Timeline for Why is the supersingular locus the zero locus of a modular form?
Current License: CC BY-SA 3.0
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| Sep 6, 2014 at 19:59 | comment | added | Felipe Voloch | Another way. If an elliptic curve over a base has a non-zero point of order $p$, then on an open set of the base, this point specializes to a non-zero point, still of order $p$, so ordinarity is open. | |
| Sep 6, 2014 at 19:54 | comment | added | Bear | I think an answer may be that supersingularity for an elliptic curve is equivalent to the fact that the Verschiebung is purely inseparable, but maybe there are other reasons. | |
| Sep 6, 2014 at 19:53 | comment | added | Bear | Ok, you know that the supersingular condition is closed because you studied that there exists the Hasse invariant. However, how could you have guessed that the Hasse invariant existed without knowing it? | |
| Sep 6, 2014 at 19:37 | comment | added | Felipe Voloch | I just proved that every finite set of points is the zero locus of a modular form. The supersingularity condition is clearly closed, so either finite or everything. | |
| Sep 6, 2014 at 19:33 | comment | added | Bear | And relating this to the supersingular locus. If you didn't know that there are only finitely many isomorphism classes of supersingular elliptic curves, could you explain why there exists the Hasse invariant that cuts out the supersingular locus? | |
| Sep 6, 2014 at 19:28 | comment | added | Bear | Thank you for your answer, but I had a more general situation in mind. Let me try to explain myself better. Consider a property of elliptic curves P, so "modular" as you say, then is there a way to know if the points of the modular curve that satisfy P is the zero locus of a modular form? | |
| Sep 6, 2014 at 18:05 | history | edited | Felipe Voloch | CC BY-SA 3.0 |
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| Sep 6, 2014 at 17:45 | history | answered | Felipe Voloch | CC BY-SA 3.0 |