Chern number of a product of complex projective lines

Solution 1:

First recall that $c(\mathbb{CP}^1) = c(T\mathbb{CP}^1) = (1 + a)^2 = 1 + 2a$, not $1 + 2a^2$. Keep in mind that $a \in H^2(\mathbb{CP}^1; \mathbb{Z})$ and is Poincaré dual to the fundamental class $[\mathbb{CP}^1] \in H_2(\mathbb{CP}^1; \mathbb{Z})$, so $\langle a, [\mathbb{CP}^1]\rangle = 1$.

As $T(\mathbb{CP}^1\times\mathbb{CP}^1) = \pi_1^*T\mathbb{CP}^1\oplus\pi_2^*T\mathbb{CP}^1$ where $\pi_i$ is the projection $\mathbb{CP}^1\times\mathbb{CP}^1 \to \mathbb{CP}^1$ onto the $i^{\text{th}}$ factor, we see that

\begin{align*} c(\mathbb{CP}^1\times\mathbb{CP}^1) &= c(T(\mathbb{CP}^1\times\mathbb{CP}^1))\\ &= c(\pi_1^*T\mathbb{CP}^1\oplus\pi_2^*T\mathbb{CP}^1)\\ &= c(\pi_1^*T\mathbb{CP}^1)c(\pi_2^*T\mathbb{CP}^1)\\ &= \pi_1^*c(T\mathbb{CP}^1)\pi_2^*c(T\mathbb{CP}^1)\\ &= \pi_1^*(1+2a)\pi_2^*(1+2a)\\ &= (1 + 2a_1)(1+2a_2)\\ &= 1 + (2a_1 + 2a_2) + (4a_1a_2) \end{align*}

where $a_i := \pi_i^*a$. Therefore $c_1(\mathbb{CP}^1\times\mathbb{CP}^1) = 2a_1 + 2a_2$ and hence

$$c_1^2(\mathbb{CP}^1\times\mathbb{CP}^1) = (2a_1 + 2a_2)^2 = 4a_1^2 + 8a_1a_2 + 4a_2^2 = 8a_1a_2$$

since $a_i^2 = (\pi_i^*a)^2 = \pi_i^*a^2$ and $a^2 \in H^4(\mathbb{CP}^1; \mathbb{Z}) = 0$. As $\langle a_1a_2, [\mathbb{CP}^1\times\mathbb{CP}^1]\rangle = 1$, $c_1^2[\mathbb{CP}^1\times\mathbb{CP}^1] = 8$.