In this paper, there are two sections.
In Section 7, we simplify the eigenvalue-based surplus shortline method for arbitrary finite polysquare translation surfaces.
This makes it substantially simpler to determine the irregularity exponents of some infinite orbits, and quicker to find the escape rate to infinity of some orbits in some infinite models.
In Section 8, our primary goal is to extend the surplus shortline method, both this eigenvalue-based version as well as the eigenvalue-free version, for application to a large class of $2$-dimensional flat dynamical systems beyond polysquares, including all Veech surfaces, and establish time-quantitative equidistribution and time-quantitative superdensity of some infinite orbits in these new systems.

§ MORE ON THE EIGENVALUE-BASED SHORTLINE METHOD

§.§ The shortline method and the edge cutting lemma

Here in part (IV) we assume that the reader is more or less familiar with the earlier parts [2, 3, 4].
The eigenvalue-based surplus shortline method has been developed in the special case of the L-surface in <cit.> and <cit.>.

For a $4$-direction billiard flow in a general finite polysquare region, we can apply unfolding introduced in [2] to convert the problem to one concerning a $1$-direction geodesic flow on some related finite polysquare translation surface.
We concentrate therefore on $1$-direction geodesic flow on a finite polysquare translation surface.

We show that the surplus shortcut-ancestor process, introduced earlier in [2, 3] in connection with the L-surface, can be adapted to every finite polysquare translation surface with $1$-direction geodesic flow.
For a quick introduction, we first illustrate the method by applying it to a $1$-direction geodesic flow with a particular quadratic irrational slope on the surface $S_2$ given in the picture on the left in Figure 7.1.1.

We compute the $2$-step transition matrix, and also determine the irregularity exponent for this particular quadratic irrational slope.
We illustrate the fact that this $2$-step transition matrix is usually very redundant.
This in turn motivates the so-called edge cutting lemma which we formulate later.

We consider $1$-direction geodesic flow on $S_2$.
Here the boundary pairings come from simple perpendicular translations.
However, the pair of vertical edges $v_2$ and the pair of horizontal edges $h_2$ around the missing square represent a slightly less straightforward form of perpendicular translation.

The surface $S_2$ has $1$ vertical street of length $1$, $3$ vertical streets of length $3$, $1$ horizontal street of length $1$, and $3$ horizontal streets of length $3$, so the street-LCM is equal to $3$.
The picture on the right in Figure 7.1.1 shows the streets where, for instance, the entries $\leftrightarrow2$ and $\updownarrow4$ in a square indicates that the square face is on the $2$-nd horizontal and $4$-th vertical street.

To apply the surplus shortline method, we have to restrict our study to slopes $\alpha$ or $\alpha^{-1}$ for which the continued fraction of $\alpha$ has the special form

\begin{equation}\label{eq7.1.1}
\alpha=[3c_0;3c_1,3c_2,3c_3,\ldots]
\end{equation}

where the digits $c_0,c_1,c_2,c_3,\ldots$ are positive integers.

For illustration we consider here the simplest slope satisfying (<ref>), namely

\begin{equation}\label{eq7.1.2}
\alpha=[3;3,3,3,\ldots]=3+\frac{1}{3+\frac{1}{3+\frac{1}{3+\cdots}}}=\frac{3+\sqrt{13}}{2},
\end{equation}

and its reciprocal $\alpha^{-1}$.
We study the long-term behavior of two particular geodesics $V(t)$ and $H(t)$ on $S_2$ that start from the origin, which is some chosen vertex of one of the square faces of $S_2$.
Crucially, this guarantees that $V(t)$ and $H(t)$ are surplus shortlines of each other.
The almost vertical geodesic $V(t)$ has slope $\alpha$, while the almost horizontal geodesic $H(t)$ has slope $\alpha^{-1}$.
Following <cit.>, we briefly elaborate on the details of the surplus shortline method in this particular case.

We distinguish the $20$ types of almost vertical units $a_i$, $1\le i\le20$, in the picture on the left in Figure 7.1.2.

We say that each of the almost vertical units

is of type $\uparrow$, and starts at the bottom edge of a square face and ends at the top edge of the same square face, while each of the almost vertical units

is of type $\nuparrow$, and starts at the bottom edge of a square face and ends at the top edge of an adjoining square face.
Of particular interest is the unit $a_{11}$ which hits the left edge of the gap at some point and continues from the corresponding point on the identified right edge of the gap.

Similarly, we distinguish the $20$ types of almost horizontal units $b_j$, $1\le j\le20$, in the picture on the right in Figure 7.1.2.

Note that the surface $S_2$ exhibits a $45$-degree reflection symmetry.
But symmetry is not necessary, and indeed totally irrelevant, for the success of the shortline method.
To emphasize this point, we have deliberately used labelling in the picture on the right in Figure 7.1.2 which is not a 45-degree reflection of the labelling in the picture on the left in Figure 7.1.2.

Applying a straightforward adaptation of the surplus shortline method discussed in <cit.>, we can determine the surplus ancestor units of $a_i$,
$1\le i\le20$.
Similarly, we can determine the surplus ancestor units of $b_j$, $1\le j\le20$.

Consider first the almost vertical unit $a_1$.
It is not difficult to see from Figure 7.1.3, which shows only the top horizontal street of $S_2$, that $a_1$ is the shortcut of an almost horizontal detour crossing of the this horizontal street, made up of a fractional almost horizontal unit $b_{11}$, full almost horizontal units $b_2$, $b_4$ and $b_6$, and then a fractional almost horizontal unit $b_1$.
We apply the delete end rule, meaning that we keep the initial fractional almost horizontal unit $b_{11}$ as a full unit and discard the final fractional almost horizontal unit $b_1$, and call $b_{11},b_2,b_4,b_6$ the ancestor units of $a_1$.

We do likewise for $a_i$, $2\le i\le20$, and can summarize in the form

\begin{align}\label{eq7.1.3}
a_1&\rightharpoonup b_{11},b_2,b_4,b_6,
\nonumber
\\
a_2&\rightharpoonup b_9,b_4,b_6,
\nonumber
\\
a_3&\rightharpoonup b_9,b_4,b_6,b_2,
\nonumber
\\
&\ \ \vdots
\nonumber
\\
a_{19}&\rightharpoonup b_3,b_{20},b_{20},b_{20},
\nonumber
\\
a_{20}&\rightharpoonup b_3,b_{20},b_{20}.
\end{align}

This leads to a transition matrix $M_1(S_2)$, where the $i$-th row captures the information in the $i$-th ancestor relation in (<ref>), with the entry on the $j$-th column displaying the multiplicity of $b_j$.

A similar exercise with the roles of the horizontal and vertical interchanged, again using the delete end rule, results in

\begin{align}\label{eq7.1.4}
b_1&\rightharpoonup a_1,a_{14},a_{10},a_2,
\nonumber
\\
b_2&\rightharpoonup a_1,a_{14},a_{10},
\nonumber
\\
b_3&\rightharpoonup a_3,a_{20},a_{16},a_4,
\nonumber
\\
&\ \ \vdots
\nonumber
\\
b_{19}&\rightharpoonup a_{19},a_{16},a_4,a_{20},
\nonumber
\\
b_{20}&\rightharpoonup a_{19},a_{16},a_4.
\end{align}

This leads to an analogous transition matrix $M_2(S_2)$.

Instead of using the delete end rule, we may also use the keep end rule, meaning that we keep the final fractional unit as a full unit and discard the initial fractional unit.
Both rules are for bookkeeping purposes only, as we do not want to count any unit twice.
Depending on which rule we use, (<ref>) and (<ref>), and hence also the matrices $M_1(S_1)$ and $M_2(S_2)$, may be a little different, but this will not affect the subsequent argument.

Composition of the two transition matrices arising from (<ref>) and (<ref>) leads to the product matrix $M_1(S_2)M_2(S_2)$.
The transpose of this product matrix happens to be the $20\times20$ matrix

\begin{equation}\label{eq7.1.5}
\bfA(S_2)=\begin{pmatrix}
1\ \ 0\ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\\
0\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1\ \ 1\ \ 2\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\\
1\ \ 1\ \ 1\ \ 0\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\\
0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1\ \ 1\ \ 2\ \ 1\ \ 4\ \ 3\\
1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 0\ \ 0\\
1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 2\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 2\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\\
0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 2\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\\
0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 2\ \ 3\ \ 5\ \ 2\ \ 2\ \ 2\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\\
0\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1\ \ 0\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\\
1\ \ 1\ \ 2\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 0\ \ 1\ \ 1\ \ 2\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\\
1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\\
2\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 2\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\\
0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 0\ \ 1\ \ 0\ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\\
1\ \ 1\ \ 2\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 2\ \ 1\ \ 0\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\\
0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1\ \ 0\ \ 1\ \ 1\ \ 0\ \ 0\\
1\ \ 1\ \ 1\ \ 1\ \ 2\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 0\ \ 4\ \ 3\\
0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\\
2\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 2\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 0\ \ 0\\
0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 0\ \ 3\ \ 2\\
1\ \ 1\ \ 1\ \ 1\ \ 2\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 1\ \ 1\ \ 1\ \ 1\ \ 2\ \ 1\ \ 1\ \ 1
\end{pmatrix}.
\end{equation}

We use the transpose, because in the edge cutting lemma, to be formulated later, we are interested in eigenvectors as column vectors of $\bfA(S_2)$.
And it helps that MATLAB, like any other linear algebra computer program, automatically computes right or column eigenvectors.

We start with the $3$ eigenvalues of (<ref>) with the largest absolute values.
By MATLAB, the largest eigenvalue is

\begin{equation}\label{eq7.1.6}
\lambda_1= \lambda_1(\bfA(S_2))=\frac{11+3\sqrt{13}}{2}=\left(\frac{3+\sqrt{13}}{2}\right)^2=\alpha^2,
\end{equation}

in view of (<ref>), with corresponding eigenvector

\begin{align}\label{eq7.1.7}
\bfv_1
\nonumber
\\
\end{align}

The second largest eigenvalue is

\begin{equation}\label{eq7.1.8}
\lambda_2= \lambda_2(\bfA(S_2))=3+2\sqrt{2}=(1+\sqrt{2})^2,
\end{equation}

with corresponding eigenvector

\begin{align}\label{eq7.1.9}
\bfv_2
\nonumber
\\
\end{align}

\begin{align}
\quad
\quad
\quad
\nonumber
\\
\quad
\quad
\nonumber
\end{align}

The third largest eigenvalue is

\begin{equation}\label{eq7.1.10}
\lambda_3= \lambda_3(\bfA(S_2))=\frac{3+\sqrt{5}}{2},
\end{equation}

with corresponding eigenvector

\begin{align}\label{eq7.1.11}
\bfv_3
\nonumber
\\
\end{align}

\begin{align}
\quad
\quad
\quad
\nonumber
\\
\quad
\quad
\nonumber
\end{align}

There are $3$ other eigenvalues which are the algebraic conjugates of $\lambda_i$, $1\le i\le3$, and these are between $0$ and $1$.
There are $14$ other eigenvalues with absolute value $1$.
We shall see that $\lambda_i$, $1\le i\le3$, are the only relevant eigenvalues of $\bfA(S_2)$, and the other $17$ eigenvalues are irrelevant.
We comment that the Jordan normal form of the matrix $\bfA(S_2)$ is simple, as the matrix has $20$ different eigenvectors.

The irregularity exponent of this special slope $\alpha$ given by (<ref>) is, by definition, equal to

Before discussing the substantial redundancy and repetition of the coordinates in (<ref>), (<ref>) and (<ref>), we make a short detour and state two general theorems.

What we are doing here with the surface $S_2$ and have done with the L-surface in [2, 3] can be adapted for any finite polysquare translation surface, and we shall give some details in the next section.
In particular, we obtain the following result.

Let $\PPP$ be a finite polysquare translation surface, with street-LCM, i.e., the least common multiple of the lengths of the horizontal and vertical streets of $\PPP$,  denoted by $\LCM(\PPP)$.
Consider a geodesic flow with a quadratic irrational slope where the ordinary continued fraction digits are all divisible by $\LCM(\PPP)$.

Using the eigenvalue-based version of the surplus shortline method developed in [2, 3], we can explicitly compute the irregularity exponent for such a slope.

Combining the irregularity exponent with the method of zigzagging introduced in <cit.>, we can also describe, for a geodesic flow on $\PPP$ with such a slope, the time-quantitative behavior of the edge cutting and face crossing numbers, as well as equidistribution relative to all convex sets.

We remark that for the slopes in Theorem <ref>, <cit.> applies, and not only guarantees superdensity, but establishes superdensity for more slopes than Theorem <ref> can provide the explicit values of the irregularity exponents.
On top of the arithmetic condition that the continued fraction digits are divisible by the street-LCM, for superdensity in Theorem 6.4.1, we also need the boundedness of the continued fraction digits given by the badly approximable slopes.
Furthermore, we need in Theorem <ref> the stronger condition of periodicity of the tail of the sequence of continued fraction digits, so that the slope is a quadratic irrational.

The irregularity exponent can be computed from the two eigenvalues with the largest absolute values of an appropriate $2d\times2d$ matrix, where $d$ is the number of square faces of the polysquare translation surface $\PPP$.

Since for every non-integrable polysquare translation surface the street-LCM is at least $2$, the surplus shortline method does not have a chance of determining the irregularity exponent for every quadratic irrational slope.

The reason why in the special case of the L-surface we are able to determine the irregularity exponent for every quadratic irrational slope is two-fold.
First, the street-LCM is equal to $2$.
More importantly, we combine the surplus and deficit versions of the shortline method according to the $\pm$-even type continued fraction of the slope.

If the street-LCM of a polysquare region or translation surface is equal to $2$, then we call it a $2$-polysquare region or translation surface.
One such surface with the simplest boundary identification via perpendicular translation is called a $2$-polysquare snake surface.
The special class of $2$-polysquare snake surfaces is surprisingly large, and Figure 7.1.4 gives an example.

We can easily characterize every $2$-polysquare snake surface by making use of the simple graph-theoretic term of path.

Given a $2$-polysquare snake surface, we can put a point at the center of every square face of the underlying polysquare region, and call it the capital of the square face.
We join two capitals by a dashed line-segment if and only if the corresponding square faces share a common edge.
In this way we obtain a non self-intersecting strictly alternating h-v-path (or v-h-path) of the capitals, where “h” stands for a horizontal edge of unit length and “v” stands for a vertical edge of unit length.
The term strictly alternating means that h is always preceded and followed by v, and v is always preceded and followed by h, unless the path stops.

We also have the converse, that every non self-intersecting strictly alternating h-v-path (or v-h-path) corresponds to a $2$-polysquare snake surface.
Indeed, we cover every vertex of the path with a unit size square
such that the vertex is the capital of the square, and
we may place some extra walls, denoted by boldface line-segments, if necessary.

It can be shown that the special treatment of the L-surface in [2, 3] can be adapted to every $2$-polysquare translation surface.
The idea, as for the L-surface, is to combine the surplus and deficit versions of the eigenvalue-based shortline method according to the $\pm$-even type continued fraction expansion of the slope.

Let $\PPP$ be a finite $2$-polysquare translation surface.
Then the LCM-divisibility condition in Theorem <ref> may be dropped, so that the conclusion of Theorem <ref> can be extended to every quadratic irrational slope.

To explain the substantial redundancy or repetition of the coordinates in the eigenvectors (<ref>), (<ref>) and (<ref>), the trick is to visualize and represent the surface $S_2$ in Figure 7.1.1 as the period of a doubly periodic infinite polysquare translation surface $S_2(\infty)$ in Figure 7.1.5 below.

The infinite polysquare translation surface $S_2(\infty)$ is essentially a square lattice of copies of $S_2$ with a noticeable difference – observe carefully the edge pairings in $S_2(\infty)$.
For convenience, we say that $S_2$ is the period surface of $S_2(\infty)$.

Following the surplus shortline method, we study two special geodesics on the infinite polysquare translation surface $S_2(\infty)$.
The first geodesic $V_*(t)$, $t\ge0$, starts from the origin, namely a vertex of one of the square faces, and has slope $\alpha$ defined in (<ref>).
The second geodesic $H_*(t)$, $t\ge0$, also starts from the origin and has reciprocal slope $\alpha^{-1}$.
Note that the projections of $V_*(t)$ and $H_*(t)$ on the period surface $S_2$
give back the geodesics $V(t)$ and $H(t)$ defined earlier.

A particle traveling on the almost vertical geodesic $V_*(t)$, $t\ge0$, of the doubly periodic infinite translation surface $S_2(\infty)$ moves from one copy of the period $S_2$ to another copy of $S_2$, back and forth, up and down, and generally wanders around.

Note that in $S_2(\infty)$, each copy of $S_2$ has $4$ immediate neighbors: we shall refer to them as the North, South, East and West neighbors.
Traveling along the geodesic $V_*(t)$, $t\ge0$, is somewhat similar to a symmetric random walk on the $2$-dimensional doubly periodic
lattice $\Zz^2$.
Escaping from a copy of $S_2$ to the North means following a unit of type $a_3$ or $a_4$ in Figure 7.1.2, while escaping to the South means a following a unit of type $a_{15}$ or $a_{16}$.
Similarly, escaping to the East means following a unit of type $a_7$, while escaping to the West means following a unit of type $a_{11}$.

