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Dec 6, 2016 at 17:30 comment added Subhajit Jana Why do you expect the normalized operator, as you have defined, to be nonzero? Shouldn't it have a zero at the special representations, i.e. $\chi_1\chi_2^{-1}=p^{\pm1}$? Most probably to get a non-vanishing holomorphic operator one normalizes by $\epsilon(0,\chi_1\chi_2^{-1})\frac{L(1,\chi_2\chi_1^{-1})}{L(0,\chi_1\chi_2^{-1})}.$
Jan 9, 2016 at 14:37 comment added Joseph Hundley A little. I didn't come across anything GL(2) (or SL(2)) specific. It seemed like this simplest case would probably be simpler and older.
Jan 9, 2016 at 14:00 vote accept Joseph Hundley
Jan 8, 2016 at 17:20 answer added paul garrett timeline score: 4
Jan 8, 2016 at 16:59 comment added user1688 Did you consult Arthur's early papers on the trace formula?
Jan 8, 2016 at 16:11 history asked Joseph Hundley CC BY-SA 3.0