Timeline for Why not add cuspidal curves in the moduli space of stable curves?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Nov 4, 2019 at 14:12 | comment | added | Yuhang Chen | @roysmith Oh, I misinterpreted your words. Thanks for the correction and the reference. | |
| Nov 4, 2019 at 3:31 | comment | added | roy smith | It was Mayer and Mumford, as Mumford states in his lecture in the notes from the 1964 Woods Hole conference, in the talk near the end of part III, titled "Further comments on boundary points". jmilne.org/math/Documents/woodshole.pdf | |
| Sep 26, 2019 at 2:56 | comment | added | Yuhang Chen | @roysmith Thanks for the reference. I will check it up. So it was Griffith and Mumford who first constructed compactified moduli space? | |
| Sep 26, 2019 at 2:36 | vote | accept | Yuhang Chen | ||
| Sep 26, 2019 at 2:35 | comment | added | Yuhang Chen | @Asvin Huh, that's easier than I thought. :) | |
| Sep 26, 2019 at 2:34 | comment | added | Asvin | @yYuhangChen good redn just means that all the fibers are smooth. | |
| Sep 26, 2019 at 2:32 | comment | added | Yuhang Chen | @Asvin Thanks for the clarification. I don't know what a "good reduction" means. But I will look it up later. | |
| Sep 26, 2019 at 1:56 | comment | added | roy smith | You might enjoy working through exercise A, pages 161-163 of Geometry of Algebraic Curves, vol. II, by Arbarello, Cornalba, and Grifiths, on exactly this example. It was mentioned already by Mayer in his IAS lectures in the Grifiths' Seminar on degeneration of algebraic varieties, around 1969, where he sketched the construction of the compactified moduli space, due originally to himself and Mumford, but unfortunately I no longer have my copy of those notes. | |
| Sep 26, 2019 at 1:33 | history | became hot network question | |||
| Sep 26, 2019 at 1:09 | comment | added | Asvin | @yYuhangChen i was basically referring to the last part of David's answer where he talks about how after adding a 6th root of t, you get an isotrivial (hence good redn) curve. | |
| Sep 26, 2019 at 0:15 | comment | added | Yuhang Chen | @Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"? | |
| Sep 26, 2019 at 0:10 | comment | added | Yuhang Chen | @Qfwfq Do you mean the cuspidal curve $y^2=x^3$ has an automorphism group $\mathbb{C^*}$, even with a marked point at the cusp? | |
| Sep 25, 2019 at 20:57 | comment | added | Qfwfq | Also, a cusp would have a $\mathbb{C}^*$-worth of automorphisms, while the usual moduli spaces of stable curves are designed to be as rigid as possible and in fact are Deligne-Mumford. | |
| Sep 25, 2019 at 19:13 | answer | added | David E Speyer | timeline score: 42 | |
| Sep 25, 2019 at 18:45 | review | Close votes | |||
| Sep 27, 2019 at 13:48 | |||||
| Sep 25, 2019 at 18:03 | comment | added | Asvin | I believe this has to do with the fact that $\mathscr M_{g,n}$ is stacky and so the valuative criterion doesn't hold on the nose. You might need to take an etale extension of your curve before you can extend the map from the generic point. In this particular case, this is related to the fact that the cuspidal singularity is unstable: After an unramified base change of $\mathbb C[[t]]$, the family will become either a node or good reduction. | |
| Sep 25, 2019 at 17:29 | history | asked | Yuhang Chen | CC BY-SA 4.0 |