We classify and enumerate all rational numbers with approximation
constant at least $\frac{1}{3}$ using hyperbolic geometry. Rational
numbers correspond to geodesics in the modular torus with both ends
in the cusp, and the approximation constant measures how far they
stay out of the cusp neighborhood in between. Compared to the
original approach, the geometric point of view eliminates the need
to discuss the intricate symbolic dynamics of continued fraction
representations, and it clarifies the distinction between the two
types of worst approximable rationals: (1) There is a plane forest
of Markov fractions whose denominators are Markov
numbers. They correspond to simple geodesics in the modular torus
with both ends in the cusp. (2) For each Markov fraction, there are
two infinite sequences of companions, which correspond to
non-simple geodesics with both ends in the cusp that do not
intersect a pair of disjoint simple geodesics, one with both ends in
the cusp and one closed.

MSC (2010).
10F20, 30F60

Key words and phrases.
Approximation constant,
Diophantine approximation,
Markov equation,
modular torus

§ INTRODUCTION

For any real number $x$, the approximation constant

\begin{equation}
\label{eq:C}
\end{equation}

measures how well $x$ can be approximated by rational
numbers $\frac{p}{q}\not=x$. Excluding $x$ from the admissible
approximants $\frac{p}{q}$ makes $C(x)$ strictly positive for all
rational numbers $x$. (This is easy to see and will be discussed
shortly.) Intuitively, the larger the approximation constant $C(x)$ of
a rational number $x\in\Q$, the more isolated the rational number $x$
is among the other rational numbers.

An irrational number $x$ is sometimes called badly approximable
if $C(x)>0$, and this is the case if and only if the sequence of
partial denominators $a_{k}$ of the continued fraction expansion

\begin{equation*}
+ \cfrac{1}{a_{1}
+\cfrac{1}{a_{2} + \ldots}}
\end{equation*}

is bounded (see, e.g., <cit.>
or <cit.>).

By contrast (as stated before) $C(x)>0$ for all rational numbers
$x=\frac{p_{0}}{q_{0}}$. Moreover, the nonzero infimum in (<ref>)
is attained for some fraction $\frac{p}{q}$ with smaller denominator,
$q\leq q_{0}$. Because, on the one hand, if $q>q_{0}$ and
$\frac{p}{q}\not=\frac{p_{0}}{q_{0}}$, then

\begin{equation*}
\frac{q}{q_{0}}\cdot\big|\,p_{0}q\;-\;q_{0}p\,\big| > 1.
\end{equation*}

On the other hand, if $q=1$ and $p$ is a closest integer to $x$
(except $x$ itself if $x\in\Z$) then

\begin{equation*}
q^{2}\cdot\big|\,x-\frac{p}{q}\,\big| = |\,x-p\,| \leq 1
\end{equation*}

with equality if and only if $x\in\Z$.

This shows not only that all rational numbers are badly approximable
in the sense that $C(x)>0$, but also that the very worst approximable
rational numbers are the integers $x\in\Z$ with $C(x)=1$. They are
followed by the half-integers $x\in \Z+\frac{1}{2}$ with
$C(x)=\frac{1}{2}$, which are best approximated by the two the nearest
integers. This can be seen as a striking confirmation of Hardy and
Wright's general observation: “From the point of view of rational
approximation, the simplest numbers are the worst”

This article is about a geometric method to classify and enumerate the
rational numbers $x$ with approximation constant
$C(x)\geq\frac{1}{3}$. Previously, the real numbers with
$C(x)>\frac{1}{3}$ were classified in terms of their continued
fraction representations by Flahive [14] for rational $x$ and
by Gurwood [17] for irrational $x$. Flahive's
classification apparently extends to include rational numbers with

All of this is closely related to the classification of Markov
irrationals and Markov forms, i.e., of irrational numbers
$x$ with Lagrange number $L(x)>\frac{1}{3}$ and of indefinite
quadratic forms $f(x,y)=ax^{2}+2bxy+cy^{2}$ with Markov
constant $M(f)>\frac{2}{3}$, where

\begin{gather}
\label{eq:Lagrange_number}
\\\intertext{and}
\label{eq:Markov_constant}
\min_{(x,y)\,\in\,\Z^{2}\setminus\{(0,0)\}}
\frac{f(x,y)}{\sqrt{-\det\,f}}\,.
\end{gather}

The monographs [1, 6, 10] and their
bibliographies are excellent resources for that theory. (Often the
Lagrange number is defined as the reciprocal value, $1/L(x)$.)

While the Lagrange number $L(x)$ and the Markov constant $M(f)$ are
invariant under the actions of $\GLTZ$, the approximation number
$C(x)$ is only invariant under the group of $\Z$-affine

\begin{equation}
\label{eq:Z-affine}
x\;\longmapsto\; \pm\,x\;+\;n\qquad (n\;\in\;\Z).
\end{equation}

In fact, the irrational numbers with $C(x)>\frac{1}{3}$ classified by
Gurwood are special representatives of the $\GLTZ$-classes of Markov
irrationals. In this article, we focus on the worst approximable
rational numbers.

The connection between continued fractions and Diophantine
approximation on the one hand, and hyperbolic geometry on the other
hand is well established [4, 11, 20]. Both
Markov forms and Markov irrationals correspond to geodesics in the
modular torus [1, 7, 8, 16, 18, 31]: Markov forms correspond to simple closed geodesics, and
Markov irrationals correspond to simple geodesics with one end in the
cusp and the other end spiraling into a simple closed geodesic. The
Lagrange number $L(x)$ and the Markov constant $M(f)$ correspond to
distances between geodesics and the Ford circles interpreted as
horocycles. This makes it possible to classify the Markov irrationals
and Markov forms using purely geometric
methods [32]. In this article, we take the same
geometric approach to classify the worst approximable rational

Figure <ref>

[anchor=south west, inner sep=0pt] at (0,0) ;
at (.5,6.5) $\dfrac{0}{1}$;
at (7.675,0.85) $\dfrac{1}{3}$;
at (11.35,2) $\dfrac{1}{2}$;
at (9.19,.33) $\dfrac{2}{5}$;

[anchor=south west, inner sep=0pt] at (0,0) ;

at (6.1,2.8) $\dfrac{2}{5}$;
at (0.85,1.1) $\dfrac{3}{8}$;
at (2.86,0.45) $\dfrac{5}{13}$;
at (9.56,0.5) $\dfrac{5}{12}$;

(1.7,-.2) node[below=-2pt] $\frac{8}{21}$ – (2.1,.2);
(2.2,-.2) node[below=-2pt] $\frac{21}{55}$ – (2.28,.065);
(2.7,-.2) node[below=-2pt] $\frac{13}{34}$ – (2.39,.1);

(3.29,-.2) node[below=-2pt] $\frac{29}{75}$ – (3.29,.05);

(8.46,-.2) node[below=-2pt] $\frac{31}{75}$ – (8.86,.05);
(8.96,-.2) node[below=-2pt] $\frac{12}{29}$ – (8.96,.12);
(9.46,-.2) node[below=-2pt] $\frac{29}{70}$ – (9.065,.045);

Ford circles scaled by $\frac{2}{3}$. Those whose highest
point is visible from $\sminfty$ belong to rational numbers with
approximation constant $\geq\frac{1}{3}$. A few of them are
indicated in the figure (compare
Figure <ref>
and Table <ref>). The bottom figure shows a detail of the
top figure.

illustrates the geometric interpretation of the approximation
constant. The extended real line $\RP^{1}=\R\cup\{\infty\}$ is the
ideal boundary of the hyperbolic plane $H^{2}$ in the half-plane
model. For an irrational number $x\in\R\setminus\Q$ and a bound $c>0$,
the approximation constant satisfies the inequality $C(x)\geq c$ if
and only if $x$ is visible from $\sminfty$ behind the Ford circles
after they are scaled by $2c$. For a rational number, $x\in\Q$, the
condition is that $x$ is visible after the Ford circle at $x$ itself
is removed, or equivalently, that the highest point of the Ford circle
at $x$ is visible

The classical approach to Diophantine approximation based on continued
fractions can also be interpreted geometrically, because continued
fractions encode the symbolic dynamics of geodesics in the Farey
triangulation. But our geometric approach goes further. In effect, we
trivialize the symbolic dynamics by considering not only one but all
ideal triangulations of the modular torus.

§ OVERVIEW

§.§ Classification of the worst approximable rational numbers

The main result of this article is the classification of the
worst approximable rational numbers in terms of Markov
fractions (see Section <ref>) and their
companions (see Section <ref>):

For a rational number $x\in\Q$, the following two statements are

* $C(x)\geq\frac{1}{3}$

* The number $x$ is either a Markov fraction or a companion of a
Markov fraction.
Furthermore, $C(x)=\frac{1}{3}$ if and only if $x$ is the first left
or right companion of a Markov fraction.

