Cohomology of Segre varieties

Question

Let $\Sigma_{n,m}$ be a Segre variety, i.e. the image of the Segre map $\mathbb{P}^n\times\mathbb{P}^m\to\mathbb{P}^{(n+1)(m+1)-1}$.

Then how can I calculate the first cohomology group of its tangent bundle, i.e. $H^1(\Sigma_{n,m},\mathcal{T})$?

Answer 1

Forget about Segre and use the Künneth formula for $T=T_{\mathbb P^n\times {\mathbb P^m}}=T_{\mathbb P^n}\boxtimes T_{\mathbb P^m}$, obtaining:$$ H^1(\mathbb{P}^n\times\mathbb{P}^m,T)\\=[ H^0(\mathbb P^n,T_{\mathbb P^n})\otimes H^1(\mathbb{P}^m,T_{\mathbb P^m})]\oplus [H^1(\mathbb P^n,T_{\mathbb P^n}) \otimes H^0((\mathbb{P}^m,T_{\mathbb P^m})]\\=[\operatorname {(whatever) }\otimes 0]\oplus [0\otimes \operatorname {(whatever)}]\\=\Large {0} $$ I have used that for any projective space $\mathbb P^N$ we have $H^i(\mathbb P^N, T_{\mathbb P^N})=0 \; \operatorname {for all}\: i\geq1$, which follows from the long cohomology exact sequence associated to the Euler exact sequence.