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fixed the problem in the short exact sequence
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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 $$ 0 \to \mathcal O_X \to \mathcal O_X(C) \to N_{C|X} \to 0 $$ 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?

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?

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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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 $\mathcal{O}_X(C)$$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?

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 $\mathcal{O}_X(C)$ 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?

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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Upper bound on the dimension of linear series on a smooth hypersurface

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 $\mathcal{O}_X(C)$ 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?