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In some recent work, the following strange-looking exponential sum arose: $$ \sum_k \sum_r \sum_s e\bigg( \frac{r \bar s^{(r)} \bar k^{(r^2+s^2)}}{r^2+s^2} \bigg). $$ Here $e(x) = e^{2\pi i x}$ as usual, and $\bar a^{(b)}$ denotes the inverse of $a$ modulo $b$. (For simplicity, let's think of the three sums as being over intervals of positive integers, excluding any integers for which the modular inverses aren't defined.)

Related versions of the sum are something like $$ \sum_k \sum_r \sum_s e\bigg( \frac{\bar s^{(r)} \bar k^{(r^2+s^2)}}{r} \bigg) $$ and $$ \sum_k \sum_r \sum_s e\bigg( \frac{\bar s^{(r)} \bar (r^2+s^2)^{(k)}(r^2+s^2)}{rk} \bigg). $$ In all these cases, we have inverses to two different moduli running around.

Is there anything in the literature that addresses these multi-modulus (Ramanujan or) Kloosterman sums? Or is there an approach for deriving estimates for them more directly from Weyl's/Deligne's estimates?

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  • $\begingroup$ 1. Are you really summing over the variables you are modding out by? 2. The simplest variable, $k$, should be the innermost sum. If $k$ is large with respect to $r^2+s^2$ you get a lot of cancellation. If $k$ is small you can view this as an incomplete Kloosterman sum and use standard analytic number theory methods for that sum (e.g. Polya-Vinagradov). Hopefully that's enough cancellation - for the $r$ and $s$ sums I can't help you. $\endgroup$
    – Will Sawin
    Jul 9, 2014 at 22:25
  • $\begingroup$ 1. Yep. 2. Fixing $k$ doesn't give me quite enough cancellation, unfortunately. $\endgroup$ Jul 9, 2014 at 22:38
  • $\begingroup$ The expressions you write are currently ambiguous. For instance, $\bar{s}^{(r)}$ is only determined up to a multiple of $r$, and this affects the first exponential. Are you going to take the representative of this residue class in $\{0,1,\dots,r-1\}$? If so, the sum is going to get rather non-algebraic, and it will be significantly more difficult to use the usual methods to handle these sums. $\endgroup$
    – Terry Tao
    Jul 10, 2014 at 0:02
  • $\begingroup$ Yes, I've simplified the expressions a bit for ¿clarity? ... the first $r\bar s^{(r)}$ is actually $r\bar s^{(r)}-s\bar r^{(s)}$, which turns out to be well-defined, or something like that. In certain ranges for $r,s$, I think the sums can be split up and the ambiguity isn't as much of an issue. But you're right of course, in the end. I was hoping to be close enough to elicit some references to relevant literature, which I could then use to address the non-simplified sums. $\endgroup$ Jul 10, 2014 at 4:04

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