To describe an arbitrary finite segment of the geodesic $V_*(t)$, $t\ge0$, of slope $\alpha$ on the infinite polysquare translation surface $S_2(\infty)$, it suffices to consider its projection on the period surface $S_2$.
The difference between number of North-escapes and the number of South-escapes gives the vertical change on $S_2(\infty)$, while the difference between the number of East-escapes and the number of West-escapes gives the horizontal change on $S_2(\infty)$.
This is how we can determine the escape rate to infinity of the special geodesic $V_*(t)$, $t\ge0$, on $S_2(\infty)$ from the behavior of its projection $V(t)$, $t\ge0$, on the period surface $S_2$.

Since the two special geodesics $V(t)$ and $H(t)$ are surplus shortlines of each other on the period surface $S_2$, the time evolution of
$V(t)$, $t\ge0$, of slope $\alpha$ and starting from the origin is described by powers of the $2$-step transition matrix $\bfA(S_2)$ given by (<ref>).
Since $\bfA(S_2)$ is diagonalizable, it is particularly easy to express the number of types of units by using the relevant eigenvalues and their eigenvectors.
Since the number of types of units is expressed as a linear combination of the powers of the eigenvalues, in case of large powers the contribution of the irrelevant eigenvalues is clearly negligible.

Recall that escaping to the North means following a unit of type $a_3$ or $a_4$ in Figure 7.1.2, and escaping to the South means following a unit of type $a_{15}$ or $a_{16}$.
It follows that for the vertical change, we need to know the $3$-rd, $4$-th, $15$-th and $16$-th coordinates of the relevant eigenvectors $\bfv_i$, $i=1,2,3$, in (<ref>), (<ref>) and (<ref>).
Escaping to the East means following a unit of type $a_7$, and escaping to the West means following a unit of type $a_{11}$.
It follows that for the horizontal change, we need to know the $7$-th and $11$-th coordinates of the relevant eigenvectors.

Recall that $v_i(j)$ denotes the $j$-th coordinate of $\bfv_i$, $i=1,2,3$ and $j=1,2,\ldots,20$.
It is easily seen from (<ref>), (<ref>) and (<ref>) that

\begin{equation}\label{eq7.1.12}
\quad
\end{equation}

\begin{equation}\label{eq7.1.13}
\quad
\end{equation}

The reader may find (<ref>) and (<ref>) a surprising coincidence.
However, there is a simple explanation why these equalities must hold, namely, that the geodesic $V_*(t)$, $t\ge0$, of $S_2(\infty)$ satisfies the conditions of <cit.> and exhibits super-slow logarithmic escape rate to infinity; see the Remark after the proof there.
Meanwhile, a violation of (<ref>) and (<ref>) would imply a power-size escape rate to infinity, exponentially larger than logarithmic escape rate to infinity.

The equalities (<ref>) and (<ref>) remain true even if in obtaining (<ref>) and (<ref>) we replace the delete end rule by the keep end rule, or mix the two rules arbitrarily.
It follows from the fact that the relevant eigenvalues and eigenvectors remain the same under these changes.

The edge cutting lemma, which we shall formulate later, is simply a far-reaching generalization of the equalities (<ref>) and (<ref>).

It can be seen from (<ref>), (<ref>) and (<ref>) that (<ref>) can be extended to include the $9$-th coordinate, so that

\begin{equation}\label{eq7.1.14}
\quad
\end{equation}

Note from Figure 7.1.2 that the entries in (<ref>) represent almost vertical units of type $\nuparrow$ on the second horizontal street of $S_2$, of length $3$.
For the first horizontal street and the third horizontal street, both also of length $3$, we have respectively the analogs

\begin{equation}\label{eq7.1.15}
\quad
\end{equation}

\begin{equation}\label{eq7.1.16}
\quad
\end{equation}

both of which hold, as can be seen from (<ref>), (<ref>) and (<ref>).

It can also be seen from (<ref>), (<ref>) and (<ref>) that (<ref>) can be extended to

\begin{equation}\label{eq7.1.17}
\quad
\end{equation}

Note from Figure 7.1.2 that the entries in (<ref>) represent almost vertical units of type $\uparrow$ or $\nuparrow$ on the third vertical street of $S_2$, of length $3$.
Note that we have chosen almost vertical units of type $\nuparrow$ that intersect the left edge of each square face.
In view of (<ref>) and (<ref>), it would have made no difference if we had chosen instead any that intersects the right edge of the same square face.
For the second vertical street and the fourth vertical street, both also of length $3$, we have respectively the analogs

\begin{equation}\label{eq7.1.18}
\quad
\end{equation}

\begin{equation}\label{eq7.1.19}
\quad
\end{equation}

both of which hold, as can be seen from (<ref>), (<ref>) and (<ref>).

The edge cutting lemma basically says that for any finite polysquare translation surface with $1$-direction geodesic flow, analogs of the equalities
(<ref>)–(<ref>) hold.
Section <ref> contains a proof of this fact.
In fact, we prove even more; see Theorem <ref>.

We conclude this section by giving a simple proof of the edge cutting lemma in the simplest special case of the L-surface.
We include it, because this simple proof already illustrates quite well the method we use in Section <ref>.

We go back to <cit.>, and consider the L-surface with a $1$-direction geodesic flow with slope $\alpha=1+\sqrt{2}$.
Figure 7.1.6 below shows the edges of the L-surface as well as the almost vertical units.

Corresponding to the matrix $\bfA(S_2)$ for the polysquare translation surface $S_2$, we have the $2$-step transition matrix

\begin{equation}\label{eq7.1.20}
\bfA=M_2^TM_1^T=\begin{pmatrix}
\end{pmatrix}.
\end{equation}

In (<ref>), we have used the lexicographic order for the almost vertical units

\begin{equation}\label{eq7.1.21}
\end{equation}

The L-surface has precisely one vertical and one horizontal street of length greater than $1$.
If $\bfv=(v_1,v_2,v_3,v_4,v_5,v_6)$ is a relevant eigenvector of $\bfA$, then the coordinates correspond to the $6$ almost vertical units in (<ref>).
It follows that the L-surface analog of the equalities (<ref>)–(<ref>) is

\begin{equation}\label{eq7.1.22}
\end{equation}

and the L-surface analog of the inequalities (<ref>)–(<ref>) is

\begin{equation}\label{eq7.1.23}
\end{equation}

These two equations together give an invariant subspace of the matrix $\bfA$.
To see this, suppose that

\begin{align}
\nonumber
\\
\nonumber
\\
\nonumber
\\
\nonumber
\\
\nonumber
\\
\nonumber
\end{align}

It is easy to see that $w_1=w_4$ follows from $v_1=v_4$.
On the other hand, we have

\begin{align}
\nonumber
\\
\nonumber
\end{align}

so that

and so $w_1+w_2=w_5+w_6$ follows from $v_1+v_2=v_5+v_6$.

The $2$-step transition matrix $\bfA$ has $6$ eigenvalues and $6$ independent eigenvectors in $\Rr^6$.
The eigenvalues, in order of decreasing absolute value, are

\begin{align}\label{eq7.1.24}
\lambda_1=3+2\sqrt{2},
\quad
\lambda_2=\frac{3+\sqrt{5}}{2},
\quad
\lambda_3=1,
\nonumber
\\
\lambda_4=1,
\quad
\lambda_5=\frac{3-\sqrt{5}}{2},
\quad
\lambda_6=3-2\sqrt{2}.
\end{align}

Suppose that the corresponding eigenvectors of $\bfA$ are respectively $\bfv_i$, $1\le i\le6$.

\begin{align}
\lambda_1\to\bfv_1
\nonumber
\\
\lambda_2\to\bfv_2
\nonumber
\\
\lambda_3\to\bfv_3
\nonumber
\\
\lambda_4\to\bfv_4
\nonumber
\\
\lambda_5\to \bfv_5
\nonumber
\\
\lambda_6\to\bfv_6
\nonumber
\end{align}

Note now that the $4$-dimensional invariant subspace of the matrix $\bfA$ given by the equations (<ref>) and (<ref>) contains the
eigenvectors $\bfv_1$ and $\bfv_2$ corresponding to the two relevant eigenvalues $\lambda_1$ and $\lambda_2$.
It also contains the eigenvectors $\bfv_5$ and $\bfv_6$ of the two conjugate eigenvalues $\lambda_5$ and $\lambda_6$.

This simple example illustrates what we call an invariant subspace argument.
In the next section we use a similar but more sophisticated version of this to prove the edge cutting lemma and more.

Finally observe that the edge cutting lemma has the following vague intuitive meaning.
If the surplus shortline method works for a $1$-direction geodesic on a finite polysquare translation surface with some fixed slope, then the edge cutting numbers of the edges in the same street are nearly the same.
We shall clarify this later.

§.§ Reducing the 2-step transition matrix: the street-spreading matrix

We see from the examples in Section <ref> that the $2$-step transition matrix of the shortline method is quite redundant.
Here we elaborate on this.
What we are interested in is the long-term behavior of the geodesics, and when the shortline method works, it suffices to consider the relevant eigenvalues, those with absolute value greater than $1$, and the corresponding eigenvectors.
For example, while the $2$-step transition matrix $\bfA(S_2)$ in (<ref>) has $20$ eigenvalues, only $3$ of them are relevant.
Moreover, each of the corresponding $20$-dimensional eigenvectors (<ref>), (<ref>) and (<ref>) has at most $8$ distinct coordinates, with some multiplicities.

Accordingly, we introduce a general reduction process that, roughly speaking, eliminates the irrelevant part of the $2$-step transition matrix $\bfA$.
We shall give a recipe of how we can read out the relevant eigenvalues and their eigenvectors from a smaller matrix called the street-spreading matrix.

We consider a vector space $W$ with basis made up of all the distinct types of almost vertical units of a finite polysquare translation surface $\PPP$, of dimension equal to twice the number of distinct square faces of $\PPP$.
We shall show that the equations in the edge cutting lemma define a subspace of $W$, and we want to show that this subspace is $\bfA$-invariant and contains all the relevant eigenvectors.
Thus the basic idea is to find such a relevant $\bfA$-invariant subspace, and a convenient basis of it, leading to the usually substantially smaller street-spreading matrix.
A proof of the edge cutting lemma will come as a byproduct of the reduction process.
Theorem <ref>, the main result, will be formulated towards the end of this section.

Let $\PPP$ be a given finite polysquare translation surface, with $d$ square faces.
Let $m$ be any fixed integer multiple of the lengths of the horizontal streets of $\PPP$, and let $n$ be any fixed integer multiple of the lengths of the vertical streets of $\PPP$.

We adopt the following convention.

A typical square face of $\PPP$, on the $i$-th horizontal street and $j$-th vertical street, is denoted by $S_{i,j}$.
For any fixed $i$, there is a finite set $J_i$ of indices $j$ such that $j\in J_i$ if and only if $S_{i,j}\subset\PPP$, so that

is the $i$-th horizontal street of $\PPP$.
For any fixed $j$, there is a finite set $I_j$ of indices $i$ such that $i\in I_j$ if and only if $S_{i,j}\subset\PPP$, so that

is the $j$-th vertical street of $\PPP$.

Our convention here assumes that if a horizontal street and a vertical street of a polysquare translation surface intersect, then the intersection is a unique square face.
This is an over-simplification of the general situation, as it is in fact possible for two or more distinct square faces of a polysquare translation surface to lie on the same horizontal street and the same vertical street simultaneously, so that the notation $S_{i,j}$ does not allow us to distinguish between two distinct square faces that lie on the $i$-th horizontal street and the $j$-th vertical street.
However, our results remain valid in the general situation, but we have chosen to adopt the present simplification and convention here as the notation is somewhat simpler and more convenient for us to study many of the polysquare translation surfaces of interest.

Indeed, the first occasion in this paper where our notation becomes inadequate is at the end of Section <ref> when we discuss the regular octagon surface, and we shall deal with the problem on a case by case basis when it arises.

Nevertheless, we make some further comments on the general case after the proof of Theorem <ref>.

We consider two types of almost vertical units in $S_{i,j}$.
Both types start from the bottom edge of $S_{i,j}$.
However, type $\uparrow_{i,j}$ ends on the top edge of $S_{i,j}$, whereas type $\nuparrow_{i,j}$ exits $S_{i,j}$ through the right edge.
We also consider two types of almost horizontal units in $S_{i,j}$.
Both types end on the right edge of $S_{i,j}$.
However, type $\rightarrow_{i,j}$ starts from the left edge of $S_{i,j}$, whereas type $\nrightarrow_{i,j}$ enters $S_{i,j}$ through the bottom edge.

We now consider a $1$-direction almost vertical geodesic $V_0$ on $\PPP$, starting at a vertex of $\PPP$ and with slope

\begin{equation}\label{eq7.2.1}
\alpha=[n;m,n,m,\ldots]=n+\frac{1}{m+\frac{1}{n+\frac{1}{m+\cdots}}},
\end{equation}

so that $n\le\alpha<n+1$.
The geodesic $V_0$ is made up of almost vertical units of type $\uparrow_{i,j}$ and $\nuparrow_{i,j}$, all with slope $\alpha$.
Thus we may consider a vector space $W$, with basis

\begin{equation}\label{eq7.2.2}
\WWWW=\{\uparrow_{i,j},\nuparrow_{i,j}:S_{i,j}\subset\PPP\}.
\end{equation}

The shortline of $V_0$ is a $1$-direction almost horizontal geodesic $H_0$ on $\PPP$, starting at the same vertex of $\PPP$ and with
slope $\alpha_1^{-1}$, where

so that $m\le\alpha_1<m+1$.
The geodesic $H_0$ is made up of almost horizontal units of type $\rightarrow_{i,j}$ and $\nrightarrow_{i,j}$, all with slope $\alpha_1^{-1}$.
Thus we may consider a vector space $W'$, with basis

The shortline of $H_0$ is the original $1$-direction almost vertical geodesic $V_0$ on $\PPP$, with slope $\alpha$.
Thus we return to the vector space $W$, with basis $\WWWW$.

Note that $V_0$ and $H_0$ are mutual shortlines.
Note the special property of $\alpha$, that its continued fraction expansion has period $2$.

Our task is to start with $\WWWW$ and identify the $2$-step ancestors in $\WWWW$ of each of the almost vertical units.
Let $\ANC$ denote one step of this ancestor process.

Step 1. We consider each of the almost vertical units in $\WWWW$, and determine its ancestors in $\WWWW'$ using the delete end rule, in the same spirit as in (<ref>).
This leads to a transition matrix $M_1$, where each row captures the information concerning the ancestors of a particular almost vertical unit, with the columns displaying the multiplicities of the individual ancestor almost horizontal units.
Indeed, $M_1^T$ is the matrix for this ancestor process $W\to W'$, where the coefficient vectors are taken as column vectors.
Note that $M_1^T$ is a $2d\times2d$ matrix.

Thus for every $i_0,j_0$ such that $S_{i_0,j_0}\subset\PPP$, using the delete end rule, we have the ancestor relationships

\begin{align}
\ANC(\{\uparrow_{i_0,j_0}\})
\cup\{\rightarrow_{i_0,j}:j\in J_{i_0}^*\}
\setminus\{\rightarrow_{i_0,j_0}\},
\label{eq7.2.3}
\\
\ANC(\{\nuparrow_{i_0,j_0}\})
\cup\{\rightarrow_{i_0,j}:j\in J_{i_0}^*\},
\label{eq7.2.4}
\end{align}

where the $*$ in $J_{i_0}^*$ denotes that each edge with indices $i_0,j$ with $j\in J_{i_0}$ is counted with multiplicity $m\vert J_{i_0}\vert^{-1}$, where $\vert J_{i_0}\vert$ denotes the number of distinct elements of the set $J_{i_0}$.
We also adopt the convention that unions and complements are taken with appropriate multiplicities.

Step 2. We consider each of the almost horizontal units in $\WWWW'$, and determine its ancestors in $\WWWW$ using the keep end rule, so not quite in the same spirit as in (<ref>).
This leads to a transition matrix $M_2$, where each row captures the information concerning the ancestors of a particular almost horizontal unit, with the columns displaying the multiplicities of the individual ancestor almost vertical units.
Indeed, $M_2^T$ is the matrix for this ancestor process $W'\to W$, where the coefficient vectors are taken as column vectors.
Note that $M_2^T$ is a $2d\times2d$ matrix.

Thus for every $i_1,j_1$ such that $S_{i_1,j_1}\subset\PPP$, using the keep end rule, we have the ancestor relationships

\begin{align}
\ANC(\{\rightarrow_{i_1,j_1}\})
\cup\{\uparrow_{i,j_1}:i\in I_{j_1}^*\}
\setminus\{\uparrow_{i_1,j_1}\},
\label{eq7.2.5}
\\
\ANC(\{\nrightarrow_{i_1,j_1}\})
\cup\{\uparrow_{i,j_1}:i\in I_{j_1}^*\},
\label{eq7.2.6}
\end{align}

where the $*$ in $I_{j_1}^*$ denotes that each edge of $I_{j_1}$ is counted with multiplicity $n\vert I_{j_1}\vert^{-1}$, where $\vert I_{j_1}\vert$ denotes the number of distinct elements of the set $I_{j_1}$.
We also adopt the convention that unions and complements are taken with appropriate multiplicities.

Step 3. Finally, we combine the two steps and end up with a $2$-step transition matrix $\bfA=(M_1M_2)^T$, of size $2d\times2d$, in the same spirit as in (<ref>).
Thus combining the ancestor relationships (<ref>)–(<ref>), we deduce that

\begin{align}
\ANC(\ANC(\{\uparrow_{i_0,j_0}\}))
\nonumber
\\
\{\nuparrow_{i_0,j_0}\}
\cup\{\uparrow_{i,j_0}:i\in I_{j_0}^*\}
\right)
\nonumber
\\
\cup\left(
\{\nuparrow_{i_0,j}:j\in J_{i_0}^*\}
\cup\{\uparrow_{i,j}:j\in J_{i_0}^*,i\in I_j^*\}
\setminus\{\uparrow_{i_0,j}:j\in J_{i_0}^*\}
\right)
\nonumber
\\
\setminus\left(
\{\nuparrow_{i_0,j_0}\}
\cup\{\uparrow_{i,j_0}:i\in I_{j_0}^*\}
\setminus\{\uparrow_{i_0,j_0}\}
\right)
\nonumber
\\
\cup\{\uparrow_{i,j}:j\in J_{i_0}^*,i\in I_j^*\}
\cup\{\nuparrow_{i_0,j}:j\in J_{i_0}^*\}
\setminus\{\uparrow_{i_0,j}:j\in J_{i_0}^*\},
\label{eq7.2.7}
\\
\ANC(\ANC(\{\nuparrow_{i_0,j_0}\}))
\nonumber
\\
\{\nuparrow_{i_0,j_0}\}
\cup\{\uparrow_{i,j_0}:i\in I_{j_0}^*\}
\right)
\nonumber
\\
\cup\left(
\{\nuparrow_{i_0,j}:j\in J_{i_0}^*\}
\cup\{\uparrow_{i,j}:j\in J_{i_0}^*,i\in I_j^*\}
\setminus\{\uparrow_{i_0,j}:j\in J_{i_0}^*\}
\right)
\nonumber
\\
=\{\nuparrow_{i_0,j_0}\}\cup\{\uparrow_{i,j}:j\in J_{i_0}^*,i\in I_j^*\}
\nonumber
\\
\cup\{\uparrow_{i,j_0}:i\in I_{j_0}^*\}
\cup\{\nuparrow_{i_0,j}:j\in J_{i_0}^*\}
\setminus\{\uparrow_{i_0,j}:j\in J_{i_0}^*\}.
\label{eq7.2.8}
\end{align}

We choose this particular convention regarding the delete end rule and keep end rule, as it makes the matrix reduction particularly simple.
For instance, if we use this particular convention for the surface $S_2$, then in the $2$-step transition matrix $\bfA$, the $14$ eigenvalues of absolute value $1$ are all the same and equal to $1$.
Indeed, we shall give a good reason later, when we discuss a simpler approach to the process, why they are all equal to $1$.

Suppose that $\lambda_1,\ldots,\lambda_s$ are the eigenvalues of the $2$-step transition matrix $\bfA$, with multiplicities $d_1,\ldots,d_s$ respectively, and where $\vert\lambda_1\vert\ge\ldots\ge\vert\lambda_s\vert$.
Then clearly $d_1+\ldots+d_s=2d$.
Furthermore, the space $\Cc^{2d}$ can be decomposed into a direct sum

where each $W_i$, $i=1,\ldots,s$, is an $\bfA$-invariant subspace of $\Cc^{2d}$ and also contains an eigenvector $\Psi_i$ corresponding to the
eigenvalue $\lambda_i$.
If $d_i=1$, then $\Psi_i$ generates $W_i$ and gives rise to a basis of $W_i$.
If $d_i>1$, then we can find a basis $\Psi_{i,j}$, $j=1,\ldots,d_i$, of $W_i$, with $\Psi_i=\Psi_{i,1}$.
Thus the collection

gives rise to a basis of $\Cc^{2d}$.

The almost vertical geodesic $V_0$ starts at a vertex of $\PPP$, and it starts with a finite succession of almost vertical units.
Let $\bfw_0$ denote the column coefficient vector of this finite collection of units with respect to the basis $\WWWW$.
Then we can find coefficients $c_{i,j}\in\Cc$, $i=1,\ldots,s$, $j=1,\ldots,d_i$, such that

\begin{equation}\label{eq7.2.9}
\bfw_0=\sum_{i=1}^s\sum_{j=1}^{d_i}c_{i,j}\Psi_{i,j}.
\end{equation}

Suppose that $\bfw_r=\bfA^r\bfw_0$.

\begin{equation}\label{eq7.2.10}
\bfw_r=\sum_{i=1}^s\sum_{j=1}^{d_i}c_{i,j}\bfA^r\Psi_{i,j}.
\end{equation}

We shall consider the special case where for every eigenvalue $\lambda_i$ of $\bfA$ with $\vert\lambda_i\vert>1$, the $\bfA$-invariant subspace $W_i$ has a basis consisting entirely of eigenvectors of $\bfA$ with eigenvalue $\lambda_i$.
Let $\lambda_i$, $i=1,\ldots,s_0$, be these eigenvalues.

are all eigenvectors of $\bfA$.
Furthermore, $\vert\lambda_i\vert\le1$, $i=s_0+1,\ldots,s$.

In this case, we have

\begin{equation}\label{eq7.2.11}
\bfw_r=\sum_{i=1}^{s_0}\sum_{j=1}^{d_i}c_{i,j}\lambda_i^r\Psi_{i,j}+O(r^{D-1}),
\end{equation}

where $D=\max\{d_1,\ldots,d_s\}$, and the error term $O(r^{D-1})$ gives an upper bound on the absolute value of the coordinates of the missing vectors.
To see this, note that the matrix $\bfA$ is similar to a matrix of the form

where a typical Jordan block $J=J(\tau,e)$ is an $e\times e$ matrix of the form

τ   1

⋱   ⋱

⋱   1

where every entry on the diagonal is equal to an eigenvalue $\tau$ of $\bfA$, every entry on the superdiagonal is equal to $1$, and every other entry is equal to $0$.

The contribution to the error term in (<ref>) comes from those eigenvalues of $\bfA$ with absolute value at most $1$.
These are related to Jordan blocks $J=J(\tau,e)$ with $\vert\tau\vert\le1$ and $e\le D$.
The matrix $J^r$ is $e\times e$ and upper-triangular, and the entry on row $i$ and column $j$, with $j\ge i$, is given by

where the binomial coefficient is equal to $0$ if $j-i>r$.
The bound $O(r^{D-1})$ follows on observing that

It is crucial to bring those eigenvalues with the largest absolute values into play.
The coefficient in (<ref>) of the eigenvector corresponding to the largest eigenvalue is non-zero, since the left hand side of (<ref>) contains terms of order of magnitude $\lambda_1^r$.
For the coefficient of the eigenvector corresponding to the second largest eigenvalue, this is less obvious.
The trick here is to start with an almost vertical unit from the chosen vertex.
If the coefficient corresponding to the second eigenvector is non-zero, then we have a good initial vector $\bfw_0$.
If the coefficient is zero, then we add the next unit, and keep on doing so but stop as soon as we get a non-zero coefficient.
Take this as our starting succession of almost vertical units, and use the corresponding good $\bfw_0$.
This process must stop after bounded time, depending only on $\PPP$ and the slope $\alpha$, as soon every basis element in $\WWWW$ will come into play.
Of the $2d$ types of almost vertical units, it is clear that at least one of them gives rise to a non-zero coefficient.

The above method in rather laborious.
Finding the eigenvalues and eigenvectors of the $2$-step transition matrix $\bfA$, of size $2d\times2d$, is not a very pleasant task.

It is clear that the expression (<ref>) is dominated by the terms arising from those eigenvalues $\lambda_i$, $i=1,\ldots,s$, with absolute values exceeding $1$.
We call these the relevant eigenvalues, as the remaining eigenvalues make little to no contribution.
We next seek a simpler way of finding these relevant eigenvalues and their corresponding eigenvectors.

For any set $\SSSS$ of almost vertical units in $\PPP$, counted with multiplicity, let $[\SSSS]$ denote the column coefficient vector of $\SSSS$ with respect to the
basis $\WWWW$.
Note that $[\SSSS]\in\Cc^{2d}$, where $d$ is the number of square faces of $\PPP$.

As the matrix $\bfA$ is the transition matrix of the $2$-step ancestor process with respect to the basis $\WWWW$, (<ref>) and
(<ref>) can be rewritten in the form

\begin{align}
=[\{\uparrow_{i,j}:j\in J_{i_0}^*,i\in I_j^*\}]
\nonumber
\\
+[\{\nuparrow_{i_0,j}:j\in J_{i_0}^*\}]
-[\{\uparrow_{i_0,j}:j\in J_{i_0}^*\}],
\label{eq7.2.12}
\\
=[\{\uparrow_{i,j}:j\in J_{i_0}^*,i\in I_j^*\}]
+[\{\uparrow_{i,j_0}:i\in I_{j_0}^*\}]
\nonumber
\\
+[\{\nuparrow_{i_0,j}:j\in J_{i_0}^*\}]
-[\{\uparrow_{i_0,j}:j\in J_{i_0}^*\}].
\label{eq7.2.13}
\end{align}

\begin{align}
\bfu_{i_0}
&=[\{\uparrow_{i,j}:j\in J_{i_0}^*,i\in I_j^*\}],
\label{eq7.2.14}
\\
\bfv_{i_0}
&=[\{\nuparrow_{i_0,j}:j\in J_{i_0}^*\}]-[\{\uparrow_{i_0,j}:j\in J_{i_0}^*\}],
\label{eq7.2.15}
\\
\bfz_{j_0}
&=[\{\uparrow_{i,j_0}:i\in I_{j_0}^*\}].
\label{eq7.2.16}
\end{align}

Then $\bfu_{i_0},\bfv_{i_0},\bfz_{j_0}\in\Cc^{2d}$, and (<ref>) and (<ref>) can be expressed in the form

\begin{align}
\label{eq7.2.17}
\\
\label{eq7.2.18}
\end{align}

Motivated by the equations (<ref>) and (<ref>), we consider $\image(\bfA-I)$, the image of the matrix $\bfA-I$ in $\Cc^{2d}$.
Note that this is an $\bfA$-invariant subspace of $\Cc^{2d}$.
Indeed, for every vector $\bfw\in\Cc^{2d}$, we have

where $\bfw'\in\Cc^{2d}$ satisfies $\bfw'=\bfA\bfw$.
It is clear from the equations (<ref>) and (<ref>) that the subspace $\image(\bfA-I)$ is generated by the elements of the form

\begin{equation}\label{eq7.2.19}
\bfu_{i_0},\bfv_{i_0},\bfz_{j_0}\in\image(\bfA-I)\subset\Cc^{2d}.
\end{equation}

All eigenvectors $\lambda\ne1$ of $\bfA$ belong to the subspace $\image(\bfA-I)\subset\Cc^{2d}$.

Suppose that $\bfw\in\Cc^{2d}$ satisfies $\bfA\bfw=\lambda\bfw$ with $\lambda\ne1$.

so that $\bfw\in\image(\bfA-I)$.

Let $\PPP$ be a finite polysquare translation surface.
Let $m$ be any fixed integer multiple of the lengths of the horizontal streets of $\PPP$, and let $n$ be any fixed integer multiple of the lengths of the vertical streets of $\PPP$.
Consider a $1$-direction almost vertical geodesic $V_0$ on $\PPP$, with slope $\alpha$ given by (<ref>).
Let $\bfA$ denote the transition matrix of the $2$-step ancestor process with respect to the basis $\WWWW$, where $\WWWW$ is given by (<ref>).
Then the following hold:

For any two distinct almost vertical units $\nuparrow_{i_0,j_1}$ in a square face $S_{i_0,j_1}$ and $\nuparrow_{i_0,j_2}$ in a square face $S_{i_0,j_2}$ on the same horizontal street of $\PPP$ and every eigenvector corresponding to a relevant eigenvalue of $\bfA$, the two entries in the eigenvector corresponding to $\nuparrow_{i_0,j_1}$ and $\nuparrow_{i_0,j_2}$ are equal.

For any two distinct pairs of almost vertical units $\uparrow_{i_1,j_0},\nuparrow_{i_1,j_0}$ in a square face $S_{i_1,j_0}$ and
$\uparrow_{i_2,j_0},\nuparrow_{i_2,j_0}$ in a square face $S_{i_2,j_0}$ on the same vertical street of $\PPP$ and every eigenvector corresponding to a relevant eigenvalue of $\bfA$, the sum of the two entries in the eigenvector corresponding to $\uparrow_{i_1,j_0},\nuparrow_{i_1,j_0}$ in the square face $S_{i_1,j_0}$ is equal to the sum of the two entries in the eigenvector corresponding to $\uparrow_{i_2,j_0},\nuparrow_{i_2,j_0}$ in the square face $S_{i_2,j_0}$.

In view of Lemma <ref>, it is enough to concentrate our attention on the subspace $\image(\bfA-I)$.
Since this subspace is generated by the vectors in (<ref>), it suffices to check that these generating vectors all satisfy (i) and (ii).
Our argument is reduced to a simple case study.

Note that $\bfu_{i_0}$ does not involve units of type $\nuparrow_{i,j}$, so (i) holds by default.
On the other hand, $\bfu_{i_0}$ satisfies (ii), due to cyclic vertical symmetry.
A similar argument applies to $\bfz_{j_0}$.

On the other hand, $\bfv_{i_0}$ satisfies (i), due to cyclic horizontal symmetry.
Clearly it also satisfies (ii), as the count for each $\nuparrow_{i_0,j}$ cancels the count for $\uparrow_{i_0,j}$.

This completes the proof of the edge cutting lemma.

Let $h=h(\PPP)$ and $v=v(\PPP)$ denote respectively the number of horizontal and vertical streets in $\PPP$.

Let $\VVV$ be the subspace of $\Cc^{2d}$ generated by $\bfu_i,\bfv_i$, $1\le i\le h$, defined in (<ref>) and (<ref>), so that the dimension of $\VVV$ is at most $2h$.

We have $\image((\bfA-I)^2)\subset\VVV$.

Since $\image((\bfA-I)^2)=(\bfA-I)(\image(\bfA-I))$, and the subspace ${\rm Im}(\bfA-I)$ is generated by the vectors in (<ref>), it suffices to show that

\begin{equation}\label{eq7.2.20}
\quad
\quad
\end{equation}

From (<ref>) and (<ref>), it is clear that both $\bfu_{i_0}$ and $\bfz_{j_0}$ involve only the almost vertical units of various types
$\uparrow_{i,j}$, so it follows from (<ref>) that the first and third statements in (<ref>) hold.
On the other hand, it follows from (<ref>)–(<ref>) that

\begin{equation}\label{eq7.2.21}
=\sum_{j\in J_{i_0}^*}\bfz_j
=[\{\uparrow_{i,j}:i\in I_j^*,j\in J_{i_0}^*\}]
\end{equation}

and the second statement in (<ref>) follows immediately.

By definition $\VVV$ is a subspace of $\image(\bfA-I)$.
Thus the following lemma is an extension of Lemma <ref>.

The vector space $\VVV$ is an $\bfA$-invariant subspace of $\Cc^{2d}$, and contains the eigenvectors corresponding to each of the eigenvalues
$\lambda\ne1$ of $\bfA$.

To prove that $\VVV$ is $\bfA$-invariant, it suffices to check that

_i∈    _i∈.

This is trivial, since we know from (<ref>) that $(\bfA-I)\bfu_i\in\VVV$ and $(\bfA-I)\bfv_i\in\VVV$.
Next suppose that $\bfw\in\Cc^{2d}$ satisfies $\bfA\bfw=\lambda\bfw$ for some $\lambda\ne1$.

so that $\bfw\in\image((\bfA-I)^2)\subset\VVV$,
where in the last step we have used Lemma <ref>.

Next we study how the matrix $\bfA$ acts on $\VVV$.
First of all, the equation (<ref>) implies

\begin{equation}\label{eq7.2.22}
\bfA\bfv_i=\bfu_i+\bfv_i.
\end{equation}

We already know that $\bfA\bfu_i\in\VVV$, so that $\bfA\bfu_i$ is a linear combination of the vectors $\bfu_i,\bfv_i$, $1\le i\le h$.
Next we shall find these coefficients explicitly, but this process is much more complicated than (<ref>), as the coefficients depend heavily on the combinatorial arrangement of the square faces of the given polysquare translation surface $\PPP$.
In fact, they depend on the given triple $(\PPP;m,n)$.
This will eventually lead us to the desired $h\times h$ street-spreading matrix $\bfS=\bfS(\PPP;m,n)$ which contains all the relevant information.

Combining (<ref>) and (<ref>), one can determine the desired coefficients by using a particular algorithm.
The abstract/formal definition of this algorithm in the general case is somewhat complicated.
As with most algorithms, the best way to learn it is to study a few concrete examples with figures.
We strongly recommend the reader to study the next two examples, after which the abstract/formal definition becomes almost self-explanatory.

The algorithm can easily be summarized in a nutshell: spread the horizontal streets vertically.
We illustrate part of the recipe by two examples.

Consider again the polysquare translation surface $S_2$ in Figure 7.1.1, and consider a $1$-direction geodesic starting from some vertex of $S_2$ with slope $\alpha$ given by (<ref>) with $m=n=3$.

We shall number the horizontal streets from top to bottom, and the vertical streets from left to right.

Consider the first horizontal street at the top, corresponding to $\bfu_1$.
We now use (<ref>), and since $J_1^*=\{2,3,4\}$, it follows that $\bfu_1$ concerns almost vertical units of type $\uparrow$ on the vertical streets $2,3,4$.
In other words, we start with the first horizontal street, and spread along every vertical street that intersects this horizontal street.
The picture on the left in Figure 7.2.2 shows a listing of all the almost vertical units of type $\uparrow$ along these vertical streets.
Applying (<ref>), we have

\begin{align}\label{eq7.2.23}
\nonumber
\\
\nonumber
\\
\end{align}

For the second horizontal street, corresponding to $\bfu_2$, highlighted in the picture on the right in Figure 7.2.2, we have $J_2^*=\{1,2,4\}$, and a similar argument gives

\begin{align}\label{eq7.2.24}
\nonumber
\\
\nonumber
\\
\end{align}

The multiple $3$ for $\uparrow_{2,1}$ in the square face $S_{2,1}$ comes from the fact that $I_1^*=\{2,2,2\}$, since $n=3$ but the vertical street has length $1$.

