Why does the vanishing of this cohomology say that?

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Consider the following proposition found in the article: Stable Vector Bundles of Rank 2 on $\mathbb{P}^{3}$ - Hartshorne.

Proposition 3.1. Let $\mathscr{E}$ be a rank 2 bundle on $\mathbb{P}^{3}$ corresponding to a curve $Y$ in $\mathbb{P}^{3}$. Then $\mathscr{E}$ is stable >>(respectively semistable) if and only if

(1) $c_{1}(\mathscr{E}) > 0$ (respectively $c_{1}(\mathscr{E}) \geq 0$), and

(2) $Y$ is not contained in any surface of degree $\leq \dfrac{1}{2}c_{1}(\mathscr{E})$ (respectively $< \dfrac{1}{2}c_{1}(\mathscr{E}))$

In proof of this proposition, he assumes that $c_{1}(\mathscr{E})$ is even. I understood the fact that $c_{1}(\mathscr{E}) > 0$. However, I tried, but I couldn't understand why $H^{0} \bigl ( \mathscr{I}_{Y}(\dfrac{1}{2}c_{1})\bigr ) = 0$ implies that $Y$ is not contained in any surface of degree $ \dfrac{1}{2}c_{1}(\mathscr{E})$ as the author states in the demonstration.

Help and suggestions will be appreciated.

Thank you.

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The cohomology exact sequence associated with the sequence of sheaves $$ 0 \to J_Y(d) \to \mathcal{O}(d) \to \mathcal{O}_Y(d) \to 0 $$ shows that a section of $J_Y(d)$ is the same as a section of $\mathcal{O}(d)$ (i.e., a degree-$d$ hypersurface) vanishing on $Y$.