Timeline for Is a holomorphic family whose fibers are all smooth locally trivial?
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| Aug 23, 2013 at 16:39 | history | edited | roy smith | CC BY-SA 3.0 |
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| Aug 23, 2013 at 8:49 | comment | added | Dan Petersen | Finally, the locus of curves with automorphisms is not really a problem. For instance, you can work with curves and abelian varieties with level $\geq 3$ structure. In general it's a bit tricky to define level structures on stable curves, but when they have compact type it's straightforward (you just use the usual definition). So you map the space of compact type curves with level structure to the corresponding cover of $A_g$, and take the Satake compactification of that. | |
| Aug 23, 2013 at 8:46 | comment | added | Dan Petersen | ... its image is at least three. But of course this does not work if $h=1$ or $g-h=1$, since then you can't really "forget" the choice of point where the two curves are attached to each other. So $M_{1,1} \times M_{g-1,1}$ maps to something of codim 2. And as you say, when $g=2$ you have products of elliptic curves, and you can't forget either marked point. | |
| Aug 23, 2013 at 8:44 | comment | added | Dan Petersen | To add to my previous comment. As you say the Satake compactification of $M_g$ is constructed in two steps: first you consider the locus of curves in $\overline M_g$ of compact type (i.e. all nodes are separating) and map that to $A_g$, and then you take the Satake compactification. The Satake boundary is small but you need to understand what happens at the loci of products of Jacobians. A codimension one stratum in $\overline M_g$ of compact type has the form $M_{h,1} \times M_{g-h,1}$ and the map from this stratum to $A_g$ will factor through $M_h \times M_{g-h}$. So the codimension of ... | |
| Aug 23, 2013 at 2:34 | history | edited | roy smith | CC BY-SA 3.0 |
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| Aug 23, 2013 at 2:27 | history | edited | roy smith | CC BY-SA 3.0 |
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| Aug 22, 2013 at 22:10 | history | edited | roy smith | CC BY-SA 3.0 |
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| Aug 22, 2013 at 22:06 | comment | added | Dan Petersen | Actually it's not going to work in genus 2 since $M_2$ is affine. But $M_3$ contains a complete curve. | |
| Aug 22, 2013 at 17:59 | history | edited | roy smith | CC BY-SA 3.0 |
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| Aug 22, 2013 at 17:53 | history | answered | roy smith | CC BY-SA 3.0 |