For the third horizontal street, corresponding to $\bfu_3$, highlighted in the picture on the left in Figure 7.2.3, we have $J_3^*=\{2,3,4\}$, and a similar argument gives

\begin{align}\label{eq7.2.25}
\nonumber
\\
\nonumber
\\
\end{align}

For the fourth horizontal street, corresponding to $\bfu_4$, highlighted in the picture on the right in Figure 7.2.3, we have $J_4^*=\{3,3,3\}$, and a similar argument gives

\begin{align}\label{eq7.2.26}
\nonumber
\\
\nonumber
\\
\end{align}

We now define the street-spreading matrix by

\begin{equation}\label{eq7.2.27}
\bfS=\begin{pmatrix}
\end{pmatrix},
\end{equation}

where the $j$-th column comes from the coefficients of $(\bfu_i+\bfv_i)$, $i=1,2,3,4$, in the expression for $(\bfA-I)\bfu_j$, $j=1,2,3,4$, as given by

Combining (<ref>)–(<ref>), we obtain

\begin{align}
\bfA\bfu_1
\nonumber
\\
\bfA\bfu_2
\nonumber
\\
\bfA\bfu_3
\nonumber
\\
\bfA\bfu_4
\nonumber
\\
\bfA\bfv_1
\nonumber
\\
\bfA\bfv_2
\nonumber
\\
\bfA\bfv_3
\nonumber
\\
\bfA\bfv_4
\nonumber
\end{align}

For any $\bfw\in\VVV$, if we write

4   2   3   3   1   0   0   0

2   6   2   0   0   1   0   0

3   2   4   3   0   0   1   0

1   0   1   4   0   0   0   1

3   2   3   3   1   0   0   0

2   5   2   0   0   1   0   0

3   2   3   3   0   0   1   0

1   0   1   3   0   0   0   1

so the $8\times8$ matrix in question is

+I   I

We shall return to this example later.

Consider the polysquare translation surface $S_3$, given in Figure 7.2.4.

This has $6$ horizontal streets of lengths $3,3,3,4,6,6$, and $9$ vertical streets of lengths $1,3,3,4,2,2,4,3,3$, as illustrated in Figure 7.2.5 where, for instance, $\leftrightarrow4$ and $\updownarrow6$ in a square face indicates that this square face is part of the $4$-th horizontal street and the $6$-th vertical street.

Consider a $1$-direction geodesic starting from some vertex of $S_3$ with slope $\alpha$ given by (<ref>) with $m=n=12$.

The polysquare translation surface $S_3$ has $25$ square faces, so any $2$-step transition matrix $\bfA$ has size $50\times50$, making it an onerous task to find any eigenvalue, even with the help of MATLAB.
It is clearly much simpler to find the $6\times6$ street-spreading matrix.

First of all, we define the vectors $\bfu_i,\bfv_i$, $i=1,2,3,4,5,6$, according to (<ref>) and (<ref>).

Consider the horizontal street corresponding to $\bfu_1$, as highlighted in the left half of Figure 7.2.6.
We see that

since $m=12$, and $\bfu_1$ concerns almost vertical units on the vertical streets $2,3,4$.
In other words, we start with the first horizontal street, and spread along every vertical street that intersects this horizontal street.
The left half of Figure 7.2.6 shows a listing of all the almost vertical units of type $\uparrow$ along these vertical streets.
Applying (<ref>), we have

\begin{align}\label{eq7.2.28}
\nonumber
\\
\nonumber
\\
\nonumber
\\
\nonumber
\\
\end{align}

We need to explain the multiples in Figure 7.2.6.
Observe the horizontal street corresponding to $\bfu_1$ consists of $3$ square faces.
As $m=12$, the correct multiplicity is $12/3=4$.
On the other hand, the vertical street containing the square face $S_{1,4}$ consists of $4$ square faces.
As $n=12$, the correct multiplicity is $12/4=3$.
Thus the multiple for that square face is $3\times4$.

We always have $\vert I_j^*\vert=n$ for every $j=1,\ldots,v$, and $\vert J_i^*\vert=m$ for every $i=1,\ldots,h$.
The first of these corresponds to the fact that every almost vertical detour crossing of $V_0$ travels a vertical distance between $n$ and $n+1$.
The second of these corresponds to the fact that every almost horizontal detour crossing of $H_0$ travels a horizontal distance between $m$ and $m+1$.

With $J_2^*=\{7,7,7,7,8,8,8,8,9,9,9,9\}$, for the horizontal street corresponding to $\bfu_2$, as highlighted in the right half of Figure 7.2.6, a similar argument gives

\begin{align}\label{eq7.2.29}
\nonumber
\\
\end{align}

With $J_3^*=\{1,1,1,1,2,2,2,2,9,9,9,9\}$, for the horizontal street corresponding to $\bfu_3$, as highlighted in Figure 7.2.7, a similar argument gives

\begin{equation}\label{eq7.2.30}
\end{equation}

With $J_4^*=\{4,4,4,5,5,5,6,6,6,7,7,7\}$, for the horizontal street corresponding to $\bfu_4$, as highlighted in Figure 7.2.8, a similar argument gives

\begin{align}\label{eq7.2.31}
\nonumber
\\
\end{align}

With $J_5^*=\{2,2,3,3,4,4,7,7,8,8,9,9\}$, for the horizontal street corresponding to $\bfu_5$, as highlighted in Figure 7.2.9, a similar argument gives

\begin{align}\label{eq7.2.32}
\nonumber
\\
\end{align}

With $J_6^*=\{3,3,4,4,5,5,6,6,7,7,8,8\}$, for the horizontal street corresponding to $\bfu_6$, as highlighted in Figure 7.2.10, a similar argument gives

\begin{align}\label{eq7.2.33}
\nonumber
\\
\end{align}

We now define the street-spreading matrix by

44   0   16   9   22   14

0   44   16   9   22   14

16   16   80   0   16   0

12   12   0   54   12   36

44   44   32   18   44   28

28   28   0   54   28   52

where the $j$-th column comes from the coefficients of $(\bfu_i+\bfv_i)$, $i=1,2,3,4,5,6$, in the expression for $(\bfA-I)\bfu_j$, $j=1,2,3,4,5,6$, as given by (<ref>)–(<ref>).

Let $\PPP$ be a finite polysquare translation surface with $d$ square faces.
Assume that $h\le v$, where $h$ is the number of horizontal streets in $\PPP$ and $v$ is the number of vertical streets in $\PPP$.
Let $m$ be any fixed integer multiple of the lengths of the horizontal streets of $\PPP$, and let $n$ be any fixed integer multiple of the lengths of the vertical streets of $\PPP$.

Consider a $1$-direction almost vertical geodesic $V_0$ on $\PPP$, with slope $\alpha$ given by (<ref>).
Let $\bfA$ denote the transition matrix of the $2$-step ancestor process with respect to the basis $\WWWW$, where $\WWWW$ is given by
(<ref>) .
Then all the relevant eigenvalues of $\bfA$ and their corresponding eigenvectors can be determined as follows:

Let $\VVV$ denote the subspace of $\Cc^{2d}$ generated by $\bfu_i,\bfv_i$, $i=1,\ldots,h$, given by (<ref>) and (<ref>).
Then $\VVV$ is an $\bfA$-invariant subspace of $\Cc^{2d}$, and contains the eigenvectors corresponding to all the relevant eigenvalues of $\bfA$.
Furthermore, there exist non-negative integers $s_{i,j}$, $i,j=1,\ldots,h$, such that for every $j=1,\ldots,h$,

\begin{equation}\label{eq7.2.34}
\end{equation}

\begin{equation}\label{eq7.2.35}
\bfA\bfv_j=\bfu_j+\bfv_j.
\end{equation}

Consider the street-spreading matrix

\begin{equation}\label{eq7.2.36}
\bfS=\begin{pmatrix}
\vdots&&\vdots\\
\end{pmatrix},
\end{equation}

as well as the matrix

\begin{equation}\label{eq7.2.37}
\bfA\vert_\VVV=\begin{pmatrix}
\bfS+I&I\\
\bfS&I
\end{pmatrix}.
\end{equation}

Suppose that $\tau$ is an eigenvalue of $\bfS$ with eigenvector $\psi$, so that $\bfS\psi=\tau\psi$.

\begin{equation}\label{eq7.2.38}
\lambda(\tau;\pm)=1+\frac{\tau\pm\sqrt{\tau^2+4\tau}}{2}
\end{equation}

are two eigenvalues of $\bfA\vert_\VVV$ with product equal to $1$, with corresponding eigenvectors

\begin{equation}\label{eq7.2.39}
\Psi(\tau;\pm)=\left(\psi,\frac{-\tau\pm\sqrt{\tau^2+4\tau}}{2}\psi\right)^T\in\Cc^{2h}.
\end{equation}

In particular, each eigenvalue of $\bfA\vert_\VVV$ comes from an eigenvalue of $\bfS$.

If $\Psi(\tau;\pm)=(a_1,a_2,a_3,a_4,b_1,b_2,b_3,b_4)^T$ is an eigenvector corresponding to an eigenvalue $\lambda(\tau;\pm)$ of $\bfA\vert_\VVV$, then $\lambda(\tau;\pm)$ is an eigenvalue of $\bfA$ with eigenvector

\begin{equation}\label{eq7.2.40}
\bfw=\sum_{i=1}^ha_i\bfu_i+\sum_{i=1}^hb_i\bfv_i.
\end{equation}

The assumption that the number $h=h(\PPP)$ of horizontal streets is less than or equal to the number $v=v(\PPP)$ of vertical streets, i.e., $h\le v$, is one for convenience, as the street-spreading matrix is smaller.
For any polysquare translation surface $\PPP$ where $h\ge v$, we can always interchange the horizontal and vertical directions.

In view of Lemma <ref>, it remains to establish (<ref>) and (<ref>).
For (<ref>), note from (<ref>) that each $\bfu_j$ involves only units of type $\uparrow$, and so (<ref>) follows immediately from (<ref>), on observing that $\bfu_i,\bfv_i$, $i=1,\ldots,h$, generate $\VVV$.
On the other hand, (<ref>) follows immediately from (<ref>).

Suppose that $\Psi$ is a column eigenvector corresponding to an eigenvalue $\lambda$ of $\bfA\vert_\VVV$.
From (<ref>), it is clear that $\bfA\vert_\VVV$ can be reduced to the matrix

I   0

by elementary row operations.
The determinant of this matrix is clearly equal to $1$.
It follows that $\det(\bfA\vert_\VVV)=1$, and so $\lambda\ne0$.
We can write

\begin{equation}\label{eq7.2.41}
\Psi=(\psi,\psi^*)^T,
\end{equation}

where $\psi$ and $\psi^*$ are both column vectors with $h$ entries.

+I   I

and so

\begin{align}\label{eq7.2.42}
\bfS\psi+\psi+\psi^*&
\nonumber
\\
\bfS\psi+\psi^*&
\end{align}

Subtracting the second equation in (<ref>) from the first, we obtain $\psi=\lambda(\psi-\psi^*)$, and so

\begin{equation}\label{eq7.2.43}
\psi^*=\frac{\lambda-1}{\lambda}\psi.
\end{equation}

Substituting this into the first equation in (<ref>), we obtain

so that

is an eigenvalue of $\bfS$ with eigenvector $\psi$.
This is equivalent to

\begin{equation}\label{eq7.2.44}
\lambda=1+\frac{\tau\pm\sqrt{\tau^2+4\tau}}{2}.
\end{equation}

This proves that any eigenvalue $\lambda$ of $\bfA\vert_\VVV$ is obtained from an eigenvalue $\tau$ of $\bfS$ by (<ref>), and establishes (<ref>).
Finally, note that

In view of (<ref>) and (<ref>), this establishes (<ref>).

Suppose that $(a_1,\ldots,a_h,b_1,\ldots,b_h)^T$ is an eigenvector corresponding to an eigenvalue $\lambda$ of $\bfA\vert_\VVV$.
Let the vector $\bfw$ be given by (<ref>).

\begin{equation}\label{eq7.2.45}
\bfA\bfw=\sum_{i=1}^hc_i\bfu_i+\sum_{i=1}^hd_i\bfv_i,
\end{equation}

where the coefficients are related by

It is clear that

and so it follows from (<ref>) and (<ref>) that $\bfA\bfw=\lambda\bfw$, so that $\lambda$ is an eigenvalue of $\bfA$ with
eigenvector $\bfw$.

As commented earlier, if two or more distinct square faces of a polysquare translation surface lie on the same horizontal street and the same vertical street simultaneously, then the notation $S_{i,j}$ does not allow us to distinguish between two distinct square faces that lie on the $i$-th horizontal street and the $j$-th vertical street.
In this case, we need to denote each square face by $S_\delta$, where $\delta=1,\ldots,d$.

We see from (<ref>) as well as Examples <ref> and <ref> that the crux of our argument concerns expressions of the form

so we need to find suitable generalizations of the expressions (<ref>)–(<ref>).

For instance, let $\HS_{i_0}$ denote the $i_0$-th horizontal street.
For any $\delta=1,\ldots,d$, let $\HS(S_\delta)$ and $\VS(S_\delta)$ denote respectively the horizontal street and the vertical street that contains the square face $S_\delta$.
Then the expression (<ref>) can be rewritten in the form

\begin{align}
=[\{\uparrow_\delta:S_\delta\subseteq\VS^*(S_\eta)\mbox{ and }S_\eta\subseteq\HS^*(S_{\delta_0})\}]
\nonumber
\\
\nonumber
\end{align}

while the expression (<ref>) can then be rewritten in the form

In both instances, we adopt the convention that every $S_\eta$ is counted and $*$ denotes counting with the appropriate multiplicity.
We also need analogous generalizations of the other identities.

We make use of this more general version in the study of one-street polysquare translation surfaces later in this section.
We also make further use of this general version in Section <ref> where we study geodesic flow on the regular octagon surface, the double-pentagon surface, polyrhombus surfaces and the infinite-halving staircase surface; see Figures 8.2.5, 8.5.2, 8.6.9 and 8.9.2 respectively.

Recall from (<ref>), (<ref>) and (<ref>) that the relevant eigenvalues of $\bfA(S_2)$ are

obtained by tedious calculation using MATLAB.
Let us instead retrieve the same information using the street-spreading matrix $\bfS$ given by (<ref>).

The eigenvalues of $\bfS$ are $\tau_1=9$, $\tau_2=4$, $\tau_3=1$ and $\tau_4=0$.
Using (<ref>), the corresponding eigenvalues of $\bfA$ are

Note that in view of Lemma <ref>, $\lambda(9;\pm)$, $\lambda(4;\pm)$ and $\lambda(1;\pm)$ are the only eigenvalues of $\bfA$ that are not equal to $1$.
Thus the multiplicity of the eigenvalue $1$ is $14$, as we have claimed earlier in the Remark following (<ref>).

Consider again the L-surface, given in Figure 7.1.6, and consider a $1$-direction geodesic starting from some vertex of the L-surface with slope
$\alpha$ given by (<ref>) with $m=n=2$.
It is easy to see from Figure 7.2.11 that

\begin{align}
\label{eq7.2.46}
\\
\label{eq7.2.47}
\end{align}

It follows from (<ref>) and (<ref>) that the street-spreading matrix is given by

2   1

2   3

with eigenvalues $\tau_1=4$ and $\tau_2=1$.
Using (<ref>) leads to the eigenvalues

including those given in (<ref>) with absolute values greater than $1$.

It is not difficult to show that geodesic flow on the surface of the unit cube is equivalent to $1$-direction geodesic flow on the polysquare translation surface $\PPP$ as shown in Figure 7.2.12 that represents the $4$-copy version of the cube surface.

Note that $\PPP$ has $6$ horizontal streets and $6$ vertical streets, all of length $4$, as shown in Figure 7.2.13, where, for instance,
$\leftrightarrow4$ and $\updownarrow6$ in a square face indicates that this square face is part of the $4$-th horizontal street and $6$-th vertical street.

Figures 7.2.12 and 7.2.13 are not very convenient for us to visualize the horizontal and vertical streets clearly.
We can re-draw these two figures, and attempt to arrange the square faces in an array where each row represents a horizontal street and each column represents a vertical street.

In Figure 7.2.14, we do precisely this, where for $i,j=1,\ldots,6$, the $i$-th row represents the $i$-th horizontal street and the $j$-th column represents the $j$-th vertical street as shown in Figure 7.2.13.
Note that this exercise is somewhat cumbersome and requires very careful edge identifications as illustrated.
We include the details here for the sake of completeness, and comment that the edge identifications are not required in the process for the determination of the street-spreading matrix.

Consider a $1$-direction geodesic on the polysquare translation surface $\PPP$ starting from some vertex of $\PPP$ with slope $\alpha$ given by (<ref>) with $m=n=4$.

The surface $\PPP$ has $24$ square faces, so any $2$-step transition matrix $\bfA$ has size $48\times48$.
It is much simpler to find the $6\times6$ street-spreading matrix.

As before, we define the matrices $\bfu_i,\bfv_i$, $i=1,\ldots,6$, according to (<ref>) and (<ref>).