For the first companions of a Markov fraction, the approximation
constant $C(x)=\frac{1}{3}$ is attained only by the associated Markov
fraction, while for every rational number with $C(x)>\frac{1}{3}$, the
approximation constant is attained by exactly two rational numbers
(see Theorems <ref>
and <ref>).

The proof of Theorem <ref> that is presented in
Section <ref> uses the characterization of Markov
fractions and their companions in terms of geodesics in the modular
torus (see Section <ref>).

Interest in classifying the worst approximable rational numbers was
initially motivated by the uniqueness conjecture for Markov
numbers [1, 24, 33]. This is the
proposition — yet to be proved or refuted — that every Markov
number is the maximum of exactly one ordered Markov triple. Indeed,
the uniqueness conjecture for Markov numbers is equivalent to the
following uniqueness conjecture for Markov fractions:

For each Markov number $q$, the interval $[0,\frac{1}{2}]$ contains
at most one Markov fraction with denominator $q$ .

One may equivalently write “exactly one” instead of “at most one”,
because the Markov fractions come in $\Z$-affine orbits and every
Markov number appears among the denominators (see
Section <ref>).

§.§ The forest of Markov fractions

The Markov fractions between zero and one are naturally associated
with the faces of a plane binary tree with the root edge and
generating rule shown in Figure <ref>
(see Lemma <ref>).

2*  $\dfrac{1}{1}$

2*  $\dfrac{0}{1}$

2*  $\dfrac{p_{2}}{q_{2}}$
1|c2*  $\dfrac{p'}{q'}$
2*  $\dfrac{p_{1}}{q_{1}}$       1|c

\begin{equation}
\label{eq:pqprime}
\begin{split}
p' \;&=\;
\frac{p_{1}\,q_{1}\,+\,p_{2}\,q_{2}}{p_{2}\,q_{1}\,-\,p_{1}\,q_{2}}\\[\medskipamount]
q' \;&=\;
\frac{q_{1}^{2}\,+\,q_{2}^{2}}{p_{2}\,q_{1}\,-\,p_{1}\,q_{2}}
\end{split}
\end{equation}

Root edge and generating rule for the tree of Markov
fractions in $[0,1]$, of which one half is shown in
Figure <ref>

For example, the first branching after the root edge produces the
Markov fraction $\frac{1}{2}$. At each node, the numbers $p'$ and
$q'$ obtained by equations (<ref>) are coprime integers, so
the Markov fraction $\frac{p'}{q'}$ is reduced.

Figure <ref>

Markov fractions in the interval
$[0,\frac{1}{2}]$. Numerical values are shown in the right

shows the first few levels of the subtree containing the Markov
fractions in the interval $[0,\frac{1}{2}]$. There is a symmetric
subtree containing the Markov fractions in $[\frac{1}{2},1]$, obtained
by a reflection and replacing each Markov fraction $x$ with
$1-x$. Likewise, the Markov fractions $x+n$ in any other integer
interval $[n,n+1]$ are arranged in a similar tree, and all these trees
form a plane binary forest. The generating rule is the same for all
trees, and the root edges are infinite parallel rays separating faces
associated with consecutive integers (see Figure <ref>).

2*  $n+1$

2*-.2pt– – – –          1|c2*  $n+\dfrac{1}{2}$

3*  $n$

2*-.2pt– – – –          1|c2*  $n-\dfrac{1}{2}$

2*  $n-1$

Roots of the forest of Markov fractions

At each node in the forest, the three Markov fractions form a
rational Markov triple
which is centered (see Definitions <ref>
and <ref>). Conversely, all centered rational
Markov triples arise in this way, except for the integer triples
$\big(n-1,n,n+1\big)\in\Z^{3}$ which correspond to pairs of
neighboring root edges.

§.§ Limits of infinite paths in the forest

The forest structure of the set of Markov fractions is useful to
study the set of its accumulation points. But first note that every
accumulation point of the set of Markov fractions is
irrational. Indeed for $\frac{a}{b}\in\Q$ and a Markov fraction
$\frac{p}{q}\not=\frac{a}{b}$, we have
$\big|\frac{p}{q}-\frac{a}{b}\big|>\frac{b^{2}}{3}$, so $\frac{a}{b}$ is not
an accumulation point.

Now consider infinite paths in the tree of Markov fractions between
$0$ and $1$, starting at the root edge. Each path corresponds to a
sequence of Markov fractions with increasing denominators, which
converges to an irrational number $x$ with Lagrange number

For example, always taking the lower branch in
Figure <ref> generates the sequence

\begin{equation*}
\frac{1}{2}\,,\; \frac{2}{5}\,,\; \frac{5}{13}\,,\; \frac{13}{34}\,,\;
\ldots
\end{equation*}

which converges to

\begin{equation*}
\phi^{-2}\;=\;\frac{1}{2}\,\big(3-\sqrt{5}\big)\;=\;2\,-\,\phi\,,
\end{equation*}

with $\phi=\frac{1}{2}(1+\sqrt{5})$, the golden ratio. This is a Markov
irrational with Lagrange number

Always taking the upper branch in
Figure <ref> generates the sequence

\begin{equation*}
\frac{2}{5}\,,\;\frac{12}{29}\,,\;\frac{70}{169}\,,\;\frac{408}{985}\,,\;\ldots
\end{equation*}

with limit $-1+\sqrt{2}$. This is another Markov irrational, with
Lagrange number

For these two examples, the limits represent the two worst
approximable classes of irrational numbers. But there are only a
countable number of $\PGLTZ$-classes of Markov irrationals and an
uncountable number of paths in an infinite binary tree, so these two
examples are exceptional. In fact, the limit $x$ is a Markov
irrational if the path turns always left or always right except for
finitely many branchings. If, on the other hand, there are infinitely
many left turns and infinitely many right turns, then the limit has
Lagrange number $L(x)=\frac{1}{3}$. Before we come back to this in
Section <ref>, let us use the Stern–Brocot tree to
label Markov fractions.

§.§ Labeling the Markov fractions in the unit interval

The denominators of Markov fractions are Markov numbers. If we take
the subtree of Markov fractions in the interval $[0,\frac{1}{2}]$
shown in Figure <ref> and replace the
fractions by their denominators alone, we obtain one of the six
symmetric subtrees of the tree of Markov numbers shown in
Figure <ref>.

[grow cyclic]
level 1=[level distance=25pt,sibling angle=120]
level 2=[level distance=25pt,sibling angle=80]
level 3=[level distance=20pt,sibling angle=60]
level 4=[level distance=15pt,sibling angle=60]
level 5=[level distance=15pt,sibling angle=60]
every node=[font=]
child foreach in 1,2,3
child foreach in 1,2
child foreach in 1,2
child foreach in 1,2
child foreach in 1,2;

(-35pt,0pt) node 1;
[rotate=120] (-35pt,0pt) node 1;
[rotate=-120] (-35pt,0pt) node 1;

(58pt,0pt) node 2;
[rotate=120] (58pt,0pt) node 2;
[rotate=-120] (58pt,0pt) node 2;

(62pt,31pt) node 5;
(62pt,-31pt) node 5;
[rotate=120] (62pt,31pt) node 5;
[rotate=120] (62pt,-31pt) node 5;
[rotate=-120] (62pt,31pt) node 5;
[rotate=-120] (62pt,-31pt) node 5;

(81pt,23pt) node 29;
(81pt,-23pt) node 29;
[rotate=120] (81pt,23pt) node 29;
[rotate=120] (81pt,-23pt) node 29;
[rotate=-120] (81pt,23pt) node 29;
[rotate=-120] (81pt,-23pt) node 29;

(56.5pt,51pt) node 13;
(56.5pt,-51pt) node 13;
[rotate=120] (56.5pt,51pt) node 13;
[rotate=120] (56.5pt,-51pt) node 13;
[rotate=-120] (56.5pt,51pt) node 13;
[rotate=-120] (56.5pt,-51pt) node 13;

(95pt,10pt) node 169;
(95pt,-10pt) node 169;
[xshift=0.5pt,yshift=0pt] [rotate=120] (95pt,10pt) node 169;
[xshift=-2pt,yshift=-3pt] [rotate=120] (95pt,-10pt) node 169;
[xshift=0pt,yshift=4pt] [rotate=-120] (95pt,10pt) node 169;
[xshift=1pt,yshift=0pt] [rotate=-120] (95pt,-10pt) node 169;

