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

Solution 1:

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 $2$).