Prove that $H^1(\mathbb P^n,T_{\mathbb P^n})=0$

Question

There is an exercise in Ravi Vakil's notes, namely exercise 21.5.Q, asking to prove that $H^1(\mathbb P^n,T_{\mathbb P^n})=0$, where $T_{\mathbb P^n}$ is the tangent bundle of the projective space. I would like a hint on how to do this. I started by looking at the Euler sequence $$0\to \mathcal O_{\mathbb P^n}\to \mathcal O_{\mathbb P^n}(1)^{(n+1)}\to T_{\mathbb P^n}\to 0,$$ so that a piece of the long exact sequence would be $$H^1(\mathbb P^n,\mathcal O_{\mathbb P^n}(1)^{(n+1)})\to H^1(\mathbb P^n,T_{\mathbb P^n})\to H^2(\mathbb P^n,\mathcal O_{\mathbb P^n})=0,$$ but I do not know whether the group on the left is $0$.

Thanks for any suggestion in this direction, or any other approach!

@MarianoSuárez-Alvarez: thank you! what you say is just functoriality, right? — Brenin, Dec 01 '13 at 21:59
It is a consequence of the fact that the functor is additive. — Mariano Suárez-Álvarez, Dec 02 '13 at 00:47
They are obtained from an additive functors by derivation. Any textbook on homological will observe this. — Mariano Suárez-Álvarez, Dec 03 '13 at 03:00

Michael Albanese · Answer 1 · 2019-03-23T15:42:06.743

As pointed out by Mariano Suárez-Álvarez in the comments, we have

$$H^1(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(1)^{n+1}) = H^1(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(1))^{n+1}.$$

Now note that

$$H^1(\mathbb{P}^n, \mathcal{O}(1)) \cong H^1(\mathbb{P}^n, \mathcal{O}(-n-1)\otimes\mathcal{O}(n+2)) \cong H^1(\mathbb{P}^n, K_{\mathbb{P}^n}\otimes\mathcal{O}(n+2)).$$

As $\mathcal{O}(n+2)$ is ample, we see that $H^1(\mathbb{P}^n, \mathcal{O}(1)) = 0$ by the Kodaira vanishing theorem. Therefore

$$H^1(\mathbb{P}^n, T_{\mathbb{P}^n}) \cong H^2(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}) = 0.$$

Prove that $H^1(\mathbb P^n,T_{\mathbb P^n})=0$

1 Answers1