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Given a smooth genus 2 curve $C$, it is canonically a two-fold cover of a $\mathbb{P}^1$, branched at six points. Allowing stable curves allows degenerations in two directions: the six points are allowed to collide (in certain ways), and $\mathbb{P}^1$ can "break" into a degenerate conic in $\mathbb{P}^2$ (here a union of two distinct lines). For instance, a smooth genus one curve with two of its points glued together is an example of the former situation, while two genus one curves joined to each other along a point is an example of the latter.

It is possible to construct explicit families (ideally, over high-dimensional bases and without appealing to stable reduction results) of stable genus 2 curves that include degenerations of both kinds? (I would even be interested in degeneration of smooth curves to the second locus of unions of elliptic curves.) One difficulty here is that, if $C$ is a union of elliptic curves, then the map from $C$ to the quotient of $C$ by its "hyperelliptic" involution (multiplication by $-1$ on each elliptic curve) is not flat at the singular point.

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Have you tried taking a linear subspace $\Lambda$ of the $\mathbb{P}^{5}$ of plane conics which avoids double lines (e.g. $\Lambda$ a general element of $\mathbb{G}(2,|\mathcal{O}_{\mathbb{P}^{2}}(2)|)$), looking at the total family $\mathcal{C} \rightarrow \Lambda,$ and taking a double cover of $\mathcal{C}$ branched over a section of the divisor that comes from the intersection of all these plane conics with a fixed plane cubic? I'm pretty sure you could arrange for the latter to be tangent to some of the conics at exactly one point, and to miss the node of most of the singular conics. –  Yusuf Mustopa Jun 26 '13 at 19:45
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@YusufMustopa: I don't think this can work, because the two-fold cover from $C$ to two $\mathbb{P}^1$'s glued together at a node is not flat, which seems to make it difficult to construct such families by such a method. (I'll add this to the original question.) –  Akhil Mathew Jun 26 '13 at 20:12
    
Perhaps more seriously, the singular fibers in the construction I wrote down do not seem to give the union of two elliptic curves with a common point. If the plane cubic in question avoids the node of the singular conic, then the branched double cover I cooked up actually has 2 nodes and not 1, and if it does hit said node, then the preimage of each line cannot be an elliptic curve. It seems I was being quite careless :) –  Yusuf Mustopa Jun 26 '13 at 20:46
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As Akhil points out, you can have both types of behavior simultaneously. For instance, consider a curve $C$ that is a union of two irreducible components $C'$ and $C''$ that intersect transversally at a single point $p$ (smooth on each component). Let $C'$ be a smooth, genus $1$ curve. Let $C''$ be a nodal genus $1$ curve with geometric genus $0$. Let $\nu':C'\to B'$ be a double cover of a genus $0$ curve ramified at $p$. Let $\nu'':C''\to B''$ be a double cover of a genus $0$ curve ramified at $p$. Then there is a unique at-worst-nodal curve $B=B'\cup B''$ of genus $0$, glued together at $\nu'(p)=\nu''(p)$. Also there is a unique morphism $\nu:C\to B$ whose restriction to $C'$ is $\nu'$ and whose restriction to $C''$ is $\nu''$.

In some sense, this already shows how to produce explicit families: deform the pointed curve $(B,D)$, where $D$ is the union of the branch points of $\nu$ (not counting $p$). This can be made very explicit: first, identify $B$ with $(B'\times\{p\})\cup (\{p\}\times B'')$ inside $B'\times B''$. Now deform this curve in a general pencil of divisors $(\mathcal{B}_s)_{s\in S}$ in $B'\times B''$. The divisor $D$ is the intersection in $B'\times B''$ of $B$ and a divisor $E$ of type $(3,3)$. So also deform $E$ in a general pencil $(\mathcal{E}_t)_{t\in T}$ of divisors of type $(3,3)$ in $B'\times B''$. Now, over the base $S\times T$, consider the family of curves $\mathcal{B}_s$ with divisors $\mathcal{D}_{s,t} = \mathcal{B}_s \cap \mathcal{E}_t$.

The only issue is that, in order to explicitly form the double cover $\mathcal{C}_{s,t}\to \mathcal{B}_t$ branched over $\mathcal{D}_{s,t}$, we need an invertible sheaf on $\mathcal{B}_t$ and an explicit isomorphism of this invertible sheaf with $\mathcal{O}_{\mathcal{B}_t}(\mathcal{D}_{s,t})$. We can form this ideal sheaf on the generic fiber. However, filling in the branched cover almost certainly will require passing to a degree $2$ cover of $S\times T$ (at least, that is my recollection).

