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.



Cohomology of the tangent sheaf of $\mathbb{P}(1,2,3)$
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 $h^0 = 5$.


