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Nov 24, 2022 at 19:21 answer added Subhajit Jana timeline score: 1
Nov 21, 2022 at 14:20 comment added user50139 @SubhajitJana can you please give more details? I am not familiar with the Kirillov model, and knowing how to obtain such a bound could be extremely helpful to me.
Nov 19, 2022 at 20:40 comment added Subhajit Jana You usually expect square-root cancellation, that is, the expected bound is $\ll_{q,\epsilon} B^{1/2+\epsilon}$ with polynomial dependency in $q$ which follows from some Voronoi formula. In this particular case, using the Kirillov model one can show the sum is $O(\sqrt{B})$ uniformly in $q$.
Nov 18, 2022 at 19:27 comment added user50139 @Kimball I guess :)
Nov 18, 2022 at 19:27 comment added user50139 @SubhajitJana, on a first approximation you may think of $h(x)=f(x/B)$ where $f$ is a fixed compactly supported function and $B$ is some parameter tending to infinity, so upper bounds should depend on $B$. Here $\chi$ is fixed, and so is $N$ (the cusp form itself is fixed) but $q$ is varying.
Nov 18, 2022 at 19:22 comment added user50139 @m34 At a first glance it seems to be the sort of thing I want, yes, thank you very much! Other remarks are still welcome, of course :)
Nov 18, 2022 at 9:48 comment added Subhajit Jana What is the support of $h$ here? Is $h$ a fixed test function? Then the sum is trivially $O(1)$. Also, you need to specify which parameters ($\pi$, $N$, $q$, $C(\chi)$, etc.) are fixed and which are varying.
Nov 18, 2022 at 6:07 comment added Kimball In an attempt to solve an unrelated problem, I was led to... - Doesn't that mean it's not unrelated?
Nov 17, 2022 at 18:18 comment added Alex M. That lemma equates the sum that the OP is studying to an even more complicated expression involving three sums and an integral. In particular, it gives no upper bound for the OP's sum.
Nov 17, 2022 at 17:33 comment added m34 Does Lemma 2.4 of arxiv.org/pdf/1707.01576.pdf do what you need?
Nov 17, 2022 at 17:16 history asked user50139 CC BY-SA 4.0