0
$\begingroup$

There is a basic question regrading the 3-dimensional hyper-Kloosterman sum which needs some help from the experts here:

For any integers $q,h \in \mathbb{N}$, how to estimate the sum: $$\sideset{_{}^{}}{^{\ast}_{}}\sum _{{x,y \bmod {qh}}} \sum_{\substack{t\bmod h \\(qt+y,h)=1 }}\, e\left (\frac{x+\overline{xy} +\overline{y+qt}+(y+qt) }{qh}\right)?$$ Here, the $\ast$ indicates that the summation is restricted to $(x,qh) = 1$ and $(y,qh) = 1$.

If $(q,h)=1$, then one may split this sum into several sub-sums of modulous $q$ and $h$; applying the bound $\sideset{_{}^{}}{^{\ast}_{}}\sum _{x,y \bmod {c}} e\left (({x+\overline{xy} +\overline{y}+y })/{c}\right)\ll c^{1+\varepsilon}$ for any $c\in \mathbb{N}$, one would quickly get the conclusion.

The question now is that if $q,h$ are not co-prime with each other, how do we get a decomposition of certain individual sub-sums, or more directly, whether we can obtain an estimate for the triple sum above?

I searched many papers; but, it seems that there is no a direct record regarding this type of triple sums. So, if any expert here knows some knowledge on this question, please show some guides or the corresponding references, many many many thanks.

Thanks in advance!

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ You want to split this into a product over prime factors $p$ of $qh$. Those that divide only one or the other can be handled as you say. Those that divide both will automatically divide the modulus with multiplicity at least $2$, so you will want to use the method of $p$-adic stationary phase for these primes. If the $p$-adic valuation of $q$ is at least the $p$-adic evaluation of $h$ then $\overline{y+ qt} = \overline{y} - q t \overline{y}^2$ and the sum in the $t$ variable is very easy to evaluate and the sum should simplify a lot from there. $\endgroup$
    – Will Sawin
    Commented Sep 26, 2023 at 11:56
  • $\begingroup$ @WillSawin Dear Sawin, thanks for timely reply. Maybe I didn't state my purpose clearly. My point of the departure is to give a bound for this sum. But, I find the known estimate for double hyper-Kloosterman sum in the post above can not be put into use. So I speculate if we have a further decomposition by changing variables. $\endgroup$
    – hofnumber
    Commented Sep 26, 2023 at 13:48
  • $\begingroup$ @WillSawin Dear professor, I also wanna ask, for fixed $h$, if we have the bound $\ll h (qh)^{1+\varepsilon} $ for this triple sum, by plugging the sum over $h$ outside, and then using the estimate $\sideset{_{}^{}}{^{\ast}_{}}\sum _{x,y \bmod {c}} e\left (({x+\overline{xy} +\overline{y}+y })/{c}\right)\ll c^{1+\varepsilon}$ for inner hyper-Kloosterman sum over $x,y$? Notice that there is still a factor of $q$ in the numerator. I am not sure if this would work. $\endgroup$
    – hofnumber
    Commented Sep 26, 2023 at 13:52
  • $\begingroup$ I don't understand your objection. I am not trying to solve the problem, just make some basic suggestions for how to start solving it. I don't see a change of variables that improves the situation much. $\endgroup$
    – Will Sawin
    Commented Sep 26, 2023 at 14:16
  • $\begingroup$ @WillSawin Dear professor, many thanks for so kindly help without any rewards! I need a quantitative bound for this sum now. The obstacle is that there exist some factor of $q$ in the numerator of the exponential sum above which may bring some troubles, I think. I don't know if one may study the Newton polyhedron nondegeneracy of the corresponding Newton polyhedron, or the works of Fouvry-Kowalski-Michel can be put into use. $\endgroup$
    – hofnumber
    Commented Sep 27, 2023 at 3:19

0

Reset to default

You must log in to answer this question.