# Uniformity of geodesic flow
in non-integrable 3-manifolds

J. Beck  Department of Mathematics, Hill Center for the Mathematical Sciences,
Rutgers University, Piscataway NJ 08854, USA<EMAIL_ADDRESS>,  W.W.L.
Chen  School of Mathematical and Physical Sciences, Faculty of Science and
Engineering, Macquarie University, Sydney NSW 2109, Australia
<EMAIL_ADDRESS>and  Y. Yang  School of Science, Beijing University of
Posts and Telecommunications, Beijing 100876, China<EMAIL_ADDRESS>
##  1\. Introduction and the splitting method

Very little is known about non-integrable dynamical systems in dimensions
greater than $2$. A perfect illustration of this sad fact is the following
open problem which is the $3$-dimensional non-integrable analogue of the
Kronecker–Weyl equidistribution theorem concerning geodesic flow on the unit
torus $[0,1)^{3}$.

A vector $\mathbf{v}=(v_{1},v_{2},v_{3})\in\mathbb{R}^{3}$ is called a
Kronecker direction in $3$-space if the coordinates $v_{1},v_{2},v_{3}$ are
linearly independent over $\mathbb{Q}$.

######  Open Problem.

Suppose that $\mathcal{M}$ is a finite polycube translation $3$-manifold. Is
it true that any half-infinite geodesic in $\mathcal{M}$ with a Kronecker
direction and which does not hit a singular point of $\mathcal{M}$ is
uniformly distributed?

The answer to the corresponding $2$-dimensional problem is given by the
Gutkin–Veech theorem, but the method does not seem to work in higher
dimensions. In $3$ dimensions, even the simpler problem of density is wide
open.

Here we introduce a new geometric method which enables us to solve the Open
Problem for infinitely many finite polycube translation $3$-manifolds which,
in the first instance, satisfy a very special condition which makes them
essentially integrable in one of the $3$ directions. We then extend our
argument in Sections 8–9 to include more finite polycube translation
$3$-manifolds.

We call our new approach the splitting method. It uses the Birkhoff ergodic
theorem, and leads to time-qualitative uniformity. It is an open problem
whether time-quantitative uniformity can be established in these situations.

To illustrate our method and ideas, we consider in the first instance the
L-solid translation $3$-manifold, the simplest non-integrable example in
$3$-dimensions. This $3$-manifold consists of $3$ atomic cubes arranged in the
shape of the letter L, with a top cube, a middle cube and a right cube,
together with face identification given by perpendicular translation. The
picture on the right in Figure 1.1 shows the $Y$-faces which are perpendicular
to the $y$-axis. The two faces $Y_{2}$ are identified with each other, and the
two faces $Y_{3}$ are identified with each other.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-1-1.pdf}\vspace{3pt}\\\
\mbox{Figure 1.1: the $X$-faces and $Y$-faces of the L-solid translation
$3$-manifold}\end{array}$

######  Theorem 1.

Any half-infinite geodesic in the L-solid translation $3$-manifold with a
Kronecker direction $\mathbf{v}\in\mathbb{R}^{3}$ is uniformly distributed
unless it hits a singularity.

Suppose that $\mathbf{v}=(v_{1},v_{2},v_{3})\in\mathbb{R}^{3}$ is a Kronecker
direction in $3$-space. Linear independence guarantees that the coordinates
$v_{1},v_{2},v_{3}$ are non-zero. It is convenient to denote the Kronecker
direction instead by the parallel vector $\mathbf{v}=(\alpha,1,\beta)$, where
$\alpha=v_{1}/v_{2}$, $\beta=v_{3}/v_{2}$ and $\alpha/\beta$ are all
irrational. For simplicity, we assume that $\alpha>0$ and $\beta>0$. We can
further assume that $\alpha<1$. If $\alpha>1$, then we simply interchange the
roles of $x$ and $y$.

We consider the discretization of the $\mathbf{v}$-flow in the L-solid
translation $3$-manifold relative to the $Y$-faces, where the flow hits the
faces $Y_{1},Y_{2},Y_{3}$. Let

$T_{\alpha,\beta}:\mathcal{Y}=Y_{1}\cup Y_{2}\cup Y_{3}\to\mathcal{Y}$ (1.1)

denote the relevant discrete transformation defined by consecutive hitting
points.

Under our assumptions on the parameters $\alpha$ and $\beta$, the
$\mathbf{v}$-flow encounters split singularities in the form of edges of the
L-solid translation $3$-manifold, highlighted in Figure 1.1 in bold. For
instance, consider a geodesic segment in the $\mathbf{v}$-direction that hits
the common edge of the faces $Y_{1}$ and $Y_{3}$. If we move this geodesic
segment marginally to the left, then it crosses the face $Y_{1}$, quickly hits
the right face $X_{1}$ and then jumps to the left face $X_{1}$. If we move
this geodesic segment marginally to the right, then it crosses the face
$X_{3}$, quickly hits the top face $Y_{3}$ and then jumps to the bottom face
$Y_{3}$.

Consider now the image under $T_{\alpha,\beta}^{-1}$ of the singular edge of
the L-solid translation $3$-manifold which is the common edge of the faces
$Y_{1}$ and $Y_{3}$. It is a line segment in the $z$-direction on the face
$Y_{2}$, as shown in Figure 1.2.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-1-2.pdf}\vspace{3pt}\\\
\mbox{Figure 1.2: a strip round the inverse image of a singular
edge}\end{array}$

Consider a narrow strip neighbourhood $S_{2}$ of this line segment, shown in
Figure 1.2 as the gray strip on the face $Y_{2}$. The image of this strip
under $T_{\alpha,\beta}$ splits along the singular edge. Our splitting method
concerns making use of such splits.

######  Remark.

Note that the front $Z$-faces on the plane $z=0$ and the corresponding back
$Z$-faces on the plane $z=1$ are pairwise identified, and none of the edges of
these faces is a singular edge, so the L-solid translation $3$-manifold is
essentially integrable in the $z$-direction.

The discretization $T_{\alpha,\beta}$ on $\mathcal{Y}$ defined by (1.1) is
area preserving. Our first goal is to prove that it is also ergodic on
$\mathcal{Y}$. Suppose, on the contrary, that it is not ergodic. Then there
exist two disjoint $T_{\alpha,\beta}$-invariant subsets $\mathcal{W}$ (white)
and $\mathcal{S}$ (silver) of $\mathcal{Y}$ such that
$\mathcal{Y}=\mathcal{W}\cup\mathcal{S}$ and xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$0<\lambda_{2}(\mathcal{W})\leqslant\lambda_{2}(\mathcal{S})<3,$ (1.2)

where $\lambda_{2}$ denotes the $2$-dimensional Lebesgue measure. We show that
this leads to a contradiction.

Consider the projection of $\mathcal{Y}$ to the unit torus $[0,1)^{2}$, given
by

$\mathcal{Y}\to[0,1)^{2}:(x,y,z)\mapsto(\\{x\\},\\{z\\}),$ (1.3)

where we take the fractional parts of the first and the third coordinates and
omit reference to the second coordinate. Then for every point $P\in[0,1)^{2}$,
there are precisely $3$ distinct points $P_{1},P_{2},P_{3}\in\mathcal{Y}$
which have projection image $P$. Let

$f_{\mathcal{W}}(P)=|\\{P_{1},P_{2},P_{3}\\}\cap\mathcal{W}|\quad\mbox{and}\quad
f_{\mathcal{S}}(P)=|\\{P_{1},P_{2},P_{3}\\}\cap\mathcal{S}|$ (1.4)

denote the number of these $3$ points that fall into $\mathcal{W}$ and
$\mathcal{S}$ respectively. We refer to $f_{\mathcal{W}}$ and
$f_{\mathcal{S}}$ as the multiplicity functions of the invariant sets
$\mathcal{W}$ and $\mathcal{S}$ respectively, and they are non-negative
integer valued functions defined on the unit torus $[0,1)^{2}$ that satisfy
the condition

$f_{\mathcal{W}}(P)+f_{\mathcal{S}}(P)=3\quad\mbox{for almost all
$P\in[0,1)^{2}$}.$ (1.5)

We also consider the corresponding projection of

$T_{\alpha,\beta}:\mathcal{Y}\to\mathcal{Y}\quad\mbox{to}\quad
T_{0}:[0,1)^{2}\to[0,1)^{2},$

which is simply the $(\alpha,\beta)$-shift on the unit torus $[0,1)^{2}$.

######  Remark.

Strictly speaking, $\mathcal{W}$ and $\mathcal{S}$ are subsets of
$\mathcal{Y}$. However, it is convenient to view the pair as a $2$-colouring
of the set $\mathcal{Y}$ with colours $\mathcal{W}$ and $\mathcal{S}$, and not
make any distinction between subset and colour.

Since $\mathbf{v}$ is a Kronecker direction, it is not difficult to show that
$T_{0}$ is ergodic. Indeed, one way is to derive it from the Kronecker–Weyl
equidistribution theorem on the unit torus. Here we give a simpler proof by
using the more elementary Kronecker density theorem which is valid in every
dimension $d\geqslant 1$. However, we restrict our discussion to the case
$d=2$.

Let $\mathbf{w}=(\alpha,\beta)\in\mathbb{R}^{2}$, where $\alpha,\beta$ are
linearly independent over $\mathbb{Q}$, and let $T_{0}$ be the
$\mathbf{w}$-shift on the unit torus $[0,1)^{2}$. If $T_{0}$ is not ergodic,
then there is a $\mathbf{w}$-shift-invariant measurable subset
$A\subset[0,1)^{2}$ such that

$0<\lambda_{2}(A)<1\quad\mbox{and}\quad A+\mathbf{w}=A,$ (1.6)

where $A+\mathbf{w}$ denotes the image of $A$ under $T_{0}$.

######  Lemma 1.1.

Suppose that the set $A\subset[0,1)^{2}$ is measurable. Then

$\int_{[0,1)^{2}}\lambda_{2}(A\cap(A+\mathbf{t}))\,\mathrm{d}\mathbf{t}=(\lambda_{2}(A))^{2}.$
(1.7)

###### Proof.

For any set $B\subset[0,1)^{2}$, let $\chi_{B}$ denote the characteristic
function of $B$. Then

$\displaystyle\int_{[0,1)^{2}}\lambda_{2}(A\cap(A+\mathbf{t}))\,\mathrm{d}\mathbf{t}=\int_{[0,1)^{2}}\left(\int_{A}\chi_{A+\mathbf{t}}(\mathbf{x})\,\mathrm{d}\mathbf{x}\right)\mathrm{d}\mathbf{t}$
$\displaystyle\quad=\int_{[0,1)^{2}}\left(\int_{A}\chi_{A-\mathbf{x}}(-\mathbf{t})\,\mathrm{d}\mathbf{x}\right)\mathrm{d}\mathbf{t}=\int_{A}\left(\int_{[0,1)^{2}}\chi_{A-\mathbf{x}}(-\mathbf{t})\,\mathrm{d}\mathbf{t}\right)\mathrm{d}\mathbf{x}$
$\displaystyle\quad=\int_{A}\lambda_{2}(A-\mathbf{x})\,\mathrm{d}\mathbf{x}=\int_{A}\lambda_{2}(A)\,\mathrm{d}\mathbf{x}=(\lambda_{2}(A))^{2}.$

The change in the order of integration is justified by Fubini’s theorem, and
this completes the proof. ∎

The identity (1.7) represents an average. Let $\varepsilon>0$ be chosen to be
sufficiently small. Then there exists $\mathbf{t}_{0}\in[0,1)^{2}$ such that

$\lambda_{2}(A\cap(A+\mathbf{t}_{0}))<(\lambda_{2}(A))^{2}+\varepsilon,$

and the Kronecker density theorem implies that there exists a positive integer
$n_{0}$ such that the multiple $n_{0}\mathbf{w}$ gets so close to
$\mathbf{t}_{0}$ modulo $[0,1)^{2}$ that

$\lambda_{2}(A\cap(A+n_{0}\mathbf{w}))<\lambda_{2}(A\cap(A+\mathbf{t}_{0}))+\varepsilon.$

Then it follows that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\lambda_{2}(A\cap(A+n_{0}\mathbf{w}))<(\lambda_{2}(A))^{2}+2\varepsilon.$
(1.8)

On the other hand, the identity on the right in (1.6) clearly gives
$A+n_{0}\mathbf{w}=A$, so (1.8) becomes
$\lambda_{2}(A)<(\lambda_{2}(A))^{2}+2\varepsilon$. However, this inequality
cannot possibly hold when $\varepsilon>0$ is sufficiently small, since
$0<\lambda_{2}(A)<1$. The contradiction establishes the ergodicity of $T_{0}$.

The Birkhoff ergodic theorem now guarantees that both $f_{\mathcal{W}}$ and
$f_{\mathcal{S}}$ are constant integer valued functions. It then follows from
(1.2) and (1.5) that

$f_{\mathcal{W}}(P)=1\quad\mbox{and}\quad f_{\mathcal{S}}(P)=2\quad\mbox{for
almost all $P\in[0,1)^{2}$}.$ (1.9)

Let us return to the narrow strip $S_{2}$ on the face $Y_{2}$, as shown in
Figure 1.2. We denote by $S_{1}$ and $S_{3}$ narrow strips on the lower faces
$Y_{1}$ and $Y_{3}$ respectively, where the images on $[0,1)^{2}$ of
$S_{1},S_{2},S_{3}$ under the projection (1.3) coincide. In other words, the
strips $S_{1},S_{2},S_{3}$ are in the same relative positions on the faces
$Y_{1},Y_{2},Y_{3}$ respectively.

Using line segments in the $x$-direction, we can divide the strips
$S_{1},S_{2},S_{3}$ into small congruent rectangles in a natural way. We refer
to them as small special rectangles of the decomposition of the strips
$S_{1},S_{2},S_{3}$. Note that we have so far not specified the dimensions of
these rectangles.

Let $R_{1},R_{2},R_{3}$ denote small special rectangles of the decomposition
of the strips $S_{1},S_{2},S_{3}$ on the faces $Y_{1},Y_{2},Y_{3}$
respectively, with the condition that their images on $[0,1)^{2}$ under the
projection (1.3) coincide, so that they are in the same relative positions on
the faces $Y_{1},Y_{2},Y_{3}$ respectively. It is clear that the image of each
of $R_{1},R_{2},R_{3}$ under $T_{\alpha,\beta}$ splits along the singular
edges $E_{1},E_{2},E_{3}$ into congruent halves, as shown in the side views
given in Figure 1.3 where the $z$-direction is suppressed. Here, for each
$\sigma=1,2,3$, the rectangle $R_{\sigma}$ splits into the left half
$R_{\sigma}^{-}$ and the right half $R_{\sigma}^{+}$, and it is clear that
their images under $T_{\alpha,\beta}$ may lie on distinct $Y$-faces.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-1-3.pdf}\vspace{3pt}\\\
\mbox{Figure 1.3: side view of L-solid translation $3$-manifold showing}\\\
\mbox{the rectangles $R_{1},R_{2},R_{3}$ and their images under
$T_{\alpha,\beta}$}\end{array}$

Figure 1.4 is a summary of the pictures in Figure 1.3. The union of the pairs

$T_{\alpha,\beta}(R_{3}^{-})\cup T_{\alpha,\beta}(R_{1}^{+}),\quad
T_{\alpha,\beta}(R_{2}^{-})\cup T_{\alpha,\beta}(R_{3}^{+}),\quad
T_{\alpha,\beta}(R_{1}^{-})\cup T_{\alpha,\beta}(R_{2}^{+})$

are rectangles the same size as the rectangles $R_{1},R_{2},R_{3}$, each split
down the middle into congruent halves by a line in the $z$-direction.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-1-4.pdf}\vspace{3pt}\\\
\mbox{Figure 1.4: side view of L-solid translation $3$-manifold showing}\\\
\mbox{the rectangles $R_{1},R_{2},R_{3}$ and their images under
$T_{\alpha,\beta}$}\end{array}$

The situation is somewhat clearer if we study the images of
$R_{1},R_{2},R_{3}$ under $T_{\alpha,\beta}^{2}$, where the second application
of $T_{\alpha,\beta}$ does not produce any further splitting, as shown in
Figure 1.5.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-1-5.pdf}\vspace{3pt}\\\
\mbox{Figure 1.5: side view of L-solid translation $3$-manifold showing}\\\
\mbox{the rectangles $R_{1},R_{2},R_{3}$ and their images under
$T_{\alpha,\beta}^{2}$}\end{array}$

Figure 1.6 is a summary of the pictures in Figure 1.5. The union of the pairs

$\begin{array}[]{l}R^{*}_{1}=T_{\alpha,\beta}^{2}(R_{3}^{-})\cup
T_{\alpha,\beta}^{2}(R_{1}^{+}),\vspace{3pt}\\\
R^{*}_{2}=T_{\alpha,\beta}^{2}(R_{2}^{-})\cup
T_{\alpha,\beta}^{2}(R_{3}^{+}),\vspace{3pt}\\\
R^{*}_{3}=T_{\alpha,\beta}^{2}(R_{1}^{-})\cup
T_{\alpha,\beta}^{2}(R_{2}^{+}),\end{array}$ (1.10)

are rectangles on the faces $Y_{1},Y_{2},Y_{3}$ respectively, the same size as
$R_{1},R_{2},R_{3}$, each split into congruent halves by a line in the
$z$-direction. Note that the images on $[0,1)^{2}$ of
$R^{*}_{1},R^{*}_{2},R^{*}_{3}$ under the projection (1.3) coincide, so that
$R^{*}_{1},R^{*}_{2},R^{*}_{3}$ are in the same relative positions on the
faces $Y_{1},Y_{2},Y_{3}$ respectively.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-1-6.pdf}\vspace{3pt}\\\
\mbox{Figure 1.6: side view of L-solid translation $3$-manifold showing}\\\
\mbox{the rectangles $R_{1},R_{2},R_{3}$ and their images under
$T_{\alpha,\beta}^{2}$}\end{array}$

We need the following measure theoretic lemma.

######  Lemma 1.2.

There exists a small special rectangle $R_{1}$ on the face $Y_{1}$ of the
L-solid translation $3$-manifold such that the following two conditions are
satisfied:

_(i)_ The set $R_{1}$ satisfies

$\frac{\lambda_{2}(R_{1}\cap\mathcal{W})}{\lambda_{2}(R_{1})}>\frac{99}{100}\quad\mbox{or}\quad\frac{\lambda_{2}(R_{1}\cap\mathcal{S})}{\lambda_{2}(R_{1})}>\frac{99}{100}.$
(1.11)

_(ii)_ There are analogues of _(i)_ for the small special rectangles $R_{2}$
and $R_{3}$ on the faces $Y_{2}$ and $Y_{3}$ respectively of the L-solid
translation $3$-manifold, defined such that the images on $[0,1)^{2}$ of
$R_{1},R_{2},R_{3}$ under the projection (1.3) coincide, as well as for the
rectangles $R^{*}_{1},R^{*}_{2},R^{*}_{3}$ defined by (1.10).

These rectangles are illustrated in Figure 1.7.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-1-7.pdf}\vspace{3pt}\\\
\mbox{Figure 1.7: the rectangles $R_{1},R_{2},R_{3}$ and
$R^{*}_{1},R^{*}_{2},R^{*}_{3}$}\end{array}$

The choice of $99/100$ in (1.11) is accidental. Note also that Lemma 1.2
somewhat resembles the Lebesgue density theorem. We include the proof in
Section 3.

Recall that we have assumed that $T_{\alpha,\beta}$ is not ergodic, and this
implies that there exist disjoint $T_{\alpha,\beta}$-invariant subsets
$\mathcal{W},\mathcal{S}\subset\mathcal{Y}$ such that
$\mathcal{Y}=\mathcal{W}\cup\mathcal{S}$ and (1.4) and (1.9) are satisfied.
Using Lemma 1.2, we are now ready to deduce a contradiction.

Lemma 1.2 ensures that each rectangle $R_{\sigma}$, $\sigma=1,2,3$, has a
dominant colour $\mathcal{C}_{\sigma}$, and each rectangle $R^{*}_{\sigma}$,
$\sigma=1,2,3$, has a dominant colour $\mathcal{C}^{*}_{\sigma}$, where each
$\mathcal{C}_{\sigma}$ or $\mathcal{C}^{*}_{\sigma}$ is white or silver. We
now use Figures 1.3, 1.5 and 1.6, and also note that since $\mathcal{W}$ and
$\mathcal{S}$ are $T_{\alpha,\beta}$-invariant, they are also
$T_{\alpha,\beta}^{2}$-invariant.