Consider the horizontal street corresponding to $\bfu_1$, and also the horizontal street corresponding to $\bfu_4$, as highlighted in the picture on the left in Figure 7.2.15.

We see that $J_1^*=J_4^*=\{2,3,5,6\}$, and both $\bfu_1$ and $\bfu_4$ concern almost vertical units on the vertical streets $2,3,5,6$.
Applying (<ref>), we have

\begin{align}\label{eq7.2.48}
\nonumber
\\
\nonumber
\\
\nonumber
\\
\nonumber
\\
\nonumber
\\
\end{align}

With $J_2^*=J_5^*=\{1,2,4,6\}$, for the horizontal street corresponding to $\bfu_2$, and also the horizontal street corresponding to $\bfu_5$, as highlighted in the picture on the right in Figure 7.2.15, a similar argument gives

\begin{align}\label{eq7.2.49}
\nonumber
\\
\nonumber
\\
\end{align}

With $J_3^*=J_6^*=\{1,3,4,5\}$, for the horizontal street corresponding to $\bfu_3$, and also the horizontal street corresponding to $\bfu_6$, as highlighted in Figure 7.2.16, a similar argument gives

\begin{align}\label{eq7.2.50}
\nonumber
\\
\nonumber
\\
\end{align}

It follows from (<ref>)–(<ref>) that the street-spreading matrix is given by

4   2   2   4   2   2

2   4   2   2   4   2

2   2   4   2   2   4

4   2   2   4   2   2

2   4   2   2   4   2

2   2   4   2   2   4

with eigenvalues $\tau_1=16$, $\tau_2=\tau_3=4$ and $\tau_4=\tau_5=\tau_6=0$.
Using (<ref>), the corresponding eigenvalues of $\bfA$ are

with the appropriate multiplicities.
The two largest eigenvalues of $\bfA$ are therefore

Furthermore, the multiplicity of the eigenvalue $1$ is $42$.

Note that

exhibit digit-halving in their continued fraction expansions, and are also the two largest eigenvalues of the $1$-step transition matrix of a
$1$-direction geodesic of slope $\alpha=2+\sqrt{5}$ on the L-surface; see <cit.>.

We comment that digit-halving is still exhibited if $\alpha$ is defined by (<ref>) with $m=n=4k$ for any positive integers $k$.
In this general case, the largest eigenvalue is the square of $[4k;4k,4k,4k,\ldots]$, and the second largest eigenvalue is the square of

We complete this section by considering two consequences of Theorem <ref> and its more general form.
The first one is fairly straightforward.

The determinant of the $2$-step transition matrix $\bfA$ is equal to $1$.

We have shown earlier that $\det(\bfA\vert_\VVV)=1$.

Let $p(x)$ be the characteristic polynomial of $\bfA\vert_\VVV$.
In view of Theorem <ref>, it is clear that the characteristic polynomial of $\bfA$ is equal to $p(x)q(x)$, where $q(x)$ has degree $2d-2h$ and its roots are all the eigenvalues of $\bfA$ that are not eigenvalues of $\bfA\vert_\VVV$.
In view of Lemma <ref>, we have $q(x)=(x-1)^{2d-2h}$.
On the other hand, it follows from (<ref>) that

where $\tau_1,\ldots,\tau_h$ are the eigenvalues of $\bfS$, and $\lambda(\tau_i;+)\lambda(\tau_i;-)=1$, $i=1,\ldots,h$.

Clearly the determinant of $\bfA$ is equal to $p(0)q(0)=1$.

To motivate the second consequence, we start with the L-surface, and obtain a new polysquare translation surface with $6$ smaller square faces, each of area $1/2$.
The trick is to draw two diagonals on each of the $3$ square faces of the L-surface.
We use the boundary pairing in the picture on the right in Figure 7.2.17, which gives the so-called diagonal subdivision surface of the L-surface, or DS-L-surface.
Of course the genus remains the same; namely, $2$.
Rotating this picture on the right anticlockwise by $45$ degrees and resizing, the DS-L-surface becomes a polysquare translation surface in the standard horizontal-vertical position.
It is easy to see that the DS-L-surface has only one horizontal street, which has length $6$ and is defined by the cycle $1,2,3,4,6,5$.
It also has only one vertical street, defined by the cycle $1,5,6,2,3,4$.
So the street size multiple of the DS-L-surface is equal to $6$.

Now <cit.> says that, applying the surplus shortline method to this particular one-street translation surface, we obtain infinitely many explicit slopes with the property that a $1$-direction geodesic with any such slope is superuniform.
However, this theorem is obtained by using <cit.>, the main result for the L-surface in that paper, with an exceptionally long proof.
So the mysterious connection between a concrete one-street translation surface and superuniformity there may seem at first sight a very deep, and perhaps accidental, result.
However, nothing is further from the truth.

If a polysquare translation surface has only one horizontal street, or only one vertical street, then we call it a one-street polysquare translation surface.

A one-street polysquare translation surface does not in general satisfy the condition that if a horizontal street and a vertical street intersect, then the intersection is a unique square face.
Thus we cannot directly apply Theorem <ref> as stated.
However, there is a general connection between such surfaces and superuniformity, and it is an extremely easy consequence of the general version of Theorem <ref>.

Indeed, we simply need the condition that the polysquare translation surface has only one horizontal street (or one vertical street), but no restriction about the other direction.
Then the street-spreading matrix $\bfS$ is a $1\times1$ matrix which gives rise to the largest eigenvalue of the $2$-step transition matrix $\bfA$.
It follows that the second largest eigenvalue is irrelevant, which implies that the surplus shortline method provides explicit quadratic irrational slopes for which the irregularity exponent is equal to $0$, so that the corresponding geodesic is superuniform.

We summarize our discussion as follows.

For every one-street polysquare translation surface, there exist infinitely many explicit quadratic irrational slopes such that every $1$-direction geodesic with any such slope is superuniform.

We conclude this section by demonstrating that one-street polysquare translation surfaces are not rare.
The idea is based on a a far-reaching generalization of the DS-L-surface.

Recall the $2$-polysquare snake surface in Figure 7.1.4.
Snake surfaces are not restricted to $2$-polysquare translation surfaces, as illustrated in Figure 7.2.18.

More precisely, let $\PPP$ be a finite polysquare translation surface.
We put a point at the center of every square face of $\PPP$, and call it the capital of the square face.
Consider the neighbor graph of $\PPP$, where the vertices are the capitals, and where two capitals are joined by an edge of the graph if the corresponding square faces share a common edge.
If the neighbor graph of $\PPP$ is a path, with no branching, then $\PPP$ is called a snake surface, as shown in Figure 7.2.18.
If the neighbor graph of $\PPP$ is a tree, then $\PPP$ is called a polysquare-tree translation surface, as shown in Figure 7.2.18 and 7.2.19.

Given any graph, the number of edges at a vertex is called the degree of the vertex.
In a path, the degree of every vertex is $2$, except at the two end-vertices where the degree is $1$.
In a tree we may have higher degrees.
For example, the neighbor tree of the polysquare-tree translation surface in Figure 7.2.19 has two vertices of degree $3$.

One-street polysquare translation surfaces are not rare, because the diagonal subdivision of every polysquare-tree translation surface has only one street.

The proof goes by induction on the number of square faces in the polysquare.
The inductive step is based on the elementary result in graph theory that every tree has a vertex of degree $1$.
Indeed, it is easy to see that a tree with $n$ vertices has precisely $n-1$ edges, and that the sum of the degrees is twice the number of edges.
Since the sum of the degrees is $2(n-1)$, this implies that there must be a vertex with degree less than $2$, and so must be equal to $1$.

In terms of the polysquare-tree region, this result in graph theory means that there is always a square face with $3$ sides not shared with neighboring square faces.
Let us call them end square faces.
We have highlighted these end square faces in Figures 7.2.18 and 7.2.19.

Figure 7.2.20 shows the smaller polysquare-tree translation surface obtained from the polysquare-tree translation surface in Figure 7.2.19 by removing one of the end square faces.
By the induction hypothesis, the diagonal subdivision of this smaller polysquare-tree translation surface is a one-street translation surface, labelled by the numbers $1$ to $38$.

Note that we carefully start the labelling from a common edge shared with the removed end square face, and keep moving in the north-east direction.
Adding back the removed end square face, Figure 7.2.21 shows an explicit one-street in the original polysquare-street translation surface, where the consecutive squares along the street are labelled with the consecutive integers from $1$ to $40$.

Note that the labelling in Figure 7.2.21 is obtained as a trivial extension of the labelling in Figure 7.2.20 by including $39$ and $40$.
This completes the inductive step.

We cannot entirely drop the tree condition for the neighbor graph, as cycles may spoil the argument.
Consider the two polysquare cycles in Figure 7.2.22.

Neither of these polysquare-cycles leads to a one-street diagonal decomposition surface, as can be shown in Figure 7.2.23.

In the picture on the left, we have a tilted street of length $8$, covering only half the tilted square faces of the polysquare.
In the picture on the right, we have a tilted street of length $4$, again covering only half the tilted square faces of the polysquare.

§.§ Application of Theorem <ref>: super-fast escape to infinity

Our goal in this section is to find explicit quadratic irrational slopes which give rise to orbits that exhibit super-fast escape rate to infinity.
More precisely, given any $\eps>0$, we shall find explicit quadratic irrational slopes such that for any orbit with any such slope, there exists an infinite sequence $T_n$, $n\ge1$, such that $T_n\to\infty$ as $n\to\infty$, and the diameter of the initial segment of length $T_n$ of the orbit exceeds $c_0T_n^{1-\eps}$, where $c_0$ is an absolute constant depending only on the given special slope.

Billiard orbits and geodesics in integrable polysquare systems behave in the same way for all quadratic irrational slopes.
In complete contrast, billiard orbits and geodesics in infinite non-integrable polysquare systems may behave in completely different ways for different quadratic irrational slopes.
We may therefore have the full spectrum from super-slow to super-fast escape rate to infinity.

We return to the problem of billiard in the $\infty$-L-strip region shown in Figure 7.3.1.
In <cit.>, it is shown that there are infinitely many quadratic irrational slopes such that a billiard orbit starting from a corner point with such a slope is dense on the $\infty$-L-strip region and also exhibits super-slow logarithmic escape rate to infinity.
In particular, any initial segment of such an orbit of length $T$, $T\ge2$, has distance at most $c_0\log T$ from the starting point, where $c_0$ is an absolute constant depending only on the given special slope.

As illustrated in Figure 7.3.2, the L-shape is the building block of the $\infty$-L-strip region.

The trick of unfolding reduces the problem of billiard on the $\infty$-L-strip region, a $4$-direction flow, to the problem of a $1$-direction geodesic flow on an appropriate infinite polysquare translation surface.
We call this surface the translation surface for $\infty$-L-strip billiard, and denote it by $\Bil(\infty;1)$.

To describe $\Bil(\infty;1)$, we first unfold each of the L-shapes $L_i$ by reflecting it horizontally and vertically to obtain $4$ reflected copies and then put them together to form a $4$-copy-$L_i$.
We then obtain $\Bil(\infty;1)$ by gluing together the infinitely many copies of these $4$-copy-$L_i$.

Clearly there is billiard flow from each L-shape $L_i$ to its two immediate neighbours $L_{i-1}$ and $L_{i+1}$.
We therefore need to identify corresponding edges of these $4$-copy versions very carefully.
A simple examination will convince the reader that the edge identifications can be as illustrated in Figure 7.3.3.
Note that a $1$-direction geodesic on $\Bil(\infty;1)$ that goes from the vertical edge $v_2^{(i+1)}$ to the vertical edge $v_2^{(i)}$ corresponds to a billiard path going from $L_{i+1}$ to $L_i$, whereas a $1$-direction geodesic on $\Bil(\infty;1)$ that goes from the vertical edge $v_3^{(i-1)}$ to the vertical edge $v_3^{(i)}$ corresponds to a billiard path going from $L_i$ to $L_{i+1}$.

Thus the motion of the infinite L-strip billiard can be described in terms of $1$-direction geodesic flow on a finite polysquare translation surface $\PPP$, the period-surface of $\Bil(\infty;1)$, illustrated in Figure 7.3.4, where we have rotated by $90$ degrees in the anticlockwise direction.

Clearly $\PPP$ has $12$ square faces, $4$ horizontal streets and $6$ vertical streets, labelled according to the picture on the right where, for instance, the information
$\leftrightarrow3$ and $\updownarrow5$ in a square face indicates that this square face is on the $3$-rd horizontal street and the $5$-th vertical street.
Note that the anticlockwise rotation by $90$ degrees makes the application of Theorem <ref> more convenient.

The boundary identification in $\PPP$ is clearly the simplest perpendicular translation.
Consider a $1$-direction geodesic on $\PPP$ with slope given by the arrow in the picture on the left in Figure 7.3.4.
If such a geodesic hits the edge $v_3$ at the top, then it jumps down vertically to the identified edge $v_3$ in the middle.
In terms of the $\infty$-L-strip billiard this means precisely that the billiard moves from $L_i$ to $L_{i+1}$, i.e., moving right.
If such a geodesic hits the edge $v_2$ in the middle, then it jumps down vertically to the  identified edge $v_2$ at the bottom.
In terms of the $\infty$-L-strip billiard this means precisely that the billiard moves from $L_i$ to $L_{i-1}$, i.e., moving left.
Thus by counting the number of edge cuttings for $v_2$ and for $v_3$, we know the total number of steps moving to the left and the total number of steps moving to the right.
Taking the difference of these two numbers, we then know precisely which particular constituent L-shape happens to contain the billiard at a given moment.

Now the edge cutting numbers are expressed in terms of powers of the relevant eigenvalues of the $2$-step transition matrix corresponding to the given quadratic irrational slope.
It makes it plausible to guess that the irregularity exponent of the period-surface $\PPP$ gives the escape rate to infinity for the infinite L-strip billiard, as long as the shortline method works.
Next we expand on this and justify this intuition.

We need to choose the parameters $m$ and $n$, which have to be integer multiples of the lengths of the horizontal and vertical streets respectively.
For this period surface, the horizontal streets have length $2$ or $4$, while the vertical streets all have length $2$.
Thus we let $m=4k$ and $n=2k$, where $k\ge1$ is any integer.
Consequently, we let

\begin{equation}\label{eq7.3.1}
\alpha_k=[2k;4k,2k,4k,\ldots]=k+\frac{\sqrt{4k^2+2}}{2}.
\end{equation}

Since the period-surface $\PPP$ has $12$ squares, the original shortline method leads to a $24\times24$ $2$-step transition matrix $\bfA$.
To find the relevant eigenvalues and eigenvectors of such a large matrix directly is clearly rather inconvenient.

Instead we apply Theorem <ref>, which leads to a much smaller $4\times4$ street-spreading matrix $\bfS$ defined by (<ref>).
We follow the notation in Section <ref>.
We consider the $\bfA$-invariant subspace $\VVV$ generated by the $8$ vectors $\bfu_i,\bfv_i$, $i=1,2,3,4$, defined by (<ref>) and
The relevant eigenvalues of $\bfA$ are then eigenvalues of the matrix $\bfA\vert_\VVV$, defined by (<ref>).

Consider the horizontal street corresponding to $\bfu_1$, as highlighted in the picture on the left in Figure 7.3.5.
Clearly $J_1=\{2,4\}$.
Using (<ref>), we have

\begin{align}\label{eq7.3.2}
\nonumber
\\
\end{align}

With $J_2=\{1,2,4,6\}$, for the horizontal street corresponding to $\bfu_2$, as highlighted in the picture on the right in Figure 7.3.5, a similar argument gives

\begin{equation}\label{eq7.3.3}
\end{equation}

With $J_3=\{1,3,5,6\}$, for the horizontal street corresponding to $\bfu_3$, as highlighted in the picture on the left in Figure 7.3.6, a similar argument gives

\begin{equation}\label{eq7.3.4}
\end{equation}

With $J_4=\{3,5\}$, for the horizontal street corresponding to $\bfu_3$, as highlighted in the picture on the right in Figure 7.3.6, a similar argument gives

\begin{equation}\label{eq7.3.5}
\end{equation}

It follows from (<ref>)–(<ref>) that the street-spreading matrix is given by

4k^2   2k^2   0   0

4k^2   4k^2   2k^2   0

0   2k^2   4k^2   4k^2

0   0   2k^2   4k^2

4   2   0   0

4   4   2   0

0   2   4   4

0   0   2   4

This has eigenvalues and corresponding eigenvectors

\begin{align}
\tau_1=8k^2,
\nonumber
\\
\tau_2=6k^2,
\nonumber
\\
\tau_3=2k^2,
\nonumber
\\
\tau_4=0,
\nonumber
\end{align}

Using the formula (<ref>), we obtain corresponding eigenvalues of $\bfA\vert_\VVV$ given by

\begin{align}
\lambda(\tau_1;\pm)
\nonumber
\\
\lambda(\tau_2;\pm)
=1+3k^2\pm k\sqrt{9k^2+6},
\nonumber
\\
\lambda(\tau_3;\pm)
=1+k^2\pm k\sqrt{k^2+2},
\nonumber
\\
\lambda(\tau_4;\pm)
\nonumber
\end{align}

The three relevant eigenvalues of $\bfA\vert_\VVV$ and corresponding eigenvectors are therefore

\begin{align}
\lambda_1
\quad\Psi_1=(1,2,2,1,b_{11},b_{12},b_{13},b_{14})^T,
\label{eq7.3.6}
\\
\lambda_2
\quad\Psi_2=(1,1,-1,-1,b_{21},b_{22},b_{23},b_{24})^T,
\label{eq7.3.7}
\\
\lambda_3
\quad\Psi_3=(1,-1,-1,1,b_{31},b_{32},b_{33},b_{34})^T,
\nonumber
\end{align}

where the values of $b_{ij}$, $i=1,2,3$, $j=1,2,3,4$, can be calculated using (<ref>).