(89pt,38pt) node 433;
(89pt,-38pt) node 433;
[xshift=-1pt,yshift=-3pt] [rotate=120] (89pt,38pt) node 433;
[xshift=2.5pt,yshift=0pt] [rotate=120] (89pt,-38pt) node 433;
[xshift=2pt,yshift=0pt] [rotate=-120] (89pt,38pt) node 433;
[xshift=0pt,yshift=3.5pt] [rotate=-120] (89pt,-38pt) node 433;

(76pt,53pt) node 194;
(76pt,-53pt) node 194;
[xshift=-2pt,yshift=-3pt] [rotate=120]  (76pt,53pt) node 194;
[xshift=2pt,yshift=1pt] [rotate=120]  (76pt,-53pt) node 194;
[xshift=2.5pt,yshift=-1pt] [rotate=-120]  (76pt,53pt) node 194;
[xshift=-1pt,yshift=4pt] [rotate=-120]  (76pt,-53pt) node 194;

(46pt,63pt) node 34;
(46pt,-63pt) node 34;
(32pt,71pt) node 34;
(32pt,-71pt) node 34;
(-79pt,8pt) node 34;
(-79pt,-8pt) node 34;

line width=0.5pt,
every node/.style=inner sep=0pt, minimum size=0pt,
line width=0pt
(0,0) node (l)
node [left=0.2] $q_{1}$
node [above=.2] $q_{2}$
node [below=.2] $q_{3}$
+(.5,0) node (r)
node [right=0.2] $q_{1}\smash{'}$;
(l) – +(120:0.5);
(l) – +(-120:0.5);
(r) – +(60:0.5);
(r) – +(-60:0.5);

The tree of Markov numbers and its generating rule

The combinatorial isomorphism with the
Stern–Brocot tree leads to the customary labeling of Markov numbers
by the rational numbers in the interval $[0,1]$. If we do not forget
about the numerators of the Markov fractions, we obtain a labeling of
the Markov fractions in the interval $[0,\frac{1}{2}]$.  For our
purposes, it makes sense to extend the Stern–Brocot tree to contain
all non-negative rational numbers and $\frac{1}{0}=\infty$. That is,
we consider the rooted trivalent plane tree with the following root
edge and generating rule:

2*  $\dfrac{1}{0}$

2*  $\dfrac{0}{1}$

2*  $\dfrac{p_{2}}{q_{2}}$
1|c2* $\dfrac{p_{1}\,+\,p_{2}}{q_{1}\,+\,q_{2}}$
2*  $\dfrac{p_{1}}{q_{1}}$       1|c

The combinatorial isomorphism of rooted plane trees leads to a

\begin{equation}
\label{eq:mu}
\begin{split}
\mu:\;\Q_{\geq 0}\;\cup\;\{\infty\}\quad
\big\{\,\text{Markov fractions}\,\big\}\;\cap\;[0,1]
\\
\frac{n}{m}\quad
\mu_{\frac{n}{m}}\;.
\end{split}
\end{equation}

The symmetries of the (extended) Stern-Brocot tree and the tree of
Markov fractions in $[0,1]$ lead to the functional equation
$\mu_{\frac{n}{m}}+\mu_{\frac{m}{n}}=1.$ For example:

The same correspondence $\mu$ also arises from the construction
involving triangle paths in the Eisenstein lattice explained in
Section <ref>.

§.§ Limits of infinite paths (continued)

The combinatorial isomorphism between the (extended) Stern–Brocot
tree and the tree of Markov fractions in the interval $[0,1]$ extends
to a bijection between infinite paths starting at the root edges of
either tree, respectively.

Each infinite path in the Stern–Brocot tree generates a sequence of
nonnegative rational numbers that converges to a finite limit or
diverges properly to $\sminfty$. The limit of a path with infinitely
many left turns and infinitely many right turns is irrational, and
each positive irrational number is the limit for precisely one such
path. Indeed, the sequence of partial denominators of the continued
fraction expansion is the run-length encoding of the sequence of left
and right turns. On the other hand, an infinite path with only
finitely many left or finitely many right turns has a rational limit,
and every positive rational number is the limit for two such
paths. The two corresponding paths in the tree of Markov fractions
have different but $\PGLTZ$-equivalent Markov irrationals as
limits. Thus, the correspondence of infinite paths defines a map

\begin{equation*}
\widehat{\mu}:\;\R_{\geq 0}\;\cup\;\{\infty\}
\quad\longrightarrow\quad
\big\{\,\text{$\PGLTZ$-classes of real numbers with $L\geq\tfrac{1}{3}$}\,\big\}.
\end{equation*}

If $x$ is rational, $\widehat{\mu}(x)$ is a class of Markov
irrationals. Otherwise, $\widehat{\mu}(x)$ is a class of irrationals
with Lagrange number $=\frac{1}{3}$. It is not difficult to see that
the restriction of $\widehat{\mu}$ to $\Q\cap[0,1]$ is a bijection
onto the set of equivalence classes of Markov irrationals. What about
the restriction of $\widehat{\mu}$ to $[0,1]\setminus\Q$? The
following two questions seem to be open:

Is every irrational number $x$ with Lagrange number
$L(x)=\frac{1}{3}$ equivalent to the limit of some infinite path in
the tree of Markov fractions between $0$ and $1$ ?

Do two different infinite paths exist in the subtree containing all
Markov fractions between $0$ and $\frac{1}{2}$ (see
Figure <ref>) such that their limits are
$\PGLTZ$-equivalent with Lagrange number $L=\frac{1}{3}$ ?

§.§ Triangle paths for Markov fractions

The Markov fractions $\mu_{\frac{n}{m}}$ can be found by a
Fibonacci-like recursion with Farey addition along triangle paths (see
Figure <ref>).

[anchor=south west, inner sep=0pt] at (0,0) ;
[anchor=south west, inner sep=0pt] at (0,0) ;
[anchor=south west, inner sep=0pt] at (0,0) ;
[anchor=south west, inner sep=0pt] at (0,0) ;
[anchor=south west, inner sep=0pt] at (0,0) ;

[anchor=north, inner sep=4pt] at (.09,.09) $\dfrac{1}{0}$;
[anchor=north, inner sep=4pt] at (1.49,.09) $\dfrac{0}{1}$;
[anchor=north, inner sep=4pt] at (2.9,.09) $\dfrac{1}{3}$;
[anchor=north, inner sep=4pt] at (4.3,.09) $\dfrac{3}{8}$;
[anchor=north, inner sep=4pt] at (5.71,.09) $\dfrac{8}{21}$;
[anchor=north, inner sep=4pt] at (7.12,.09) $\dfrac{21}{55}$;

[anchor=south, inner sep=4pt] at (0.8,1.3) $\dfrac{1}{1}$;
[anchor=south, inner sep=4pt] at (2.2,1.3) $\dfrac{1}{2}$$\,=\mu_{\frac{1}{1}}$;
[anchor=south, inner sep=4pt] at (3.61,1.3) $\dfrac{2}{5}$$\,=\smash{\mu_{\frac{1}{2}}}$;
[anchor=south, inner sep=4pt] at (5.01,1.3) $\dfrac{5}{13}$$\,=\smash{\mu_{\frac{1}{3}}}$;
[anchor=south, inner sep=4pt] at (6.42,1.3) $\dfrac{13}{34}$$\,=\smash{\mu_{\frac{1}{4}}}$;
[anchor=south, inner sep=4pt] at (7.82,1.3) $\dfrac{34}{89}$$\,=\smash{\mu_{\frac{1}{5}}}$;

[anchor=south west, inner sep=0pt] at (0,0);

[anchor=south west, inner sep=0pt] at (0,0);
[anchor=south west, inner sep=0pt] at (0,0);
[anchor=south west, inner sep=0pt] at (0,0);

[anchor=north, inner sep=4pt] at (.09,.09) $\dfrac{1}{0}$;
[anchor=north, inner sep=4pt] at (1.49,.09) $\dfrac{0}{1}$;
[anchor=north, inner sep=4pt] at (2.9,.09) $\dfrac{1}{3}$;
[anchor=north, inner sep=4pt] at (4.3,.09) $\dfrac{3}{8}$;
[anchor=north, inner sep=4pt] at (5.11,1.3) $\dfrac{5}{13}$;
[anchor=north, inner sep=4pt] at (6.42,1.3) $\dfrac{17}{44}$;
[anchor=north, inner sep=4pt] at (7.82,1.3) $\dfrac{46}{119}$;

