# Periodic paths on the pentagon, double pentagon and golden L

Diana Davis and Samuel Lelièvre

###### Abstract

We give a tree structure on the set of all periodic directions on the golden
L, which gives an associated tree structure on the set of periodic directions
for the pentagon billiard table and double pentagon surface. We use this to
give the periods of periodic directions on the pentagon and double pentagon.
We also show examples of many periodic billiard trajectories on the pentagon,
which are strikingly beautiful, and we describe some of their properties.
Finally, we give conjectures and future directions based on experimental
computer evidence.

##  1 Main results, introduction and background

###  1.1 Periodic trajectories in polygons: existence, abundance, enumeration

One problem in billiards is whether every polygonal billiard table has a
periodic path. This question is open and of great interest even for triangles,
where Richard Schwartz showed that every triangle whose largest angle is less
than $100^{\circ}$ has a periodic path, and this result was recently extended
by four authors to $112.3^{\circ}$ [S08, TGMM18]. From Howard Masur’s result
we know that every rational polygon has a periodic path, and in fact countably
many [Ma86]. That every _regular_ polygon has a periodic path is clear; a path
connecting midpoints of any two edges extends to a periodic path.

One specific family of polygons is those that tile the plane by reflections:
the square, rectangle, equilateral triangle, hexagon, half-equilateral
triangle, and isosceles right triangle. For those, because of the possibility
to develop to the plane, it is easy to describe the countable set of
directions in which there are periodic paths, and to enumerate those periodic
paths, as we recall in §1.3 for the case of the square.

In a sense, the next interesting polygons to look at are non-tiling regular
polygons and non-tiling rational triangles, where existence is known but
enumeration is new. Here we enumerate periodic paths on the regular pentagon,
which allows us to produce pictures of striking beauty. Several examples of
such paths appear below.

> Figure 1: Some periodic paths on the regular pentagon. These are the
> trajectories 1000-short, 1231-long, 102-short, and 133-short, respectively
> (see §2.2).

###  1.2 Main results

In this paper, we use the relationship between regular pentagons and the
_golden L_ , a right-angled, L-shaped translation surface with edge lengths 1,
the golden ratio $\phi$, and its inverse $1/\phi$ (Definition 2.1). We give a
description of all saddle connection vectors on the golden L in a quaternary
_tree_ analogous to the Farey tree; this tree is our main tool.

######  Theorem 2.11.

Let $A^{*}$ be the set of all finite words on the alphabet
$A=\\{\sigma_{0},\sigma_{1},\sigma_{2},\sigma_{3}\\}$ of linear maps taking
the positive quadrant to its four partitioning sectors (Definition 2.5). Then
the set of all vectors $\sigma\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$, for $\sigma\in A^{*}$, gives the set of short
cylinder vectors in the positive quadrant.

A periodic direction on the golden L is a direction with slope in
$\mathbf{Q}[\sqrt{5}]$ [AS09]. There are, of course, many ways to express such
a slope as a vector. We use the long saddle connection vector in each periodic
direction; this is the vector obtained from our algorithm to generate the
tree.

######  Theorem 2.22.

Corresponding to any periodic direction vector $\mathbf{v}$ in the first
quadrant on the golden L is a unique sequence $a_{v}=(k_{0},k_{1},...,k_{n})$
of sectors as in Definition 2.19, such that ${\bf
v}=\ell_{v}\sigma_{k_{0}}\ldots\sigma_{k_{n}}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$ for some length $\ell_{v}$. Then the long saddle
connection vector in the direction $\mathbf{v}$ has $\ell_{v}=1$ and is thus
$\sigma_{k_{0}}\ldots\sigma_{k_{n}}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$. Furthermore, the long saddle connection vector is
of the form $[a+b\phi,c+d\phi]$ for $a,b,c,d\in\mathbf{N}$.

We use the saddle connection vector in this canonical representation to
determine the combinatorial period of the associated double pentagon
trajectory:

######  Theorem 3.15.

In each periodic direction, there are two periodic trajectories, whose
corresponding direction in the golden L has long saddle connection vector
$[a+b\phi,c+d\phi]$ as above. The short trajectory in this direction has
holonomy vector $\mathbf{v}=[a+b\phi,c+d\phi]$ in the golden L and
combinatorial period $2(a+b+c+d)$ in the double pentagon, and the long
trajectory in this direction has holonomy vector $\phi\mathbf{v}$ in the
golden L and combinatorial period $2(a+2b+c+2d)$ in the double pentagon.

This result is directly analogous to the corresponding result on the square
billiard table, where a trajectory with slope $p/q$ has associated holonomy
vector $[q,p]$ and period $2(p+q)$; see §3.2.

###  1.3 Analogous known results for the square

For a rational billiard table, the set of periodic directions is countable and
dense [Ma86]. In this paper, we describe the structure of this set for the
pentagon and the golden L. The structure of the set of periodic directions is
well understood for the square torus and square billiard table. We summarize
those results here, as we will extend them to the golden L surface, the double
pentagon surface and the pentagon billiard table.

######  Theorem 1.1.

For the square torus and square billiard table:

  1. 1.

The periodic directions on the square torus and billiard table are those with
rational slope.

  2. 2.

A trajectory with slope $a/b$ (in lowest terms) on the square torus has
combinatorial period $a+b$, and on the square billiard table has combinatorial
period $2(a+b)$.

###### Proof.

We assume $a,b\in\mathbf{N}$ and that $a/b$ is in lowest terms, and we assume
as usual that the trajectory does not hit a vertex.

  1. 1.

A trajectory with rational slope meets only a finite number of points on the
edges of the table, so it must be periodic; unfolding a periodic path yields a
trajectory with rational slope.

  2. 2.

Developing the square torus to a grid of squares, we can see that a line of
slope $a/b$ on the square grid repeats itself after going up $a$ squares (down
if $a<0$) and to the right $b$ squares (left if $b<0$). The number of edges
crossed is thus $a+b$. Unfolding the square billiard table to the square torus
requires $4$ copies of the table inside each torus, which is equivalent to
adding grid lines at half integers, so the path repeats after $2(a+b)$ edge
crossings of these lines.

∎

For more discussion and for illustrations, see [D17], §4.

###  1.4 Previous work on the pentagon, double pentagon and golden L

William Veech studied billiards in regular polygons, including the regular
pentagon [V92]. Curt McMullen characterized the L-shaped polygons that are
lattice (Veech) surfaces; one such surface is the golden L, an example given
in that paper [Mc03, Theorem 9.2 and Figure 4].

In [DFT11], the first author, Dmitry Fuchs and Sergei Tabachnikov explored
periodic directions on the double pentagon surface and pentagon billiard
table, as we do here. There is significant overlap between that paper and the
current paper, but we have a different emphasis. In particular, in that paper
the hyperbolic plane in the Poincaré model is used as the basis for all of the
direction computations on the pentagon, and the results on periodic directions
are given in these terms. By contrast, in the current paper we use translation
surfaces, in particular the golden L, as the basis for our direction
computations. Several of our results are the same $-$ our Theorem 2.11
characterizes periodic directions using a Farey-like tree based on the golden
L, as does their Theorem 1 using the hyperbolic plane; their matrices $X_{i}$
in §2.2 are the same as our matrices $\sigma_{i}$ in Definition 2.5; their
generating transition diagrams in equation (1) and “enhanced graphs” in Figure
14 are the same as our transition diagrams (under a permutation on edge
labels) in Lemma 3.6; their Theorems 7 and 10 combine to give a result
analogous to our Combinatorial Period Theorem 3.15, but our methods for that
Theorem are more direct.

The following results are unique to our paper: Theorems 2.11 and 2.22
characterizing saddle connections on the golden L, Theorem 2.15 about the
symmetry of the tree of directions, Corollary 3.18 about the structure of
cutting sequences on the double pentagon, §3.3-3.4 and Theorem 3.26 about the
Veech group of the necklace surface and the applications to the symmetries of
trajectories on the pentagon billiard, Corollary 3.27 giving the proportion of
paths with and without rotational symmetry, Theorem 3.29 and Proposition 3.33
giving details about how different types of paths arise, Propositions 4.1 and
4.2 about the periods that arise in the double pentagon and the pentagon
billiard, and §4.2 about non-equidistribution of certain periodic billiard
paths. Furthermore, the pictures from our `Sage` program (see Appendix B)
allow us to look at periodic trajectories on the pentagon for the first time,
which are strikingly beautiful and afford us additional insights about the
system.

The second author, Jayadev Athreya and Jon Chaika studied slope gaps on the
golden L, which is the same surface that we use here, but with an entirely
different focus [ACL15].

### Acknowledgements

This project began at the Oberwolfach workshop “Flat Surfaces and Dynamics on
Moduli Space” in 2014. We are grateful to “Dynamics and Geometry in the
Teichmüller Space” at CIRM in 2015, “Flat Surfaces and Dynamics of Moduli
Space” at BIRS in Oaxaca in 2016, “Teichmüller Space, Polygonal Billiard,
Interval Exchanges” at CIRM in 2017, the Billiards Cluster at Tufts University
in 2017, and “Teichmüller Dynamics, Mapping Class Groups and Applications” at
Institut Fourier in 2018, which allowed for our continued collaborations.

All of our computations were done in `Sage`, many using `CoCalc` [S+18, CC18].
We thank Pierre Arnoux, Pascal Hubert, Sergei Tabachnikov, Ronen Mukamel,
Barak Weiss, Corinna Ulcigrai, Curt McMullen and John Smillie for productive
conversations about this project.

##  2 The double pentagon and the golden L

A classical construction relates billiards on a rational polygon to a linear
flow on a translation surface, by reflecting the billiard table across its
edges until each edge is matched with an opposite parallel edge of the same
length. It is easier to study linear flows than billiard paths, because the
direction of the flow does not change.

We first unfold the pentagon to its corresponding flat surface, called the
_necklace_ , which is tiled by 10 copies of the pentagon. Veech gave a
presentation of the necklace in [V92, Figure 2.3]. The necklace is a 5-fold
cover of the _double pentagon_. For previous work on the double pentagon, see
[D13]. Since the double pentagon is simpler, we consider trajectories there
instead of on the necklace. Then we perform an affine automorphism of the
double pentagon to transform it into the golden L. The golden L is easier to
work with than the double pentagon, because it is a right-angled surface.

###  2.1 From the pentagon to the necklace, double pentagon and golden L

We unfold the pentagon billiard table across its edges, labeling each image
edge with the same label as the original billiard table edge, until each edge
has an oppositely-oriented, parallel edge with the same label. This requires
10 copies of the pentagon (the order of the dihedral group $D_{5}$), which can
be arranged in a circle or _necklace_ , shown in the middle of Figure 2. The
edge identifications, which are inherited from the unfolding of the pentagon
table, are as labeled in the figure.

Figure 2: Left: the regular pentagon billiard table. Center: the unfolding of
the pentagon billiard table into the necklace translation surface. The number
1-5 (along with the color) indicates which edge of the billiard table was
unfolded to get that edge, and the letter indicates which of the copies are
glued to each other. Right: The double pentagon translation surface.

######  Definition 2.1.

The golden ratio $\phi=\frac{1+\sqrt{5}}{2}\approx 1.618\ldots$ is the
positive solution of $x^{2}=x+1$. We denote by $\bar{\phi}$ its inverse, i.e.
$\bar{\phi}=1/\phi=\phi-1.$ Note that $\bar{\phi}$ satisfies the equation
$x^{2}=1-x$.

The _golden L_ is an L-shaped translation surface with length-to-thickness
ratio the golden ratio. It is built out of a $1\times 1$ square with
$1\times\bar{\phi}$ and $\bar{\phi}\times 1$ rectangles glued to it along
their length-$1$ sides (see the left side of Figure 3, and Figure 5). The
_double pentagon_ is made of two regular pentagons, one of which is a
reflected copy of the other, with opposite parallel edges identified.

A _cylinder_ is a maximal parallel family of periodic trajectories. The golden
L and double pentagon have two cylinders in each periodic direction: a _long
cylinder_ and a _short cylinder_. In Figure 3, the long and short cylinders in
the horizontal direction are shaded dark and light grey, respectively. The two
cylinders have the same modulus (aspect ratio) and have length ratio $\phi$.

We now turn to the double pentagon, and start by giving the affine equivalence
with the golden L.

