2 Geodesics in the L-solid manifold

We begin the proof of Theorem 1.1 with a discussion of the special case when the polycube $3$ -manifold in question is obtained as a translation $3$ -manifold from the L-solid region shown in Figure 3. Here the three unit cubes of the region are called the top cube, the middle cube and the right cube. The street-LCM of the resulting L-solid $3$ -manifold is $2$ . We consider geodesics inside this manifold and, in particular, those with directions given by (1-1)–(1-2).

Figure 3

The L-solid region.

For convenience, we refer to the L-solid region or the L-solid $3$ -manifold simply as the L-solid.

We introduce a convenient labeling of the faces of the L-solid; see Figures 4 and 5.

The picture on the left in Figure 4 shows the three front faces of the L-solid, with $y=0$ . The front top square face $Y_{3}$ has vertices $(0,0,1)$ , $(1,0,1)$ , $(0,0,2)$ and $(1,0,2)$ . We denote this fact by

$$ \begin{align*} Y_3=\operatorname{\mathrm{SQ}}\{(0,0,1),(1,0,1),(0,0,2),(1,0,2)\}. \end{align*} $$

The front middle and front right square faces are denoted respectively by

$$ \begin{align*} Y_{1}&=\operatorname{\mathrm{SQ}}\{(0,0,0),(1,0,0),(0,0,1),(1,0,1)\},\\ Y_{2}&=\operatorname{\mathrm{SQ}}\{(1,0,0),(2,0,0),(1,0,1),(2,0,1)\}. \end{align*} $$

On the other hand, the picture on the right in Figure 4 shows the three back faces of the L-solid, with $y=1$ . The back top, back middle and back right square faces are denoted respectively by

$$ \begin{align*} Y_{6}&=\operatorname{\mathrm{SQ}}\{(0,1,1),(1,1,1),(0,1,2),(1,1,2)\},\\ Y_{4}&=\operatorname{\mathrm{SQ}}\{(0,1,0),(1,1,0),(0,1,1),(1,1,1)\},\\ Y_{5}&=\operatorname{\mathrm{SQ}}\{(1,1,0),(2,1,0),(1,1,1),(2,1,1)\}. \end{align*} $$

These six square faces are perpendicular to the y-axis, justifying use of the letter Y.

We see from Figure 5 that between the front and back faces, there are four faces

$$ \begin{align*} X_{1}&=\operatorname{\mathrm{SQ}}\{(0,0,0),(0,1,0),(0,0,1),(0,1,1)\},\\ X_{2}&=\operatorname{\mathrm{SQ}}\{(0,0,1),(0,1,1),(0,0,2),(0,1,2)\},\\ X_{3}&=\operatorname{\mathrm{SQ}}\{(2,0,0),(2,1,0),(2,0,1),(2,1,1)\},\\ X_{4}&=\operatorname{\mathrm{SQ}}\{(1,0,1),(1,1,1),(1,0,2),(1,1,2)\}, \end{align*} $$

on the boundary of the L-solid that are perpendicular to the x-axis, another four faces

$$ \begin{align*} Z_{1}&=\operatorname{\mathrm{SQ}}\{(0,0,0),(1,0,0),(0,1,0),(1,1,0)\},\\ Z_{2}&=\operatorname{\mathrm{SQ}}\{(1,0,0),(2,0,0),(1,1,0),(2,1,0)\},\\ Z_{3}&=\operatorname{\mathrm{SQ}}\{(0,0,2),(1,0,2),(0,1,2),(1,1,2)\},\\ Z_{4}&=\operatorname{\mathrm{SQ}}\{(1,0,1),(2,0,1),(1,1,1),(2,1,1)\}, \end{align*} $$

on the boundary of the L-solid that are perpendicular to the z-axis, and two inside faces

$$ \begin{align*} X_{5}&=\operatorname{\mathrm{SQ}}\{(1,0,0),(1,1,0),(1,0,1),(1,1,1)\},\\ Z_{5}&=\operatorname{\mathrm{SQ}}\{(0,0,1),(1,0,1),(0,1,1),(1,1,1)\}. \end{align*} $$

Figure 4

Labeling the front and back faces of the L-solid.

Figure 5

Labeling the rest of the faces of the L-solid.

Let $k\geqslant 1$ be an arbitrary but fixed integer. Assume that a segment of a geodesic $\mathcal {L}_{k}$ starts from the origin $\mathbf {0}=(0,0,0)$ and ends at a point $C=(1,x,y)$ on the inside square face $X_{5}$ , and in between hits the square face $Z_{3}$ on k separate occasions; see Figure 6, which shows the special case $k=1$ .

Figure 6

First detour crossing in the L-solid: $k=1$ .

Let B denote the point on the square face $Z_{3}$ that $\mathcal {L}_{k}$ hits on the last occasion before it bounces down to the point $B^{\prime }$ and continues towards the point C.

Let A denote the point on the inside square face $Z_{5}$ that $\mathcal {L}_{k}$ hits on the first occasion, and assume that $A=(x,y,1)$ , so that its coordinates form a permutation of the coordinates of $C=(1,x,y)$ with the same quantities x and y. It is easy to see that $B=(2kx,2ky,2)$ and $B^{\prime }=(2kx,2ky,0)$ . The geometric fact that the two vectors $\mathbf {0} A$ and $B^{\prime }C$ are parallel gives rise to the equations

$$ \begin{align*} \frac{1-2kx}{x}=\frac{x-2ky}{y}=\frac{y-0}{1}. \end{align*} $$

The first equality reduces to $y=x^{2}$ . Substituting this into the right-hand side and then equating with the left-hand side, we conclude that

(2-1) $$ \begin{align} y=x^{2} \quad \mbox{and}\quad x^{3}+2kx-1=0. \end{align} $$

Note that the cubic polynomial $x^{3}+2kx-1$ is strictly increasing and has precisely one root $\alpha _{k}$ satisfying

(2-2) $$ \begin{align} \frac{1}{2k+1}<\alpha_{k}<\frac{1}{2k}. \end{align} $$

Of course, (2-1)–(2-2) represent the special case of (1-1) when $h=2$ , the street-LCM of the L-solid.

We now take $x=\alpha _{k}$ and $y=\alpha ^{2}_{k}$ , and consider the segment of the geodesic $\mathcal {L}_{k}$ with direction given by $\mathbf {v}_{0}=(\alpha _{k},\alpha ^{2}_{k},1)$ illustrated in Figure 6. Starting from the origin $\mathbf {0}=(0,0,0)$ , this geodesic segment exhibits up-and-down zigzagging, and finally ends its journey at the point $C=(1,\alpha _{k},\alpha ^{2}_{k})$ on the middle square face $X_{5}$ . It represents a left-to-right detour crossing inside a tower-like three-dimensional street, where the latter is the union of the middle cube and the top cube. Observe that with $0<\alpha _{k}<1$ , the geodesic $\mathcal {L}_{k}$ goes to the right faster than it goes to the back, and so crosses from one X-face to another faster than crossing from one Y-face to another.

The straight-line segment joining the two endpoints $\mathbf {0}$ and $C=(1,\alpha _{k},\alpha ^{2}_{k})$ of this detour crossing is called the shortcut of this detour crossing.

The first extension of this particular zigzagging segment of the geodesic $\mathcal {L}_{k}$ with direction given by $\mathbf {v}_{0}=(\alpha _{k},\alpha ^{2}_{k},1)$ starts from the point $C=(1,\alpha _{k},\alpha ^{2}_{k})$ on the square face $X_{5}$ and goes to the point $D=(2,2\alpha _{k},2\alpha ^{2}_{k})$ on the right face $X_{3}$ ; see Figure 7. This zigzagging extension hits the square face $Z_{4}$ on $2k$ separate occasions, at the points $(1+i\alpha _{k}-\alpha _{k}^{3},\alpha _{k}+i\alpha ^{2}_{k}-\alpha _{k}^{4},1)$ , $1\leqslant i\leqslant 2k$ , before ending at the point D on the square face $X_{3}$ .

Figure 7

Second detour crossing in the L-solid: $k=1$ .

This zigzagging second segment of the geodesic $\mathcal {L}_{k}$ from C to D with direction given by $\mathbf {v}_{0}$ represents a left-to-right detour crossing inside a three-dimensional street, which is simply the right cube. Again the straight-line segment joining the two endpoints C and D of this detour crossing is called the shortcut of this detour crossing.

These two shortcuts $\mathbf {0} C$ and $CD$ , with endpoints $\mathbf {0}=(0,0,0)$ , $C=(1,\alpha _{k},\alpha ^{2}_{k})$ and $D=(2,2\alpha _{k},2\alpha ^{2}_{k})$ , are clearly collinear. In fact, the point C is the midpoint of the line segment $\mathbf {0} D$ .

It is easy to see that this collinearity of the shortcuts is preserved as we continue and take the third, fourth and subsequent segments of the geodesic $\mathcal {L}_{k}$ starting from the origin with direction given by $\mathbf {v}_{0}=(\alpha _{k},\alpha ^{2}_{k},1)$ . This collinearity means precisely that these consecutive shortcuts together form another geodesic $\mathcal {L}^{*}_{k}$ starting from the origin, but it has a new direction given by $\mathbf {v}_{1}=(1,\alpha _{k},\alpha ^{2}_{k})$ obtained by a permutation of the coordinates of $\mathbf {v}_{0}$ . We refer to this new geodesic $\mathcal {L}^{*}_{k}$ as the shortline of the original geodesic $\mathcal {L}_{k}$ . Formally, $\mathbf {S}(\mathcal {L}_{k})=\mathcal {L}^{*}_{k}$ , where $\mathbf {S}$ denotes the shortline operation. In the special case $k=1$ , Figure 8 is basically Figures 6 and 7 put together. It shows the two parts of the zigzagging $\mathcal {L}_{k}$ from $\mathbf {0}$ to D, with the point C separating the two parts. It also shows the corresponding shortline segment of $\mathcal {L}^{*}_{k}$ , indicated by the dashed arrow, from $\mathbf {0}$ to D.

Figure 8

Detour crossing and its shortline in the L-solid: $k=1$ .

The crucial fact is that the geodesic $\mathcal {L}_{k}$ and its shortline $\mathcal {L}^{*}_{k}$ hit every square face $X_{i}$ , $1\leqslant i\leqslant 5$ , at precisely the same points, like C and D in Figures 7 and 8. We refer to this observation as the X-face hitting property of the infinite geodesic $\mathcal {L}_{k}$ and its shortline $\mathcal {L}^{*}_{k}$ .

The vector $\mathbf {v}_{1}=(1,\alpha _{k},\alpha ^{2}_{k})$ is clearly obtained from the vector $\mathbf {v}_{0}=(\alpha _{k},\alpha ^{2}_{k},1)$ by a left shift in the cyclic permutation of the coordinates

$$ \begin{align*} 1\to\alpha_{k}\to\alpha^{2}_{k}\to1. \end{align*} $$

Applying a second left shift in this cyclic permutation, we obtain in turn a new vector $\mathbf {v}_{2}=(\alpha ^{2}_{k},1,\alpha _{k})$ . It is easy to see that the shortline of the geodesic $\mathcal {L}^{*}_{k}$ is a new geodesic $\mathcal {L}^{**}_{k}$ that starts at the origin and has direction given by $\mathbf {v}_{2}$ . Formally, $\mathbf {S}(\mathcal {L}^{*}_{k})=\mathcal {L}^{**}_{k}$ . Note that $\mathcal {L}^{*}_{k}$ consists of front-to-back detour crossings, and $\mathcal {L}^{**}_{k}$ is the union of the corresponding shortcuts. Observe that with $0<\alpha _{k}<1$ , the geodesic $\mathcal {L}^{*}_{k}$ goes to the back faster than it goes up, and so crosses from one Y-face to another faster than crossing from one Z-face to another.

Again the crucial fact is that the geodesic $\mathcal {L}^{*}_{k}$ and its shortline $\mathcal {L}^{**}_{k}$ hit every square face $Y_{i}$ , $1\leqslant i\leqslant 6$ , at precisely the same points. We refer to this observation as the Y-face hitting property of the infinite geodesic $\mathcal {L}^{*}_{k}$ and its shortline $\mathcal {L}^{**}_{k}$ .

Applying a third left shift in the cyclic permutation, we return to the original direction given by $\mathbf {v}_{0}=(\alpha _{k},\alpha ^{2}_{k},1)$ . It is easy to see that the shortline of the geodesic $\mathcal {L}^{**}_{k}$ is the original geodesic $\mathcal {L}_{k}$ that starts at the origin and has direction vector $\mathbf {v}_{0}$ . Formally, $\mathbf {S}(\mathcal {L}^{**}_{k})=\mathcal {L}_{k}$ . Note that $\mathcal {L}^{**}_{k}$ consists of bottom to top detour crossings, and $\mathcal {L}_{k}$ is the union of the corresponding shortcuts. Observe that with $0<\alpha _{k}<1$ , the geodesic $\mathcal {L}^{**}_{k}$ goes up faster than it goes to the right, and so crosses from one Z-face to another faster than crossing from one X-face to another.

Again the crucial fact is that the geodesic $\mathcal {L}^{**}_{k}$ and its shortline $\mathcal {L}_{k}$ hit every square face $Z_{i}$ , $1\leqslant i\leqslant 5$ , at precisely the same points. We refer to this observation as the Z-face hitting property of the infinite geodesic $\mathcal {L}^{**}_{k}$ and its shortline $\mathcal {L}_{k}$ .

The remarkable face hitting properties explain why we focus on geodesics with direction given by one of the vectors

(2-3) $$ \begin{align} \mathbf{v}_{0}=(\alpha_{k},\alpha^{2}_{k},1), \quad \mathbf{v}_{1}=(1,\alpha_{k},\alpha^{2}_{k}), \quad \mathbf{v}_{2}=(\alpha^{2}_{k},1,\alpha_{k}). \end{align} $$

We prove uniformity of such geodesics in the L-solid by applying an adaptation of an area magnification process via shortlines, developed in [Reference Beck, Chen and Yang1, Section 6.3]. To make the present paper self-contained, we explain this magnification process in full detail in Sections 3 and 4.

In [Reference Beck, Chen and Yang1, Section 6.3], we establish time-quantitative density. In fact, we establish a nearly optimal form of density of such geodesics in the L-solid. To explain this, let $\eta>0$ be an arbitrarily small but fixed positive number. We say that a half-infinite geodesic $\mathcal {L}$ in the L-solid is $\eta $ -nearly superdense if there exists an effectively computable explicit threshold $N_{0}(\eta )$ such that, for every integer $n\geqslant N_{0}(\eta )$ , the initial segment of $\mathcal {L}$ with length $n^{2+\eta }$ intersects every axis-parallel cube of side length $1/n$ in the L-solid.

The following result is [Reference Beck, Chen and Yang1, Theorem 6.1.2].

In fact, we need only a straightforward corollary of the above result, obtained by choosing any fixed value of $\eta $ in the interval $0<\eta <1$ .

Our first goal is to establish uniformity of these special geodesics in the L-solid. Unfortunately, our proof does not give any error term, so what we can prove here is time-qualitative uniformity. The proof consists of four major steps:

1. (i) preparation for the magnification process;

2. (ii) the magnification process;

3. (iii) grids and iteration; and

4. (vi) conclusion via time-quantitative density B.

We cover these in the next four sections.

3 Preparation for the magnification process

We use Birkhoff’s well-known pointwise ergodic theorem concerning measure-preserving systems $(X,\mathcal {A},\mu ,T)$ . The triple $(X,\mathcal {A},\mu )$ is a measure space, where X is the underlying space, $\mathcal {A}$ is a $\sigma $ -algebra of sets in X, while $\mu $ is a nonnegative $\sigma $ -additive measure on X with $\mu (X)<\infty $ , and $T:X\to X$ is a measure-preserving transformation, so that $T^{-1}A\in \mathcal {A}$ and $\mu (T^{-1}A)=\mu (A)$ for every $A\in \mathcal {A}$ . Here we simply apply ergodic theory, and do not expect the reader to have any serious expertise in the subject. Knowledge of Lebesgue integral and basic measure theory suffices.

Let $L^{1}(X,\mathcal {A},\mu )$ denote the space of measurable and integrable functions in the measure space $(X,\mathcal {A},\mu )$ . Then Birkhoff’s pointwise ergodic theorem says that for every function $f\in L^{1}(X,\mathcal {A},\mu )$ , the limit

(3-1) $$ \begin{align} \lim_{m\to\infty}\frac{1}{m}\sum_{j=0}^{m-1}f(T^{j}x)=f^{*}(x) \end{align} $$

exists for $\mu $ -almost every $x\in X$ , where $f^{*}\in L^{1}(X,\mathcal {A},\mu )$ is a T-invariant measurable function satisfying the condition

$$ \begin{align*} \int_{X}f\,{d}\mu=\int_{X}f^{*}\,{d}\mu. \end{align*} $$

A particularly important special case is when T is ergodic, when every measurable T-invariant set $A\in \mathcal {A}$ is trivial in the precise sense that $\mu (A)=0$ or $\mu (A)=\mu (X)$ . This is equivalent to the assertion that every measurable T-invariant function is constant $\mu $ -almost everywhere.

If T is ergodic, then (3-1) simplifies to

(3-2) $$ \begin{align} \lim_{m\to\infty}\frac{1}{m}\sum_{j=0}^{m-1}f(T^{j}x)=\int_{X}f\,{d}\mu, \end{align} $$

and the right-hand side of (3-1) is the same constant for $\mu $ -almost every $x\in X$ .

The remarkable intuitive interpretation of (3-2) is that the time average on the left-hand side is equal to the space average on the right-hand side.

Unfortunately, Birkhoff’s theorem does not give the speed of convergence in (3-1) or (3-2).

Next we explain how ergodicity and Birkhoff’s theorem are used in the proof. Recall the labeling of the square faces of the L-solid shown in Figures 4 and 5. Consider the five square faces $X_{i}$ , $1\leqslant i\leqslant 5$ , each with area $1$ . Boundary identification in the L-solid gives $X_{1}=X_{3}$ and $X_{2}=X_{4}$ , so that we simply have the three square faces $X_{1},X_{2},X_{5}$ . We define our underlying measure space as the set $X_{0}=X_{1}\cup X_{2}\cup X_{5}$ with the usual area, or two-dimensional Lebesgue measure, denoted by . Then $X_{0}$ is in fact a compact flat surface, that is, a polysquare surface, of area $3$ , so that .

We use the special direction $\mathbf {v}_{1}=(1,\alpha _{k},\alpha ^{2}_{k})$ , where $\alpha _{k}$ is a root of the cubic equation $x^{3}+2kx-1=0$ satisfying (2-2).

The $\mathbf {v}_{1}$ -flow in the L-solid defines a -preserving transformation on $X_{0}$ in the natural way as illustrated in Figure 9. For instance, the point $C\in X_{5}$ is mapped to the point $D\in X_{3}=X_{1}$ via the $\mathbf {v}_{1}$ -flow. Similarly, the $\mathbf {v}_{1}$ -flow maps almost every point of the measure-space $X_{0}$ to another point of $X_{0}$ ; here we ignore the singularities. Let $T=T_{\mathbf {v}_{1}}:X_{0}\to X_{0}$ denote this -preserving transformation. In particular, $T(C)=D$ in Figure 9.

Figure 9

Transformation $T=T_{\mathbf {v}_{1}}$ , where $\mathbf {v}_{1}=CD$ .

The major part of our argument is to prove that this particular transformation $T=T_{\mathbf {v}_{1}}$ is ergodic. In other words, we need to show that if $S\subset X_{0}$ is a measurable T-invariant set, then or . Once ergodicity is established, it is relatively straightforward to derive uniformity via Birkhoff’s theorem, at least for almost every starting point.

We prove ergodicity by contradiction. Assume to the contrary that there is a nontrivial measurable T-invariant set $S_{0}\subset X_{0}$ such that . We then derive a contradiction by using a version of the magnification process via shortlines developed in [Reference Beck, Chen and Yang1, Sections 6.2–6.3]. Note that our assumption ensures that the complement $S_{0}^{c}=X_{0}\setminus S_{0}$ is another nontrivial measurable T-invariant set.

Removing a set of measure zero, we can clearly assume that for every point $x\in S_{0}$ , $T^{j}x$ is well defined for every $j\geqslant 1$ .

Given a point $z\in X_{0}$ and a radius $0<r<1/2$ , let $D(z;r)$ denote the circular disk of radius r and centered at z. Then $D(z;r)$ has area $r^{2}\pi $ . Note that $D(z;r)\subset X_{0}$ , due to the fact that $X_{0}$ is a compact flat surface. Lebesgue’s density theorem then says that, for almost every $z\in S_{0}$ ,

and for almost every $z\in S_{0}^{c}=X_{0}\setminus S_{0}$ ,

Let $n\geqslant 1$ be an integer, and divide each of the square faces $X_{1},X_{2},X_{5}$ into $n^{2}$ congruent squares of area $n^{-2}$ in the standard way. We refer to them as small n-squares. Thus, we have $3n^{2}$ small n-squares in $X_{0}$ .

Lemma 3.1. Suppose that the real number $\tau $ satisfies

(3-3)

Let $\varepsilon>0$ be arbitrarily small but fixed. Then there exists a finite integer-valued threshold $n=n(S_{0};\varepsilon )$ such that there exist at least $(\tau -\varepsilon )n^{2}$ small n-squares Q with

(3-4)

Proof. Since $S_{0}^{c}=X_{0}\setminus S_{0}$ is Lebesgue measurable, given any $\delta>0$ , there exist finitely many disjoint axis-parallel rectangles such that their union V is $\delta $ -close to $S_{0}^{c}=X_{0}\setminus S_{0}$ in the sense of the measure of the symmetric difference, so that

It follows that

Since V is a finite union of disjoint axis-parallel rectangles, there clearly exists an integer-valued threshold $n=n(V;\delta )$ such that the union $V_{1}$ of the small n-squares Q contained in V has measure at least

Let $\mathcal {B}$ denote the set of n-squares Q that do not satisfy (3-4), so that

Then

so that

$$ \begin{align*} \vert\mathcal{B}\vert\leqslant\frac{\delta n^2}{\varepsilon}=\frac{\varepsilon n^2}{3}, \end{align*} $$

if we take $\delta =\varepsilon ^{2}/3$ . We may assume without loss of generality that $0<\varepsilon <1$ . Deleting the n-squares $Q\in \mathcal {B}$ , we see that $V_{1}$ contains at least

$$ \begin{align*} (\tau-2\delta)n^2-\frac{\varepsilon n^2}{3}=\bigg(\tau-\frac{2\varepsilon^2}{3}-\frac{\varepsilon}{3}\bigg)n^2>(\tau-\varepsilon)n^2 \end{align*} $$

n-squares Q that satisfy (3-4). Finally, note that V and $\delta $ depend on $S_{0}$ and $\varepsilon $ .

We now continue our discussion under the assumptions of Lemma 3.1.

Let $\mathcal {Q}$ denote the set of small n-squares Q that satisfy the bound (3-4). For notational convenience, we may write

(3-5) $$ \begin{align} \mathcal{Q}=\{Q_{i}:i\in I\}. \end{align} $$

Then, in view of Lemma 3.1, we have

(3-6) $$ \begin{align} \vert I\vert=\vert\mathcal{Q}\vert\geqslant(\tau -\varepsilon)n^{2}. \end{align} $$

For every $i\in I$ in (3-5), let $\chi _{i}$ denote the characteristic function of the subset $S_{0}\cap Q_{i}$ of $X_{0}$ , so that

$$ \begin{align*} \chi_i(x)=\begin{cases} 1\quad&\mbox{if }x\in S_{0}\cap Q_{i},\\ 0\quad&\mbox{otherwise}. \end{cases} \end{align*} $$

Applying Birkhoff’s theorem in the form (3-1) to each function $\chi _{i}$ , $i\in I$ , we have

(3-7) $$ \begin{align} \lim_{m\to\infty}\frac{1}{m}\sum_{j=0}^{m-1}\chi_{i}(T^{j}x)=\chi^{*}_{i}(x), \end{align} $$

for $\mu $ -almost every $x\in X_{0}$ , where each $\chi ^{*}_{i}(x)$ is some T-invariant measurable function satisfying the condition

(3-8)

Combining (3-4) and (3-8), for every $i\in I$ , we have

(3-9)

Using the left-hand side of (3-7), we deduce that $\chi ^{*}_{i}(x)$ is nonnegative $\mu $ -almost everywhere. It then follows from (3-9) that

(3-10)

Taking the minimum of the function

$$ \begin{align*} g(x)=\frac{1}{\vert I\vert}\sum_{i\in I}\chi^*_i(x) \end{align*} $$

over $x\in S_{0}$ , or getting arbitrarily close to that, (3-10) implies the existence of some $x_{0}\in S_{0}$ such that

(3-11) $$ \begin{align} 0\leqslant\frac{1}{\vert I\vert}\sum_{i\in I}\chi^{*}_{i}(x_{0})<\frac{\varepsilon}{(3-\tau)n^{2}}, \end{align} $$

where the factor $3-\tau $ in the denominator on the right-hand side is the measure of $S_{0}$ given by (3-3).

It now follows from (3-11) that there exists a subset $I_{0}\subset I$ satisfying $\vert I_{0}\vert \geqslant \vert I\vert /2$ such that

(3-12) $$ \begin{align} 0\leqslant\chi^{*}_{i}(x_{0})\leqslant\frac{2\varepsilon}{(3-\tau)n^{2}} \quad\mbox{for every }i\in I_{0}. \end{align} $$

Note that in view of (3-6), we have

(3-13) $$ \begin{align} \vert I_{0}\vert\geqslant\frac{(\tau-\varepsilon)n^{2}}{2}. \end{align} $$

On the other hand, combining (3-7) and (3-12), we see that

(3-14) $$ \begin{align} \lim_{m\to\infty}\frac{1}{m}\sum_{j=0}^{m-1}\chi_{i}(T^{j}x_{0}) =\chi^{*}_{i}(x_{0}) \leqslant\frac{2\varepsilon}{(3-\tau)n^{2}} \quad\mbox{for every }i\in I_{0}. \end{align} $$

Since every point in $S_{0}$ is nonpathological, $T^{j}x_{0}$ is well defined for every $j\geqslant 1$ . It also follows from (3-14) that there exists a finite threshold $m_{0}$ such that for every integer $m\geqslant m_{0}$ ,

(3-15) $$ \begin{align} \frac{1}{m}\sum_{j=0}^{m-1}\chi_{i}(T^{j}x_{0}) <\frac{3\varepsilon}{(3-\tau)n^{2}} \quad\mbox{for every }i\in I_{0}. \end{align} $$

We recall that $\chi _{i}$ is the characteristic function of the set $S_{0}\cap Q_{i}$ . Since $x_{0}$ is in the nontrivial T-invariant set $S_{0}$ , the infinite sequence $T^{j}x_{0}$ , $j\geqslant 0$ , never enters the set $Q_{i}\setminus S_{0}$ . Let $\chi ^{\star \star }_{i}$ denote the characteristic function of the small n-square $Q_{i}$ , so that

$$ \begin{align*} \chi^{\star\star}_i(x)=\begin{cases} 1\quad&\mbox{if }x\in Q_{i},\\ 0\quad&\mbox{otherwise}. \end{cases} \end{align*} $$

Then (3-15) is equivalent to the assertion that for every integer $m\geqslant m_{0}$ ,

(3-16) $$ \begin{align} \frac{1}{m}\sum_{j=0}^{m-1}\chi^{\star\star}_{i}(T^{j}x_{0}) <\frac{3\varepsilon}{(3-\tau)n^{2}} \quad\mbox{for every }i\in I_{0}. \end{align} $$

Note that every small n-square $Q_{i}$ has area $n^{-2}$ . Then (3-16) implies that for every integer $m\geqslant m_{0}$ ,

(3-17)

Choosing a sufficiently small $\varepsilon>0$ , this gives the message that the small n-square $Q_{i}$ is grossly undervisited by the sequence $T^{j}x_{0}$ , $0\leqslant j\leqslant m-1$ .

Indeed, since , the term

(3-18)

represents the expected value of the number of the points $T^{j}x_{0}\in Q_{i}$ , $0\leqslant j\leqslant m-1$ . If $0<\tau <3$ is fixed and $\varepsilon>0$ is sufficiently small, then the factor

$$ \begin{align*} \frac{9\varepsilon}{3-\tau} \end{align*} $$

in (3-17) justifies the term grossly undervisited. Here we assume that the expected value (3-18) is large, which is clearly possible, since $m\geqslant m_{0}$ can be arbitrarily large.

Note that every $Q_{i}$ , where $i\in I_{0}$ , is grossly undervisited. In view of (3-13), these represent a positive proportion of the total number $3n^{2}$ of small n-squares in $X_{0}$ .

Suppose that every small n-square $Q_{i}$ , $i\in I_{0}$ , is divided into

(3-19) $$ \begin{align} \frac{6\varepsilon m}{(3-\tau)n^{2}} \end{align} $$

convex parts with equal area; for notational simplicity, let us assume here that the quantity (3-19) is an integer. We refer to these as the tiny convex parts. Since

$$ \begin{align*} \frac{6\varepsilon m}{(3-\tau)n^2}=2\frac{9\varepsilon}{(3-\tau)}\frac{m}{3n^2}, \end{align*} $$

it follows from (3-17) and (3-19) that at least half of these tiny convex parts of $Q_{i}$ are empty, that is, they do not contain any element of the sequence $T^{j}x_{0}$ , $0\leqslant j\leqslant m-1$ . We refer to them as the empty tiny convex parts of $Q_{i}$ , $i\in I_{0}$ .

In the next section, we give an explicit construction of these tiny convex parts of $Q_{i}$ , $i\in I_{0}$ , which have the same area.

4 The magnification process

We consider iterated area magnification of convex sets on the faces by using the three X–Y–Z-face hitting properties and the $3$ -periodicity of the shortline process. We elaborate on this.

Figure 10 illustrates area magnification via tilted parallel projection.

Figure 10

Magnifying the area via tilted parallel projection.

In order to visualize the parallel projection a little better, we have included an extra copy of three unit square faces, indicated by the dashed line squares, on the plane $y=1$ in cartesian $3$ -space that can be identified with the square faces $Y_{4},Y_{5},Y_{6}$ of the L-solid. We start with a parallelogram $S_{0}=ABCD$ on the square face $X_{3}$ . Using the $\mathbf {v}_{1}$ -flow indicated by the arrow, where $\mathbf {v}_{1}=(1,\alpha _{k},\alpha ^{2}_{k})$ is the direction vector of the shortline $\mathcal {L}^{*}_{k}$ of the geodesic $\mathcal {L}_{k}$ , we project the parallelogram on to the plane $y=1$ . For simplicity, suppose that it is projected on to unit squares identified with $Y_{4}\cup Y_{5}$ , as shown. This tilted parallel projection maps the parallelogram $S_{0}$ to a new parallelogram $S_{1}=A^{\prime }B^{\prime }C^{\prime }D^{\prime }$ , and the area of $S_{1}$ is $1/\alpha _{k}$ times the area of $S_{0}$ . We say that in Figure 10, the image of $S_{0}=ABCD$ splits, in the sense that the image parallelogram $S_{1}=A^{\prime }B^{\prime }C^{\prime }D^{\prime }$ is located on more than one square face. We aim to avoid splitting as much as possible.

But we do not stop here. We want to describe a full $3$ -cycle $S_{0}\to S_{1}\to S_{2}\to S_{3}$ as illustrated in Figure 11, where $S_{2}$ is on a Z-face and $S_{3}$ returns to an X-face.

Figure 11

Illustrating the $3$ -cycle of the magnification process.

In general, we start with a parallelogram $S_{0}$ on some square face $X_{i}$ of the L-solid that is contained in a plane $x=x_{0}^{*}$ in cartesian $3$ -space, where $x_{0}^{*}$ is an integer.

As a first step, let us project $S_{0}$ by the vector $\mathbf {v}_{1}=(1,\alpha _{k},\alpha _{k}^{2})$ to a parallelogram $S_{1}$ on a plane $y=y_{1}^{*}$ , where $y_{1}^{*}$ is an integer, with the point $(x_{0}^{*},c_{1},c_{2})\in S_{0}$ at the center projected to the point $(c_{3},y_{1}^{*},c_{4})\in S_{1}$ . If $\mathbf {v}_{1}=(1,\alpha _{k},\alpha _{k}^{2})$ projects a point $(x_{0}^{*},c_{1}+y_{0},c_{2}+z_{0})\in S_{0}$ to the point $(c_{3}+x_{1},y_{1}^{*},c_{4}+z_{1})\in S_{1}$ , then a simple calculation shows that

$$ \begin{align*} x_1=-\frac{y_0}{\alpha_k}=\frac{z_1-z_0}{\alpha_k^2}. \end{align*} $$

Hence, $x_{1}=-\alpha _{k}^{-1}y_{0}$ and $z_{1}=z_{0}-\alpha _{k}y_{0}$ , and so

(4-1) $$ \begin{align} \begin{pmatrix} x_{1}\\ z_{1} \end{pmatrix} =\begin{pmatrix} -\alpha_{k}^{-1}&0\\ -\alpha_{k}&1 \end{pmatrix} \begin{pmatrix} y_{0}\\ z_{0} \end{pmatrix}. \end{align} $$

As a second step, let us project $S_{1}$ by the vector $\mathbf {v}_{2}=(\alpha _{k}^{2},1,\alpha _{k})$ to a parallelogram $S_{2}$ on a plane $z=z_{2}^{*}$ , where $z_{2}^{*}$ is an integer, with the point $(c_{3},y_{1}^{*},c_{4})\in S_{1}$ at the center projected to the point $(c_{5},c_{6},z_{2}^{*})\in S_{2}$ . If $\mathbf {v}_{2}=(\alpha _{k}^{2},1,\alpha _{k})$ projects a point $(c_{3}+x_{1},y_{1}^{*},c_{4}+z_{1})\in S_{1}$ to the point $(c_{5}+x_{2},c_{6}+y_{2},z_{2}^{*})\in S_{2}$ , then a simple calculation shows that

$$ \begin{align*} \frac{x_2-x_1}{\alpha_k^2}=y_2=-\frac{z_1}{\alpha_k}. \end{align*} $$

Hence, $x_{2}=x_{1}-\alpha _{k} z_{1}$ and $y_{2}=-\alpha _{k}^{-1}z_{1}$ , and so

(4-2) $$ \begin{align} \begin{pmatrix} x_{2}\\ y_{2} \end{pmatrix} =\begin{pmatrix} 1&-\alpha_{k}\\ 0&-\alpha_{k}^{-1} \end{pmatrix} \begin{pmatrix} x_{1}\\ z_{1} \end{pmatrix}. \end{align} $$

As a final step, let us project $S_{2}$ by the vector $\mathbf {v}_{0}=(\alpha _{k},\alpha _{k}^{2},1)$ to a parallelogram $S_{3}$ on a plane $x=x_{3}^{*}$ , where $x_{3}^{*}$ is an integer, with the point $(c_{5},c_{6},z_{2}^{*})\in S_{2}$ at the center projected to the point $(x_{3}^{*},c_{7},c_{8})\in S_{3}$ . If $\mathbf {v}_{0}=(\alpha _{k},\alpha _{k}^{2},1)$ projects a point $(c_{5}+x_{2},c_{6}+y_{2},z_{2}^{*})\in S_{2}$ to the point $(x_{3}^{*},c_{7}+y_{3},c_{8}+z_{3})\in S_{3}$ , then simple calculation shows that

$$ \begin{align*} -\frac{x_2}{\alpha_k}=\frac{y_3-y_2}{\alpha_k^2}=z_3. \end{align*} $$

Hence, $y_{3}=y_{2}-\alpha _{k} x_{2}$ and $z_{3}=-\alpha _{k}^{-1}x_{2}$ , and so

(4-3) $$ \begin{align} \begin{pmatrix} y_{3}\\ z_{3} \end{pmatrix} =\begin{pmatrix} -\alpha_{k}&1\\ -\alpha_{k}^{-1}&0 \end{pmatrix} \begin{pmatrix} x_{2}\\ y_{2} \end{pmatrix}. \end{align} $$

Combining (4-1)–(4-3), we see that this three-step magnification process has resulted in the transition

$$ \begin{align*} \begin{pmatrix} y_3\\ z_3 \end{pmatrix} =\begin{pmatrix} -\alpha_k&1\\ -\alpha_k^{-1}&0 \end{pmatrix} \begin{pmatrix} 1&-\alpha_k\\ 0&-\alpha_k^{-1} \end{pmatrix} \begin{pmatrix} -\alpha_k^{-1}&0\\ -\alpha_k&1 \end{pmatrix} \begin{pmatrix} y\\ z \end{pmatrix}. \end{align*} $$

The transition matrix in question is

$$ \begin{align*} \mathcal{A}=\begin{pmatrix} -\alpha_k&1\\ -\alpha_k^{-1}&0 \end{pmatrix} \begin{pmatrix} 1&-\alpha_k\\ 0&-\alpha_k^{-1} \end{pmatrix} \begin{pmatrix} -\alpha_k^{-1}&0\\ -\alpha_k&1 \end{pmatrix} =\begin{pmatrix} 2-\alpha_k^3&\alpha_k^2-\alpha_k^{-1}\\ \alpha_k^{-2}-\alpha_k&1 \end{pmatrix}. \end{align*} $$

The eigenvalues of this matrix are the roots of the quadratic equation

$$ \begin{align*} \det\begin{pmatrix} 2-\alpha_k^3-\Lambda&\alpha_k^2-\alpha_k^{-1}\\ \alpha_k^{-2}-\alpha_k&1-\Lambda \end{pmatrix} =0, \end{align*} $$

or $\Lambda ^{2}-(3-\alpha _{k}^{3})\Lambda +\alpha _{k}^{-3}=0$ , with solutions

(4-4) $$ \begin{align} \Lambda_{i}=\frac{(3-\alpha_{k}^{3})\pm\mathrm{i}\sqrt{4\alpha_{k}^{-3}-(3-\alpha_{k}^{3})^{2}}}{2}, \quad i=1,2, \end{align} $$

and corresponding eigenvectors

(4-5) $$ \begin{align} \mathbf{W}_{i}=\bigg(\frac{\Lambda_{i}-1}{\alpha_{k}^{-2}-\alpha_{k}},1\bigg), \quad i=1,2. \end{align} $$

Note that the two eigenvalues are complex conjugates of each other, as are the first coordinates of the two eigenvectors. It follows that the two vectors

(4-6) $$ \begin{align} \mathbf{u}_{1}=\mathbf{W}_{1}+\mathbf{W}_{2} \quad\mbox{and}\quad \mathbf{u}_{2}=\frac{1}{\mathrm{i}}(\mathbf{W}_{1}-\mathbf{W}_{2}) \end{align} $$

have real coordinates. Indeed, $\mathbf {u}_{1}=(u_{1,1},u_{1,2})$ , where

(4-7) $$ \begin{align} u_{1,1}=\frac{2-\alpha_{k}^{3}}{\alpha_{k}^{-2}-\alpha_{k}} \quad\mbox{and}\quad u_{1,2}=2, \end{align} $$

and $\mathbf {u}_{2}=(u_{2,1},u_{2,2})$ , where

(4-8) $$ \begin{align} u_{2,1}=\frac{\sqrt{4\alpha_{k}^{-3}-(3-\alpha_{k}^{3})^{2}}}{\alpha_{k}^{-2}-\alpha_{k}} \quad\mbox{and}\quad u_{2,2}=0, \end{align} $$

if we ensure that $u_{2,1}$ is positive.

These precise formulas motivate us to start the area magnification process with a small parallelogram $S_{0}$ on an X-face that lies on a plane $x=x_{0}^{*}$ in cartesian $3$ -space, and with some very special properties attached to it. Omitting reference to the x-coordinate, we ensure that the nonparallel sides of $S_{0}$ are parallel to $\mathbf {u}_{1}$ and $\mathbf {u}_{2}$ , with lengths, respectively,

(4-9) $$ \begin{align} \delta\vert\mathbf{u}_{1}\vert=\delta(u_{1,1}^{2}+u_{1,2}^{2})^{1/2} \quad\mbox{and}\quad \delta\vert\mathbf{u}_{2}\vert=\delta u_{2,1}, \end{align} $$

where $\delta>0$ is a small fixed real number.

The three-step magnification process transforms $S_{0}$ into a parallelogram $S_{3}$ with the nonparallel sides parallel to the vector

$$ \begin{align*} \mathbf{u}_1^{(1)} =\mathcal{A}\mathbf{u}_1 =\mathcal{A}(\mathbf{W}_1+\mathbf{W}_2) =\mathcal{A}\mathbf{W}_1+\mathcal{A}\mathbf{W}_2 =\Lambda_1\mathbf{W}_1+\Lambda_2\mathbf{W}_2, \end{align*} $$

with real coordinates and of length

$$ \begin{align*} \delta\vert\Lambda_1\mathbf{W}_1+\Lambda_2\mathbf{W}_2\vert, \end{align*} $$

and parallel to the vector

$$ \begin{align*} \mathbf{u}_2^{(1)} =\mathcal{A}\mathbf{u}_2 =\frac{1}{\mathrm{i}}\mathcal{A}(\mathbf{W}_1-\mathbf{W}_2)r =\frac{1}{\mathrm{i}}(\mathcal{A}\mathbf{W}_1-\mathcal{A}\mathbf{W}_2) =\frac{1}{\mathrm{i}}(\Lambda_1\mathbf{W}_1-\Lambda_2\mathbf{W}_2), \end{align*} $$

with real coordinates and of length

$$ \begin{align*} \delta\vert\Lambda_1\mathbf{W}_1-\Lambda_2\mathbf{W}_2\vert. \end{align*} $$

Indeed, this three-step magnification process repeated $r\geqslant 1$ times transforms $S_{0}$ into a parallelogram $S_{3r}$ on some X-face with the nonparallel sides parallel to the vector

(4-10) $$ \begin{align} \mathbf{u}_{1}^{(r)} =\Lambda_{1}^{r}\mathbf{W}_{1}+\Lambda_{2}^{r}\mathbf{W}_{2}, \end{align} $$

with real coordinates and of length

(4-11) $$ \begin{align} \delta\vert\Lambda_{1}^{r}\mathbf{W}_{1}+\Lambda_{2}^{r}\mathbf{W}_{2}\vert, \end{align} $$

and parallel to the vector

(4-12) $$ \begin{align} \mathbf{u}_{2}^{(r)} =\frac{1}{\mathrm{i}}(\Lambda_{1}^{r}\mathbf{W}_{1}-\Lambda_{2}^{r}\mathbf{W}_{2}), \end{align} $$

with real coordinates and of length

(4-13) $$ \begin{align} \delta\vert\Lambda_{1}^{r}\mathbf{W}_{1}-\Lambda_{2}^{r}\mathbf{W}_{2}\vert. \end{align} $$

We clearly wish to understand how far the parallelogram $S_{3r}$ differs from the original parallelogram $S_{0}$ . In particular, we wish to show that it does not look like a long needle, with one side substantially longer than the other.

Assume that k is a fixed large integer.

First we estimate the area of $S_{3r}$ . It is clear, since $u_{2,2}=0$ , that

(4-14) $$ \begin{align} \operatorname{\mathrm{area}}(S_{0}) =\delta^{2}\vert u_{1,2}\vert\vert u_{2,1}\vert =(4+o_{k}(1))\delta^{2}\alpha_{k}^{1/2}. \end{align} $$

It then follows from the magnification process that

(4-15) $$ \begin{align} \operatorname{\mathrm{area}}(S_{3r}) =\alpha_{k}^{-3r}\operatorname{\mathrm{area}}(S_{0}) =(4+o_{k}(1))\delta^{2}\alpha_{k}^{-3r}\alpha_{k}^{1/2}. \end{align} $$

Next, we estimate the maximum side length of $S_{3r}$ . Note first of all that it clearly follows from (4-4) that

$$ \begin{align*} \vert\Lambda_i\vert=\alpha_k^{-3/2}, \quad i=1,2. \end{align*} $$

Combining this with (4-5), we see that

$$ \begin{align*} \vert\mathbf{W}_i\vert\leqslant2, \quad i=1,2. \end{align*} $$

It follows, on noting that the side lengths of $S_{3r}$ are given by (4-11) and (4-13), that the maximum side length of $S_{3r}$ satisfies

(4-16) $$ \begin{align} \operatorname{\mathrm{max-sidelength}}(S_{3r}) \leqslant\delta(\vert\Lambda_{1}\vert^{r}\vert\mathbf{W}_{1}\vert+\vert\Lambda_{2}\vert^{r}\vert\mathbf{W}_{2}\vert) \leqslant4\delta\alpha_{k}^{-3r/2}. \end{align} $$

Combining (4-15) and (4-16), we deduce that

(4-17) $$ \begin{align} \frac{(\operatorname{\mathrm{max-sidelength}}(S_{3r}))^{2}}{\operatorname{\mathrm{area}}(S_{3r})}\leqslant\frac{4+o_{k}(1)}{\alpha_{k}^{1/2}}. \end{align} $$

This shows that uniformly $S_{3r}$ does not look like a long needle.

Recall that towards the end of the previous section, we wish to divide each small n-square $Q_{i}$ , $i\in I_{0}$ , into

(4-18) $$ \begin{align} \frac{6\varepsilon m}{(3-\tau)n^{2}} \end{align} $$

convex parts with equal area, referred to as the tiny convex parts, and that for notational simplicity, we assume that the quantity (4-18) is an integer. Recall also that at least half of these tiny convex parts of $Q_{i}$ do not contain any element of the sequence $T^{j}x_{0}$ , $0\leqslant j\leqslant m-1$ , and that we refer to them as the empty tiny convex parts of $Q_{i}$ , $i\in I_{0}$ . We are now ready to give the explicit construction.

Since every empty tiny convex part of $Q_{i}$ , $i\in I_{0}$ , has relatively large area, we can employ the area magnification process on any of them. Of course the magnified image of an empty set is also empty. We show that there exists at least one such empty tiny convex part of some $Q_{i}$ , $i\in I_{0}$ , which avoids image-splitting during the entire area magnification process.

We define the tiny convex parts as congruent parallelograms on an X-face such that one side is parallel to $\mathbf {u}_{1}$ with length $\delta \vert \mathbf {u}_{1}\vert $ , and the other side is parallel to $\mathbf {u}_{2}$ with length $\delta \vert \mathbf {u}_{2}\vert $ ; see (4-5)–(4-9). Thus, we consider a parallelogram lattice

(4-19) $$ \begin{align} \Omega=\Omega(\mathbf{u}_{1},\mathbf{u}_{2};\delta) \end{align} $$

on the plane, where any fundamental parallelogram $\mathcal {P}\in \Omega $ , an atom of $\Omega $ , has sides $\delta \mathbf {u}_{1}$ and $\delta \mathbf {u}_{2}$ . The concrete value of the parameter $\delta>0$ will be specified later.

It is a minor irritation that the parallelogram lattice $\Omega $ in (4-19) may not tile a unit size square precisely. However, we show later that, with an appropriate choice of the parameters, this technical issue has a totally negligible effect.

The whole magnification process consists of a large number of consecutive $3$ -cycles. It makes the calculations more convenient if we write the total number of consecutive $3$ -cycles as $\ell -O(1)$ , where the value of the constant $O(1)$ will be specified later.

We choose the parameters $\delta $ and $\ell $ such that

(4-20) $$ \begin{align} \delta\vert\Lambda^{\ell}_{1}\vert=\delta\vert\Lambda^{\ell}_{2}\vert =\delta\alpha_{k}^{-3\ell/2}\asymp1. \end{align} $$

This will be made more precise later.

The area of any fundamental parallelogram in the lattice $\Omega =\Omega (\mathbf {u}_{1},\mathbf {u}_{2};\delta )$ defined in (4-19) is chosen to be precisely equal to $(3-\tau )/6\varepsilon m$ . Combining this with (4-14), the area of any fundamental parallelogram in the lattice $\Omega =\Omega (\mathbf {u}_{1},\mathbf {u}_{2};\delta )$ is equal to

(4-21) $$ \begin{align} (4+o_{k}(1))\delta^{2}\alpha_{k}^{1/2}=\frac{3-\tau}{6\varepsilon m}. \end{align} $$

Since the area of an arbitrary small n-square $Q_{i}$ , $i\in I_{0}$ , is equal to $n^{-2}$ , it follows that the number of the fundamental parallelograms $\mathcal {P}$ in $\Omega =\Omega (\mathbf {u}_{1},\mathbf {u}_{2};\delta )$ that intersect $Q_{i}$ is between $5\varepsilon m/(3-\tau )n^{2}$ and $7\varepsilon m/(3-\tau )n^{2}$ , where the number would be precisely $6\varepsilon m/(3-\tau )n^{2}$ in the case of perfect tiling, assuming that m is sufficiently large.

We now take the fundamental parallelograms $\mathcal {P}$ of $\Omega =\Omega (\mathbf {u}_{1},\mathbf {u}_{2};\delta )$ in $Q_{i}$ as the tiny convex parts mentioned above. Then it follows that there are at least

$$ \begin{align*} \frac{5\varepsilon m}{(3-\tau)n^2}=\frac{5}{3}\frac{9\varepsilon}{(3-\tau)}\frac{m}{3n^2} \end{align*} $$

fundamental parallelograms in $Q_{i}$ . It then follows from (3-17) that a proportion of at least $1-(3/5)=2/5$ of the fundamental parallelograms $\mathcal {P}$ in $Q_{i}$ are empty, that is, they do not contain any element of the sequence $T^{j}x_{0}$ , $0\leqslant j\leqslant m-1$ . We refer to them as the empty fundamental parallelograms $\mathcal {P}$ in $Q_{i}$ , $i\in I_{0}$ . There are at least

$$ \begin{align*} \frac{2\varepsilon m}{(3-\tau)n^2} \end{align*} $$

of them. Combining this with (3-13), there are at least

(4-22) $$ \begin{align} \frac{(\tau-\varepsilon)n^{2}}{2}\frac{2\varepsilon m}{(3-\tau)n^{2}}=\frac{\varepsilon(\tau-\varepsilon)m}{3-\tau} \end{align} $$

empty fundamental parallelograms $\mathcal {P}$ in $X_{0}$ .

Since , noting that the area of a fundamental parallelogram is given by (4-21), there are

(4-23) $$ \begin{align} \frac{3}{(3-\tau)/{6\varepsilon m}}\pm\varepsilon^{2}m=\bigg(\frac{18\varepsilon}{3-\tau}\pm\varepsilon^{2}\bigg)m \end{align} $$

fundamental parallelograms of the lattice $\Omega =\Omega (\mathbf {u}_{1},\mathbf {u}_{2};\delta )$ in $X_{0}$ , assuming m is large enough. Here, the error term $\pm \varepsilon ^{2}m$ comes from the contribution near the border.

Comparing (4-22) and (4-23), we conclude that the ratio of empty fundamental parallelograms $\mathcal {P}$ in $X_{0}$ compared to the total number in $X_{0}$ is at least

(4-24) $$ \begin{align} \frac{\tau-\varepsilon}{20}, \end{align} $$

assuming that $\varepsilon>0$ is sufficiently small.

Our basic idea is that each such grossly undervisited small n-square $Q_{i}$ has many pairwise disjoint convex empty sets $\mathcal {P}\in \Omega =\Omega (\mathbf {u}_{1},\mathbf {u}_{2};\delta )$ with the same nonnegligible area, for which we can apply the area magnification process via shortlines. In the magnification process, we work with the concrete half-infinite geodesic that starts from the special point $x_{0}\in S_{0}$ , and of course we use its shortlines.

We want to avoid the annoying technical problem of image-splitting illustrated in Figure 10. Since there are many candidates for convex empty sets, we have a good chance of showing that at least one of them is splitting-free in the precise sense that it never splits during the entire magnification process.

The details of the magnification process are rather complicated, and the reader may wish to jump ahead right now and read the conclusion of Section 5 summarized in the last paragraph of the section and then Section 6. This is the quickest way to get a rough idea on how the proof works. After that the reader can come back to Section 5 and study the technical details of the magnification process.

Let $1\leqslant r\leqslant \ell $ be an arbitrary integer. We use r as a running parameter to describe a general $3$ -cycle in the magnification process of $\ell -O(1)$ consecutive $3$ -cycles.

Starting from a fundamental parallelogram $\mathcal {P}$ of $\Omega (\mathbf {u}_{1},\mathbf {u}_{2};\delta )$ , and carrying out r consecutive $3$ -cycles of the area magnification process via shortlines, we obtain a new parallelogram $\mathcal {P}^{(r)}$ on an X-face, or X-faces, such that one side is the vector $\delta \mathbf {u}^{(r)}_{1}$ , and the other side is the vector $\delta \mathbf {u}^{(r)}_{2}$ ; see (4-10)–(4-13).

Of course, we are particularly interested in empty fundamental parallelograms $\mathcal {P}\in \Omega (\mathbf {u}_{1},\mathbf {u}_{2};\delta )$ .

5 Grids and iteration

We first establish the following geometric result which we use a few times.

Lemma 5.1 (‘area lemma’)

Let $\Omega $ be an arbitrary parallelogram lattice in $\mathbb {R}^{2}$ such that each fundamental parallelogram has diameter $\eta>0$ and area $\Delta $ . Then a straight-line segment S with length $L>0$ intersects at most

$$ \begin{align*} \frac{2\eta(L+2\eta)}{\Delta} \end{align*} $$

fundamental parallelograms in $\Omega $ .

On the other hand, let A be a convex set in $\mathbb{R}^2$ lying within an aligned square U of area 1.Then the boundary $\partial A$ of A intersects at most

$$ \begin{align*} 8\cdot\frac{2\eta(\sqrt{2}+2\eta)}{\Delta} \end{align*} $$

fundamental parallelograms in $\Omega$ .

Proof. For any two points $P,Q\in \mathbb {R}^{2}$ , let $\rho (P,Q)$ denote the usual Euclidean distance between P and Q. For any $\eta>0$ , consider the $\eta $ -neighborhood of S given by

$$ \begin{align*} S(\eta)=\{P\in\mathbb{R}^2:\rho(P,Q)\leqslant\eta\mbox{ for some }Q\in S\}. \end{align*} $$

It is easy to see that $\operatorname {\mathrm {area}}(S(\eta ))\leqslant 2\eta (L+2\eta )$ . Note next that every fundamental parallelogram in $\Omega $ that intersects S is completely contained in $S(\eta )$ . Since each such fundamental parallelogram has area $\Delta $ , a simple calculation shows that the number of fundamental parallelograms in $\Omega $ that intersect S does not exceed

$$ \begin{align*} \frac{\operatorname{\mathrm{area}}(S(\eta))}{\Delta}\leqslant\frac{2\eta(L+2\eta)}{\Delta}. \end{align*} $$

The second assertion follows from the inequality

$$ \begin{align*} \operatorname{\mathrm{area}}((\partial A)(\eta))\le\operatorname{\mathrm{area}}((\partial U)(\sqrt{2}\eta)), \end{align*} $$

where $\partial U$ denotes the boundary of U.

This completes the proof.

Throughout, we assume that $k\geqslant 8$ , so that $\alpha _{k}<1/16$ .

We need some notation.

First of all, we rewrite (4-20) in the form

(5-1) $$ \begin{align} \delta=\frac{\alpha_{k}^{3\ell/2}}{c_{1}}, \end{align} $$

where the constant $c_{1}>0$ will be specified later. We also write

$$ \begin{align*} \ell=r+(r-1)+(r-2)+\cdots+c_2=\frac{r(r+1)}{2}-\frac{(c_2-1)c_2}{2}, \end{align*} $$

where the constant $c_{2}\geqslant 10$ will be specified later. More generally, for every integer $i=0,1,2,\ldots ,r-c_{2}$ , we write

(5-2) $$ \begin{align} \ell_{i}=\sum_{h=0}^{i}(r-h)=\frac{r(r+1)}{2}-\frac{(r-i-1)(r-i)}{2}, \end{align} $$

so that, in particular,

(5-3) $$ \begin{align} \ell_{0}=r, \quad \ell_{1}=r+(r-1), \quad \ell_{r-c_{2}-1}=\ell-c_{2} \quad\mbox{and}\quad \ell_{r-c_{2}}=\ell, \end{align} $$

and let the integer $q_{i}$ satisfy the inequalities

(5-4) $$ \begin{align} \frac{c_{1}}{(r-i)^{4}}\alpha_{k}^{-3(\ell-\ell_{i})/2}-1<q_{i}\leqslant\frac{c_{1}}{(r-i)^{4}}\alpha_{k}^{-3(\ell-\ell_{i})/2}. \end{align} $$

In particular, the integer $q=q_{0}$ satisfies the inequalities

(5-5) $$ \begin{align} \frac{c_{1}}{r^{4}}\alpha_{k}^{-3(\ell-r)/2}-1<q\leqslant\frac{c_{1}}{r^{4}}\alpha_{k}^{-3(\ell-r)/2}. \end{align} $$

The calculations here and below makes use of the key fact that we can choose the value of the parameter $\ell $ as large as possible. It follows that $r\approx \sqrt {2\ell }$ can also be arbitrarily large.

We are now ready to describe an iterative process that allows us to construct grids that contain empty fundamental parallelograms that are also nonsplitting during the magnification process.

Preparatory step. We begin our process with a $q\times q$ grid $\mathcal {G}$ of fundamental parallelograms $\mathcal {P}\in \Omega $ . Since each fundamental parallelogram $\mathcal {P}\in \Omega $ has sides given by the vectors $\delta \mathbf {u}_{1}$ and $\delta \mathbf {u}_{2}$ , a $q\times q$ grid $\mathcal {G}$ consists of $q^{2}$ fundamental parallelograms $\mathcal {P}\in \Omega $ and is itself a larger parallelogram with sides given by the vectors $q\delta \mathbf {u}_{1}$ and $q\delta \mathbf {u}_{2}$ .

Consider a parallelogram lattice $\mathcal {H}$ that contains an X-face, with grid fundamental parallelograms given by $q\times q$ grids $\mathcal {G}$ with sides given by the vectors $q\delta \mathbf {u}_{1}$ and $q\delta \mathbf {u}_{2}$ .

Remark 5.2. Here we use the term grid fundamental parallelograms to distinguish these larger parallelograms from the fundamental parallelograms $\mathcal {P}\in \Omega $ .

It follows easily from (4-7)–(4-9) that the diameter of each grid fundamental parallelogram in $\mathcal {H}$ does not exceed

(5-6) $$ \begin{align} 8q\delta\leqslant\frac{8\alpha_{k}^{3r/2}}{r^{4}}\leqslant\frac{1}{2r^{4}}, \end{align} $$

in view of (5-1), (5-5) and the assumption on k. This is extremely small when r is large, and ensures that there are many $q\times q$ grids $\mathcal {G}$ that are completely contained in an X-face of $X_{0}$ . It also follows easily from (4-14) that each grid fundamental parallelogram in $\mathcal {H}$ has area $\Delta =(4+o_{k}(1))q^{2}\delta ^{2}\alpha _{k}^{1/2}$ . On the other hand, an X-face has four boundary edges, each with unit length. Applying Lemma 5.1, we conclude that the number of grid fundamental parallelograms in $\mathcal {H}$ that intersect the boundary of any of the three X-faces in $X_{0}$ does not exceed

$$ \begin{align*} 3\cdot4\cdot\frac{16q\delta(1+16q\delta)}{(4+o_k(1))q^2\delta^2\alpha_k^{1/2}}=\frac{48+o_k(1)}{q\delta\alpha_k^{1/2}}. \end{align*} $$

The total number of fundamental parallelograms $\mathcal {P}\in \Omega $ is roughly equal to

$$ \begin{align*} \frac{3q^2}{(4+o_k(1))q^2\delta^2\alpha_k^{1/2}}, \end{align*} $$

so it follows from (4-24) and (5-6) that the total number of empty fundamental parallelograms $\mathcal {P}\in \Omega $ contained in $q\times q$ grids $\mathcal {G}$ that do not intersect the boundary of any of the three X-faces in $X_{0}$ is at least

$$ \begin{align*} &\frac{3q^{2}}{(4+o_{k}(1))q^{2}\delta^{2}\alpha_{k}^{1/2}}\cdot\frac{\tau-\varepsilon}{20}-\frac{(48+o_{k}(1))q^{2}}{q\delta\alpha_{k}^{1/2}}\\ &\quad=\frac{3}{(4+o_{k}(1))q^{2}\delta^{2}\alpha_{k}^{1/2}}\Bigg(\frac{\tau - \epsilon}{20}-(64+o_{k}(1))q\delta\Bigg)q^{2}. \end{align*} $$

Thus, the average proportion of empty fundamental parallelograms in a $q\times q$ grid $\mathcal {G}$ that does not intersect the boundary of any of the three X-faces in $X_{0}$ is at least

$$ \begin{align*} \frac{\tau-\varepsilon}{20}-(64+o_k(1))q\delta. \end{align*} $$

Note that it follows from (5-6) that

$$ \begin{align*} (64+o_k(1))q\delta\leqslant\frac{4}{r^4}<\frac{\tau}{100r^2} \end{align*} $$

for large r.

We now choose a $q\times q$ grid $\mathcal {G}_{0}$ that is completely contained in an X-face of $X_{0}$ and contains the maximum number of empty fundamental parallelograms $\mathcal {P}\in \Omega $ . Clearly the proportion of empty fundamental parallelograms $\mathcal {P}\in \mathcal {G}_{0}$ is at least

(5-7) $$ \begin{align} \tau_{0}=\frac{\tau-\varepsilon}{20}-\frac{\tau}{100r^{2}}. \end{align} $$

Initial step. Starting with $\mathcal {G}_{0}$ and applying the first $3$ -cycle of magnifications

$$ \begin{align*} X_0 \mathop{\longrightarrow}\limits^{\mathbf{v}_1}Y_0 \mathop{\longrightarrow}\limits^{\mathbf{v}_2}Z_0 \mathop{\longrightarrow}\limits^{\mathbf{v}_0}X_0, \end{align*} $$

where

$$ \begin{align*} Y_0=Y_1\cup Y_2\cup Y_3 \quad\mbox{and}\quad Z_0=Z_1\cup Z_2\cup Z_5 \end{align*} $$

as shown in Figures 4 and 5, we obtain a magnified $q\times q$ grid $\mathcal {G}_{0}^{(1)}$ . Employing this argument r times in succession, we obtain a magnified $q\times q$ grid $\mathcal {G}_{0}^{(r)}=\mathcal {G}_{0}^{(\ell _{0})}$ at the end of r consecutive $3$ -cycles of magnifications. At each magnification, the diameter of the parallelogram grows by a factor $\alpha _{k}^{-1/2}$ , so that the diameter of the parallelogram at each stage of the $3r$ -step magnification process, assuming that no splitting takes place, never exceeds

(5-8) $$ \begin{align} 8q\delta\alpha_{k}^{-3r/2}\leqslant\frac{8}{r^{4}}\leqslant1, \end{align} $$

in view of (5-6), as long as $r\geqslant 2$ . If splitting occurs at any step of the magnification process, the bound (5-8) on the diameter guarantees that the image splits into at most four convex parts, each contained in a square face. For convenience, we call these convex square-parts.

Remark 5.3. Note that the L-solid has only $14$ square faces in total. This number is reduced to nine with face identification. Thus, many distinct convex square-parts may fall into the same square face. However, we need to exercise care and treat each separately.

Since there are $3r$ steps in the magnification process, it follows that the $q\times q$ grid $\mathcal {G}_{0}^{(r)}$ splits into $R\leqslant 4^{3r}$ convex square-parts.

Let A be a convex square-part of $\mathcal {G}_{0}^{(r)}$ , and suppose that it is contained in the unit square face U.

Suppose first of all that A contains a vertex $P_{0}$ of $\mathcal {G}_{0}^{(r)}$ . We consider a parallelogram lattice $\mathcal {H}_{0}$ , where $P_{0}$ is one of the vertices and a grid fundamental parallelogram in $\mathcal {H}_{0}$ has sides given by the vectors $q_{1}\delta \mathbf {u}_{1}^{(r)}$ and $q_{1}\delta \mathbf {u}_{2}^{(r)}$ . We say that a grid fundamental parallelogram in $\mathcal {H}_{0}$ is good if it is completely inside A, and bad if it intersects the boundary of A. Those grid fundamental parallelograms that do not intersect A are irrelevant to our discussion.

Meanwhile, it follows from (4-16) that the diameter $\eta $ of each grid fundamental parallelogram in $\mathcal {H}_{0}$ satisfies

$$ \begin{align*} \eta\leqslant8q_1\delta\alpha_k^{-3r/2}, \end{align*} $$

and it follows from (4-15) that each grid fundamental parallelogram in $\mathcal {H}_{0}$ has area

$$ \begin{align*} \Delta=(4+o_k(1))q_1^2\delta^2\alpha_k^{-3r}\alpha_k^{1/2}. \end{align*} $$

Applying Lemma 5.1, we conclude that there are at most

(5-9) $$ \begin{align} 8\cdot\frac{16q_{1}\delta\alpha_{k}^{-3r/2}(\sqrt{2}+16q_{1}\delta\alpha_{k}^{-3r/2})}{(4+o_{k}(1))q_{1}^{2}\delta^{2}\alpha_{k}^{-3r}\alpha_{k}^{1/2}} \leqslant\frac{64+o_{k}(1)}{q_{1}\delta\alpha_{k}^{-3r/2}\alpha_{k}^{1/2}} \end{align} $$

bad grid fundamental parallelograms in $\mathcal {H}_{0}$ that intersect the convex square-part A of $\mathcal {G}_{0}^{(r)}$ .

Suppose next that A does not contain a vertex $P_{0}$ of $\mathcal {G}_{0}^{(r)}$ . Then A is small, and we can consider an arbitrary parallelogram lattice $\mathcal {H}_{0}$ with sides given by the vectors $q_{1}\delta \mathbf {u}_{1}^{(r)}$ and $q_{1}\delta \mathbf {u}_{2}^{(r)}$ . An analogous argument shows that the estimate (5-9) remains valid.

There are at most $4^{3r}$ convex square-parts of $\mathcal {G}_{0}^{(r)}$ . It follows that the total number of bad grid fundamental parallelograms that intersect any convex square-part of $\mathcal {G}_{0}^{(r)}$ is at most

$$ \begin{align*} 4^{3r}\cdot\frac{64+o_k(1)}{q_1\delta\alpha_k^{-3r/2}\alpha_k^{1/2}}, \end{align*} $$

and so the total number of fundamental parallelograms $\mathcal {P}^{(r)}\in \mathcal {G}_{0}^{(r)}$ contained in bad grid fundamental parallelograms in $\mathcal {G}_{0}^{(r)}$ is at most

$$ \begin{align*} 4^{3r}\cdot\frac{(64+o_k(1))q_1^2}{q_1\delta\alpha_k^{-3r/2}\alpha_k^{1/2}}. \end{align*} $$

Now the total number of fundamental parallelograms $\mathcal {P}^{(r)}\in \mathcal {G}_{0}^{(r)}$ is equal to $q^{2}$ . On the other hand, since the magnified image of an empty set is empty, the proportion of empty fundamental parallelograms $\mathcal {P}^{(r)}\in \mathcal {G}_{0}^{(r)}$ is at least $\tau _{0}$ , which is given by (5-7). Hence, the total number of empty fundamental parallelograms $\mathcal {P}^{(r)}\in \mathcal {G}_{0}^{(r)}$ contained in good grid fundamental parallelograms in $\mathcal {G}_{0}^{(r)}$ is at least

$$ \begin{align*} q^2\tau_0-4^{3r}\cdot\frac{(64+o_k(1))q_1^2}{q_1\delta\alpha_k^{-3r/2}\alpha_k^{1/2}}. \end{align*} $$

Thus, the average proportion of empty fundamental parallelograms in a good $q_{1}\times q_{1}$ grid fundamental parallelogram in $\mathcal {G}_{0}^{(r)}$ is at least

$$ \begin{align*} \tau_0-\frac{4^{3r}}{q^2}\cdot\frac{(64+o_k(1))q_1^2}{q_1\delta\alpha_k^{-3r/2}\alpha_k^{1/2}}. \end{align*} $$

Note that it follows from (5-1) and (5-3)–(5-5) that

$$ \begin{align*} \frac{4^{3r}}{q^2}\cdot\frac{(64+o_k(1))q_1^2}{q_1\delta\alpha_k^{-3r/2}\alpha_k^{1/2}} =\frac{(64+o_k(1))}{(r-1)^4}\alpha_k^{-2}(4\alpha_k^{1/2})^{3r}r^8 <\frac{\tau}{100(r-1)^2} \end{align*} $$

for large r.

We now choose a $q_{1}\times q_{1}$ grid $\mathcal {G}_{1}$ which is a good grid fundamental parallelogram in $\mathcal {G}_{0}^{(\ell _{0})}=\mathcal {G}_{0}^{(r)}$ and which contains the maximum number of empty fundamental parallelograms $\mathcal {P}^{(\ell _{0})}=\mathcal {P}^{(r)}$ which are images of $\mathcal {P}\in \mathcal {G}_{0}$ . Clearly the proportion of empty fundamental parallelograms $\mathcal {P}^{(\ell _{0})}=\mathcal {P}^{(r)}\in \mathcal {G}_{1}$ is at least

(5-10) $$ \begin{align} \tau_{1}=\tau_{0}-\frac{\tau}{100(r-1)^{2}}. \end{align} $$

Furthermore, since the image $\mathcal {P}^{(\ell _{0})}=\mathcal {P}^{(r)}$ of the fundamental parallelogram $\mathcal {P}$ under the $3\ell _{0}=3r$ magnifications is completely inside a convex square-part of $\mathcal {G}_{0}^{(\ell _{0})}=\mathcal {G}_{0}^{(r)}$ , $\mathcal {P}$ never splits during the $3\ell _{0}=3r$ magnifications. In particular, these empty fundamental parallelograms are nonsplitting.

Inductive step. Suppose, in general, that $\mathcal {G}_{i}$ is $q_{i}\times q_{i}$ grid which is a good grid fundamental parallelogram, contained in a convex square-part of $\mathcal {G}_{0}^{(\ell _{i-1})}=\mathcal {G}_{i-1}^{(r-i+1)}$ , and that the proportion of empty and nonsplitting fundamental parallelograms $\mathcal {P}^{(\ell _{i-1})}\in \mathcal {G}_{i}$ is at least

(5-11) $$ \begin{align} \tau_{i}=\tau_{i-1}-\frac{\tau}{100(r-i)^{2}}, \end{align} $$

where $\mathcal {P}^{(\ell _{i-1})}$ represents the image of $\mathcal {P}$ under $3\ell _{i-1}$ magnifications, remaining empty and nonsplitting during the entire process $\mathcal {G}_{0}\to \mathcal {G}_{i}$ .

Starting with $\mathcal {G}_{i}$ and applying the first $3$ -cycle of magnifications

$$ \begin{align*} X_0 \mathop{\longrightarrow}\limits^{\mathbf{v}_1}Y_0 \mathop{\longrightarrow}\limits^{\mathbf{v}_2}Z_0 \mathop{\longrightarrow}\limits^{\mathbf{v}_0}X_0, \end{align*} $$

we now obtain a magnified $q_{i}\times q_{i}$ grid $\mathcal {G}_{i}^{(1)}$ . Employing this argument $r-i$ times in succession, we obtain a magnified $q_{i}\times q_{i}$ grid $\mathcal {G}_{i}^{(r-1)}=\mathcal {G}_{0}^{(\ell _{i})}$ at the end of $r-i$ consecutive $3$ -cycles of magnifications. The diameter of the grid $\mathcal {G}_{i}$ is clearly bounded by $8q_{i}\delta \alpha _{k}^{-3\ell _{i-1}/2}$ . At each magnification, the diameter of the parallelogram grows by a factor $\alpha _{k}^{-1/2}$ , so that the diameter of the parallelogram at each stage of the present $3(r-i)$ -step magnification process, assuming that no splitting takes place, never exceeds

(5-12) $$ \begin{align} 8q_{i}\delta\alpha_{k}^{-3\ell_{i-1}/2}\alpha_{k}^{-3(r-i)/2} =8q_{i}\delta\alpha_{k}^{-3\ell_{i}/2} \leqslant\frac{8}{(r-i)^{4}}\leqslant1, \end{align} $$

in view of (5-1) and (5-4), as long as $r-i\geqslant 2$ . If splitting occurs at any step of the magnification process, the bound (5-12) on the diameter guarantees that the image splits into at most four convex square-parts, each contained in a square face. Since there are $3(r-i)$ steps in the magnification process, it follows that the $q_{i}\times q_{i}$ grid $\mathcal {G}_{i}^{(r-i)}$ splits into $R_{i}\leqslant 4^{3(r-i)}$ convex square-parts.

Let A be a convex square-part of $\mathcal {G}_{i}^{(r-i)}$ , and suppose that it is contained in the unit square face U. We may suppose that A contains a vertex $P_{0}$ of $\mathcal {G}_{0}^{(r)}$ , as the argument in the alternative case requires only minor modifications. We consider a parallelogram lattice $\mathcal {H}_{i}$ , where $P_{0}$ is one of the vertices and a grid fundamental parallelogram in $\mathcal {H}_{i}$ has sides given by the vectors $q_{i+1}\delta \mathbf {u}_{1}^{(\ell _{i})}$ and $q_{i+1}\delta \mathbf {u}_{2}^{(\ell _{i})}$ . We say that a grid fundamental parallelogram in $\mathcal {H}_{i}$ is good if it is completely inside A, and bad if it intersects the boundary of A. Those grid fundamental parallelograms that do not intersect A are irrelevant to our discussion.

Meanwhile, it follows from (4-16) that the diameter $\eta $ of each grid fundamental parallelogram in $\mathcal {H}_{i}$ satisfies

$$ \begin{align*} \eta\leqslant8q_{i+1}\delta\alpha_k^{-3\ell_i/2}, \end{align*} $$

and it follows from (4-15) that each grid fundamental parallelogram in $\mathcal {H}_{0}^{(r)}$ has area

$$ \begin{align*} \Delta=(4+o_k(1))q_{i+1}^2\delta^2\alpha_k^{-3\ell_i}\alpha_k^{1/2}. \end{align*} $$

Applying Lemma 5.1, we conclude that there are at most

$$ \begin{align*} 8\cdot\frac{16q_{i+1}\delta\alpha_k^{-3\ell_i/2}(\sqrt{2}+16q_{i+1}\delta\alpha_k^{-3\ell_i/2})} {(4+o_k(1))q_{i+1}^2\delta^2\alpha_k^{-3\ell_i}\alpha_k^{1/2}} \leqslant\frac{64+o_k(1)}{q_{i+1}\delta\alpha_k^{-3\ell_i/2}\alpha_k^{1/2}} \end{align*} $$

bad grid fundamental parallelograms in $\mathcal {H}_{i}$ that intersect the convex square-part A of $\mathcal {G}_{i}^{(r-i)}$ .