[anchor=south, inner sep=4pt] at (0.8,1.3) $\dfrac{1}{1}$;
[anchor=south, inner sep=4pt] at (2.2,1.3) $\dfrac{1}{2}$;
[anchor=south, inner sep=4pt] at (2.2,1.3) $\dfrac{1}{2}$;
[anchor=south, inner sep=4pt] at (3.51,1.3) $\dfrac{2}{5}$;
[anchor=south, inner sep=4pt] at (4.3,2.52) $\dfrac{7}{18}$;
[anchor=south, inner sep=4pt] at (5.71,2.52) $\dfrac{12}{31}$;
[anchor=south, inner sep=4pt] at (7.12,2.52) $\dfrac{29}{75}$;
[anchor=south, inner sep=4pt] at (8.53,2.52) $\dfrac{75}{194}$$\smash{\,=\mu_{\frac{2}{5}}}$;

[anchor=south west, inner sep=0pt] at (0,0);

[anchor=south west, inner sep=0pt] at (0,0);
[anchor=south west, inner sep=0pt] at (0,0);

[anchor=north, inner sep=4pt] at (.09,.09) $\dfrac{1}{0}$;
[anchor=north, inner sep=4pt] at (1.49,.09) $\dfrac{0}{1}$;
[anchor=north, inner sep=4pt] at (2.9,.09) $\dfrac{1}{3}$;
[anchor=north, inner sep=4pt] at (3.71,1.3) $\dfrac{2}{5}$;
[anchor=north, inner sep=4pt] at (5.01,1.3) $\dfrac{7}{17}$;
[anchor=north, inner sep=4pt] at (6.42,1.3) $\dfrac{19}{46}$;
[anchor=north, inner sep=4pt] at (7.32,2.51) $\dfrac{31}{75}$;
[anchor=north, inner sep=4pt] at (8.53,2.51) $\dfrac{105}{254}$;

[anchor=south, inner sep=4pt] at (0.8,1.3) $\dfrac{1}{1}$;
[anchor=south, inner sep=4pt] at (2.1,1.3) $\dfrac{1}{2}$;
[anchor=south, inner sep=4pt] at (2.9,2.51) $\dfrac{3}{7}$;
[anchor=south, inner sep=4pt] at (4.3,2.51) $\dfrac{5}{12}$;
[anchor=south, inner sep=4pt] at (5.56,2.51) $\dfrac{12}{29}$;
[anchor=south, inner sep=4pt] at (6.42,3.72) $\dfrac{43}{104}$;
[anchor=south, inner sep=4pt] at (7.85,3.72) $\dfrac{74}{179}$;
[anchor=south, inner sep=4pt] at (9.3,3.72) $\dfrac{179}{433}$$\smash{\,=\mu_{\frac{3}{5}}}$;

[anchor=south west, inner sep=0pt] at (0,0);
[anchor=south west, inner sep=0pt] at (0,0);
[anchor=south west, inner sep=0pt] at (0,0);
[anchor=south west, inner sep=0pt] at (0,0);

[anchor=south west, inner sep=0pt] at (0,0);

[anchor=north, inner sep=4pt] at (.09,.09) $\dfrac{1}{0}$;
[anchor=north, inner sep=4pt] at (1.49,.09) $\dfrac{0}{1}$;
[anchor=north, inner sep=4pt] at (2.9,.09) $\dfrac{1}{3}$;
[anchor=north, inner sep=4pt] at (3.71,1.3) $\dfrac{2}{5}$$\smash{\,=\mu_{\frac{1}{2}}}$;
[anchor=north, inner sep=4pt] at (5.15,1.3) $\dfrac{7}{17}$;
[anchor=north, inner sep=4pt] at (5.86,2.51) $\dfrac{12}{29}$$\smash{\,=\mu_{\frac{2}{3}}}$;
[anchor=north, inner sep=4pt] at (7.32,2.51) $\dfrac{41}{99}$;
[anchor=north, inner sep=4pt] at (8.0,3.72) $\dfrac{70}{169}$$\smash{\,=\mu_{\frac{3}{4}}}$;
[anchor=north, inner sep=4pt] at (9.55,3.72) $\dfrac{239}{577}$;

[anchor=south, inner sep=4pt] at (0.8,1.3) $\dfrac{1}{1}$;
[anchor=south, inner sep=4pt] at (2.1,1.3) $\dfrac{1}{2}$;
[anchor=south, inner sep=4pt] at (2.9,2.51) $\dfrac{3}{7}$;
[anchor=south, inner sep=4pt] at (4.1,2.51) $\dfrac{5}{12}$;
[anchor=south, inner sep=4pt] at (5.01,3.72) $\dfrac{17}{41}$;
[anchor=south, inner sep=4pt] at (6.22,3.72) $\dfrac{29}{70}$;
[anchor=south, inner sep=4pt] at (7.12,4.93) $\dfrac{99}{239}$;
[anchor=south, inner sep=4pt] at (8.53,4.93) $\dfrac{169}{408}$;
[anchor=south, inner sep=4pt] at (9.94,4.93) $\dfrac{408}{985}$$\smash{\,=\mu_{\frac{4}{5}}}$;

[]Triangle paths of the Markov fractions $\mu_{\frac{n}{m}}$
with $1\leq n<m\leq 5$.

The following theorem defines this construction more precisely,
extending the snake graph construction for Markov
numbers [27].

Let $m\geq 0$ and $n\geq 0$ be coprime integers, let
$z_{m,n}=m+n\,\omega$ with $\omega=\frac{1}{2}\,(1+i\sqrt{3})$,
and let $\alpha$ be the oriented straight line segment from $0$ to
$z_{m,n}$. Let $\tau_{0}\;=\;[0,\;1,\;\omega]$, and unless $(m,n)$
equals $(1,0)$ or $(0,1)$, let

\begin{equation*}
\tau_{1}\;=\;[1,\;\omega,\;1+\omega],
\quad\ldots,\quad
\tau_{N}=[z_{m,n}-1,\;z_{m,n}-\omega,\; z_{m,n}]
\end{equation*}

be the finite sequence of triangles in the Eisenstein lattice
$\Z+\omega\,\Z$ that $\alpha$ intersects after
$\tau_{0}$. Recursively label the vertices of these triangles as

* Label the vertices $0$, $1$, and $\omega$ of $\tau_{0}$ with
$\frac{1}{0}$, $\frac{0}{1}$, and $\frac{1}{1}$, respectively.

* If the vertices of $\tau_{0},\ldots,\tau_{j-1}$ are already
labeled, label the remaining vertex of triangle $\tau_{j}$ with
$\frac{p_{1}+p_{2}}{q_{1}+q_{2}}$, where $\frac{p_{1}}{q_{1}}$
and $\frac{p_{2}}{q_{2}}$ are the labels at the vertices shared
with triangle $\tau_{j-1}$.

Then the label at $z_{m,n}$ is the Markov fraction $\mu_{\frac{n}{m}}$.

In Section <ref>, this construction is derived by
following geodesics in the Farey triangulation.

In Figure <ref>, the point $0$
(labeled $\frac{1}{0}$) is connected to the points $z_{m,n}$
(labeled with the Markov fractions $\mu_{\frac{n}{m}}$) not by
straight line segments as described in
Theorem <ref>, but by polygonal curves that are
determined as follows. Define a hyperbolic metric on each triangle
strip by equipping each triangle with the hyperbolic Beltrami–Klein
metric defined by its circumcircle <cit.>. In this
metric, the polygonal curves shown in Figures <ref>
and <ref> are hyperbolic geodesics connecting
their ideal endpoints. We use this construction because it works
equally well for companions (see Figure <ref>),
whereas the straight line segments would have vertices in their
interior. More importantly, Markov fractions and their companions
correspond to certain geodesics in the modular torus with both ends
in the cusp, and the polygonal curves shown in
Figures <ref> and <ref> are
natural representations of these geodesics in the following sense:
The canonical ideal triangulation of the modular torus induces a
euclidean metric on it [12], and if one uses this
euclidean metric to lift the hyperbolic geodesics to the euclidean
plane, one obtains the polygonal curves shown in the figures.

§.§ Companions of Markov fractions

Table <ref>

The first few right companions,
$\gamma^{+}_{2},\,\gamma^{+}_{3},\,\ldots$, of the Markov
fractions $\frac{p}{q}$ with denominator $q<1000$. The limits are
Markov irrationalities, i.e., irrational numbers $x$ with Lagrange
number $L(x)\;>\;\tfrac{1}{3}$. The fractions in the first row are
equal to $F_{2k-2}/F_{2k}$, and the denominators in the second row
are equal to the Pell numbers $P_{2k}$ (see
Remark <ref> and compare
Figure <ref>).

lists the first few right companions,
for each Markov fraction $\frac{p}{q}=\gamma^{+}_{1}(\frac{p}{q})$
with denominator $q<1000$ (see Definition <ref>).  The
corresponding left companions $\gamma^{-}_{k}(\frac{p}{q})$ are
symmetrically located:

\begin{equation}
\label{eq:companions_symmetric}
\gamma^{-}_{k}\,\Big(\frac{p}{q}\Big)
\,+\,
\gamma^{+}_{k}\,\Big(\frac{p}{q}\Big)
\;=\;2\;\frac{p}{q}\,.
\end{equation}

The sequences of
right and left companions converge to $\frac{p}{q}\pm \delta_{q}$,
respectively, with

\begin{equation}
\delta_{q}\;=\;\frac{3}{2}\;-\;\sqrt{\,\frac{9}{4}\;-\;\frac{1}{q^{2}}\,}\;.
\end{equation}

The closed interval

\begin{equation}
\label{eq:Ipq}
\end{equation}

all companions of $\frac{p}{q}$, but neither any other Markov
fractions nor their companions, but the endpoints of the interval are
limit points of the set of Markov fractions (see Lemma <ref>).

Figure <ref>

[anchor=south west, inner sep=0pt] (fig1) at (0,12)

(a) at (0.4, 12.45)$\frac{0}{1}$;
(b) at (6.92, 12.45)$\frac{1}{3}$;
(c) at (7.73, 12.45)$\frac{3}{8}$;
(d) at (7.95, 12.45)$\frac{5}{13}$;
(e) at (8.23, 12.45)$\frac{2}{5}$;
(f) at (8.51, 12.45)$\frac{12}{29}$;
(g) at (10.2, 12.45)$\frac{1}{2}$;
[-Stealth, thick, densely dotted] (7.74, 12.15) |- (0.4, 11.8) – (0.4, 11.3);
[-Stealth, thick, densely dotted] (7.935, 12.15) |- (10.2, 11.8) – (10.2, 11.3);

[anchor=south west, inner sep=0pt] (fig1) at (0,6)

(h) at (0.4, 6.45)$\frac{3}{8}$;
(i) at (6.465, 6.45)$\frac{8}{21}$;
(j) at (7.31, 6.45)$\frac{21}{55}$;
(k) at (7.575, 6.45)$\frac{34}{89}$;
(l) at (7.86, 6.45)$\frac{13}{34}$;
(m) at (8.24, 6.45)$\frac{507}{1325}$;
(n) at (10.2, 6.45)$\frac{5}{13}$;
[-Stealth, thick, densely dotted] (7.34, 6.15) |- (0.4, 5.8) – (0.4, 5.3);
[-Stealth, thick, densely dotted] (7.56, 6.15) |- (10.2, 5.8) – (10.2, 5.3);

[anchor=south west, inner sep=0pt] (fig1) at (0,0)

(o) at (0.4, 0.45)$\frac{21}{55}$;
(p) at (6.4, 0.45)$\frac{55}{144}$;
(q) at (7.17, 0.45)$\frac{144}{377}$;
(r) at (7.56, 0.45)$\frac{233}{610}$;
(s) at (7.95, 0.45)$\frac{89}{233}$;
(t) at (8.45, 0.45)$\frac{23763}{62210}$;
(u) at (10.2, 0.45)$\frac{34}{89}$;

Markov fractions and their companions. Markov fractions are
marked by vertical lines and connected to their companions by
semicircles. The only companions that are visually discernible at
the scales of these figures are the first six right companions of
$\frac{0}{1}$ (see Table <ref>). The vertical lines
and semicircles project to simple geodesics in the modular torus
with both ends in the cusp.

indicates a striking self-similarity in the arrangement of Markov
fractions and their companions.

§.§ Triangle paths for companions

The left and right companions of a Markov fraction can also be found
by a Fibonacci-like recursion along triangle paths (see
Figure <ref>).

[anchor=south west, inner sep=0pt] at (0,0) ;
[anchor=south west, inner sep=0pt] at (0,0) ;
[anchor=south west, inner sep=0pt] at (0,0) ;

[anchor=center, inner sep=4pt] at (0,1.3) (a);

[anchor=north, inner sep=4pt] at (.09,.09) $\dfrac{1}{0}$;
[anchor=north, inner sep=4pt] at (1.49,.09) $\dfrac{0}{1}$;
[anchor=north, inner sep=4pt] at (2.9,.09) $\dfrac{1}{3}$;
[anchor=north, inner sep=4pt] at (3.71,1.3) $\dfrac{2}{5}$;
[anchor=north, inner sep=4pt] at (5.01,1.3) $\dfrac{7}{17}$;
[anchor=north, inner sep=4pt] at (6.42,1.3) $\dfrac{19}{46}$;
[anchor=north, inner sep=4pt] at (7.32,2.51) $\dfrac{31}{75}$;
[anchor=north, inner sep=4pt] at (8.53,2.51) $\dfrac{105}{254}$;
[anchor=north, inner sep=4pt] at (9.95,2.51) $\dfrac{284}{687}$;

[anchor=south, inner sep=4pt] at (0.8,1.3) $\dfrac{1}{1}$;
[anchor=south, inner sep=4pt] at (2.1,1.3) $\dfrac{1}{2}$;
[anchor=south, inner sep=4pt] at (2.9,2.51) $\dfrac{3}{7}$;
[anchor=south, inner sep=4pt] at (4.3,2.51) $\dfrac{5}{12}$;
[anchor=south, inner sep=4pt] at (5.56,2.51) $\dfrac{12}{29}$;
[anchor=south, inner sep=4pt] at (6.42,3.72) $\dfrac{43}{104}$;
[anchor=south, inner sep=4pt] at (7.85,3.72) $\dfrac{74}{179}$;
[anchor=south, inner sep=4pt] at (9.25,3.72) $\dfrac{179}{433}$;
[anchor=south, inner sep=4pt] at (10.65,3.72) $\dfrac{463}{1120}$;

[anchor=south west, inner sep=0pt] at (0,0) ;
[anchor=south west, inner sep=0pt] at (0,0) ;
[anchor=south west, inner sep=0pt] at (0,0) ;

[anchor=center, inner sep=4pt] at (0,1.3) (b);

[anchor=north, inner sep=4pt] at (.09,.09) $\dfrac{1}{0}$;
[anchor=north, inner sep=4pt] at (1.49,.09) $\dfrac{0}{1}$;
[anchor=north, inner sep=4pt] at (2.9,.09) $\dfrac{1}{3}$;
[anchor=north, inner sep=4pt] at (4.3,.09) $\dfrac{3}{8}$;
[anchor=north, inner sep=4pt] at (5.11,1.3) $\dfrac{5}{13}$;
[anchor=north, inner sep=4pt] at (6.42,1.3) $\dfrac{17}{44}$;
[anchor=north, inner sep=4pt] at (7.82,1.3) $\dfrac{46}{119}$;
[anchor=north, inner sep=4pt] at (8.73,2.51) $\dfrac{75}{194}$;
[anchor=north, inner sep=4pt] at (9.95,2.51) $\dfrac{254}{657}$;

[anchor=south, inner sep=4pt] at (0.8,1.3) $\dfrac{1}{1}$;
[anchor=south, inner sep=4pt] at (2.2,1.3) $\dfrac{1}{2}$;
[anchor=south, inner sep=4pt] at (3.51,1.3) $\dfrac{2}{5}$;
[anchor=south, inner sep=4pt] at (4.3,2.52) $\dfrac{7}{18}$;
[anchor=south, inner sep=4pt] at (5.71,2.52) $\dfrac{12}{31}$;
[anchor=south, inner sep=4pt] at (6.92,2.52) $\dfrac{29}{75}$;
[anchor=south, inner sep=4pt] at (7.85,3.72) $\dfrac{104}{269}$;
[anchor=south, inner sep=4pt] at (9.25,3.72) $\dfrac{179}{463}$;
[anchor=south, inner sep=4pt] at (10.65,3.72) $\dfrac{433}{1120}$;

[](a) Triangle paths for $\mu_{\frac{1}{2}}=\frac{2}{5}$ and the
first and second right companions
$\gamma^{+}_{2}(\frac{2}{5})=\frac{31}{75}$ and
(b) Same for the first and second left companions
$\gamma^{-}_{2}(\frac{2}{5})=\frac{29}{75}$ and

Roughly speaking, while the Markov fractions are obtained from
triangle paths of primitive lattice vectors, the companions are
obtained from triangle paths of imprimitive lattice vectors. To avoid
lattice points in their interior, the paths are bent slightly to
the the left or to the right, which leads to right or left companions,
respectively. More precisely:

