Stiefel-Whitney Numbers of $\mathbb{R}P^2\times \mathbb{R}P^2$

Question

I'd like to calculate the Stiefel-Whitney numbers of $\mathbb{R}P^2\times\mathbb{R}P^2,$ but don't know how to. My first instinct was to say that the tangent bundle is isomorphic to the product of tangent bundles over $\mathbb{R}P^2$ and then do this directly. However, I don't know if this is true, and I don't know how to compute the cohomology ring $H^{*}(\mathbb{R}P^2\times\mathbb{R}P^2; \mathbb{Z}_2)$ easily.

Answer 1

Let $p_1, p_2$ be the projections $M \times N \to M, M \times N \to N$; then it is true that $T(M\times N) \cong p_1^* TM \oplus p_2^* TN$.

You can calculate that cohomology ring via the Kunneth theorem, since your coefficients are in a field. It gives you the tensor product of the rings: $H^*(X \times Y; F) \cong H^*(X;F) \otimes_F H^*(Y; F)$. Because $H^*(\Bbb{RP}^2; \Bbb Z_2) \cong \Bbb Z_2[x]/(x^3)$, you get $H^*(\Bbb{RP}^2 \times \Bbb{RP}^2;\Bbb Z_2) \cong \Bbb Z_2[x,y]/(x^3,y^3)$, where $|x|=|y|=1$.

Now, we know $w(\Bbb{RP}^2) = (1+x)^3 = 1+x+x^2 \in H^*(\Bbb{RP}^2;\Bbb Z_2)$. Pulling back, we get that

$w(p_1^*(T\Bbb{RP}^2)) = p_1^*w(\Bbb{RP}^2) = 1+x+x^2$, and $w(p_2^*(T\Bbb{RP}^2)) = 1+y+y^2$. Because $w(E \oplus E') = w(E)w(E')$, we see that $$w(\Bbb{RP}^2 \times \Bbb{RP}^2) = 1+(x+y)+(x^2+xy+y^2)+(x^2y+xy^2)+x^2y^2.$$

Taking products of these terms, we see that the Whitney numbers corresponding to the partitions $(1,1,1,1), (2,1,1)$ and $(3,1)$ are all zero, and the Whitney numbers corresponding to $(2,2)$ and $(4)$ are both one.