Counting lattice points inside a parallelepiped

Question

The problem I am about to state is in three dimensions and does not follow from Davenport's theorem. Its two-dimensional version is an immediate consequence of Pick's theorem.

Consider the lattice $\mathbb{Z}^3$. I will now define some properties of a parallelepiped $P$, for which we want to count the lattice points in the interior.

Let $M, B>0$ be two real numbers, where $M>>B$. Think of $M$ as a large number and $B$ as a constant.

• Base: the base of the parallelepiped $P$ will be a square $S$ of side-length $M$ (when I say square, I mean the boundary along with the interior). The square $S$ can be placed arbitrarily in $\mathbb{R}^3$. The only constraint is that S does not contain any points of $\mathbb{Z}^3$. Finally, let $u$ be a unit vector normal to $S$.
• Small slant-vector: let $v\in \mathbb{Z}^3$ be a lattice-vector with norm $\|v\|_2\le B$. Suppose that the inner product $v\cdot u \ge 0$.

The parallelepiped is defined as $P:=\{x+tv\ :\ x\in S, \ t\in[0,1]\}$.

Note that $vol(P)=M^2 (v\cdot u)$. The goal is to prove (or disprove) that

$$\left | \frac{|\mathbb{Z}^3 \cap P|}{M^2} - v \cdot u \right| \to 0 $$

as the real number $M\to \infty$. In other words, $vol(P)$ is a "first-order" approximation for $|\mathbb{Z}^3 \cap P|$.

Comments:

1. Observe that if we drop the assumption that $S$ does not contain any lattice point or the assumption that $v\in \mathbb{Z}^3$, then the statement does not hold.
2. Motivation: This problem comes from the theory of elasticity in physics. It is a folklore result in this field that the macroscopic continuous equations can be derived as a limit of a simple atomic model. However, I could not find a rigorous proof, so I want to prove it. The above (hopefully true) fact is needed for the proof.

Answer 1

I have not filled in absolutely all the details, but hopefully this is enough to be convincing.

Let's let $M$ take positive integer values, and let's consider the parallelepiped: $$P'=\{x+Mtv\mid x\in S, t\in [0,1]\}.$$ Because $v$ is a lattice vector and $S$ contains no lattice points, the number of lattice points in $P'$ is $M$ times the number of lattice points in $P$.

The volume of $P'$ is $M^3 (u\cdot v)$. For each lattice point $x$ in $P'$, consider a unit cube centred at $x$. "Most" of the lattice points in $P'$ are going to have the property that their entire cube lies in $P'$. More precisely, the difference between the volume of $P'$ and the sum of the volumes of the cubes centred at lattice points inside $P'$ is going to be bounded by something proportional to the area of the sides of $P'$ (because only lattice points near one of the boundary walls won't simply have their whole cube contained in $P'$). This area is of course quadratic in $M$. Another way to say this is that that the union of the unit cubes centred at the lattice points of $P'$ is a reasonable approximation to $P'$, with the difference in the two volumes being bounded by something quadratic in $M$.

Thus, the volume of $P'$ and the sum of the volumes of the cubes around the lattice points in $P'$ differ by an amount bounded by $cM^2$ for some $c$ depending on $S$ and $v$. The sum of the volumes of the unit cubes is, of course, the number of lattice points in $P'$.

Thus, $|M^3 (u\cdot v) - M|P\cap \mathbb Z^3|| < cM^2$. Dividing through by $M^3$, we get the desired result.