Recall that for the classical Kloosterman sum $$ K(a,b,p^t):= \sum_{x \in (\mathbb{Z}/ p^t \mathbb{Z})^* } \psi \left(\frac{ax+bx^{-1}}{p^t} \right),$$ where $\psi(x)=e^{2\pi ix}$, $a,b,t$ are natural numbers and $p$ is a fixed prime number. We have the Weil bound, i.e. $$ \vert K(a,b,p^t) \rvert \leq (t+1) \sqrt{(\gcd(a,b,p^t))} \sqrt{p^t}.$$ Now if we consider the following short Kloosterman sum $$ K'(a,b,p^t):= \sum_{x \in A} \psi \left(\frac{ax+bx^{-1}}{p^t} \right),$$ where $A:=\left \{ x \rvert x \equiv 1 \mod p^m,\;x \in (\mathbb{Z}/ p^t \mathbb{Z})^* \right \}.$ Here m is a fixed positive integer and we can also assume that $t>>m$. My question is that can we also achieve an analogy of Weil bound for the above short Kloosterman sum $K'(a,b,p^t)$? Hopefully, I expect the following bound $$ \vert K'(a,b,p^t) \rvert \leq A_m \cdot (t+1) \sqrt{(\gcd(a,b,p^t))} \sqrt{p^t}.$$ Here $A_m$ is a positive constant only depend on the choice of $p$ and $m$.

Since we assume that $t>>m$, then the number of elements in the finite set $A$ is greater than $\sqrt{p^t}$. Then from the “short Kloosterman sums" in Wikipedia, we may achieve an analogy of Weil bound for the above short Kloosterman sum. However, I cannot find any references. So any ideas or references for the bound of above short Kloosterman sum are welcome.


If $x=1+p^my$ ($1\le y\le p^{t-m}$) then $$\psi \left(\frac{ax+bx^{-1}}{p^t} \right)=\psi \left(\frac{a(1+p^my)+b(1+p^my)^{-1}}{p^t} \right)=\psi \left(\frac{a+b}{p^t} \right)\psi \left(\frac{f(y)}{p^{t-m}} \right),$$ where $f(y)=ay+b(-y+y^2p^m-y^3p^{2m}+\cdots)$. The sum $$\sum_{y=1}^{p^{n}} \psi \left(\frac{f(y)}{p^{n}} \right)$$ can be calculated explicitly, because $f(y)$ has a nice form (almost all coefficients are divisible by $p$), see Lemma 2.1 in Generalized Twisted Kloosterman Sum Over ℤ[i] by S. Varbanets. The main idea is to take $y=y_0+y_1p^{n-1}$, where $1\le y_0\le p^{n-1}$, $1\le y_1\le p.$ The sum becomes linear over $y_1$. Apply this idea twice.

For $n\ge 1$ $$f(y)\equiv f(y_0)+f'(y_0)y_1p^{n-1}\equiv f(y_0)+(a-b)y_1p^{n-1}\pmod{ p^n}.$$ So $$S_n(a,b)=\sum_{y=1}^{p^n}e_{p^n}(f(y))=\sum_{y_0=1}^{p^{n-1}}\sum_{y_1=1}^{p}e_{p^n}(f(y_0)+(a-b)y_1p^{n-1})=p\delta_p(a-b)\sum_{y_0=1}^{p^{n-1}}e_{p^n}(f(y_0)),$$ where $e_N(x)=e^{2\pi ix/N}$ and $$\delta_q(x)=\begin{cases} 1,& \text{ if }q\mid x;\\ 0,& \text{ if }q\nmid x. \end{cases}$$ This sum does not vanish if $a\equiv b\pmod{p }$. We may also assume that $a\equiv b\not \equiv0\pmod{p }$ because otherwise original sum can be simplified: for $a=pa_1$, $b=pb_1$ $$\sum_{x \in (\mathbb{Z}/ p^n \mathbb{Z})^*}e_{p^n}(ax+bx^{-1})=p\sum_{x \in (\mathbb{Z}/ p^{n-1} \mathbb{Z})^*}e_{p^{n-1}}(a_1x+b_1x^{-1}).$$ Let $a=b+p^\alpha a_1$, $\alpha\ge 1$, $(a_1,p)=1$. Then
$f(y)=a_1p^\alpha y+b(y^2p^m-y^3p^{2m}+\ldots)$, and $$S_n(a,b)=p \sum_{y=1}^{p^{n-1}}e_{p^{n-1}}(f(y)p^{-1}).$$

If $\alpha\ge m$ then $$S_n(a,b)=p \sum_{y=1}^{p^{n-1}}e_{p^{n-m}}(g(y))=p^m \sum_{y=1}^{p^{n-m}}e_{p^{n-m}}(g(y)),$$ where $$g(y)=a_1p^{\alpha-m}y+b(y^2-y^3p^{m}+\ldots),$$ and one can apply Lemma 2.1 from the cited article.

If $\alpha<m$ then $$S_n(a,b)=p^\alpha \sum_{y=1}^{p^{n-\alpha}}e_{p^{n-\alpha}}(g(y)),$$ where $$g(y)=a_1y+b(y^2p^{m-\alpha}-y^3p^{2m-\alpha}+\ldots).$$ Again $y=y_0+y_1p^{n-\alpha-1}$, $1\le y_0\le p^{n-\alpha-1}$, $1\le y_1\le p$ $$g(y)\equiv g(y_0)+g'(y_0)y_1p^{n-\alpha-1}\equiv g(y_0)+a_1y_1p^{n-\alpha-1}\pmod{ p^{n-\alpha}}.$$ $$S_n(a,b)= \sum_{y=1}^{p^{n-\alpha}}e_{p^{n-\alpha}}(g(y))= \sum_{y_0=1}^{p^{n-\alpha-1}}\sum_{y_1=1}^{p}e_{p^{n-\alpha}}(g(y_0)+a_1y_1p^{n-\alpha-1})=0,$$ because sum over $y_1$ vanishes.

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  • $\begingroup$ Thanks a lot! Is there any similar result for $p$ equals 2? $\endgroup$ – JACK Feb 21 at 15:41
  • $\begingroup$ I think maybe we cannot apply Lemma 2.1 in your reference. Since Lemma 2.1 requires that the coefficient of $y^2$ should be relatively prime to $p$. However, the polynomial in your calculation does not satisfy this condition.@Alexey Ustinov $\endgroup$ – JACK Feb 21 at 21:35
  • $\begingroup$ @JACK You may repeat the same calculations for $p=2$. Lemma 2.1 not always applied directly. I've added more details in my answer. $\endgroup$ – Alexey Ustinov Feb 22 at 2:46
  • $\begingroup$ Thanks! @Alexey Ustinov I think for $p$ equals 2 case, we may not repeat the same calculations since Lemma 2.1 cannot apply to this case. So maybe we need some modifications. $\endgroup$ – JACK Feb 23 at 2:09
  • $\begingroup$ @JACK For $p=2$ take $y=y_0+y_1p^{n-\alpha-2}$. It works for $m>1$. For $m=1$ take $y=y_0+y_1p^{n-\alpha-3}$. $\endgroup$ – Alexey Ustinov Feb 23 at 2:38

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