# Cohomology of the tangent sheaf of $\mathbb{P}(1,2,3)$

Using the exact sequence

$$0\mapsto\mathcal{O}_{\mathbb{P}^{2}}\rightarrow\mathcal{O}_{\mathbb{P}^{2}}(1)^{\oplus 3}\rightarrow T_{\mathbb{P}^{2}}\mapsto 0$$

it is easy to compute $H^{1}(\mathbb{P}^{2},T_{\mathbb{P}^{2}}) = H^{2}(\mathbb{P}^{2},T_{\mathbb{P}^{2}}) = 0$ while $h^{0}(\mathbb{P}^{2},T_{\mathbb{P}^{2}}) = 8$.

On the singular variety $\mathbb{P}(1,2,3)$ by $T_{\mathbb{P}(1,2,3)}$ I mean $\mathcal{H}om(\Omega_{\mathbb{P}(1,2,3)},\mathcal{O}_{\mathbb{P}(1,2,3)})$.

Is there an analogous way (or a completely different way) of computing the cohomology groups of $T_{\mathbb{P}(1,2,3)} = \mathcal{H}om(\Omega_{\mathbb{P}(1,2,3)},\mathcal{O}_{\mathbb{P}(1,2,3)})$ ?

If it helps $\mathbb{P}(1,2,3)$ can be embedded in $\mathbb{P}^{6}$ as a singular Del Pezzo surface of degree six.

If you think about $P(1,2,3)$ as about stack then there is an analogue of the Euler sequence $$0 \to O \to O(1) \oplus O(2) \oplus O(3) \to T \to 0.$$ It allows to compute $h^1 = h^2 = 0$ and $h^0 = 5$.
• Just to add one comment: away from small characteristic (2 and 3), the computation "should" be the same whether you work with $\mathbb{P}(1,2,3)$ as a stack or as a coarse moduli space. There is an action of $\mathbb{G}_m$ on $\mathbb{A}^3\setminus\{0\}$, and the affine, geometric quotient $q:(\mathbb{A}^3\setminus\{0\})\to \mathbb{P}(1,2,3)$ gives a way of realizing $\Theta$ as the invariant part of $q_* \widetilde{T}$, for $\widetilde{T}$ defined as Sasha suggests on $\mathbb{A}^3\setminus\{0\}$. Since $q$ is affine, compute for $\widetilde{T}$ and take $\mathbb{G}_m$-invariants. – Jason Starr Feb 11 '14 at 19:09