I'm having truble with Exercise 14B of Milnor/Stesheff Characteristic classes: prove that the total Chern class of a comple bundle is mapped to the Stiefel-Whitney class by the coefficient homomorphism $H^*(B,\mathbb{Z})\to H^*(B,\mathbb{Z}_2)$. I'm aware of this question/answer where a very different approach is used in the answer but would like to understand what one is supposed to do in the Milnor's exercise.

My effort:

To show that odd dimensional SW classes of a complex bundle $\xi$ are zero, we use the fact that there is a bundle map from $\xi$ to the canonical bundle over the complex Grassmanian which has no odd-dimensional cells: naturality of SW classes implies then that $w_{2k+1}(\xi)=0$. (I hope that this is correct.)

For the top-class $c_n$ which equals the Euler class, it is shown in & 9.2 that the coefficient homomorphism maps it to the top SW class.

However, I don't see how to use some kind of an induction. If $E_0$ is the complement of the zero section, there is a commutative diagram $$\require{AMScd} \begin{CD} H^{2j}(B) @>{\pi^*}>> H^{2j}(E_0)\\ @VVV @VVV \\ H^{2j}(B, \mathbb{Z}_2) @>{\pi^*}>> H^{2j}(E_0, \mathbb{Z}_2)\\ \end{CD} $$ where the horizontal lines are isomorphisms for $j<n$. The upper horizontal arrow maps the Chern class $c_j(\xi)$ to the Chern class $c_j(\xi_0)$ of the "orthogonal complement bundle". By induction, the right hand side vertical arrow maps the Chern class of $\xi_0$ to its SW class. However, how to prove that the SW class $w_{2j}(\xi)$ is mapped to $w_{2j}(\xi_0)$? (The lower horizontal map is not induced by a bundle homomorphism..)

Edit: Is this the right idea? The projection $\pi: E_0\to B$ induces a pullback complex $n$-bundle $\pi^*\xi$ (with base space $E_0$) which can be naturally decomposed as the sum of a trivial line bundle $\epsilon^1$ and $\xi_0$ (The fiber of $\epsilon^1$ over $e_0$ is generated by multiples of $e_0$ in its fiber over $\pi(e_0)$). Then using $w_{2j}(\pi^*\xi)=w_{2j}(\epsilon^1\oplus\xi_0)=w_{2j}(\xi_0)$ and the fact that $\pi^*$ maps $w_{2j}(\xi)$ to $w_{2j}(\pi^* \xi)$ by naturality, we conclude that $\pi^*$ maps $w_{2j}(\xi)$ to $w_{2j}(\xi_0)$. This, together with the induction above, completes the proof.

Is this correct?

  • $\begingroup$ Would using the axiomatic definition of the SW classes be easier? $\endgroup$
    – anomaly
    May 11, 2015 at 5:57
  • $\begingroup$ @anomaly I also thought about this but don't see how to proceed.. $\endgroup$ May 11, 2015 at 6:34

1 Answer 1


This is everything perfectly alright. However notice two things: you have to assume $n>1$, which is fine since for $n=1$ we are in the already known case of the top class i.e. the euler class (but I guess you did that as the start of the induction). And the other thing is that what you do is actually not too different from the splitting principle, it is actually almost the same. What you try to do in the splitting principle, is first to find a space where you can split of a line bundle. Here you do the same just you did not care about the resulting space to be compact (in case $B$ were for example). So if you would compactify your space $E_0$ by projectivizing (i.e. quotiening out scalar multiplication), you would get precisely the same.

As a last comment about my first comment: I guess this actually depends on how you interpret the induction. Either you start with line bundles and then go from $n$-bundles to $n+1$-bundles or you start with an $n$-bundle and then try to reduce it to an $n-1$-bundle to apply your knowledge of the euler class.

That's it.

  • $\begingroup$ Thank you! The induction I had in mind was wrt. the number "dimension of the bundle minus index of the characteristic class." (As far as I understood, this quantity is used in the inductive definition of Chern classes in Milnor, although it is not explicitely stated.) $\endgroup$ May 11, 2015 at 6:33
  • $\begingroup$ If you had time, could you please also check a similar question here? Regards math.stackexchange.com/questions/1291272/… $\endgroup$ May 21, 2015 at 14:43

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