Edit. I can be a little more explicit. For a general pencil $S$ as above, there are precisely two singular members of the pencil, i.e., there are two $k$-rational points, $0,\infty \in S$, such that for $S^* = S\setminus\{0,\infty\}$, the restricted morphism $\pi^*:\mathcal{B}^* \to S^*$ is smooth and projective. Moreover, there are sections: the two base points of the pencil. So $\mathcal{B}^*\to S^*$ is the projectivization of a rank $2$, locally free sheaf on $S^*$. Note that $S^*$ is isomorphic to $\mathbb{G}_m$, and thus has trivial Picard group. Therefore, $\text{Pic}(\mathcal{B}^*)$ is a free Abelian group of rank 1, with an ample generator $\mathcal{A}$ that has degree $1$ on the geometric fibers of $\pi^*$. In particular, the pullback from $B'\times B''$ of $\mathcal{O}(3,3)$ is isomorphic to $\mathcal{A}^{\otimes 6}$. Therefore, over $S^*\times T$, the ample invertible sheaf we are trying to "halve" is $\mathcal{A}^{\otimes 6}\boxtimes \mathcal{O}_T(1).$ Thus, over each of the two standard open affines in $T$, there is a square root of $\mathcal{O}_{\mathcal{B}^*\times T}(\mathcal{D}^*)$. Hence, over each open affine in $T$, there is also the associated double cover of $\nu^*:\mathcal{C}^* \to \mathcal{B}^*\times T$ branched over $\mathcal{D}^*$. Gluing these together gives a branched cover over all of $\mathcal{B}^*\times T$.

The "discriminant divisor" $\Delta$ of $\mathcal{D} \to S\times T$ is the divisor over which the fibers of the Cartier divisors $\mathcal{B} \times T$ and $S \times \mathcal{E}$ in $(S\times T)\times(B'\times B'')$ are tangent (one can prove that $\Delta$ is a Cartier divisor using sheaves of relative differentials and the "norm" of the finite flat map $\mathcal{D}\to S \times T$). If $T$ is a sufficiently general pencil, then $\Delta$ is a smooth Cartier divisor away from finitely many points of $S\times T$, none of which is the special point $(0,0)$ parameterizing $(B,D)$, and $\Delta$ intersects $\{0\}\times T$ transversally at $(0,0)$. Thus, after deleting finitely many points from $S\times T$ (but not the special point), we may assume that $\Delta \cup \{0,\infty\}\times T$ is a simple normal crossings divisor in $S\times T$.

The family of curves, $\mathcal{C}^*\to S^*\times T$ is a family of smooth curves on the complement of $\Delta$. The fibers over geometric points of $\Delta$ are irreducible curves of arithmetic genus that have a single ordinary double point.

The family of stable curves $\mathcal{C}^*$ over $S^*\times T$ does not extend to $S\times T$. However, it does extend to codimension $1$ points after a ramified base change of order $2$. More precisely, let $u:\widetilde{S}\to S$ be the degree $2$, finite flat morphism that is &eeacute;tale over $S^*$ and is ramified at $0$ and $\infty$. Then the pullback of $\mathcal{C}^*$ to $\widetilde{S}^*\times T$ does extend to codimension $1$ points of $\widetilde{S}\times T$.

Moreover, the pullback $\widetilde{\Delta}$ of $\Delta$ to $\widetilde{S}\times T$ is still transverse to $\{0,\infty\}\times T$ at the special point $(0,0)$, and away from all but finitely many points. Thus, deleting those finitely many points, the full discriminant locus, $\widetilde{\Delta} \cup (\{0,\infty\} \times T)$, is a simple normal crossings divisor in $\widetilde{S}\times T$. Thus, now we can apply the main theorem of de Jong - Oort, "Extending families of curves", to conclude that the family of stable curves extends to a family $\mathcal{C}$ over all of $\widetilde{S}\times T$ (less the finitely many points we deleted).

Since $\widetilde{S}\times T$ is $\mathbb{P}^1\times \mathbb{P}^1$, I think this is about as explicit as possible. One thing to note: for the morphism $\widetilde{\nu}:\widetilde{C}\to \widetilde{B}$, the morphism does fail to be flat over the codimension $2$ locus of "nodes" of $\widetilde{B}$ that projects isomorphically to $\{0,\infty\}\times T$ inside $\widetilde{S}\times T$. So the coherent sheaf $\widetilde{\nu}_*\mathcal{O}_{\widetilde{C}}/\mathcal{O}_{\widetilde{B}}$ is not an invertible sheaf: it is a pure, torsion-free sheaf of rank $1$. It should not be that difficult to work out this sheaf. This would make the family $\widetilde{\mathcal{C}}$ even more explicit.