The image of the rectangle $R_{1}$ under $T_{\alpha,\beta}$ splits along the
edge $E_{1}$ into congruent halves. The image of $R_{1}^{-}$, the left half of
$R_{1}$, under $T_{\alpha,\beta}^{2}$ forms the left half of $R^{*}_{3}$. The
image of $R_{1}^{+}$, the right half of $R_{1}$, under $T_{\alpha,\beta}^{2}$
forms the right half of $R^{*}_{1}$. Thus

$\mathcal{C}_{1}=\mathcal{C}^{*}_{1}=\mathcal{C}^{*}_{3}.$ (1.12)

The image of the rectangle $R_{2}$ under $T_{\alpha,\beta}$ splits along the
edge $E_{2}$ into congruent halves. The image of $R_{2}^{-}$, the left half of
$R_{2}$, under $T_{\alpha,\beta}^{2}$ forms the left half of $R^{*}_{2}$. The
image of $R_{2}^{+}$, the right half of $R_{2}$, under $T_{\alpha,\beta}^{2}$
forms the right half of $R^{*}_{3}$. Thus

$\mathcal{C}_{2}=\mathcal{C}^{*}_{2}=\mathcal{C}^{*}_{3}.$ (1.13)

The image of the rectangle $R_{3}$ under $T_{\alpha,\beta}$ splits along the
edge $E_{3}$ into congruent halves. The image of $R_{3}^{-}$, the left half of
$R_{3}$, under $T_{\alpha,\beta}^{2}$ forms the left half of $R^{*}_{1}$. The
image of $R_{3}^{+}$, the right half of $R_{3}$, under $T_{\alpha,\beta}^{2}$
forms the right half of $R^{*}_{2}$. Thus

$\mathcal{C}_{3}=\mathcal{C}^{*}_{1}=\mathcal{C}^{*}_{2}.$ (1.14)

Combining (1.12)–(1.14), we conclude
$\mathcal{C}_{1}=\mathcal{C}_{2}=\mathcal{C}_{3}$. This means that the
dominant colour in the rectangles $R_{1},R_{2},R_{3}$ is the same colour
$\mathcal{C}$.

Thus we can find measurable subsets $R_{\sigma}^{(0)}\subset R_{\sigma}$,
$\sigma=1,2,3$, satisfying

$R_{\sigma}^{(0)}\cap\mathcal{C}=R_{\sigma}^{(0)}\quad\mbox{and}\quad\frac{\lambda_{2}(R_{\sigma}^{(0)})}{\lambda_{2}(R_{\sigma})}=\frac{9}{10},$

and such that their images on $[0,1)^{2}$ under the projection (1.3) coincide.
This clearly contradicts (1.9), and establishes the ergodicity of
$T_{\alpha,\beta}$.

The Birkhoff ergodic theorem now implies that for any fixed Kronecker
direction $\mathbf{v}\in\mathbb{R}^{3}$, a half-infinite geodesic in the
L-solid translation $3$-manifold with almost any starting point is uniformly
distributed. This establishes a weaker form of Theorem 1. In Section 4, we
extend ergodicity to unique ergodicity and complete the proof.

Theorem 1 is best possible in the sense that there is no uniform distribution
if the direction $\mathbf{v}=(v_{1},v_{2},v_{3})\in\mathbb{R}^{3}$ of a half-
infinite geodesic is not a Kronecker direction, so that the coordinates
$v_{1},v_{2},v_{3}$ are linearly dependent over $\mathbb{Q}$. In this case, we
do not even have density. Indeed, the classical Kronecker density theorem
implies that even the modulo one projection of the geodesic orbit to the unit
torus $[0,1)^{3}$ is not dense.

##  2\. Splitting diagram and splitting permutation

The key part in the proof of ergodicity in the case of geodesics in the
L-solid translation $3$-manifold is illustrated by the splitting diagram given
in Figure 2.1, which is a simpler version of Figure 1.6.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-2-1.pdf}\vspace{3pt}\\\
\mbox{Figure 2.1: splitting diagram of the L-solid translation
$3$-manifold}\end{array}$

Here we label the top, middle and right cubes by the numbers $1,2,3$
respectively. Then for every $\sigma=1,2,3$, $Y_{\sigma}$ is the bottom
$Y$-face of cube $\sigma$, $R_{\sigma}$ is a small special rectangle on
$Y_{\sigma}$, and the image of $R_{\sigma}$ under $T_{\alpha,\beta}$ splits
along the edge $E_{\sigma}$, and we then study the images of the congruent
halves $R_{\sigma}^{-}$ and $R_{\sigma}^{+}$ of $R_{\sigma}$ under
$T_{\alpha,\beta}^{2}$. Furthermore, for every $\sigma=1,2,3$, the small
special rectangle $R_{\sigma}$, and so also its images under
$T_{\alpha,\beta}^{2}$, has predominantly the colour $\mathcal{C}_{\sigma}$.

The label $3\sim 1$ on the face $Y_{1}$ indicates that
$T_{\alpha,\beta}^{2}(R_{3}^{-})\cup T_{\alpha,\beta}^{2}(R_{1}^{+})$ is,
apart from a set of zero measure, a rectangle on the face $Y_{1}$, and
$\mathcal{C}_{3}=\mathcal{C}_{1}$. The label $2\sim 3$ on the face $Y_{2}$
indicates that $T_{\alpha,\beta}^{2}(R_{2}^{-})\cup
T_{\alpha,\beta}^{2}(R_{3}^{+})$ is, apart from a set of zero measure, a
rectangle on the face $Y_{2}$, and $\mathcal{C}_{2}=\mathcal{C}_{3}$. The
label $1\sim 2$ on the face $Y_{3}$ indicates that
$T_{\alpha,\beta}^{2}(R_{1}^{-})\cup T_{\alpha,\beta}^{2}(R_{2}^{+})$ is,
apart from a set of zero measure, a rectangle on the face $Y_{3}$, and
$\mathcal{C}_{1}=\mathcal{C}_{2}$. Furthermore, we can compress this
information into a single cyclic permutation $3\to 1\to 2\to 3$. We call this
the splitting permutation of the L-solid translation $3$-manifold.

Our argument in Section 1 can therefore be adapted to study geodesic flow on a
class of finite polycube translation $3$-manifolds where each member is
essentially the cartesian product of a finite polysquare translation surface
and the unit torus $[0,1)$, provided that the corresponding splitting
permutation is a single cyclic permutation.

We now examine a few more examples.

######  Example 2.1.

Consider the $5$-cube staircase translation $3$-manifold, with splitting
diagram given in Figure 2.2.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-2-2.pdf}\vspace{3pt}\\\
\mbox{Figure 2.2: splitting diagram of the $5$-cube staircase translation
$3$-manifold}\end{array}$

The splitting permutation

$4\to 1\to 5\to 2\to 3\to 4$ (2.1)

is given in the form of a single cyclic permutation.

Let us add $2$ cubes to this $3$-manifold and consider a $7$-cube staircase
translation $3$-manifold, with splitting diagram given in Figure 2.3.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-2-3.pdf}\vspace{3pt}\\\
\mbox{Figure 2.3: splitting diagram of the $7$-cube staircase translation
$3$-manifold}\end{array}$

The splitting permutation

$4\to 1\to 5\to 6\to 2\to 3\to 7\to 4$ (2.2)

is also given in the form of a single cyclic permutation.

Note that to go from (2.1) to (2.2), we insert $7$ after $3$, and insert $6$
after $5$. The insertion corresponds to the following observations after
adding cubes $6$ and $7$:

$\circ$ The image $T_{\alpha,\beta}^{2}(R_{3}^{-})$ moves from $Y_{5}$ to
$Y_{6}$.

$\circ$ The image $T_{\alpha,\beta}^{2}(R_{5}^{-})$ moves from $Y_{3}$ to
$Y_{7}$.

$\circ$ The image $T_{\alpha,\beta}^{2}(R_{\sigma}^{-})$ is unchanged for
$\sigma=1,2$ and $\sigma=4$.

$\circ$ The image $T_{\alpha,\beta}^{2}(R_{\sigma}^{+})$ is unchanged for
$\sigma=1,2,3,4,5$.

$\circ$ The images $T_{\alpha,\beta}^{2}(R_{6}^{-})$ and
$T_{\alpha,\beta}^{2}(R_{7}^{-})$ are on $Y_{3}$ and $Y_{5}$ respectively.

$\circ$ The images $T_{\alpha,\beta}^{2}(R_{6}^{+})$ and
$T_{\alpha,\beta}^{2}(R_{7}^{+})$ are on $Y_{7}$ and $Y_{6}$ respectively.

It can be shown that if we go from the $s$-cube staircase translation
$3$-manifold to the $(s+2)$-cube staircase translation $3$-manifold, where $s$
is odd, then the following assertions are valid:

$\circ$ The image $T_{\alpha,\beta}^{2}(R_{s-2}^{-})$ moves from $Y_{s}$ to
$Y_{s+1}$.

$\circ$ The image $T_{\alpha,\beta}^{2}(R_{s}^{-})$ moves from $Y_{s-2}$ to
$Y_{s+2}$.

$\circ$ The image $T_{\alpha,\beta}^{2}(R_{\sigma}^{-})$ is unchanged for
$\sigma=1,\ldots,s-3$ and $\sigma=s-1$.

$\circ$ The image $T_{\alpha,\beta}^{2}(R_{\sigma}^{+})$ is unchanged for
$\sigma=1,\ldots,s$.

$\circ$ The images $T_{\alpha,\beta}^{2}(R_{s+1}^{-})$ and
$T_{\alpha,\beta}^{2}(R_{s+2}^{-})$ are on $Y_{s-2}$ and $Y_{s}$ respectively.

$\circ$ The images $T_{\alpha,\beta}^{2}(R_{s+1}^{+})$ and
$T_{\alpha,\beta}^{2}(R_{s+2}^{+})$ are on $Y_{s+2}$ and $Y_{s+1}$
respectively.

This leads to the insertion of $s+2$ after $s-2$ and the insertion of $s+1$
after $s$ in the splitting permutation. Indeed, this argument repeatedly
inductively then shows that for every $s$-cube staircase translation
$3$-manifold, where $s$ is odd, the splitting permutation is a single cyclic
permutation.

######  Example 2.2.

A $(3,2)$-snake translation surface is a finite polysquare translation surface
where the square faces form a finite zigzagging path from left to right, every
street has length $3$, apart from the two streets at the end which have length
$2$. A $(3,2)$-snake translation $3$-manifold is then the cartesian product of
a $(3,2)$-snake translation surface and the unit torus $[0,1)$. Figure 2.4 is
the splitting diagram of such a $3$-manifold with $7$ cubes.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-2-4.pdf}\vspace{3pt}\\\
\mbox{Figure 2.4: splitting diagram of a $7$-cube $(3,2)$-snake translation
$3$-manifold}\end{array}$

The splitting permutation

$3\to 2\to 4\to 6\to 5\to 7\to 1\to 3$ (2.3)

is given in the form of a single cyclic permutation.

Let us add $2$ cubes to this $3$-manifold and consider a $9$-cube
$(3,2)$-snake translation $3$-manifold, with splitting diagram given in Figure
2.5.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-2-5.pdf}\vspace{3pt}\\\
\mbox{Figure 2.5: splitting diagram of a $9$-cube $(3,2)$-snake translation
$3$-manifold}\end{array}$

The splitting permutation

$3\to 2\to 4\to 6\to 5\to 9\to 7\to 8\to 1\to 3$ (2.4)

is also given in the form of a single cyclic permutation.

Note that to go from (2.3) to (2.4), we insert $9$ after $5$, and insert $8$
after $7$. Here the general case is more complicated. As we go from left to
right, any vertical street along the way may go up or down, and the different
choices lead to somewhat different splitting diagrams.

######  Example 2.3.

Figure 2.6 shows the splitting diagram of a polycube translation $3$-manifold
which is the cartesian product of a finite polysquare translation surface with
a hole and the unit torus $[0,1)$. Here the finite polysquare translation
surface has edge identification via perpendicular translation, and has $5$
horizontal and $6$ vertical streets.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-2-6.pdf}\vspace{3pt}\\\
\mbox{Figure 2.6: splitting diagram of a finite polycube translation
$3$-manifold}\\\ \mbox{with a hole}\end{array}$

The splitting permutation

$8\to 2\to 5\to 6\to 1\to 7\to 3\to 9\to 11\to 10\to 4\to 8$

is given in the form of a single cyclic permutation.

The next two examples concern cases where the splitting permutation is not a
single cyclic permutation. This can fail in more than one way.

######  Example 2.4.

Here we consider a very simple case, where we add an extra cube to the L-solid
translation $3$-manifold to form a $4$-cube L-solid translation $3$-manifold.
Figure 2.7 shows the splitting diagram.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-2-7.pdf}\vspace{3pt}\\\
\mbox{Figure 2.7: splitting diagram of a $4$-cube L-solid translation
$3$-manifold}\end{array}$

The splitting permutation is given by $4\to 1\to 2\to 4$ and $3\to 3$ which is
clearly not a single cyclic permutation. There is a stationary point $3$.

######  Example 2.5.

Consider a polycube translation $3$-manifold which is the cartesian product of
a finite polysquare translation surface with $6$ square faces and the unit
torus $[0,1)$. Figure 2.8 shows the splitting diagram.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-2-8.pdf}\vspace{3pt}\\\
\mbox{Figure 2.8: splitting diagram of a $6$-cube solid polysquare translation
$3$-manifold}\end{array}$

Here the splitting permutation

$3\to 2\to 6\to 3\quad\mbox{and}\quad 4\to 5\to 1\to 4$

is a product of two disjoint cyclic permutations.

For the class of polycube translation $3$-manifolds which are cartesian
products of finite polysquare translation surfaces and the unit torus $[0,1)$,
the question of which of these have splitting permutations which are cyclic
has a rather simple answer. Cyclic splitting permutation means a single cycle
in the cycle decomposition, and this happens if and only if the boundary edge
identification of the finite polysquare translation surface reduces the number
of vertices to $1$. Indeed, this reduction to $1$ vertex is precisely the same
process by which the colours in the splitting points are proved to be equal
and we end up with a single colour. And vice versa.

In Section 5, we prove the following result.

######  Theorem 2.

Suppose that a polycube translation $3$-manifold $\mathcal{M}$ is the
cartesian product of a polysquare translation surface $\mathcal{P}$ with $d$
square faces and the unit torus $[0,1)$. Suppose further that the splitting
permutation contains a cycle of length greater than $d/2$. Then any half-
infinite geodesic in $\mathcal{M}$ with a Kronecker direction
$\mathbf{v}\in\mathbb{R}^{3}$ is uniformly distributed unless it hits a
singularity.

It is clear that the restriction that we have imposed on the polycube
translation $3$-manifolds is a very severe one. It is an interesting open
problem to study the corresponding problem for general finite polycube
translation $3$-manifolds.

##  3\. Proof of Lemma 1.2

Recall that we denote the Kronecker direction by
$\mathbf{v}=(\alpha,1,\beta)$, where $\alpha$, $\beta$ and $\alpha/\beta$ are
all irrational, and also assume that $0<\alpha<1$ and $\beta>0$.

The irrational number $\alpha$ has infinite continued fraction

$\alpha=[0;a_{1},a_{2},a_{3},\ldots]=\frac{1}{a_{1}+\frac{1}{a_{2}+\frac{1}{a_{3}+\ldots}}},$
(3.1)

where $a_{1},a_{2},a_{3},\ldots$ are positive integers. For any
$k=1,2,3,\ldots,$ the $k$-th convergent

$\frac{p_{k}}{q_{k}}=\frac{p_{k}(\alpha)}{q_{k}(\alpha)}=[0;a_{1},a_{2},a_{3},\ldots,a_{k}],$

where the integers $p_{k}$ and $q_{k}$ are coprime, is defined in terms of a
finite initial segment of the continued fraction (3.1), with denominator
$q_{k}=q_{k}(\alpha)$. It is well known that

$\|q\alpha\|\geqslant\|q_{k}\alpha\|=|q_{k}\alpha-p_{k}|,\quad
q=1,\ldots,q_{k+1}-1,$ $\|q_{k+1}\alpha\|<\|q_{k}\alpha\|,$ (3.2)

and xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\frac{1}{q_{k+1}+q_{k}}\leqslant\|q_{k}\alpha\|\leqslant\frac{1}{q_{k+1}}.$

These show that the convergents give the best rational approximations to
$\alpha$.

For any integer $k\geqslant 1$, let $\mathcal{A}_{k}(\alpha)$ be the partition
of $[0,1)$ with $q_{k+1}$ division points

$\\{\ell\alpha\\},\quad\ell=0,1,2,\ldots,q_{k+1}-1.$ (3.3)

We need information on the gaps between neighboring partition points, and
these can be described in terms of continued fractions. We also use the famous
$3$-distance theorem [4, 5, 6, 7, 8, 9] in its strong form.

######  Lemma 3.1 ($2$-distance theorem).

The distance between any pair of neighboring partition points (3.3) of
$\mathcal{A}_{k}(\alpha)$ can only take the values

$\|q_{k}\alpha\|\quad\mbox{and}\quad\|q_{k}\alpha\|+\|q_{k+1}\alpha\|,$

precisely $q_{k+1}-q_{k}$ and $q_{k}$ times respectively.

For every $i=1,\ldots,q_{k+1}-1$, consider the short special interval

$J_{k}(i)=J_{k}(\alpha;i)=\left(\\{i\alpha\\}-\frac{\|q_{k}\alpha\|}{2},\\{i\alpha\\}+\frac{\|q_{k}\alpha\|}{2}\right),$
(3.4)

with length xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$|J_{k}(\alpha;i)|=\|q_{k}\alpha\|.$

In view of $\|q_{k}\alpha\|+\|q_{k+1}\alpha\|<2\|q_{k}\alpha\|$, a simple
consequence of (3.2), it is clear that these intervals (i) are pairwise
disjoint, (ii) are contained in the open interval $(0,1)$, and (iii) have
total length greater than $1/3$.

The irrational number $\beta$ has infinite continued fraction

$\beta=[b_{0};b_{1},b_{2},b_{3},\ldots]=b_{0}+\frac{1}{b_{1}+\frac{1}{b_{2}+\frac{1}{b_{3}+\ldots}}},$
(3.5)

where $b_{0}$ is a non-negative integer and $b_{1},b_{2},b_{3},\ldots$ are
positive integers. For any $h=1,2,3,\ldots,$ the $h$-th convergent

$\frac{p^{\prime}_{h}}{q^{\prime}_{h}}=\frac{p^{\prime}_{h}(\beta)}{q^{\prime}_{h}(\beta)}=[b_{0};b_{1},b_{2},b_{3},\ldots,b_{h}],$

where the integers $p^{\prime}_{h}$ and $q^{\prime}_{h}$ are coprime, is
defined in terms of a finite initial segment of the continued fraction (3.5),
with denominator $q^{\prime}_{h}=q^{\prime}_{h}(\beta)$.

For any integer $h\geqslant 1$, let $\mathcal{B}_{h}(\beta)$ be the partition
of $[0,1)$ with $q^{\prime}_{h+1}$ division points

$\\{\ell\beta\\},\quad\ell=0,1,2,\ldots,q^{\prime}_{h+1}-1.$ (3.6)

In view of Lemma 3.1, the distance between any pair of neighboring partition
points (3.6) of $\mathcal{B}_{h}(\beta)$ can only take the values

$\|q^{\prime}_{h}\beta\|\quad\mbox{and}\quad\|q^{\prime}_{h}\beta\|+\|q^{\prime}_{h+1}\beta\|.$
(3.7)

For every integer $j=1,\ldots,q^{\prime}_{h+1}-1$, consider the short special
interval

$J^{\prime}_{h}(j)=J^{\prime}_{h}(\beta;j)=\left(\\{j\beta\\}-\frac{\|q^{\prime}_{h}\beta\|}{2},\\{j\beta\\}+\frac{\|q^{\prime}_{h}\beta\|}{2}\right),$
(3.8)

with length xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$|J^{\prime}_{h}(\beta;j)|=\|q^{\prime}_{h}\beta\|.$

It is clear that these intervals (i) are pairwise disjoint, (ii) are contained
in the open interval $(0,1)$, and (iii) have total length greater than $1/3$.

Using the short special intervals above, we construct some small special
rectangles.

Let the denominator $q_{k}=q_{k}(\alpha)$ of the $k$-th convergent of $\alpha$
be large, and let the denominator $q^{\prime}_{h}=q^{\prime}_{h}(\beta)$ of
the $h$-th convergent of $\beta$ be substantially larger, both to be made
precise later.

The faces $Y_{1},Y_{2},Y_{3}$ of the L-solid translation $3$-manifold have
bottom left vertex

$\mathbf{p}_{1}=(0,1,0),\quad\mathbf{p}_{2}=(0,0,0),\quad\mathbf{p}_{3}=(1,0,0)$

respectively. For every $\sigma=1,2,3$ and every
$j=1,\ldots,q^{\prime}_{h+1}-q_{k+1}$, we consider the small special rectangle

$R_{\sigma,k,h}(\alpha;1;\beta;j)=\\{\mathbf{p}_{\sigma}+(x,0,z)\in
Y_{\sigma}:(x,z)\in J_{k}(\alpha;1)\times J^{\prime}_{h}(\beta;j)\\}$

on the face $Y_{\sigma}$. For every such $\sigma$ and $j$, and for every
$i=0,1,\ldots,q_{k+1}-2$, we also consider the image of this small special
rectangle under $T_{\alpha,\beta}^{i}$, given by

$R^{\dagger}_{\sigma,k,h}(\alpha;1+i;\beta;j+i)=T_{\alpha,\beta}^{i}(R_{\sigma,k,h}(\alpha;1;\beta;j))$
(3.9)

which is not necessarily in $Y_{\sigma}$ but can be anywhere in $\mathcal{Y}$,
and is splitting free, in view of the properties (i)–(iii) of the short
special intervals. In particular, the lack of splitting implies that they are
congruent rectangles, with the same area

$\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1+i;\beta;j+i))=\lambda_{1}(J_{k}(\alpha;1))\lambda_{1}(J^{\prime}_{h}(\beta;1)),$
(3.10)

where $\lambda_{1}$ and $\lambda_{2}$ denote respectively $1$\- and
$2$-dimensional Lebesgue measures. Thus they are also small special
rectangles.

Next, recall from Section 1 and (1.9) that the underlying set $\mathcal{Y}$
has a non-trivial decomposition into a disjoint union
$\mathcal{Y}=\mathcal{W}\cup\mathcal{S}$ of $T_{\alpha,\beta}$-invariant
measurable subsets $\mathcal{W}$ and $\mathcal{S}$, with
$\lambda_{2}(\mathcal{W})=1$ and $\lambda_{2}(\mathcal{S})=2$. Let
$\varepsilon>0$ be arbitrarily small but fixed. Then there exists a subset
$\mathcal{W}_{1}=\mathcal{W}_{1}(\varepsilon;\mathcal{W})\subset\mathcal{Y}$
which is a finite union of disjoint rectangles and such that the symmetric
difference
$\mathcal{W}\bigtriangleup\mathcal{W}_{1}=(\mathcal{W}\setminus\mathcal{W}_{1})\cup(\mathcal{W}_{1}\setminus\mathcal{W})$
has measure xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\lambda_{2}(\mathcal{W}\bigtriangleup\mathcal{W}_{1})<\varepsilon.$ (3.11)

Note also that the $\mathcal{Y}$-complement
$\mathcal{W}_{1}^{\,c}=\mathcal{Y}\setminus\mathcal{W}_{1}$ of
$\mathcal{W}_{1}$ satisfies

$\mathcal{W}\bigtriangleup\mathcal{W}_{1}=\mathcal{S}\bigtriangleup\mathcal{W}_{1}^{\,c}.$
(3.12)

For any pair $(\sigma,j)$ with $\sigma=1,2,3$ and
$j=1,\ldots,q^{\prime}_{h+1}-q_{k+1}$, we call the collection

$R^{\dagger}_{\sigma,k,h}(\alpha;1+i;\beta;j+i),\quad i=0,1,\ldots,q_{k+1}-2,$
(3.13)

of small special rectangles the $T_{\alpha,\beta}$-power chain of the pair
$(\sigma,j)$. Let $\delta=\delta(\varepsilon)>0$ be fixed. We say that this
$T_{\alpha,\beta}$-power chain (3.13) is defective if every member of the
chain is bad in the sense that at least one of the following two properties
holds:

(1) The member intersects the boundary of a rectangle of the set
$\mathcal{W}_{1}$.

(2) The inequality

$\frac{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1+i;\beta;j+i)\cap(\mathcal{W}\bigtriangleup\mathcal{W}_{1}))}{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1+i;\beta;j+i))}\geqslant\delta$
(3.14)

holds.

To show that defective $T_{\alpha,\beta}$-power chains form a small minority,
let $\Delta$ denote the total number of pairs $(\sigma,j)$ satisfying
$\sigma=1,2,3$ and $j=1,\ldots,q^{\prime}_{h+1}-q_{k+1}$ such that the
$T_{\alpha,\beta}$-power chain (3.13) is defective. Then the total number
$\Delta(q_{k+1}-1)$ of small special rectangles (3.9) with

$\sigma=1,2,3,\quad i=0,1,\ldots,q_{k+1}-2,\quad
j=1,\ldots,q^{\prime}_{h+1}-q_{k+1}$ (3.15)

in defective $T_{\alpha,\beta}$-power chains satisfies

$\Delta(q_{k+1}-1)\leqslant\Lambda+\Omega,$ (3.16)

where $\Lambda$ denotes the total number of small special rectangles (3.9)
with (3.15) that intersect the boundary of a rectangle in the set
$\mathcal{W}_{1}$, and $\Omega$ denotes the total number of small special
rectangles (3.9) with (3.15) such that the inequality (3.14) holds.

Consider a face $Y_{\sigma}$, where a typical point is of the form
$\mathbf{p}_{\sigma}+(x,0,z)$. Note that the intervals $J_{k}(\alpha;1+i)$,
$i=0,1,\ldots,q_{k+1}-2$, are disjoint, and for each fixed $i$, the intervals
$J^{\prime}_{h}(\beta;j+i)$, $j=1,\ldots,q^{\prime}_{h+1}-q_{k+1}$, are
disjoint. It follows that on $Y_{\sigma}$, any line segment in the
$x$-direction intersects less than $q_{k+1}$ of the small special rectangles,
while any line segment in the $z$-direction intersects less than
$q^{\prime}_{h+1}$ of the small special rectangles. This means that the
boundary of any rectangle in the set $\mathcal{W}_{1}$ intersects less than
$2(q_{k+1}+q^{\prime}_{h+1})$ small special rectangles. Since
$\mathcal{W}_{1}$ contains only finitely many rectangles, it follows that
there exists a constant $c_{1}=c_{1}(\varepsilon;\mathcal{W})$ such that
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\Lambda\leqslant c_{1}(q_{k+1}+q^{\prime}_{h+1}).$ (3.17)

Next, note that properties (i)–(iii) imply the trivial bounds

$\lambda_{1}(J_{k}(\alpha;1))\geqslant\frac{1}{3(q_{k+1}-1)}\quad\mbox{and}\quad\lambda_{1}(J^{\prime}_{k}(\beta;1))\geqslant\frac{1}{3(q^{\prime}_{h+1}-1)},$

so that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1+i;\beta;j+i))\geqslant\frac{1}{9(q_{k+1}-1)(q^{\prime}_{h+1}-1)}.$
(3.18)

Combining (3.11), (3.14) and (3.18), we deduce that

$\varepsilon>\lambda_{2}(\mathcal{W}\bigtriangleup\mathcal{W}_{1})\geqslant\frac{\Omega\delta}{9(q_{k+1}-1)(q^{\prime}_{h+1}-1)}.$

This inequality can be simplified to

$\Omega<\varepsilon^{1/2}(q_{k+1}-1)(q^{\prime}_{h+1}-1)$ (3.19)

if we choose the parameter $\delta$ to satisfy

$\delta=\delta(\varepsilon)=9\varepsilon^{1/2}.$ (3.20)

Combining (3.16), (3.17) and (3.19), we deduce that

$\Delta<\frac{c_{1}(q_{k+1}+q^{\prime}_{h+1})}{q_{k+1}-1}+\varepsilon^{1/2}(q^{\prime}_{h+1}-1).$

Since there are precisely $3(q^{\prime}_{h+1}-q_{k+1})$
$T_{\alpha,\beta}$-power chains, it then follows that as long as (3.20) holds,
the proportion of defective $T_{\alpha,\beta}$-power chains is at most

$\frac{c_{1}(q_{k+1}+q^{\prime}_{h+1})}{3(q^{\prime}_{h+1}-q_{k+1})(q_{k+1}-1)}+\frac{\varepsilon^{1/2}(q^{\prime}_{h+1}-1)}{3(q^{\prime}_{h+1}-q_{k+1})},$
(3.21)

We now repeat the argument for the reverse flow.

For any integer $k\geqslant 1$, we consider the partition
$\mathcal{A}_{k}(-\alpha)$ of $[0,1)$ with $q_{k+1}$ division points
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\\{-\ell\alpha\\},\quad\ell=0,1,2,\ldots,q_{k+1}-1.$

For every $i=1,\ldots,q_{k+1}-1$, we consider the special short interval
$J_{k}(\alpha;-i)$, defined by replacing $i$ by $-i$ in (3.4).

For every $\sigma=1,2,3$ and every $j=q_{k+1}+1,\ldots,q^{\prime}_{h+1}$, we
consider the small special rectangle

$R_{\sigma,k,h}(\alpha;-1;\beta;j-2)=\\{\mathbf{p}_{\sigma}+(x,0,z)\in
Y_{\sigma}:(x,z)\in J_{k}(\alpha;-1)\times J^{\prime}_{h}(\beta;j-2)\\}$

on the face $Y_{\sigma}$. For every $\sigma$ and $j$, and for every
$i=0,1,\ldots,q_{k+1}-2$, we also consider the image of this small special
rectangle under $T_{\alpha,\beta}^{-i}$, given by

$R^{\dagger}_{\sigma,k,h}(\alpha;-1-i;\beta;j-2-i)=T_{\alpha,\beta}^{-i}(R_{\sigma,k,h}(\alpha;-1;\beta;j-2))$

which is not necessarily in $Y_{\sigma}$ but can be anywhere in $\mathcal{Y}$,
and is splitting free. Thus these are also small special rectangles, with the
same area given by (3.10).

As before, for any pair $(\sigma,j)$ with $\sigma=1,2,3$ and
$j=q_{k+1}+1,\ldots,q^{\prime}_{h+1}$, we say that the
$T_{\alpha,\beta}^{-1}$-power chain of the pair $(\sigma,j)$, given by

$R^{\dagger}_{\sigma,k,h}(\alpha;-1-i;\beta;j-2-i),\quad
i=0,1,\ldots,q_{k+1}-2,$ (3.22)

is defective if every member of the chain is bad in the sense that at least
one of the following two properties holds:

(1) The member intersects the boundary of a rectangle of the set
$\mathcal{W}_{1}$.

(2) The inequality

$\frac{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;-1-i;\beta;j-2-i)\cap(\mathcal{W}\bigtriangleup\mathcal{W}_{1}))}{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;-1-i;\beta;j-2-i))}\geqslant\delta$

holds, where $\delta$ is defined as before.

An analogous argument then shows that the proportion of defective
$T_{\alpha,\beta}^{-1}$-power chains is at most (3.21).

######  Lemma 3.2.

Suppose that the proportion (3.21) both of defective $T_{\alpha,\beta}$-power
chains (3.13) and of defective $T_{\alpha,\beta}^{-1}$-power chains (3.22) is
sufficiently small. Then there exists an integer $j_{0}$ satisfying
$q_{k+1}+1\leqslant j_{0}\leqslant q^{\prime}_{h+1}-q_{k+1}$ such that for
every $\sigma=1,2,3$,

_(i)_ the $T_{\alpha,\beta}$-power chain (3.13) with $j=j_{0}$ is not
defective; and

_(ii)_ the $T_{\alpha,\beta}^{-1}$-power chain (3.22) with $j=j_{0}$ is not
defective.

It follows from (i) that for every $\sigma=1,2,3$, there exists an integer
$i_{0}$ satisfying $0\leqslant i_{0}\leqslant q_{k+1}-2$ such that the small
special rectangle

$R^{\dagger}_{\sigma,k,h}(\alpha;1+i_{0};\beta;j_{0}+i_{0})$

does not intersect the boundary of a rectangle of the set $\mathcal{W}_{1}$
and also satisfies the inequality xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\frac{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1+i_{0};\beta;j_{0}+i_{0})\cap(\mathcal{W}\bigtriangleup\mathcal{W}_{1}))}{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1+i_{0};\beta;j_{0}+i_{0}))}<\delta.$
(3.23)

The former implies that

$R^{\dagger}_{\sigma,k,h}(\alpha;1+i_{0};\beta;j_{0}+i_{0})\subset\mathcal{W}_{1}\quad\mbox{or}\quad
R^{\dagger}_{\sigma,k,h}(\alpha;1+i_{0};\beta;j_{0}+i_{0})\subset\mathcal{W}_{1}^{c}.$
(3.24)

If the first condition in (3.24) holds, then it follows from (3.23) that

$\frac{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1+i_{0};\beta;j_{0}+i_{0})\cap\mathcal{W})}{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1+i_{0};\beta;j_{0}+i_{0}))}\geqslant
1-\delta.$ (3.25)

If the second condition in (3.24) holds, then it follows from (3.12) and
(3.23) that

$\frac{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1+i_{0};\beta;j_{0}+i_{0})\cap\mathcal{S})}{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1+i_{0};\beta;j_{0}+i_{0}))}\geqslant
1-\delta.$ (3.26)

Since $T_{\alpha,\beta}$ is measure-preserving and the subsets $\mathcal{W}$
and $\mathcal{S}$ are $T_{\alpha,\beta}$-invariant, it follows that each
condition (3.25) or (3.26), if true, extends to the whole
$T_{\alpha,\beta}$-power chain. Thus in particular, one of the inequalities

$\frac{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1;\beta;j_{0})\cap\mathcal{W})}{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1;\beta;j_{0}))}\geqslant
1-\delta\quad\mbox{or}\quad\frac{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1;\beta;j_{0})\cap\mathcal{S})}{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;1;\beta;j_{0}))}\geqslant
1-\delta$

holds. Similarly, it follows from (ii) that one of the inequalities

$\frac{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;-1;\beta;j_{0}-2)\cap\mathcal{W})}{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;-1;\beta;j_{0}-2))}\geqslant
1-\delta\quad\\!\mbox{or}\quad\\!\frac{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;-1;\beta;j_{0}-2)\cap\mathcal{S})}{\lambda_{2}(R^{\dagger}_{\sigma,k,h}(\alpha;-1;\beta;j_{0}-2))}\geqslant
1-\delta$

holds. For $\sigma=1,2,3$, we now take

$R_{\sigma}=R^{\dagger}_{\sigma,k,h}(\alpha;-1;\beta;j_{0}-2)=R_{\sigma,k,h}(\alpha;-1;\beta;j_{0}-2)$

on the face $Y_{\sigma}$, and take

$R^{*}_{\sigma}=R^{\dagger}_{\sigma,k,h}(\alpha;1;\beta;j_{0})=R_{\sigma,k,h}(\alpha;1;\beta;j_{0})$

on the face $Y_{\sigma}$.

To complete the proof of Lemma 1.2, it remains to specify the parameters.

To satisfy the condition (1.11) and its analogues, we need $1-\delta>99/100$,
so that $\delta<1/100$. In view of (3.20), we need the parameter
$\varepsilon>0$ to satisfy

$\varepsilon<\frac{1}{810000}.$

The parameter $\varepsilon>0$ also needs to be sufficiently small to ensure
that the proportion (3.21) of defective power chains is small. Note that that
we can guarantee that

$\frac{q_{k+1}+q^{\prime}_{h+1}}{3(q^{\prime}_{h+1}-q_{k+1})}\leqslant
1\quad\mbox{and}\quad\frac{q^{\prime}_{h+1}-1}{3(q^{\prime}_{h+1}-q_{k+1})}\leqslant
1$

by ensuring that the two continued fraction denominators $q_{k+1}$ and
$q^{\prime}_{h+1}$ satisfy the condition $q^{\prime}_{h+1}\geqslant
c_{2}q_{k+1}$ for some suitably large absolute constant $c_{2}$. Recall next
that the constant $c_{1}$ depends only on $\varepsilon$ and $\mathcal{W}$. It
follows that we can choose the continued fraction denominator $q_{k+1}$ to be
sufficiently large to ensure that the factor

$\frac{c_{1}}{q_{k+1}-1}$

is as small as we please. This completes the proof of Lemma 1.2.

##  4\. Extending ergodicity to unique ergodicity

The term unique ergodicity refers to the extension from uniformity for half-
infinite geodesics with almost every starting point to uniformity for half-
infinite geodesics with every starting point and that do not hit any
singularity and become undefined. The basic ideas and pioneering results are
due to Bogoliuboff and Krylov [1] in 1937; see also Furstenberg [2, Sections
3.2–3.3].

The proof of Theorem 1 is by contradiction and consists of two parts. Part 1
is a general argument based on Krylov–Bogoliuboff theory, and the key idea
here is to reformulate the problem in terms of invariant Borel measures and to
use functional analysis. Part 2 is an ad hoc argument based on the special
property that the $\mathbf{v}$-flow in the L-solid translation $3$-manifold,
when reduced modulo one to geodesic flow in the unit torus $[0,1)^{3}$,
exhibits uniformity.

For convenience, we denote by $\mathcal{M}$ the L-solid translation
$3$-manifold.

######  Part 1.

Suppose, on the contrary, that there exist a half-infinite geodesic
$\mathcal{L}(\mathbf{v};\mathbf{p}_{0};t)$, $t\geqslant 0$, with arc-length
parametization, where $\mathbf{v}\in\mathbb{R}^{3}$ is a Kronecker direction
and $\mathbf{p}\in\mathcal{M}$ is the starting point, and a non-zero
continuous function $f_{0}$ on $\mathcal{M}$ for which uniformity fails. Then
the infinite sequence of integrals

$\frac{1}{n}\int_{0}^{n}f_{0}(\mathcal{L}(\mathbf{v};\mathbf{p}_{0};t))\,\mathrm{d}t,\quad
n=1,2,3,\ldots,$

does not converge to xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\frac{1}{3}\int_{\mathcal{M}}f_{0}\,\mathrm{d}\lambda_{3},$ (4.1)

where $\frac{1}{3}\lambda_{3}$ is a probability measure on $\mathcal{M}$, so
that there exists an infinite strictly increasing sequence
$h_{0}<h_{1}<h_{2}<h_{3}<\ldots$ of positive integers such that the limit

$\lim_{n\to\infty}\frac{1}{h_{n}}\int_{0}^{h_{n}}f_{0}(\mathcal{L}(\mathbf{v};\mathbf{p}_{0};t))\,\mathrm{d}t$

exists but is not equal to (4.1).

######  Lemma 4.1.

Under these assumptions, there exists an ergodic measure-preserving system
$(\mathcal{M},\mathcal{B},\nu,\mbox{$\mathbf{v}$-flow})$, where $\mathcal{B}$
denotes the Borel $\sigma$-algebra in $\mathcal{M}$, and $\nu$ denotes a new
$\mathbf{v}$-flow-invariant Borel probability measure, such that

$\lim_{n\to\infty}\frac{1}{h_{n}}\int_{0}^{h_{n}}f_{0}(\mathcal{L}(\mathbf{v};\mathbf{p}_{0};t))\,\mathrm{d}t=\int_{\mathcal{M}}f_{0}\,\mathrm{d}\nu\neq\frac{1}{3}\int_{\mathcal{M}}f_{0}\,\mathrm{d}\lambda_{3}.$
(4.2)

###### Proof.

We introduce, for every integer $n=1,2,3,\ldots,$ the particular normalized
length-counting measure $\nu_{n}$ on Borel sets $B$ in $\mathcal{M}$ defined
by

$\nu_{n}(B)=\frac{1}{h_{n}}\int_{0}^{h_{n}}\chi_{B}(\mathcal{L}(\mathbf{v};\mathbf{p}_{0};t))\,\mathrm{d}t,$

where the characteristic function $\chi_{B}$ satisfies

$\chi_{B}(\mathbf{p})=\left\\{\begin{array}[]{ll}1,&\mbox{if $\mathbf{p}\in
B$},\\\ 0,&\mbox{if $\mathbf{p}\not\in B$}.\end{array}\right.$

We then apply a general theorem in functional analysis which says that the
space of Borel probability measures on any compact set is compact in the weak-
star topology, which means that for every continuous function $f$ on the
compact space, as $n\to\infty$,

$\mu_{n}\to\mu\quad\mbox{if and only if}\quad\int f\,\mathrm{d}\mu_{n}\to\int
f\,\mathrm{d}\mu.$ (4.3)

######  Remark.

This compactness theorem is a non-trivial result. The standard proof is based
on the Riesz representation theorem and can be found in most textbooks on
functional analysis.

Thus the space $\mathfrak{M}=\mathfrak{M}(\mathcal{M})$ of Borel probability
measures on $\mathcal{M}$ is compact, so there exists a subsequence
$\nu_{n_{m}}$ of the sequence $\nu_{n}$ and a Borel probability measure
$\nu_{\infty}$ such that $\nu_{n_{m}}\to\nu_{\infty}$ as $m\to\infty$, and it
follows from (4.3) that

$\lim_{m\to\infty}\int_{\mathcal{M}}f\,\mathrm{d}\nu_{n_{m}}=\int_{\mathcal{M}}f\,\mathrm{d}\nu_{\infty}\quad\mbox{for
every continuous function $f$ on $\mathcal{M}$}.$ (4.4)

Let $t_{1}>0$ be arbitrary but fixed. For any continuous function $f$ defined
on $\mathcal{M}$, let $f_{t_{1}}$ denote the function obtained from $f$
writing $f(t+t_{1})=f_{t_{1}}(t)$. Then

$\displaystyle\int_{\mathcal{M}}f_{t_{1}}\,\mathrm{d}\nu_{n_{m}}$
$\displaystyle=\frac{1}{h_{n_{m}}}\int_{0}^{h_{n_{m}}}f_{t_{1}}(\mathcal{L}(\mathbf{v};\mathbf{p}_{0};t))\,\mathrm{d}t=\frac{1}{h_{n_{m}}}\int_{0}^{h_{n_{m}}}f(\mathcal{L}(\mathbf{v};\mathbf{p}_{0};t+t_{1}))\,\mathrm{d}t$
$\displaystyle=\frac{1}{h_{n_{m}}}\int_{0}^{h_{n_{m}}}f(\mathcal{L}(\mathbf{v};\mathbf{p}_{0};t))\,\mathrm{d}t$
$\displaystyle\qquad+\frac{1}{h_{n_{m}}}\int_{0}^{t_{1}}(f(\mathcal{L}(\mathbf{v};\mathbf{p}_{0};t+h_{n_{m}}))-f(\mathcal{L}(\mathbf{v};\mathbf{p}_{0};t)))\,\mathrm{d}t$
$\displaystyle=\int_{\mathcal{M}}f\,\mathrm{d}\nu_{n_{m}}+E(n_{m}),$

where xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$|E(n_{m})|\leqslant\frac{2t_{1}\sup|f|}{h_{n_{m}}}.$

Since $f$ is continuous on the compact set $\mathcal{M}$, the factor $\sup|f|$
exists and is finite. Thus $E(n_{m})\to 0$ as $m\to\infty$, so that

$\lim_{m\to\infty}\int_{\mathcal{M}}f_{t_{1}}\,\mathrm{d}\nu_{n_{m}}=\lim_{m\to\infty}\int_{\mathcal{M}}f\,\mathrm{d}\nu_{n_{m}}.$
(4.5)

Combining (4.4) and (4.5), we conclude that

$\int_{\mathcal{M}}f_{t_{1}}\,\mathrm{d}\nu_{\infty}=\int_{\mathcal{M}}f\,\mathrm{d}\nu_{\infty},$

and this establishes the $\mathbf{v}$-flow-invariance of the limit measure
$\nu_{\infty}$. Furthermore, combining (4.2) and (4.4) with $f=f_{0}$, we
conclude that

$\int_{\mathcal{M}}f_{0}\,\mathrm{d}\nu_{\infty}\neq\frac{1}{3}\int_{\mathcal{M}}f_{0}\,\mathrm{d}\lambda_{3}.$
(4.6)

This completes the proof. ∎

Let $\mathfrak{M}_{1}\subset\mathfrak{M}$ denote the set of Borel probability
measures $\mu$ on $\mathcal{M}$ that are $\mathbf{v}$-flow-invariant and
satisfy the requirement

$\int_{\mathcal{M}}f_{0}\,\mathrm{d}\mu=\int_{\mathcal{M}}f_{0}\,\mathrm{d}\nu_{\infty}.$

Then $\nu_{\infty}\in\mathfrak{M}_{1}$, so that $\mathfrak{M}_{1}$ is a non-
empty closed subset of $\mathfrak{M}$, and so $\mathfrak{M}_{1}$ is compact.

With some appropriate $\nu^{*}\in\mathfrak{M}_{1}$, we can extend the above to
a measure-preserving system
$(\mathcal{M},\mathcal{B},\nu^{*},\mbox{$\mathbf{v}$-flow})$ which is ergodic.
To prove this, we use the almost trivial fact that $\mathfrak{M}_{1}$ is
convex. The general Krein–Millman theorem in functional analysis implies that
the non-empty compact and convex set $\mathfrak{M}_{1}$ is spanned by its
extremal points. It is a well known result in ergodic theory that the extremal
points in the space of measures are the ergodic measures; see Furstenberg [2,
Proposition 3.4]. Thus we can choose the measure $\nu^{*}\in\mathfrak{M}_{1}$
above to be such an extremal point.

######  Part 2.

Consider the measure-preserving system
$(\mathcal{M},\mathcal{B},\nu^{*},\mbox{$\mathbf{v}$-flow})$ which is ergodic.
We now apply the flow version of the Birkhoff ergodic theorem. For a
continuous function $f$ on $\mathcal{M}$ and $\nu^{*}$-almost every starting
point $\mathbf{p}\in\mathcal{M}$, we have

$\lim_{n\to\infty}\frac{1}{n}\int_{0}^{n}f(\mathcal{L}(\mathbf{v};\mathbf{p};t))\,\mathrm{d}t=\int_{\mathcal{M}}f\,\mathrm{d}\nu^{*}.$
(4.7)

We return to the continuous function $f_{0}$ at the beginning of Part 1 for
which uniformity fails, and consider the Borel set

$Y=\left\\{\mathbf{p}\in\mathcal{M}:\lim_{n\to\infty}\frac{1}{n}\int_{0}^{n}f_{0}(\mathcal{L}(\mathbf{v};\mathbf{p};t))\,\mathrm{d}t=\int_{\mathcal{M}}f_{0}\,\mathrm{d}\nu^{*}\right\\}.$
(4.8)

Then it follows from (4.7) and (4.8) that

$\nu^{*}(Y)=1.$ (4.9)

Meanwhile, the weaker form of Theorem 1 gives uniformity for
$\lambda_{3}$-almost every starting point $\mathbf{p}\in\mathcal{M}$, and so

$\lim_{n\to\infty}\frac{1}{n}\int_{0}^{n}f_{0}(\mathcal{L}(\mathbf{v};\mathbf{p};t))\,\mathrm{d}t=\frac{1}{3}\int_{\mathcal{M}}f_{0}\,\mathrm{d}\lambda_{3}.$
(4.10)

Combining (4.6) with $\nu^{*}$ replacing $\nu_{\infty}$, (4.7) in the special
case $f=f_{0}$ and (4.10), we conclude that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\lambda_{3}(Y)=0.$

This implies that for every $\delta>0$, there exists an infinite collection of
disjoint axis-parallel rectangular boxes $\mathfrak{R}_{i}$, $i=1,2,3,\ldots,$
such that

$\sum_{i=1}^{\infty}\lambda_{3}(\mathfrak{R}_{i})<\delta\quad\mbox{and}\quad
Y\subset\bigcup_{i=1}^{\infty}\mathfrak{R}_{i},$ (4.11)

and it follows from (4.9) that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\nu^{*}\left(\bigcup_{i=1}^{\infty}\mathfrak{R}_{i}\right)=1.$ (4.12)

Comparing (4.11) and (4.12), we see that there exists a positive integer
$i_{0}$ such that

$\frac{\lambda_{3}(\mathfrak{R}_{i_{0}})}{\nu^{*}(\mathfrak{R}_{i_{0}})}<\delta.$
(4.13)

Clearly, for $\nu^{*}$-almost every starting point $\mathbf{p}\in\mathcal{M}$,
we have

$\lim_{n\to\infty}\frac{1}{n}\int_{0}^{n}\chi_{\mathfrak{R}_{i_{0}}}(\mathcal{L}(\mathbf{v};\mathbf{p};t))\,\mathrm{d}t=\nu^{*}(\mathfrak{R}_{i_{0}}),$
(4.14)

where $\chi_{\mathfrak{R}_{i_{0}}}$ is the characteristic function of
$\mathfrak{R}_{i_{0}}$.

We next make use of the crucial fact that the $\mathbf{v}$-flow in
$\mathcal{M}$ reduces modulo one to $\mathbf{v}$-flow in the unit torus
$[0,1)^{3}$. Since $\mathcal{M}$ consists of $3$ cubes, every point in the
unit torus $[0,1)^{3}$ has $3$ pre-images in $\mathcal{M}$. We can clearly
assume that $\mathfrak{R}_{i_{0}}$ is contained in one of these $3$ cubes.
Write $\mathfrak{R}_{i_{0}}=\mathfrak{R}(1)$, and then let $\mathfrak{R}(2)$
and $\mathfrak{R}(3)$ denote axis-parallel rectangular boxes in the other $2$
cubes with the property that $\mathfrak{R}(1),\mathfrak{R}(2),\mathfrak{R}(3)$
reduce modulo one to the same axis-parallel rectangular box in the unit torus
$[0,1)^{3}$.

Since $\mathbf{v}$ is a Kronecker direction, the continuous form of the
Kronecker–Weyl equidistribution theorem applies for geodesic flow in direction
$\mathbf{v}$ in the unit torus $[0,1)^{3}$. It follows that

$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{3}\int_{0}^{n}\chi_{\mathfrak{R}(i)}(\mathcal{L}(\mathbf{v};\mathbf{p};t))\,\mathrm{d}t=\lambda_{3}(\mathfrak{R}(1)\cup\mathfrak{R}(2)\cup\mathfrak{R}(3))=3\lambda_{3}(\mathfrak{R}_{i_{0}})$
(4.15)

holds for every starting point $\mathbf{p}\in\mathcal{M}$ where the half-
infinite geodesic does not hit a singularity.

Recall that (4.14) holds for $\nu^{*}$-almost every starting point
$\mathbf{p}\in\mathcal{M}$. Let $\mathbf{p}^{*}$ be such a starting point.
Combining (4.13) and (4.14) gives

$\frac{\lambda_{3}(\mathfrak{R}_{i_{0}})}{\delta}<\lim_{n\to\infty}\frac{1}{n}\int_{0}^{n}\chi_{\mathfrak{R}_{i_{0}}}(\mathcal{L}(\mathbf{v};\mathbf{p}^{*};t))\,\mathrm{d}t.$
(4.16)

On the other hand, the right hand side of (4.16) can be bounded from above by
using (4.15) with $\mathbf{p}=\mathbf{p}^{*}$, and we have

$\lim_{n\to\infty}\frac{1}{n}\int_{0}^{n}\chi_{\mathfrak{R}_{i_{0}}}(\mathcal{L}(\mathbf{v};\mathbf{p}^{*};t))\,\mathrm{d}t\leqslant
3\lambda_{3}(\mathfrak{R}_{i_{0}}).$

Choosing $\delta=1/3$ clearly leads to a contradiction, and this completes the
proof of Theorem 1.

##  5\. Proof of Theorem 2

The proof of Theorem 2 is an adaptation of the proof of Theorem 1, where the
only extra idea is one more application of ergodicity.

The geodesic flow in $\mathcal{P}\times[0,1)$ under consideration is
$\mathbf{v}$-flow, with $\mathbf{v}=(\alpha,1,\beta)$, where $\alpha$, $\beta$
and $\alpha/\beta$ are all irrational. For simplicity, we assume that
$\alpha>0$ and $\beta>0$, and like before, we further assume that $\alpha<1$.

For the $\sigma$-th atomic cube of $\mathcal{P}\times[0,1)$, where
$\sigma=1,\ldots,d$, we denote by $X_{\sigma}$ and $Y_{\sigma}$ the left
$X$-face and the bottom $Y$-face respectively, and denote by $E_{\sigma}$ the
common edge of the right $X$-face and the top $Y$-face, as shown in Figure
5.1.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-5-1.pdf}\vspace{3pt}\\\
\mbox{Figure 5.1: the faces $X_{\sigma},Y_{\sigma}$ and the edge $E_{\sigma}$
of the $\sigma$-th atomic cube,}\\\ \mbox{and the narrow strip $S_{\sigma}$ on
$Y_{\sigma}$}\end{array}$

We consider the discretization of the $\mathbf{v}$-flow in polycube
translation $3$-manifold $\mathcal{P}\times[0,1)$ relative to the $Y$-faces,
where the flow hits the faces $Y_{1},\ldots,Y_{d}$. Let

$T_{\alpha,\beta}:\mathcal{Y}=Y_{1}\cup\ldots\cup Y_{d}\to\mathcal{Y}$ (5.1)

denote the relevant discrete transformation defined by consecutive hitting
points. Note that the projection of the $\mathbf{v}$-flow to the $xy$-plane
gives rise to a flow on the polysquare translation surface $\mathcal{P}$ with
slope $\alpha^{-1}$.

We need to prove that $T_{\alpha,\beta}$ is ergodic. As before, we proceed by
contradiction. Assume, on the contrary, that $T_{\alpha,\beta}$ is not
ergodic. Then there exist two $T_{\alpha,\beta}$-invariant subsets
$\mathcal{W}$ (white) and $\mathcal{S}$ (silver) such that
$\mathcal{Y}=\mathcal{W}\cup\mathcal{S}$ and

$1\leqslant\lambda_{2}(\mathcal{W})\leqslant\lambda_{2}(\mathcal{S})\leqslant
d-1,$ (5.2)

where $\lambda_{2}$ denotes the $2$-dimensional Lebesgue measure. We show that
this leads to a contradiction.

Again, we consider the multiplicity functions $f_{\mathcal{W}}$ and
$f_{\mathcal{S}}$ of the invariant sets $\mathcal{W}$ and $\mathcal{S}$
respectively. More precisely, consider the projection of $\mathcal{Y}$ to the
unit torus $[0,1)^{2}$, given by (1.3). Then for every point $P\in[0,1)^{2}$,
there are precisely $d$ points $P_{1},\ldots,P_{d}\in\mathcal{Y}$ which have
projection image $P$. Let

$f_{\mathcal{W}}(P)=|\\{P_{1},\ldots,P_{d}\\}\cap\mathcal{W}|\quad\mbox{and}\quad
f_{\mathcal{S}}(P)=|\\{P_{1},\ldots,P_{d}\\}\cap\mathcal{S}|$ (5.3)

denote the number of these $d$ points that fall into $\mathcal{W}$ and
$\mathcal{S}$ respectively, so that

$f_{\mathcal{W}}(P)+f_{\mathcal{S}}(P)=d\quad\mbox{for almost all
$P\in[0,1)^{2}$}.$ (5.4)

We also consider the corresponding projection of

$T_{\alpha,\beta}:\mathcal{Y}\to\mathcal{Y}\quad\mbox{to}\quad
T_{0}:[0,1)^{2}\to[0,1)^{2},$

which is simply the $(\alpha,\beta)$-shift on the unit torus $[0,1)^{2}$.

For each $\sigma=1,\ldots,d$, there is a line segment on the face $Y_{\sigma}$
which is the image under $T_{\alpha,\beta}^{-1}$ of the edge $E_{\sigma}$ of
$\mathcal{P}\times[0,1)$ of the $\sigma$-th atomic cube. It is possible that
this edge $E_{\sigma}$ is a singular edge of $\mathcal{P}\times[0,1)$.
Consider a narrow strip neighbourhood $S_{\sigma}$ of this line segment, as
illustrated in light gray in the picture on the right in Figure 5.1. As in the
proof of Lemma 1.2, we can think of the $x$-coordinate of every point in
$S_{\sigma}$ to fall modulo one into an interval of the type

$J_{k}(\alpha;-1)=\left(\\{-\alpha\\}-\frac{\|q_{k}\alpha\|}{2},\\{-\alpha\\}+\frac{\|q_{k}\alpha\|}{2}\right).$

In other words, if $\mathbf{p}_{\sigma}$ represents the bottom left vertex of
the face $Y_{\sigma}$ of the translation $3$-manifold $\mathcal{M}$, then

$S_{\sigma}=\\{\mathbf{p}_{\sigma}+(x,0,z)\in Y_{\sigma}:(x,z)\in
J_{k}(\alpha;-1)\times[0,1)\\}.$

It is straightforward to extend Lemma 1.2 to the case of a polycube
translation $3$-manifold which is the cartesian product of a polysquare
translation surface with $d$ faces with the unit torus. However, to establish
Theorem 2, we need something stronger than this extension, and this is
facilitated by making a small change to the definition of the small special
rectangles.

For every integer $j=1,\ldots,q^{\prime}_{h+1}-1$, we replace the short
special interval (3.8) by the possibly longer interval

$J^{\prime}_{h}(j)=J^{\prime}_{h}(\beta;j)=\left(\\{j\beta\\}-\frac{\mathfrak{d}_{h}(\beta;j;-)}{2},\\{j\beta\\}+\frac{\mathfrak{d}_{h}(\beta;j;+)}{2}\right),$
(5.5)

where $\mathfrak{d}(\beta;j;-)$ and $\mathfrak{d}(\beta;j;+)$ denote the
distances of $\\{j\beta\\}$ to its immediate left neighbour and to its
immediate right neighbour respectively in $\mathcal{B}_{h}(\beta)$. Clearly
each of $\mathfrak{d}(\beta;j;-)$ and $\mathfrak{d}(\beta;j;+)$ has value
equal to one of the values in (3.7).

The small special rectangles $R_{\sigma,k,h}$ and $R^{\dagger}_{\sigma,k,h}$
as well as the $T_{\alpha,\beta}$-power chains and the
$T^{-1}_{\alpha,\beta}$-power chains are then defined with the modification
(5.5) in place. Note in particular that this modification is in the
$z$-direction which is integrable, so all the small special rectangles and
their images remain splitting free as before.

Let $I\subset[0,1)$ be an interval of the form (5.5), and let

$R_{\sigma}=\\{(x,y,z)\in S_{\sigma}:z\in I\\},\quad\sigma=1,\ldots,d,$ (5.6)

denote small special rectangles of the decomposition of the strips
$S_{1},\ldots,S_{d}$ on the faces $Y_{1},\ldots,Y_{d}$ respectively with the
condition that their images on $[0,1)^{2}$ under the projection (1.3)
coincide, so that they are in the same relative positions on the faces
$Y_{1},\ldots,Y_{d}$ respectively. It is clear that the image of each of
$R_{1},\ldots,R_{d}$ under $T_{\alpha,\beta}$ splits along the edges
$E_{1},\ldots,E_{d}$ into congruent halves. For each $\sigma=1,\ldots,d$, the
rectangle $R_{\sigma}$ splits into the left half $R_{\sigma}^{-}$ and the
right half $R_{\sigma}^{+}$, and it is clear that their images under
$T_{\alpha,\beta}$ may lie on distinct $Y$-faces. On the other hand, for every
$\sigma=1,\ldots,d$, there is a small special rectangle in $Y_{\sigma}$ of the
form

$R^{*}_{\sigma}=T_{\alpha,\beta}^{2}(R_{\rho^{\prime}}^{-})\cup
T_{\alpha,\beta}^{2}(R_{\rho^{\prime\prime}}^{+}),$

which is the union of the image under $T_{\alpha,\beta}^{2}$ of the left half
$R_{\rho^{\prime}}^{-}$ of some small special rectangle $R_{\rho^{\prime}}$
and the image under $T_{\alpha,\beta}^{2}$ of the right half
$R_{\rho^{\prime\prime}}^{+}$ of some small special rectangle
$R_{\rho^{\prime\prime}}$. Here the two small special rectangles
$R_{\rho^{\prime}}$ and $R_{\rho^{\prime\prime}}$ may be identical or
different. However, in view of the restriction imposed on the small special
rectangles $R_{1},\ldots,R_{d}$ that their images under the projection (1.3)
coincide, we deduce that the images of the small special rectangles
$R^{*}_{1},\ldots,R^{*}_{d}$ under the projection (1.3) also coincide, apart
possibly from sets of zero measure caused by the singular edges of
$\mathcal{P}\times[0,1)$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-5-2.pdf}\vspace{3pt}\\\
\mbox{Figure 5.2: the small special rectangles $R_{\sigma}$ and
$R^{*}_{\sigma}$ on the face $Y_{\sigma}$,}\\\ \mbox{and the projectied view
with the $z$-coordinate removed}\end{array}$

It is clear that the rectangles $R_{1},\ldots,R_{d}$ and
$R^{*}_{1},\ldots,R^{*}_{d}$ have the same area.

