6
$\begingroup$

Let $E=\xi-\eta$ be a virtual vector bundle over a compact base $B$, which we may assume is a CW complex. A quick and dirty way to define the total Stiefel-Whitney class $w(E)\in H^\ast(B;\mathbb{Z}/2\mathbb{Z})$ would be to say that, morally, $E\oplus\eta\cong \xi$, and so we can use the Whitney product formula $w(E\oplus\eta)=w(E)w(\eta)$ to get $$w(E) = w(\xi)w(\eta)^{-1},$$ the element $w(\eta)\in H^\ast(B;\mathbb{Z}/2\mathbb{Z})$ being invertible since $B$ is compact.

I am interested in a more righteous definition, along the lines of Whitney's original definition for honest bundles in terms of obstructions to finding linearly independent sections over skeleta. I think that such a definition exists (maybe in terms of sectioning a certain Hom-bundle) but I wasn't able to find a reference. Hence I ask:

Can the Stiefel-Whitney classes of a virtual vector bundle be defined obstruction theoretically, and if so, is there a good reference in the literature describing this construction?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

Given $\eta$, there is a trivial bundle $N$ such that $\eta\subset N$. Then $-\eta\sim N/\eta$, and the classes of $-\eta$ can be defined as the loci where a certain number of constant sections of $N$ stops being independent modulo $\eta$. This can certainly be restated more formally in terms of obstructions.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you for your answer! Still, I would like to know how this restatement goes. Do you know of any references? $\endgroup$
    – Mark Grant
    Commented Jan 14, 2014 at 12:23
  • $\begingroup$ No, sorry; I take it as folklore :) You can use the ordinary Stiefel bundle associated with $N/\eta$, or you can probably define some kind of relative Stiefel manifolds (frames in $\mathbb{R}^N$ independent modulo a subspace), or something else like that. I guess this depends on the problem you are trying to solve. $\endgroup$
    – Alex Degtyarev
    Commented Jan 14, 2014 at 13:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .