Recently, in my research I bumped onto gonal morphisms. At the moment, my knowledge is based upon some things I read on the internet. Before stating my questions, I added some definitions/facts that might motivate the questions below.

By a curve, I mean a smooth projective connected curve over $\mathbf{C}$. A non-constant morphism $\pi:X\longrightarrow \mathbf{P}^1$ is gonal if $\deg \pi$ is minimal. The gonality of a curve $X$, denoted by $\gamma_X$, is the degree of a gonal morphism $\pi:X\longrightarrow \mathbf{P}^1$. Thus, for example, a curve of genus $g\geq 2$ is hyperelliptic iff it is $2$-gonal.

The hyperelliptic map of a hyperelliptic curve is unique. (Of course, here by unique we mean unique up to composition with an isomorphism of the projective line.)

Edit: In the questions below, we consider the set of gonal morphisms of a curve modulo the action of Aut$(\mathbf{P}^1)$.

Fact 1. For any curve $X$ of genus $g\geq 2$, we have that $\gamma_X \leq [\frac{g+3}{2}]$.

Fact 2. For any integer $\gamma \geq 2$, the closure of the locus of $\gamma$-gonal curves in the moduli space $\mathcal{M}_g$ of smooth curves of genus $g\geq 2$ is irreducible of dimension $2g-5+2\gamma$.

Fact 3. For any prime number $p$ and integer $g\geq 2$ such that $g\geq (p-1)^2$, Accola showed that any $p$-gonal curve of genus $g$ has a unique gonal morphism.

I can't prove these facts, but I do remember where I got them from. So if necessary I could give the references.

Question 1. Let $X$ be a $\gamma$-gonal curve of genus $g\geq 2$. Is the set of gonal morphisms for $X$ modulo the action of Aut$(\mathbf{P}^1)$ finite?

I expect the answer to this question to be negative if $g-\gamma$ is small. In view of Fact 3, I would like to propose the following question.

Question 2a. Fix $\gamma\geq 3$. Does there exist an integer $g_\gamma$ such that for any $g\geq g_\gamma$ and any $\gamma$-gonal curve $X$ of genus $g$, the gonal morphism for $X$ is unique?

Question 2b. Fix $\gamma\geq 3$. Does there exist an integer $g_\gamma$ such that for any $g\geq g_\gamma$ and any $\gamma$-gonal curve X of genus $g$, the set of gonal morphisms for $X$ is finite?

Question 3. Does there exist a positive integer $g_0$ with the following property? For any $g\geq g_0$ and curve $X$ of genus $g$, the set of gonal morphisms of $X$ is finite?

Question 4. Do there exist curves with infinitely many gonal morphisms? (Edit: In hindsight, this question is the same as Question 1.)

I think it's clear that these questions aren't unrelated. They are all related to the set of gonal morphisms associated to a curve. It would be wonderful to know when this set is finite.

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    $\begingroup$ About question 4: a smooth plane quartic $C$ has gonality 3 and has infinitely many $g^1_3$, corresponding to projecting to ${\mathbb P}^1$ from a point of $C$. $\endgroup$
    – rita
    Aug 20, 2011 at 12:16
  • $\begingroup$ thank you for this answer. What is $g_3^1$ stand for? I came across this notation a couple of times. Could you give a definition? I'm guessing $g_3$ means gonality three? What does the 1 on top mean? $\endgroup$ Aug 20, 2011 at 12:18
  • $\begingroup$ I do not understand the difference between Q.1 and Q.4. If you do not divide by the action of $Aut({\mathbb P}^1$ then you always get infinitely many gonal morphisms. $\endgroup$
    – rita
    Aug 20, 2011 at 12:38
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    $\begingroup$ No. This will only be true for general curves, i.e. curves corresponding to some non-empty Zariski open subset of the moduli space $M_g$. $\endgroup$
    – naf
    Aug 22, 2011 at 13:13
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    $\begingroup$ as an example generalizing rita's and confirming ulrich's comment, consider a smooth plane quintic, which has genus 6 and gonality 4, and project from points of the curve. $\endgroup$
    – roy smith
    Aug 31, 2011 at 19:08

1 Answer 1


Extending Rita's example, if $X$ is, say, a double cover of a curve of genus $3$, then $X$ can have arbitrarily large genus and it has gonality (at most) $6$. Moreover it has infinitely many $g^1_6$ (BTW $g^r_d$ means a linear system of degree $d$ and dimension $r$, so a $g^1_d$ is a map to $\mathbb{P}^1$ of degree $d$). So the answer to all of your questions is no.

Here is something that can be done. If you have two maps of degree $d$ from $X$ to $\mathbb{P}^1$, then you get a map from $X$ to $\mathbb{P}^1 \times \mathbb{P}^1$. If this map is injective, then the genus of $X$ is at most $(d-1)^2$ (or something like that). So if the genus is large, there must be a map $X \to Y$ such that any map of degree $d$ from $X$ to $\mathbb{P}^1$ factors through $Y$. If $d$ is prime, this cannot happen, hence your fact 3.

  • $\begingroup$ A $g^1_d$ is not exactly a map to $\mathbb{P}^1$ of degree $d$ (since it can have basepoints)... But nice answer! $\endgroup$
    – naf
    Aug 20, 2011 at 12:49
  • $\begingroup$ What is your curve $Y$ in the second paragraph? $\endgroup$ Aug 20, 2011 at 13:16
  • $\begingroup$ @Ariyan: The image of $X$ in $\mathbb{P}^1\times\mathbb{P}^1$, if the map is not injective, which will happen if the genus of $X$ is large enough. If you want to consider all maps of degree $d$ then you need to prove that there is a $Y$ that works simultaneously for all of them. $\endgroup$ Aug 20, 2011 at 13:38
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    $\begingroup$ In your example the gonality is precisely 6 for $g$ large enough. Assume the curve has a $g^1_d$ with $d<6$. Then you get a birational map $C\to D\subset {\mathbb P}^1\times X=:S$. By construction $DK_S=8−2d\le 8$ and by the index theorem $2D^2\le(2+d)^2\le 49$. So $p_a(D)\ge g$ is bounded. $\endgroup$
    – rita
    Aug 20, 2011 at 18:29
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    $\begingroup$ Sorry for the messy notation. $C$ is a curve of genus $g$ which is a double cover of a plane quartic $X$, $D$ is the image of $C$ via the map $f\times g$, where $f\colon C\to X$ is the double cover and $g\colon C\to {\mathbb P}^1$ is the $g^1_d$. The Hodge index theorem is applied to $D$ and $\{pt\}\times X+{\mathbb P}^1\times\{pt\}$. $\endgroup$
    – rita
    Aug 20, 2011 at 20:07

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