Given a positive number $\eta>0$, we say that a collection
$(R_{1},\ldots,R_{d})$ of the form (5.6) is $\eta$-nice if for each
$\sigma=1,\ldots,d$, the set $R_{\sigma}$ satisfies

$\frac{\lambda_{2}(R_{\sigma}\cap\mathcal{W})}{\lambda_{2}(R_{\sigma})}>1-\eta\quad\mbox{or}\quad\frac{\lambda_{2}(R_{\sigma}\cap\mathcal{S})}{\lambda_{2}(R_{\sigma})}>1-\eta,$

and the set $R^{*}_{\sigma}$ satisfies

$\frac{\lambda_{2}(R^{*}_{\sigma}\cap\mathcal{W})}{\lambda_{2}(R^{*}_{\sigma})}>1-\eta\quad\mbox{or}\quad\frac{\lambda_{2}(R^{*}_{\sigma}\cap\mathcal{S})}{\lambda_{2}(R^{*}_{\sigma})}>1-\eta.$

We have the following stronger form of Lemma 1.2.

######  Lemma 5.1.

Let $\varepsilon>0$ and $\eta>0$ be arbitrarily small and fixed. Then there
exist sufficiently large integers $k$ and $h$ such that for every
$\sigma=1,\ldots,d$, the union of all rectangles
$R_{\sigma,k,h}(\alpha;1;\beta;j)$ in the strip $S_{\sigma}$ arising from
$\eta$-nice collections $(R_{1},\ldots,R_{d})$ has area at least
$(1-\varepsilon)\lambda_{2}(S_{\sigma})$.

###### Idea of proof.

There are two critical changes to the proof of Lemma 1.2, and each contributes
to part of the value of $\varepsilon$.

First of all, by modifying the small rectangles in the $z$-direction, we
ensure that for each $\sigma=1,\ldots,d$, the inequality

$\lambda_{2}\left(\bigcup_{j=1}^{q^{\prime}_{h+1}-q_{k+1}}R_{\sigma,k,h}(\alpha;1;\beta;j)\right)\geqslant\left(1-q_{k+1}(\|q^{\prime}_{h}\beta\|+\|q^{\prime}_{h+1}\beta\|)\right)\lambda_{2}(S_{\sigma})$

holds. With sufficiently large values of $h$ and $k$, the factor
$q_{k+1}(\|q^{\prime}_{h}\beta\|+\|q^{\prime}_{h+1}\beta\|)$ can be made
arbitrarily close to $0$. This means that the small special rectangles in the
strip $S_{\sigma}$ make up nearly all of $S_{\sigma}$.

Next, the conclusion of Lemma 3.2 can be significantly extended. It can be
shown that by choosing $h$ and $k$ suitably large, we can guarantee a
proportion arbitrarily close to $1$ of indices $j_{0}$ satisfying
$q_{k+1}+1\leqslant j_{0}\leqslant q^{\prime}_{h+1}-q_{k+1}$ such that for
every $\sigma=1,\ldots,d$, the $T_{\alpha,\beta}$-power chain (3.13) with
$j=j_{0}$ is not defective, and the $T_{\alpha,\beta}^{-1}$-power chain (3.22)
with $j=j_{0}$ is not defective. ∎

We may assume, without loss of generality, that $\sigma=1$ is part of the
cycle of length $d_{0}>d/2$ of the splitting permutation.

Suppose that there exist two different small special rectangles
$R^{\prime}_{1}$ and $R^{\prime\prime}_{1}$ in the strip $S_{1}$ that arise
from $\eta$-nice collections $(R^{\prime}_{1},\ldots,R^{\prime}_{d})$ and
$(R^{\prime\prime}_{1},\ldots,R^{\prime\prime}_{d})$ respectively such that
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\frac{\lambda_{2}(R^{\prime}_{1}\cap\mathcal{W})}{\lambda_{2}(R^{\prime}_{1})}>1-\eta\quad\mbox{and}\quad\frac{\lambda_{2}(R^{\prime\prime}_{1}\cap\mathcal{S})}{\lambda_{2}(R^{\prime\prime}_{1})}>1-\eta.$

Applying the splitting method argument at the end of Section 1 to
$R^{\prime}_{1}$, we conclude that there are at least $d_{0}$ sets among
$R^{\prime}_{1},\ldots,R^{\prime}_{d}$ which are overwhelmingly in
$\mathcal{W}$, so that $f_{\mathcal{W}}\geqslant d_{0}>d/2$. Meanwhile,
applying the same argument to $R^{\prime\prime}_{1}$, we conclude that there
are at least $d_{0}$ sets among
$R^{\prime\prime}_{1},\ldots,R^{\prime\prime}_{d}$ which are overwhelmingly in
$\mathcal{S}$, so that $f_{\mathcal{S}}\geqslant d_{0}>d/2$. This is clearly a
contradiction, since $f_{\mathcal{W}}+f_{\mathcal{S}}=d$.

Hence the strip $S_{1}$ is overwhelmingly monochromatic. We may assume,
without loss of generality, that the strip $S_{1}$ is overwhelmingly in
$\mathcal{W}$. In fact, a proportion exceedingly $(1-\varepsilon)(1-\eta)$ of
$S_{1}$ is in $\mathcal{W}$. We shall show that this eventually leads to a
contradiction.

We return to the translation $3$-manifold
$\mathcal{M}=\mathcal{P}\times[0,1)$, with $\lambda_{3}$-preserving flow in
the direction $\mathbf{v}=(\alpha,1,\beta)$. We have assumed that this flow is
not ergodic, so that there exists a decomposition
$\mathcal{M}=\mathcal{W}_{3}\cup\mathcal{S}_{3}$ into two $\mathbf{v}$-flow-
invariant and disjoint subsets $\mathcal{W}_{3}$ and $\mathcal{S}_{3}$ such
that

$1\leqslant\lambda_{3}(\mathcal{W}_{3})=f_{\mathcal{W}}\leqslant d-1,\quad
1\leqslant\lambda_{3}(\mathcal{S}_{3})=f_{\mathcal{S}}\leqslant d-1,\quad
f_{\mathcal{W}}+f_{\mathcal{S}}=d.$

######  Remark.

In view of the relationship between the discrete $T_{\alpha,\beta}$-invariance
and the continuous $\mathbf{v}$-flow-invariance, one can visualize a close
relationship between the pair $(\mathcal{W},\mathcal{S})$ and the pair
$(\mathcal{W}_{3},\mathcal{S}_{3})$, in that

$f_{\mathcal{W}}(P)=\lambda_{3}(\mathcal{W}_{3})\quad\mbox{and}\quad
f_{\mathcal{S}}(P)=\lambda_{3}(\mathcal{S}_{3})\quad\mbox{for almost all
$P\in[0,1)^{2}$}.$

For convenience, we may write $f_{\mathcal{W}}=\lambda_{3}(\mathcal{W}_{3})$
and $f_{\mathcal{S}}=\lambda_{3}(\mathcal{S}_{3})$.

Let $\chi=\chi_{\mathcal{W}_{3}}$ denote the characteristic function of the
invariant subset $\mathcal{W}_{3}\subset\mathcal{M}$, so that
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\chi(\mathbf{s})=\left\\{\begin{array}[]{ll}1,&\mbox{if
$\mathbf{s}\in\mathcal{W}_{3}$},\\\ 0,&\mbox{if
$\mathbf{s}\not\in\mathcal{W}_{3}$}.\end{array}\right.$

Writing $\mathbf{s}=(x,y,z)$ and using Fubini’s theorem, we have

$\int_{\mathcal{M}}\chi(\mathbf{s})\,\mathrm{d}\mathbf{s}=\int_{\mathcal{P}}\left(\int_{[0,1)}\chi(x,y,z)\,\mathrm{d}z\right)\mathrm{d}\lambda_{2}(x,y),$

where the inner integral xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\psi(x,y)=\int_{[0,1)}\chi(x,y,z)\,\mathrm{d}z$ (5.7)

is well defined in the Lebesgue sense for almost every $(x,y)\in\mathcal{P}$.

######  Lemma 5.2.

Suppose that the $\mathbf{v}$-flow in $\mathcal{M}$ is not ergodic. Then for
almost every $(x,y)\in\mathcal{P}$, we have xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\frac{1}{d}\leqslant\psi(x,y)=\frac{f_{\mathcal{W}}}{d}\leqslant\frac{d-1}{d}.$
(5.8)

###### Sketch of proof.

Recall that the projection of the $\mathbf{v}$-flow in the translation
$3$-manifold $\mathcal{M}$ to the $xy$-plane is a geodesic flow with slope
$\alpha^{-1}$ on the polysquare translation surface $\mathcal{P}$. Meanwhile,
note that $\mathcal{M}=\mathcal{P}\times[0,1)$ is a cartesian product where
the second factor, the unit torus $[0,1)$, is in the integrable $z$-direction.
Since $\mathcal{W}_{3}\subset\mathcal{M}$ is a $\mathbf{v}$-flow-invariant
set, it follows that the function $\psi(x,y)$ in (5.7) on $\mathcal{P}$ is
invariant under the geodesic flow with slope $\alpha^{-1}$ on $\mathcal{P}$.
As the slope $\alpha^{-1}$ is irrational, it follows from the Gutkin–Veech
theorem [3, 10] that this geodesic flow on $\mathcal{P}$ is ergodic. This
implies that the function $\psi(x,y)$ is constant almost everywhere on
$\mathcal{P}$. The result follows. ∎

Lemma 5.2 implies that $f_{\mathcal{S}}/d\geqslant 1/d$ part of $S_{1}$ is in
$\mathcal{S}$, a contradiction. This gives ergodicity in Theorem 2. Unique
ergodicity can be established by a straightforward adaptation of the argument
in Section 4.

We also establish the following simple consequence for use later.

Let $\partial_{X}\mathcal{P}$ denote the union of the $d$ edges of the
polysquare surface $\mathcal{P}$ which are in the $x$-direction.

######  Lemma 5.3.

Suppose that the $\mathbf{v}$-flow in $\mathcal{M}$ is not ergodic. Then for
almost every point $(x,y)\in\partial_{X}\mathcal{P}$, the inequalities (5.8)
hold.

###### Sketch of proof.

Since the $\mathbf{v}$-flow is linear, the integral (5.7) is well defined for
almost every $(x,y)\in\partial_{X}\mathcal{P}$. On the other hand, for every
$x^{*}\in[0,1)$, there are precisely $d$ points
$(x_{\sigma},y_{\sigma})\in\partial_{X}\mathcal{P}$, $\sigma=1,\ldots,d$, such
that $x_{\sigma}\equiv x^{*}\bmod{1}$. ∎

##  6\. A stronger result

The next step in our investigation is to remove the somewhat artificial
hypothesis in Theorem 2 concerning the splitting permutation.

######  Theorem 3.

Suppose that a polycube translation $3$-manifold $\mathcal{M}$ is the
cartesian product of a polysquare translation surface $\mathcal{P}$ with $d$
square faces and the unit torus $[0,1)$. Then any half-infinite geodesic in
$\mathcal{M}$ with a Kronecker direction $\mathbf{v}\in\mathbb{R}^{3}$ is
uniformly distributed unless it hits a singularity.

As before for Theorem 2, we assume that $\mathbf{v}=(\alpha,1,\beta)$, where
$\alpha$, $\beta$ and $\alpha/\beta$ are all irrational, and that $0<\alpha<1$
and $\beta>0$. We also adopt the notation and terminology in Section 5. Again,
we assume that the discrete transformation $T_{\alpha,\beta}$ given by (5.1)
is not ergodic. This leads to a partition
$\mathcal{Y}=\mathcal{W}\cup\mathcal{S}$ into a disjoint union of two
$T_{\alpha,\beta}$-invariant measurable subsets $\mathcal{W}$ and
$\mathcal{S}$ such that (5.2)–(5.4) are valid. Since $\mathcal{W}$ is
measurable, given any arbitrarily small $\varepsilon>0$, to be fixed later,
there exists a set $\mathcal{W}_{1}=\mathcal{W}_{1}(\varepsilon;\mathcal{W})$
which is a finite union of disjoint rectangles aligned with the edges of the
faces $Y_{1},\ldots,Y_{d}$ such that the symmetric difference

$\mathcal{W}\bigtriangleup\mathcal{W}_{1}=(\mathcal{W}\setminus\mathcal{W}_{1})\cup(\mathcal{W}_{1}\setminus\mathcal{W})$

satisfies the inequality xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\lambda_{2}(\mathcal{W}\bigtriangleup\mathcal{W}_{1})<\varepsilon.$ (6.1)

Given any positive integer $H$, we can divide the unit interval $[0,1)$ in the
usual way into $H$ equal subintervals

$I(\tau)=\left[\frac{\tau-1}{H},\frac{\tau}{H}\right),\quad\tau=1,\ldots,H.$

For each $\tau=1,\ldots,H$ and $\sigma=1,\ldots,d$, let $I_{\sigma}(\tau)$ be
the subinterval on the part of $\partial_{X}\mathcal{P}$ corresponding to the
face $Y_{\sigma}$ such that $I_{\sigma}(\tau)\equiv I(\tau)\bmod{1}$, and
consider the strip $I_{\sigma}(\tau)\times\\{y_{\sigma}\\}\times[0,1)$ on the
face $Y_{\sigma}$, where $y_{\sigma}$ is the common value of the
$y$-coordinates of the points on $Y_{\sigma}$.

Since the set $\mathcal{W}_{1}$ is a finite union of disjoint aligned
rectangles, it follows that as long as the threshold $H=H(\mathcal{W}_{1})$ is
chosen sufficiently large, then at least half of the indices $\tau=1,\ldots,H$
are safe in the sense that for every $\sigma=1,\ldots,d$, the strip
$I_{\sigma}(\tau)\times\\{y_{\sigma}\\}\times[0,1)$ on the face $Y_{\sigma}$
does not contain any $z$-parallel edge of any aligned rectangle in the finite
union $\mathcal{W}_{1}$. Let $\mathcal{H}\subset\\{1,\ldots,H\\}$ denote the
collection of safe indices. Then it is clear that $|\mathcal{H}|\geqslant
H/2$.

Let $\tau^{*}\in\mathcal{H}$ satisfy the property

$\displaystyle\lambda_{2}\left((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap\bigcup_{\sigma=1}^{d}(I_{\sigma}(\tau^{*})\times\\{y_{\sigma}\\}\times[0,1))\right)$
$\displaystyle\quad=\min_{\tau\in\mathcal{H}}\lambda_{2}\left((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap\bigcup_{\sigma=1}^{d}(I_{\sigma}(\tau)\times\\{y_{\sigma}\\}\times[0,1))\right).$

For convenience, write

$I(\tau^{*})=\mathcal{I}_{0}=\mathcal{I}_{0}(\mathcal{W}_{1})\subset[0,1),\quad\mbox{so
that}\quad|\mathcal{I}_{0}|=H^{-1},$ (6.2)

and for $\sigma=1,\ldots,d$, write

$\mathcal{I}_{\sigma}=I_{\sigma}(\tau^{*}),\quad\mbox{so
that}\quad\mathcal{I}_{\sigma}\equiv\mathcal{I}_{0}\bmod{1}.$

We have the following intermediate results.

######  Property A.

For every $\sigma=1,\ldots,d$ and every $x\in\mathcal{I}_{\sigma}$, consider
the $z$-parallel cross section of $\mathcal{W}_{1}$ on the face $Y_{\sigma}$
given by

$\operatorname{cs}_{\sigma}(x;\mathcal{W}_{1})=\\{z\in[0,1):(x,y_{\sigma},z)\in\mathcal{W}_{1}\cap
Y_{\sigma}\\}\subset[0,1).$

Then $\operatorname{cs}_{\sigma}(x;\mathcal{W}_{1})$ is independent of the
choice of $x\in\mathcal{I}_{\sigma}$.

###### Sketch of proof.

Consider the set

$\operatorname{cs}_{\sigma}(\mathcal{W}_{1})=\\{(x,z)\in\mathcal{I}_{\sigma}\times[0,1):(x,y_{\sigma},z)\in\mathcal{W}_{1}\cap
Y_{\sigma}\\}.$

Note that $\mathcal{I}_{\sigma}=I_{\sigma}(\tau^{*})$, where $\tau^{*}$ is a
safe index. This means that for any aligned rectangle $\mathfrak{R}$ in the
finite union $\mathcal{W}_{1}$, either

$\operatorname{cs}_{\sigma}(\mathcal{W}_{1})\cap\mathfrak{R}=\emptyset\quad\mbox{or}\quad\operatorname{cs}_{\sigma}(\mathcal{W}_{1})\cap\mathfrak{R}=\mathcal{I}_{\sigma}\times\mathfrak{J}$

for some subinterval $\mathfrak{J}\subset[0,1)$. The result follows almost
immediately. ∎

######  Remark.

For every $\sigma=1,\ldots,d$ and almost every $x\in\mathcal{I}_{\sigma}$, the
$z$-parallel cross section of $\mathcal{W}$ on the face $Y_{\sigma}$ given by

$\operatorname{cs}_{\sigma}(x;\mathcal{W})=\\{z\in[0,1):(x,y_{\sigma},z)\in\mathcal{W}\cap
Y_{\sigma}\\}\subset[0,1)$

is well defined and measurable, but is not independent of the choice of
$x\in\mathcal{I}_{\sigma}$.

######  Property B.

For every $\sigma=1,\ldots,d$, we have

$\lambda_{2}((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap(\mathcal{I}_{\sigma}\times\\{y_{\sigma}\\}\times[0,1)))<\frac{2\varepsilon}{H}.$
(6.3)

###### Proof.

We have

$\displaystyle\lambda_{2}((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap(\mathcal{I}_{\sigma}\times\\{y_{\sigma}\\}\times[0,1)))=\lambda_{2}((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap(I_{\sigma}(\tau^{*})\times\\{y_{\sigma}\\}\times[0,1)))$
$\displaystyle\quad\leqslant\lambda_{2}\left((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap\bigcup_{\sigma=1}^{d}(I_{\sigma}(\tau^{*})\times\\{y_{\sigma}\\}\times[0,1))\right)\leqslant\frac{\lambda_{2}(\mathcal{W}\bigtriangleup\mathcal{W}_{1})}{|\mathcal{H}|}<\frac{2\varepsilon}{H},$

where the second last inequality comes from the fact that the minimum over
$\mathcal{H}$ does not exceed the average and the fact that

$\bigcup_{\tau\in\mathcal{H}}\bigcup_{\sigma=1}^{d}(I_{\sigma}(\tau)\times\\{y_{\sigma}\\}\times[0,1))\subset\mathcal{Y}$

is a disjoint union so that

$\displaystyle\sum_{\tau\in\mathcal{H}}\lambda_{2}\left((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap\bigcup_{\sigma=1}^{d}(I_{\sigma}(\tau)\times\\{y_{\sigma}\\}\times[0,1))\right)$
$\displaystyle\quad\leqslant\lambda_{2}((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap\mathcal{Y})=\lambda_{2}(\mathcal{W}\bigtriangleup\mathcal{W}_{1}).$

This completes the proof. ∎

In view of Property A, for every $\sigma=1,\ldots,d$, we can choose
$x_{\sigma}\in\mathcal{I}_{\sigma}$ arbitrarily, and consider the $z$-parellel
cross section

$U_{\sigma}=\operatorname{cs}_{\sigma}(x_{\sigma};\mathcal{W}_{1}).$

We need some auxiliary results.

We consider Lebesgue measurable subsets $U_{\sigma}\subset[0,1)$,
$\sigma=1,\ldots,d$, of the unit torus $[0,1)$. In particular, we make the
assumption that $0<\lambda_{1}(U_{1})<1$, where $\lambda_{1}$ denotes
$1$-dimensional Lebesgue measure. Furthermore, for any real number
$u\in\mathbb{R}$, we consider the $u$-translated copy of $U_{1}$, given by

$u+U_{1}=\\{\\{u+z\\}:z\in U_{1}\\}.$

######  Lemma 6.1.

The set of values $u_{0}\in[0,1)$ for which the inequalities

$\lambda_{1}(U_{\sigma}\bigtriangleup(u_{0}+U_{1}))\geqslant\frac{1}{32d^{2}}\lambda_{1}(U_{1})(1-\lambda_{1}(U_{1})),\quad\sigma=1,\ldots,d,$

hold simultaneously has Lebesgue measure at least $1/2$.

For every real number $z\in\mathbb{R}$, denote by

$\|z\|=\min_{n\in\mathbb{Z}}|z-n|$

the distance of $z$ from the nearest integer.

######  Lemma 6.2.

Let $\mathbf{v}=(\alpha,1,\beta)\in\mathbb{R}^{3}$ be a Kronecker vector. For
every $\varepsilon_{1}>0$, there exists an infinite sequence

$1\leqslant
m_{1}(\varepsilon_{1})<m_{2}(\varepsilon_{1})<m_{3}(\varepsilon_{1})<\ldots$

of positive integers such that

$\|m_{j}(\varepsilon_{1})\alpha\|<\varepsilon_{1},\quad j=1,2,3,\ldots,$ (6.4)

and the sequence $\\{m_{j}(\varepsilon_{1})\beta\\}$, $j=1,2,3,\ldots,$ is
uniformly distributed in $[0,1)$.

###### Proof.

Since $\alpha$, $\beta$ and $\alpha/\beta$ are all irrational, the
Kronecker–Weyl equidistribution theorem implies that the sequence

$(\\{j\alpha\\},\\{j\beta\\}),\quad j=1,2,3,\ldots,$

is uniformly distributed in the unit square $[0,1)^{2}$. For any
$\varepsilon_{1}>0$, let $m_{1},m_{2},m_{3},\ldots$ be the subsequence of
$1,2,3,\ldots$ such that

$\\{m_{j}\alpha\\}\in[0,\varepsilon_{1})\cup(1-\varepsilon_{1},1),\quad
j=1,2,3,\ldots.$

The conclusion follows easily. ∎

Since the sequence $\\{m_{j}(\varepsilon_{1})\beta\\}$, $j=1,2,3,\ldots,$ is
uniformly distributed in $[0,1)$, it follows that for any $\delta>0$, there
exists a subsequence $m_{j_{1}},m_{j_{2}},m_{j_{3}},\ldots$ such that for each
term $m_{j_{t}}=m_{j_{t}}(\varepsilon_{1};\delta)$, the inequality

$\|m_{j_{t}}(\varepsilon_{1};\delta)\beta-u_{0}\|<\delta$

holds, where $u_{0}\in[0,1)$ is given by Lemma 6.1 and fixed. Clearly there
exist two successive convergents of $\alpha$, with denominators
$q_{k}(\alpha)$ and $q_{k+1}(\alpha)$ such that

$q_{k}(\alpha)\leqslant m_{j_{t}}(\varepsilon_{1};\delta)<q_{k+1}(\alpha).$
(6.5)

We work with the denominator $q_{k}=q_{k}(\alpha)$ of a particular convergent
of $\alpha$ that satisfies the inequalities (6.5), and can choose
$m_{j_{t}}(\varepsilon_{1};\delta)$ large enough so that $q_{k}$ is
substantially larger than $|\mathcal{I}_{0}|^{-1}=H$, where $\mathcal{I}_{0}$
is defined by (6.2).

Write $N=m_{j_{t}}(\varepsilon_{1};\delta)$, so that

$q_{k}(\alpha)\leqslant N<q_{k+1}(\alpha).$

For every $\sigma=1,\ldots,d$ and $i=1,\ldots,q_{k+1}-1$, consider the
$J_{k}$-type $z$-parallel strip

$S_{\sigma,k}(i)=\\{(x,y_{\sigma},z)\in Y_{\sigma}:x\in J_{k}(i)\bmod{1}\\},$

where the short special interval $J_{k}(i)$ is given by (3.4). Furthermore,
consider the half $J_{k}$-type $z$-parallel strips

$\displaystyle S^{-}_{\sigma,k}(i)=\\{(x,y_{\sigma},z)\in Y_{\sigma}:x\in
J^{-}_{k}(i)\bmod{1}\\},$ $\displaystyle
S^{+}_{\sigma,k}(i)=\\{(x,y_{\sigma},z)\in Y_{\sigma}:x\in
J^{+}_{k}(i)\bmod{1}\\},$

where

$J^{-}_{k}(i)=\left(\\{i\alpha\\}-\frac{\|q_{k}\alpha\|}{2},\\{i\alpha\\}\right)\quad\mbox{and}\quad
J^{+}_{k}(i)=\left(\\{i\alpha\\},\\{i\alpha\\}+\frac{\|q_{k}\alpha\|}{2}\right).$

Since $q_{k}$ is large compared to $|\mathcal{I}_{0}|^{-1}$, the $z$-parallel
strip $\mathcal{I}_{1}\times\\{y_{1}\\}\times[0,1)\subset Y_{1}$ contains a
significant number of strips of the form $S_{1,k}(i)$, where
$i=1,\ldots,q_{k+1}-1$.

######  Lemma 6.3.

For every $i=1,\ldots,q_{k+1}-1$, at least one of the two
$T_{\alpha,\beta}$-power chains

$\displaystyle T_{\alpha,\beta}^{n}S^{-}_{1,k}(i),\quad n=1,\ldots,N,$ (6.6)
$\displaystyle T_{\alpha,\beta}^{n}S^{+}_{1,k}(i),\quad n=1,\ldots,N,$ (6.7)

is splitting free.

###### Proof.

For any fixed $i=1,\ldots,q_{k+1}-1$, consider the $T_{\alpha,\beta}$-power
chain

$T_{\alpha,\beta}^{n}S_{1,k}(i),\quad n=1,\ldots,N.$

Either no member of this chain splits, or there is a smallest
$n_{0}=1,\ldots,N$ satisfying $i+n_{0}\geqslant q_{k+1}$, in view of the
$2$-distance theorem, such that $T_{\alpha,\beta}^{n_{0}}S_{1,k}(i)$ splits.
In the latter case, we have

$\left(\\{(i+n_{0})\alpha\\}-\frac{\|q_{k}\alpha\|}{2},\\{(i+n_{0})\alpha\\}+\frac{\|q_{k}\alpha\|}{2}\right)\cap\\{0,1\\}\neq\emptyset.$
(6.8)

Suppose first that the intersection (6.8) is $\\{0\\}$. Then

$0<\\{(i+n_{0})\alpha\\}<\frac{\|q_{k}\alpha\|}{2}<\\{(i+n_{0})\alpha\\}+\frac{\|q_{k}\alpha\|}{2}<1.$

Any point $(x,y,z)\in T_{\alpha,\beta}^{n_{0}}S^{+}_{1,k}(i)$ must satisfy

$x\in\left(\\{(i+n_{0})\alpha\\},\\{(i+n_{0})\alpha\\}+\frac{\|q_{k}\alpha\|}{2}\right)\bmod{1}.$
(6.9)

Since the interval in (6.9) is contained in $(0,1)$, it follows that
$T_{\alpha,\beta}^{n_{0}}S^{+}_{1,k}(i)$ does not split. We can now prove by
induction that for every $n=n_{0}+1,\ldots,N$, any point $(x,y,z)\in
T_{\alpha,\beta}^{n}S^{+}_{1,k}(i)$ must satisfy

$x\in\left(\\{(i+n)\alpha\\},\\{(i+n)\alpha\\}+\frac{\|q_{k}\alpha\|}{2}\right)\bmod{1}.$
(6.10)

The inequality $0<n-n_{0}<q_{k+1}-1$ and the $2$-distance theorem now ensure
that the interval in (6.10) is contained in $(0,1)$. It follows that
$T_{\alpha,\beta}^{n}S^{+}_{1,k}(i)$ does not split. We thus conclude that the
$T_{\alpha,\beta}$-power chain (6.7) does not split.

Suppose next that the intersection (6.8) is $\\{1\\}$. Then a similar argument
shows that the $T_{\alpha,\beta}$-power chain (6.6) does not split. ∎

We denote by xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$T_{\alpha,\beta}^{n}S^{\dagger}_{1,k}(i),\quad n=1,\ldots,N,$ (6.11)

one of the $T_{\alpha,\beta}$-power chain (6.6) and (6.7) that is splitting
free.

It is clear from the properties of the short special intervals $J_{k}(i)$,
$i=1,\ldots,q_{k+1}-1$, that the sets $S^{\dagger}_{1,k}(i)$,
$i=1,\ldots,q_{k+1}-1$, that are contained in
$\mathcal{I}_{1}\times\\{y_{1}\\}\times[0,1)\subset Y_{1}$ together occupy
more than $1/10$ of $\mathcal{I}_{1}\times\\{y_{1}\\}\times[0,1)$, and so have
total area exceeding $1/10H$. For each of these sets, consider the last
element of the chain (6.11), of the form xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$T_{\alpha,\beta}^{N}S^{\dagger}_{1,k}(i)\subset Y_{\sigma}\quad\mbox{for some
$\sigma=1,\ldots,d$}.$

Suppose that $Y_{\sigma_{0}}$ contains the maximum number of such images, and
let $\mathfrak{I}(\sigma_{0})$ denote the collection of indices
$i=1,\ldots,q_{k+1}-1$ such that
$T_{\alpha,\beta}^{N}S^{\dagger}_{1,k}(i)\subset Y_{\sigma_{0}}$ and
$S^{\dagger}_{1,k}(i)\subset\mathcal{I}_{1}\times\\{y_{1}\\}\times[0,1)$. Then

$\bigcup_{i\in\mathfrak{I}(\sigma_{0})}T_{\alpha,\beta}^{N}S^{\dagger}_{1,k}(i)\subset
Y_{\sigma_{0}}\quad\mbox{and}\quad\lambda_{2}\left(\bigcup_{i\in\mathfrak{I}(\sigma_{0})}T_{\alpha,\beta}^{N}S^{\dagger}_{1,k}(i)\right)\geqslant\frac{1}{10dH}.$
(6.12)

######  Lemma 6.4.

We have

$\lambda_{2}\left((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap\bigcup_{i\in\mathfrak{I}(\sigma_{0})}T_{\alpha,\beta}^{N}S^{\dagger}_{1,k}(i)\right)<2\varepsilon_{1}+20\varepsilon
d\lambda_{2}\left(\bigcup_{i\in\mathfrak{I}(\sigma_{0})}T_{\alpha,\beta}^{N}S^{\dagger}_{1,k}(i)\right),$
(6.13)

and xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\lambda_{2}\left((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap\bigcup_{i\in\mathfrak{I}(\sigma_{0})}S^{\dagger}_{1,k}(i)\right)<20\varepsilon
d\lambda_{2}\left(\bigcup_{i\in\mathfrak{I}(\sigma_{0})}S^{\dagger}_{1,k}(i)\right).$
(6.14)

###### Proof.

For every $i\in\mathfrak{I}(\sigma_{0})$, the condition
$S^{\dagger}_{1,k}(i)\subset\mathcal{I}_{1}\times\\{y_{1}\\}\times[0,1)$,
together with (6.4) and $N=m_{j_{t}}(\varepsilon_{1};\delta)$, ensure that the
total area of the images in (6.12) that fall outside the subset
$\mathcal{I}_{\sigma_{0}}\times\\{y_{\sigma_{0}}\\}\times[0,1)\subset
Y_{\sigma_{0}}$ does not exceed $2\varepsilon_{1}$. It then follows from (6.3)
with $\sigma=\sigma_{0}$ and the second inequality in (6.12) that

$\displaystyle\lambda_{2}\left((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap\bigcup_{i\in\mathfrak{I}(\sigma_{0})}T_{\alpha,\beta}^{N}S^{\dagger}_{1,k}(i)\right)$
$\displaystyle\quad\leqslant\lambda_{2}\left(\left(\bigcup_{i\in\mathfrak{I}(\sigma_{0})}T_{\alpha,\beta}^{N}S^{\dagger}_{1,k}(i)\right)\setminus(\mathcal{I}_{\sigma_{0}}\times\\{y_{\sigma_{0}}\\}\times[0,1))\right)$
$\displaystyle\quad\qquad+\lambda_{2}((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap(\mathcal{I}_{\sigma_{0}}\times\\{y_{\sigma_{0}}\\}\times[0,1)))$
$\displaystyle\quad<2\varepsilon_{1}+\frac{2\varepsilon}{H}\leqslant
2\varepsilon_{1}+20\varepsilon
d\lambda_{2}\left(\bigcup_{i\in\mathfrak{I}(\sigma_{0})}T_{\alpha,\beta}^{N}S^{\dagger}_{1,k}(i)\right),$

establishing (6.13).

To establish (6.14), note from the second inequality in (6.12) that

$\lambda_{2}\left(\bigcup_{i\in\mathfrak{I}(\sigma_{0})}S^{\dagger}_{1,k}(i)\right)\geqslant\frac{1}{10dH}.$
(6.15)

It then follows from (6.3) with $\sigma=1$ and (6.15) that

$\displaystyle\lambda_{2}\left((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap\bigcup_{i\in\mathfrak{I}(\sigma_{0})}S^{\dagger}_{1,k}(i)\right)\leqslant\lambda_{2}((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap(\mathcal{I}_{1}\times\\{y_{1}\\}\times[0,1)))$
$\displaystyle\quad<\frac{2\varepsilon}{H}\leqslant 20\varepsilon
d\lambda_{2}\left(\bigcup_{i\in\mathfrak{I}(\sigma_{0})}S^{\dagger}_{1,k}(i)\right).$

This completes the proof. ∎

######  Lemma 6.5.

Let $x_{1}$ be the $x$-coordinate of a point of $S^{\dagger}_{1,k}(i)$ for
some $i\in\mathfrak{I}(\sigma_{0})$. Then xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$N\beta+\operatorname{cs}_{1}(x_{1};\mathcal{W})=\operatorname{cs}_{\sigma_{0}}(x_{\sigma_{0}};\mathcal{W}),$

where xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\|N\beta-u_{0}\|<\delta$ (6.16)

and $x_{\sigma_{0}}$ is the $x$-coordinate of a point of
$T_{\alpha,\beta}^{N}S^{\dagger}_{1,k}(i)$ satisfying
$\\{x_{\sigma_{0}}\\}=\\{x_{1}+N\alpha\\}$.

###### Sketch of proof.

Note that $T_{\alpha,\beta}^{N}$ gives rise to a shift in the $z$-direction by
$N\beta\bmod{1}$ and a shift in the $x$-direction by $N\alpha\bmod{1}$. Note
also that $\mathcal{W}$ is invariant under $T_{\alpha,\beta}$. We have earlier
chosen $N$ to satisfy (6.16) and $\|N\alpha\|<\varepsilon_{1}$. ∎

We are interested in the set

$\mathcal{X}_{0}=\left\\{x:(x,y_{1},z)\in\bigcup_{i\in\mathfrak{I}(\sigma_{0})}S^{\dagger}_{1,k}(i)\mbox{
for some $z\in[0,1)$}\right\\}$ (6.17)

of the $x$-coordinates of points on $Y_{1}$ and the set

$\mathcal{X}_{N}=\left\\{x:(x,y_{\sigma_{0}},z)\in\bigcup_{i\in\mathfrak{I}(\sigma_{0})}T_{\alpha,\beta}^{N}S^{\dagger}_{1,k}(i)\mbox{
for some $z\in[0,1)$}\right\\}$ (6.18)

of the $x$-coordinates of points in $Y_{\sigma_{0}}$.

Let $x_{1}\in\mathcal{X}_{0}$ and $x_{\sigma_{0}}\in\mathcal{X}_{N}$ be chosen
arbitrarily. Applying Lemma 6.1 with $\sigma=\sigma_{0}$ to the sets
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$U_{1}=\operatorname{cs}_{1}(x_{1};\mathcal{W}_{1})\quad\mbox{and}\quad
U_{\sigma_{0}}=\operatorname{cs}_{\sigma_{0}}(x_{\sigma_{0}};\mathcal{W}_{1}),$
(6.19)

we deduce that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\lambda_{1}(U_{\sigma_{0}}\bigtriangleup(u_{0}+U_{1}))\geqslant\frac{1}{32d^{2}}\lambda_{1}(U_{1})(1-\lambda_{1}(U_{1})).$

By the continuity of the Lebesgue measure, for any fixed $\eta>0$, if
$\delta>0$ in (6.16) is sufficiently small, then

$\lambda_{1}(U_{\sigma_{0}}\bigtriangleup(N\beta+U_{1}))\geqslant\frac{1}{32d^{2}}\lambda_{1}(U_{1})(1-\lambda_{1}(U_{1}))-\eta.$

Choosing xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\eta=\frac{1}{64d^{2}}\lambda_{1}(U_{1})(1-\lambda_{1}(U_{1})),$

we deduce that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\lambda_{1}(U_{\sigma_{0}}\bigtriangleup(N\beta+U_{1}))\geqslant\frac{1}{64d^{2}}\lambda_{1}(U_{1})(1-\lambda_{1}(U_{1})),$
(6.20)

provided that $\delta>0$ is sufficiently small.

######  Lemma 6.6.

We have

$\lambda_{1}(\\{x\in\mathcal{X}_{N}:\lambda_{1}(\operatorname{cs}_{\sigma_{0}}(x;\mathcal{W})\bigtriangleup\operatorname{cs}_{\sigma_{0}}(x;\mathcal{W}_{1}))>300\varepsilon
d\\})<\frac{1}{10}\lambda_{1}(\mathcal{X}_{N}),$ (6.21)

provided that $\varepsilon_{1}$ is sufficiently small. We also have

$\lambda_{1}(\\{x\in\mathcal{X}_{0}:\lambda_{1}(\operatorname{cs}_{1}(x;\mathcal{W})\bigtriangleup\operatorname{cs}_{1}(x;\mathcal{W}_{1}))>200\varepsilon
d\\})<\frac{1}{10}\lambda_{1}(\mathcal{X}_{0}).$ (6.22)

###### Proof.

First, note that

$\int_{\mathcal{X}_{N}}\lambda_{1}(\operatorname{cs}_{\sigma_{0}}(x;\mathcal{W})\bigtriangleup\operatorname{cs}_{\sigma_{0}}(x;\mathcal{W}_{1}))\,\mathrm{d}x=\lambda_{2}\left((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap\bigcup_{i\in\mathfrak{I}(\sigma_{0})}T_{\alpha,\beta}^{N}S^{\dagger}_{1,k}(i)\right).$

Combining this with (6.13) and noting

$\lambda_{2}\left(\bigcup_{i\in\mathfrak{I}(\sigma_{0})}T_{\alpha,\beta}^{N}S^{\dagger}_{1,k}(i)\right)=\lambda_{1}(\mathcal{X}_{N}),$

we conclude that

$\int_{\mathcal{X}_{N}}\lambda_{1}(\operatorname{cs}_{\sigma_{0}}(x;\mathcal{W})\bigtriangleup\operatorname{cs}_{\sigma_{0}}(x;\mathcal{W}_{1}))\,\mathrm{d}x<2\varepsilon_{1}+20\varepsilon
d\lambda_{1}(\mathcal{X}_{N})<30\varepsilon d\lambda_{1}(\mathcal{X}_{N})$

provided that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\varepsilon_{1}<5\varepsilon d\lambda_{1}(\mathcal{X}_{N}).$ (6.23)

The inequality (6.21) follows immediately from this.

