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Here by $P^n$ I mean $CP^n$, and what I want to do is to calculate the number of global sections of the holomorphic tangent bundle of $CP^n$.

If $n=1$, it is well known that $h^0(P^1, TP^1)=h^o(P^1,\mathcal{O}_{P^1}(2))=3$.

If $n>1$, I did some calculation in local coordinates, and find out that $h^0(P^n, TP^n) = n(n+1)$.

I am not sure if this is the correct answer and wonder if anyone else has calculated this before.

Besides, does anybody know the value of $h^1(P^n, TP^n)$? Even the $n=2$ case is enough for me. Many thanks!

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I removed the LaTeX from the title and made it more descriptive. – Harry Gindi Mar 25 '10 at 17:09
Yes, I realize in general that this module may not have a dimension in any sense of the word, but for complex projective space, the structure sheaf is made up of only constant functions by Liouville's theorem, so we can identify $\mathcal{O}_{\mathbb{CP}^n}(\mathbb{CP}^n)$ with $\mathbb{C}$, which is a field, which means that the module of global sections is free and has a well-defined dimension. – Harry Gindi Mar 25 '10 at 17:19
that is, global sections of the tangent bundle! – Harry Gindi Mar 25 '10 at 17:34
up vote 5 down vote accepted

The dimension of $H^0(\mathbb P^n,T\mathbb P^n)$ is $(n+1)^2-1$ and $h^1(\mathbb P^n, T \mathbb P^n)=0$. Using the Euler sequence (see for instance Griffiths-Harris, Principles of Algebraic Geometry) you can reduce the computation of these guys to the computation of the comology of $\mathcal O_{\mathbb P^n}$ and $\mathcal O_{\mathbb P^n}(1)$.

For the dimension of the space of holomorphic vector fields on $\mathbb P^n$ it is perhaps easier to realize that $$Aut(\mathbb P^n) = PSL(n+1, \mathbb C)$$ and its Lie algebra is $$\mathfrak{sl}(n+1,\mathbb C)$$.

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Thanks for the answer! I must have made some mistake in my calculation. One more question: what is the reason of $h^1(P^n, TP^n)=0$, is it guaranteed by some vanishing theorem? And how to see it intuitively (say in the language of Cech cohomology)? – Hanxiong Zhang Mar 25 '10 at 3:28
Look at the Euler sequence. You will see that $h^1(TP^n)$ vanishes because $h^1(O(1))=0$ (induction on $n$) and $h^2(O)=0$ ($P^n$ has no holomorphic $2$-forms). – Jorge Vitório Pereira Mar 25 '10 at 3:43
I can see that $h^1(P^1,\mathcal{O}_{P^1}(1)) = 0$, how to proceed with induction to prove $h^1(P^n,\mathcal{O}_{P^n}(1)) = 0$ for $n=2,3...$? – Hanxiong Zhang Mar 25 '10 at 4:03

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