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](https://arxiv.org/pdf/1701.06579.pdf), 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



Special divisors on smooth plane curves
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.


