Timeline for Does X(13) have potentially good reduction at 13?
Current License: CC BY-SA 3.0
9 events
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| Mar 18, 2016 at 22:19 | comment | added | znt | goes precisely via this sort of construction. With $J(p)/J_0(p)$ the story is more complicated. For a start there will be tame quadratic twists of Steinberg forms (e.g. $J_0(p)$ itself) so it doesn't have pot good redn. And then you see some dihedral chars induced from chars of order $p^2-1$ so you need bigger tame extensions. Maybe there are more too, I don't carry the classification of low conductor supercuspidals around in my head :-/ | |
| Mar 18, 2016 at 22:16 | comment | added | znt | In fact it has good reduction over $\mathbb{Q}(\zeta_p)^+$ (because the tame level is 1). There are two ways of thinking about this. First you can write down a semistable model for $X_1(p)$ over $\mathbb{Q}(\zeta_p)$ and compare with $X_0(p)$ via a vanishing cycle computation as is done in KM. Alternatively you can invoke local Langlands and see that the forms contributing have level $p$ and non-trivial character so are principal series with one unramified and one tame character, to get the same result. Of course both approaches are "the same" in the sense that the proof of local-global... | |
| Mar 18, 2016 at 12:01 | comment | added | François Brunault | According to Katz-Mazur, Deligne and Rapoport showed that for $p$ prime, the abelian variety $J_1(p)/J_0(p)$ has good reduction over $\mathbf{Q}(\zeta_p)$. I wonder what can be said about $J(p)/J_0(p)$. | |
| Mar 17, 2016 at 8:42 | comment | added | Laurent Berger | @znt on a mac it actually works the same as with an ipad/iphone: press and hold a key and all the possible modifications of the letter will eventually be displayed for you to choose from | |
| Mar 16, 2016 at 23:10 | history | edited | znt | CC BY-SA 3.0 |
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| Mar 16, 2016 at 21:54 | comment | added | Bobby Grizzard | On a mac I believe option+c does the trick. | |
| Mar 16, 2016 at 21:44 | comment | added | François Brunault | Thanks for answering my question. For the cedilla the simplest is copy&paste, but I'm not really sensible to it | |
| Mar 16, 2016 at 19:05 | comment | added | znt | PS I'm new here -- how do I get the cedilla on Francois using a US keyboard? | |
| Mar 16, 2016 at 19:02 | history | answered | znt | CC BY-SA 3.0 |