# Non-hyperelliptic curves of genus at least two

A hyperelliptic curve can be understood as the set of points satisfying an equation of the form $$\displaystyle z^2 = f(x,y),$$ where $f(x,y)$ is a binary form of degree $d = 2g+2$. In this case, $g$ can be shown to be the genus of the curve.

It turns out that when $g=1$, the curve is an elliptic curve and in fact all curves of genus 1 curves over $\mathbb{C}$. However, this is not the case for higher genus; in particular, most curves of genus $g \geq 2$ are not hyperelliptic,

My questions are:

1) Is there a description, for a fixed genus $g$, an infinite family of non-hyperelliptic curves of genus $g$? By that I mean an equation similar to the one above that characterizes all curves in the family.

2) Bhargava recently proved that a proportion tending to 100 percent as $g \rightarrow \infty$ of hyperelliptic curves have no rational points and morever, the failure is accounted for by the Brauer-Manin obstruction. I believe that it is conjectured that all curves either satisfy the Hasse principle or if it fails the Hasse principle, the failure is accounted for by the Brauer-Manin obstruction. Since 'most' genus $g \geq 2$ curves are not hyperelliptic, Bhargava's theorem does not account for the situation for all genus $g$ curves. What is the best known result on the density of general curves of fixed genus $g$ with no rational points?

Thanks for any insight.

• – Felipe Voloch Jul 5 '14 at 3:15
• If $f(x,y)$ is not squarefree the genus is not as you claim. – joro Jul 5 '14 at 7:27

Perhaps a couple of things should be pointed out.

1) All curves of genus $g=2$ are hyperelliptic.

2) Your equation $z^2=f(x, y)$ is not the equation of an hyperelliptic curve. If you take $f(x, y)$ to be a binary form of degree $2g+2$ then the (projective) equation of the curve is $$z^2 y^{2g} = f(x, y),$$ where the discriminant $\Delta (f) \neq 0$.

3) A natural generalization of hyperelliptic curves are superelliptic curves, i.e, curves which have affine equation $y^n=f(x)$.

Perhaps, one can attempt to investigate what Bhargava type statements mean for superelliptic curves first or at least fix the "type" of equation. For example, take non-hyperelliptic curves of genus 3, which are ternary quartics.

• Regarding your point (2), I had assumed that this was an equation in weighted projective space, where $z$ has weight $g+1$ and $x$ and $y$ have weight 1. Then I think that one does get a hyperelliptic curve. Personally, I prefer to take two affine curves $z^2 = f(x,1)$ and $w^2=f(1,y)$ and glue them together, but that's a matter of taste. – Joe Silverman Jul 7 '14 at 0:00

Here is an example for question (1). Consider curves of the form $y^3=f(x, z)$, where $f$ is separable of degree $d$. These have genus $d-2$ by Riemann-Hurwitz. But if $d>4$, they cannot be hyperelliptic, for if they were they would map to $\mathbb{P}^1\times \mathbb{P}^1$ via the obvious $3:1$ cover of $\mathbb{P}^1$ (project to $x$) as well as the $2:1$ cover coming from hyperellipticity, with image a $(3,2)$ curve. The map is birational onto its image, which has arithmetic genus $2$. But the geometric genus must be less than or equal to the arithmetic genus, hence $$d-2\leq 2.$$

• A $(3,2)$ curve has genus $2$, I think. So this works for any $g>2$. – Will Sawin Jul 7 '14 at 4:04
• Oops, you're right of course. The easy way to see it is to degenerate to a line arrangement. – Daniel Litt Jul 7 '14 at 7:14