2
$\begingroup$

Let $m\geq 2$ be a positive integer and consider the Kloosterman sums $$ \mathrm{Kl}(a,b,m)=\sum_{\substack{1\leq x\leq m\\ \gcd(x,m)=1}}\exp\left(\frac{2\pi i}{m}(ax+b\bar{x})\right), $$ where $\bar{x}$ is the inverse of $x$ modulo $m$. The following upper bound is well-known $$ \mathrm{Kl}(a,b,m)\leq \tau(m)\gcd(a,b,m)^{1/2}m^{1/2},\qquad\qquad\qquad (1) $$ where $\tau(m)=\sum_{d\mid m}d$ is the divisor function.

However it seems that Estermann proved slightly better upper bound (see Kaisa Matomaki's paper). When $32\nmid m$ we have $$ \mathrm{Kl}(a,b,m)\leq 2^{\omega(m)}\gcd(a,b,m)^{1/2}m^{1/2},\qquad \qquad\qquad(2)$$ where $\omega(m)$ is the number of distinct prime divisors of $m$. When $32\mid m$ the bound holds with an additional factor $\sqrt{2}$ on the right hand side. Clearly we have $2^{\omega(m)}\leq \tau(m)$ and so (2) is much better than (1). The surprising point for me is that even Estermann in his paper formulated (1) rather than (2). My questions now are the following

  1. First I would like to know whether (2) is correct and if it is correct, I would like to have a reference stating (2) as a theorem.
  2. If (2) is correct why in most litterateur authors use (1). Is there any technical reason why (1) is used most of the time.
$\endgroup$
  • 2
    $\begingroup$ I think (2) is technically harder to state, e.g. because of the case distinction $32\mid m$ vs. $32\nmid m$. In analytic number theory, there is not much of a difference between $2^{\omega(m)}$ and $\tau(m)$. For example, they have the same maximal order and essentially the same moments. Of course a textbook should emphasize (2) as well, because in certain situations the improvement over (1) might be substantial. $\endgroup$ – GH from MO Apr 7 '15 at 18:41
  • 1
    $\begingroup$ Your divisor function is unusual in that as it is it has order $O(n\log\log n)$. $\endgroup$ – Fan Zheng Apr 7 '15 at 19:05
2
$\begingroup$
  1. Yes, and it follows from Estermann article.

  2. Usually it is enough to estimate $\tau(m)$ as $m^\varepsilon.$

$\endgroup$
  • $\begingroup$ I agree. See also my comment to the OP's post. $\endgroup$ – GH from MO Apr 7 '15 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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