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Jun 7, 2011 at 6:43 answer added Marc Palm timeline score: 1
Jun 6, 2011 at 20:20 comment added HYL Thank you for your response, I didn't think of twisting the character.
Jun 6, 2011 at 20:17 vote accept HYL
Jun 6, 2011 at 18:45 comment added Kevin Buzzard For 1. I am a bit confused. Classical cuspidal eigenforms (not modular forms) give rise to irreducible subrepresentations of the space you're interested in, but there are other irreducible subrepresentations not coming from classical eigenforms -- namely those coming from Maass forms. Does this answer 2. for you? For 1. why can't you just twist to reduce to the situation where $\psi$ is trivial on the positive reals?
Jun 6, 2011 at 17:42 answer added paul garrett timeline score: 12
Jun 6, 2011 at 13:37 history asked HYL CC BY-SA 3.0