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.
Best Answer
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 $h^0 = 5$.