With $m$, $n$, $\omega$, $z_{m,n}$ , and $\alpha$ as in
Theorem <ref>, let $k$ be an integer $\geq 2$, so
that the oriented line segment $k\alpha$ from $0$ to
$k\,z_{m,n}=z_{km,kn}$ contains $k-1$ vertices of the Eisenstein
lattice in its interior. Bend this line segment slightly to the left
or to the right while keeping the endpoints fixed to obtain an
oriented curve $\beta^{+}$ or $\beta^{-}$, respectively, that avoids
lattice points in the interior. Or, to be more precise and more
definite, let $\beta^{\pm}$ be the curve

\begin{equation*}
\beta^{\pm}:\;[0,1]\rightarrow\C,\quad
\beta^{\pm}(t)\;=\;(t\,\pm\,i\,\epsilon\,t\,(1-t))\,k\,z_{m,n},
\end{equation*}

where $\epsilon>0$ is chosen small enough so that there are no
lattice points between the line segment $k\alpha$ and $\beta^{\pm}$
or in the interior of $\beta^{\pm}$. As in
Theorem <ref>, label the vertices $0$, $1$, and
$\omega$ with $\frac{1}{0}$, $\frac{0}{1}$, and $\frac{1}{1}$,
respectively, and label the vertices of the triangles
$ \tilde\tau^{\pm}_{0},\,\ldots,\,\tilde\tau^{\pm}_{N}$ crossed by
$\beta^{\pm}$ recursively, so that in each triangle of the sequence
the as yet unlabeled vertex is labeled with the Farey sum of the
already labeled vertices.

Then the label at $k\,z_{m,n}$ is $\gamma^{\pm}_{k}\big(\frac{p}{q}\big)$,
the $(k-1)$st left or right companion of $\frac{p}{q}=\mu_{\frac{n}{m}}$.

Like the analogous construction for Markov fractions (see
Theorem <ref>), this is derived by following
geodesics in the Farey triangulation (see
Section <ref>).

§.§ Hyperbolic geometry

In the upper half-space model of the the hyperbolic plane,

\begin{equation*}
H^{2}\;=\;\{\,z\in\C\;|\;\im\,z\; >\; 0\}\quad
\text{with metric}\quad
ds=\frac{|dz|}{\im z},
\end{equation*}

we associate a horocycle $h(p,q)$ with every pair $(p,q)$ of real
numbers, not both zero:

* If $q\not=0$, $h(p,q)$ is the horocycle centered at
$\frac{p}{q}$ with euclidean diameter $\frac{1}{q^{2}}$.

* $h(p,0)$ is the horizontal horocycle centered at $\sminfty$ with equation
$\im z = p^{2}$.

This establishes a $\PSLTR$-equivariant
bijection between the space

\begin{equation*}
(\R^{2}\setminus\{(0,0)\})/\{\pm 1\}
\end{equation*}

of lax vectors (in Conway's terminology [9]) and
the space of horocycles in the hyperbolic plane <cit.>,

The following observation translates between Diophantine approximation
and hyperbolic geometry:

The signed distance $d(g_{x},\,h(p,q))$ of the vertical
geodesic $g_{x}$ in $H^{2}$ joining $x\in\R$ and $\sminfty$ and a
horocycle $h(p,q)$ with $q\not=0$ and $\frac{p}{q}\not=x$ satisfies

\begin{equation}
\label{eq:dhg}
\end{equation}

See Figure <ref>.

[label= below:$x\vphantom{\frac{p}{q}}$] (x) at (0,0);
[label= below:$\frac{p}{q}$] (pq) at (1.5,0);

(-1,0) – (3.5,0);
(x) – +(0,3.5) node[pos=0.75, left] $g_{x}$;
(pq) – +(0,3.5);
($(pq)+(0,1)$) circle (1) node[right, xshift=9.5mm, yshift=2mm] $h(p,q)$;

[dashed] ($(pq)+(1.5,1.5)$) arc (0:338:1.5);

[dashed] ($(pq)+(0,2)$) – (3.25,2) node[right]$\im z=\dfrac{1}{q^{2}}$;
[dashed] ($(pq)+(0,3)$) – (3.25,3) node[right]$\im z=2\,
\Big|\,x-\dfrac{p}{q}\,\Big|$;

(0,1.5) arc (90:0:1.5);
[very thick] (0,1.5) arc (90:67.4:1.5) node[midway, above]
[very thick] ($(pq)+(0,2)$) – ($(pq)+(0,3)$) node[midway,
right] $d$;

in x, pq

[black] () circle (1pt);

Signed distance $d=d(g_{x},h(p,q))$ between the vertical
geodesic $g_{x}$ and the horocycle $h(p,q)$.

If $p$ and $q$ are coprime integers, then the horocycle $h(p,q)$ is a
Ford circle. The previous lemma relates the approximation
number $C(x)$ to the minimal signed distance between the geodesic
$g_{x}$ and a Ford circle $h(p,q)$ not centered at either $\sminfty$
or $x$ (see also Figure <ref>):

For $x\in\R$,

\begin{equation}
\label{eq:C_and_d}
\log(2C(x))\;=\;
\inf_{\frac{p}{q}\;\in\;\Q\,\setminus\,\{x\}}\;
d(g_{x}, h(p,q)).
\end{equation}

In particular, the following two statements are equivalent:

* $C(x)\geq\frac{1}{3}$

* For all Ford circles $h(p,q)$ with

\begin{equation}
\label{eq:dhg_ineq}
\end{equation}

The hyperbolic plane $H^{2}$ is the universal cover of the modular
torus $M$ (see Section <ref>), and the Ford circles
project to the boundary of the maximal cusp neighborhood in the
modular torus. The geometric proof of the classification
Theorem <ref> for worst approximable rational numbers in
Section <ref> also proves the following classification
theorem of geodesics in $M$ that have both ends in the
cusp but otherwise stay furthest away from the cusp:

Let $g$ be a geodesic in the modular torus $M$ with both ends in the
cusp. Then the following statements are equivalent:

* Away from the ends, the signed distance between $g$ and the
cusp is $\geq\log\frac{2}{3}$.

* $g$ does not intersect all simple closed geodesics in $M$.

* $g$ does not intersect all simple geodesics in $M$ with both
ends in the cusp.

If one and hence all of these conditions hold, then either

* $g$ is a simple geodesic. In this case there is a unique
simple closed geodesic that does not intersect $g$, and infinitely
many simple geodesics with both ends in the cusp that do not
intersect $g$.

* $g$ has self-intersections, and then there are a unique closed
geodesic $c$ in $M$ and a unique simple geodesic $\tilde{g}$ in $M$
with both ends in the cusp that do not intersect $g$.

In case (a), the geodesic $g$ in $M$ is the projection of a vertical
geodesic $g_{x}$ in $H^{2}$ for some Markov fraction $x$. In case (b),
the geodesic $g$ in $M$ is the projection of a vertical geodesic
$g_{x}$ in $H^{2}$ for some companion $x$ of a Markov fraction
$\tilde{x}$, and the geodesic $\tilde{g}$ in $M$ is the projection of
the vertical geodesic $g_{\tilde{x}}$ in $H^{2}$.

§ THE ALGEBRAIC POINT OF VIEW

This section treats the two types of worst approximable rational
numbers, the Markov fractions and their companions, from the algebraic
point of view. Geometrically minded readers may want to look at
Section <ref> first, which provides a complementary
geometric point of view.

§.§ Markov numbers and Markov triples

For reference and to fix notation, we begin with a brief review of
some basic definitions and facts about Markov numbers and Markov
triples [1, 6, 10].

A Markov triple $(q_{1},q_{2},q_{3})$ is a positive integer
solution of Markov's equation

\begin{equation}
\label{eq:Markov}
x^{2}\,+\,y^{2}\,+\,z^{2}\;=\;3 xyz\,.
\end{equation}

Equivalently, a triple of positive integers
$(q_{1},q_{2},q_{3})\in{(\Z_{>0})}^{3}$ is a Markov triple if and only if

\begin{equation}
\label{eq:Markov_q}
\frac{q_{1}}{q_{2}\,q_{3}}
\;+\;
\frac{q_{2}}{q_{3}\,q_{1}}
\;+\;\frac{q_{3}}{q_{1}\,q_{2}}
\;=\; 3.
\end{equation}

A Markov number is a positive integer that is an element of
some Markov triple. For example, $(5,1,2)$ and $(1,5,2)$ are two of
the six Markov triples containing the Markov numbers $1$, $2$, and

Since Markov's equation (<ref>) is quadratic in each
variable, there are two solutions for each variable if the other two
variables are fixed. Thus, if $(q_{1},q_{2},q_{3})$ is a Markov
triple, then so are its neighbors

\begin{equation*}
\end{equation*}

where the $q_{i}'$ are determined by Vieta's formulas for quadratic
polynomials, i.e.,

\begin{equation}
\label{eq:qiprime}
\;=\;3\,q_{j}\,q_{k}\,-\,q_{i}
\end{equation}

with $\{i,j,k\}=\{1,2,3\}$. This neighbor relation turns the set of
Markov triples into a plane binary tree. The Markov numbers correspond
to the regions into which the tree separates the plane (see
Figure <ref>).