Next, note that

$\int_{\mathcal{X}_{0}}\lambda_{1}(\operatorname{cs}_{1}(x;\mathcal{W})\bigtriangleup\operatorname{cs}_{1}(x;\mathcal{W}_{1}))\,\mathrm{d}x=\lambda_{2}\left((\mathcal{W}\bigtriangleup\mathcal{W}_{1})\cap\bigcup_{i\in\mathfrak{I}(\sigma_{0})}S^{\dagger}_{1,k}(i)\right).$

Combining this with (6.14) and noting

$\lambda_{2}\left(\bigcup_{i\in\mathfrak{I}(\sigma_{0})}S^{\dagger}_{1,k}(i)\right)=\lambda_{1}(\mathcal{X}_{0}),$

we conclude that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\int_{\mathcal{X}_{0}}\lambda_{1}(\operatorname{cs}_{1}(x;\mathcal{W})\bigtriangleup\operatorname{cs}_{1}(x;\mathcal{W}_{1}))\,\mathrm{d}x<20\varepsilon
d\lambda_{1}(\mathcal{X}_{0}).$

The inequality (6.22) follows immediately from this. ∎

It now follows from Lemma 6.6 that there exist
$x_{1}^{\star}\in\mathcal{X}_{0}$ and
$x_{\sigma_{0}}^{\star}\in\mathcal{X}_{N}$ satisfying
$\\{x_{\sigma_{0}}^{\star}\\}=\\{x_{1}^{\star}+N\alpha\\}$ such that

$\lambda_{1}(\operatorname{cs}_{\sigma_{0}}(x_{\sigma_{0}}^{\star};\mathcal{W})\bigtriangleup\operatorname{cs}_{\sigma_{0}}(x_{\sigma_{0}}^{\star};\mathcal{W}_{1}))\leqslant
300\varepsilon d,$ (6.24)

provided that (6.23) holds, and

$\lambda_{1}(\operatorname{cs}_{1}(x_{1}^{\star};\mathcal{W})\bigtriangleup\operatorname{cs}_{1}(x_{1}^{\star};\mathcal{W}_{1}))\leqslant
200\varepsilon d.$ (6.25)

Combining (6.19) and (6.20), we have

$\displaystyle\frac{1}{64d^{2}}\lambda_{1}(U_{1})(1-\lambda_{1}(U_{1}))$
$\displaystyle\leqslant\lambda_{1}(\operatorname{cs}_{\sigma_{0}}(x_{\sigma_{0}};\mathcal{W}_{1})\bigtriangleup(N\beta+\operatorname{cs}_{1}(x_{1};\mathcal{W}_{1})))$
$\displaystyle=\lambda_{1}(\operatorname{cs}_{\sigma_{0}}(x_{\sigma_{0}}^{\star};\mathcal{W}_{1})\bigtriangleup(N\beta+\operatorname{cs}_{1}(x_{1}^{\star};\mathcal{W}_{1}))),$
(6.26)

where the last step is justified by Property A. Next, by the triangle
inequality

$\lambda_{1}(A\bigtriangleup C)\leqslant\lambda_{1}(A\bigtriangleup
B)+\lambda_{1}(B\bigtriangleup C),$

valid for measurable sets $A,B,C\subset[0,1)$, we have

$\displaystyle\lambda_{1}(\operatorname{cs}_{\sigma_{0}}(x_{\sigma_{0}}^{\star};\mathcal{W}_{1})\bigtriangleup(N\beta+\operatorname{cs}_{1}(x_{1}^{\star};\mathcal{W}_{1})))$
$\displaystyle\quad\leqslant\lambda_{1}(\operatorname{cs}_{\sigma_{0}}(x_{\sigma_{0}}^{\star};\mathcal{W}_{1})\bigtriangleup\operatorname{cs}_{\sigma_{0}}(x_{\sigma_{0}}^{\star};\mathcal{W}))$
$\displaystyle\quad\qquad+\lambda_{1}(\operatorname{cs}_{\sigma_{0}}(x_{\sigma_{0}}^{\star};\mathcal{W})\bigtriangleup(N\beta+\operatorname{cs}_{1}(x_{1}^{\star};\mathcal{W})))$
$\displaystyle\quad\qquad+\lambda_{1}((N\beta+\operatorname{cs}_{1}(x_{1}^{\star};\mathcal{W}))\bigtriangleup(N\beta+\operatorname{cs}_{1}(x_{1}^{\star};\mathcal{W}_{1})))$
$\displaystyle\quad\leqslant\lambda_{1}(\operatorname{cs}_{\sigma_{0}}(x_{\sigma_{0}}^{\star};\mathcal{W}_{1})\bigtriangleup\operatorname{cs}_{\sigma_{0}}(x_{\sigma_{0}}^{\star};\mathcal{W}))$
$\displaystyle\quad\qquad+\lambda_{1}((N\beta+\operatorname{cs}_{1}(x_{1}^{\star};\mathcal{W}))\bigtriangleup(N\beta+\operatorname{cs}_{1}(x_{1}^{\star};\mathcal{W}_{1}))),$
(6.27)

where the last step is justified by Lemma 6.5.

Combining (6.24)–(6), we now conclude that if (6.23) holds, then

$\frac{1}{64d^{2}}\lambda_{1}(U_{1})(1-\lambda_{1}(U_{1}))\leqslant
500\varepsilon d.$ (6.28)

It remains to specify the parameters $\varepsilon$ and $\varepsilon_{1}$.

The assumption that the discrete transformation $T_{\alpha,\beta}$ is not
ergodic also implies that the $\mathbf{v}$-flow in $\mathcal{M}$ is not
ergodic. Thus Lemma 5.3 implies that

$\frac{1}{d}\leqslant\lambda_{1}(\operatorname{cs}_{1}(x_{1}^{\star};\mathcal{W}))\leqslant\frac{d-1}{d}.$

Combining this with (6.25), we deduce that

$\frac{1}{d}-200\varepsilon
d\leqslant\lambda_{1}(\operatorname{cs}_{1}(x_{1}^{\star};\mathcal{W}_{1}))\leqslant\frac{d-1}{d}+200\varepsilon
d.$

Choosing $\varepsilon$ to satisfy xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$200\varepsilon d\leqslant\frac{1}{2d},$ (6.29)

and noting that $U_{1}=\operatorname{cs}_{1}(x_{1}^{\star};\mathcal{W}_{1})$,
we deduce that

$\frac{1}{2d}\leqslant\lambda_{1}(U_{1})\leqslant 1-\frac{1}{2d}.$

Hence xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\frac{1}{64d^{2}}\lambda_{1}(U_{1})(1-\lambda_{1}(U_{1}))\geqslant\frac{1}{64d^{2}}\left(\frac{1}{2d}\right)^{2}.$
(6.30)

Clearly (6.28) and (6.30) contradict each other if

$\varepsilon<\frac{1}{128000d^{5}}.$

This is stronger than the restriction (6.29).

Finally, note from (6.17) and (6.18) that

$\lambda_{1}(\mathcal{X}_{N})=\lambda_{1}(\mathcal{X}_{0})\leqslant|\mathcal{I}_{1}|=|\mathcal{I}_{0}|=\frac{1}{H},$

where $H=H(\mathcal{W}_{1})$ is chosen sufficiently large. Here
$\mathcal{W}_{1}$ depends on $\varepsilon$, in view of (6.1). Clearly
$\varepsilon_{1}$ can be chosen sufficiently small in terms of $\varepsilon$
so that the condition (6.23) is satisfied.

This gives ergodicity in Theorem 3. Unique ergodicity can then be established
by a straightforward adaptation of the argument in Section 4.

##  7\. Proof of Lemma 6.1

For every $\sigma=1,\ldots,d$, let

$\chi_{U_{\sigma}}(x)=\left\\{\begin{array}[]{ll}1,&\mbox{if $x\in
U_{\sigma}$},\\\ 0,&\mbox{if $x\not\in U_{\sigma}$},\end{array}\right.$

and xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\chi_{u+U_{1}}(x)=\left\\{\begin{array}[]{ll}1,&\mbox{if $x\in u+U_{1}$},\\\
0,&\mbox{if $x\not\in u+U_{1}$},\end{array}\right.$

denote respectively the characteristic functions of $U_{\sigma}$ and
$u+U_{1}$. Then it is not difficult to see that

$\lambda_{1}(U_{\sigma}\bigtriangleup(u_{0}+U_{1}))=\int_{0}^{1}(\chi_{U_{\sigma}}(x)-\chi_{u+U_{1}}(x))^{2}\,\mathrm{d}x.$
(7.1)

On the other hand, the $n$-th Fourier coefficient of $\chi_{U_{\sigma}}$ and
of $\chi_{u+U_{1}}$ are respectively

$a_{n;\sigma}=\int_{0}^{1}\chi_{U_{\sigma}}(x)\mathrm{e}^{-2\pi\mathrm{i}nx}\,\mathrm{d}x$

and xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\int_{0}^{1}\chi_{u+U_{1}}(x)\mathrm{e}^{-2\pi\mathrm{i}nx}\,\mathrm{d}x=\mathrm{e}^{-2\pi\mathrm{i}nu}\int_{0}^{1}\chi_{U_{1}}(x-u)\mathrm{e}^{-2\pi\mathrm{i}n(x-u)}\,\mathrm{d}x=\mathrm{e}^{-2\pi\mathrm{i}nu}a_{n;1}.$

Applying the Parseval formula to the right hand side of (7.1) then gives

$\lambda_{1}(U_{\sigma}\bigtriangleup(u_{0}+U_{1}))=\sum_{n\in\mathbb{Z}}|a_{n;\sigma}-\mathrm{e}^{-2\pi\mathrm{i}nu}a_{n;1}|^{2}.$
(7.2)

Suppose that $n\in\mathbb{Z}$ is non-zero. The complex valued function
$\mathrm{e}^{-2\pi\mathrm{i}nu}a_{n;1}$ is periodic in $u\in\mathbb{R}$ with
period $|n|^{-1}$. Indeed, its values over a period forms a circle

$\operatorname{circle}(n;1)=\\{\mathrm{e}^{-2\pi\mathrm{i}nu}a_{n;1}:0\leqslant
u<|n|^{-1}\\}$

on the complex plane, as shown on the left in Figure 7.1.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-7-1.pdf}\vspace{3pt}\\\
\mbox{Figure 7.1: $\operatorname{circle}(n;1)$ and
$\operatorname{disk}(n;\sigma;r)$}\end{array}$

Let $\operatorname{disk}(n;\sigma;r)$ denote the circular disk on the complex
plane centred at $a_{n;\sigma}$ and with radius $r>0$, as shown on the right
in Figure 7.1. The intersection of $\operatorname{circle}(n;1)$ and
$\operatorname{disk}(n;\sigma;r)$, if not empty, is a circular arc within
$\operatorname{disk}(n;\sigma;r)$ with length clearly less than the
circumference $2\pi r$ of the disk.

We let $r=\eta|a_{n;1}|$, where the parameter $\eta\in(0,1)$ will be
determined later. Now partition $\operatorname{circle}(n;1)$ into two parts,
one disjoint from $\operatorname{disk}(n;\sigma;\eta|a_{n;1}|)$ and the other
contained in $\operatorname{disk}(n;\sigma;\eta|a_{n;1}|)$. It is then clear
that the inequality

$|a_{n;\sigma}-\mathrm{e}^{-2\pi\mathrm{i}nu}a_{n;1}|^{2}\geqslant\eta^{2}|a_{n;1}|^{2}$
(7.3)

holds for $u\in[0,1)$, with the possible exception of a subset

$\operatorname{bad}(n;\sigma;\eta)\subset[0,1)\quad\mbox{with}\quad\lambda_{1}(\operatorname{bad}(n;\sigma;\eta))\leqslant\eta.$
(7.4)

For every $n\in\mathbb{Z}\setminus\\{0\\}$, every $\sigma=1,\ldots,d$ and
every $u\in[0,1)$, let

$F_{n;\sigma;\eta}(u)=\left\\{\begin{array}[]{ll}|a_{n;1}|^{2},&\mbox{if
$u\in\operatorname{bad}(n;\sigma;\eta)$},\\\ 0,&\mbox{if
$u\not\in\operatorname{bad}(n;\sigma;\eta)$}.\end{array}\right.$ (7.5)

Then for every $\sigma=1,\ldots,d$, it follows from (7.4) and (7.5) that

$\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\int_{0}^{1}F_{n;\sigma;\eta}(u)\,\mathrm{d}u\leqslant\eta\sum_{n\in\mathbb{Z}\setminus\\{0\\}}|a_{n;1}|^{2}.$
(7.6)

If, further, we let

$\operatorname{violator}(\sigma;\eta)=\left\\{u\in[0,1):\sum_{n\in\mathbb{Z}\setminus\\{0\\}}F_{n;\sigma;\eta}(u)\geqslant\frac{1}{2}\sum_{n\in\mathbb{Z}\setminus\\{0\\}}|a_{n;1}|^{2}\right\\},$
(7.7)

then it follows from (7.5) and (7.6) that

$\lambda_{1}(\operatorname{violator}(\sigma;\eta))\leqslant 2\eta.$ (7.8)

If $u\in[0,1)\setminus\operatorname{violator}(\sigma;\eta)$, then using (7.5)
and (7.7), we have

$\displaystyle\sum_{\begin{subarray}{c}{n\in\mathbb{Z}\setminus\\{0\\}}\\\
{F_{n;\sigma;\eta}(u)=0}\end{subarray}}|a_{n;1}|^{2}$
$\displaystyle=\sum_{n\in\mathbb{Z}\setminus\\{0\\}}|a_{n;1}|^{2}-\sum_{\begin{subarray}{c}{n\in\mathbb{Z}\setminus\\{0\\}}\\\
{F_{n;\sigma;\eta}(u)\neq 0}\end{subarray}}|a_{n;1}|^{2}$
$\displaystyle=\sum_{n\in\mathbb{Z}\setminus\\{0\\}}|a_{n;1}|^{2}-\sum_{n\in\mathbb{Z}\setminus\\{0\\}}F_{n;\sigma;\eta}(u)>\frac{1}{2}\sum_{n\in\mathbb{Z}\setminus\\{0\\}}|a_{n;1}|^{2}.$
(7.9)

Recalling that the condition $F_{n;\sigma;\eta}(u)=0$ implies the inequality
(7.3), it then follows from (7) that

$\sum_{\begin{subarray}{c}{n\in\mathbb{Z}\setminus\\{0\\}}\\\
{F_{n;\sigma;\eta}(u)=0}\end{subarray}}|a_{n;\sigma}-\mathrm{e}^{-2\pi\mathrm{i}nu}a_{n;1}|^{2}\geqslant\eta^{2}\sum_{\begin{subarray}{c}{n\in\mathbb{Z}\setminus\\{0\\}}\\\
{F_{n;\sigma;\eta}(u)=0}\end{subarray}}|a_{n;1}|^{2}>\frac{\eta^{2}}{2}\sum_{n\in\mathbb{Z}\setminus\\{0\\}}|a_{n;1}|^{2}.$

We therefore deduce that for every $\sigma=1,\ldots,d$, the inequality

$\sum_{n\in\mathbb{Z}\setminus\\{0\\}}|a_{n;\sigma}-\mathrm{e}^{-2\pi\mathrm{i}nu}a_{n;1}|^{2}>\frac{\eta^{2}}{2}\sum_{n\in\mathbb{Z}\setminus\\{0\\}}|a_{n;1}|^{2}$
(7.10)

holds for every $u\in[0,1)\setminus\operatorname{violator}(\sigma;\eta)$,
where the bound (7.8) holds.

To evaluate the sum xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\sum_{n\in\mathbb{Z}\setminus\\{0\\}}|a_{n;1}|^{2}=\sum_{n\in\mathbb{Z}}|a_{n;1}|^{2}-|a_{0;1}|^{2},$
(7.11)

we note that the Parseval formula gives

$\sum_{n\in\mathbb{Z}}|a_{n;1}|^{2}=\int_{0}^{1}\chi_{U_{1}}^{2}(x)\,\mathrm{d}x=\int_{0}^{1}\chi_{U_{1}}(x)\,\mathrm{d}x=\lambda_{1}(U_{1}),$
(7.12)

while by definition we have

$|a_{0;1}|^{2}=\left(\int_{0}^{1}\chi_{U_{1}}(x)\,\mathrm{d}x\right)^{2}=(\lambda_{1}(U_{1}))^{2}.$
(7.13)

From (7.11)–(7.13), we deduce that

$\sum_{n\in\mathbb{Z}\setminus\\{0\\}}|a_{n;1}|^{2}=\lambda_{1}(U_{1})(1-\lambda_{1}(U_{1})).$
(7.14)

Combining (7.2), (7.10) and (7.14), we now deduce that for every
$\sigma=1,\ldots,d$, the inequality xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\lambda_{1}(U_{\sigma}\bigtriangleup(u_{0}+U_{1}))>\frac{\eta^{2}}{2}\lambda_{1}(U_{1})(1-\lambda_{1}(U_{1}))$

holds for every $u\in[0,1)\setminus\operatorname{violator}(\sigma;\eta)$,
where the bound (7.8) holds. Finally, let $\eta=(4d)^{-1}$. Then it follows
from (7.8) that

$\lambda_{1}\left(\bigcup_{\sigma=1}^{d}\operatorname{violator}(\sigma;\eta)\right)\leqslant
2\eta d=\frac{1}{2}.$

This completes the proof of Lemma 6.1.

##  8\. Kronecker–Weyl polycube translation $3$-manifolds

Both Theorems 2 and 3 concern finite polycube translation $3$-manifolds that
are the cartesian products of finite polysquare translation surfaces with the
unit torus $[0,1)^{2}$. These special polycube translation $3$-manifolds all
have one direction which is integrable, and can therefore be considered as
degenerate cases in the class of all finite polycube translation
$3$-manifolds.

We are currently unable to answer in full generality the question raised in
the Open Problem in Section 1. However, we are nevertheless able to answer
this question in the affirmative for infinitely many finite polycube
translation $3$-manifolds that do not have any direction which can be
considered integrable. The construction of such examples of finite polycube
translation $3$-manifolds comes from the notion of split-covering.

Before we describe this construction, we consider a simple example.

######  Example 8.1.

Consider the unit torus $[0,1)^{3}$ as a basic building block. Then any half-
infinite geodesic in $[0,1)^{3}$ with a Kronecker direction is uniformly
distributed. The L-solid translation $3$-manifold $\mathcal{M}_{0}$ can be
viewed as a $3$-fold split-covering of $[0,1)^{3}$, and Theorem 1 states that
any half-infinite geodesic in $\mathcal{M}_{0}$ with a Kronecker direction is
uniformly distributed unless it hits a singularity.

Let us now take $\mathcal{M}_{0}$ as a basic building block, and consider next
a rather special $4$-fold split-covering of $\mathcal{M}_{0}$ where the $4$
copies of $\mathcal{M}_{0}$ are placed side-by-side in the $x$-direction, as
shown in Figure 8.1.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-8-1.pdf}\vspace{3pt}\\\
\mbox{Figure 8.1: $4$ L-solid translation $3$-manifolds}\end{array}$

Clearly if we simply glue the $4$ copies of $\mathcal{M}_{0}$ together, then
the resulting polycube translation $3$-manifold $\mathcal{M}$ is integrable in
the $x$-direction, and can be considered degenerate. To ensure that the
resulting polycube translation $3$-manifold $\mathcal{M}$ is not degenerate,
we convert some of the $X$-faces into walls, as shown in the examples in
Figure 8.2.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-8-2.pdf}\vspace{3pt}\\\
\mbox{Figure 8.2: non-degenerate $4$-fold split-coverings}\\\ \mbox{of the
L-solid translation $3$-manifolds}\end{array}$

We claim that any half-infinite geodesic in either example of $\mathcal{M}$ in
Figure 8.2 with a Kronecker direction is uniformly distributed unless it hits
a singularity.

To give a brief glimpse of our method and help formulate our result, let us
consider the example of $\mathcal{M}$ on the right hand side, and ignore the
atomic cubes at the front, as indicated in Figure 8.3.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-8-3.pdf}\vspace{3pt}\\\
\mbox{Figure 8.3: ignoring parts of a non-degenerate $4$-fold split-
covering}\\\ \mbox{of the L-solid translation $3$-manifolds}\end{array}$

Removing the front atomic cubes altogether, we obtain the collection of atomic
cubes as shown in Figure 8.4. Here we see a $4$-fold split-covering in the
$x$-direction with an $X$-street of length $4$, while immediately below are
$4$ $X$-streets of length $1$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-8-4.pdf}\vspace{3pt}\\\
\mbox{Figure 8.4: a very special $4$-fold split-covering}\end{array}$

We shall show that this is the crucial part of $\mathcal{M}$, and that the
remaining atomic cubes of $\mathcal{M}$ are totally irrelevant.

######  Theorem 4.

Let $\mathcal{P}$ be a finite polysquare translation surface, and let
$\mathcal{M}_{0}=[0,1)\times\mathcal{P}$ denote the cartesian product of the
unit torus $[0,1)$ with $\mathcal{P}$, where the $x$-direction is the
direction of the unit torus. For any fixed positive integer $s$, let
$\mathcal{M}$ denote an $s$-fold split-covering of $\mathcal{M}_{0}$, where
the $s$ copies of $\mathcal{M}_{0}$ are placed side-by-side in the
$x$-direction and glued together, and where some $X$-faces are replaced by
walls.

Suppose that $\mathcal{M}$ contains an $s\times 2\times 1$ array of atomic
cubes, where the top row of $s$ atomic cubes gives rise to an $X$-street of
length $s$, and where any $X$-face of any atomic cube in the bottom row is a
wall. Then any half-infinite geodesic in $\mathcal{M}$ with a Kronecker
direction is uniformly distributed unless it hits a singularity.

######  Remark.

Note that in Figure 8.4, we can take the bottom row of atomic cubes and place
them at the top instead. Note that in Figure 8.5, we must ensure that the
bottom $Y$-face of any atomic cube in the bottom row is identified with the
top $Y$-face of the corrresponding atomic cube in the top row.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-8-5.pdf}\vspace{3pt}\\\
\mbox{Figure 8.5: an alternative very special $4$-fold split-
covering}\end{array}$

######  Remark.

In Theorem 4, we can also allow $\mathcal{M}$ to contain an $s\times 2\times
1$ array of atomic cubes, where the bottom row of $s$ atomic cubes gives rise
to an $X$-street of length $s$, and where any $X$-face of any atomic cube in
the top row is a wall.

######  Example 8.2.

Consider a finite polysquare translation surface $\mathcal{P}$ of $9$ faces as
shown in Figure 8.6, where identified edges are obtained from each other by
perpendicular translation, and where $2$ of the $y$-perpendicular edges and
$2$ of the $z$-perpendicular edges are walls.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-8-6.pdf}\vspace{3pt}\\\
\mbox{Figure 8.6: a finite polysquare translation surface}\end{array}$

Let $\mathcal{M}_{0}=[0,1)\times\mathcal{P}$ denote the polycube translation
$3$-manifold obtained from the cartesian product of the unit torus $[0,1)$
with $\mathcal{P}$, as shown in the picture on the left in Figure 8.7.
Corresponding to the $2$ $y$-perpendicular edges of $\mathcal{P}$, the
$3$-manifold $\mathcal{M}_{0}$ has $2$ $Y$-faces that are walls. Corresponding
to the $2$ $z$-perpendicular edges of $\mathcal{P}$, the $3$-manifold
$\mathcal{M}_{0}$ has $2$ $Z$-faces that are walls.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-8-7.pdf}\vspace{3pt}\\\
\mbox{Figure 8.7: a very special $5$-fold split-covering}\end{array}$

Next, we construct a $5$-fold split-covering of $\mathcal{M}_{0}$ with $5$
copies of $\mathcal{M}_{0}$ placed side-by-side in the $x$-direction and glued
together, and where some $X$-faces are replaced by walls, as shown in the
picture on the right in Figure 8.7. It is clear from this picture that the
resulting polycube translation $3$-manifold $\mathcal{M}$ contains a $5\times
2\times 1$ array of atomic cubes of the type described in the statement of
Theorem 4. Those $X$-faces that are walls and are not in the special $5\times
2\times 1$ array of atomic cubes are irrelevant. The $Y$-faces and $Z$-faces
present in $\mathcal{M}$ correspond precisely to those present in
$\mathcal{M}_{0}$.

We say that a finite polycube translation $3$-manifold $\mathcal{M}$ is
Kronecker–Weyl if every half-infinite geodesic in $\mathcal{M}$ with a
Kronecker direction is uniformly distributed unless it hits a singularity.
Using $s$-fold split-covering and noting the irrelevance of whether many
$X$-faces are walls or otherwise, we see that we can construct exponentially
many such polycube translation $3$-manifolds. On the other hand, we are a long
way from a full solution of the Open Problem in Section 1.

##  9\. Proof of Theorem 4

The proof of Theorem 4 goes very much along the lines of that of Theorem 1.
However, extra ideas are needed at various stages of the argument.

Suppose that the period polycube translation $3$-manifold $\mathcal{M}_{0}$
contains precisely $d$ distinct atomic cubes. Let $Y^{(1)},\ldots,Y^{(d)}$
denote the bottom $Y$-faces of these atomic cubes, and let
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\mathcal{Y}_{0}=Y^{(1)}\cup\ldots\cup Y^{(d)}.$

Then the $s$-fold split-covering $\mathcal{M}$ contains precisely $sd$
distinct atomic cubes, where each $Y$-face $Y^{(\gamma)}$,
$\gamma=1,\ldots,d$, of $\mathcal{M}_{0}$ leads to $s$ distinct $Y$-faces
$Y^{(\gamma)}_{\sigma}$, $\sigma=1,\ldots,s$. Let
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\mathcal{Y}=\bigcup_{\gamma=1}^{d}\bigcup_{\sigma=1}^{s}Y^{(\gamma)}_{\sigma}.$

As before, we consider the discretization of the $\mathbf{v}$-flow in
$\mathcal{M}$ relative to the $Y$-faces, where $\mathbf{v}=(\alpha,1,\beta)$
is a Kronecker direction, and let

$T_{\alpha,\beta}:\mathcal{Y}\to\mathcal{Y}$

denote the relevant discrete transformation defined by consecutive hitting
points. We need to show that this area-preserving map is ergodic on
$\mathcal{Y}$. Suppose, on the contrary, that it is not ergodic. Then there
exist two disjoint $T_{\alpha,\beta}$-invariant subsets $\mathcal{W}$ and
$\mathcal{S}$ of $\mathcal{Y}$ such that
$\mathcal{Y}=\mathcal{W}\cup\mathcal{S}$ and

$0<\lambda_{2}(\mathcal{W})\leqslant\lambda_{2}(\mathcal{S})<sd,$ (9.1)

where $\lambda_{2}$ denotes the $2$-dimensional Lebesgue measure. We show that
this leads to a contradiction.

The natural projection of $\mathcal{Y}$ to $\mathcal{Y}_{0}$ can be
represented in the form

$\mathcal{Y}\to\mathcal{Y}_{0}:(x,y,z)\mapsto(\\{x\\},y,z),$ (9.2)

indicating that, for every $\gamma=1,\ldots,d$, the images of the $Y$-faces
$Y^{(\gamma)}_{\sigma}$, $\sigma=1,\ldots,s$, coincide and are all equal to
$Y^{(\gamma)}$, and the image of each point in $Y^{(\gamma)}_{\sigma}$ is in
the same relative position in $Y^{(\gamma)}$. Thus for every point
$P\in\mathcal{Y}_{0}$, there are precisely $s$ distinct points
$P_{1},\ldots,P_{s}\in\mathcal{Y}$ which have projection image $P$. Let

$f_{\mathcal{W}}(P)=|\\{P_{1},\ldots,P_{s}\\}\cap\mathcal{W}|\quad\mbox{and}\quad
f_{\mathcal{S}}(P)=|\\{P_{1},\ldots,P_{s}\\}\cap\mathcal{S}|$

denote the number of these $s$ points that fall into $\mathcal{W}$ and
$\mathcal{S}$ respectively. They are non-negative integer valued functions
defined on $\mathcal{Y}_{0}$ that satisfy the condition

$f_{\mathcal{W}}(P)+f_{\mathcal{S}}(P)=s\quad\mbox{for almost all
$P\in\mathcal{Y}_{0}$}.$ (9.3)

We also consider the corresponding projection of

$T_{\alpha,\beta}:\mathcal{Y}\to\mathcal{Y}\quad\mbox{to}\quad
T_{0}=T_{0,\alpha,\beta}:\mathcal{Y}_{0}\to\mathcal{Y}_{0},$ (9.4)

which is simply the discretization of the $\mathbf{v}$-flow in
$\mathcal{M}_{0}$ relative to the $Y$-faces.

Theorem 3 establishes the ergodicity of $T_{0}$. The Birkhoff ergodic theorem
then guarantees that both $f_{\mathcal{W}}$ and $f_{\mathcal{S}}$ are constant
integer valued functions. It then follows from (9.1) and (9.3) that there
exist two integers $s_{1}$ and $s_{2}$ satisfying

$1\leqslant s_{1},s_{2}\leqslant s-1\quad\mbox{and}\quad s_{1}+s_{2}=s$ (9.5)

such that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$f_{\mathcal{W}}(P)=s_{1}\quad\mbox{and}\quad
f_{\mathcal{S}}(P)=s_{2}\quad\mbox{for almost all $P\in\mathcal{Y}_{0}$}.$
(9.6)

We then derive a contradiction by establishing a suitable analogue of Lemma
1.2.

For a Kronecker direction $\mathbf{v}=(\alpha,1,\beta)$, where $\alpha$,
$\beta$ and $\alpha/\beta$ are all irrational, we can clearly assume that
$\alpha>0$ and $\beta>0$. In the simple case of Theorem 1 and Lemma 1.2, the
parameter $\beta$ is concerned with a direction which is considered to be
integrable, so its value causes no difficulty. Furthermore, since we can
interchange the other two directions, we can assume that $0<\alpha<1$.

Here, the parameter $\beta$ is no longer concerned with a direction which is
considered to be integrable, so its value can potentially cause difficulties.
Meanwhile, the value of the parameter $\alpha$ can also cause some
inconvenience. To formulate and establish a suitable analogue of Lemma 1.2, we
need to first understand these difficulties.

######  Remark on the value of $\alpha$.

Assume that the $z$-direction is integrable, so that the parameter $\beta$
does not cause any difficulties. Figure 9.1 shows the singularities of the
$\mathbf{v}$-flow, in the special case when $0<\alpha<1$, in a $4\times
2\times 1$ array of atomic cubes of the type described in the statement of
Theorem 4 in connection with a $4$-fold split-covering, with the $z$-direction
suppressed. The dashed line segment in each bottom atomic cube indicates that
a line segment in the $z$-direction on the bottom $Y$-face of the atomic cube
is taken by the flow to the singular edge at the top of the wall that is the
right $X$-face of the atomic cube.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-9-1.pdf}\vspace{3pt}\\\
\mbox{Figure 9.1: the case when $0<\alpha<1$}\end{array}$

The situation is a bit more complicated when $\alpha>1$, as illustrated in
Figure 9.2.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-9-2.pdf}\vspace{3pt}\\\
\mbox{Figure 9.2: the case when $2<\alpha<3$}\end{array}$

In each case, we start with $4$ small rectangles

$R_{1}=R_{1}^{-}\cup R_{1}^{+},\quad R_{2}=R_{2}^{-}\cup R_{2}^{+},\quad
R_{3}=R_{3}^{-}\cup R_{3}^{+},\quad R_{4}=R_{4}^{-}\cup R_{4}^{+}.$

Each is on the bottom $Y$-face of a bottom atomic cube, and they are in the
same relative position within their own $Y$-faces. Each is split under
$T_{\alpha,\beta}$ by the singular edge at the top of the wall that forms the
right $X$-face of its own atomic cube.

Although we draw the same conclusion, for instance, that the image
$T_{\alpha,\beta}^{2}(R_{1}^{-})$ of the left half $R_{1}^{-}$ of $R_{1}$ and
the image $T_{\alpha,\beta}^{2}(R_{2}^{+})$ of the right half $R_{2}^{+}$ of
$R_{2}$ together essentially form a small rectangle on some $Y$-face, the
identity of this $Y$-face is not so straightforward as in the case
$0<\alpha<1$ earlier.

This is not a major issue, but we shall fix this at the same time as we deal
with the real issue concerning the value of $\beta$. The idea is that of short
transportation, illustrated in Figure 9.3, where we keep our attention on the
flow much closer to the singularities.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-9-3.pdf}\vspace{3pt}\\\
\mbox{Figure 9.3: fixing the problem with the value of $\alpha$}\end{array}$

######  Remark on the value of $\beta$.

The parameter $\beta$ deals with the part of the flow in the $z$-direction.
Figure 9.1 shows the singularities of the $\mathbf{v}$-flow, in the special
case when $\beta>0$ is small, in a $4\times 2\times 1$ array of atomic cubes
of the type described in the statement of Theorem 4 in connection with a
$4$-fold split-covering. The dashed line segment in each bottom atomic cube
indicates that a line segment in the $z$-direction on the bottom $Y$-face of
the atomic cube is taken by the flow to the singular edge at the top of the
wall that is the right $X$-face of the atomic cube. As the value of $\beta$ is
small, we can find suitable small rectangles such that their images under
$T_{\alpha,\beta}$ and $T_{\alpha,\beta}^{2}$ stay within this $4\times
2\times 1$ array of atomic cubes.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-9-4.pdf}\vspace{3pt}\\\
\mbox{Figure 9.4: small values of $\beta$}\end{array}$

The situation is different if $\beta>0$ is not so small, as there is the
possibility that the flow from a small rectangle crosses the back $Z$-face of
the bottom atomic cube before reaching the wall that is the right $X$-face of
the atomic cube, as illustrated in Figure 9.5. In this case, the images of the
small rectangle under $T_{\alpha,\beta}$ and $T_{\alpha,\beta}^{2}$ may no
longer stay within this $4\times 2\times 1$ array of atomic cubes. What then
happens depends on whether some $X$-faces outside this array are walls or not.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-9-5.pdf}\vspace{3pt}\\\
\mbox{Figure 9.5: larger values of $\beta$}\end{array}$

Figure 9.6 shows the view from the top of what may happen, with small
$\alpha>0$, and we do not have the appropriate composition of the image
rectangles. For instance, the images $T_{\alpha,\beta}^{2}(R_{1}^{-})$ and
$T_{\alpha,\beta}^{2}(R_{1}^{+})$ now form a small rectangle. It seems
complicated to eradicate the situation if $\beta>1$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-9-6.pdf}\vspace{3pt}\\\
\mbox{Figure 9.6: larger values of $\beta$}\end{array}$

Figure 9.3 gives a hint on a possible remedy of the situation for the value of
the parameter $\alpha$ if the value of the parameter $\beta$ does not cause
any serious difficulties. It turns out that essentially the same idea can also
deal with any difficulties arising from the value of the parameter $\beta$.

Short transportation means concentrating on what happens much closer to the
singularities. However, to do so, we need to move into the interior of the
atomic cubes and not just the $Y$-faces. Thus we move outside the set
$\mathcal{Y}$ and have to abandon the discretization of the $\mathbf{v}$-flow
in $\mathcal{M}$ relative to the $Y$-faces.

A simpler approach is to magnify the polycube translation $3$-manifold
$\mathcal{M}$ instead. More precisely, given any Kronecker direction
$\mathbf{v}=(\alpha,1,\beta)$ with $\alpha>0$ and $\beta>0$, let $\mathscr{N}$
be an integer satisfying

$\mathscr{N}>\max\\{\alpha,1+2\beta\\}.$ (9.7)

We now magnify the polycube translation $3$-manifold $\mathcal{M}$ by a factor
$\mathscr{N}$ in each direction, so that any atomic cube in $\mathcal{M}$ is
magnified to volume $\mathscr{N}^{3}$, and any face in $\mathcal{M}$ is
magnified to area $\mathscr{N}^{2}$. We then split each magnified atomic cube
into $\mathscr{N}^{3}$ atomic cubes of unit volume, so that each magnified
face of $\mathcal{M}$ is split into $\mathscr{N}^{2}$ faces of unit area. We
shall abuse notation and denote the resulting polycube translation
$3$-manifold also by $\mathcal{M}$. This new $\mathcal{M}$ has
$\mathscr{N}^{3}sd$ atomic cubes, with $\mathscr{N}^{3}sd$ distinct $Y$-faces
that make up a new set $\mathcal{Y}$. We can perform a similar exercise to
obtain a new polycube translation $3$-manifold $\mathcal{M}_{0}$ with
$\mathscr{N}^{3}d$ atomic cubes, with $\mathscr{N}^{3}d$ distinct $Y$-faces
that make up a corresponding new set $\mathcal{Y}_{0}$. Clearly the new
$\mathcal{M}$ is an $s$-fold split-covering of the new $\mathcal{M}_{0}$.

We can consider the corresponding new mappings given in (9.4), and repeat our
previous argument. Then the inequalities (9.1) are replaced by the
inequalities

$0<\lambda_{2}(\mathcal{W})\leqslant\lambda_{2}(\mathcal{S})<\mathscr{N}^{3}sd.$

Crucially, the assertions (9.5) and (9.6) remain valid.

The special $s\times 2\times 1$ array of atomic cubes is now replaced by an
$s\mathscr{N}\times 2\mathscr{N}\times\mathscr{N}$ array of atomic cubes.
Suppose that the coordinates of the points are in

$[0,s\mathscr{N}]\times[0,2\mathscr{N}]\times[0,\mathscr{N}].$ (9.8)

Then the walls correspond precisely to the collection of points

$\\{(x,y,z):x\in\\{0,\mathscr{N},\ldots,s\mathscr{N}\\}\mbox{ and
}y,z\in[0,\mathscr{N}]\\}.$

For the special case when $s=4$ and $\mathscr{N}=4$, this array is shown in
Figure 9.3 with the $z$-direction suppressed.

It is also convenient to write

$\mathscr{B}_{\sigma}=[(\sigma-1)\mathscr{N},\sigma\mathscr{N}]\times[0,2\mathscr{N}]\times[0,\mathscr{N}],\quad\sigma=1,\ldots,s.$
(9.9)

Their union gives the $s$-fold split-covering of
$[0,\mathscr{N}]\times[0,2\mathscr{N}]\times[0,\mathscr{N}]\subset\mathcal{M}_{0}$
that corresponds to $\mathcal{M}$ being the $s$-fold split-covering of
$\mathcal{M}_{0}$.

We are ready to formulate a suitable analogue of Lemma 1.2. To do so, we need
to find small special rectangles $R_{1},\ldots,R_{s}$ on suitable Y-faces
$Y_{1},\ldots,Y_{s}$ as well as small special rectangles
$R^{*}_{1},\ldots,R^{*}_{s}$ on suitable Y-faces $Y^{*}_{1},\ldots,Y^{*}_{s}$.

######  Remark.

We comment in advance that while in Lemma 1.2, we have

$\mathcal{Y}=Y_{1}\cup Y_{2}\cup Y_{3}=Y^{*}_{1}\cup Y^{*}_{2}\cup Y^{*}_{3},$

here we have xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$Y_{1}\cup\ldots\cup Y_{s}\neq Y^{*}_{1}\cup\ldots\cup Y^{*}_{s},$ (9.10)

and both sides of (9.10) are proper subsets of $\mathcal{Y}$.

Let us now choose a suitable $Y$-face $Y_{1}$ in $\mathscr{B}_{1}$. We shall
make use of the coordinate system given by (9.8) and (9.9).

Figure 9.3 suggests that the $y$-coordinate of any point on $Y_{1}$ should be
equal to $\mathscr{N}-1$, making it a $y$-distance $1$ below the top of the
vertical wall separating $\mathscr{B}_{1}$ and $\mathscr{B}_{2}$. Figure 9.3
also suggests that $Y_{1}$ should contain a line segment $S_{1}$ in the
$z$-direction of length $1$ and which is mapped by the $\mathbf{v}$-flow to
the top edge of this wall. Clearly the $x$-coordinate of any point of $S_{1}$
must be equal to $\mathscr{N}-\alpha\in(0,\mathscr{N})$. This explains the
requirement that $\mathscr{N}>\alpha$ in (9.7).

Now let $\mathbf{p}_{1}=(x_{1},y_{1},z_{1})$ be the coordinates of the bottom
left vertex of $Y_{1}$. The above argument shows that
$x_{1}=[\mathscr{N}-\alpha]$ and $y_{1}=\mathscr{N}-1$. We shall take

$z_{1}=0,$ (9.11)

to be explained later.

We next determine the $Y$-faces $Y_{2},\ldots,Y_{s}$ in
$\mathscr{B}_{2},\ldots,\mathscr{B}_{s}$ and containing the line segments
$S_{2},\ldots,S_{s}$ in the $z$-direction respectively, so that the images of
$Y_{1},\ldots,Y_{s}$ under the modulo $\mathscr{N}$ analog of the natural
projection (9.2) coincide, as do the images of $S_{1},\ldots,S_{s}$.

The argument thus far is summarized in the bottom half of Figure 9.7, where
the $y$-direction is suppressed.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-9-7.pdf}\vspace{3pt}\\\
\mbox{Figure 9.7: a process equivalent to short transportation}\end{array}$

Consider next the image $S^{*}_{2}=S_{1}+(2\alpha,2,2\beta)$ of $S_{1}$ that
corresponds to two applications of the map $T_{\alpha,\beta}$ if we ignore the
singularity at the top of the wall.

The $y$-coordinate of any point on $S^{*}_{2}$ is clearly $\mathscr{N}+1$. The
$x$-coordinate is equal to $\mathscr{N}+\alpha\in(\mathscr{N},2\mathscr{N})$.
The $z$-coordinate lies in the interval
$(2\beta,1+2\beta)\subset(0,\mathscr{N})$. Hence $S^{*}_{2}$ lies on two
adjoining $Y$-faces in $\mathscr{B}_{2}$, with total length $1$, and explains
the choice (9.11) as well as the condition $\mathscr{N}>1+2\beta$ in (9.7).

Now let $Y^{*}_{2}$ denote the $Y$-face that contains the longer half of the
line segment $S^{*}_{2}$, with bottom left vertex
$\mathbf{p}^{*}_{2}=(x^{*}_{2},y^{*}_{2},z^{*}_{2})$. Then
$x^{*}_{1}=[\mathscr{N}+\alpha]+1$, $y^{*}_{2}=\mathscr{N}+1$ and

$z^{*}_{2}=\left\\{\begin{array}[]{ll}[2\beta],&\mbox{if
$\\{2\beta\\}<1/2$},\\\ 1+[2\beta],&\mbox{if
$\\{2\beta\\}>1/2$}.\end{array}\right.$