# Is the set of hyperelliptic curves with a K-point closed?

I am actually interested in the same question for more general kinds of curves, but I will be specific.

Let $$K$$ be a field and $$\overline{K}$$ be an algebraic closure of $$K$$. Let's say that a "hyperelliptic curve" is a smooth projective $$K$$-curve $$C$$ of genus $$\ge 2$$ such that there is a degree $$2$$ morphism $$C_{\overline{K}} \to \mathbb{P}^1_{\overline{K}}$$. Let $$M_g$$ be the (coarse) moduli space of hyperelliptic curves of genus $$g$$. Given $$a \in M_g(K)$$ let $$C_a$$ be the corresponding hyperelliptic $$K$$-curve. Let $$X$$ be the set of $$a \in M_g(K)$$ such that $$C_a$$ has a $$K$$-point.

Now suppose that $$K$$ is a local field or maybe $$\mathbb{C}((t))$$ and equip $$M_g(K)$$ with the resulting topology. Is $$X$$ a closed subset of $$M_g(K)$$? This seems like it should be true.

• I don't understand this very well: I think on the coarse moduli space, all the twists of a hyperelliptic curve (so of the form $dy^2 = f(x)$ for varying d) correspond to the same point and having $K$ point depends on which twist you take (?). So the question doesn't seem to be well defined as is. Perhaps you want to say that some twist has a $K$ point? – Asvin Oct 29 '20 at 1:00
• @Asvin But there's always a twist with a rational point, just take your favorite $x_0,y_0\in K$ and set $d=f(x_0)/y_0^2$. Maybe one could instead work on the moduli stack, instead of the moduli space? – Joe Silverman Oct 29 '20 at 1:30
• Does the same thing always happen for the coarse moduli space of genus $g$ curves? If that's the case then I probably need to work with something that is either more sophisticated or less sophisticated. – Erik Walsberg Oct 29 '20 at 3:10
• I think there will always exist some points for which stuff like this happens. It's related to the aut group of the curve being non trivial, I am not sure how "non generic" a non trivial automorphism group is. – Asvin Oct 29 '20 at 6:33

In the "more sophisticated" direction, we can ask a similar question about the moduli stack $$\mathscr{M}_g$$ of hyperelliptic curves of genus $$g$$. If $$K$$ is a topological field, there is a natural topology on the set $$\vert\mathscr{M}_g(K)\vert$$ of isomorphism classes of of genus $$g$$ hyperelliptic curves over $$K$$: a subset $$\Omega$$ of $$\vert\mathscr{M}_g(K)\vert$$ is open if for every family $$f:C\to S$$ of genus $$g$$ hyperelliptic curves over a $$K$$-variety $$S$$, the set $$\Omega(f):=\left\{s\in S(K)\mid C_s\in\Omega\right\}$$ is open in $$S(K)$$. If you take for $$\Omega$$ the set of curves with a rational point, then $$\Omega(f)$$ is just the image of $$f(K):C(K)\to S(K)$$.

Assume now that $$K$$ is a valued field, with completion $$\widehat{K}$$. Here is what I know:

1. If $$K$$ is henselian, the map $$f(K)$$ is open because $$f$$ is smooth, so $$\Omega(f)$$ is open.
2. If $$K$$ is a local field (i.e. locally compact) then $$f(K)$$ is topologically proper (because $$f$$ is proper) and in particular closed, so $$\Omega(f)$$ is closed.
3. For henselian $$K$$ it is not true in general that $$f(K)$$ is a closed map. However, if $$\widehat{K}/K$$ is a separable extension (e.g. if $$K$$ is complete, or has characteristic zero) then $$f(K)$$ has closed image. This follows from the "strong approximation property", see [3], Theorem 1.3.

All this works for other moduli problems, or when $$K$$ is a field with an archimedean absolute value, in which case "henselian" means "algebraically closed or real closed", and "local" means $$\mathbb{R}$$ or $$\mathbb{C}$$.

Of course, in case a fine moduli scheme $$M$$ exists, the meaning of "open" (resp. "closed") is the naive one, as formulated in the question. This is the case for instance for the moduli $$U_g$$ of curves of genus $$g\geq3$$ without nontrivial automorphisms.

For general facts on topologizing points of stacks, see:
[1] L. Moret-Bailly, Problèmes de Skolem sur les champs algébriques, Compositio Math. 125(1) (2001), 1–30; doi:10.1023/A:1002686625404.
[2] K. Cesnavicius, Topology on cohomology of local fields, Forum of Mathematics (2015) https://doi.org/10.1017/fms.2015.18
For the strong approximation property as used above:
[3] L. Moret-Bailly, An extension of Greenberg’s theorem to general valuation rings, Manuscripta Math (2011), doi:10.1007/s00229-011-0510-5.

• Thanks! This is useful – Erik Walsberg Oct 29 '20 at 20:41