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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.

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up vote 6 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.

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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 $w_{n-1}(\tau)$? – Dustin Tran 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. – Dustin Tran May 2 '12 at 2:25

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