$H^i(X, \mathcal{O}_X(H))$ for $X$ a cubic fourfold

Question

$\newcommand{\oh}{\mathcal{O}}$Let $X \subset \mathbb{P}^5$ be a cubic fourfold. I'd like to known the cohomology $H^i(X, \mathcal{O}_X(H))$ where $i : H \subset X$ is the ample divisor class.

One idea I had was to use the short exact sequence $ 0 \to \mathcal{O}_X(-H) \to \mathcal{O}_X \to i_* \mathcal{O}_H \to 0$ associated to the divisor $H \subset X$. Now tensor this by $\mathcal{O}_X(H)$ to get the sequence $0 \to \mathcal{O}_X \to \mathcal{O}_X(H) \to i_* \mathcal{O}_H \to 0$.

I think we have $i_*\oh_H \otimes \oh_X(H) \cong i_*\oh_H$ at the end because $i_* \oh_H \otimes \oh_X(H) \cong i_*(\oh_H \otimes i^*\oh_X(H))$ by projection formula, and the pullback of a structure sheaf is a structure sheaf, but there can be no functions on $H$ with poles on all of $H$, so $i^* \oh_X(H) \cong \oh_H$ and thus $i_*\oh_H \otimes \oh_X(H) \cong i_*\oh_H$ (I'm not 100% sure this is right though).

Now take the long exact sequence in cohomology of the short exact sequence above. We have $H^i(\oh_X)=\mathbb{C},0$ for $i=0, \neq 0$ respectively, but I'm unsure how to find $H^i(i_* \oh_H)$.

Answer 1

Writing $P=\mathbb{P}^5$, one has an exact sequence, $0\to O_P(-2)\to O_P(1)\to O_X(H)\to 0$. Taking cohomologies, one gets $h^0(O_X(H))=6$ and $h^i=0, i>0$.