# Bounds on imaginary parts of partial Kloosterman sums?

For a prime $p$ and integers $a,m$, $0<a,m<p$ define the (partial Kloosterman) sum $$S_p(a,m) = \sum_{0<k<m} \exp\left(\frac{2\pi\mathrm{i}}{p}(a x + x^{-1})\right),$$ where $x^{-1}$ is the multiplicative inverse of $x$ modulo $p$. I know $\Im S_p(a,p)=0$ and that there is the Weil bound $$|\Re S_p(a,p)| \leq 2\sqrt{p}.$$

Is there a similar bound on $\Im S_p(a,m)$ that holds for all $m=1,\ldots,p$? I imagine this question might have a known answer in the literature, but I couldn't find anything. Conjectured bounds would also be interesting to me.

A related question: is the Weil bound true for partial sums as well, i.e., is it true that $$\max_m |\Re S_p(a,m)|\leq 2\sqrt{p}?$$

If I pretend that $(x+a x^{-1})/p$ is uniformly distributed in $[0,1]$, then the distribution of real and imaginary parts of each term is the same, but all imaginary terms cancel. So intuitively only half as many imaginary terms contribute to $\max_m |\Im S_p(a,m)|$ as there are real terms that altogether contribute to make up $2\sqrt{p}$. So I would expect some bound of the form $$\max_m |\Im S_p(a,m)| \leq 2\sqrt{p/2},$$ but I am not really convinced this is right (for example: should it be $2\sqrt{p/2}$ because $p$ is halved or $\sqrt{p}$ because the number of terms is halved but $p$ is the same?).

• It's certainly $O(p^{1/2 + o(1)})$ by the usual trick of writing the characteristic function of $[0,m]$ as a linear combination of complex exponentials. I suspect the imaginary part of a partial sum is not quite $O(p^{1/2})$ as you hope. – Noam D. Elkies Nov 28 '14 at 0:25

The partial sums, normalized by $\sqrt{p}$, are unbounded, as one varies $a$ over all invertible classes modulo $p$n and lets $p$ go to infinity. This follows from results in this paper, which also has more precise information on the distribution of the partial sums, including real and imaginary parts.
(By the way, it is more usual to write $ax+x^{-1}$ in the phase of the Kloosterman sum, instead of $x+ax^{-1}$; this emphasizes that it is, as a function of $a$, the discrete Fourier transform of $e_p(x^{-1})$).
• If I computed right, the first case of $\left|{\rm Im} \, S_p(a,m)\right| > \sqrt p$ is ${\rm Im} \, S_{53}(24,14) = 7.2877\ldots = 1.0010\ldots \sqrt{53}$, and there is no case of $\left|{\rm Im} \, S_p(a,m)\right| > \sqrt{2p}$ for $p<2000$, though the ratios are inching closer to $\sqrt 2$, and are bound to exceed $\sqrt 2$ eventually. – Noam D. Elkies Nov 28 '14 at 6:09