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I'm imagining there's a way to relate the pontryagin classes of $T(M\times N)$ to the pontryagin classes of $M$ and those of $N$, but I haven't been able to find a helpful reference. Could someone explain this or direct me to a good source? I am interested specifically in $CP^2\times CP^2$ but also in the general case. Thanks!

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up vote 3 down vote accepted

Given $M_1$ and $M_2$, let $\pi_i:M_1\times M_2\rightarrow M_i$ be the canonical projection map.

Then we have $T(M_1\times M_2) \cong \pi_1^*(TM_1)\oplus \pi_2^*(TM_2)$.

The Pontryagin classes satisfy the Whitney sum formula, at least mod 2 torsion. So, \begin{align*} p(T(M_1\times M_2)) &= p(\pi_1^*(TM_1)\oplus \pi_2^*(TM_2)) \\\ &= p(\pi_1^*(TM_1))\cup p(\pi_2^*(TM_2)) \\\ &= \pi_1^*(p(TM_1))\cup \pi_2^*(p(TM_2))\end{align*}

(All of the equalites are mod 2-torsion)

This same type calculation works for the Chern classes, Stiefel-Whitney classes, and the Euler class assuming they're defined.

In the case of $\mathbb{C}P^2$, the cohomology ring has no 2-torsion, so these calcuations go through just fine. Writing $H^*(\mathbb{C}P^2) \cong \mathbb{Z}[x]/x^3$ with $x$ of degree $2$, we have $p_1 = 3x^2$. By the Kunneth formula, we have $H^*(\mathbb{C}P^2\times \mathbb{C}P^2) \cong \mathbb{Z}[x,y]/x^3,y^3$.

It follows that $p(T(\mathbb{C}P^2\times \mathbb{C}P^2)) = 1 + 3x^2 + 3y^2 + 9x^2y^2$.

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