Recall that by counting the number of edge cuttings for $v_2$ and for $v_3$, we know the total number of steps of the billiard orbit moving to the left and the total number of steps of the billiard orbit moving to the right.
The difference between these two numbers then gives us information on the movement of the billiard orbit.

We now proceed to find the edge cutting numbers of $v_2$ and $v_3$.

The eigenvector of $\bfA$ corresponding to the eigenvector $\Psi_1$ of $\bfA\vert_\VVV$ is given by

We first find the number of almost vertical units counted here that cut the edge $v_2$; see Figure 7.3.4.

Concerning the contribution from those units counted by $\bfu_1$, it is clear from the picture on the left in Figure 7.3.5 that we have $2k^2$ units of type $\uparrow_{2,2}$ and $2k^2$ units of type $\uparrow_{2,4}$, making a total of $4k^2$.

Concerning the contribution from those units counted by $\bfu_2$, it is clear from the picture on the right in Figure 7.3.5 that we have $k^2$ units of type $\uparrow_{2,2}$ and $k^2$ units of type $\uparrow_{2,4}$, making a total of $2k^2$.

Concerning the contribution from those units counted by $\bfu_3$ and $\bfu_4$, it is clear from Figure 7.3.6 that there is none.

To study the contribution from those units counted by $\bfv_1,\ldots,\bfv_4$, we appeal to (<ref>) and Figure 7.2.1.
There is clearly no contribution from $\bfv_1,\bfv_3,\bfv_4$, while for $\bfv_2$, we have positive contributions from $\nuparrow_{2,1}$ and
$\nuparrow_{2,2}$, and negative contributions from $\uparrow_{2,2}$ and $\uparrow_{2,4}$, each counted with the multiplicity of $J_2^*$, and so there is perfect cancellation.

Thus for the edge $v_2$, we have a total count of $4k^2+2(2k^2)=8k^2$.

We next find the number of almost vertical units counted here that cut the edge $v_3$; see Figure 7.3.4.

Concerning the contribution from those units counted by $\bfu_1$ and $\bfu_2$, it is clear from Figure 7.3.5 that there is none.

Concerning the contribution from those units counted by $\bfu_3$, it is clear from the picture on the left in Figure 7.3.6 that we have $k^2$ units of type $\uparrow_{4,3}$ and $k^2$ units of type $\uparrow_{4,5}$, making a total of $2k^2$.

Concerning the contribution from those units counted by $\bfu_4$, it is clear from the picture on the right in Figure 7.3.6 that we have $2k^2$ units of type
$\uparrow_{4,3}$ and $2k^2$ units of type $\uparrow_{4,5}$, making a total of $4k^2$.

There is clearly no contribution from $\bfv_1,\bfv_2,\bfv_3$, while for $\bfv_4$, we have positive contributions from $\nuparrow_{4,3}$ and $\nuparrow_{4,5}$, and negative contributions from $\uparrow_{4,3}$ and $\uparrow_{4,5}$, each counted with the multiplicity of $J_4^*$, and so there is perfect cancellation.

Thus for the edge $v_3$, we have a total count of $2(2k^2)+4k^2=8k^2$.

Since the two counts are the same, it follows that the eigenvalue $\lambda_1$ does not contribute to the difference between the edge cutting numbers of $v_2$ and $v_3$.

The perfect cancellation is not surprising at all.
Indeed, the high powers corresponding to the largest eigenvalue represents the main term of the problem.
Any lack of cancellation here would violate the uniformity of the geodesic.
With a quadratic irrational slope, this would specifically violate the Gutkin–Veech theorem [9, 16, 18] that guarantees uniformity of any
$1$-direction geodesic in any polysquare translation surface with irrational slope.
See also [5, 6].

The eigenvector of $\bfA$ corresponding to the eigenvector $\Psi_2$ of $\bfA\vert_\VVV$ is given by

We first find the number of almost vertical units counted here that cut the edge $v_2$.
Recall that the contribution from $\bfu_1$ and $\bfu_2$ are $4k^2$ and $2k^2$ respectively, while there is no contribution from $\bfu_3$ and
$\bfu_4$, and also no contribution from $\bfv_1,\ldots,\bfv_4$.
Thus for the edge $v_2$, we have a total count of $4k^2+2k^2=6k^2$.

We next find the number of almost vertical units counted here that cut the edge $v_3$.
Recall that the contribution from $\bfu_3$ and $\bfu_4$ are $2k^2$ and $4k^2$ respectively, while there is no contribution from $\bfu_1$ and
$\bfu_2$, and also no contribution from $\bfv_1,\ldots,\bfv_4$.
Thus for the edge $v_3$, we have a total count of $-2k^2-4k^2=-6k^2$.

The difference, in absolute value, is therefore $12k^2$.
Let $c_2=c_{2,1}$ in (<ref>).
Then it follows that the second eigenvalue $\lambda_2$ contributes $12c_2\lambda_2^rk^2$ to the difference between the edge cutting numbers of $v_2$ and $v_3$.

So the deviation from the starting point comes from the second largest eigenvalue, and this has the order of magnitude $\lambda_2^r$ compared to the order of magnitude $\lambda_1^r$ of the main term.
Choosing $T=\lambda_1^r$, we have $\lambda_2^r\asymp T^{\kappa_0}$, where

\begin{equation}\label{eq7.3.8}
\kappa_0=\kappa_0(k)=\frac{\log\lambda_2}{\log\lambda_1}=\frac{2\log k+\log6}{2\log k+\log8}+o(1),
\end{equation}

in view of (<ref>) and (<ref>), is the irregularity exponent of a $1$-direction geodesic of slope (<ref>) on the period-surface in Figure 7.3.4.

Clearly $\kappa_0=\kappa_0(k)\to1$ as $k\to\infty$.
So we have just established $T^{\kappa_0}=T^{1-\eps}$ size super-fast escape rate to infinity for the infinite L-strip billiard with the explicit class of quadratic irrational slopes in (<ref>) where the parameter $k\ge1$ is any integer.

It is easy to see that the irregularity exponent $\kappa_0=\kappa_0(k)$ in (<ref>) is precisely the escape rate to infinity of this infinite billiard.
Indeed, the exponent of the escape rate to infinity cannot be larger than the expression (<ref>) coming from the two largest eigenvalues.

Consider the Ehrenfest wind-tree model with $a=b=1/2$, rescaled so that we have square faces of unit area.
Consider a $4$-direction billiard trajectory in the infinite region in Figure 7.3.7.
Here the building block of this infinite polysquare translation surface is an L-shape.

As usual, this infinite billiard model is equivalent to a $1$-direction geodesic flow on an infinite polysquare translation surface that we denote by $\Bil(\infty;2)$, with the index $2$ here indicating that this is double periodic.
To construct $\Bil(\infty;2)$, we take one of the L-shape building blocks, and unfold the $4$-direction billiard flow on it to a $1$-direction geodesic flow on a $4$-copy version of it, obtained by reflecting horizontally and vertically.
Note that each $4$-copy L-shape has a right-neighbor, a left-neighbor, a down-neighbor and an up-neighbor in $\Bil(\infty;2)$, and we need appropriate edge identification which amount to a more complicated version of Figure 7.3.3.
The period-surface $\Bil(2)$ of $\Bil(\infty;2)$ is shown in the picture on the left in Figure 7.3.8.

The horizontal and vertical streets of $\Bil(2)$ are indicated in the picture on the right in Figure 7.3.8 where, for instance, the entries
$\leftrightarrow3$ and $\updownarrow4$ in a square face indicates that the square face is on the $3$-rd horizontal street and the $4$-th vertical street.
It is easy to see that $\Bil(2)$ has $6$ horizontal streets and $6$ vertical streets.

The boundary identification in $\Bil(2)$ is clearly the simplest perpendicular translation.
Consider a $1$-direction geodesic in $\Bil(2)$ with slope given by the arrow in the picture on the left in Figure 7.3.8.
If such a geodesic hits the edge $h_2$ or $h_4$ at the top, then it jumps down vertically to the identified edge $h_2$ or $h_4$ in the middle.
In terms of the Ehrenfest wind-tree billiard this means precisely that the billiard moves from one L-shape to the next above, i.e., moving up.
If such a geodesic hits the edge $h_1$ or $h_3$ in the middle, then it jumps down vertically to the  identified edge $h_1$ or $h_3$ at the bottom.
In terms of the Ehrenfest wind-tree billiard this means precisely that the billiard moves from one L-shape to the next below, i.e., moving down.
Thus by counting the number of edge cuttings for $h_2,h_4$ and for $h_1,h_3$, we know the total number of steps moving up and the total number of steps moving down.
Taking the difference of these two numbers, we then know precisely the vertical location of the billiard at a given moment.

We need to choose the parameters $m$ and $n$, which have to be integer multiples of the lengths of the horizontal and vertical streets respectively.
For this period surface, the horizontal and vertical streets all have length $2$.
Thus we let $m=2k$ and $n=2k$, where $k\ge1$ is any integer.
Consequently, we let

\begin{equation}\label{eq7.3.9}
\alpha_k=[2k;2k,2k,2k,\ldots]=k+\sqrt{k^2+1}.
\end{equation}

We apply Theorem <ref>, leading to a $6\times6$ street-spreading matrix $\bfS$ defined by (<ref>).
We follow the notation in Section <ref>.
We consider the $\bfA$-invariant subspace $\VVV$ generated by the $12$ vectors $\bfu_i,\bfv_i$, $i=1,\ldots,6$, defined by (<ref>) and (<ref>).
The relevant eigenvalues of $\bfA$ are then eigenvalues of the matrix $\bfA\vert_\VVV$, defined by (<ref>).

Consider the horizontal street corresponding to $\bfu_1$, as highlighted in the picture on the left in Figure 7.3.9.
Clearly $J_1=\{2,4\}$.
Using (<ref>), we have

\begin{align}\label{eq7.3.10}
\nonumber
\\
\end{align}

With $J_2=\{1,2\}$, for the horizontal street corresponding to $\bfu_2$, as highlighted in the picture in the middle in Figure 7.3.9, a similar argument gives

\begin{equation}\label{eq7.3.11}
\end{equation}

With $J_3=\{4,6\}$, for the horizontal street corresponding to $\bfu_3$, as highlighted in the picture on the right in Figure 7.3.9, a similar argument gives

\begin{equation}\label{eq7.3.12}
\end{equation}

With $J_4=\{1,3\}$, for the horizontal street corresponding to $\bfu_4$, as highlighted in the picture on the left in Figure 7.3.10, a similar argument gives

\begin{equation}\label{eq7.3.13}
\end{equation}

With $J_5=\{5,6\}$, for the horizontal street corresponding to $\bfu_5$, as highlighted in the picture in the middle in Figure 7.3.10, a similar argument gives

\begin{equation}\label{eq7.3.14}
\end{equation}

With $J_6=\{3,5\}$, for the horizontal street corresponding to $\bfu_5$, as highlighted in the picture in the middle in Figure 7.3.10, a similar argument gives

\begin{equation}\label{eq7.3.15}
\end{equation}

It follows from (<ref>)–(<ref>) that the street-spreading matrix is given by

2k^2   k^2   k^2   0   0   0

k^2   2k^2   0   k^2   0   0

k^2   0   2k^2   0   k^2   0

0   k^2   0   2k^2   0   k^2

0   0   k^2   0   2k^2   k^2

0   0   0   k^2   k^2   2k^2

2   1   1   0   0   0

1   2   0   1   0   0

1   0   2   0   1   0

0   1   0   2   0   1

0   0   1   0   2   1

0   0   0   1   1   2

This has non-zero eigenvalues and corresponding eigenvectors

\begin{align}
\tau_1=4k^2,
\nonumber
\\
\tau_2=3k^2,
\nonumber
\\
\nonumber
\\
\tau_3=k^2,
\nonumber
\\
\nonumber
\\
\tau_4=0,
\nonumber
\end{align}

Using the formula (<ref>), we obtain corresponding eigenvalues of $\bfA\vert_\VVV$ given by

\begin{align}
\lambda(\tau_1;\pm)
\nonumber
\\
\lambda(\tau_2;\pm)
\nonumber
\\
\lambda(\tau_3;\pm)
\nonumber
\\
\lambda(\tau_4;\pm)
\nonumber
\end{align}

The three relevant eigenvalues of $\bfA\vert_\VVV$ and corresponding eigenvectors are therefore

\begin{align}
\lambda_1
\quad\Psi_1=(1,1,1,1,1,1,b_{11},\ldots,b_{16})^T,
\label{eq7.3.16}
\\
\lambda_2
\quad\Psi_{2,1}=(-1,0,-1,1,0,1,b_{21}^{(1)},\ldots,b_{26}^{(1)})^T,
\label{eq7.3.17}
\\
\quad\Psi_{2,2}=(0,-1,1,-1,1,0,b_{21}^{(2)},\ldots,b_{26}^{(2)})^T,
\nonumber
\\
\lambda_3
\quad\Psi_{3,1}=(1,0,-1,-1,0,1,b_{31}^{(1)},\ldots,b_{36}^{(1)})^T,
\nonumber
\\
\quad\Psi_{3,2}=(0,1,-1,-1,1,0,b_{31}^{(2)},\ldots,b_{36}^{(2)})^T,
\nonumber
\end{align}

where the values of $b_{1j},b_{2j}^{(1)},b_{2j}^{(2)},b_{3j}^{(1)},b_{3j}^{(2)}$, $j=1,\ldots,6$, can all be calculated using (<ref>).

We now proceed to find the edge cutting numbers of $h_2,h_4$ and $h_1,h_3$.

As in Example <ref>, the eigenvalue $\lambda_1$ does not contribute to their difference, as any lack of cancellation would violate the Gutkin–Veech theorem that guarantees uniformity.

The eigenvector of $\bfA$ corresponding to the eigenvector $\Psi_{2,1}$ of $\bfA\vert_\VVV$ is given by

\begin{equation}\label{eq7.3.18}
\end{equation}

The eigenvector of $\bfA$ corresponding to the eigenvector $\Psi_{2,2}$ of $\bfA\vert_\VVV$ is given by

\begin{equation}\label{eq7.3.19}
\end{equation}

We first find the number of almost vertical units counted here that cut the edges $h_2$ and $h_4$; see Figure 7.3.8.
The count from each of $\bfu_i$, $i=1,\ldots,6$ are

On the other hand, it is easy to show that the total contribution from $\bfv_i$, $i=1,\ldots,6$ is $0$.
Thus corresponding to the eigenvector $\Psi_{2,1}$ and (<ref>), the total count is $3k^2$.
Corresponding to the eigenvector $\Psi_{2,2}$ and (<ref>), the total count is $0$.

We next find the number of almost vertical units counted here that cut the edges $h_1$ and $h_3$; see Figure 7.3.8.
The count from each of $\bfu_i$, $i=1,\ldots,6$ are

Like before, the total contribution from $\bfv_i$, $i=1,\ldots,6$ is $0$.
Thus corresponding to the eigenvector $\Psi_{2,1}$ and (<ref>), the total count is $-3k^2$.
Corresponding to the eigenvector $\Psi_{2,2}$ and (<ref>), the total count is $0$.

It follows that the second eigenvalue $\lambda_2$ contributes $6c_{2,1}\lambda_2^rk^2$ to the difference between the edge cuttings numbers of $h_2,h_4$ and $h_1,h_3$.

So the deviation from the starting point comes from the second largest eigenvalue, and this has the order of magnitude $\lambda_2^r$ compared to the order of magnitude $\lambda_1^r$ of the main term.
Choosing $T=\lambda_1^r$, we have $\lambda_2^r\asymp T^{\kappa_0}$, where

\begin{equation}\label{eq7.3.20}
\kappa_0=\kappa_0(k)=\frac{\log\lambda_2}{\log\lambda_1}=\frac{2\log k+\log3}{2\log k+\log4}+o(1),
\end{equation}

in view of (<ref>) and (<ref>), is the irregularity exponent of a $1$-direction geodesic of slope (<ref>) on the period-surface in Figure 7.3.8.

Clearly $\kappa_0=\kappa_0(k)\to1$ as $k\to\infty$.
So we have just established $T^{\kappa_0}=T^{1-\eps}$ size super-fast escape rate to infinity for the infinite L-strip billiard with the explicit class of quadratic irrational slopes in (<ref>) where the parameter $k\ge1$ is any integer.

It is easy to see that the irregularity exponent $\kappa_0=\kappa_0(k)$ in (<ref>) is precisely the escape rate to infinity of this infinite billiard.
Indeed, the exponent of the escape rate to infinity cannot be larger than the expression (<ref>) coming from the two largest eigenvalues.

This example also highlights that different quadratic irrational slopes can lead to $1$-direction geodesics that exhibit vastly different escape rates to infinity.
To explain this, we shall consider an infinite polysquare translation surface $\PPP$ where there is an absolute bound $\ell$ such that the length of any horizontal or vertical street of $\PPP$ is at most $\ell$.
This is called an $\ell$-square-maze translation surface, and $1$-direction geodesics on square-maze translation surfaces are studied in <cit.>.