The root Markov triple $(1,1,1)$ and its neighbors $(2,1,1)$,
$(1,2,1)$, and $(1,1,2)$ are called singular. All other Markov
triples are called non-singular and consist of three different
Markov numbers. For each Markov triple, there is a unique path to the
root. At each step along this path, the largest Markov number of the
triple is replaced according to the rule (<ref>).

§.§ Markov fractions and rational Markov triples

Markov fractions are one of the two types of worst approximable
rational numbers (see Theorems <ref>
and <ref>). Like the Markov numbers,
Markov fractions are defined in terms of triples:

(i) A triple of rational numbers $(x_{1},x_{2},x_{3})$ is a
rational Markov triple if $x_{k}=\frac{p_{k}}{q_{k}}$ for
some Markov triple $(q_{1},q_{2},q_{3})$ and some integers
$p_{1},p_{2},p_{3}$ satisfying the equations

\begin{align}
\label{eq:p1p2}
\label{eq:p2p3}
\end{align}

(ii) A Markov fraction is a rational number $x_{k}$ that is an
element of some rational Markov triple $(x_{1},x_{2},x_{3})$.

A rational Markov triple $(x_{1},x_{2},x_{3})$ is
strictly increasing,

\begin{equation*}
\end{equation*}

because Markov numbers are positive and equations (<ref>)
and (<ref>) are equivalent to

\begin{equation}
\label{eq:p1p2_alt}
\frac{p_{2}}{q_{2}}\,-\,\frac{p_{1}}{q_{1}}\;=\;\frac{q_{3}}{q_{1}\,q_{2}}
\qquad\text{and}\qquad
\frac{p_{3}}{q_{3}}\,-\,\frac{p_{2}}{q_{2}}\;=\;\frac{q_{1}}{q_{2}\,q_{3}}.
\end{equation}

Equations (<ref>), (<ref>),
and (<ref>) also yield the relation

\begin{equation}
\label{eq:p1p3}
\end{equation}

with $q_{2}'$ defined by (<ref>), which will
be used in the proof of Lemma <ref>.

Since the Markov numbers $q_{k}$ of a Markov triple
$(q_{1},q_{2},q_{3})$ are pairwise coprime, equations (<ref>)
and (<ref>) imply that each pair $p_{k}$, $q_{k}$ is also
coprime, so the fractions $\frac{p_{k}}{q_{k}}$ are reduced. This
means that the Markov triple $(q_{1}, q_{2}, q_{3})$ in
Definition <ref> is uniquely determined by the
rational Markov triple $(x_{1},x_{2},x_{3})$.
Further, if $(x_{1},x_{2},x_{3})$ is a rational Markov triple, then for
every $n\in\Z$,

\begin{equation*}
\end{equation*}

is also a rational Markov triple with the same Markov triple
$(q_{1},q_{2},q_{3})$ of denominators, and all rational Markov triples
with this denominator triple are obtained in this way. Since
$(-x_{3},-x_{2},-x_{1})$ is also a rational Markov triple, the group
of $\Z$-affine transformations (<ref>) acts on the set of
rational Markov triples, and hence on the set of Markov fractions.

There is another $\Z$-action on the set of rational Markov
triples: Using equations (<ref>) and (<ref>),
it is easy to check that

\begin{equation}
\label{eq:Zshift}
\quad\text{and}\quad
\end{equation}

are also rational Markov triples, with cyclically permuted
denominators. So the action of the cyclic permutation group
$A_{3}\simeq\Z_{3}$ on the set of Markov triples $(q_{1},q_{2},q_{3})$
lifts to a free action of $\Z$ on the set of rational Markov triples
$\big(\frac{p_{1}}{q_{1}},\frac{p_{2}}{q_{2}},\frac{p_{3}}{q_{3}}\big)$. Accordingly,
the tree of Markov numbers (see Figure <ref>) corresponds
to the forest of Markov fractions constructed in
Lemma <ref>, a subtree of which is shown in
Figure <ref>.

To see that there exists a rational Markov triple
for each Markov triple $(q_{1},q_{2},q_{3})$, note that the
coprimality of Markov triples implies the existence of $p_{1}$ and
$p_{2}$ satisfying (<ref>) (and $p_{2}$ and $p_{3}$
satisfying (<ref>), and $p_{1}$ and $p_{3}$
satisfying (<ref>)). By the following lemma, there exists a
suitable $p_{3}$ (or $p_{1}$ or $p_{2}$, respectively) to complete
the rational Markov triple.

Let $(q_{1},q_{2},q_{3})$ be a Markov

* If $p_{1},p_{2}\in\Z$ satisfy equation (<ref>), then
$ \frac{p_{2}^{2}\,+\,1}{q_{2}}\,\in\,\Z, $ and

\begin{equation}\label{eq:p3}
p_{3}\;\coloneqq\; q_{1}\cdot\,\frac{p_{2}^{2}\,+\,1}{q_{2}}\,-\,p_{1}\,p_{2}
\end{equation}

satisfies equation (<ref>).

* If $p_{2},p_{3}\in\Z$ satisfy equation (<ref>), then
$ \frac{p_{2}^{2}\,+\,1}{q_{2}}\,\in\,\Z, $ and

\begin{equation}\label{eq:p1}
p_{1}\;\coloneqq\; p_{2}\,p_{3}\,-\,q_{3}\cdot\frac{p_{2}^{2}\,+\,1}{q_{2}}
\end{equation}

satisfies equation (<ref>).

* If $p_{1},p_{3}\in\Z$ satisfy equation (<ref>), then

\begin{equation}\label{eq:p2}
\end{equation}

is an integer and satisfies equations (<ref>) and (<ref>).

In any case, therefore,
is a rational Markov triple, which is, moreover, uniquely determined
by $\big(\frac{p_{1}}{q_{1}},\frac{p_{2}}{q_{2}}\big)$,
$\big(\frac{p_{2}}{q_{2}},\frac{p_{3}}{q_{3}}\big)$, or
$\big(\frac{p_{1}}{q_{1}},\frac{p_{3}}{q_{3}}\big)$, respectively.

For a proof see Section <ref>. The integrality
$\frac{p_{2}^{2}\,+\,1}{q_{2}}$ implies the following necessary condition for
Markov numbers, which is well known <cit.> and
was proved in a similar way <cit.>:

If $q$ is a Markov number, then $-1$ is a quadratic residue modulo

It will be useful to extend the distinction between singular and
non-singular Markov triples (see Section <ref>) to
rational Markov triples, and to have a special term to single out the
rational Markov triples with denominators $(1,1,1)$. Also, since we
are especially interested in rational Markov triples in which the
second denominator is largest (see
Theorem <ref>), it makes sense to
introduce a special term for those, too:

A rational Markov triple
is called

* singular if the Markov triple $(q_{1},q_{2},q_{3})$
is singular, i.e., $\{q_{1},q_{2},q_{3}\}\subset\{1,2\}$,

* integral if $q_{1}=q_{2}=q_{3}=1$, and

* centered if

\begin{equation}
\label{eq:centered}
\end{equation}

Integral rational Markov triples are of the form $(n-1,n,n+1)$ with
$n\in\Z$. They are integral, centered, and the only centered rational
Markov triples for which (<ref>) is satisfied with
equality. All non-integral singular Markov triple have one of the
forms $(n,n+\frac{1}{2},n+1)$, $(n-2,n,n+\frac{1}{2})$ or
$(n-\frac{1}{2},n,n+2)$ for $n\in\Z$, of which precisely those of the
first form are centered.

The notion of a centered rational Markov triple is closely related
to Rockett & Szüsz's concept of a complete Markov
triple <cit.>: If
is a centered rational Markov triple with
$\frac{p_{k}}{q_{k}}\in[0,\frac{1}{2}]$ then
$(q_{2}, p_{2}; q_{1}, p_{1}; q_{3}, p_{3})$ is a complete Markov
triple. Compare also Figure <ref>
with <cit.>. However, the idea to interpret the
pairs $(q_{i}, p_{i})$ as rational numbers $\frac{p_{i}}{q_{i}}$
does not seem to be obvious in the context of [28].