######  Definition 2.2.

Let $P=\begin{bmatrix}1&\cos(\pi/5)\\\ 0&\sin(\pi/5)\end{bmatrix}$, so
$P^{-1}=\begin{bmatrix}1&\cos(\pi/5)\\\ 0&\sin(\pi/5)\end{bmatrix}^{-1}$.

######  Lemma 2.3.

The matrix $P$ takes the golden L surface to the double pentagon surface, and
its inverse $P^{-1}$ takes the double pentagon to the golden L. In particular,
they take long cylinder vectors to long cylinder vectors, and the same for
short cylinder vectors.

###### Proof.

We can cut up the golden L on the left side of Figure 3, along the dark lines
shown, and reassemble it by translation, respecting edge identifications, as
shown with the dashed arrows, into the sheared version of the double pentagon
outlined in the same picture. Then we can shear the double pentagon to the
right, obtaining the double regular pentagon on the right side of Figure 3.

We construct the shearing matrix that takes the golden L to the double
pentagon, column by column. The vectors $[1,0]$ and $[0,1]$ are shown as thick
arrows in the golden L on the left side of Figure 3. The unit horizontal edge
$[1,0]$ stays as a unit horizontal. The vertical edge is mapped to a diagonal
of the pentagon, also of unit length, and making an angle of $\pi/5$ with the
positive horizontal. ∎

######  Corollary 2.4.

A direction $\mathbf{v}$ is periodic on the double pentagon if and only if the
direction $P^{-1}\mathbf{v}$ is periodic on the golden L.

###### Proof.

The saddle connection in direction $\mathbf{v}$ on the double pentagon can be
cut-and-pasted into a saddle connection in direction $P^{-1}\mathbf{v}$ on the
golden L. ∎

Figure 3: The cut-and-paste, shear, and vertical compression between the
golden L and the double pentagon. For a similar picture and more discussion,
see [Mc03, Figure 4].

We are interested in linear flows on these surfaces, especially periodic
trajectories. A linear flow is defined by a nonzero vector. By the $4$-fold
rotational symmetry of the golden L, we may reduce our attention to flows
whose direction $\theta$ satisfies $\theta\in[0,\pi/2)$. By the $10$-fold
rotational symmetry of the double pentagon, we may reduce our attention to
flows whose direction $\theta$ satisfies $\theta\in[0,\pi/5)$. The linear map
$P$ between the golden L and the double pentagon takes these sectors to each
other.

With this correspondence, we take advantage of the best of both worlds: we
consider trajectories on both sides, thinking of their vectors on the golden L
side (where vectors are easier), and thinking of their cutting sequences on
the double pentagon side (where cutting sequences are nice).

###  2.2 The tree of saddle connection vectors for the golden L

As stated in the introduction, we wish to study the directions of periodic
trajectories on the golden L. Periodic directions on the golden L are all of
those with slope in $\mathbf{Q}[\sqrt{5}]$. This is quite special; for
example, periodic directions on the double regular heptagon are a proper
subset of the field that they generate [AS09].

Periodic trajectories come in cylinders (Definition 2.1), and have a naturally
associated vector, the holonomy vector of this cylinder. In each direction of
a periodic trajectory on the golden L, there are two cylinders of periodic
trajectories, one short and one long, with length ratio $\phi$. The horizontal
and vertical directions are examples of this. Examples for pentagon billiard
trajectories in each of the two cylinders are in Figure 4.

> Figure 4: Two pairs of short and long billiard trajectories. For each pair,
> the trajectory is in the same direction. Notice that within each pair, the
> “empty spots” in one tend to correspond to “dense spots” in the other. This
> is because at an empty spot in the left pentagon, no core curve of a short
> cylinder crosses there, so the core curves of the long cylinder are more
> likely to cross that area in the right pentagon, and vice-versa. These have
> tree words 23 and 130, respectively.

In each periodic direction, there are also two lengths of saddle connection,
with length ratio $\phi$. The long saddle connection vector coincides with the
short cylinder vector, and the sum of the long and short saddle connection
vectors coincides with the long cylinder vector. Again, the horizontal and
vertical directions are examples of this.

Our approach to enumerating periodic directions is to take advantage of the
affine self-similarities of the golden L (Veech’s “hidden symmetries”). We
single out four such affine symmetries in the following definition.

######  Definition 2.5.

Let $\Sigma=\left\\{\left[\begin{smallmatrix}x\\\
y\end{smallmatrix}\right]:x>0,y\geq 0\right\\}$, the positive cone (first
quadrant). For $i=0,1,2,3$, define $\Sigma_{i}$ and $\sigma_{i}$, as follows.

$\sigma_{0}=\left[\begin{smallmatrix}1&\phi\\\ 0&1\end{smallmatrix}\right]$,
and $\Sigma_{0}=\sigma_{0}\Sigma=\left\\{\left[\begin{smallmatrix}x\\\
y\end{smallmatrix}\right]:0\leq y<\bar{\phi}x\right\\}$.

$\sigma_{1}=\left[\begin{smallmatrix}\phi&\phi\\\
1&\phi\end{smallmatrix}\right]$, and
$\Sigma_{1}=\sigma_{1}\Sigma=\left\\{\left[\begin{smallmatrix}x\\\
y\end{smallmatrix}\right]:\bar{\phi}x\leq y<x\right\\}$.

$\sigma_{2}=\left[\begin{smallmatrix}\phi&1\\\
\phi&\phi\end{smallmatrix}\right]$, and
$\Sigma_{2}=\sigma_{2}\Sigma=\left\\{\left[\begin{smallmatrix}x\\\
y\end{smallmatrix}\right]:x\leq y<\phi x\right\\}$.

$\sigma_{3}=\left[\begin{smallmatrix}1&0\\\ \phi&1\end{smallmatrix}\right]$,
and $\Sigma_{3}=\sigma_{3}\Sigma=\left\\{\left[\begin{smallmatrix}x\\\
y\end{smallmatrix}\right]:\phi x\leq y\right\\}$.

######  Lemma 2.6.

We have the following facts about the $\sigma_{i}$ and $\Sigma_{i}$:

  1. 1.

The first quadrant $\Sigma$ is partitioned into $\Sigma_{0}$, $\Sigma_{1}$,
$\Sigma_{2}$, $\Sigma_{3}$.

  2. 2.

The matrices $\sigma_{i}$ generate the Veech group.

###### Proof.

The first claim is clear by construction. For the second, notice that
$\sigma_{0}\cdot\sigma_{3}^{-1}\cdot\sigma_{0}\cdot\sigma_{3}^{-1}\cdot\sigma_{0}=\rho=\left[\begin{smallmatrix}0&-1\\\
1&0\end{smallmatrix}\right]$, the matrix for a quarter-turn rotation. Together
$\rho$ and $\sigma_{0}$ generate the Veech group for the golden L, the
$(2,5,\infty)$ triangle group, a maximal discrete subgroup of
$SL_{2}(\mathbf{R})$, which therefore must be the full Veech group of the
golden L. ∎

Figure 5: The golden L, and the vectors that are the column vectors for the
matrices $\sigma_{i}$ in Definition 2.5, which bound the associated sectors
$\Sigma_{i}$ in the first quadrant. The slopes of the middle three dividing
lines are $\bar{\phi}$, $1$, and $\phi$.

We will use these matrices to generate all of the saddle connection vectors.

######  Remark 2.7.

The matrices $\sigma_{0}$ and $\sigma_{3}$ alone generate the Veech group, but
we choose to use all four $\sigma_{i}$ because they generate the positive
semigroup of the Veech group.

######  Definition 2.8.

Let $\Lambda$ (respectively $\Lambda_{i}$) be the intersection of the set of
_short cylinder vectors_ (equivalently, the set of long saddle connection
vectors) with $\Sigma$ (respectively $\Sigma_{i}$).

######  Lemma 2.9.

The $\Lambda_{i}$s form a partition of $\Lambda$, and
$\Lambda_{i}=\sigma_{i}\Lambda$.

###### Proof.

The first statement is clear by construction.

We have $\sigma_{i}\Lambda\subseteq\Lambda\cap\Sigma_{i}$ because
$\Lambda\subseteq\Sigma$, so the images satisfy
$\sigma_{i}\Lambda\subseteq\sigma_{i}\Sigma=\Sigma_{i}$. Since long saddle
connection vectors go to long saddle connection vectors, for any
$v\in\Lambda$, $\sigma_{i}v$ is a long saddle connection vector that lies in
$\Sigma$, so it is also in $\Lambda$.

For the reverse inclusion $\sigma_{i}\Lambda\supseteq\Lambda\cap\Sigma_{i}$,
let $v\in\Lambda\cap\Sigma_{i}$. Then $\sigma_{i}^{-1}v$ is also a long saddle
connection vector, and it is in $\Sigma$, so it is in $\Lambda$, and so
$v=\sigma_{i}(\sigma_{i}^{-1})v\in\Lambda_{i}$. ∎

######  Definition 2.10.

Let $A=\\{\sigma_{0},\sigma_{1},\sigma_{2},\sigma_{3}\\}$, and let $A^{*}$ be
the set of all finite words on the alphabet $A$, including the empty word. For
an element $\sigma=\sigma_{k_{1}},\sigma_{k_{2}},\ldots,\sigma_{k_{n}}$ of
$A^{*}$, let $m(\sigma)=\sigma_{k_{n}}\cdots\sigma_{k_{2}}\sigma_{k_{1}}$ be
the product of the matrices comprising $\sigma$, in order of application.

######  Theorem 2.11 (Tree Theorem).

The set of all vectors $m(\sigma)\cdot\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$, for $\sigma\in A^{*}$, gives the entire set
$\Lambda$. Furthermore, leaving out the words that start with $\sigma_{0}$, we
get a bijection between the set of
$m(\sigma)\cdot\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$ and
$\Lambda$.

###### Proof.

Notice that, for vectors in $\Sigma$, all of the $\sigma_{i}$s send them to
longer vectors (except $\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$, which is sent to itself). Thus the
$\sigma_{i}^{-1}$s send vectors in $\Lambda_{i}$ to shorter vectors in
$\Lambda$ (except $\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$,
which is sent to itself). Since $\Lambda$ is discrete, the image eventually
lands on the shortest vector, $\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$. Working backwards, starting with
$\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$ and applying the
$\sigma_{i}$s, gives the result. Since
$\sigma_{0}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]=\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$, and this is the only time when
$\sigma_{i}\mathbf{v}=\mathbf{v}$ for $\mathbf{v}\in\Lambda$, we omit words
starting with $\sigma_{0}$ and get the desired bijection. ∎

Theorem 2.11 gives a way of generating all saddle connection vectors on the
golden L (Figure 6). For a different presentation of the same construction,
see [DFT11, §2.2 and Theorem 4].

Figure 6: (a) Golden L long saddle connection vectors from the first six
levels of the tree (3075 points). (b) All of the long saddle connection
vectors in $[0,100]^{2}$, which requires 61 levels of the tree; see Corollary
2.13 (6007 points). Points are color coded by level; the color scale is
different in the two pictures.

######  Proposition 2.12.

At depth $k\geq 1$ in the tree, the shortest vectors are
$\left[\begin{smallmatrix}k\phi\\\ 1\end{smallmatrix}\right]$ and
$\left[\begin{smallmatrix}1\\\ k\phi\end{smallmatrix}\right]$.

###### Proof.

Beginning with $\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$,
applying the matrices $\sigma_{1},\sigma_{2},\sigma_{3}$ yields the vectors
$\left[\begin{smallmatrix}\phi\\\ 1\end{smallmatrix}\right]$,
$\left[\begin{smallmatrix}1\\\ \phi\end{smallmatrix}\right]$,
$\left[\begin{smallmatrix}\phi\\\ \phi\end{smallmatrix}\right]$, respectively.
The latter is longer than the first two, so we discard it. Then the result
follows from the following claims for any
$\mathbf{v}=\left[\begin{smallmatrix}a\\\ b\end{smallmatrix}\right]$ in the
positive cone:

  1. 1.

$|\sigma_{0}\mathbf{v}|<|\sigma_{1}\mathbf{v}|$ and
$|\sigma_{3}\mathbf{v}|<|\sigma_{2}\mathbf{v}|$.

  2. 2.

$|\sigma_{0}\left[\begin{smallmatrix}a\\\
b\end{smallmatrix}\right]|<|\sigma_{3}\left[\begin{smallmatrix}a\\\
b\end{smallmatrix}\right]|$ if $a>b$ and
$|\sigma_{0}\left[\begin{smallmatrix}a\\\
b\end{smallmatrix}\right]|>|\sigma_{3}\left[\begin{smallmatrix}a\\\
b\end{smallmatrix}\right]|$ if $a<b$.

Both of these follow from multiplying out the matrices and considering the
lengths of the resulting vectors. By (1), the shortest vector will always be a
product of $\sigma_{0}$ and $\sigma_{3}$ applied to $\mathbf{v}$. By (2),
applying ${\sigma_{0}}^{k}$ or ${\sigma_{3}}^{k}$ results in a shorter vector
than a product including both matrices, and which one of these is shorter
depends on the location of the vector. ∎

######  Corollary 2.13.

To obtain all of the vectors in the square $[0,N]^{2}$, it is necessary to go
to depth $N/\phi$ in the tree.

For example, in Figure 6, depth $61=\lfloor 100/\phi\rfloor$ is required to
get everything in $[0,100]^{2}$.

######  Definition 2.14.

We refer to the _tree of periodic directions_ as the tree given by the
construction in Theorem 2.11: The base node of the tree is vector
$\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$, the first level is
given by $\sigma_{i}\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$
for $i\in\\{1,2,3\\}$, and the $(n+1)^{\text{st}}$ level for $n\geq 1$ is
given by
$\sigma_{j_{n}}\cdots\sigma_{j_{1}}\sigma_{i}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$ for $i\in\\{1,2,3\\},j_{k}\in\\{0,1,2,3\\}$. The
_tree word_ corresponding to a periodic direction is given by the of
subscripts of the transformations $\sigma_{i}$ in the order of application,
e.g. tree word 120 for the direction
$\sigma_{0}\sigma_{2}\sigma_{1}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$.

$1$$2$$3$$0$$1$$2$$3$$0$$1$$2$$3$$0$$1$$2$$3$$10$$11$$12$$13$$20$$21$$22$$23$$30$$31$$32$$33$
Figure 7: The first two levels in the tree of periodic directions on the
regular pentagon. The short trajectory corresponding to the core curve of the
cylinder is shown for each.

The short pentagon billiard trajectory for each direction in the tree is in
Figure 7. The picture hints that the tree is left-right symmetric:

######  Theorem 2.15.

The tree is symmetric: to get an identical trajectory, for the first digit of
the tree word we do the swap $1\leftrightarrow 3$, and for all the rest of the
digits, we do the swap $0\leftrightarrow 3$ and $1\leftrightarrow 2$.

We need the following results, which follow from matrix multiplication.

######  Lemma 2.16.

If $A=\left[\begin{smallmatrix}a&b\\\ c&a\end{smallmatrix}\right]$, $B=A^{T}$,
and $S=\left[\begin{smallmatrix}0&1\\\ 1&0\end{smallmatrix}\right]$, then

  1. 1.

$SAS=B$ and $SBS=A$ and $AS=SB$ and $SA=BS$.

  2. 2.

$A\left[\begin{smallmatrix}x\\\
y\end{smallmatrix}\right]=SB\left[\begin{smallmatrix}y\\\
x\end{smallmatrix}\right]$.

###### Proof of Theorem 2.15.

Observe that $v_{2}:=\sigma_{2}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$ satisfies $Sv_{2}=v_{2}$, and
$v_{1}:=\sigma_{1}\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$ and
$v_{3}:=\sigma_{3}\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$
satisfy $Sv_{1}=v_{3}$ and $Sv_{3}=v_{1}$. This proves the result for words of
length 1. Since we can apply Lemma 2.16 to $(A,B)=(\sigma_{0},\sigma_{3})$ and
to $(A,B)=(\sigma_{1},\sigma_{2})$, the result follows by induction.

The above shows that the directions on the golden L corresponding to a tree
word and its swapped word are symmetric across the line $y=x$. Applying the
matrix $P$ to directions that are symmetric on the golden L results in
directions that are symmetric across the long diagonal of the double pentagon,
so they stay symmetric when lifted to the necklace, and when folded into a
pentagon billiard trajectory. ∎

######  Remark 2.17.

On the golden L or on the double pentagon, the central representative of a
cylinder of periodic trajectories passes through two regular Weierstrass
points. In the double pentagon, these are the midpoints of the edges; in the
golden L these are the centers of the rectangles and square, and the midpoints
of the short sides of the rectangles. After lifting to the necklace and
folding to the pentagon billiard table, both points are mapped to the midpoint
of the same edge, or of all of the edges in the case of trajectories with
rotational symmetry. One can observe in the pictures that anytime a trajectory
hits the midpoint of a given edge, it does so twice.

###  2.3 Directions to saddle connection vectors on the golden L

Theorem 2.11 gives us all of the saddle connection vectors for the golden L.
In this section, we answer a related question: _Given a periodic direction on
the golden L, what is the saddle connection vector in that direction?_

######  Remark 2.18.

Given a rational slope $a/b$ in lowest terms on the square torus, we can
easily see that the associated saddle connection vector in that direction is
$[b,a]$. However, suppose that we are given a direction such as
$\left[1618/3141,8776/72785\right]$ on the square torus. It is clear that this
direction is periodic, because it corresponds to a rational slope. However, in
order to get the slope in lowest terms, we need to perform the Euclidean
algorithm on the two fractions. What follows is an analogous process for
periodic directions on the golden L.

######  Definition 2.19.

We define an iterated process on a direction vector $\mathbf{v}$ in the first
quadrant:

Let $\mathbf{v}_{0}=\mathbf{v}$. For $i\geq 0$, let
$\mathbf{v}_{i+1}=\sigma_{k_{i}}^{-1}\mathbf{v}_{i}$ where
$\mathbf{v}_{i}\in\Sigma_{k_{i}}$. Let $a_{v}=\\{k_{0},k_{1},\ldots\\}$.

Here $\mathbf{v}_{0}$ is the original direction vector, which is in some
sector $\Sigma_{k_{0}}$. The operation $\sigma^{-1}_{k_{0}}$ stretches sector
$\Sigma_{k_{0}}$ into the entire first quadrant, and
$\sigma^{-1}_{k_{0}}\mathbf{v}_{0}$ is the image vector of $\mathbf{v}_{0}$
under this operation. We iterate this process, repeatedly stretching whichever
sector our vector lies in, into the entire first quadrant. The sequence
$a_{v}$ is the _itinerary_ of sectors that the vector lies in at each step of
the process. This is analogous to the continued fraction algorithm. If our
vector is in a periodic direction, eventually one of its images will land on
the boundary of the sector, so the process stops:

######  Proposition 2.20.

If $\mathbf{v}$ is in a saddle connection direction, then

  1. 1.

The sequence $a_{v}$ is eventually constant equal to $0$, starting from some
index $k$, and

  2. 2.

For this value of $k$,
$\mathbf{v}_{k}=\left[\begin{smallmatrix}\ell_{\mathbf{v}}\\\
0\end{smallmatrix}\right]$ for some ${\ell_{v}}\in\mathbf{R}$.

###### Proof.

By Theorem 2.11, every saddle connection vector can be obtained by applying a
finite number of $\sigma_{i}$’s to $\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$. Thus, a finite number of repeated applications of
the appropriate $\sigma_{i}^{-1}$ will transform any saddle connection vector
to $\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$. Since
$\mathbf{v}$ is in a saddle connection direction, it is a constant multiple of
a saddle connection vector, so a finite number of repeated applications of the
appropriate $\sigma_{i}^{-1}$’s will transform it to a constant multiple of
$\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$. After this finite
number of applications, the transformed vector is
${\ell_{\mathbf{v}}}\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$
in $\Sigma_{0}$, and
$\sigma_{0}({\ell_{\mathbf{v}}}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right])={\ell_{\mathbf{v}}}\sigma_{0}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]={\ell_{\mathbf{v}}}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$, so it stays in $\Sigma_{0}$. ∎

We can use this process to give the saddle connection vectors in a particular
direction:

######  Proposition 2.21.

Given a periodic direction $\mathbf{v}$ on the golden L, the long and short
saddle connection vectors in that direction are $\mathbf{v}/\ell_{\mathbf{v}}$
and $\mathbf{v}\bar{\phi}/{\ell_{\mathbf{v}}}$, respectively, where
$\ell_{\mathbf{v}}$ is as in Proposition 2.20.

###### Proof.

The maps $\sigma_{i}$ are linear. The horizontal long saddle connection vector
is $\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$, so if we end
with $\left[\begin{smallmatrix}\ell_{v}\\\ 0\end{smallmatrix}\right]$ it means
we started from $\ell_{v}$ times a long saddle connection vector. The short
saddle connection vector is $\bar{\phi}$ times the long saddle connection
vector. ∎

######  Theorem 2.22.

Corresponding to any periodic direction vector $\mathbf{v}$ is a unique tree
word $k_{0}k_{1}\cdots k_{n}$, which is an itinerary of sectors as in
Definition 2.19, such that
$\mathbf{v}=\ell_{\mathbf{v}}\sigma_{k_{0}}\ldots\sigma_{k_{n}}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$ for some length $\ell_{\mathbf{v}}$. Then the
saddle connection vector in the direction $\mathbf{v}$ is
$\sigma_{k_{0}}\ldots\sigma_{k_{n}}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$. Furthermore, the saddle connection vector is of
the form $[a+b\phi,c+d\phi]$ for $a,b,c,d\in\mathbf{N}$.

###### Proof.

That the saddle connection vector is
$\sigma_{k_{0}}\ldots\sigma_{k_{n}}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$ is clear by construction. Multiplying out the
product of matrices $\sigma_{k_{i}}$ with $\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$ yields a vector consisting of a sum of positive
integers and positive integer multiples of powers of $\phi$. Applying the
identity $\phi^{2}=1+\phi$ reduces each coordinate to the sum of a positive
integer and a positive integer multiple of $\phi$. Said another way, the
vector $\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$ and each
matrix $\sigma_{i}$ each have all of their entries in the ring
$\mathbf{Z}[\phi]$, so their products do as well. ∎

##  3 Combinatorial periods of pentagon and double pentagon trajectories

###  3.1 Cutting sequences on the double pentagon

######  Definition 3.1.

Given a translation surface decomposed into polygons with labeled edges, and
an oriented bi-infinite linear trajectory $\tau$ on this translation surface,
let $c(\tau)$ be the bi-infinite sequence of labels of the edges that the
trajectory crosses. If the trajectory hits a corner, it stops; in this case,
the sequence is not bi-infinite; otherwise it is.

######  Definition 3.2.

The _transition_ $ab$ is _allowed_ in sector $\Sigma_{i}$ if some trajectory
in $\Sigma_{i}$ cuts through edge $a$ and then through edge $b$.

For $i=0,1,2,3$, the transition diagram $\mathfrak{D}_{i}$ is a directed graph
whose vertices are edge labels of the double pentagon (the numbers
$1,\ldots,5$), with an arrow from edge label $a$ to edge label $b$ if and only
if the transition $ab$ is allowed in $\Sigma_{i}$. We say that a sequence is
_admissible_ in diagram $\mathfrak{D}_{i}$ if it is given by a path following
arrows on $\mathfrak{D}_{i}$.

######  Lemma 3.3.

For a trajectory with $\theta\in[0,\pi/5)$ on the double pentagon, the
corresponding transition diagram is:

${\mathbf{1}}$${\mathbf{2}}$${\mathbf{3}}$${\mathbf{4}\ }$${\mathbf{5}}$

###### Proof.

This is easily checked by hand, using Figure 8. To do this, use one of the
pentagons with its edge labels (thin), and ignore the internal line segments
and their bold labels. ∎

######  Definition 3.4.

For $i=0,1,2,3$, define $\tilde{\sigma_{i}}=P\sigma_{i}P^{-1}$.

These matrices do the same job as the $\sigma_{i}$’s on the golden L, now on
the double pentagon.

######  Lemma 3.5.

The image of the double pentagon under each $\tilde{\sigma_{i}}$ is as shown
in Figure 8.

> Figure 8: The image of the double pentagon under each
> $\tilde{\sigma_{i}}^{-1}$: top left $\tilde{\sigma_{0}}^{-1}$, bottom left
> $\tilde{\sigma_{1}}^{-1}$, top right $\tilde{\sigma_{2}}^{-1}$, bottom right
> $\tilde{\sigma_{3}}^{-1}$.

