Ideal sheaf on a surface

Question

Let $S\subset\mathbb{P}^n$ a smooth complex projective surface. I consider the exact sequence $$0\rightarrow I_S\rightarrow\mathcal{O}_{\mathbb{P}^n}\rightarrow\mathcal{O}_S\rightarrow 0,$$ where $I_S$ is the ideal sheaf of $S$ and with $\mathcal{O}_S$ I denote $i_*\mathcal{O}_S$ ($i :S\hookrightarrow\mathbb{P}^n$ the inclusion), so it is the extension by zero outside $S$ of the sheaf $\mathcal{O}_S$.

From this exact sequence I obtain this other exact sequence $$0\rightarrow I_S(k)\rightarrow\mathcal{O}_{\mathbb{P}^n}(k)\rightarrow\mathcal{O}_S(k)\rightarrow 0.$$ $I_S(k)$ is the sheaf of polynomials of degree $k$ that vanish on $S$.

My question is, how can i look at $\mathcal{O}_S(k)$? Is it correct to say that it is equal to $\mathcal{O}_S(kH)$ where $H$ is an hyperplane section of $S$? It should be actually a sheaf on $\mathbb{P}^n$.

My aim is to calculate $h^0(I_S(k))=dimH^0(I_S(k))$, so i need to know $h^0(\mathcal{O}_S(k))$ and to know it i'd like to use Riemann-Roch, that's why i thought to introduce a section $H$ on $S$.

Answer 1

Yes, $\mathcal O_S(k)$ is precisely $\mathcal O_S(kH)$.

Answer 2

You can somewhat think of $\mathcal O_S(k)$ as either a sheaf on $S$ or on $\mathbb P^n$, since computing its cohomology on either space gives the same answer. This is the content of Lemma III.2.10 of Hartshorne. $\mathcal O(1)$ is the associated sheaf to a hyperplane in $\mathbb P^n$, and $\mathcal O_S(1) = i^\ast \mathcal O(1)$ is associated to a hyperplane section $S\cap H$. Your approach seems reasonable to me!