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Is it possible to construct a nonisotrivial family of genus four curves $X \rightarrow S$, with the following properties:

(1) $S$ is a complete curve;

(2) All the fibers are smooth;

(3) The generic fiber lies on a singular quadratic in $\mathbb{P}^3$.

It is not possible to construct such a family if we replace (3) by "all the fibers...". But the problem is that there could be smooth hyperelliptic reductions.

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Nitpick: You haven't ruled out the constant family, or more generally isotrivial families. I assume you want to impose that the map $S \to M_4$ be nonconstant. –  David Speyer Apr 3 '12 at 3:11
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There are two similar and simpler questions: (1) You could ask for a complete family of genus $3$ curves, most of which are degree 4 curves in $P^2$ and some of which are hyperelliptic (2) you could ask for a complete family of genus $4$ curves, most of which lie on a nonsingular quadratic in $P^3$ and some of which are hyperelliptic. Do you already know the answers to these? –  David Speyer Apr 3 '12 at 3:13
    
I don't know if this will interest you, but there are non isotrivial families of curves of genus $6$ over a smooth complete curve - a special case of the so-called Kodaira fibrations. See Astérisque 127, Exp. X, Prop. 1. –  Damian Rössler Apr 3 '12 at 8:09
    
Thanks for the comment. Concerning the genus three curves, there is such a family like that. I'm not sure of the genus four family. But that seems to be easier than the original question. A naive idea to the original question would be take a complete surface in $M_4$, and the intersection with the Petri locus gives the desired family. But I've just found out yesterday that we don't know if such a complete surface exists. –  xuehang Apr 4 '12 at 15:05
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I know one very explicit construction of a complete curve in $M_4$, which I learned from Chris Zaal's thesis. You start by fixing a single curve $X$ of genus two. One can find a complete curve $S$ and two disjoint nonconstant sections of the trivial family $X \times S \to S$. Choosing a double cover branched over the two sections (possibly after replacing $S$ by a finite étale base change) gives a nonconstant family of genus four curves over $S$. I don't know if the resulting genus four curves will lie on a singular quadratic though. I can give more details if you want. –  Dan Petersen Apr 4 '12 at 15:20
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1 Answer

No, there is no such $S$:

EDIT: (BIG) GAP BELOW I compute limits of certain linear series in the Hurwitz scheme, and then I make claims about limits of other (bigger) linear systems taken over curves in $\mathcal{M}_3$. However, the limit curves in the Hurwitz scheme are not stable (the first of them is analyzed in Donagi's paper quoted below in Example 2.10 (iii) ).

The trigonal construction (see Donagi's "the fibers of the prym map" - sadly the arxiv version is without the pretty pictures) gives you a 1-1 correspondence between:

  • Smooth genus $4$ curves $C$, a non trivial 2-torsion point $\alpha\in JC[2]$, and a base point free $g^1_3$ on $C$.

  • Smooth genus $3$ curve $X$ -- which is the Prym curve of the pair $(C,\alpha)$ -- with a $g^1_4$, such that no fiber of a the $g^1_4$ is of the form $2(p+q)$.

Assuming that this $g^1_4$ is not a canonical pencil, it is a complete linear system $|K_X+\beta|$. One then verifies that there is a natural isomorphism between the unramified double cover of $C$ associated with $\alpha$, and

$$\overline{X\times_{|K_X+\beta|} X\setminus\mathrm{diagonal}}/(x,y)\sim(y,x)\cong \Theta_X \cap (\Theta_X+\beta),$$

which means that the inverse of the trigonal construction on $|K_X-\beta|$ gives $C$ with a different $g^1_3$. Which means that under the 1-1 correspondence above, $C$s sitting on a singular quadric correspond to the case where the $g^1_4$ is canonical.

