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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.

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    $\begingroup$ 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? $\endgroup$ – Asvin Oct 29 '20 at 1:00
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    $\begingroup$ @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? $\endgroup$ – Joe Silverman Oct 29 '20 at 1:30
  • $\begingroup$ 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. $\endgroup$ – Erik Walsberg Oct 29 '20 at 3:10
  • $\begingroup$ 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. $\endgroup$ – Asvin Oct 29 '20 at 6:33
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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.

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

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