###### Proof.

We obtain these diagrams via direct computation: For example,
${\sigma_{0}}^{-1}$ sends $\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$ to $\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$ in the golden L, so it also preserves the
horizontal direction in the double pentagon. ${\sigma_{0}}^{-1}$ sends
$\left[\begin{smallmatrix}0\\\ 1\end{smallmatrix}\right]$ to
$\left[\begin{smallmatrix}-\phi\\\ 1\end{smallmatrix}\right]$ in the golden L,
meaning that it twists to the left by one cylinder length. The other three
diagrams are obtained in the same way. ∎

######  Lemma 3.6.

For a trajectory $\tau$ with angle $0\leq\theta<\pi/5$, if we transform $\tau$
by $\tilde{\sigma_{i}}$, the induced effect on the associated cutting sequence
is as in the following diagrams:

$\mathfrak{D}$: ${\mathbf{1}}$${\mathbf{2}}$${\mathbf{3}}$${\mathbf{4}\
}$${\mathbf{5}}$ $\mathbf{1}\mathbf{2}\mathbf{3}\mathbf{4}\mathbf{5}$

$\mathfrak{D}(0)$:
${\mathbf{2}}$${\mathbf{1}}$${\mathbf{4}}$${\mathbf{3}}$${\mathbf{5}}$$\scriptstyle{23}$$\scriptstyle{32}$$\scriptstyle{4}$$\scriptstyle{4}$
$\mathbf{2}\mathbf{1}23\mathbf{4}\mathbf{3}4\mathbf{5}$

$\mathfrak{D}(1)$:
${\mathbf{3}}$${\mathbf{5}}$${\mathbf{1}}$${\mathbf{4}}$${\mathbf{2}}$$\scriptstyle{4}$$\scriptstyle{432}$$\scriptstyle{4}$$\scriptstyle{23}$$\scriptstyle{234}$$\scriptstyle{3}$$\scriptstyle{32}$$\scriptstyle{3}$
$\mathbf{3}4\mathbf{5}432\mathbf{1}23\mathbf{4}3\mathbf{2}$

$\mathfrak{D}(2)$:
${\mathbf{4}}$${\mathbf{2}}$${\mathbf{5}}$${\mathbf{1}}$${\mathbf{3}}$$\scriptstyle{3}$$\scriptstyle{34}$$\scriptstyle{3}$$\scriptstyle{432}$$\scriptstyle{43}$$\scriptstyle{2}$$\scriptstyle{234}$$\scriptstyle{2}$
$\mathbf{4}3\mathbf{2}34\mathbf{5}432\mathbf{1}2\mathbf{3}$

$\mathfrak{D}(3)$:
${\mathbf{1}}$${\mathbf{3}}$${\mathbf{2}}$${\mathbf{5}}$${\mathbf{4}}$$\scriptstyle{2}$$\scriptstyle{2}$$\scriptstyle{34}$$\scriptstyle{43}$
$\mathbf{1}2\mathbf{3}\mathbf{2}34\mathbf{5}\mathbf{4}$

On the left, we use the usual presentation of such “generation” transition
diagrams in the literature. On the right, we use a simplified form that
emphasizes the one-dimensional nature of a path on a diagram. (This was also
done in [DFT11, §2.3].)

###### Proof.

Given any periodic trajectory, its cutting sequence is a periodic word in the
alphabet $\\{1,2,3,4,5\\}$ describing the edges it crosses. We can deform this
trajectory into a piecewise-linear zigzag path that connects midpoints of
successive edges crossed. The original linear trajectory and the zigzag path
cross the same image edges for trajectories with $\theta\in[0,\pi/5)$; this
can easily be checked by hand.

Now we associate to each node label $k$ of the transition diagram
$\mathfrak{D}(i)$ the image edge that passes through the midpoint of edge $k$,
or that coincides with edge $k$. Using Figure 8, we label the arrow between
these image edges with the other image edges that must be crossed between
original edges $k$ and $k\pm 1$, the sign depending on if they are upper
arrows or lower arrows in the diagram; the edges that are crossed are also
easily checked by hand. ∎

######  Definition 3.7.

A _decorated cutting sequence_ is a cutting sequence that includes the extra
information of the arrows from the transition diagram; in particular, it
indicates which symbol is next. To obtain a decorated cutting sequence from a
cutting sequence, we add an exponent between every pair of symbols, which is
the second symbol of the pair: the periodic cutting sequence $e_{1}e_{2}\cdots
e_{k}$ becomes ${e_{1}}^{e_{2}}{e_{2}}^{e_{3}}\cdots{e_{k}}^{e_{1}}$.

######  Definition 3.8.

Define the substitution rules $r_{i}$ for $i=0,1,2,3$ on a decorated cutting
sequence:

$r_{0}=\begin{cases}{1}^{2}\to{2}^{1}\\\ {2}^{1}\to{1}^{2}\\\
{2}^{3}\to{1}^{2}{2}^{3}{3}^{4}\\\ {3}^{2}\to{4}^{3}{3}^{2}{2}^{1}\\\
{3}^{4}\to{4}^{3}\\\ {4}^{3}\to{3}^{4}\\\ {4}^{5}\to{3}^{4}{4}^{5}\\\
{5}^{4}\to{5}^{4}{4}^{3}\end{cases}r_{1}=\begin{cases}{1}^{2}\to{3}^{4}{4}^{5}\\\
{2}^{1}\to{5}^{4}{4}^{3}\\\ {2}^{3}\to{5}^{4}{4}^{3}{3}^{2}{2}^{1}\\\
{3}^{2}\to{1}^{2}{2}^{3}{3}^{4}{4}^{5}\\\ {3}^{4}\to{1}^{2}{2}^{3}{3}^{4}\\\
{4}^{3}\to{4}^{3}{3}^{2}{2}^{1}\\\ {4}^{5}\to{4}^{3}{3}^{2}\\\
{5}^{4}\to{2}^{3}{3}^{4}\end{cases}r_{2}=\begin{cases}{1}^{2}\to{4}^{3}{3}^{2}\\\
{2}^{1}\to{2}^{3}{3}^{4}\\\ {2}^{3}\to{2}^{3}{3}^{4}{4}^{5}\\\
{3}^{2}\to{5}^{4}{4}^{3}{3}^{2}\\\ {3}^{4}\to{5}^{4}{4}^{3}{3}^{2}{2}^{1}\\\
{4}^{3}\to{1}^{2}{2}^{3}{3}^{4}{4}^{5}\\\ {4}^{5}\to{1}^{2}{2}^{3}\\\
{5}^{4}\to{3}^{2}{2}^{1}\end{cases}r_{3}=\begin{cases}{1}^{2}\to{1}^{2}{2}^{3}\\\
{2}^{1}\to{3}^{2}{2}^{1}\\\ {2}^{3}\to{3}^{2}\\\ {3}^{2}\to{2}^{3}\\\
{3}^{4}\to{2}^{3}{3}^{4}{4}^{5}\\\ {4}^{3}\to{5}^{4}{4}^{3}{3}^{2}\\\
{4}^{5}\to{5}^{4}\\\ {5}^{4}\to{4}^{5}\end{cases}$ (1)

######  Lemma 3.9.

Applying the substitution rules (1) is equivalent to implementing the
transition diagrams in Lemma 3.6.

In other words, applying the substitution rules (1) is equivalent to taking a
decorated cutting sequence $c(\tau)$ on the double pentagon, where the
associated cutting sequence is admissible on diagram $\mathfrak{D}$, running
the cutting sequence through diagram $\mathfrak{D}(i)$ in Lemma 3.6,
collecting the letters to create a new cutting sequence, and then turning the
cutting sequence into a decorated cutting sequence as described in Definition
3.7.

###### Proof.

For each diagram, we write down the node label, with the arrow label as
exponent. ∎

######  Remark 3.10.

The literature on cutting sequences and transition diagrams contains a variety
of different choices about how to symbolically represent a path on a
transition diagram, and how to represent the operation that transforms one
cutting sequence into another cutting sequence. John Smillie and Corinna
Ulcigrai [SU10] used a “dual” transition diagram where arrow labels become
nodes, and nodes become arrow labels. The first author, Irene Pasquinelli and
Corinna Ulcigrai [DPU18] used substitutions on arrows, without using nodes or
arrow labels.

The symbol $a^{b}$ in the notation we use represents an arrow from $a$ to $b$,
so the substitutions given in Definition 3.8 are also substitutions on arrows.
The current opinion of these and the current authors is that a direct
substitution rule on arrows is the best way of expressing this type of
operator.

Finally, we note that in the substitution rules in Definition 3.8, we have
combined “derivation” and “normalization” (see [DPU18, §7]) into one step.
Namely, given a cutting sequence that corresponds to a path on transition
diagram $\mathfrak{D}$, running it through any of the substitutions $r_{i}$
yields a cutting sequence that also corresponds to a path on transition
diagram $\mathfrak{D}$, in a single step.

######  Definition 3.11.

For a given decorated periodic cutting sequence $\tau$ on the double pentagon,
we count how many of the symbols are associated to each of the four pairs of
arrows in the transition diagram. Specifically, define

$a_{\tau}=\\#\\{{1}^{2},{2}^{1}\\},\qquad
d_{\tau}=\\#\\{{2}^{3},{3}^{2}\\},\qquad
b_{\tau}=\\#\\{{3}^{4},{4}^{3}\\},\qquad c_{\tau}=\\#\\{{4}^{5},{5}^{4}\\}.$

The variable names are assigned in this non-alphabetical order because they
turn out to correspond to the coefficients in the vector
$\begin{bmatrix}a+b\phi\\\ c+d\phi\end{bmatrix}$; see Lemma 3.20.

######  Lemma 3.12.

The effects of each $r_{i}$ on the lengths of cutting sequences, depending on
the values of $a_{\tau},b_{\tau},c_{\tau},d_{\tau}$, are as follows:

$\displaystyle r_{0}$
$\displaystyle:[a_{\tau},b_{\tau},c_{\tau},d_{\tau}]\to[a_{\tau}+d_{\tau},b_{\tau}+c_{\tau}+d_{\tau},c_{\tau},d_{\tau}]$
$\displaystyle r_{1}$
$\displaystyle:[a_{\tau},b_{\tau},c_{\tau},d_{\tau}]\to[b_{\tau}+d_{\tau},a_{\tau}+b_{\tau}+c_{\tau}+d_{\tau},a_{\tau}+d_{\tau},b_{\tau}+c_{\tau}+d_{\tau}]$
$\displaystyle r_{2}$
$\displaystyle:[a_{\tau},b_{\tau},c_{\tau},d_{\tau}]\to[b_{\tau}+c_{\tau},a_{\tau}+b_{\tau}+d_{\tau},b_{\tau}+d_{\tau},a_{\tau}+b_{\tau}+c_{\tau}+d_{\tau}]$
$\displaystyle r_{3}$
$\displaystyle:[a_{\tau},b_{\tau},c_{\tau},d_{\tau}]\to[a_{\tau},b_{\tau},b_{\tau}+c_{\tau},a_{\tau}+b_{\tau}+d_{\tau}].$

###### Proof.

Given a short periodic trajectory $\tau$ and its corresponding cylinder vector
$v$, and cutting sequence $s$, given $i\in\\{0,1,2,3\\}$, let
$v^{\prime}=\sigma_{i}(v)$ and $\tau^{\prime}$ the short periodic trajectory
in direction $v^{\prime}$, and $s^{\prime}$ the corresponding cutting
sequence.

Staring at the definition of the $r_{i}$’s in Definition 3.8, we just need to
count, for each symbol ${1}^{2}$, ${2}^{1}$, ${2}^{3}$, ${3}^{2}$, ${3}^{4}$,
${4}^{3}$, ${4}^{5}$, ${5}^{4}$, how many of its appearances on the right hand
side come from each type on the left-hand side.

For example, for $r_{0}$, letters ${1}^{2}$ and ${2}^{1}$ come from letters
${1}^{2}$, ${2}^{1}$, ${2}^{3}$, ${3}^{2}$, consequently,
$a_{\tau^{\prime}}=a_{\tau}+d_{\tau}$. ∎

###  3.2 Periods of periodic double pentagon trajectories

In this section we answer a simple question: Given a periodic direction on the
double pentagon surface or on the pentagonal billiard table, what are the
corresponding periods? The analogous result is well known for the square; a
trajectory with slope $p/q$ on the square torus has period $p+q$, and a
trajectory with slope $p/q$ on the square billiard table has period $2(p+q)$
(Theorem 1.1).

It turns out that the answer for the double pentagon is the same as for the
square, if we set up our definitions correctly. First we must establish
periodic directions:

######  Lemma 3.13.

Periodic directions in the double pentagon and in the pentagonal billiard
table are exactly the images of the directions in $\mathbf{Q}[\sqrt{5}]$ under
the matrix $P$.

