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Let $X$ be a smooth plane projective curve of degree $6$ and genus $10$ (over complex numbers). Then my question is the following :

Question : Is it possible that there exists a special divisor $D$ of degree $9$ on $X$ admitting exactly $4$ independent sections?

Observations : $(i)$ From Clifford's theorem: we have, $h^0(\mathcal O_X(D)) -1 =3\leq \frac{\text{deg}(D)}{2} =4.5$. Therefore, this theorem says that such divisor may exists on $X$.

$(ii)$ Note that, if such a divisor exists on $X$, then it belongs to $G^3_9(X)$. Since, $\rho(10,3,9)=-6<0$, we can't guarantee the non-emptiness of $G^3_9(X)$.

Any insight or remark from anyone is welcome.

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No, this is not possible. Let $H$ be the divisor of a line. Use the base point free pencil trick to get an exact sequence $$0\rightarrow H^0(D-H)\rightarrow H^0(D)^2\rightarrow H^0(D+H)$$ Since $\deg(D+H)=15$, we have by Riemann-Roch $h^0(D+H)\leq 7$, hence $h^0(D-H)\geq 1$. Thus $D\equiv H+E$ with $E\geq 0$ of degree 3. But by Serre duality, $h^0(H+E)=h^0(2H-E)$. Now $E$ imposes independent conditions on conics, so $h^0(2H-E)=h^0(2H)-3=3$, contradicting $h^0(D)=4$.

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    $\begingroup$ The system $\lvert 2H\rvert$ is cut down by conics in $\mathbb{P}^2$, so $\lvert 2H-E\rvert$ is cut down by conics passing through $E$. Now any length 3 subscheme of $\mathbb{P}^2$ imposes 3 conditions on conics — see Lemma 0.1.1 in Beltrametti-Sommese Zero cycles and $k$-th order embeddings of smooth projective surfaces, INDAM Symposia Mathematica, 32,(1992), 33-48. $\endgroup$
    – abx
    Commented Mar 16, 2021 at 20:27
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    $\begingroup$ What is your question if $\deg(D)=8$? You can have $h^0=3$, and certainly not 4 — the exact sequence would give you a $g^1_2$. $\endgroup$
    – abx
    Commented Mar 17, 2021 at 16:53
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    $\begingroup$ $E$ cannot be a $g^1_2$, $X$ would be hyperelliptic. $\endgroup$
    – abx
    Commented Mar 17, 2021 at 17:37
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    $\begingroup$ If it is 4, $D+p$ for any $p\in X$ contradicts my previous answer. $\endgroup$
    – abx
    Commented Mar 17, 2021 at 18:26
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    $\begingroup$ Yes, this is what I mean. $\endgroup$
    – abx
    Commented Mar 17, 2021 at 19:27

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