Timeline for Why is the cuspidal spectrum discrete?
Current License: CC BY-SA 3.0
11 events
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| Apr 4, 2016 at 17:45 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Apr 11, 2011 at 5:55 | vote | accept | Marc Palm | ||
| Apr 10, 2011 at 18:27 | comment | added | GH from MO | @pm: Maass forms don't live on the compactified surface. | |
| Apr 10, 2011 at 16:09 | answer | added | GH from MO | timeline score: 6 | |
| Apr 10, 2011 at 15:02 | history | edited | Marc Palm | CC BY-SA 3.0 |
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| Apr 10, 2011 at 14:52 | comment | added | Marc Palm | So is it true, that Maass forms live on the compactified surface? I will try to make the question more specific. | |
| Apr 10, 2011 at 13:31 | comment | added | Kevin Buzzard | I'll offer a vague answer to what seems to me to be a vague question. N represents a loop around a cusp. If you're perpendicular to $\pi$ then you satisfy some boundedness criterion at the cusps, and adding the cusps to a modular curve makes it compact, so now you're looking at functions on a compact space and then spectral theory works very well and you get discrete decompositions. Is this any help or are you looking for something totally different? | |
| Apr 10, 2011 at 11:57 | history | edited | Marc Palm | CC BY-SA 3.0 |
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| Apr 10, 2011 at 11:08 | history | edited | Marc Palm | CC BY-SA 3.0 |
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| Apr 10, 2011 at 9:33 | history | edited | Marc Palm |
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| Apr 10, 2011 at 9:21 | history | asked | Marc Palm | CC BY-SA 3.0 |