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For questions about Stiefel-Whitney classes which are characteristic classes associated to real vector bundles.

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Using Stiefel-Whitney class to build new principal bundles

I'm reading this paper and at the beginning of the second section, he states many results that aren't clear to me. Consider a principal $SO(3)$-bundle $P\rightarrow R^2\times \Sigma$, where $\Sigma$ ...
3
votes
2answers
273 views

Is Cohen immersion conjecture (theorem) known for vector bundles?

R. Cohen proved the immersion conjecture in a 1985 Annals paper: Cohen, Ralph L., The immersion conjecture for differentiable manifolds, Ann. Math. (2) 122, 237-328 (1985). ZBL0592.57022. Any smooth ...
7
votes
0answers
231 views

Different definitions of Stiefel-Whitney classes

It is quite easy to show that different definitions of the Stiefel-Whitney classes agree by showing that they satisfy the well-known axioms. Nevertheless I have been asking myself wether one can prove ...
16
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2answers
519 views

What is the $\mathbb{Z}_2$ cohomology of an oriented grassmannian?

Let $\operatorname{Gr}(k, n)$ and $\operatorname{Gr}^+(k, n)$ denote the unoriented and oriented grassmannians respectively. The $\mathbb{Z}_2$ cohomology of the unoriented grassmannian is $$H^*(\...
10
votes
2answers
497 views

Is the top Stiefel-Whitney number of a topological manifold the Euler characteristic mod two?

Recall that the Stiefel-Whitney classes of a smooth manifold are defined to be those of its tangent bundle - this definition doesn't extend to topological manifolds as they don't have a tangent bundle....
17
votes
1answer
1k views

What are the possible Stiefel-Whitney numbers of a five-manifold?

On a compact five-manifold, the Stiefel-Whitney number $w_2w_3$ can be nonzero. An example is the manifold $SU(3)/SO(3)$, and also another example is a $\mathbb{CP}^2$ bundle over a circle where the ...
3
votes
2answers
420 views

Milnor's proof of cohomology of BO(n)

In Milnor/Stasheff characteristic classes there is the proof that $H^*(BO(n);\mathbb{Z}_2)$ is the polynomial ring on the first n Stiefel-Whitney classes. I understand the part that the latter ring is ...
8
votes
0answers
212 views

Dixmier-Douady class is the third integral Stiefel-Whitney class

Let $M$ be (say smooth) manifold. From the short exact sequence of groups $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}_2 \to 0$ (where the first map is multiplication by $2$) one obtain long exact ...
5
votes
0answers
139 views

Stiefel-Whitney Class associated to Quadratic Form

I am trying to understand a bit of Milnor's paper "Algebraic K-Theory and Quadratic Forms". I guess, first things first. Why did Milnor come up with such an invariant? For any number field (such as ...
4
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0answers
151 views

Obstruction to the existence of lifting of the classifying map

Let $E$ be an $n$-plane bundle over CW complex $X$. Then $E$ is a pullback of tautological bundle $\gamma_n$ over $BO(n)$ i.e. $E=f^*(\gamma_n)$. This $f$ is called classyfing map. One can show that ...
5
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2answers
261 views

Two set of axioms for Stiefel-Whitney classes

Let $E \to X$ be a vector bundle. We can associate to $E$ several invariants: among them are the Stiefel-Whitney classes $w_i(E) \in H^i(X;\mathbb{Z}_2)$. These classes may be defined using the axioms:...
8
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3answers
537 views

Stiefel-Whitney total class with prescribed zeros

First things first, I am aware of the existence of this topic. It's related, but old and my question hasn't been discussed there. So I hope it's not wrong to start a new topic. I'm currently ...
7
votes
2answers
714 views

Second Stiefel-Whitney class is a square

I'm interested in examples of manifolds which are orientable and such that the second Stiefel-Whitney class is a square. (Of course the second Stiefel-Whitney class should be non-zero.) An easy ...
7
votes
2answers
497 views

Vector bundle over an oriented manifold with non-vanishing w_2w_3

I am looking for an example of an oriented rank 5 (or lower) real vector bundle $V$ over an oriented manifold such that the cup product $w_2(V) w_3(V)$ of Stiefel-Whitney classes does not vanish. It ...