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I am trying to show that if $C\subseteq\mathbb{P}^3$ is a smooth curve of degree $11$ and genus $15$, then $C$ is contained in a cubic surface $S\subseteq\mathbb{P}^3$. The idea is to use the exact sequence of sheaves $$ 0 \to \ \mathcal{I}_{C}\to \mathcal{O}_{\mathbb{P}^3} \to \mathcal{O}_{C} \to 0 $$ where $\mathcal{I}_{C}$ is the ideal sheaf of $C$. Twisting the exact sequence by 3, we get $$ 0 \to \ \mathcal{I}_{C}(3)\to \mathcal{O}_{\mathbb{P}^3}(3) \to \mathcal{O}_{C}(3)\to 0 $$ Taking the long exact sequence in cohomology and using $H^{i}(\mathcal{O}_{\mathbb{P}^3}(3))=0$ for $i=1, 2$, we have $$ 0 \to \ H^{0}(\mathcal{I}_{C}(3))\to H^{0}(\mathcal{O}_{\mathbb{P}^3}(3)) \to H^{0}(\mathcal{O}_{C}(3))\to H^1(\mathcal{I}_{C}(3))\to 0\to H^1(\mathcal{O}_{C}(3))\to H^2(\mathcal{I}_{C}(3))\to 0 $$ Now, by Riemann-Roch, $\dim H^0(\mathcal{O}_{C}(3))- \dim H^{1}(\mathcal{O}_{C}(3)) = 3\cdot 11 + 1 - 15 = 19$. So if $\dim H^{1}(\mathcal{O}_{C}(3))=0$, then $\dim H^0(\mathcal{O}_{C}(3))=19$. On the other hand, $\dim H^{0}(\mathcal{O}_{\mathbb{P}^{3}}(3)) = 20$. But then, the above exact sequence (and basic linear algebra) shows that $\dim H^{0}(\mathcal{I}_{C}(3)) \geq 20-19 = 1$, so there is a degree $3$ form vanishing along $C$. In other words, the curve $C$ is contained in a cubic surface, as desired.

So my question is:

Why is $H^{1}(\mathcal{O}_{C}(3))=0$?

The above exact sequence shows that $H^{1}(\mathcal{O}_{C}(3))=H^{2}(\mathcal{I}_{C}(3))$. So maybe it is easier to show that the latter is $0$?

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  • $\begingroup$ I realize that $H^{1}(\mathcal{O}_{C}(3))$ may not be $0$. If that is possible, is there a way to complete the argument to show that $C$ still lies on a cubic surface? $\endgroup$ – Prism Oct 1 '16 at 17:50
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Okay I think I figured out the answer! If $H^{1}(\mathcal{O}_{C})\neq 0$, then $\mathcal{O}_{C}(3)$ corresponds to a special effective divisor $D$ on $C$, and so by Clifford's theorem, we have $$ \dim H^{0}(D) - 1 \leq \frac{1}{2} \deg(D) = \frac{1}{2}\cdot 33 < 17 $$ So $\dim H^{0}(C, \mathcal{O}_{C}(3))<18$, and the argument above applies again to show that $C$ is contained in a cubic surface. I would very much appreciate if someone checked this answer!

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  • $\begingroup$ Can't you use Riemann--Roch? Deg $D = 33$, so $h^1(D) = h^0(K -D)$, and $K-D$ has negative degree. Or have I gotten the numbers muddled? $\endgroup$ – tracing Oct 2 '16 at 14:12
  • $\begingroup$ @tracing I think you are right! In this case, $\deg K = 2\cdot 15 - 2 = 28$ while $\deg D = 33$, so $K-D$ indeed has negative degree. This is much simpler than appealing to Clifford's theorem. Thank you very much! $\endgroup$ – Prism Oct 2 '16 at 19:29

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