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\begin{center}
\rule{0.95\textwidth}{2pt}
\\[3pt]
\large \sc
Lattice points in the 3D torus
\\[3pt]
\rule{0.95\textwidth}{2pt}
\end{center}

\subsubsection*{\color{darkgreen}Introduction}

\noindent
$\bullet$ This talk is based on a joint work with my former student {\sf Dulcinea Raboso}.

Lattice point problems are very classic. They can be traced back to the first half of the 19th century.


\noindent
$\bullet$ {\sf C.F. Gauss} (1837) circle problem,  {\sf P.G.L. Dirichlet} (1849) divisor problem.

\

The general situation is that we take a domain $\Omega\subset\R^d$, we scale it to $\Omega_R=R\Omega$ with a large constant $R$, and we approximate the number of integral points in $\Omega_R$ by the volume $|\Omega_R|= |\Omega|R^d$, 
\[
 |\Omega_R\cap \Z^d|= |\Omega|R^d+O\big(R^\alpha\big).
\]
{\bf Question:} What is the best (minimal) value of $\alpha$?

\

\noindent
{\bf Example} (circle problem){\bf.}
$\Omega =$\hspace{1pt}unit circle, $d=2$.
\[
 \#\big\{
 (n,m)\in\Z^2\;:\; n^2+m^2\le R^2
 \big\}
 =
 \sum_{n\le R^2} r_2(n)
 =
 \pi R^2 +O\big(R^{\alpha_2}\big).
\]
$\bullet$ {\sf W. Sierpinski} 1906,  proved that $\alpha_2=2/3$, the so-called ``trivial'' exponent, is valid. Actually, it is not trivial, let us say that it is easy once one knows a harmonic analytic approach.

\noindent
$\bullet$ {\sf G.H. Hardy} 1915,  conjectured that any $\alpha_2>1/2$ is valid. Still open.

\noindent
$\bullet$ {\sf M.N. Huxley} 2003,  proved that $\alpha_2>131/208$ is valid (for ``general''  regular convex plane regions).

\begin{center}
 \begin{tabular}{c}
\includegraphics[scale=0.5]{./images/seg_circle.eps}
\end{tabular}
\end{center}
In a century we have walked approximately 2 steps out of 9 towards the conjecture. And the improvements are really small. For instance, the last but one best known result was $46/73$ and $46/73-131/208<1/3000$.
This is a classic and hard problem in analytic number theory.



\subsubsection*{\color{darkgreen}Usual approach}


The basic idea is using Poisson summation formula
\[
 |\Omega_R\cap \Z^d|
 = 
 \sum_{\vec{n}\in\Z^d}
 \chi_{\Omega_R}(\vec{n})
 \underset{\text{\color{blue}formal}}{=} 
 \sum_{\vec{n}\in\Z^d}
 \widehat{\chi}_{\Omega_R}(\vec{n})
\]
The last series does not converges absolutely and even it is not expected to converge at all for large $d$. Here the hat indicates the Fourier transform
\[
 \widehat{f}(\xi)= \int_{\R^d} f(\vec{x})e^{-2\pi i\vec{\xi}\cdot\vec{x}}\; d\vec{x}.
\]
To avoid the convergence problems one notes that details of size $\delta$ require frequencies at least comparable to $\delta^{-1}$.

\begin{center}
 \begin{tabular}{c}
\includegraphics[scale=0.5]{./images/checker1.eps}
\end{tabular}
\qquad
 \begin{tabular}{c}
\includegraphics[scale=0.5]{./images/checker2.eps}
\end{tabular}
\end{center}

\noindent
The effect in the formula is like losing the volume of a band of width $\delta$
\[
 |\Omega_R\cap \Z^d|
 \underset{\text{\color{blue}approx.}}{=} 
 \sum_{\|\vec{n}\|<\delta^{-1}}
 \widehat{\chi}_{\Omega_R}(\vec{n})
 +O\big(R^{d-1}\delta\big). 
\]
It is obvious that $|\widehat{\chi}_{\Omega_R}(\vec{0})|= |\Omega|R^d$, then 
\[
\boxed{
 |\Omega_R\cap \Z^d|
 \underset{\text{\color{blue}approx.}}{=} 
 |\Omega|R^d+
 \sum_{0\ne \|\vec{n}\|<\delta^{-1}}
 \widehat{\chi}_{\Omega_R}(\vec{n})
 +O\big(R^{d-1}\delta\big)}. 
\]
If we take $\delta$ small the last term is small but then we need good methods on exponential sums (perhpas too good) to deal with the middle term that oscillates.



