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.



On emptiness of certain $G^r_d(X)$ on a smooth plane curve
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 $h^0(D)=4$.


