# hyperelliptic stable genus four curve

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.
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