Timeline for Why is a general curve automorphism-free?
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18 events
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| Feb 12, 2011 at 2:39 | comment | added | roy smith | and the really classic paper of hurwitz: Hurwitz, A. (1893), "Über algebraische Gebilde mit Eindeutigen Transformationen in sich", Mathematische Annalen 41 (3): 403–442, i recommend looking at it even if your german is minimal. | |
| Feb 12, 2011 at 2:30 | comment | added | roy smith | here is the classic paper of harry rauch. ams.org/journals/bull/1962-68-04/S0002-9904-1962-10818-0/… | |
| Feb 12, 2011 at 2:22 | comment | added | roy smith | F. Oort — Singularities of moduli schemes. S ́em. P. Dubreil 29 (1975/76). Lect. Notes Math. 586, Springer - Verlag 1977; pp. 61 - 76. | |
| Feb 12, 2011 at 1:31 | comment | added | jlk | @roy smith: Thanks! I had not seen the Cornalba paper before. | |
| Feb 11, 2011 at 16:28 | comment | added | roy smith | and if one does the same for A(g) and gets maximum dimension less than 3g-3, then of course this locus does not contain Jacobians. This may be done in a paper of Oort in seminaire dubreil. | |
| Feb 11, 2011 at 16:27 | comment | added | roy smith | Here is a paper of Cornalba which enumerates all subloci in M(g) of curves with automorphisms. www-dimat.unipv.it/cornalba/papers/autonew.pdf | |
| Feb 11, 2011 at 15:31 | comment | added | roy smith | The excellent answers above depend on specialization, to singular, hyperelliptic, or curves with prime order automorphisms. For an approach that does not use any specialization, over C, you might use that an automorphism is induced by a non euclidean symmetry of a fundamental polygon in the upper half plane covering the curve. This imposes symmetry conditions on the polygon that should not be generally true. Equivalently the associated polygon group should be its own normalizer in the group of non euclidean motions,(Siegel). This metric approach has been generalized recently by McKernan. | |
| Feb 11, 2011 at 8:13 | comment | added | jlk | @roy smith: Thanks for the response! The generic reducedness idea might work. A direct computation should show that the completed local ring of $M_g$ at a general hyperelliptic curve is reduced (though maybe the computation is essentially Torsten Ekedahl's answer). I'd have to think more about the argument using the Jacobian.... | |
| Feb 11, 2011 at 6:23 | comment | added | roy smith | yes you are right. I was trying to come up with a new idea. Another approach is that automorphisms give singularities of M(g), so this would seem to amount to the generic reducedness of M(g), which might follow from the injectivity of the infinitesimal Torelli map in general. No I guess that has the same flaw. My apologies. So i guess I'm back to the earlier answers, namely, finiteness of the automorphism group, plus bounds on the dimension of the subvariety along which an automorphism can persist. This argument by Hurwitz is on p.276 of Griffiths-Harris. | |
| Feb 11, 2011 at 6:15 | comment | added | jlk | @Felipe Voloch: Thanks for the answer! | |
| Feb 11, 2011 at 5:29 | comment | added | jlk | @roy smith: I am not quite sure I follow your argument. Don't you need to rule out the possibility that the Schottky locus is contained in the locus of ppav's with an extra automorphism? | |
| Feb 11, 2011 at 3:34 | comment | added | roy smith | I guess it suffices to show a general ppav has none except the minus map. (by the Jacobian functor from curves to ppav's.) | |
| Feb 11, 2011 at 1:59 | comment | added | Felipe Voloch | @Dan: That is a good simple way to prove it, but I understood the OP specifically did not want to do that (going to the boundary and consider singular curves). | |
| Feb 10, 2011 at 16:56 | comment | added | Dan Petersen | It seems to me that this can be done very painlessly if one constructs a curve without automorphisms instead in the Deligne-Mumford boundary of $M_g$. To find an automorphism-free curve of genus three you could take a general curve of genus two and glue together any two points of it which are not conjugate under the hyperelliptic involution and not Weierstrass points. Then you could just iterate: by gluing together two arbitrary points of an automorphism-free curve of genus g you get an automorphism-free curve of genus g+1. | |
| Feb 10, 2011 at 7:42 | comment | added | jlk | @Felipe Voloch: I did mean "general" in the sense of algebraic geometry. | |
| Feb 10, 2011 at 7:28 | comment | added | Felipe Voloch | @Mariano: I am assuming he meant general in the algebraic geometry sense, i.e., in a non-empty Zariski open set of moduli. But it is not hard to show that curves with automorphisms form a Zariski closed subset of moduli (using the finiteness of the automorphism group), so once you know there is one, you conclude that most don't have automorphisms. | |
| Feb 10, 2011 at 6:53 | comment | added | Mariano Suárez-Álvarez | Why is it enough to exhibit one curve? A semicontinuity argument? | |
| Feb 10, 2011 at 6:48 | history | answered | Felipe Voloch | CC BY-SA 2.5 |