The problem connected to the Ehrenfest wind-tree model billiard with $a=b=1/2$ that we have considered here leads to a problem of $1$-direction geodesic flow on
Now $\Bil(\infty;2)$ has infinite horizontal and vertical streets, so is not a square-maze translation surface.
However, if we consider streets in $\Bil(\infty;2)$ at $45$ degrees to the horizontal and vertical axes, then the situation becomes very different.
First of all, observe such a street in the Ehrenfest wind-tree model billiard as shown in Figure 7.3.11.

The street indicated is contained in $4$ constituent L-shapes.
If we consider the analogous problem of $1$-direction geodesic flow in $\Bil(\infty;2)$, then the image of this street is shown in Figure 7.3.12, made up of the $4$-copy versions of these $4$ constituent L-shapes.
We have in fact shown that $\Bil(\infty;2)$ rotated by $45$ degrees becomes a square-maze translation surface where every horizontal and vertical street has length $12$.

<cit.> asserts that for an $\ell$-square-maze translation surface, there are infinitely many numbers $\alpha$ of the form

where $a\ge\ell!$ is divisible by $\ell!$, such that there exist $1$-direction geodesics with slope $\alpha$ that exhibit time-quantitative density and super-slow logarithmic escape rate to infinity.
For a square-maze translation surface where every street has length precisely $\ell$, the condition on $a$ can be relaxed to include all integers $a\ge\ell$ that are divisible by $\ell$.

It follows that for the rotated $\Bil(\infty;2)$, there exist infinitely many numbers $\alpha_k$ of the form

\begin{equation}\label{eq7.3.21}
\alpha_k=[12k;12k,12k,12k,\ldots]=12k+\frac{1}{12k+\frac{1}{12k+\frac{1}{12k+\cdots}}},
\end{equation}

where $k\ge1$ is an integer, such that there exist $1$-direction geodesics with slope $\alpha_k$ that exhibit super-slow logarithmic escape rate to infinity.

Now note that slopes $\alpha_k$ of the form (<ref>) make up a subset of those slopes $\alpha_k$ of the form (<ref>).
It follows that slopes $\alpha_k$ of the form (<ref>) give rise to $1$-direction geodesics on $\Bil(\infty;2)$ that exhibit super-fast escape rate to infinity.

It is easy to show that a line with slope $\alpha_k$, after a clockwise rotation of $45$ degrees, now has slope equal to

\begin{equation}\label{eq7.3.22}
\alpha_k^*=\frac{\alpha_k-1}{\alpha_k+1}.
\end{equation}

Thus we have shown that there are infinitely many numbers $\alpha_k$ of the form (<ref>) such that there exist $1$-direction geodesics of slope $\alpha_k$ in $\Bil(\infty;2)$ that exhibit super-fast escape rate to infinity, and also $1$-direction geodesics of slope $\alpha_k^*$ in
$\Bil(\infty;2)$ that exhibit super-slow logarithmic escape rate to infinity.
Note that in view of (<ref>) and (<ref>), both $\alpha_k$ and $\alpha_k^*$ are quadratic irrationals.

§.§ More on the escape rate to infinity

In this section, we consider some more complicated infinite billiards.

Consider C-wall obstacles, as shown in Figure 7.4.1.

More precisely, the building block of this infinite polysquare region is a $3\times3$ square, with walls on three sides of the middle square face as shown.

As usual, this infinite billiard model is equivalent to a $1$-direction geodesic flow on an infinite polysquare translation surface that we denote by $\Bil(\infty;2;\mathrm{C})$.
Here the index $2$ indicates that this is double periodic, and the letter $\mathrm{C}$ refers to the common shape of the obstacles.
To construct $\Bil(\infty;2;\mathrm{C})$, we take one of the building blocks, and unfold the $4$-direction billiard flow on it to a $1$-direction geodesic flow on a $4$-copy version of it, obtained by reflecting horizontally and vertically.
Note that each $4$-copy version has a right-neighbor, a left-neighbor, a down-neighbor and an up-neighbor in $\Bil(\infty;2;\mathrm{C})$, and we need appropriate edge identification for gluing them together.

The period-surface $\Bil(2;\mathrm{C})$ of $\Bil(\infty;2;\mathrm{C})$ is shown in the picture on the left in Figure 7.4.2.
The horizontal and vertical streets of $\Bil(2;\mathrm{C})$ are indicated in the picture on the right in Figure 7.4.2 where, for instance, the entries
$\leftrightarrow3$ and $\updownarrow5$ in a square face indicates that the square face is on the $3$-rd horizontal street and the $5$-th vertical street.
It is easy to see that $\Bil(2;\mathrm{C})$ has $8$ horizontal streets of length $3$, and $2$ horizontal streets of length $6$.
It also has $8$ vertical streets of length $3$, $2$ vertical streets of length $4$, and $2$ vertical streets of length $2$.
Thus the street-LCM is equal to  $12$.

We shall consider an almost vertical geodesic $V_0$ in $\Bil(2;\mathrm{C})$ with slope $\alpha$ given by (<ref>) with $m=n=12$.
As $\Bil(2;\mathrm{C})$ has $36$ square faces, the $2$-step transition matrix $\bfA$ is $72\times72$ which is mildly inconvenient.
Instead, we shall determine the street-spreading matrix $\bfS$ which, at size $10\times10$, is considerably smaller.

We follow the notation in Section <ref>.
We consider the $\bfA$-invariant subspace $\VVV$ generated by the $20$ vectors $\bfu_i,\bfv_i$, $i=1,\ldots,10$, defined by (<ref>) and (<ref>).
The relevant eigenvalues of $\bfA$ are then eigenvalues of the matrix $\bfA\vert_\VVV$, defined by (<ref>).

With $J_1=J_4=\{1,3,9\}$, for the horizontal streets corresponding to $\bfu_1,\bfu_4$ as highlighted in the picture on the left in Figure 7.4.3, we have, using (<ref>),

\begin{align}
\nonumber
\\
\nonumber
\\
\nonumber
\\
\nonumber
\\
\nonumber
\\
\label{eq7.4.1}
\end{align}

With $J_2=J_5=\{5,7,10\}$, for the horizontal streets corresponding to $\bfu_2,\bfu_5$ as highlighted in the picture on the right in Figure 7.4.3, we have, using (<ref>),

\begin{align}
\nonumber
\\
\nonumber
\\
\label{eq7.4.2}
\end{align}

With $J_6=J_9=\{2,4,9\}$ and $J_7=J_{10}=\{6,8,10\}$, for the horizontal streets corresponding to $\bfu_6,\bfu_7,\bfu_9,\bfu_{10}$ highlighted in Figure 7.4.4, we have

\begin{align}
\nonumber
\\
\nonumber
\\
\label{eq7.4.3}
\\
\nonumber
\\
\nonumber
\\
\label{eq7.4.4}
\end{align}

With $J_3=\{1,3,5,7,11,12\}$ and $J_8=\{2,4,6,8,11,12\}$, for the horizontal streets corresponding to $\bfu_3,\bfu_8$ highlighted in Figure 7.4.5, we have

\begin{align}
\nonumber
\\
\label{eq7.4.5}
\\
\nonumber
\\
\label{eq7.4.6}
\end{align}

It follows from (<ref>)–(<ref>) that the street-spreading matrix is given by

\begin{equation}\label{eq7.4.7}
\bfS=\begin{pmatrix}
\end{pmatrix}.
\end{equation}

This has non-zero eigenvalues and corresponding eigenvectors given by

\begin{align}
\tau_1=144,
\psi_1=(1,1,2,1,1,1,1,2,1,1)^T,
\nonumber
\\
\tau_2=112,
\psi_2=(-1,1,0,-1,1,-1,1,0,-1,1)^T,
\nonumber
\\
\tau_3=96,
\psi_3=(-1,-1,-2,-1,-1,1,1,2,1,1)^T,
\nonumber
\\
\tau_4=64,
\psi_4=(1,-1,0,1,-1,-1,1,0,-1,1)^T,
\nonumber
\\
\tau_5=48,
\psi_5=(1,1,-4,1,1,1,1,-4,1,1)^T.
\nonumber
\end{align}

Note that the street-spreading matrix $\bfS$ given by (<ref>) corresponds to the choice of parameters $m=n=12$.
Switching to $m=n=12k$, where $k\ge1$ is any integer, we obtain the new street-spreading matrix $\bfS(k)=k^2\bfS$ simply by multiplying the matrix
$\bfS$ by $k^2$.
Then of course the eigenvalues are also multiplied by $k^2$, but the eigenvectors remain the same.
Naturally the $2$-step transition matrix $\bfA$ is modified to $\bfA(k)$.

We next determine some of the eigenvalues of $\bfA(k)$ using (<ref>).
The largest eigenvalue of $\bfA(k)\vert_\VVV$ is

\begin{equation}\label{eq7.4.8}
\lambda_1(k)
\end{equation}

while the second largest eigenvalue of $\bfA(k)\vert_\VVV$ is

\begin{equation}\label{eq7.4.9}
\lambda_2(k)
\end{equation}

with eigenvector of the form

Let us return to Figures 7.4.1 and 7.4.2.
It is clear that the billiard moves to the up-neighbor if the $1$-direction geodesic hits any of the edges $h_1,\ldots,h_6$, and moves to the
down-neighbor if the $1$-direction geodesic hits any of the edges $h_7,\ldots,h_{12}$.
It is also clear that the billiard moves to the right-neighbor if the $1$-direction geodesic hits any of the edges $v_7,\ldots,v_{12}$, and moves to the
left-neighbor if the $1$-direction geodesic hits any of the edges $v_1,\ldots,v_6$.

We now proceed to find the edge cutting numbers of $h_1,\ldots,h_6$ and $h_7,\ldots,h_{12}$.

As in the example in Section <ref>, the eigenvalue $\lambda_1$ does not contribute to their difference, as any lack of cancellation would violate the Gutkin–Veech theorem that guarantees uniformity.

The eigenvector of $\bfA(k)$ corresponding to the eigenvector $\Psi_2$ of $\bfA(k)\vert_\VVV$ is given by

\begin{align}\label{eq7.4.10}
\nonumber
\\
\end{align}

We first find the number of almost vertical units counted here that cut the edges $h_1,\ldots,h_6$; see Figure 7.4.2.
The counts from $\bfu_i$, $i=1,\ldots,10$, are

\begin{equation}\label{eq7.4.11}
\begin{array}{lllll}
\bfu_1\mapsto44k^2,\ {}
&\bfu_2\mapsto44k^2,\ {}
&\bfu_3\mapsto32k^2,\ {}
&\bfu_4\mapsto44k^2,\ {}
\\
\bfu_6\mapsto12k^2,\ {}
&\bfu_7\mapsto12k^2,\ {}
&\bfu_8\mapsto0,\ {}
&\bfu_9\mapsto12k^2,\ {}
\end{array}
\end{equation}

On the other hand, it is not difficult to show that each of $\bfv_1,\ldots\bfv_{10}$ contributes precisely zero.
Thus for the edges $h_1,\ldots,h_6$, we have a total count of

We next find the number of almost vertical units counted here that cut the edges $h_7,\ldots,h_{12}$; see Figure 7.4.2.
The counts from $\bfu_i$, $i=1,\ldots,10$, are

\begin{equation}\label{eq7.4.12}
\begin{array}{lllll}
\bfu_1\mapsto12k^2,\ {}
&\bfu_2\mapsto12k^2,\ {}
&\bfu_3\mapsto0,\ {}
&\bfu_4\mapsto12k^2,\ {}
\\
\bfu_6\mapsto44k^2,\ {}
&\bfu_7\mapsto44k^2,\ {}
&\bfu_8\mapsto32k^2,\ {}
&\bfu_9\mapsto44k^2,\ {}
\end{array}
\end{equation}

Again, it is not difficult to show that each of $\bfv_1,\ldots\bfv_{10}$ contributes precisely zero.
Thus for the edges $h_7,\ldots,h_{12}$, we have a total count of

Thus there is perfect cancellation in the vertical direction.

We next proceed to find the edge cutting numbers of $v_1,\ldots,v_6$ and $v_7,\ldots,v_{12}$; see Figure 7.4.2.

Again, the eigenvalue $\lambda_1$ does not contribute to their difference, as any lack of cancellation would violate the Gutkin–Veech theorem that guarantees uniformity.
So we concentrate our attention on the contribution from the eigenvector $\Psi_2$ of $\bfA\vert_\VVV$.
Here, note that none of $\bfu_1,\ldots,\bfu_{10}$ makes any non-zero contribution, as only units of type $\nuparrow$ can contribute to the count.
For the contributions from $\bfv_1,\ldots,\bfv_{10}$, we use (<ref>).

We first find the number of almost vertical units counted in (<ref>) that cut the edges $v_1,\ldots,v_6$; see Figure 7.4.2.
Since $\bfv_3$ and $\bfv_8$ do not feature in (<ref>), it is not necessary to any counting for them.
Clearly we have a count of $4k$ for each of $\bfv_1,\bfv_4,\bfv_6,\bfv_9$, and none for the rest, making a total of $-16\tau_2^*k$.

We next find the number of almost vertical units counted in (<ref>) that cut the edges $v_7,\ldots,v_{12}$; see Figure 7.4.2.
Clearly we have a count of $4k$ for each of $\bfv_2,\bfv_5,\bfv_7,\bfv_{10}$, and none for the rest, making a total of $16\tau_2^*k$.

The difference, in absolute value, is therefore $32\tau_2^*k$.
Let $c_2=c_{2,1}$ in (<ref>) and (<ref>).
It follows that the second eigenvalue $\lambda_2(k)$ contributes $32c_2\tau_2^*\lambda_2^r(k)k$ to the difference between the edge cuttings numbers of $v_1,\ldots,v_6$ and $v_7,\ldots,v_{12}$.

So the horizontal deviation from the starting point comes from the second largest eigenvalue, with order of magnitude $\lambda_2^r$ compared to the order of magnitude $\lambda_1^r$ of the main term.
Choosing $T=\lambda_1^r$, we have $\lambda_2^r\asymp T^{\kappa_0}$, where

\begin{equation}\label{eq7.4.13}
\kappa_0=\kappa_0(k)=\frac{\log\lambda_2}{\log\lambda_1}=\frac{2\log k+\log112}{2\log k+\log144}+o(1),
\end{equation}

in view of (<ref>) and (<ref>), is the irregularity exponent of a $1$-direction geodesic of slope

\begin{equation}\label{eq7.4.14}
\alpha_k=[12k;12k,12k,12k,\ldots]=6k+\sqrt{36k^2+1}=\sqrt{\lambda_1(k)}
\end{equation}

on the period-surface $\Bil(2;\mathrm{C})$ in Figure 7.4.2.

Clearly $\kappa_0=\kappa_0(k)\to1$ as $k\to\infty$.
So we have just established $T^{\kappa_0}=T^{1-\eps}$ size super-fast escape rate to infinity for this infinite billiard with the explicit class of quadratic irrational slopes in (<ref>) where the parameter $k\ge1$ is any integer.

It is easy to see that the irregularity exponent $\kappa_0=\kappa_0(k)$ in (<ref>) is precisely the escape rate to infinity of this infinite billiard.
The exponent of the escape rate to infinity cannot be larger than the expression (<ref>) coming from the two largest eigenvalues.

Note that this super-fast escape rate to infinity comes from horizontal deviation.
As observed earlier, we have perfect cancellation in the vertical direction for the two largest eigenvalues.
So let us investigate what the third eigenvalue gives.

The third largest eigenvalue of $\bfA(k)\vert_\VVV$ is

with eigenvector of the form

The eigenvector of $\bfA(k)$ corresponding to the eigenvector $\Psi_3$ of $\bfA(k)\vert_\VVV$ is given by

\begin{align}
\nonumber
\\
\nonumber
\\
\nonumber
\end{align}

We first find the number of almost vertical units counted here that cut the edges $h_1,\ldots,h_6$; see Figure 7.4.2.
The counts from $\bfu_i$, $i=1,\ldots,10$, are given by (<ref>).
Again, it is not difficult to show that each of $\bfv_1,\ldots\bfv_{10}$ contributes precisely zero.
Thus for the edges $h_1,\ldots,h_6$, we have a total count of

We next find the number of almost vertical units counted here that cut the edges $h_7,\ldots,h_{12}$; see Figure 7.4.2.
The counts from $\bfu_i$, $i=1,\ldots,10$, are given by (<ref>).
Again, it is not difficult to show that each of $\bfv_1,\ldots\bfv_{10}$ contributes precisely zero.
Thus for the edges $h_7,\ldots,h_{12}$, we have a total count of

The difference, in absolute value, is therefore $392\tau_3^*k^2$.
Let $c_3=c_{3,1}$ in (<ref>) and (<ref>).
It follows that the third eigenvalue $\lambda_3(k)$ contributes $392c_3\tau_3^*\lambda_3^r(k)k^2$ to the difference between the edge cuttings numbers of $h_1,\ldots,h_6$ and $h_7,\ldots,h_{12}$.

So the vertical deviation from the starting point comes from the third largest eigenvalue, with order of magnitude $\lambda_3^r$ compared to the order of magnitude $\lambda_1^r$ of the main term.

Consider the Ehrenfest wind-tree model with $a=b=1/3$, rescaled so that we have square faces of unit area.
Consider a $4$-direction billiard trajectory in the infinite region in Figure 7.4.6.
Here the building block of this infinite polysquare region is a $3\times3$ square, with walls on all sides of the middle square face as shown.

