Timeline for Definition of CM modular form
Current License: CC BY-SA 3.0
7 events
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| Oct 21, 2014 at 21:07 | history | edited | Olivier | CC BY-SA 3.0 |
Corrected typo
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| Feb 14, 2013 at 20:46 | comment | added | user30035 | OK I'm happy :-) If $K$ is any quadratic extension of $F$, totally real or totally imaginary or otherwise, and if $\chi$ is a grossencharacter of $K$ then of course you can automorphically induce $\chi$ up to $GL(2)/F$ by standard converse theorems and Hecke/Tate. The point is that if $K$ isn't CM then you have far less choice about what you can do at infinity because of units. Kevin | |
| Feb 14, 2013 at 8:49 | history | edited | Olivier | CC BY-SA 3.0 |
added 119 characters in body
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| Feb 14, 2013 at 8:48 | comment | added | Olivier | Dear wccanard (!), I did write that $k$ should be greater than 2, but now I realize I did not repeat the condition when passing to $F$. Thanks for pointing this out. | |
| Feb 13, 2013 at 20:49 | comment | added | user30035 | Olivier -- your assertion about $K$ being CM is false in the generality that you write it, in the totally real case (at least at the time I am writing this comment); there are for example Hilbert modular forms of parallel weight 1, and also automorphic representations of Artin type attached to non-holomorphic $\pi$s, which are isomorphic to a quadratic twist of themselves where the associated quadratic extension is not $CM$. In fact in the totally real case $K$ may be neither totally real nor totally imaginary. | |
| Feb 13, 2013 at 10:31 | comment | added | Olivier | Let me add a reference. You can read Motives and automorphic forms: the potentially abelian case, available on L.Fargues webpage. This is a modern exploration of the topic (which contains much much more than the answer to your question). | |
| Feb 13, 2013 at 9:29 | history | answered | Olivier | CC BY-SA 3.0 |