###### Proof.

A periodic direction in the double pentagon corresponds to a saddle connection
vector $\mathbf{v}_{P}$ on the double pentagon. The image
$P^{-1}\mathbf{v}_{P}$ of this vector on the golden L is a saddle connection
on the golden L; see Figure 3. By Theorem 2.22, saddle connection vectors on
the golden L have components in $\mathbf{Z}[\phi]$, so their slopes are in
$\mathbf{Q}[\sqrt{5}]$. Going in the other direction, any direction in
$\mathbf{Q}[\sqrt{5}]$ is a saddle connection direction on the golden L, for
some saddle connection vector $\mathbf{v}_{L}$, and the image
$P\mathbf{v}_{L}$ of this vector on the double pentagon is a saddle connection
on the double pentagon, and hence a periodic direction. ∎

######  Theorem 3.14.

Given a vector $\mathbf{v}\in\Sigma$ parallel to a periodic trajectory on the
golden L, there are two cylinders of periodic trajectories on the L, one short
and one long, in this direction. The short one is in $\Lambda$, and the long
one is in $\phi\Lambda$. The one in $\Lambda$ is $\ell^{-1}\mathbf{v}$, where
$\ell=\ell_{\mathbf{v}}$ from Proposition 2.20. It lives in
$\mathbf{Z}[\phi]^{2}$, i.e. it has coordinates $[a+b\phi,c+d\phi]$ with
$a,b,c,d$ nonnegative integers and $(a,b)\neq(0,0)$.

###### Proof.

This is a rephrasing of Proposition 2.21. ∎

We now give an expression for the period of a trajectory on the double
pentagon, corresponding to a given periodic direction:

######  Theorem 3.15 (Combinatorial Period Theorem).

Given a vector $\mathbf{v}=[a+b\phi,c+d\phi]\in\Lambda$, the corresponding
short trajectory has holonomy vector $\mathbf{v}$ and the long trajectory has
holonomy vector $\phi\mathbf{v}$. The combinatorial periods of the
corresponding trajectories in the double pentagon are $2(a+b+c+d)$ and
$2(a+2b+c+2d)$, respectively.

######  Example 3.16.

To determine the (periodic) word corresponding to a given direction
$\mathbf{v}$, we use the (finite) sequence of sectors
$a_{\mathbf{v}}=\\{k_{0},k_{1},\ldots,k_{n}\\}$ to work backwards. The word
corresponding to the direction $[1,0]$ is $w_{n}={1}^{2}{2}^{1}$ (see Figure
8). We run this word through the transition diagram corresponding to $k_{n}$,
to obtain the previous word $w_{n-1}$, and so on. For example, if
$a_{\mathbf{v}}=\\{1,2,0\\}$, to get the short trajectories, we run
$w_{3}={1}^{2}{2}^{1}$ through $r_{1}$, run the result through $r_{2}$, and
run the result of that through $r_{0}$, obtaining the sequences

$\displaystyle w_{3}$ $\displaystyle={1}^{2}{2}^{1}$ $\displaystyle w_{2}$
$\displaystyle={3}^{4}{4}^{5}{5}^{4}{4}^{3}$ $\displaystyle w_{1}$
$\displaystyle={5}^{4}{4}^{3}{3}^{2}{2}^{1}{1}^{2}{2}^{3}{3}^{2}{2}^{1}{1}^{2}{2}^{3}{3}^{4}{4}^{5}$
$\displaystyle w_{0}$
$\displaystyle={5}^{4}{4}^{3}{3}^{4}{4}^{3}{3}^{2}{2}^{1}{1}^{2}{2}^{1}{1}^{2}{2}^{3}{3}^{4}{4}^{3}{3}^{2}{2}^{1}{1}^{2}{2}^{1}{1}^{2}{2}^{3}{3}^{4}{4}^{3}{3}^{4}{4}^{5}$

in turn.

We also use $a_{\mathbf{v}}$ to find the direction. We compute
$\sigma_{0}\sigma_{2}\sigma_{1}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$, remembering that $\phi^{2}=\phi+1$.

$\displaystyle\sigma_{0}\sigma_{2}\sigma_{1}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$ $\displaystyle=\begin{bmatrix}1&\phi\\\
0&1\end{bmatrix}\begin{bmatrix}\phi&1\\\
\phi&\phi\end{bmatrix}\begin{bmatrix}\phi&\phi\\\
1&\phi\end{bmatrix}\begin{bmatrix}1\\\ 0\end{bmatrix}$
$\displaystyle=\begin{bmatrix}1&\phi\\\
0&1\end{bmatrix}\begin{bmatrix}\phi&1\\\
\phi&\phi\end{bmatrix}\begin{bmatrix}\phi\\\ 1\end{bmatrix}$
$\displaystyle=\begin{bmatrix}1&\phi\\\
0&1\end{bmatrix}\begin{bmatrix}\phi+2\\\ 2\phi+1\end{bmatrix}$
$\displaystyle=\begin{bmatrix}4\phi+4\\\ 2\phi+1\end{bmatrix}.$

We show the intermediary steps above because they give us the direction
corresponding to each word: $w_{3}=12$ corresponds to direction
$\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$,
$w_{2}=3454$ corresponds to direction $\left[\begin{smallmatrix}\phi\\\
1\end{smallmatrix}\right]$,
$w_{1}=543212321234$ corresponds to direction
$\left[\begin{smallmatrix}\phi+2\\\ 2\phi+1\end{smallmatrix}\right]$, and
$w_{0}=5434321212343212123434$ corresponds to direction
$\left[\begin{smallmatrix}4\phi+4\\\ 2\phi+1\end{smallmatrix}\right]$.

We have given undecorated sequences for ease of reading and counting. The
reader can check that at each step, the length of the period is twice the sum
of $a,b,c$, and $d$ in the vector $\left[\begin{smallmatrix}a\phi+b\\\
c\phi+d\end{smallmatrix}\right]$.

Now we develop some preliminary results that will help us to prove Theorem
3.15.

######  Lemma 3.17.

Every decorated periodic cutting sequence consists of equal numbers of $a^{b}$
and $b^{a}$ for each $a,b\in\\{1,2,3,4,5\\}$.

###### Proof of Lemma 3.17 by contradiction.

If the cutting sequence contained $n$ ${a}^{b}$s and fewer than $n$
${b}^{a}$s, this would mean that the associated path on the transition diagram
goes at least once from $a$ to $b$, without going back from $b$ to $a$,
contradicting the fact that the path returns to where it started. ∎

###### Proof of Lemma 3.17 by induction.

Every (short) periodic sequence can be obtained via a series of substitutions
$r_{i}$ (see Equation (1)) applied to the periodic sequence
$({1}^{2}{2}^{1})$. For each $i$, $r_{i}({a}^{b})$ is the same as
$r_{i}({b}^{a})$ but in the reverse order and with the base and exponent
reversed, so that each ${c}^{d}$ in the $i^{\text{th}}$ position in
$r_{i}({a}^{b})$ is matched with ${d}^{c}$ in the $i^{\text{th}}$ position
from the end in $r_{i}({b}^{a})$. Since the original sequence
$({1}^{2}{2}^{1})$ consists of pairs of this form, and each substitution
preserves this property, every cutting sequence consists of pairs of this
form. ∎

The inductive proof gives an additional result:

######  Corollary 3.18.

Every periodic cutting sequence has several “palindrome” properties: It reads
the same in the reverse order if

  1. 1.

the base and exponent are reversed, or if

  2. 2.

bases and exponents are treated the same, or if

  3. 3.

we ignore the exponents and remove the first digit (using the order given by
applying the substitutions (1) to $1^{2}2^{1}$ or $3^{4}4^{3}$).

For example, the periodic cutting sequence $543212321234$ from Example 3.16
can be rewritten in decorated form as
${5}^{4}{4}^{3}{3}^{2}{2}^{1}{1}^{2}{2}^{3}{3}^{2}{2}^{1}{1}^{2}{2}^{3}{3}^{4}{4}^{5}$,
showing these properties.

######  Lemma 3.19.

The matrices $\sigma_{i}$ have the following effects on a vector in
$\mathbf{R}^{2}$ with entries in $\mathbf{Z}[\phi]^{2}$:

$\displaystyle\sigma_{0}\begin{bmatrix}a+b\phi\\\ c+d\phi\end{bmatrix}$
$\displaystyle=\begin{bmatrix}(a+d)+(b+c+d)\phi\\\ c+d\phi\end{bmatrix},$
$\displaystyle\sigma_{1}\begin{bmatrix}a+b\phi\\\ c+d\phi\end{bmatrix}$
$\displaystyle=\begin{bmatrix}(b+d)+(a+b+c+d)\phi\\\
(a+d)+(b+c+d)\phi\end{bmatrix},$
$\displaystyle\sigma_{2}\begin{bmatrix}a+b\phi\\\ c+d\phi\end{bmatrix}$
$\displaystyle=\begin{bmatrix}(b+c)+(a+b+d)\phi\\\
(b+d)+(a+b+c+d)\phi\end{bmatrix},$
$\displaystyle\sigma_{3}\begin{bmatrix}a+b\phi\\\ c+d\phi\end{bmatrix}$
$\displaystyle=\begin{bmatrix}a+b\phi\\\ (b+c)+(a+b+d)\phi\end{bmatrix}.$

###### Proof.

This is easily checked by multiplying out the matrices on the left hand side
(each matrix $\sigma_{i}$ is given in Definition 2.5), and applying the
identity $\phi^{2}=1+\phi$. ∎

######  Lemma 3.20.

Let $\tau$ be a short periodic trajectory on the double pentagon and
$\mathbf{v}=\begin{bmatrix}a+b\phi\\\ c+d\phi\end{bmatrix}$ be the
corresponding short saddle connection vector. Let
$a_{\tau},b_{\tau},c_{\tau},d_{\tau}$ be the number of pairs of types of
letters in the cutting sequence as described in Definition 3.11. Then
$(a_{\tau},b_{\tau},c_{\tau},d_{\tau})=(2a,2b,2c,2d)$.

###### Proof.

Corresponding to the vector $\mathbf{v}$ is a unique sequence
$a_{\mathbf{v}}=(k_{0},k_{1},...,k_{n})$ of sectors as in Definition 2.19 such
that $\mathbf{v}=\sigma_{k_{0}}...\sigma_{k_{n}}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$. Correspondingly,
$\tau=r_{k_{0}}...r_{k_{n}}({1}^{2}{2}^{1})$. The proof is by induction on
$n$, and the induction step consists in observing that the effect of the
$\sigma_{i}$’s on $(a,b,c,d)$ and of the $r_{i}$’s on
$(a_{\tau},b_{\tau},c_{\tau},d_{\tau})$ match, by comparing Lemmas 3.12 and
3.19. ∎

Now we are ready to prove the main theorem.

###### Proof of the Combinatorial Period Theorem 3.15.

The proof is by induction on depth in the tree.

For the short trajectory, the base case is the vector
$\mathbf{v}=\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]=\left[\begin{smallmatrix}1+0\phi\\\
0+0\phi\end{smallmatrix}\right]$, corresponding to cutting sequence
${1}^{2}{2}^{1}$. The period is twice the sum of the coefficients, so the
result holds.

Lemma 3.20 shows that each application of the $r_{i}$’s and the $\sigma_{i}$’s
has an equivalent effect on the cutting sequence letters and on the vector
coefficients, so the relationship is preserved.

For the long trajectory, we start with cutting sequence $3^{4}4^{3}$, and the
combinatorics matches just as with the short trajectory. ∎

At the end of this section in §3.6, we give an extended proof of the
Combinatorial Period Theorem 3.15 that introduces some new ideas.

###  3.3 The L-necklace, the Five Ls, and their Veech group

As discussed in §2.1, the _Necklace_ , which we will now call the _P-necklace_
, is a translation surface made of $10$ pentagons glued edge-to-edge, which is
the smallest translation surface that is a cover of the pentagonal billiard
table. It is a 5-fold cover of the double pentagon surface. Here, we introduce
a different presentation of the same topological surface: the _L-necklace_ is
a sheared version of the necklace, made of $5$ copies of the golden $L$ (see
Figure 9).

> Figure 9: The $P$-necklace and the $L$-necklace, which are affine images of
> each other. Each edge is glued to the parallel facing edge; for labeling,
> see Figure 2. The dotted rectangles show the underlying structure of golden
> $L$s.

We study the necklace because a cutting sequence on the necklace can be
projected down to a cutting sequence on the double pentagon via a 5-to-1 edge
label map. Also, the necklace is the smallest translation surface that covers
the pentagonal billiard table; the necklace can be folded into the table via a
10-to-1 edge label map.

It is useful to reassemble the $L$-necklace into the underlying golden $L$s,
as in Figure 10. We call this presentation the _Five $L$s_.

> Figure 10: A rearrangement of the $L$-necklace into the $5$ golden $L$s. The
> thin edge labels are as in Figure 2 and the bold labels are used in
> Definition 3.21.

######  Definition 3.21.

We define a $4$-tuple $(a,b,c,d)$ to keep track of the gluings of the Five
$L$s. If $b=0$, it means that edge $\mathbf{b}$ is glued to the vertically
downward edge, and $4e$ is glued to $3b$, and so on for all five edges in the
position of $\mathbf{b}$. If $c=1$ it means that $\mathbf{c}$ is glued to
$1a$, and so on. We also define the surfaces $f(a,b,c,d)$ which are gluings of
the Five Ls as described in Definition 3.21.

For the Five $L$s, $f(a,b,c,d)=f(1,3,2,4).$ This is easy to check.

######  Definition 3.22.

Let $T$ be the horizontal shear $\left[\begin{smallmatrix}1&\phi\\\
0&1\end{smallmatrix}\right]=\sigma_{0}$ and let $R$ be the counter-clockwise
quarter-turn $\left[\begin{smallmatrix}0&-1\\\ 1&0\end{smallmatrix}\right]$.
$T$ and $R$ generate the Veech group $\Gamma$ of the golden L. We analyze $T$
and $R$ with respect to their effects on the surfaces $f(a,b,c,d)$. $T\cdot
f(a,b,c,d)$ records the gluings that result from applying the shear to the
surface with gluings $(a,b,c,d)$, and similarly for $R\cdot f(a,b,c,d)$.

######  Lemma 3.23.

$T\cdot f(a,b,c,d)=f(a,b-a,c,d-a-c)$ and $R\cdot f(a,b,c,d)=f(-d,c,-b,a)$.
Furthermore, we can eliminate redundancies modulo $5$ with the following
reductions, where the quotients are all taken modulo $5$:

If $a\neq 0$, $f(a,b,c,d)=f(a/a,b/a,c/a,d/a)=f(1,b/a,c/a,d/a)$.

If $a=0$, then $f(a,b,c,d)=f(a/d,b/d,c/d,d/d)=f(0,b/d,c/d,1)$. Note that it is
not possible to have $a=0$ and $d=0$, because then the Five Ls are not
connected.

###### Proof.

The first two equalities are a simple exercise in translation surfaces. The
reductions are obtained by relabeling the Five Ls. ∎

######  Proposition 3.24.

The $\Gamma$-orbit of the Five Ls consists of six surfaces (Figure 11).

###### Proof.

We compute the images of the Five Ls under the generating set $\langle
T,R\rangle$ of $\Gamma$. The gluings and relations in Figure 11 are obtained
again by a simple calculation in translation surfaces. ∎

${f(1,3,2,4)}$${f(1,4,2,2)}$${f(1,0,2,0)}$${f(0,2,0,1)}$${f(1,2,2,1)}$${f(1,1,2,3)}$$R$$T$$T$$T$$T$$R$$T$$R$$R$$T$

> Figure 11: The effects of the shear $T$ and the rotation $R$ on the gluings
> of the Five Ls. This is the Schreier graph for the action of Hecke 5 on the
> necklace and its images, for generating set $\langle T,R\rangle$.

We name the five nodes that are cyclically related by $T$ to be $A,B,C,D,E$
and the node $f(0,2,0,1)$ to be $F$, as shown in Figure 12.

###  3.4 Symmetries of trajectories via the group structure

There are two kinds of trajectories on the pentagonal billiard table: those
that _are_ rotationally symmetric (Figure 13), and those that are _not_
rotationally symmetric, for which there are thus five congruent paths that
differ only by rotation (Figure 14). In this section we prove this result
algebraically, using the group structure of the Five Ls.

To each periodic direction, we associate a starting horizontal trajectory,
with cutting sequence either $12$ (short) or $34$ (long) on the double
pentagon, and a tree word. For example, the tree word $120$ means that we
apply the product $\sigma_{0}\sigma_{2}\sigma_{1}$ to
$\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$ in order to get the
associated direction. In order to extract information about where we end up in
the graph depending on its tree word, we express the $\sigma_{i}$ in terms of
$T$ and $R$ and re-draw the graph in Figure 12.

######  Lemma 3.25.

Expressing each $\sigma_{i}$ in terms of $T$ and $R$ and following them around
the directed graph in Figure 11 yields the directed graph in Figure 12, which
tells us how the $\sigma_{i}$s permute the surfaces $A$, $B$, $C$, $D$, $E$
and $F$.

###### Proof.

We have the following relationships:

$\displaystyle\sigma_{0}$ $\displaystyle=T$ $\displaystyle\sigma_{1}$
$\displaystyle=TRT$ $\displaystyle\sigma_{2}$ $\displaystyle=TRTRT$
$\displaystyle\sigma_{3}$ $\displaystyle=TRTRTRT$

Starting at each node in the graph in Figure 11, we determine the effect of
each $\sigma_{i}$ by following the arrows corresponding to $T$ and $R$. This
yields the directed graph in Figure 12. ∎

ABCDEF Figure 12: The location in the Veech group of the Five Ls, depending on
the $\sigma_{i}$. Start at node $A$ and follow arrows _backwards_ according to
the tree word for a given direction, read _backwards_ , to determine its
associated symmetry. The colors are as follows: dotted black is $\sigma_{0}$,
thick green is $\sigma_{1}$, thin red is $\sigma_{2}$, and dashed blue is
$\sigma_{3}$. Landing on $F$ indicates only reflection symmetry; landing on
the other five indicates rotation symmetry as well.

######  Theorem 3.26.

Given a trajectory and the tree word associated to its direction, start at $A$
in the diagram in Figure 12, and follow the arrows _backwards_ according to
the tree word, read _backwards_. If it ends on $A$ through $E$, the trajectory
has order $5$ rotational symmetry; if it lands on $F$, the trajectory does not
have rotational symmetry.

###### Proof.

Every periodic direction has an associated long saddle connection vector. A
finite product $\sigma_{i}$s, for example $\sigma_{0}\sigma_{2}\sigma_{1}$ for
tree word 120, which is an element of the Veech group of the golden L, takes
$\left[\begin{smallmatrix}1\\\ 0\end{smallmatrix}\right]$ to this saddle
connection vector. By Proposition 3.24, that element takes the surface to one
of $A$, $B$, $C$, $D$, $E$ or $F$.

Thus the inverse, e.g.
$(\sigma_{0}\sigma_{2}\sigma_{1})^{-1}={\sigma_{1}}^{-1}{\sigma_{2}}^{-1}{\sigma_{0}}^{-1}$,
of the element takes the saddle connection vector to a horizontal vector.
Applying this inverse corresponds to reading the tree word backwards (021 for
our example) and following the arrows backwards, since we are using
${\sigma_{i}}^{-1}$. We look at the two cylinders of trajectories on the Five
Ls corresponding to this horizontal trajectory.

For $A$, $B$, $C$, $D$ and $E$, $a=1$ and $c=2$, so the Five Ls are arranged
horizontally as in Figure 10. Thus the horizontal cylinders are five times as
wide as in one golden L, and the horizontal trajectory traverses all of them
before repeating. This means that the pentagon trajectory traverses all five
copies of the double pentagon in the necklace, corresponding to all five
orientations (rotations by $2\pi/5$), one after the other, so the trajectory
has order 5 rotational symmetry, and its length is five times as long as in
the double pentagon.

For $F$, $a=0$ and $c=0$, so the Five Ls are arranged vertically, so a
horizontal cylinder is the same as in a single golden L. Thus the golden $L$
trajectory lifts to five distinct translated cylinders of trajectories, one
for each orientation, which are images of each other under rotation by
$2\pi/5$. So trajectories in case $F$ fail to individually have rotational
symmetry. ∎

> Figure 13: Examples of periodic paths on the pentagon with the full symmetry
> of $D_{5}$. These trajectories are 101-short, 12-long, 1000-long and
> 23-long, respectively, corresponding to the core curve of the cylinder.
> Using a non-core curve breaks the bilateral symmetry and results in a
> picture with only (usually very subtle) rotational symmetry.

######  Corollary 3.27.

Of periodic orbits in the pentagon billiard table, 5/6 have both rotational
symmetry and reflection symmetry (see Figure 13), and 1/6 have only reflection
symmetry (see Figure 14).

###### Proof.

The stochastic matrix for the graph in Figure 12 is:

$\left[\begin{matrix}0&0&1/4&1/4&1/4&1/4\\\ 1&0&0&0&0&0\\\
0&1/4&0&1/4&1/4&1/4\\\ 0&1/4&1/4&1/4&1/4&0\\\ 0&1/4&1/4&1/4&0&1/4\\\
0&1/4&1/4&0&1/4&1/4\end{matrix}\right].$

Powers of this matrix quickly approach the matrix all of whose entries are
$1/6$. Thus a path of length $n$ asymptotically visits nodes equally, so $1/6$
of paths end on vertex $F$, and $5/6$ of them end on the other vertices. ∎

> Figure 14: Examples of periodic paths on the pentagon with only bilateral
> symmetry. These are the short trajectories with tree words 121, 1112, 1223
> and 2011, respectively, corresponding to the core curve of the cylinder.
> Using a non-core curve breaks the bilateral symmetry and results in a
> picture with no symmetry at all.

Our Theorem 3.26 gives a method for following the tree word around a directed
graph to determine the symmetry type of a trajectory. Compare this to the
approach in [DFT11], which uses a congruence condition on the saddle
connection vector to determine the symmetry:

######  Theorem 3.28 ([DFT11], Theorem 11).

Given a periodic direction on the pentagon billiard table with long saddle
connection vector $[a+b\phi,c+d\phi]$, let $L=2(a+b+c+d)$. The period of the
corresponding pentagon billiard trajectory is: $\begin{cases}L&\mbox{if
}(d-b)+2(c-a)\equiv 0\\\ 5L&\mbox{if }(d-b)+2(c-a)\not\equiv
0\end{cases}\pmod{5}.$

###  3.5 Symmetries of trajectories via geometry

In this section, we develop geometric methods for understanding billiard paths
using the necklace, to give yet another characterization of billiard path
symmetry (Theorem 3.29).

Given a trajectory on the double pentagon, each time the trajectory crosses an
edge, we fold along that edge, and thereby obtain the corresponding trajectory
on the pentagon billiard table. It turns out that, for periodic trajectories,
there are two possible outcomes to this process.

######  Theorem 3.29.

For a periodic trajectory $p$ on the double pentagon surface corresponding to
the core curve of a cylinder (i.e. bouncing off of the midpoint of an edge),
two possibilities arise:

  1. 1.

The preimage on the necklace is the union of five periodic trajectories, which
each project one-to-one to $p$. In this case, the billiard trajectory has the
same length as $p$, and only reflection symmetry (see Figure 14 and Figure
15(a)).

  2. 2.

The preimage on the necklace is a single periodic trajectory, whose length is
five times that of $p$. In this case, the billiard trajectory also has length
five times that of $p$, and it has both reflection and rotation symmetries
(see Figure 13 and Figure 15(b)).

> Figure 15: Trajectories on the double pentagon lift to two distinct types of
> trajectories on the necklace. (a) The vertical trajectory lifts to five
> trajectories of the same length. (b) The horizontal trajectory lifts to a
> single trajectory five times as long.

###### Proof.

Without loss of generality, we may assume that $p$ is the representative of
its homotopy class that passes through the midpoint of two sides
$s,t\in\\{1,2,3,4,5\\}$. Removing the midpoint of $s$, the trajectory is an
open segment on the double pentagon. When we lift this segment to the
necklace, it gives us five corresponding segments, which join midpoints of
sides parallel to $s$. When we add in the midpoints, there are two
possibilities, illustrated by the vertical and horizontal trajectories,
respectively:

  1. 1.

