4
$\begingroup$

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?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

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 $4$.

Improve this answer
$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$
    – HARRY
    Commented Feb 16, 2021 at 18:25
  • 1
    $\begingroup$ 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. $\endgroup$
    – Jef
    Commented Feb 16, 2021 at 18:36
  • $\begingroup$ in that case we need to have $h^0(A) >2$ when $ \text{deg}(A)>4$, but that need not necessarily happen $\endgroup$
    – HARRY
    Commented Feb 16, 2021 at 18:49
  • 1
    $\begingroup$ 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 $\endgroup$
    – Jef
    Commented Feb 16, 2021 at 20:25
  • 1
    $\begingroup$ @HARRY You can always choose $p$ not a base point to ensure the dimension goes down by exactly one. $\endgroup$
    – Will Sawin
    Commented Feb 18, 2021 at 0:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .