whitney classes do).

Nonetheless, I am wondering if there is a more general or abstract framework that allows one to define the Stiefel-Whitney classes in the spirit of Chern-Weil. For example, maybe this is done through a more abstract definition of a connection/curvature.

Improve this question
$\endgroup$
9
  • $\begingroup$ Might be worth looking nLabwards, e.g., ncatlab.org/nlab/show/… or ncatlab.org/nlab/show/… or math.ntnu.no/~stacey/Mathforge/nForum/… . $\endgroup$
    – David Corfield
    Commented Jan 18, 2011 at 17:06
  • $\begingroup$ I don't have an answer to this question, but I have a guess. In general, if you want to use De Rham theory to recover integral invariants of a manifold, a smart thing to do is look at distributional differential forms (I first learned of the literature on this from MO, but I can't remember where). So I would start by defining a distributional connection on a vector bundle $E$ (rather, on its frame bundle) to be a distributional $E$-valued 1-form and going from there. $\endgroup$
    – Paul Siegel
    Commented Jan 18, 2011 at 18:52
  • 2
    $\begingroup$ You might want to take a look at "Stiefel-Whitney currents" by Harvey and Zweck [MR1731064 (2001c:53104)]. $\endgroup$
    – 680
    Commented Jan 19, 2011 at 12:07
  • 1
    $\begingroup$ @David and 680: thanks for the links! $\endgroup$
    – Eric O. Korman
    Commented Jan 20, 2011 at 23:37
  • 2
    $\begingroup$ @diverietti: I think the confusion is about real curves as opposed to complex curves ;-) $\endgroup$
    – Tobias Hartnick
    Commented Apr 11, 2011 at 14:41

2 Answers 2

Reset to default
6
$\begingroup$

There exist flat manifolds (i.e., closed Riemannian manifolds with vanishing sectional curvature) which do not admit any spin or spin${}^c$-structure. But since the existence of spin, resp. of spin${}^c$-structures is detected by the second Stiefel-Whitney class, I am strongly led to believe that there is no way of defining these à la Chern-Weil. The reason is that the phrase 'à la Chern-Weil' entails for me that you use in an essential way the curvature tensor, but in these examples it vanishes.

Improve this answer
$\endgroup$
5
$\begingroup$

I want to say the short answer is no.

But in certain contexts, we can get things that are analogous. For example if you take a principal $B$-bundle $Q$ over $M$ and then suppose you can have a "nice" :) central extension of your lie group $B$ by $$1 \to \underline{\mathbb{C}^*}_M \to \tilde{B} \to B \to 1,$$where $\underline{\mathbb{C}^*}_M$ is the sheaf of smooth functions into $\mathbb{C}^*$, then you can define a cohomology class in $H^1(M, \underline{B})$ by seeing how well you can lift your bundle to a $\tilde{B}$-bundle. Now by the central extension, we would have $$H^1(M, \underline{B}) \overset{\sim}{=} H^2(M , \underline{\mathbb{C}^*}_M)$$ and then by the exponential sequence you would have $$H^1(M, \underline{B}) \overset{\sim}{=} H^2(M , \underline{\mathbb{C}^*}_M) \overset{\sim}{=} H^3(M, \mathbb{Z}).$$

And so what I am saying (since principal bundles have associated vector bundles) that your vector bundle in the right conditions could give you a degree three cohomology class in the integers. I'm sure you could pluck your Z/2 coefficients out of this (I don't know why you would want to be so rigid, nor am I claiming you are asking for that). Then there is an actual geometric interpretation of this integer related to a certain curvature form in this construction that I am not quite ready to add to this answer yet :)

Improve this answer
$\endgroup$
1
  • $\begingroup$ By the way, I'm fairly certain that Weil developed some of these methods. $\endgroup$
    – cheyne
    Commented Feb 26, 2013 at 15:42

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