I encounter a hyper-Kloosterman sum which needs some help from the experts here:

For any integers $q,s \in \mathbb{N}^+$(which may not be necessarily co-prime with each other), how to bound the sum: $$\sideset{_{}^{}}{^{\ast}_{}}\sum _{\substack{x,y \bmod {q}}\\ (y+s,q)=1} \, e\left (\frac{x+\overline{xy} +\overline{y+s}+y }{q}\right)?$$

It is conjectured that the upper-bound should be $O(q^{1+\varepsilon})$, but it looks very challenging to prove this for me.

Note: It is known that one has the estimate $\sideset{_{}^{}}{^{\ast}_{}}\sum _{x,y \bmod {q}} e\left (({x+\overline{xy} +\overline{y}+y })/{q}\right)\ll q^{1+\varepsilon}$, but I don't know how to estimate the sum above with the parameter $s$ involved.

Let $$f (x_1,x_2)=x_1+\overline{x_1x_2} +\overline{x_2+s}+x_2\in \mathbb{F}_q[x_1,x_2,\overline{x_1x_2}]$$ be a Laurent polynomial. One the other hand, one can consider the Newton polyhedron $\Delta(f)$ and verify that that the Laurent polynomial $f$ is non-degenerate by, e.g., Adolphson-Sperber's works. However, I'am really not familiar with algebraic geometry.

I searched Friedlander-Iwaniec's paper "Incomplete Kloosterman Sums and a Divisor Problem" (see https://sci-hub.wf/10.2307/1971175), and Yitang Zhang's paper "Bounded gaps between primes" (see https://annals.math.princeton.edu/wp-content/uploads/annals-v179-n3-p07-s.pdf). It seems that there is no a direct record regarding this type of double exponential sums.

So, if any expert here knows some knowledge on this question, please show some guides or the corresponding references, many many thanks.

Thanks in advance!


1 Answer 1


You can easily reduce to the case that $q$ is a prime power.

For $q$ is a prime, one approach is to use the method of trace functions. The sum can be written as the sum over $y$ of $Kl_2(\overline{y})$ times $e( y+ \overline{y+s})$. Both of these are geometrically irreducible trace functions and they are not geometrically isomorphic (one arises from a rank 2 sheaf, one from rank 1) so they are quasi-orthogonal, which gives a $\sqrt{q}$ of cancellation when they are summed against each other. Together with the $\sqrt{q}$ of cancellation from the Kloosterman sum, this gives what you want. (In fact one can give an explicit bound of $5q+1$ or something like that if one really wants.)

Alternately for $q$ prime, we can introduce the variable $z = \overline{y+s}$ $$\sideset{_{}^{}}{^{\ast}_{}}\sum _{\substack{x,y , z \bmod {q}}\\ z (y+s)=1 } \, e\left (\frac{x+\overline{xy} +z+y }{q}\right)$$ and detect the equation $z(y+s)=1$ by additive characters

$$ \frac{1}{q} \sum_{t\bmod q} \sideset{_{}^{}}{^{\ast}_{}}\sum _{\substack{x,y , z \bmod {q}}} \, e\left (\frac{x+\overline{xy} +z+y +tzy +tzs-t}{q}\right)$$

The $t=0$ terms are easy to handle, with cancellation in the sum over $z$ and an ordinary hyper-Kloosterman sum in the other variables. The $t\neq 0$ terms are an exponential sum over a torus in the sense of Adolphson-Sperber and Denef-Loeser. One can probably check cancellation using their results.

However, for $q$ a prime power, I don't think you will always get square-root cancellation. For simplicity, assume $q=p^n$ with $n$ even.

The sum in $x$ is a Kloosterman sum, which vanishes if $y$ is not a perfect square mod $q$ and otherwise is $\sum_{ \substack{ z \bmod q}{z^2 =\overline{y}}} e\left(\frac{2z}{q} \right)$ and so changing variables to $z$ gives

$$\sqrt{q} \sideset{_{}^{}}{^{\ast}_{}}\sum _{\substack{ z \bmod {q}}} e \left( \frac{2z+ \overline{ \overline{z}^2+s} + \overline{z}^2}{q} \right)= \sqrt{q} \sideset{_{}^{}}{^{\ast}_{}}\sum _{\substack{ z \bmod {q}}} e \left( \frac{2z+ z^2 \overline{ 1 +z^2 s} + \overline{z}^2}{q} \right)$$

The ratio of stationary phase will express this as a sum over the critical points of the function $2z + \frac{z^2}{1+z^2 s} + z^{-2}$, i.e. over the zeroes modulo $p^{n/2}$ of the derivative $$2 + \frac{2z (1+z^2 s) - 2z^3 s}{ (1+z^2 s)^2} - 2 z^{-3} = 2\frac{ (z^3-1) (1+z^2 s)^2+ z}{ z^3 (1+z^2 s)^2}$$ If the function has a critical point of order at least $3$, i.e. if its derivative has a zero of order at least $2$, then locally near the point the exponential sum will look like an exponential of a power higher than $2$ which does not admit square-root cancellation. This will indeed happen if the discriminant of the polynomial $ (z^3-1) (1+z^2 s)^2+ z$ vanishes mod a high power of $p$, which can happen as this discriminant is a nonconstant polynomial of $s$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.