The Pólya–Vinogradov inequality asserts that a non-principal Dirichlet character $\chi$ with modulus equal to $q$ satisfies
$$\displaystyle \left \lvert \sum_{N < n < N+M} \chi(n) \right \rvert = O \left(\sqrt{q} \log q \right)$$
for all positive integers $N$, $M$.
My question concerns changing the summation condition to run over Eisenstein integers having bounded norm, and the character to a primitive cubic character given by the cubic residue symbol of some element $w \in \mathbb{Z}[\zeta_3]$, where $\zeta_3$ is a primitive third root of unity. Specifically, I want to understand the sum
$$\displaystyle \left \lvert \sum_{A < N(z) < A + B} \left(\frac{z}{w} \right)_3 \right \rvert$$
where $\left(\frac{\cdot}{\cdot} \right)_3$ is the cubic residue symbol over the Eisenstein integers and $N(\cdot)$ is the norm on $\mathbb{Z}[\zeta_3]$. Can a bound of $O(\sqrt{N(w)} \log N(w))$ be obtained as in the case of running over rational integers?
Best Answer
No.
Such a bound would imply a similar bound on $$\displaystyle \left \lvert \sum_{N(z) = M} \left(\frac{z}{w} \right)_3 \right \rvert.$$
If $M$ is a product of distinct primes $p_1,\dots p_n$ congruent to $1$ mod $3$, the norms of primes $\pi_1,\dots,\pi_n$ then $$\sum_{N(z) = M} \left(\frac{z}{w} \right)_3 = \left( \left(\frac{1}{w} \right)_3 + \left(\frac{-1 }{w} \right)_3 \right) \left( \left(\frac{1}{w} \right)_3 + \left(\frac{\omega }{w} \right)_3 + \left(\frac{\omega^2 }{w} \right)_3 \right) \prod_{i=1}^n \left( \left(\frac{\pi_i}{w} \right)_3 + \left(\frac{\overline{\pi_i} }{w} \right)_3 \right)$$
so if we choose $w$ such that $\left(\frac{-1 }{w} \right)_3 =\left(\frac{\omega }{w} \right)_3 =1$ and then choose $n$ different primes $\pi$ such that $ \left(\frac{\pi_i}{w} \right)_3 =\left(\frac{\overline{\pi_i} }{w}\right)_3$ then this will have size $6 \cdot 2^n$.
Taking $n$ sufficiently large, we contradict any bound like the one you request.
To sketch a proof of a bound, let's first understand $$\left \lvert \sum_{z} \phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) \left(\frac{z}{w} \right)_3 \right \rvert$$ where $\phi$ is a smooth bump function. The trivial bound is $O(RT)$ as we are summing over an annulus of radius $R$ and thickness $T$. Another bound is provided by integrating the Fourier transform of $\phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) $ against the Fourier transform of $\left(\frac{z}{w} \right)_3 $. Since the Fourier transform of $\left(\frac{z}{w} \right)_3 $ is a bunch of Gauss sums, the same at every point, the bound we get this way will be $\sqrt{N(w)}$ times the $L^1$ norm of the Fourier transform of $\phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) $.
For $T$ small, the Fourier transform of $\phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) $ will look like the integral of a phase over a circle, i.e. a Bessel function, thus, viewed as a function of a complex variable $x$, approximately constant for $|x| < R^{-1}$ and proportial to $1/\sqrt{x}$ for larger values of $x$. The bump function will give us a rapidly decreasing cutoff at a radius of $1/T$.
Since the value at $0$ should be $RT$, the Fourier transform will take the value $\approx RT$ for $|x| <R^{-1}$ and $\approx R^{1/2} T / |x|^{1/2}$ for $ R^{-1} < |x| < T^{-1}$. The $L^1$ norm of this is $$ \int_{ R^{-1} }^{T^{-1} } (R^{1/2} T/ r^{1/2} ) \cdot 2 \pi r \cdot dr \approx R^{1/2} T (T^{-1})^{3/2} = (R/T)^{1/2} $$ so we get
$$\left \lvert \sum_{z} \phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) \left(\frac{z}{w} \right)_3 \right \rvert \ll \min ( RT, (R N(w) /T)^{1/2} )$$
Now in the sharp cutoff case we can express the characteristic function interval of radius $\sqrt{A}$ and thickness $B/ \sqrt{A}$ as a sum of smoothed characteristic functions of the same radius and dyadically decreasing thickness. So the bound we get will be obtained by summing the bounds for (dyadically) different values of $T$ from $0$ to $B/\sqrt{A}$, and thus will be comparable to the bound for the worst case of $T$.
The worst case occurs when the two terms in the minimum are equal, $ RT= (R N(w) /T)^{1/2}$, or $T = (N(w)/ R)^{1/3}$, giving a bound of $$ R^{2/3} N(w)^{1/3} = (A N(w))^{1/3},$$ in other words, by working out the details, I believe it should be possible to prove a bound of the form
$$\displaystyle \left \lvert \sum_{A < N(z) < A + B} \left(\frac{z}{w} \right)_3 \right \rvert \ll (A N(w))^{1/3}$$
This gives cancellation for $A \gg N(w)^{1/2}$, so we still obtain cancellation in intervals of greater than square-root size, which makes sense as this is possible in the classical setting (https://arxiv.org/abs/1508.00512), but not as much cancellation for larger intervals, which also makes sense as the boundaries in this setting are less smooth than in the classical one.