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Let $K$ be a field equipped with a non-Archimedean absolute value, for example $K=\mathbb{C}((t))$. An Abelian variety $A$ over $K$ is called maximally degenerate if it admits an analytic uniformisation by a torus, i.e. if there exists an isomorphism of analytic varieties $(\mathbb{G}_m^n)^{an}/\Lambda \cong A^{an}$ where $\Lambda$ is a lattice (a free subgroup of $\mathbb{G}_m^n$ which maps injectively into its tropicalization). Being maximally degenerate can more generally be defined as having a skeleton in the sense of Berkovich of dimension coinciding with the dimension of the variety.

Elliptic curves that are maximally degenerate are characterized as those having j-invariant of absolute value $> 1$. How can one describe the set of points in the moduli space of curves of genus $g \geq 2$, $\mathcal{M}_g(K)$, which corresponds to curves which have a maximally degenerate Jacobian? Is this set semi-algebraic?

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  • $\begingroup$ The moduli space in question in an open subset of the generic fiber of a projective (formal) scheme $\bar M$ over $\mathcal{O}_K$ (the Deligne-Mumford compactification). The special fiber $\bar M_k$ parametrizes stable curves over the residue field $k$, and there is a closed subset $Z\subseteq \bar M_k$ parametrizing those all of whose components have genus zero. My guess is that the locus you are looking at should be the tube (=preimage under the specialization map) of $Z$ (or rather its intersection with $M$). It is a non-quasicompact open subset. $\endgroup$ – Piotr Achinger Apr 26 '17 at 12:40
  • $\begingroup$ @PiotrAchinger so how this would manifest in the case of elliptic curves for example? how does one see that the subset parametrizing stable curves with only genus 0 components? (I presume this is standard, but could you please give an idea?) by the way, is there a reason that you have the intuition that this particular set of stable curves should be picked out, and not curves with rational irreducible components for example? $\endgroup$ – Dima Sustretov Apr 26 '17 at 13:16
  • $\begingroup$ In the case of elliptic curves, $|j|<1$ means that the curve reduces to a curve of $j$-invariant zero over $k$, so what you're getting is the (punctured) tube of the point zero on the $j$-line. But now I'm confused about the case $|j|>1$, when the point specializes to $\infty$ on the compactified $j$-line. $\endgroup$ – Piotr Achinger Apr 26 '17 at 13:30
  • $\begingroup$ @PiotrAchinger I have confused the inequality when speaking about the $j$-invariant: it is $|j| > 1$ that corresponds to maximal degeneration $\endgroup$ – Dima Sustretov Apr 26 '17 at 17:00

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