We get five disjoint loops on the necklace, each one going through midpoints
of sides that are lifts of $s$, each one passing through the midpoint of an
edge that is the unfolded image of a different edge of the billiard table.
Thus, when the necklace is folded to yield the billiard table, these map to
five trajectories that are rotations of each other, each one passing twice
through the midpoint of exactly one edge. Thus such trajectories have
reflection symmetry across the perpendicular bisector to that edge, but no
rotation symmetry.

  2. 2.

We get a single loop of five times the length of $p$, which passes through the
midpoints of all of the edges of the necklace that are lifts of $s$. Thus,
when the necklace is folded to yield the billiard table, these fold to a
trajectory that passes through the midpoint of each edge twice, and has
reflection symmetry across the perpendicular bisector to every edge; thus,
such trajectories have rotation symmetry as well.

∎

Examples of pentagon billiard trajectories that lift to a necklace trajectory
five times as long as the double pentagon trajectory are in Figure 13. By
construction, they have rotational symmetry, so along with the bilateral
symmetry, they have the full symmetry of the dihedral group $D_{5}$. Examples
of pentagon billiard trajectories that lift to five copies of the double
pentagon trajectory on the necklace are in Figure 14. Such paths have only
bilateral symmetry.

######  Lemma 3.30.

Vectors in $\Lambda$ are primitive, in the sense that the only common factors
of $a+b\phi$ and $c+d\phi$ are units in $\mathbf{Z}[\phi]$.

###### Proof.

Since $\left[\begin{smallmatrix}a+b\phi\\\
c+d\phi\end{smallmatrix}\right]=\sigma\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]$ for some $\sigma$ in the Veech group of the golden
L, $\left[\begin{smallmatrix}a+b\phi\\\ c+d\phi\end{smallmatrix}\right]$ is
the first column of a matrix with determinant 1. Thus we have the Biezout
relation

$(a+b\phi)\cdot x+(c+d\phi)\cdot y=1.$

Therefore any common divisor of $a+b\phi$ and $c+d\phi$ divides 1, so it is a
unit. ∎

######  Lemma 3.31.

A primitive periodic orbit on the double pentagon has a primitive cutting
word.

Here “primitive periodic orbit” means that we traverse exactly one period of
the trajectory, and “primitive cutting word” means that the word does not
consist of repeats of a shorter word.

###### Proof.

Suppose the cutting word is not primitive. Then the corresponding holonomy
vector $[a+b\phi,c+d\phi]$ is not primitive as a vector
$[a,b,c,d]\in\mathbf{R}^{4}$. Since by Lemma 3.30 vectors $[a+b\phi,c+d\phi]$
are always primitive, this is impossible. ∎

######  Lemma 3.32.

For a pentagon billiard, the central trajectory in a cylinder of periodic
trajectories has half the length of the other trajectories in that cylinder if
and only if the trajectory is traveling parallel to an edge of the pentagon.

###### Proof.

This occurs if and only if it is possible to fold the necklace in half such
that the pieces of trajectory sit on top of each other with the same
orientation. This, in turn, occurs only if the trajectory is parallel to the
edge (and only if the trajectory passes through midpoints of edges). ∎

In the introduction of [DFT11], the authors mention that for a periodic
billiard in the pentagon, “if the period is odd, nearby parallel trajectories
have the length and the period twice as large.” Lemma 3.32 shows that the
_only_ such trajectories are parallel to an edge.

######  Proposition 3.33.

The length of a periodic billiard sequence corresponding to a trajectory with
rotational symmetry is five times the length of the corresponding cutting word
for the double pentagon, except for the horizontal trajectory, when it is half
of this.

###### Proof.

This follows directly from Lemma 3.32. ∎

###  3.6 Extended proof of the Combinatorial Period Theorem 3.15

######  Definition 3.34.

Let $B=\\{{1}^{2},{2}^{1},{2}^{3},{3}^{2},{3}^{4},{4}^{3},{4}^{5},{5}^{4}\\}$
and let $B^{*}$ be the set of all finite words in $B$. Note that our periodic
sequences consist of a language within $B^{*}$ of words that are admissible,
i.e. each symbol is compatible with the one before it (e.g.
$\ldots{a}^{b}{b}^{c}\ldots$).

We define some maps:

$\displaystyle m:\mathbf{N}[\phi]^{2}$ $\displaystyle\to\mathbf{N}^{4}$
$\displaystyle n:\\{B^{*}\\}\to\mathbf{N}^{4}$
$\displaystyle\begin{bmatrix}a+b\phi\\\ c+d\phi\end{bmatrix}$
$\displaystyle\mapsto\begin{bmatrix}a\\\ b\\\ c\\\ d\end{bmatrix}$
$\displaystyle\ \ \ \ \ \ \ \ \ w\mapsto\begin{bmatrix}a_{w}\\\ b_{w}\\\
c_{w}\\\ d_{w}\end{bmatrix}$

Note that $m$ is invertible, so its inverse
$m^{-1}:\mathbf{N}^{4}\to\mathbf{N}[\phi]^{2}$ is also defined.

If $\mathbf{v}=\begin{bmatrix}a+b\phi\\\ c+d\phi\end{bmatrix}\in\Lambda$,
define its complexity as $c(\mathbf{v})=a+b+c+d$.

Given a periodic word $w\in B^{*}$, define $\ell(w)$ to be its length.

###### Extended proof of Theorem 3.15.

Now Theorem 3.15 can be rephrased as: _If $\mathbf{v}$ is the vector in
$\Lambda$ corresponding to a periodic trajectory with associated cutting
sequence $w$, then $\ell(w)=2c(\mathbf{v})$._

The proof is by induction on the depth in the tree, and the base case is the
same as in the previous proof: it is true for the vector $\begin{bmatrix}1\\\
0\end{bmatrix}$ and for the associated trajectory ${1}^{2}{2}^{1}$.

Define $\hat{\Lambda}=m(\Lambda)$, which is the set of vectors
$\begin{bmatrix}a\\\ b\\\ c\\\ d\end{bmatrix}$ such that
$\begin{bmatrix}a+b\phi\\\ c+d\phi\end{bmatrix}\in\Lambda$.

Now for each $i=0,1,2,3$, define
$\hat{\sigma_{i}}:\hat{\Lambda}\to\hat{\Lambda}$ such that
$\hat{\sigma_{i}}=m^{-1}\circ\sigma_{i}\circ m$. These express the effects of
the matrices $\sigma_{i}$, just in $4$-vector coefficient coordinates instead
of $2$-vector coordinates.

These are linear maps, and we can compute them explicitly from Lemma 3.19:

$\hat{\sigma_{0}}=\begin{bmatrix}1&0&0&1\\\ 0&1&1&1\\\ 0&0&1&1\\\
0&0&0&1\end{bmatrix},\ \hat{\sigma_{1}}=\begin{bmatrix}0&1&0&1\\\ 1&1&1&1\\\
1&0&0&1\\\ 0&1&1&1\end{bmatrix},\ \hat{\sigma_{2}}=\begin{bmatrix}0&1&1&0\\\
1&1&0&1\\\ 0&1&0&1\\\ 1&1&1&1\end{bmatrix},\
\hat{\sigma_{3}}=\begin{bmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&1&1&0\\\
1&1&0&1\end{bmatrix}$.

Now notice that the statement of Lemma 3.12 can be expressed as:
$n(r_{i}(w))=\hat{\sigma_{i}}(n(w))$. This gives the induction step. ∎

######  Remark 3.35.

Each of the matrices $\hat{\sigma_{i}}$ is a cone contraction of the positive
cone of $\mathbf{R}^{4}$. We can visualize this as taking a tetrahedron to
four contracted tetrahedra inside it. Iterating this construction yields a
fractal-like object (see Figure 16). `Sage` code for doing this is in Appendix
A. John Smillie and Corinna Ulcigrai made a very similar picture associated to
the regular octagon surface that they studied in [SU10], which did not appear
in their paper but which they shared with us after seeing the picture below.

Figure 16: The fifth-level polytope obtained via the cone contractions
described in Remark 3.35.

######  Question 3.36.

What is the fractal dimension of this object, and what properties does it
have? Is it related to the fractal sine wave [Mc15, Figure 3]?

###  3.7 Summary

We now state everything we know about periodic trajectories on the double
pentagon surface and the pentagon billiard table, and say how to determine
this information from the direction:

######  Theorem 3.37.

For a given periodic direction on the golden L, we can determine everything
about the associated double pentagon and pentagon billiard trajectories, as
follows.

Find the long saddle connection vector $\mathbf{v}=[a+b\phi,c+d\phi]$, using
Theorem 2.22.

  1. 1.

The combinatorial period of the trajectory on the double pentagon surface is
$2(a+b+c+d)$.

  2. 2.

Transform $\mathbf{v}$ into a saddle connection vector on the double pentagon
surface (using the matrix $P$ from Definition 2.2). The Euclidean length of
the trajectory on the double pentagon surface is the length of the resulting
vector $P\mathbf{v}$.

Find the tree word associated to $\mathbf{v}$, such as $120$. This tells us
the product of $\sigma_{i}$s that takes $[1,0]$ to $\mathbf{v}$. Starting at
$A$ in the graph in Figure 12, follow _backward_ the arrows associated to each
$\sigma_{i}$ in the product, read _backward_ , and determine the ending node
in the graph.

  1. 3.

If the ending node is $A$, $B$, $C$, $D$, or $E$, then the trajectory has the
full symmetry of the dihedral group of the regular pentagon, and the
combinatorial period, and the euclidean length, of the pentagon billiard
trajectory are five times that of the associated double pentagon trajectory
computed above (except for the two trajectories parallel to the edges of the
table through midpoints, in which case they are 2.5 times that above).

  2. 4.

On the other hand, if the ending node is $F$, then the trajectory has only
bilateral symmetry, and its combinatorial period, and the euclidean length, of
the pentagon billiard trajectory are the same as the associated double
pentagon trajectory computed above.

###### Proof.

  1. Part (1) is Theorem 3.15; part (2) is Proposition 2.21, and part (3) is Theorem 3.26 and Proposition 3.33, and (4) is Theorem 3.26.

∎

##  4 Applications and future directions

Our `Sage` program (see Appendix B) computes the periodic trajectory for the
golden L surface, the double pentagon surface and the pentagon billiard table,
associated to any tree word. As a result of exploring these trajectories, we
have additional results, conjectures and questions.

###  4.1 Combinatorial periods achieved on the surface and table

By symmetry, all periodic trajectories on the double pentagon have periods
that are multiples of 2, and all periodic trajectories with rotational
symmetry on the pentagon billiard table have periods that are multiples of 10.
It turns out that in both cases, _every_ such number does in fact arise as a
periodic trajectory period, via simple constructions.

######  Proposition 4.1.

Every positive even number arises as a double pentagon trajectory period.

###### Proof.

A horizontal trajectory has combinatorial period 2, and the trajectory with
tree word $10^{n}$ has period $2n+2$. Geometrically, we start with the
period-4 trajectory whose tree word is 1, and do repeated horizontal Dehn
twists; each twist increases the period by 2 (see Figure 17). Algebraically,
by Theorem 2.22, the trajectory corresponding to tree word $1\cdot 0^{n}$ has
corresponding long saddle connection vector
$\sigma_{0}^{n}\sigma_{1}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]=\left[\begin{smallmatrix}(n+1)\phi\\\
1\end{smallmatrix}\right]$, so by Theorem 3.15, the period is
$2((n+1)+1)=2(n+2)$. ∎

> Figure 17: Short periodic paths on the double pentagon, with combinatorial
> period 2, 4, 6, 8, 10, respectively, corresponding to tree words 0, 1, 10,
> 100, 1000, respectively. Continuing this construction leads to paths with
> every even period.

######  Proposition 4.2.

Every positive multiple of 10 arises as the period of a rotationally symmetric
pentagon billiard trajectory.

###### Proof.

Given $k\in\mathbf{N}$, the trajectory corresponding to the following tree
word has period $10k$:

  1. 1.

If $k\not\equiv 3$ (mod 5), 1 followed by $k-2$ 0s, short trajectory.

  2. 2.

If $k\equiv 3$ (mod 10), 1 followed by $(k-3)/2$ 0s, long trajectory.

  3. 3.

If $k\equiv 8$ (mod 10), 2 followed by $(k-2)/2$ 0s, short trajectory.

These follow from using the directed graph in Figure 12 and the Combinatorial
Period Theorem 3.15. For case (1), if $k\not\equiv 3$ (mod 5), then
$k-2\not\equiv 1$ (mod 5). We start at $A$ and follow $\sigma_{0}$ backwards
$k-2$ times, landing on $A,B,C$ or $D$. Then we follow $\sigma_{1}$ backwards,
landing on $D,A,E$ or $A$, respectively, all of which correspond to a path
with rotational symmetry. (Notice that if $k\equiv 3$, this process results in
landing on $F$, yielding a path without rotational symmetry.)

