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Sorry to disturb. I have a question involving the three-dimensional Kloosterman sum, which needs some help from the experts here.

For any $\alpha, \beta, \gamma \in \mathbb{Z}$ and $q\in \mathbb{N}^+$, how can we estimate the sum $$ \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{x,y,z \bmod q} \left( \frac{x+\overline{xy}+\alpha y\overline{z}+\beta z+\gamma \overline{z}}{q}\right)?$$

I searched Pitt's paper(Theorem 3, https://sci-hub.wf/10.1215/S0012-7094-95-07711-4), Friedlander-Iwaniec's peper (Theorem2, https://sci-hub.wf/10.2307/1971175), and Michel's paper (Section 6, https://www.doc88.com/p-9052834716480.html?r=1), but find it seems they all don't match the type we are concerned above.

It can be sure some stuff involving $l$-adic cohomology can be put into use; see for example Adolphson-Sperber's paper (https://www.researchgate.net/profile/Steven-Sperber-2/publication/38390736_Exponential_sums_and_Newton_polyhedra/links/558c275708ae591c19d9efe8/Exponential-sums-and-Newton-polyhedra.pdf) or Denef-Loeser's paper (https://webusers.imj-prg.fr/~francois.loeser/inv91.pdf). However, I am not really familiar with the $l$-adic cohomology, and can not really figure out the Newton polyhedron $\Delta_\infty(f)$ of a Laurent polynomial $$f(x_1,x_2,x_3)=x_1+\overline{x_1x_2}+\alpha x_2\overline{x_3}+\beta x_3+\gamma \overline{x_3}\in \mathbf{F}_q[x_1,x_2,x_3, \overline{x_1x_2x_3}]$$ to verify whether or not $f$ is non-degenerate with respect to $\Delta_\infty(f)$.

So, if some expert has seen this type of sum in the question or leans something how to show the non-degenerateness, please give some comments or guide a reference.

Great thanks in advance!

$\mathbf{\text{Edit:}}$ If $(\alpha,q)=1,$ one sees that the triple sum turns out to be $$q\sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{x \bmod q} KL_2(\alpha \overline{x-\gamma}) KL_2(\beta x),$$ for which it seems there is still no record in the literature. This however can be compared with eq.(6.2) in https://www.doc88.com/p-9052834716480.html?r=1. But unfortunately, the available result is that for the type $\sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{x \bmod q} KL_2(\alpha \overline{x-\gamma}) KL_2(\beta \overline{x})$.

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    $\begingroup$ Maybe I am missing something, but can't the y-sum be executed to simplify the expression? In any case if q is a prime power it should be possible to evaluate this in elementary fashion using stationary phase $\endgroup$ Jun 25, 2022 at 11:45
  • $\begingroup$ @WindomEarle Sorry, there is a typo; have re-edited. Thanks! $\endgroup$
    – hofnumber
    Jun 25, 2022 at 12:29
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    $\begingroup$ Do you mean to sum mod $q$ (and not mod $c$)? $\endgroup$
    – Will Sawin
    Jun 25, 2022 at 16:38
  • $\begingroup$ @WillSawin Sorry, have edited. Many thanks. $\endgroup$
    – hofnumber
    Jun 25, 2022 at 18:16

1 Answer 1

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The proof 1 of Remark 6.2 in Lectures on applied $\ell$-adic cohomology works without modification for an arbitrary rational linear transformation and thus applies to $$\sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{x \bmod q} KL_2(\alpha \overline{x-\gamma}) KL_2(\beta x).$$ Thus for $q$ prime and $\alpha$ not divisible by $q$,

$$ \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{x,y,z \bmod q} \left( \frac{x+\overline{xy}+\alpha y\overline{z}+\beta z+\gamma \overline{z}}{q}\right) = O(q^{3/2}).$$

Here the rational linear transformation is $\beta x \mapsto \alpha (x-\gamma)^{-1}$, i.e. $x \mapsto \alpha (\beta^{-1} x- \gamma)^{-1}$ which is never the identity (the only case for which the bound fails).

I looked through a few works of Fouvry-Kowalski-Michel and didn't see a statement that could be quoted entirely off-the-sheaf here, without reasoning such as in that proof.

Probably the Newton polyhedron nondegeneracy argument can be made to work also. If $\alpha,\beta, \gamma$ are nonzero, it's the polyhedron with the vertices $(1,0,0), (-1,-1,0), (0,1,-1), (0,0,1),(0,0,-1)$, which I got by reading off the degrees of the monomials in your formula. One should add $(0,0,0)$ and then remove any points in the convex hull of the others but, when you do that $(0,0,0)$ is the only point in the convex hull of the others. This polyhedron has five vertices, six triangular two-dimensional faces, and nine edges, and one just has to check the derivative nonvanishing criterion for each of those (one doesn't have to check for the volume since it includes zero).

Since the degree of each monomial is one of the vertices, for each face $\tau$ the restricted polynomial $f_\tau$ will be a sum of one monomial for each vertex of the face. Since all the faces are points, lines, or triangles this will be a sum of one, two, or three monomials and the derivative turns into an algebraic criterion in the degrees of the monomials which is automatically satisfied in all large characteristics.

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    $\begingroup$ "off-the-sheaf" -- was that a pun or a typo, lol. $\endgroup$ Jun 26, 2022 at 2:11
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    $\begingroup$ @AnuragSahay A typo, sadly. $\endgroup$
    – Will Sawin
    Jun 26, 2022 at 2:14
  • $\begingroup$ @WillSawin Dear prof. Sawin, much obliged. This is what I would like to expect. Yes, my puzzle if whether or not the matrix like $$\gamma =\begin{pmatrix}0&1\\1&1\end{pmatrix}$$ can be applied to Remark 6.2 of Michel's paper to get the general bound of $q^{3/2}$. Thanks for telling me this works!! $\endgroup$
    – hofnumber
    Jun 26, 2022 at 5:45

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