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Thanks. I agree with that the double cover issue seems to be tricky. I'm not really sure how to do this in families. For instance, the hope was to get a family of genus two curves over the family of conics in $\mathbb{P}^2$ allowing degenerations of the conic, but an obstacle is that this is not a Zariski locally trivial $\mathbb{P}^1$-bundle (as the projectivization of the sections of the canonical bundle must be). –  Akhil Mathew Jun 27 '13 at 3:23
    
Great. I need to think a bit more to understand this, but I think this is exactly what I was looking for. –  Akhil Mathew Jun 28 '13 at 3:09
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An explicit example of such a family with $1$-dimensional basis can be found in my paper On surfaces of general type with $p_g=q=1, K^2=3$. The relevant statement is Proposition 6.1, that I will restate here for the reader's convenience.

Proposition. There exists exactly one irreducible family of minimal surfaces $S$ of general type with $p_g=q=1$, $K^2=3$ and a rational pencil $|G|$ of curves of genus $2$, and this family is parametrized by the coarse moduli space of elliptic curves. Moreover:

  1. the pencil $|G|$ is base point free, hence it defines a fibration $f \colon S \to \mathbb{P}^1$ whose general fibre is a smooth genus $2$ curve;
  2. $|G|$ is the only genus $2$ pencil on $S$;
  3. $|G|$ contains $13$ singular elements; six of these are genus $1$ curves with an ordinary double point, the other seven consist of two smooth elliptic curves intersecting transversally at a single point.

As far as I know this family was first constructed by Xiao Gang in his book Surfaces fibrées en courbes de genre deux (Lecture Notes in Mathematics 1137, Springer 1985), see in particular Chapter 3. The costruction follows from the classification of non-isotrivial genus $2$ fibrations $f \colon S \to C$ ($S$ smooth surface, $C$ smooth curve) whose associated Jacobian fibration has a fixed part $E \times C$.

In Xiao's book one can also find many other examples, see in particular the table at page $52$. The family of surfaces in the Proposition corresponds to the case $d=4$ in that table. These surfaces are explicitly mentioned in the Corollaire $4$, page $51$, where they are called fibrations $f(E, 4)$.

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Thanks for these references. –  Akhil Mathew Jun 28 '13 at 3:10
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I disagree with the premise of your question, but maybe I just do not understand it. When points collide (in the world of stable curves) a new component "bubbles off" - in other words, there is no difference between points colliding and the $\mathbb P^1$ breaking. So I think that if you want to understand degenerations of your curve in terms of what happens to the branch configuration "downstairs" then I think you should be looking at the theory of admissible covers.

If your 6-pointed $\mathbb P^1$ breaks into two pieces with three markings on each, then an admissible double cover branched at these points will have branching also at the node, so you get two elliptic curves glued at a point. If instead you have one piece with two markings and one with four, then the inverse image of the 2-pointed component is a $\mathbb P^1$ and the inverse image of the 4-pointed component is genus one curve, and these are now glued to each other at two points. This curve is not stable, but we can stabilize it, and we get an elliptic curve glued to itself.

Finally, what kind of explicit families are you looking for? Do you want to write them down in coordinates? And what is a "high-dimensional base" here -- a base of dimension more than 3 (= the dimension of the moduli space of genus two curves) seems a bit silly, right?

Edit. I just remembered a reference that may be useful for you: there is a paper of Avritzer and Lange in which they go through in detail why a stable hyperelliptic curve is the same as a an admissible double cover of a stable $(2g+2)$-pointed genus zero curve.

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I believe that, if the $\mathbb{P}^1$ is not allowed to break (but if branch points can collide), the only part of $\overline{M}_2$ will be reached is the "hyperelliptic" locus (the locus where the canonical bundle is generated by global sections), which leaves out in particular the locus I'm most interested in (unions of elliptic curves). I would like to write down families in coordinates, and I'm not too picky about "high" (say $\geq 2$). I wasn't familiar with the theory of admissible covers though, and it seems quite relevant. Thanks. –  Akhil Mathew Jun 26 '13 at 20:19
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