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Let $X$ be a stable hyperelliptic genus four curve (meaning on the moduli space it lies in the closure of the hyperelliptic locus) and $L$ a line bundle on $X$. Suppose that the pair $(X,L)$ comes from the degeneration of a $\mathfrak{g}_3^1$ on a smooth curve. This means the following. There is a DVR $R$ and a stable curve $\mathcal{X} \to Spec R$ and a line bundle $\mathcal{L}$ on $\mathcal{X}$. The special fiber of $\mathcal{X}$ is $X$ and the restriction of $\mathcal{L}$ to the special fiber is $L$. The generic fiber is a smooth genus four curve $Y$ and the restriction of $\mathcal{L}$ to $Y$ satisfies $h^0(Y, \mathcal{L}|_Y) \geq 2$.

The question is: can we take $\mathcal{X}$ to be hyperelliptic, i.e. the generic fiber $Y$ is also hyperellitpic (and smooth)?

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There's something I don't understand. If you say $\mathfrak{g}^1_3$ it means a linear system of dimension 1 and degree 3, so $h^0=2$. Probably you should re-write the question. As it stands it is difficult to understand what you want. – IMeasy Apr 16 '13 at 19:27
In fact, on any smooth genus four curve, a degree three line bundle has at most two linearly independent global sections. So $h^0=2$ and $ h^0\geq 2$ make no difference. – marker Apr 17 '13 at 2:39
ok, that's right – IMeasy Apr 18 '13 at 13:52
My suggestion is to do a case-by-case analysis. Using the theory of admissible covers, you can explicitly list all the dual graphs that can arise from a limit of a hyperelliptic curve. I went through a couple of these and proved that every "smoothable" $\mathfrak{g}^1_3$ is a limit of a "hyperelliptic" $\mathfrak{g}^1_3$. – Jason Starr Apr 18 '13 at 15:16
Yeah, this is what I did for a couple of cases. And that's how I made such an expectation. – marker Apr 18 '13 at 19:58

All smooth genus four curves have two $g^1_3$. Since the unviersal picard fibration over the moduli space of stable genus four curves is proper, yes you can assume that the generic fiber is hyperelliptic.

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This argument is not correct. Although all smooth genus $4$ curves have "at least two" $\mathfrak{g}^1_3$s, or more precisely, a length $2$ subscheme of $\text{Pic}^3_X$ inside $W^1_3(X)$, nonetheless, every smooth hyperelliptic curve has infinitely many $\mathfrak{g}^1_3$s. More precisely, for a smooth, hyperelliptic curve $X$, $W^1_3(X)$ is a copy of $X$ (embedded by the Abel map, and then translated). All of these $\mathfrak{g}^1_3$s come from the hyperelliptic $\mathfrak{g}^1_2$ by adding a base point. – Jason Starr Apr 18 '13 at 15:08

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