Cohomology of tangent sheaf of a hypersurface

Let $$X\subset\mathbb{P}^n$$ be an irreducible and reduced hypersurface of degree $$d$$. How can one explicitly compute the dimension of the vector spaces $$H^0(X,T_X),H^1(X,T_X),H^2(X,T_X)$$? Here $$T_X$$ is the tangent sheaf of $$X$$. For instance $$h^0(X,T_X)$$ gives the dimension of the automorphism group of $$X$$.

Use the normal sequence $$0 \to T_X \to T_{\mathbb{P}^n}\vert_X \to N_{X/\mathbb{P}^n} \to 0,$$ exact sequences $$0 \to \mathcal{O}_{\mathbb{P}^n} \to \mathcal{O}_{\mathbb{P}^n}(d) \to i_*N_{X/\mathbb{P}^n} \to 0$$ (we identify here $$N_{X/\mathbb{P}^n}$$ with $$\mathcal{O}_X(d)$$ and denote by $$i$$ the embedding $$X \to \mathbb{P}^n$$) and $$0 \to T_{\mathbb{P}^n}(-d) \to T_{\mathbb{P}^n} \to i_*(T_{\mathbb{P}^n}\vert_X) \to 0,$$ and Borel-Bott-Weil Theorem to compute cohomology on $$\mathbb{P}^n$$.
• I think that the first exact sequence you wrote holds only if $X$ is smooth. What if $X$ is singular? Mar 5 '19 at 18:57
• This is right (I missed the absence of the smoothness assumption in the question). In the singular case the first sequence is not exact in the right term, but its cokernel is not so hard to control. It is isomorphic to $\mathcal{O}_Z(d)$, where $Z \subset \mathbb{P}^n$ is the subscheme defined by $\{ \partial F/\partial x_0 = \partial F/\partial x_1 = \dots \partial F/\partial x_n = 0 \}$, where $F$ is the equation of $X$. Mar 5 '19 at 19:31