Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to understand the relationship between Chern classes and Stiefel-Whitney classes, and I came upon this problem (14-E) in Milnor-Stasheff's Characteristic Classes.

We are asked to define the Stiefel-Whitney classes in the same way as was constructed for Chern classes, using mod 2 coefficients: $w_n(\xi) := e(\xi)$ mod 2 and $w_i(\xi):=(\pi_0^*)^{-1}w_i(\xi_0)$ for $i<n$, where $\pi_0^*: H^{2i}(B)\to H^{2i}(E_0)$ is an isomorphism for $i<n$.

In the problem, it states:

In this approach there is some difficulty in showing that $w_{n-1}(\xi_0)$ belongs to the image of $\pi_0^*.$ It suffices to show that $w_{n-1}(\xi_0)$ restricts to zero in each fiber $F_0$, or equivalently, that the tangent bundle $\tau$ of the $(n-1)$-sphere satisfies $w_{n-1}(\tau)=0$.

This is easy to prove since $e(\tau)=0$ or $2$ depending on odd or even dimensions, so $e(\tau)$ mod $2=0=w_{n-1}(\tau).$ But why does it suffice to prove this? That is, I don't necessarily understand why this proves "$w_{n-1}(E_0)$ belongs to the image of $\pi_0^*$." I'm misunderstanding a step in this logic that is perhaps quite trivial.

share|improve this question
add comment

1 Answer 1

up vote 5 down vote accepted

The map $\pi_0^*$ fits into a long exact sequence including $$ H^{n-1}(B) \to H^{n-1}(E_0) \to H^n(E, E_0)$$ So, we want to check that $w_{n-1}(\xi_0)$ maps to zero under $H^{n-1}(E_0) \to H^n(E, E_0)$. The class in $H^n(E, E_0)$ is determined by its restrictions to the fibers $H^n(F, F_0)$ which is isomorphic via the corresponding long exact sequence to $H^{n-1}(F_0)$, thus we need to see where $w_{n-1}(\xi_0)$ maps to under the restriction $H^{n-1}(E_0)\to H^{n-1}(F_0)$. At this point, by the definition of $\xi_0$ and naturality, it should be easy to see that $w_{n-1}(\xi_0)$ maps to $w_{n-1}(\tau)$ where $\tau$ is the tangent bundle of the $(n-1)$-sphere.

share|improve this answer
    
Thanks for the information! It makes a lot of sense, but I'm stuck at the last part. For the last statement, are we somehow identifying $F_0$ as $TS^{n-1}$ to obtain the final map to $w_{n-1}(\tau)$? –  Riem May 2 '12 at 1:15
    
$F_0 = \mathbb R^n - \{0\}$ deformation retracts onto $S^{n-1}$, and if we take the pullback bundle of $\xi_0$ along the inclusion $S^{n-1} \subset F_0 = \mathbb R^n - \{0\} \subset E_0$ (remember $E_0$ is the base space of the bundle $\xi_0$), then we obtain $TS^{n-1}$. –  Justin Young May 2 '12 at 1:23
    
Ah, that makes sense. What confused me was what corresponded to the total space and base space for $\xi_0$. Thanks. –  Riem May 2 '12 at 2:25
add comment

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.