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Let $X$ be a smooth degree $d$ hypersurface in $\mathbb{P}^3$ and $d \ge 5$. Let $C \subset X$ be a reduced curve such that $C$ is not a complete intersection curve in $\mathbb{P}^3$. Is it true that the dimension of the linear series $H^0(\mathcal{O}_X(C))$ is at most $2$? The motivation of this question is as follows:

As far as I understand $h^0(N_{C|X})$ computes the dimension of the Hilbert scheme of curves in $X$. Since $C$ is not a complete intersection and $X$ is a surface, there can exist at most one dimensional family of curves deforming $C$ in $X$ which would implying that $h^0(N_{C|X}) \le 1$. We now use the long exact sequence associated to the short exact sequence, $$ 0 \to \mathcal O_X \to \mathcal O_X(C) \to N_{C|X} \to 0 $$ along with the fact that $H^1(\mathcal{O}_X)=0$ since it is a smooth hypersurface in $\mathbb{P}^3$ to conclude the above result. Is there some mistake in this logic?

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  • $\begingroup$ There a problem with the display of the short exact sequence $\endgroup$
    – Naga Venkata
    Commented Dec 17, 2012 at 11:34
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    $\begingroup$ I think the point is that it's possible that $h^0(N_{C|X})\ge 2$. $\endgroup$
    – J.C. Ottem
    Commented Dec 17, 2012 at 12:50
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    $\begingroup$ This is a nitpick, but when talking about a codimension-1 subvariety in P^3, it seems a little superfluous to call it a "hypersurface". Why not just say "surface"? $\endgroup$
    – user5117
    Commented Dec 17, 2012 at 18:58

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I think your claim is not true.

Let $H$ be the hyperplane section of $X$ and assume that $C$ is not linearly equivalent to a multiple of $H$. Then $h^0(X, C+rH)\to\infty$ for $r\to\infty$ (Rieman-Roch + Serre duality). On the other hand, $C+rH$ is not linearly equivalent to a multiple of $H$, so it is not a complete intersection.

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