# Stiefel-Whitney numbers for product bundle

I'm reading Milnor's "characteristic classes" and I want to compute Stiefel-Whitney numbers of $P^2 \times P^2$ (product of projective spaces) for one of the problems,

I know how Stiefel-Whitney classes of a product bundle are related to the two previous bundles :

$w_k(\zeta \times \eta)=\sum_{i+j=k}w_i(\zeta) \times w_j(\eta)$

but I still can't do the computation, can anyone help me with this, thank you.

Here is what I've done:

by kunneth theorem we have: $H^{*}(P^2 \times P^2 ; Z_2) = Z_2[a,b]/(a^3,b^3)$

now by the above relation for stiefel-whitney classes of product bundle we can write this classes for $P^2 \times P^2$ in terms of $a$ and $b$ :

$w_4 = a^2 b^2 , w_3=a^2b+ab^2, w_2=a^2+ab+b^2, w_1=a+b, w_0=1$

now I should compute operation of fundamental homology class of $P^2 \times P^2$ on each of these elements $w_2^2, w_1^2w_2,w_1w_3, ...$ in $H^4(P^2 \times P^2;Z_2)$ , I've got stuck in this part

• Where are you getting stuck in the computation? For example, can you compute $H^\ast(P^2\times P^2, \mathbb{Z}/2)$? – Jason DeVito Mar 25 '13 at 17:04
• @JasonDeVito I added my computations to the post. – Mehdi Mar 25 '13 at 19:14
• I see - thanks for updating the post. I'll start typing up an answer now (if no one else beats me to it). – Jason DeVito Mar 25 '13 at 19:18

Now that you have all the Stiefel-Whitney classes written down, the hard part is over. To compute Stiefel-Whitney numbers, recall that these are, by definition, obtained in the following way.

Start with a partition of $4$, that is, a sum of a bunch of positive numbers which give $4$. Here are all five of the options: $1+1+1+1,\, 1+1+2,\, 1+3,\, 2+2,\,$ and $4$.

For each choice, form the corresponding product of Stiefel-Whitney class \begin{align*} 1+1+1+1 &\leftrightarrow w_1 \cup w_1 \cup w_1 \cup w_1 \\ 1+1+2 &\leftrightarrow w_1\cup w_1\cup w_2\\ 1+3 &\leftrightarrow w_1\cup w_3\\ 2+2 &\leftrightarrow w_2\cup w_2 \\ 4&\leftrightarrow w_4\end{align*}

The point of a partition is that all the cup products on the right land in $H^4(P^2\times P^2;\mathbb{Z}/2)$. Since every manifold has an orientation class mod $2$, we can pair the element on the right with the orientation class and get a number mod $2$ out. These numbers mod $2$ are the Stiefel-Whitney numbers.

By Poincare duality, the orientation class is the dual of the unique element in $H^4(P^2\times P^2,\mathbb{Z}/2)$, that is, it's the dual of $a^2 b^2$. Hence, computing all the Stiefel-Whitney numbers is the same as computing all the above cup products (using the relations $a^3 = b^3 = 0$), and then counting, mod $2$, the number of occurrences of $a^2 b^2$.

Doing this (while supressing the cup product sign) gives \begin{align*} (w_1)^4 &= (a+b)^4 & &= 0\\ (w_1)^2 w_2 &= (a+b)^2(a^2 + b^2 + ab) & &= 0 \\ w_1 w_3 &= (a+b)(ab^2 + a^2 b) & &= 0\\ (w_2)^2 &= (a^2+b^2+ab)^2 & &= a^2 b^2\\ w_4 &= a^2b^2 & &= a^2 b^2.\end{align*}

(Note that the computations are considerable eased by noting we're working mod $2$ so $(a+b)^2 = a^2 + b^2$.)

From this calculation, we see that three of the Stiefel-Whitney numbers are $0$ (mod $2$) while the other two are $1$ (mod $2$).

• If some could fix my aligning on the right side of the second align, that'd be great. More importantly, I'm not sure why it's not correct to begin with, so link to a $\LaTeX$ lesson wouldn't hurt either ;-) – Jason DeVito Mar 25 '13 at 19:48
• Thank you for your detailed answer, using Poincare duality was really nice. Unfortunately, I don't know how to fix the align. – Mehdi Mar 25 '13 at 19:57
• I have a stupid question, because I see these Poincaré duality arguments all the time and I don't really understand: It is my understanding that Poincaré duality in this case yields an isomorphism $H^4 \to H_0$ by capping a cohomology class of degree four with the fundamental class. Why then do we say that the orientation class is dual to the element in $H^4$? Doesn't the orientation class live in $H_4$, not $H_0$? – anon Dec 6 '17 at 16:04
• @anon: Looking back on this 4 years later, I no longer see the connection with Poincare duality. All I need is that $a^2 b^2$ is dual to the unique non-zero element of $H_4(P^2\times P^2;\mathbb{Z}/2\mathbb{Z})$ (in the linear algebraic sense of dual spaces), which just requires the universal coefficieint theorem. – Jason DeVito Dec 6 '17 at 18:28
• Thanks Jason, this makes a lot more sense! – anon Dec 6 '17 at 19:06