As usual, this infinite billiard model is equivalent to a $1$-direction geodesic flow on an infinite polysquare translation surface that we denote by $\Bil(\infty;2;\mathrm{W})$.
Here the index $2$ indicates that this is double periodic, and the letter $\mathrm{W}$ refers to the wind-tree model.
To construct $\Bil(\infty;2;\mathrm{W})$, we take one of the building blocks, and unfold the $4$-direction billiard flow on it to a $1$-direction geodesic flow on a $4$-copy version of it, obtained by reflecting horizontally and vertically.
Note that each copy of the period surface has a right-neighbor, a left-neighbor, a down-neighbor and an up-neighbor in $\Bil(\infty;2;\mathrm{W})$,
and we need appropriate edge identification for gluing them together.

The period-surface $\Bil(2;\mathrm{W})$ of $\Bil(\infty;2;\mathrm{W})$ is shown in the picture on the left in Figure 7.4.7.
The horizontal and vertical streets of $\Bil(2;\mathrm{W})$ are indicated in the picture on the right in Figure 7.4.7 where, for instance, the entries
$\leftrightarrow3$ and $\updownarrow5$ in a square face indicates that the square face is on the $3$-rd horizontal street and the $5$-th vertical street.
It is easy to see that $\Bil(2;\mathrm{W})$ has $10$ horizontal streets, of which $8$ are of length $3$ and $2$ are of length $4$.
It also has $10$ vertical streets, of which $8$ are of length $3$ and $2$ are of length $4$.

Consider now a $1$-direction geodesic starting from some vertex of $\Bil(2;W)$ with slope $\alpha$ given by (<ref>) with $m=n=12$.

With $J_1=J_4=\{1,3,9\}$ and $J_2=J_5=\{5,7,10\}$, for the horizontal streets corresponding to $\bfu_1,\bfu_2,\bfu_4,\bfu_5$ highlighted in Figure 7.4.8, we have

\begin{align}
\nonumber
\\
\nonumber
\\
\label{eq7.4.15}
\\
\nonumber
\\
\nonumber
\\
\label{eq7.4.16}
\end{align}

With $J_6=J_9=\{2,4,9\}$ and $J_7=J_{10}=\{6,8,10\}$, for the horizontal streets corresponding to $\bfu_6,\bfu_7,\bfu_9,\bfu_{10}$ highlighted in Figure 7.4.9, we have

\begin{align}
\nonumber
\\
\nonumber
\\
\label{eq7.4.17}
\\
\nonumber
\\
\nonumber
\\
\label{eq7.4.18}
\end{align}

With $J_3=\{1,3,5,7\}$ and $J_8=\{2,4,6,8\}$, for the horizontal streets corresponding to $\bfu_3,\bfu_8$ highlighted in Figure 7.4.10, we have

\begin{align}
\nonumber
\\
\label{eq7.4.19}
\\
\nonumber
\\
\label{eq7.4.20}
\end{align}

It follows from (<ref>)–(<ref>) that the street-spreading matrix is given by

\begin{equation}\label{eq7.4.21}
\bfS=\begin{pmatrix}
\end{pmatrix}.
\end{equation}

This has non-zero eigenvalues and corresponding eigenvectors given by

\begin{align}
\tau_1=144,
\psi_1=(3,3,4,3,3,3,3,4,3,3)^T,
\nonumber
\\
\tau_2=112,
\psi_{2,1}=(0,-1,-1,0,-1,1,0,1,1,0)^T,
\nonumber
\\
\psi_{2,2}=(-1,0,-1,-1,0,0,1,1,0,1)^T,
\nonumber
\\
\tau_3=64,
\psi_3=(1,-1,0,1,-1,-1,1,0,-1,1)^T,
\nonumber
\\
\tau_4=16,
\psi_4=(1,1,-4,1,1,1,1,-4,1,1)^T.
\nonumber
\end{align}

Note that the street-spreading matrix $\bfS$ given by (<ref>) corresponds to the choice of parameters $m=n=12$.
Switching to $m=n=12k$, where $k\ge1$ is any integer, we obtain the new street-spreading matrix $\bfS(k)=k^2\bfS$ simply by multiplying the matrix $\bfS$ by $k^2$.
Then of course the eigenvalues are also multiplied by $k^2$, but the eigenvectors remain the same.
Naturally the $2$-step transition matrix $\bfA$ is modified to $\bfA(k)$.

We next determine some of the eigenvalues of $\bfA(k)$ using (<ref>).
The largest eigenvalue of $\bfA(k)$ is

\begin{equation}\label{eq7.4.22}
\lambda_1(k)
\end{equation}

while the second largest eigenvalue of $\bfA(k)$ is

\begin{equation}\label{eq7.4.23}
\lambda_2(k)
\end{equation}

with eigenvectors of the form

\begin{align}
\Psi_{2,1}
\nonumber
\\
\Psi_{2,2}
\nonumber
\end{align}

Let us return to Figures 7.4.6 and 7.4.7.
It is clear that the billiard moves to the up-neighbour if the $1$-direction geodesic hits any of the edges $h_1,\ldots,h_6$, and moves to the
down-neighbor if the $1$-direction geodesic hits any of the edges $h_7,\ldots,h_{12}$.
It is also clear that the billiard moves to the right-neighbor if the $1$-direction geodesic hits any of the edges $v_7,\ldots,v_{12}$, and moves to the
left-neighbor if the $1$-direction geodesic hits any of the edges $v_1,\ldots,v_6$.

We now proceed to find the edge cutting numbers of $h_1,\ldots,h_6$ and $h_7,\ldots,h_{12}$.
As before, it is easily shown that each of $\bfv_1,\ldots\bfv_{10}$ contributes precisely zero.

As in Examples <ref> and <ref>, the eigenvalue $\lambda_1$ does not contribute to their difference, as any lack of cancellation would violate the
Gutkin–Veech theorem that guarantees uniformity.

The eigenvector of $\bfA(k)$ corresponding to the eigenvector $\Psi_{2,1}$ of $\bfA\vert_\VVV$ is given by

\begin{align}\label{eq7.4.24}
\nonumber
\\
\end{align}

The eigenvector of $\bfA(k)$ corresponding to the eigenvector $\Psi_{2,2}$ of $\bfA\vert_\VVV$ is given by

\begin{align}\label{eq7.4.25}
\nonumber
\\
\end{align}

We first find the number of almost vertical units counted here that cut the edges $h_1,\ldots,h_6$; see Figure 7.4.7.
The counts from $\bfu_i$, $i=1,\ldots,10$, are

_1↦44k^2,     _2↦44k^2,     _3↦48k^2,     _4↦44k^2,     _5↦44k^2,

_6↦12k^2,     _7↦12k^2,     _8↦0,     _9↦12k^2,     _10↦12k^2.

Thus corresponding to the eigenvector $\Psi_{2,1}$ and (<ref>), the total count is $-112k^2$.
Corresponding to the eigenvector $\Psi_{2,2}$ and (<ref>), the total count is also $-112k^2$.

We next find the number of almost vertical units counted here that cut the edges $h_7,\ldots,h_{12}$; see Figure 7.4.7.
The counts from $\bfu_i$, $i=1,\ldots,10$, are

_1↦12k^2,     _2↦12k^2,     _3↦0,     _4↦12k^2,     _5↦12k^2,

_6↦44k^2,     _7↦44k^2,     _8↦48k^2,     _9↦44k^2,     _10↦44k^2.

Thus corresponding to the eigenvector $\Psi_{2,1}$ and (<ref>), the total count is $112k^2$.
Corresponding to the eigenvector $\Psi_{2,2}$ and (<ref>), the total count is also $112k^2$.

The difference, in absolute value, is therefore $224k^2$ in each case.
It follows that the second eigenvalue $\lambda_2(k)$ contributes $224(c_{2,1}+c_{2,2})\lambda_2^r(k)k^2$ to the difference between the edge cuttings numbers of $h_1,\ldots,h_6$ and $h_7,\ldots,h_{12}$.

Naturally, we can choose a starting vector $\bfw_0$ for the geodesic in question such that $c_{2,1}+c_{2,2}\ne0$.
This is clearly possible, since $c_{2,1}+c_{2,2}=0$ always would violate the linear independence of the eigenvectors $\Psi_{2,1}$ and $\Psi_{2,2}$.

So the deviation from the starting point comes from the second largest eigenvalue, with order of magnitude $\lambda_2^r$ compared to the order of magnitude
$\lambda_1^r$ of the main term.
Choosing $T=\lambda_1^r$, we have $\lambda_2^r\asymp T^{\kappa_0}$, where

\begin{equation}\label{eq7.4.26}
\kappa_0=\kappa_0(k)=\frac{\log\lambda_2}{\log\lambda_1}=\frac{2\log k+\log112}{2\log k+\log144}+o(1),
\end{equation}

in view of (<ref>) and (<ref>), is the irregularity exponent of a $1$-direction geodesic of slope

\begin{equation}\label{eq7.4.27}
\alpha_k=[12k;12k,12k,12k,\ldots]=6k+\sqrt{36k^2+1}=\sqrt{\lambda_1(k)}
\end{equation}

on the period-surface $\Bil(2;\mathrm{W})$ in Figure 7.4.8.

Again $\kappa_0=\kappa_0(k)\to1$ as $k\to\infty$.
So we have just established $T^{\kappa_0}=T^{1-\eps}$ size super-fast escape rate to infinity for this infinite billiard with the explicit class of quadratic irrational slopes in (<ref>) where the parameter $k\ge1$ is any integer.

It is easy to see that the irregularity exponent $\kappa_0=\kappa_0(k)$ in (<ref>) is precisely the escape rate to infinity of this infinite billiard.
The exponent of the escape rate to infinity cannot be larger than the expression (<ref>) coming from the two largest eigenvalues.

In view of the examples in Sections <ref> and <ref> so far, it seems plausible to conjecture that every periodic infinite polysquare billiard with at least one infinite street exhibits super-fast escape rate to infinity for some concrete infinite class of quadratic irrational slopes.
It would be nice to prove this conjecture in general.

We complete this section with a hybrid example.

Our final example is modified from the Ehrenfest wind-tree model.
Consider a $2$-direction billiard trajectory in the region in Figure 7.4.11.
Here the building block of this infinite polysquare region is a $3\times3$ square, with the middle square missing and with two vertical walls.
Note that there is no billiard in the vertical direction, so this is a hybrid problem.
The trajectory keeps on going up vertically, or going down vertically, one way but not both.

As usual, this infinite hybrid model is equivalent to a $1$-direction geodesic flow on an infinite polysquare translation surface that we denote by $\Bil(\infty;2;\mathrm{H})$.
Here the index $2$ indicates that this is double periodic, and the letter $\mathrm{H}$ refers to the hybrid model.
To construct $\Bil(\infty;2;\mathrm{H})$, we take one of the building blocks, and unfold the $2$-direction billiard flow on it to a $1$-direction geodesic flow on a $2$-copy version of it, obtained by reflecting vertically.
Note that each $2$-copy version has a right-neighbor, a left-neighbor, a down-neighbor and an up-neighbor in $\Bil(\infty;2;\mathrm{H})$, and we need appropriate edge identification for gluing them together.

The period-surface $\Bil(2;\mathrm{H})$ of $\Bil(\infty;2;\mathrm{H})$ is shown in the picture on the left in Figure 7.4.12.

The horizontal and vertical streets of $\Bil(2;\mathrm{H})$ are indicated in the picture on the right in Figure 7.4.12 where, for instance, the entries
$\leftrightarrow3$ and $\updownarrow4$ in a square face indicates that the square face is on the $3$-rd horizontal street and the $4$-th vertical street.
It is easy to see that $\Bil(2;\mathrm{H})$ has $5$ horizontal streets, of which $4$ are of length $3$ and $1$ is of length $4$.
It also has $6$ vertical streets, of which $4$ are of length $3$ and $2$ are of length $2$.

Consider now a $1$-direction geodesic starting from some vertex of $\Bil(2;H)$ with slope $\alpha$ given by (<ref>) with $m=n=12$.

With $J_1=J_4=\{1,2,5\}$, for the horizontal streets corresponding to $\bfu_1,\bfu_4$ highlighted in the picture on the left in Figure 7.4.13, we have

\begin{align}
\nonumber
\\
\label{eq7.4.28}
\end{align}

With $J_2=J_5=\{3,4,6\}$, for the horizontal streets corresponding to $\bfu_2,\bfu_5$ highlighted in the picture on the right in Figure 7.4.13, we have

\begin{align}
\nonumber
\\
\label{eq7.4.29}
\end{align}

With $J_3=\{1,2,3,4\}$, for the horizontal street corresponding to $\bfu_3$ highlighted in Figure 7.4.14, we have

\begin{align}\label{eq7.4.30}
\nonumber
\\
\end{align}

It follows from (<ref>)–(<ref>) that the street-spreading matrix is given by

\begin{equation}\label{eq7.4.31}
\bfS=\begin{pmatrix}
\end{pmatrix}.
\end{equation}

This has non-zero eigenvalues and corresponding eigenvectors given by

\begin{align}
\tau_1=144,
\psi_1=(3,3,4,3,3)^T,
\nonumber
\\
\tau_2=112,
\psi_2=(-1,1,0,-1,1)^T,
\nonumber
\\
\tau_3=16,
\psi_3=(1,1,-4,1,1)^T.
\nonumber
\end{align}

Note that the street-spreading matrix $\bfS$ given by (<ref>) corresponds to the choice of parameters $m=n=12$.
Switching to $m=n=12k$, where $k\ge1$ is any integer, we obtain the new street-spreading matrix $\bfS(k)=k^2\bfS$ simply by multiplying the matrix $\bfS$ by $k^2$.
Then of course the eigenvalues are also multiplied by $k^2$, but the eigenvectors remain the same.
Naturally the $2$-step transition matrix $\bfA$ is modified to $\bfA(k)$.

We next determine some of the eigenvalues of $\bfA(k)$ using (<ref>).
The largest eigenvalue of $\bfA(k)$ is

while the second largest eigenvalue of $\bfA(k)$ is

with eigenvector of the form

\begin{align}
\Psi_2
\nonumber
\end{align}

As observed earlier, there is no billiard in the vertical direction, so we now proceed to study its horizontal behaviour.
We thus proceed to find the edge cutting numbers of $v_1,v_2,v_3$ and $v_4,v_5,v_6$; see Figure 7.4.12.

It can be shown that the eigenvalue $\lambda_1$ does not contribute to their difference, although the Gutkin–Veech theorem does not apply here.
So we concentrate our attention on the contribution from the eigenvector $\Psi_2$ of $\bfA\vert_\VVV$.

The eigenvector of $\bfA(k)$ corresponding to the eigenvector $\Psi_2$ of $\bfA\vert_\VVV$ is given by

\begin{equation}\label{eq7.4.32}
\end{equation}

Here, note that none of $\bfu_1,\ldots,\bfu_5$ makes any non-zero contribution, as only units of type $\nuparrow$ can contribute to the count.
For the contributions from $\bfv_1,\ldots,\bfv_5$, we use (<ref>).

We first find the number of almost vertical units counted in (<ref>) that cut the edges $v_1,v_2,v_3$; see Figure 7.4.12.
Since $\bfv_3$ does not feature in (<ref>), it is not necessary to any counting for it.
Clearly we have a count of $4k$ for each of $\bfv_1,\bfv_4$, and none for $\bfv_2,\bfv_5$, making a total of $-8\tau_2^*k$.

We next find the number of almost vertical units counted in (<ref>) that cut the edges $v_4,v_5,v_6$; see Figure 7.4.12.
Clearly we have a count of $4k$ for each of $\bfv_2,\bfv_5$, and none for $\bfv_1,\bfv_4$, making a total of $8\tau_2^*k$.

The difference, in absolute value, is therefore $16\tau_2^*k$.
Thus the second eigenvalue $\lambda_2(k)$ contributes $16c_2\tau_2^*\lambda_2^r(k)k$ to the difference between the edge cuttings numbers of $v_1,v_2,v_3$ and $v_4,v_5,v_6$.

Again the horizontal deviation from the starting point comes from the second largest eigenvalue, with order of magnitude $\lambda_2^r$ compared to the order of magnitude $\lambda_1^r$ of the main term.

§.§ Geodesics with arbitrary starting points

In Sections <ref> and <ref>, we have exhibited examples of $1$-direction geodesics in infinite polysquare translation surfaces that start from a vertex and which exhibit super-fast escape rate to infinity.
In Example <ref>, we have also shown that there are $1$-direction geodesics in square-maze translation surfaces that start from a vertex and which exhibit super-slow escape rate to infinity.

We now describe a method by which we can relax the restriction that the starting point of the geodesic has to be a vertex of a polysquare translation surface, so that only the angle $\alpha$ of the geodesic matters, and can still establish results on the escape rate to infinity as before.

Note, first of all, that a $1$-direction geodesic modulo one is equivalent to a torus line in the unit square.
To help us visualize the situation even more clearly, we can replace the unit square by $\Rr^2$ and the torus line by a line on $\Rr^2$.

Next, note that we are considering slopes $\alpha$ that are badly approximable numbers.
Such numbers satisfy the following two properties.

Suppose that $\alpha$ is badly approximable.
Then there exists a constant $C^*=C^*(\alpha)>0$ such that for every real number $\ell\ge1$, every segment of length $\ell$ of a straight line of slope
$\alpha$ has distance at most $C^*/\ell$ from the nearest integer lattice point in $\Zz^2$.

Suppose that $\alpha$ is badly approximable.
Then there exists a constant $C^{**}=C^{**}(\alpha)>0$ such that for any real number $\ell\ge1$, any tilted rectangle with one side starting from the origin, with slope $\alpha$ and length $\ell$, and with the perpendicular side of length $C^{**}/\ell$, does not contain any integer lattice point in $\Zz^2$ except the origin.

The idea is to consider geodesics that are parallel to our given geodesic and which share some of its characteristics.