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Exercise 4-A of Milnor and Stasheff's book Characteristic Classes reads:

Show that the Stiefel–Whitney classes of a Cartesian product are given by $$w_k(\xi\times\eta) = \sum^k_{i=0} w_i(\xi)\times w_{k-i}(\eta)$$

This does not make sense to me because if $\xi\rightarrow B_1$ and $\eta\rightarrow B_2$ then $\xi\times\eta\rightarrow B_1\times B_2$, so $w_k(\xi\times\eta)\in H^k(B_1\times B_2;\mathbb{Z}_2)$. While $w_i(\xi)\times w_{k-i}(\eta)=(w_i(\xi),w_{k-i}(\eta))\in H^i(B_1;\mathbb{Z}_2)\times H^{k-i}(B_2;\mathbb{Z}_2)$.

So $w_k(\xi\times\eta)$ and $\sum^k_{i=0} w_i(\xi)\times w_{k-i}(\eta)$ lie in different sets, so how can I be asked to prove that they are equal?

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Do you know Künneth theorem? – Grigory M Feb 4 '13 at 8:22
They're using the conventions of the Kunneth Theorem in this formula as they describe it in the appendix. You're mis-interpreting their terminology. – Ryan Budney Feb 4 '13 at 8:23
@Ryan Budney So do they want me to show that the image of $w_k(\xi\times\eta)$ under the Kunneth isomorphism is equal to $\sum^k_{i=0} w_i(\xi)\otimes w_{k-i}(\eta)$ – Moss Feb 4 '13 at 8:34
That's it. The formula sort of follows for free from the Whitney axiom regarding direct sums -- it's almost the same axiom. – Ryan Budney Feb 4 '13 at 8:48
@Sebastian Perhaps you should post an answer to your question in order that it doesn't keep getting bumped up to the main page by the community user. – user38268 Feb 4 '13 at 11:31

I think you did not understand the $\times$ symbol at here. As others pointed out it is the cross product. So you have $$ w_{k}(\epsilon\times \delta)=w_{k}(\pi_{1}^{*}(\epsilon)\oplus\pi_{2}^{*}(\delta)) $$ By axiom 4 on page 38 we have $$ w_{k}(\pi_{1}^{*}(\epsilon)\oplus\pi_{2}^{*}(\delta))=\sum^{k}_{i=1}w_{i}(\pi_{1}^{*}(\epsilon))\cup w_{k-i}(\pi_{2}^{*}(\delta)) $$ While by the definition of cross product we have $$a\times b=\pi_{1}^{*}(a)\cup \pi_{2}^{*}(b)$$ This proved the original statement. Here the definition of cross product follows from Hatcher, page 218. Note as others pointed out the original statement is a mere restatement of Kunneth theorem with base ring be a field.

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In the last displayed formula, $p_2^*$ should be $\pi_2^*$. – Michael Albanese Jan 10 at 8:07
Thanks! Typo fixed. – Bombyx mori Jan 12 at 14:01

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