There are at most $4^{3(r-i)}$ convex square-parts of $\mathcal {G}_{i}^{(r-i)}$ . It follows that the total number of bad grid fundamental parallelograms that intersect any convex square-part of $\mathcal {G}_{i}^{(r-i)}$ is at most

$$ \begin{align*} 4^{3(r-i)}\cdot\frac{64+o_k(1)}{q_{i+1}\delta\alpha_k^{-3\ell_i/2}\alpha_k^{1/2}}, \end{align*} $$

and so the total number of fundamental parallelograms $\mathcal {P}^{(\ell _{i})}\in \mathcal {G}_{i}^{(r-i)}$ contained in bad grid fundamental parallelograms in $\mathcal {G}_{i}^{(r-i)}$ is at most

$$ \begin{align*} 4^{3(r-i)}\cdot\frac{(64+o_k(1))q_{i+1}^2}{q_{i+1}\delta\alpha_k^{-3\ell_i/2}\alpha_k^{1/2}}. \end{align*} $$

Now the total number of fundamental parallelograms $\mathcal {P}^{(\ell _{i})}\in \mathcal {G}_{i}^{(r-i)}$ is equal to $q_{i}^{2}$ . On the other hand, since the magnified image of an empty set is empty, the proportion of empty fundamental parallelograms $\mathcal {P}^{(\ell _{i})}\in \mathcal {G}_{i}^{(r-i)}$ is at least $\tau _{i}$ given by (5-11). Hence, the total number of empty fundamental parallelograms $\mathcal {P}^{(\ell _{i})}\in \mathcal {G}_{i}^{(r-i)}$ contained in good grid fundamental parallelograms in $\mathcal {G}_{i}^{(r-i)}$ is at least

$$ \begin{align*} q_i^2\tau_i-4^{3(r-i)}\cdot\frac{(64+o_k(1))q_{i+1}^2}{q_{i+1}\delta\alpha_k^{-3\ell_i/2}\alpha_k^{1/2}}. \end{align*} $$

It follows that the average proportion of empty fundamental parallelograms in a good $q_{i+1}\times q_{i+1}$ grid fundamental parallelogram in $\mathcal {G}_{i}^{(r-i)}$ is at least

$$ \begin{align*} \tau_i-\frac{4^{3(r-i)}}{q_i^2}\cdot\frac{(64+o_k(1))q_{i+1}^2}{q_{i+1}\delta\alpha_k^{-3\ell_i/2}\alpha_k^{1/2}}. \end{align*} $$

Note that it follows from (5-1), (5-2) and (5-4) that

$$ \begin{align*} \frac{4^{3(r-i)}}{q_{i}^{2}}\cdot\frac{(64+o_{k}(1))q_{i+1}^{2}}{q_{i+1}\delta\alpha_{k}^{-3\ell_{i}/2}\alpha_{k}^{1/2}} &=\frac{(64+o_{k}(1))}{(r-i-1)^{4}}\alpha_{k}^{-2}(4\alpha_{k}^{1/2})^{3(r-i)}(r-i)^{8}\\ &<\frac{\tau}{100(r-i-1)^{2}} \end{align*} $$

for large $r-i$ . Since $r-i\geqslant c_{2}$ , we can choose $c_{2}$ sufficiently large so that

$$ \begin{align*} \alpha_k^{-2}(4\alpha_k^{1/2})^{3(r-i)}(r-i)^8\leqslant1 \end{align*} $$

whenever $r-i\geqslant c_{2}$ , for instance.

We can now choose a $q_{i+1}\times q_{i+1}$ grid $\mathcal {G}_{i+1}$ which is a good grid fundamental parallelogram in $\mathcal {G}_{0}^{(\ell _{i})}=\mathcal {G}_{i}^{(r-i)}$ and which contains the maximum number of empty fundamental parallelograms $\mathcal {P}^{(\ell _{i})}$ which are images of $\mathcal {P}\in \mathcal {G}_{0}$ . Clearly the proportion of empty fundamental parallelograms $\mathcal {P}^{(\ell _{i})}\in \mathcal {G}_{i+1}$ is at least

$$ \begin{align*} \tau_{i+1}=\tau_i-\frac{\tau}{100(r-i-1)^2}. \end{align*} $$

Furthermore, since the image $\mathcal {P}^{(\ell _{i})}$ of the fundamental parallelogram $\mathcal {P}$ under the $3\ell _{i}$ magnifications is completely inside a convex square-part of $\mathcal {G}_{0}^{(\ell _{i})}=\mathcal {G}_{i}^{(r-i)}$ , $\mathcal {P}$ never splits during the $3\ell _{i}$ magnifications from the very beginning. In particular, these empty fundamental parallelograms are nonsplitting.

Last step. The final step of this iterative process concerns $i=r-c_{2}-1$ , with $\ell _{i}=\ell -c_{2}$ . We conclude that we can choose an $(r-c_{2})\times (r-c_{2})$ grid $\mathcal {G}_{r-c_{2}}$ which is a good grid fundamental parallelogram in $\mathcal {G}_{0}^{(\ell -c_{2})}=\mathcal {G}_{r-c_{2}-1}^{(c_{2}+1)}$ , and the proportion of empty and parallelograms $\mathcal {P}^{(\ell -c_{2})}\in \mathcal {G}_{r-c_{2}}$ is at least

(5-13) $$ \begin{align} \tau_{r-c_{2}}=\tau_{r-c_{2}-1}-\frac{\tau}{100c_{2}^{2}}. \end{align} $$

Furthermore, since the image $\mathcal {P}^{(\ell -c_{2})}$ of the fundamental parallelogram $\mathcal {P}$ under the $3(\ell -c_{2})$ magnifications is completely inside a convex square-part of $\mathcal {G}_{0}^{(\ell -c_{2})}$ , $\mathcal {P}$ never splits during the $3(\ell -c_{2})$ magnifications from the very beginning. In particular, these empty fundamental parallelograms are nonsplitting.

Combining (5-7), (5-10), (5-11) and (5-13), we see that

$$ \begin{align*} \tau_{r-c_2} =\frac{\tau-\varepsilon}{20}-\frac{\tau}{100}\sum_{h=c_2}^r\frac{1}{h^2}>\frac{\tau-\varepsilon}{20}-\frac{\tau}{50} >\frac{\tau-2\varepsilon}{40} >0 \end{align*} $$

if $\varepsilon <\tau /2$ . Note also from (5-3) and (5-4) that

$$ \begin{align*} q_{r-c_2}>\frac{c_1}{c_2^4}-1=1 \end{align*} $$

if we specify that $c_{1}=2c_{2}^{4}$ .

Thus, we have shown that there exists an empty parallelogram $\mathcal {P}_{0}\in \Omega $ such that the $3(\ell -c_{2})$ -step magnification process gives rise to an empty parallelogram $\mathcal {P}_{0}^{(\ell -c_{2})}$ , and there is no splitting during the entire process of $\ell -c_{2}$ consecutive $3$ -cycles of magnifications.

7 Unique ergodicity: proving Theorems 6.3 and 1.1

Unique ergodicity refers to the extension of Theorems 6.1 and 6.2 where we replace uniformity for almost every starting point with uniformity for every nonpathological starting point. The pioneering results on unique ergodicity are due to Furstenberg [Reference Furstenberg3, Sections 3.2–3.3]. The reader, however, does not need to know those sections. For the sake of completeness, we have included here the necessary arguments.

Proof of Theorem 6.3. The proof goes by contradiction, and consists of two parts. In the first part, we simply follow Furstenberg’s argument. The surprising basic idea here is a reformulation of the problem in terms of T-invariant Borel measures and the use of functional analysis. The second part is an ad hoc argument based on some special simultaneous Diophantine approximation properties of the given direction vectors $\mathbf {v}_{i}$ , $i=0,1,2$ , associated to real algebraic numbers of degree $3$ .

The first part of the argument is summarized in the following lemma.

Lemma 7.1. Suppose that there exist a nonpathological starting point $y_{0}\in X_{0}$ and a Jordan measurable set $J_{0}\subset X_{0}$ for which uniformity fails, so that the infinite sequence

(7-1) $$ \begin{align} \frac{1}{m}\sum_{j=0}^{m-1}\chi_{J_{0}}(T^{j}y_{0}), \quad m\geqslant1, \end{align} $$

where $\chi _{J_{0}}$ is the characteristic function of $J_{0}$ , does not converge to . Then there exists an ergodic measure-preserving system $(X_{0},\mathcal {B},\nu ,T)$ , where $\mathcal {B}$ is the Borel $\sigma $ -algebra on $X_{0}$ and $\nu $ is a new T-invariant Borel probability measure, such that

(7-2)

Proof. In view of the assumption, there exists an infinite subsequence

$$ \begin{align*} 0\leqslant h_0< h_1<h_2<h_3<\cdots \end{align*} $$

of the positive integers such that the limit

$$ \begin{align*} \lim_{m\to\infty}\frac{1}{h_m}\sum_{j=0}^{h_m-1}\chi_{J_0}(T^jy_0), \end{align*} $$

exists, but is not equal to .

We now repeat and adapt some ideas in [Reference Furstenberg3, Sections 3.2–3.3]. For every integer $m\geqslant 1$ , we introduce the normalized counting measure $\nu _{m}$ , defined for every Borel set $B\subset X_{0}$ by

(7-3) $$ \begin{align} \nu_{m}(B)=\frac{1}{h_{m}}\sum_{j=0}^{h_{m}-1}\chi_{B}(T^{j}y_{0}), \end{align} $$

where $\chi _{B}$ is the characteristic function of B.

We now make use of a general theorem in functional analysis which says that the space of Borel probability measures on any compact set is compact in the so-called weak-star topology. The latter means that

$$ \begin{align*} \mu_m\to\mu \quad\mbox{if and only if } \int f\,{d}\mu_m\to\int f\,{d}\mu, \end{align*} $$

where f runs over all continuous functions on the compact space.

This compactness theorem is a nontrivial result. The standard proof is based on the Riesz representation theorem.

Let $\mathcal {M}$ denote the set of Borel probability measures $\mu $ on $X_{0}$ . By the general theorem, $\mathcal {M}$ is compact. Let $\mathcal {M}_{1}\subset \mathcal {M}$ denote the set of those Borel probability measures $\mu $ on $X_{0}$ that are T-invariant and such that . It is obvious that $\mathcal {M}_{1}$ is a closed subset of $\mathcal {M}$ and therefore compact.

The compactness of $\mathcal {M}$ implies that there is a subsequence $\nu _{m_{i}}$ of the sequence $\nu _{m}$ defined by (7-3) such that $\nu _{m_{i}}\to \nu _{\infty }$ as $i\to \infty $ , where $\nu _{\infty }$ is a Borel probability measure on $X_{0}$ . It easily follows from (7-3) that $\nu _{\infty }$ is T-invariant. Indeed, writing $y_{1}=Ty_{0}$ , we have

$$ \begin{align*} \nu_{m}(T^{-1}B)&=\frac{1}{h_{m}}\sum_{j=0}^{h_{m}-1}\chi_{B}(T^{j}y_{1})=\frac{1}{h_{m}}\sum_{j=1}^{h_{m}}\chi_{B}(T^{j}y_{0}) \nonumber \\ &=\frac{1}{h_{m}}\sum_{j=0}^{h_{m}-1}\chi_{B}(T^{j}y_{0})+\frac{\chi_{B}(T^{h_{m}}y_{0})-\chi_{B}(y_{0})}{h_{m}}\\[-2pt] &=\nu_{m}(B)+\frac{\chi_{B}(T^{h_{m}}y_{0})-\chi_{B}(y_{0})}{h_{m}}, \end{align*} $$

and

$$ \begin{align*} \bigg\vert\frac{\chi_B(T^{h_m}y_0)-\chi_B(y_0)}{h_m}\bigg\vert\leqslant\frac{1}{h_m}\to0 \quad\mbox{as }m\to\infty. \end{align*} $$

Moreover, the limit measure $\nu _{\infty }$ clearly satisfies the requirement in (7-2), implying that $\nu _{\infty }\in \mathcal {M}_{1}$ , and so $\mathcal {M}_{1}$ is a nonempty compact set.

To find an appropriate $\nu \in \mathcal {M}_{1}$ which guarantees that the measure-preserving system $(X_{0},\mathcal {B},\nu ,T)$ is ergodic, we use the almost trivial fact that $\mathcal {M}_{1}$ is convex. The well-known Krein–Milman theorem in functional analysis implies that the nonempty convex set $\mathcal {M}_{1}$ is spanned by its extremal points. It is a well-known general result in ergodic theory that the extremal points are precisely the ergodic T-invariant measures; see [Reference Furstenberg3, Proposition 3.4]. Thus, we can choose our measure $\nu \in \mathcal {M}_{1}$ to be such an extremal point, and this completes the proof.

The remainder of the proof is concerned with showing that the conclusion of Lemma 7.1 leads to a contradiction by using some very special number-theoretic properties of the cubic algebraic number $\alpha _{k}$ which defines the direction vector.

