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Let $M_{3,1}$ be the (coarse, non-compactified) moduli space of genus $3$ curves with a marked point over a field $k$ of characteristic zero. Throwing away the hyperelliptic curves, take the open subset $M_{3,1}^{nh}$ of non-hyperelliptic curves, which under the canonical embedding correspond to smooth plane quartics. The variety $M_{3,1}^{nh}$ has a stratification: $$M_{3,1}^{nh}= M_{3,1}^{generic} \sqcup M_{3,1}^{bitangent} \sqcup M_{3,1}^{flex}\sqcup M_{3,1}^{hyperflex} .$$ Here the superscript denotes the behaviour of the marked point $P$ under the canonical embedding. (For example, $P$ is a hyperflex if and only if $4P$ is a canonical divisor.) It is known by work of Looijenga ('Cohomology of M_3 and M_{3,1}') that each stratum is unirational.

Let $J \rightarrow M_{3,1}^{nh}$ be the universal Jacobian over $M_{3,1}^{nh}$. For $D$ in $$\{generic, bitangent, flex, hyperflex\},$$ write $J^D \rightarrow M_{3,1}^D$ for the pullback to the corresponding stratum.

Question. Is $J^D$ a unirational variety?

I'm especially interested in the hyperflex case. Any hints or references would be appreciated.

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    $\begingroup$ Jacobian of a genus 3 curve is birational to its symmetric cube via the Abel-Jacobi map. Thus J itself is unirational because it is dominated by a rational variety parametrizing four points on $\mathbb{P}^2$ and a degree four curve passing through these points. The hyperflex condition is a linear condition on the configuration above, so the same argument seems to show that $J^{hyperflex}$ is also unirational. $\endgroup$
    – Evgeny Shinder
    Commented Jan 2, 2020 at 22:35
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    $\begingroup$ Furthermore, since 4 general points in $\mathbb{P}^2$ can be sent to $[0:0:1]$, $[0:1:0]$, $[1:0:0]$, $[1:1:1]$ by $PGL_3$-action, the universal Jacobians we consider are birational to quotients of projective spaces (degree four equations with some linear conditions on the coefficients) by a linear action of the symmetric group $S_3$. I think one can then show that these moduli spaces are in fact RATIONAL, not just unirational. $\endgroup$
    – Evgeny Shinder
    Commented Jan 3, 2020 at 12:21
  • $\begingroup$ Thanks for the comments, they make a lot of sense! $\endgroup$
    – Jef
    Commented Jan 5, 2020 at 14:48

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