# On degree and section of a line bundle on a smooth plane quintic

Let $$X$$ be a smooth plane projective quintic curve (over $$\mathbb C$$). Then we know that it has gonality $$4$$. Assume that it has genus $$g(X)=6$$. Then my question is the following:

Is it necessarily true that for every line bundle $$A$$ on $$X$$ with $$h^0(A) \geq 2$$ one has $$\text{deg}(A)\geq h^0(A)+2$$?

Gonality $$4$$ means minimum degree of line bundles with atleast $$2$$ sections is $$4$$. On the other hand we have for any line bundle with atleast $$2$$ sections and with $$h^1(A) \geq 2$$, $$\text{deg}(A) \geq 2h^0(A)-2$$. But then it's not quite clear to me that how these two facts on gonality and clifford index (and may be Riemann-Roch) give us an affirmative answer to the question. Maybe I'm missing something obvious. Does there exist a more direct proof in the literature?

This is true, and can be shown by an induction argument on $$h^0(A)$$. If $$h^0(A)=2$$, then $$\deg(A)\geq 4$$ since the gonality of $$X$$ is $$4$$. If $$h^0(A)>2$$, let $$p\in X$$ be a point in the support of an effective divisor representing $$A$$. Since $$\mathrm{H}^0(X,A(-p))\subset \mathrm{H}^0(X,A)$$ is a subspace of codimension $$\leq 1$$, we obtain the inequality $$h^0(A(-p))\geq h^0(A)-1 .$$ This implies that $$h^0(A(-p))\geq 2$$, so using the induction hypothesis and the inequality again we obtain $$\deg(A)-1 = \deg(A(-p)) \geq h^0(A(-p))+2\geq h^0(A)+1 .$$ Add $$1$$ to both sides.

Remark: this argument only uses that the gonality of $$X$$ is $$4$$.

• thank you very much for the proof. I have one trivial doubt: while applying induction hypothesis to $h^0(A(-p))$ aren't we required to make sure that we are at the $k-1$th stage I.e $h^0(A(-p)) < h^0(A)$? but codimension $\leq1$ means they could be equal. Commented Feb 16, 2021 at 18:25
• I guess you are right. Maybe it's better to induct on $\deg A$ then? One first needs to treat the case of $\deg A = 4$, which is easy using what I've written above.
– Jef
Commented Feb 16, 2021 at 18:36
• in that case we need to have $h^0(A) >2$ when $\text{deg}(A)>4$, but that need not necessarily happen Commented Feb 16, 2021 at 18:49
• What if you take a point p as in my proof, so then either $h^0(A(-p))<h^0(A)$ (in which case we can apply the induction hypothesis), or replace A by A(-p) and repeat this procedure. In other words, you're allowed to remove multiple points
– Jef
Commented Feb 16, 2021 at 20:25
• @HARRY You can always choose $p$ not a base point to ensure the dimension goes down by exactly one. Commented Feb 18, 2021 at 0:26