Since the measure-preserving system $(X_{0},\mathcal {B},\nu ,T)$ given by Lemma 7.1 is ergodic, it follows from Birkhoff’s ergodic theorem that for every Borel set $B\in \mathcal {B}$ and for $\nu $ -almost every $y\in X_{0}$ , we have

(7-4) $$ \begin{align} \lim_{m\to\infty}\frac{1}{m}\sum_{j=0}^{m-1}\chi_{B}(T^{j}y)=\nu(B). \end{align} $$

Let W be an arbitrarily large but fixed positive integer. We claim that there exists a nonempty open square $Q=Q(W)\subset X_{0}$ such that

(7-5)

To prove (7-5), we choose $B=J_{0}$ in (7-1), and consider the set

(7-6) $$ \begin{align} Y=\bigg\{y\in X_{0}:\lim_{m\to\infty}\frac{1}{m}\sum_{j=0}^{m-1}\chi_{J_{0}}(T^{j}y)=\nu(J_{0})\bigg\}. \end{align} $$

We already know from Theorem 6.2 that for -almost every $y\in X_{0}$ , we have

(7-7)

where denotes the two-dimensional Lebesgue measure. Combining (7-2), (7-4), (7-6) and (7-7), we conclude that

(7-8)

Let $\delta>0$ be arbitrarily small but fixed. Since , there exists an infinite sequence $R_{i}$ , $i\geqslant 1$ , of open squares such that

(7-9)

By (7-8) and (7-9), we have

(7-10) $$ \begin{align} \sum_{i=1}^{\infty}\nu(R_{i})\geqslant1. \end{align} $$

It follows from (7-9) and (7-10) that there exists an integer $i_{0}\geqslant 1$ such that

(7-11)

Choosing $\delta =1/W$ in (7-11), the inequality (7-5) follows with the choice $Q=R_{i_{0}}$ .

We next take advantage of the special direction vector $\mathbf {v}_{1}=(1,\alpha _{k},\alpha ^{2}_{k})$ , where $\alpha _{k}$ is a cubic algebraic number given by (2-1) and (2-2). We apply some general results from Diophantine approximation, summarized in the next two lemmas. We use the standard notation $\Vert x\Vert $ to denote the distance of a real number x from the nearest integer.

The first lemma concerns badly approximable linear forms; see [Reference Schmidt10, Ch. II and Theorem 4A].

Lemma 7.2. Let $m\geqslant 1$ be an integer, and let $\gamma _{1},\ldots ,\gamma _{m}$ be any m numbers in a real algebraic number field of degree $m+1$ such that $1,\gamma _{1},\ldots ,\gamma _{m}$ are linearly independent over the rationals. Write $\mathbf {v}=(1,\gamma _{1},\ldots ,\gamma _{m})\in \mathbb {R}^{m+1}$ . Then there exists a constant $C>0$ , depending at most on m and $\gamma _{1},\ldots ,\gamma _{m}$ , such that

(7-12) $$ \begin{align} \bigg\|\sum_{i=1}^{m}n_{i}\gamma_{i}\bigg\|\geqslant\frac{C}{(\max_{1\leqslant i\leqslant m}\vert n_{i}\vert)^{m}} \end{align} $$

for every $\mathbf {n}=(n_{1},\ldots ,n_{m})\in \mathbb {Z}^{m}$ with $\mathbf {n}\ne \mathbf {0}$ .

We comment that (7-12) follows relatively easily from the definition of the norm in an algebraic number field.

The second lemma is Mahler’s transference theorem in the relevant special case; see [Reference Mahler7] or [Reference Cassels2, Ch. 5 and Theorem 2].

Lemma 7.3. A necessary and sufficient condition that there exists a constant $C_{1}>0$ such that

$$ \begin{align*} \bigg\|\sum_{i=1}^mn_i\gamma_i\bigg\|\Big(\max_{1\leqslant i\leqslant m}\vert n_i\vert\Big)^m\geqslant C_1 \end{align*} $$

for every $\mathbf {n}=(n_{1},\ldots ,n_{m})\in \mathbb {Z}^{m}$ with $\mathbf {n}\ne \mathbf {0}$ , is that there exists another constant $C_{2}>0$ such that

$$ \begin{align*} \Big(\max_{1\leqslant j\leqslant m}\| n\gamma_j\|\Big)^{m} \vert n\vert\geqslant C_2 \end{align*} $$

for every $n\in \mathbb {Z}$ with $n\ne 0$ .

Let $\gamma _{1}=\alpha _{k}$ and $\gamma _{2}=\alpha _{k}^{2}$ . It follows from Lemma 7.2 that there exists a constant $C=C(\alpha _{k})>0$ , depending only on $\alpha _{k}$ , such that

$$ \begin{align*} \| n_1\alpha_k+n_2\alpha_k^2\|\geqslant\frac{C}{(\max\{\vert n_1\vert, \vert n_2\vert\})^2} \end{align*} $$

for every $\mathbf {n}=(n_{1},n_{2})\in \mathbb {Z}^{2}$ with $\mathbf {n}\ne \mathbf {0}$ . It then follows from Lemma 7.3 that there exists another constant $C^{\prime }=C^{\prime }(\alpha _{k})>0$ , depending only on $\alpha _{k}$ , such that

(7-13) $$ \begin{align} (\max\{\|n\alpha_{k}\|,\|n\alpha_{k}^{2}\|\})^{2}\vert n\vert\geqslant C^{\prime} \end{align} $$

for every $n\in \mathbb {Z}$ with $n\ne 0$ .

Consider the two-dimensional vector $\mathbf {w}_{1}=(\alpha _{k},\alpha ^{2}_{k})$ . Note that the transformation $T=T_{\mathbf {v}_{1}}:X_{0}\to X_{0}$ modulo one reduces to the $\mathbf {w}_{1}$ -shift

(7-14) $$ \begin{align} \mathbf{x}\to\mathbf{x}+\mathbf{w}_{1}\bmod{[0,1)^{2}} \end{align} $$

on the flat unit torus $[0,1)^{2}$ .

Applying (7-4) in the case when B is the nonempty open square $Q=Q(W)$ , we deduce that for $\nu $ -almost every $z\in X_{0}$ , we have

(7-15) $$ \begin{align} \lim_{m\to\infty}\frac{1}{m}\sum_{j=0}^{m-1}\chi_{Q}(T^{j}z)=\nu(Q), \end{align} $$

where $\chi _{Q}$ is the characteristic function of Q. Let $z=z_{0}$ in $X_{0}$ satisfy (7-15). Then there exists a threshold $m_{0}$ such that for every $m\geqslant m_{0}$ , we have

(7-16) $$ \begin{align} \frac{1}{m}\sum_{j=0}^{m-1}\chi_{Q}(T^{j}z_{0})=\frac{\nu(Q)}{2}. \end{align} $$

Choosing m sufficiently large, we can guarantee that $1/\sqrt {m}$ is much smaller than the side length of the square Q. We divide Q into small congruent subsquares $R_{i}$ , $i\in I$ , of side length $1/r$ , where $m=C^{\prime }r^{2}$ and $C^{\prime }>0$ is the constant in (7-13). Thus, if we ignore, for notational simplicity, those small squares $R_{i}$ that intersect the boundary of Q, then with negligible error if m is sufficiently large, we have

(7-17)

We claim that

(7-18) $$ \begin{align} T^{-j_{1}}R_{i}\cap T^{-j_{2}}R_{i}=\emptyset, \quad i\in I,\ 0\leqslant j_{1}<j_{2}\leqslant m-1. \end{align} $$

Suppose, to the contrary, that there exist $i\in I$ and integers $0\leqslant j_{1}<j_{2}\leqslant m-1$ such that $T^{-j_{1}}R_{i}$ intersects $T^{-j_{2}}R_{i}$ . Let

$$ \begin{align*} \mathbf{u}\in T^{-j_1}R_i\cap T^{-j_2}R_i. \end{align*} $$

Then both $T^{j_{1}}\mathbf {u}$ and $T^{j_{2}}\mathbf {u}$ are elements of the same small square $R_{i}$ of side length $1/r$ . Applying the modulo one reduction in (7-14), we see that $\mathbf {u}+j_{1}\mathbf {w}_{1}$ and $\mathbf {u}+j_{2}\mathbf {w}_{1}$ are $1/r$ -close to each other on the flat torus $[0,1)^{2}$ , and so $(j_{2}-j_{1})\mathbf {w}_{1}$ and $\mathbf {0}=(0,0)$ are $1/r$ -close to each other on the flat torus $[0,1)^{2}$ . Thus,

(7-19) $$ \begin{align} (\max\{\|(j_{2}-j_{1})\alpha_{k}\|,\|(j_{2}-j_{1})\alpha_{k}^{2}\|\})^{2}\leqslant\frac{1}{r^{2}}=\frac{C^{\prime}}{m}. \end{align} $$

On the other hand, applying (7-13) with $n=j_{2}-j_{1}$ , we obtain

$$ \begin{align*} (\max\{\|(j_2-j_1)\alpha_k\|,\|(j_2-j_1)\alpha_k^2\|\})^2\geqslant\frac{C'}{j_2-j_1}>\frac{C'}{m}, \end{align*} $$

contradicting (7-19). This establishes (7-18).

We are ready to complete the proof of Theorem 6.3. Combining (7-16) and (7-17), we have, for every $m\geqslant m_{0}$ ,

(7-20) $$ \begin{align} \frac{\nu(Q)m}{2} \leqslant\sum_{j=0}^{m-1}\chi_{Q}(T^{j}z_{0}) =\sum_{i\in I}\sum_{j=0}^{m-1}\chi_{R_{i}}(T^{j}z_{0}) \leqslant\vert I\vert, \end{align} $$

where the last inequality is a consequence of (7-18).

Combining (7-5), (7-17), (7-19) and (7-20), we conclude that

But this is absurd, since $2/C^{\prime }$ is a constant depending only on $\alpha _{k}$ , and W can be arbitrarily large. This completes the proof of Theorem 6.3. □

Finally, we discuss how we can extend Theorems 6.1–6.4 concerning the L-solid to any polycube $3$ -manifold with one-direction geodesic flow, and establish Theorem 1.1.

Let $\mathcal {P}$ be an arbitrary polycube $3$ -manifold. Our task is to extend the proof of Theorem 6.1 given in the special case of the L-solid with street-LCM equal to $2$ to the case of $\mathcal {P}$ with one-direction geodesic flow. We know that $\mathcal {P}$ has X-streets, Y-streets and Z-streets, and that the street-LCM of $\mathcal {P}$ is the least common multiple of the lengths of all the streets.

We can clearly adapt Sections 3–5, as the discussion in these sections does not use any special property of the L-solid that is not present in an arbitrary polycube $3$ -manifold. For notational convenience, and for easy visualization, we chose to work earlier with the simplest concrete special case of the L-solid.

To generalize the argument in Section 6, we need a time-quantitative density result for any polycube $3$ -manifold. The following result is [Reference Beck, Chen and Yang1, Theorem 6.4.2], and is a straightforward generalization of time-quantitative density A in Section 2.

Let $\eta>0$ be an arbitrarily small but fixed positive number. We say that a half-infinite geodesic $\mathcal {L}$ in $\mathcal {P}$ is $\eta $ -nearly superdense if there exists an effectively computable explicit threshold $N_{0}(\mathcal {P};\eta )$ such that, for every integer $n\geqslant N_{0}(\mathcal {P};\eta )$ , the initial segment of $\mathcal {L}$ with length $n^{2+\eta }$ intersects every axis-parallel cube of side length $1/n$ in $\mathcal {P}$ .

Lemma 7.4. Let $\mathcal {P}$ be an arbitrary polycube $3$ -manifold, and let $h=h(\mathcal {P})$ denote the street-LCM of $\mathcal {P}$ . Let $\eta>0$ be fixed. There exists a threshold $K_{0}(\mathcal {P};\eta )$ such that for every integer $k\geqslant K_{0}(\mathcal {P};\eta )$ , any half-infinite geodesic in $\mathcal {P}$ with direction given by one of the vectors in (1-2), where $\alpha _{k}$ is a root of the cubic equation $x^{3}+hkx-1=0$ in the interval

$$ \begin{align*} \frac{1}{hk+1}<\alpha_k<\frac{1}{hk}, \end{align*} $$

is $\eta $ -nearly superdense in $\mathcal {P}$ .

Using this density result, it is fairly straightforward to generalize the argument in Section 6 to establish an analog of Theorem 6.1 for any polycube $3$ -manifold. Finally, repeating the arguments in the derivation of Theorems 6.2–6.4 from Theorem 6.1, we can complete the proof of Theorem 1.1.