Let $X\subset \mathbb P^n (n>3)$ be a smooth hypersurface defined by degree $d$ polynomial $F$, and let $R=k[x_0,\cdots,x_n]/(\partial F_0,\cdots,\partial F_n)$ be a graded ring (sometimes called Jacobian ring).
I want to prove that:
$$R_d\cong H^1(X,\mathcal T_X)$$
where $R_d$ is the degree $d$ piece of $R$ and $\mathcal T_X$ is the tangent bundle of $X$.
Here is my approach:
I already see there is a short exact sequence \begin{equation} 0\to H^0(\mathcal T_{\mathbb P^n}|_X)\to H^0(\mathcal O_X(d))\to H^1(\mathcal T_X)\to 0 \end{equation}
So need to show $$ H^0(\mathcal O_X(d))/H^0(\mathcal T_{\mathbb P^n}|_X)\cong R_d $$
But I have no idea why this is true.
Here is (a sketch of) how I get the exact sequence:
It comes from the normal bundle sequence: $$ 0\to \mathcal T_X \to \mathcal T_{\mathbb P^n}|_X \to \mathcal N_{X/{\mathbb P^n}}\to 0 $$ and $\mathcal N_{X/{\mathbb P^n}}=\mathcal O(d)$. It induces long exact sequence on cohomology, hence remains to show $H^0(\mathcal T_X)=0$ and $H^1(\mathcal T_{\mathbb P^n}|_X)=0$. The second one comes from Euler sequence. For the first one, we note that $X$ is smooth implies $\partial_i F$ forms a regular sequence, and hence $\Sigma h_i \partial_i F=0$ implies all $h_i=0$. Let $h_i$ be a section of $\mathcal O_X(1)$, we can extend it to $\mathbb P^n$. We still use $h_i$ to denote the extension. $h_i$ lies in $H^0(\mathbb P^n,\mathcal O(1))$ hence is a linear polynomial. And $\Sigma h_i \partial_i F$ vanishes on $X$ means $\Sigma h_i \partial_i F \in (F)$ hence we can assume $\Sigma h_i \partial_i F=\Sigma x_i \partial_i F$. By $\partial_i F$ forms a regular sequence we conclude $h_i=x_i$. Therefore, any global section of $\mathcal O_X(1)^{n+1}$which has no contribution to the normal direction of $X$ also has no contribution to the tangent direction of $\mathbb P^n$ (hence also no contribution to tangent direction of $X$). Remains to see if any section of $\mathcal T_X$ can be extended to $\mathcal O_X(1)^{n+1}$, which can be done by Euler sequence.