I will state a very specific case: genus 5. Though it's particular, it admits a generalization to $M_g$, and I think reflects the nature of a general stratification of $M_g$.

It is known that if you have a genus five curve $C$ we've got the following disjoint familes:

-$C$ is hyperelliptic; meaning degree 2 map $\phi:C\rightarrow\mathbb{P}^1$.

-$C$ has a degree 3 map to $\phi:C\rightarrow\mathbb{P}^1$ (equivalently, it is a plane quintic with a node).

-$C$ has a degree 4 map to $\phi:C\rightarrow\mathbb{P}^1$.

Now, how are these facts related to the degree of the equation whose zero locus is $C$?.

Let me see If I'm reading off accurately the meaning of the situation above. We're saying implicitly that there exists a space where the degree of the equation defining $C$ is 2. There exists another space where the a degree 4 equation defines $C$, so on a so forth. does it sound all right?.

  • $\begingroup$ In projective spaces deg < 3 => genus = 0, deg < 4 => genus <2, deg < 5 => genus < 4. If you are not talking about projective spaces, then what degree are you talking about (the non connected and free part of the Picard group has to be 1-dimensional of course) ? $\endgroup$ – David Lehavi Feb 25 '10 at 13:01
  • $\begingroup$ David- I think degree in this case refers to the number of preimages a generic point in P^1 has under the map. $\endgroup$ – Ben Webster Feb 25 '10 at 14:13
  • $\begingroup$ That's right. I meant by degree "d", a d-branched cover of P^1. $\endgroup$ – Csar Lozano Huerta Feb 25 '10 at 16:32
  • $\begingroup$ I guess the tag "algebraic" is actually "ag.algebraic-geometry". I think I can retag it, but I never tried and I don't know if in this case it would be considered a good netiquette. $\endgroup$ – Andrea Ferretti Feb 25 '10 at 16:49

$M_5$ is pretty big. I'll do $M_3$, which should get across the same points.

In $M_3$, there are two strata:

The hyperelliptic locus, $H$, consists of those curves which have a degree $2$ map to $\mathbb{P}^1$.

The planar biquadrics, $B$, consist of curves which can be embedded in $\mathbb{P}^2$ with degree $4$. Such a curve has many different degree $3$ maps to $\mathbb{P}^1$; projection from any of its points gives such a map.

Every curve in $M_3$ is in precisely one of $B$ and $H$.

There are several points I want to make:

(a) The topology on $M_3$ is not the disjoint union of $B$ and $H$. Rather, $B$ is dense and open in $M_3$, and $H$ is a hypersurface.

(b) There is no notion of "the defining equation" of a curve. The curves in $B$ can be embedded in $\mathbb{P}^2$ by degree $4$ equations; but they can also be embedded in $\mathbb{P}^3$ with various larger degrees and they can be embedded in the plane with higher degrees if you allow nodes.

The issue is more dramatic for curves in $B$. They can be embedded in $\mathbb{P}^1 \times \mathbb{P}^1$, with degree $(2,4)$. To embed them in $\mathbb{P}^2$, you have to allow nodes (or worse singularities); I'm not sure what the lowest degree you can get away with is there.

(c) It is interesting to see how a curve in $B$ degenerates to a hyperelliptic curve. Let $F_t$ be a family of degree $4$ curves in $\mathbb{P}^2$, parameterized by $t$ in a disc $\Delta$. Let $C \to \Delta \setminus \{ 0 \}$ be the corresponding family of abstract curves, and suppose the limit in $M_3$ as $t \to 0$ is a hyperelliptic curve. If we normalize things properly, we can take $F_0 = Q^2$ for a conic $Q$. Let $F_t = Q^2 + t G + \cdots$.

Let $\Delta'$ be the branched cover $t=u^2$ of $\Delta$. Then we can complete the family $C' \to \Delta' \setminus \{ 0 \}$ to a family $\tilde{C} \to \Delta'$ whose fiber over $0$ is the hyperelliptic curve in question.

The family $\tilde{C}$ maps to $\mathbb{P}^2$. For $u \neq 0$, the curve $C'_u$ maps to $\{ F_{u^2}=0 \}$. At $u=0$, the hyperelliptic curve double covers $Q$, ramified over the $8$ points $\{ Q=G=0 \}$.

(d) Similarly, we can take a family of genus $3$ curves with degree $3$ maps to $\mathbb{P}^1$ and take the limit of that family, in the sense of stable maps. I'm finding this limit a bit difficult. I think you get a curve with two components, $X \cup Y$, glued along a single node; where $X$ is hyperelliptic of genus $3$ and double covers $\mathbb{P}^1$ while $Y$ is genus $0$ and maps isomorphically to $\mathbb{P}^1$.

The case of genus 5 will be similar. There will be a generic situation, which in this case is degree $8$ curves in $\mathbb{P}^4$. (And those curves do, indeed, have degree $4$ maps to $\mathbb{P}^1$.) The other strata will be contained in the closure of it.

There is no one notion of "the defining equation" or "the degree" of an abstract curve. However, given a particular family of curves mapping to $\mathbb{P}^k$, we can take the limit in the sense of stable maps and get a stable map, one of whose components will be the limiting abstract curve.

  • 1
    $\begingroup$ In b - the genus of a (2,8) curve on P^1 x P^1 is 7. I think you mean bidegree (2,4) for genus 3. (Looking at the "three degree four polynomials" A, B, C involved in the definition of such a curve, the branch points of the projection are defined by the vanishing of the discrimant A^2 - BC). In d - is it clear the stable map has to degenerate? A hyperelliptic curve also may be mapped 3:1 to P^1. $\endgroup$ – mdeland Feb 26 '10 at 18:56
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    $\begingroup$ David - in your case (d), it is easiest to look at it as a limit in the Hurwitz scheme, where what happens is exactly what you described, but for "combinatorial" reasons: Look at the underlying M_0,n (the n points are marking simple branch points); when 2 branch points collide, you "sprout" an "infinitesimal" component, and the two points "move" to that component. Now you have to analyze the various 3-fold covers of two P^1's joined at a node, with given branch points, simple ramification over these points, and arbitrary ramification pattern over the node - but this is simple combinatorics $\endgroup$ – David Lehavi Feb 27 '10 at 19:25
  • $\begingroup$ You're right, (2,4) it is. I'll edit. $\endgroup$ – David E Speyer Feb 27 '10 at 22:45
  • $\begingroup$ I Fixed a typo. $\endgroup$ – David E Speyer Apr 2 '10 at 12:57

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