We now turn to the case where $C$ is hyperelliptic. One should work carefully on the Horwitz schemes on both sides of the correspondence (essentially using the bigonal construction dictionary from Donagi's paper, and using the fact that the intermediate Abelian varieties in the two sides of the bigonal construction are dual), but the upshot is that on the $X$ side we would have a $g^1_4$ on the prym of a hyperelliptic curve, which splits to two cases (denoting by $H_C$ the hyperelliptic class on $C$):

  • $\alpha+H_C$ is the sum of two Wierstrass points $w_1,w_2$, in which case $X$ is hyperelliptic. One may identify $|H_X|$ and $|H_C|$, such that the residual points to $w_1, w_2$ correspond to the Weierstrass points of $X$ (there is an exercise about it in ACGH, in the chapter on Prym varieties).

  • $\alpha+2H_C$ is the sum of four Weirstrass points $w_1,\ldots, w_4$, in which case $X$ is reducible: it is the "gluing" of genus $1$ and a genus $2$ curve. Similarly to above (but with the dualizing system on the $X$ side and the canonical system on the $C$ side), $w_1, \ldots w_4$ corresponds to the ramification points of the $g^1_2$ on the genus $1$ component, and the residual $6$ points correspond to the ramification points of the $g^1_2$ on the genus $2$ component.

Note that the first case here becomes closer to the second if we glue a rational tail to the genus $3$ curve. Indeed, the resulting curve is not longer semi stable, but we work with Hurwitz schemes, so by and large, not much harm is done.

Armed with all this, we now take the curve $S$, throw the hyperelliptic points, lift it from $\mathcal{M}_4$ to $\mathcal{R}_4$ (i.e. adding a choice of a non trivial 2-torsion point), apply the construction to get a family of genus $3$ curves together with canonical pencils, and lift the curve from $\mathcal{M}_3$ to $\mathcal{M}_3(2)$, (adding full level structure). We now have coordinates on the universal canonical system (e.g. by fixing the "first" four bitangents). I.e. we have a non projective curve $S'$, with

  • A surface $\tilde{X}\subset \mathbb{P}^2\times S'$: the universal $X$.

  • A curve $P\subset \mathbb{P}^2\times S'$: the universal point such that for every $s\in S'$ the projection of $\tilde{X}_s$ from $p_s$ to $\mathbb{P}^2$ is the fiber's $g^1_4$.

We can now close $S', \tilde{X}$ and $P$. If you believe the analysis of the hyperelliptic cases, then the limits in both cases are two double lines (representing the two components of $X$, which may be either genus $0$ and $3$, or genus $1$ and $2$), and where the limit of $P$ over this fiber does not hit any of the lines.

I.e. we now have a surface of bidegree $(4, d_1)$, and a curve $(1, d_2)$ inside $\mathbb{P}^2\times\overline{S'}$ which do not intersect; which is impossible.

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Thank you very much for your answer. One consequence of this nonexistence is that there is no projective surface in the $M_4$ (fine moduli). Otherwise the intersection of this surface and the closure Petri locus will give a family like this. Is this right? –  xuehang Feb 20 '13 at 3:26
    
I suspected thos is what you are after, but i dont see it , care to elaboraye. ? –  David Lehavi Feb 20 '13 at 6:02
    
Also, chances are thar if you csn complete the argument, my answer is wrong.... –  David Lehavi Feb 20 '13 at 6:35
    
The nonexistence of a surface in $M_4$ was my original motivation of this question. My idea is that the closure of the Petri locus (the locus of curves which lies on the singular cone) is an ample divisor in $M_4$. If we had a surface in $M_4$, then it must intersect with the Petri locus. This will (after possibly desingularization of the intersection) give a family of curves with the property described in the question. Previously I thought I need this nonexistence for my paper, but now I realize that I don't need this anymore... –  xuehang Feb 20 '13 at 19:11
    
My skepticism here is because I tried several similar arguments a few years back, and they all collapsed (it was years back - so I don't remember if I tried the Petri locus, and I don't remember if the degenerations I used were the same): - Is the Petri locus is ample in $\overline{\mathcal{M}_4}$ or in $\mathcal{M}_4$ ? - I edited my response above as there is a gap in my argument (you have to verify that the limits of the universal dualizing systems are the same ones as in the Hurwitz schemes: a) this looks like a lot of work, and b) I'm not sure it's correct. –  David Lehavi Feb 20 '13 at 22:24
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