looking close at an example of Moduli space of curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:28:12Z http://mathoverflow.net/feeds/question/16376 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16376/looking-close-at-an-example-of-moduli-space-of-curves looking close at an example of Moduli space of curves Csar Lozano Huerta 2010-02-25T06:43:29Z 2010-04-02T12:57:23Z <p>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 $M_g$.</p> <p>It is known that if you have a genus five curve $C$ we've got the following disjoint familes:</p> <p>-$C$ is hyperelliptic; meaning degree 2 map $\phi:C\rightarrow\mathbb{P}^1$.</p> <p>-$C$ has a degree 3 map to $\phi:C\rightarrow\mathbb{P}^1$ (equivalently, it is a plane quintic with a node).</p> <p>-$C$ has a degree 4 map to $\phi:C\rightarrow\mathbb{P}^1$.</p> <p>Now, how are these facts related to the degree of the equation whose zero locus is $C$?. </p> <p>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?. </p> http://mathoverflow.net/questions/16376/looking-close-at-an-example-of-moduli-space-of-curves/16469#16469 Answer by David Speyer for looking close at an example of Moduli space of curves David Speyer 2010-02-26T03:05:39Z 2010-04-02T12:57:23Z <p>$M_5$ is pretty big. I'll do $M_3$, which should get across the same points.</p> <p>In $M_3$, there are two strata: </p> <p>The hyperelliptic locus, $H$, consists of those curves which have a degree $2$ map to $\mathbb{P}^1$. </p> <p>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. </p> <p>Every curve in $M_3$ is in precisely one of $B$ and $H$.</p> <p>There are several points I want to make:</p> <p>(a) The topology on $M_3$ is <strong>not</strong> the disjoint union of $B$ and $H$. Rather, $B$ is dense and open in $M_3$, and $H$ is a hypersurface.</p> <p>(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.</p> <p>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.</p> <p>(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 <code>$C \to \Delta \setminus \{ 0 \}$</code> 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 $F_t = Q^2 + t G + \cdots$.</p> <p>Let $\Delta'$ be the branched cover $t=u^2$ of $\Delta$. Then we can complete the family <code>$C' \to \Delta' \setminus \{ 0 \}$</code> to a family $\tilde{C} \to \Delta'$ whose fiber over $0$ is the hyperelliptic curve in question. </p> <p>The family $\tilde{C}$ maps to $\mathbb{P}^2$. For $u \neq 0$, the curve $C'_u$ maps to <code>$\{ F_{u^2}=0 \}$</code>. At $u=0$, the hyperelliptic curve double covers $Q$, ramified over the $8$ points <code>$\{ Q=G=0 \}$</code>.</p> <p>(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 <em>think</em> 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 $\mathbb{P}^1$.</p> <hr> <p>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. </p> <p>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.</p>