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Let $X$ be a smooth, plane projective curve of degree $6$ and genus $10$ (over complex numbers).

Question : Is it possible that there exists a special divisor $\Delta$ of degree $10$ on $X$ such that it has exactly $5$ independent sections?

Observations : $(1)$ From Clifford's Theorem, we have $h^0(\mathcal O_X(\Delta)) -1 =4 \leq \frac {\text{deg}(\Delta)}{2}=5$ and therefore it says this can happen.

$(2)$ If the curve $X$ is general, then from Theorem-$A$ of this paper, we have this is not possible.(Please correct me if I am wrong). This is also true from Brill-Noether theorem for general curves.

Can we say anything more in this situation when the curve $X$ is not necessarily general and if we know what $\Delta^2, K_X$ are?

Any insight from anyone is welcome

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1 Answer 1

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Such a divisor cannot exist. Let $H$ be the divisor of a line. By the base-point free pencil trick, we have an exact sequence $$0\rightarrow H^0(\Delta -H)\rightarrow H^0(\Delta)^2\rightarrow H^0(\Delta +H)\,;$$since $\deg (\Delta +H)=16$, we have $h^0(\Delta +H)\leq 8$, hence $h^0(\Delta -H)\geq 2$. Then $D:=\Delta -H$ is a $g^1_4$, thus base-point free since $X$ is not trigonal. Now applying again the base-point free pencil trick, we get an exact sequence $$ 0\rightarrow H^0(H-D)\rightarrow H^0(H)^2\rightarrow H^0(\Delta )$$ which tells us $h^0(H-D)>0$. Thus $D\equiv H-p-q$, for some points $p,q$ in $X$; but this implies $h^0(D)=1$, a contradiction.

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  • $\begingroup$ From $\text{deg}(\Delta+H) =16$ aren't we getting by Clifford's theorem that $h^0(\Delta+H) -1 \leq \frac{16}{2}=8$, i.e $h^0(\Delta+H) \leq 9$ and hence $h^0(\Delta-H) \geq 1$ ? Please correct me if this is wrong. $\endgroup$
    – User
    Mar 8, 2021 at 13:34
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    $\begingroup$ This is correct, but what do you do with that? For my argument I need $h^0(\Delta -H)\geq 2$. $\endgroup$
    – abx
    Mar 8, 2021 at 14:03
  • $\begingroup$ can we say something if we know that $\Delta^2=8$? $\endgroup$
    – User
    Mar 8, 2021 at 15:28
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    $\begingroup$ What do you call $\Delta^2$? $\endgroup$
    – abx
    Mar 8, 2021 at 15:56
  • $\begingroup$ by $\Delta^2$ I mean its self intersection number. $\endgroup$
    – User
    Mar 8, 2021 at 15:58

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