The direction vector
${\sigma_{0}}^{k-2}\sigma_{1}\left[\begin{smallmatrix}1\\\
0\end{smallmatrix}\right]=\left[\begin{smallmatrix}(k-1)\phi\\\
\phi\end{smallmatrix}\right]$, so since the path has rotational symmetry, the
period is $10k$, as desired. The other cases are similar.

∎

Based on computer evidence, we conjecture the following:

######  Conjecture 4.3.

Every even number arises as the period of an asymmetric periodic billiard
trajectory on the pentagon, other than 2, 12, 14 and 18.

We have verified this up to period 4000.

######  Conjecture 4.4.

For trajectories with only reflection symmetry, the growth rate is linear:
There are asymptotically $n/4$ trajectories that have period $2n$.

###  4.2 Non-equidistribution of periodic billiard trajectories

A given periodic trajectory consists of a finite number of line segments, and
thus by definition does not equidistribute. However, we can ask if a sequence
of longer and longer periodic trajectories equidistributes in the limit. Most
long trajectories do “equidistribute” in this sense, but some do not (Figure
18).

> Figure 18: Two periodic paths on the pentagon: (a) The short path
> corresponding to tree word 1000000000000000000000000000000000002, with
> combinatorial length 1460 and geometric length $\sim$909\. (b) The short
> path corresponding to tree word 12121, with combinatorial length 1470 and
> geometric length $\sim$964\. Both are drawn with the same line thickness,
> and correspond to the core curve of the cylinder.

######  Remark 4.5.

The only way to get non-equidistributed periodic paths like Figure 18(a) is
with long strings of horizontal ($\sigma_{0}$) or vertical ($\sigma_{3}$) Dehn
twists. Any initial product of $\sigma_{i}$s determines how much trajectory is
in each cylinder, the product of $\sigma_{0}$ or $\sigma_{3}$ twists the
horizontal or vertical cylinder to equidistribute the trajectory within the
cylinder, and then any ending product of $\sigma_{i}$s transforms the cylinder
direction out of horizontal. See Equation (2) and Figure 19. We thank Barak
Weiss for suggesting this construction.

$\text{tree word}=\overbrace{1}^{\text{controls amount of trajectory in each
cylinder}}\underbrace{00000000000000000000000000000000000}_{\text{horizontal
Dehn twists}}\overbrace{2}^{\text{transforms cylinder direction}}$ (2)

> Figure 19: The method behind Figure 18(a), shown for a simpler example: The
> short periodic paths on the double pentagon corresponding to tree words 1,
> 100000, and 1000002, respectively, illustrating Remark 4.5 and Equation (2).
> To get the billiard trajectory picture, fold the third double pentagon
> picture above so that the two pentagons align, and then superimpose four
> more copies of it on top, in the four other orientations.

In the family of periodic trajectories with tree words of the form
$1000\ldots$ (shown in Figure 17), the ratio of the “trajectory concentration”
(length of trajectory per unit area) in the small and large horizontal
cylinder is 0; in a nearly equidistributed trajectory, it is near 1. A
trajectory corresponding to a tree word of the form $A\ 0^{n}$ equidistributes
within each horizontal cylinder (converges to Lebesgue measure) as
$n\to\infty$. We conjecture that:

######  Conjecture 4.6.

The set of ratios of trajectory concentration in the horizontal cylinders for
a trajectory correspoding to $A\ 0^{n}$ forms a Cantor set and accumulates
near equidistribution.

The double pentagon and the necklace are Veech surfaces, and thus exhibit
_optimal dynamics_ : every trajectory is either periodic or equidistributed.
As we saw in Figure 18, it is possible to find periodic billiard trajectories
that are arbitrarily dense, but _not_ equidistributed. In fact, we can find
such periodic trajectories that never visit some part of the table _at all_.
Figure 20 shows examples of two families we have found that exhibit this
surprising behavior.

> Figure 20: Representatives (with $n=20$) of the families of short
> trajectories with tree words of the form $123^{n}1$ and $10^{n}1$,
> respectively, which all miss part of the table. Another family is very
> similar to the family on the left, and has tree words of the form
> $303^{n}1$.

###  4.3 More conjectures and questions

######  Question 4.7.

Some billiard trajectories appear to have “holes,” for example the one on the
right in Figure 4. What can we say about the holes?

######  Question 4.8.

(Curtis McMullen) If you pick an edge midpoint and a periodic direction, how
can you tell if it will give a short or a long trajectory?

######  Question 4.9.

So far, the structure of the set of periodic directions and periodic
trajectories is only well understood for the rectangle and regular pentagon
billiard tables, and possibly the regular hexagon. What is its structure for
other tables, such as the regular heptagon and octagon?

######  Question 4.10.

What is the relationship between geometric vs. combinatorial lengths of saddle
connection vectors, and what are the statistics of their distribution?

######  Question 4.11.

Is there a substitution to go from the short to the long billiard word in a
given periodic direction?

Julien Cassaigne, Pascal Hubert, and Serge Troubetzkoy studied the bounce
language of billiards in polygons and gave formulas for the complexity of the
language of squares, and of triangles that tile the plane by reflection
[CHT02]. Relatedly, we believe that:

######  Conjecture 4.12.

For the subword length $n$ sufficiently large, the complexity of each
individual aperiodic word is $15n+10$.

Let $p(n)$ be the complexity of the billiard language of a billiard table. At
the end of [CHT02, §1], the authors remark that “It would be interesting to
know if the limit $\lim_{n\to\infty}\frac{p(n)}{n^{3}}$ exists in the case of
Veech polygons.” Our computational evidence suggests that:

######  Conjecture 4.13.

For the regular pentagon billiard table, we have:
$\displaystyle\lim_{n\to\infty}\frac{p(n)}{n^{3}}=\frac{10}{\pi^{2}}.$

######  Question 4.14.

Corollary 3.27 shows that, at a given depth $n$ in the tree, paths
equidistribute with respect to landing on elements of the Veech group in the
graph in Figure 12. Is the same true if we measure with respect to
combinatorial length instead of tree word length?

######  Question 4.15.

Can we understand height functions for slopes in $\mathbf{Q}[\sqrt{5}]$?

######  Question 4.16.

Describe the polytope that results from applications of
$\widehat{\sigma_{i}}$s (see Question 3.36).

######  Question 4.17.

We know the slopes of periodic directions in the golden L model (see [Mc02,
Theorem A.1] and [ACL15]). What are the slopes in the pentagon model? Do they
form a field?

######  Question 4.18.

What is the language of the bounce sequences, of sides of the pentagon that
the trajectory hits? Note that this is different from the language of cutting
sequences on the double pentagon surface, because the labels of corresponding
edges are different. It would be interesting to study valid bounce sequences,
in the method of Smillie and Ulcigrai [SU10].

######  Question 4.19.

(See [L74]) There is a simple test to see if a matrix is in the Veech group of
the square torus: it must have integer entries and determinant 1. Is there are
similarly simple characterization for matrices in the Veech group of the
double pentagon or golden L?

## References

  * [AS09] Pierre Arnoux, Thomas A. Schmidt, Veech surfaces with non-periodic directions in the trace field Journal of Modern Dynamics 3 (2009), no. 4, 611–629. doi: 10.3934/jmd.2009.3.611
  * [ACL15] Jayadev Athreya, Jon Chaika, Samuel Lelièvre, The gap distribution of slopes on the golden L. Contemporary Mathematics 631 (2015): 47–62.
  * [CHT02] Julien Cassaigne, Pascal Hubert, Serge Troubetzkoy, Complexity and growth for polygonal billiards, Annales de l’institut Fourier, Volume 52 (2002) no. 3, 835–847.
  * [D13] Diana Davis, Cutting sequences, regular polygons, and the Veech group, Geometriae Dedicata, Volume 162, Issue 1 (2013), 231–261.
  * [D17] Diana Davis, Lines in positive genus: An introduction to flat surfaces, in Dynamics done with your bare hands, EMS Series of Lecture Notes (2017): arXiv:1507.02571.
  * [DFT11] Diana Davis, Dmitry Fuchs, Sergei Tabachnikov, Periodic trajectories in the regular pentagon, Moscow Mathematical Journal, Volume 11, Issue 3 (2011), 439–461.
  * [DPU18] Diana Davis, Irene Pasquinelli, Corinna Ulcigrai, Cutting sequences on Bouw-Möller surfaces: an $\mathcal{S}$-adic characterization. To appear in Annales Scientifiques de l’Ecole Normale Supérieure. arXiv:1509.03905 [math.DS] (2018).
  * [L74] Armin Leutbecher, Über die Heckeschen Gruppen $G(\lambda)$. II. Mathematische Annalen 211, 63–86 (1974).
  * [Ma86] Howard Masur, Closed trajectories for quadratic differentials with an application to billiards, Duke Mathematical Journal. Volume 53 (1986), 307–313.
  * [Mc02] Curtis T. McMullen, Teichmüller curves of infinite complexity. Acta Mathematica 191, 191–223 (2003).
  * [Mc03] Curtis T. McMullen, Billiards and Teichmüller curves on Hilbert modular surfaces. Journal of the American Mathematical Society 16 (2003), 857–885.
  * [Mc15] Curtis T. McMullen, Cascades in the dynamics of measured foliations. Annales scientifiques de l’ENS 48, fascicule 1, 1–39 (2015)
  * [S08] Richard Schwartz, Obtuse Triangular Billiards II: 100 Degrees Worth of Periodic Trajectories. Journal of Experimental Mathematics, Volume 18, Issue 2 (2008), 137–171.
  * [S+18] W. A. Stein et al., _Sage Mathematics Software_ , The Sage Development Team, 2018, http://www.sagemath.org.
  * [CC18] SageMath, Inc., _CoCalc Online Computational Mathematics_ , 2018. https://cocalc.com/.
  * [SU10] John Smillie, Corinna Ulcigrai, Geodesic flow on the Teichm$\mathrm{\ddot{u}}$ller disk of the regular octagon, cutting sequences and octagon continued fractions maps, Dynamical Numbers: Interplay between Dynamical Systems and Number Theory, Contemporary Mathematics, Volume 532 (2010), 29–65.
  * [TGMM18] George Tokarsky, Jacob Garber, Boyan Marinov, Kenneth Moore, One hundred and twelve point three degree theorem, preprint (2018), arXiv:1808.06667.
  * [V92] William Veech, The Billiard in a Regular Polygon, Geometric and functional analysis 2.3 (1992): 341–380.

##  Appendix A Code for fractal polytope

The following code defines the matrices from 3.6, rescaled to preserve the
standard simplex in $\mathbf{R}^{4}$ (the intersection of the positive cone
with the hyperplane where coordinates sum to one).

It then defines a matrix $V$ to send that simplex to a regular tetrahedron in
$\mathbf{R}^{3}$ for the sake of visualization.

M = MatrixSpace(AA, 4)mm = [M([1/sum(r)*r for r in m.rows()]) for m in [ M((1,
0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1)), M((0, 1, 0, 1, 1, 1, 1, 1, 1,
0, 0, 1, 0, 1, 1, 1)), M((0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1)),
M((1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1))]]C = Polyhedron(rays=[(1,
0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)])r2, r3 = AA(2).sqrt(),
AA(3).sqrt()V = matrix(AA, 4, 3, ( 1, 0, -r2/4, -1/2, r3/2, -r2/4, -1/2,
-r3/2, -r2/4, 0, 0, 3/4*r2))W = Words([0, 1, 2, 3])def f(n): return
sum(p.plot(frame=False, point=False, wireframe=(0.5, 0.5, 0.5),
polygon=hue(k/4^n), aspect_ratio=(1, 1, 1)) for k, p in
enumerate(Polyhedron(prod(mm[i] for i in w)*V) for w in
W.iterate_by_length(n)))

The fifth iterate of the fractal construction, as shown in Figure 16, can then
be visualized using:

f5 = f(5)f5.show(viewer=’threejs’) # or viewer=’tachyon’ or viewer=’jmol’

##  Appendix B Code for pentagon billiard trajectory pictures

The `Sage` code would add several pages to the printed form of this article,
so we include it after the `\end{document}` at the end of the LaTeX source
file, available on arXiv.