\subsubsection*{\color{darkgreen}Our result}


The region that we consider is  the 3D torus
\[
\mathbb{T}=
\Big\{(x,y,z)\in\R^3\;:\;
\big(\rho'-\sqrt{x^2+y^2}\big)^2+z^2\le\rho^2\Big\}
\]
where $0<\rho<\rho'$ are fixed constants.
The torus has two circles of points with vanishing (Gaussian) curvature. It plays a role in the asymptotic.

\begin{center}
 \begin{tabular}{c}
\includegraphics[scale=0.65]{./images/torus1.eps}
\end{tabular}
\quad
 \begin{tabular}{c}
\includegraphics[scale=0.65]{./images/torus3.eps}
\end{tabular}
\quad
 \begin{tabular}{c}
\includegraphics[scale=0.65]{./images/torus2.eps}
\end{tabular}
\end{center}

We have proved
\[
\boxed{
 |\mathbb{T}_R\cap \Z^d|= |\mathbb{T}|R^3
 +C(R)R^{3/2}
+O\big(R^\alpha\big)}
\qquad
\text{for any $\alpha>4/3$},
\]
where $C(R)$ is a periodic function (say a constant in arithmetic progressions).
\[
|\mathbb{T}|R^3=2\pi^2\rho^2\rho'R^3=\text{\small volume term},
\quad
 C(R)R^{3/2}=\text{\small secondary main term.}
\]


This problem is easier than the circle problem because we have walked approximately 2 steps out of 6 towards the conjecture that in this case claims the result for any $\alpha>1$.
\begin{center}
 \begin{tabular}{c}
\includegraphics[scale=0.5]{./images/seg_torus.eps}
\end{tabular}
\end{center}




\subsubsection*{\color{darkgreen}Our method}

Following the scheme mentioned before, we apply
\[
 |\Omega_R\cap \Z^d|
 \underset{\text{\color{blue}approx.}}{=} 
 |\Omega|R^d+
 \sum_{0\ne \|\vec{n}\|<\delta^{-1}}
 \widehat{\chi}_{\Omega_R}(\vec{n})
 +O\big(R^{d-1}\delta\big)
\]
with $\Omega=\mathbb{T}$, $d=3$ and $\delta= R^{-2/3}$, in this way $O\big(R^{d-1}\delta\big)=O\big(R^{4/3}\big)$.
Then the problem boils down to estimate
\[
 \sum_{0\ne \|\vec{n}\|<R^{2/3}}
 \widehat{\chi}_{\mathbb{T}_R}(\vec{n}).
\]

A difference with respect to the classic scheme (for convex regions) is that when we apply the stationary phase approximation to study $\widehat{\chi}_{\mathbb{T}_R}(\vec{n})$, the method collapses when $\vec{n}$ is the normal at a point of vanishing curvature of $\mathbb{T}_R$. In our case, we have to separate the vertical vectors,
\[
 \sum_{0<n<R^{2/3}}
 \widehat{\chi}_{\mathbb{T}_R}(0,0,n)\quad\rightarrow\quad
 C(R)R^{3/2}+\text{negligible terms}
\]
and we get the secondary main term.

For bounding the rest of the sum we elaborate a trick introduced in~1963 for the sphere problem by the celebrated exponential sum estimator \raisebox{-3pt}{\includegraphics[scale=0.5]{./images/caligr2.eps}} and independently and simultaneously  by one of the most conspicuous Chinese number theorists
\begin{center}
 \begin{tabular}{c}
\includegraphics[scale=0.5]{./images/caligr.eps}
\end{tabular}
\qquad
\begin{tabular}{c}
(Chen Jingrun).
\end{tabular}
\end{center}

The rough idea is that the torus has cylindrical symmetry and it is preserved in the Fourier transform side, then we have to deal with oscillatory sums of the form
\[
 \sum_m\sum_n\sum_l 
 f\big(\sqrt{n^2+m^2},l\big)
\]
When $m$ is frozen, the derivative of $\sqrt{x^2+m^2}$ is like~$1$. If we glue the variables $k=n^2+m^2$ we have something like $\sqrt{x}$ that has a smaller derivative
\[
 \sum_m\sum_n\sum_l 
 f\big(\sqrt{n^2+m^2},l\big)
=
 \sum_k\sum_l 
 r_2(k)
 f\big(\sqrt{k},l\big).
\]
In analytic terms this is, in principle,  good because sums with less oscillation are easier to control. But we have to pay prize with a chaotic arithmetic coefficient $r_2(k)$ that escapes to  analytic methods. 

Although $r_2(k)$ is chaotic, it is close to be bounded, in fact its quadratic mean grows as a logarithm.  Then it is not a fatal loss to use  Cauchy-Schwarz inequality to get rid of it
\[
\Big|
\sum_{k\le K}\sum_l 
 r_2(k)
 f\big(\sqrt{k},l\big)
\Big|^2 
 \underset{\text{\color{blue}approx.}}{\le} 
 K
\sum_{k\le K}\Big|
\sum_{l} 
 f\big(\sqrt{k},l\big)
 \Big|^2.
\]
Opening the square, and changing the order of summation we get
\[
\sum_{l_1} 
\sum_{l_1} 
\sum_k
 f\big(\sqrt{k},l_1\big)
 \bar{f}\big(\sqrt{k},l_2\big)
\]
and the inner sum is easier to study because of the control on the oscillation.

\

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\


The preprint ``Lattice points in the 3-dimensional torus'' by F. Chamizo and D. Raboso will be posted on the arXiv soon.

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