We can now state the main result about the approximation constant of a
Markov fraction, and the best approximating rationals that achieve it,
as follows:

(i) Every Markov fraction is the middle element of a
unique centered rational Markov triple.

(ii) If
is a centered rational Markov triple, then the best approximants
of $\frac{p_{2}}{q_{2}}$ are precisely $\frac{p_{1}}{q_{1}}$
and $\frac{p_{3}}{q_{3}}$ , i.e.,

\begin{equation*}
\left(
\frac{p_{2}}{q_{2}}
\right)
\;=\;
\left(
\frac{p_{2}}{q_{2}}-\frac{p_{1}}{q_{1}}
\right)
\;=\;
\left(
\frac{p_{3}}{q_{3}}-\frac{p_{2}}{q_{2}}
\right)\,,
\end{equation*}

and for every rational number $\frac{a}{b}$ not contained in the

\begin{equation*}
\left(
\frac{p_{2}}{q_{2}}
\right)
b^{2}\cdot \left| \frac{p_{2}}{q_{2}}-\frac{a}{b} \right|.
\end{equation*}

In particular, this implies

\begin{equation}
\label{eq:Cp2q2}
\left(
\frac{p_{2}}{q_{2}}
\right)
\;=\; \frac{q_{1}\,q_{3}}{q_{2}}\;>\;\frac{1}{3}\,.
\end{equation}

For a proof of (i) see Section <ref>,
for a proof of (ii) see Section <ref>.

Since $\big(\frac{0}{1},\frac{2}{5},\frac{1}{2}\big)$ and
$\big(\frac{2}{5},\frac{12}{29},\frac{1}{2}\big)$ are centered
rational Markov triples,

* the best approximants of $\frac{2}{5}$ are
$\frac{0}{1}$ and $\frac{1}{2}$, and $C\big(\tfrac{2}{5}\big)=\tfrac{2}{5}$,

* the best approximants of $\frac{12}{29}$ are
$\frac{2}{5}$ and $\frac{1}{2}$, and $C\big(\tfrac{12}{29}\big)=\tfrac{10}{29}$.

Figure <ref> illustrates the
geometric interpretation of these examples.

The plane forest of centered rational Markov triples (see
Section <ref> and Figure <ref>) is
constructed in the following lemma. It is used in
Section <ref> for the proof
Theorem <ref> (i) and in
Section <ref> to establish the geometric
characterization of rational Markov triples.

* If
is a rational Markov triple, then so are its children

\begin{equation}
\label{eq:children}
\left(\,
\frac{p_{1}}{q_{1}},\;
\frac{p_{3}'}{q_{3}'},\;
\frac{p_{2}}{q_{2}}\,
\right)
\quad\text{and}\quad
\left(\,
\frac{p_{2}}{q_{2}},\;
\frac{p_{1}'}{q_{1}'},\;
\frac{p_{3}}{q_{3}}\,
\right)
\end{equation}

and its parents

\begin{equation}
\label{eq:parents}
\left(\,
\frac{p_{2}'}{q_{2}'},\;
\frac{p_{1}}{q_{1}},\;
\frac{p_{3}}{q_{3}}\,
\right),
\qquad
\left(\,
\frac{p_{1}}{q_{1}},\;
\frac{p_{3}}{q_{3}},\;
\frac{p_{2}'}{q_{2}'}\,+\,3\,
\right)
\end{equation}

where $q_{1}'$, $q_{2}'$, and $q_{3}'$ are defined
by (<ref>), and $p_{1}'$, $p_{2}'$, and $p_{3}'$
defined by

\begin{align}
\label{eq:p1prime}
\label{eq:p2prime}
\label{eq:p3prime}
\end{align}

In particular, $p_{k}'\in\Z$ for $k\in\{1,2,3\}$.

* A rational Markov triple is a child of each of its parents and
a parent of each of its children.

* Both children of a centered rational Markov triple are
centered, and exactly one parent of a non-singular centered
rational Markov triple is centered.

Thus, the parent-child relationship induces on the set of
non-integer centered rational Markov triples the structure of a
plane binary forest, with singular root triples $(n,
n+\frac{1}{2}, n+1)$ as shown in Figure <ref>
and the generating rule (<ref>).

§.§.§ Proof of Lemma <ref> (Completion of rational Markov triples)

To see part (i) of Lemma <ref>, first note that
$q_{2}$ divides $p_{2}^{2}+1$ because $q_{2}$ and $q_{1}$ are coprime

\begin{equation*}
\overset{\eqref{eq:p1p2}}{=}\;(p_{1}q_{2}+q_{3})^{2}+q_{1}^{2}
\overset{\,\eqref{eq:Markov}\,}{=}\;
\end{equation*}

A straightforward calculation yields equation (<ref>):

\begin{equation*}
\overset{\eqref{eq:p3}}{=}\;
\overset{\eqref{eq:p1p2}}{=}\; q_{1}
\end{equation*}

Statement (ii) can be proved in the same way.

For the integrality statement of (iii), consider the identity

\begin{equation*}
\overset{\eqref{eq:p1p3}}{=}\;
\overset{\eqref{eq:qiprime}}{=}\;
\end{equation*}

So $q_{2}'$ divides $(p_{1}q_{1}+p_{3}q_{3})$ because $q_{1}$ and
$q_{2}'$ are coprime as elements of the Markov triple

To verify equations (<ref>) and (<ref>), use the
equivalent forms (<ref>): With

\begin{equation*}
\frac{p_{2}}{q_{2}}
\;\overset{\eqref{eq:p2}}{=}\;
\frac{p_{1}\,q_{1}\,+\,p_{3}\,q_{3}}{q_{2}\,q_{2}'}
\;\overset{\eqref{eq:qiprime}}{=}\;
\frac{p_{1}\,q_{1}\,+\,p_{3}\,q_{3}}{q_{1}^{2}\,+\,q_{3}^{2}}
\end{equation*}

we obtain

\begin{equation*}
\frac{p_{2}}{q_{2}}\,-\,\frac{p_{1}}{q_{1}}\;
\overset{\eqref{eq:p2}}{=}\;
\frac{(p_{3}\,q_{1}\,-\,p_{1}\,q_{3})\,q_{3}}{(\,q_{1}^{2}\,+\,q_{3}^{2}\,)\,q_{1}}\;
\overset{\eqref{eq:p1p3}}{=}\;
\frac{q_{2}'\,q_{3}}{(\,q_{1}^{2}\,+\,q_{3}^{2}\,)\,q_{1}}
\overset{\eqref{eq:qiprime}}{=}\;
\frac{q_{3}}{q_{1}\,q_{2}}\,,
\end{equation*}

proving (<ref>). An analogous calculation
proves (<ref>).

Finally, since it has been established that
is a rational Markov triple, the uniqueness of the completion
follows from equations (<ref>).

§.§.§ Proof of Lemma <ref> (Forest of centered rational Markov triples)

Parts (i) and (ii) of Lemma <ref> can be proved
by applying the Completion-Lemma <ref>
appropriately. For example, to see that the first child is a rational
Markov triple, let
$\frac{\tilde{p}_{1}}{\tilde{q}_{1}}=\frac{p_{1}}{q_{1}}$ and
$\frac{\tilde{p}_{3}}{\tilde{q}_{3}}=\frac{p_{2}}{q_{2}}$. Then

\begin{equation*}
\tilde{p}_{3}\,\tilde{q}_{1}\,-\,\tilde{p}_{1}\,\tilde{q}_{3}
\;=\;q_{3}\;=:\;\tilde{q}_{2}'\,,
\end{equation*}

and Lemma <ref> (iii) says that

\begin{equation*}
\left(\,
\frac{\tilde{p}_{1}}{\tilde{q}_{1}},\;
\frac{\tilde{p}_{2}}{\tilde{q}_{2}},\;
\frac{\tilde{p}_{3}}{\tilde{q}_{3}}\,
\right)
\left(\,
\frac{p_{1}}{q_{1}},\;
\frac{p_{3}'}{q_{3}'},\;
\frac{p_{2}}{q_{2}}\,
\right)
\end{equation*}

is a rational Markov triple.

Statement (iii) follows from the same properties of Markov triples and
their neighbors that produce the tree of Markov
numbers [6]: If $q_{2}\geq\max\{q_{1},q_{3}\}$, then
$q_{1}'>q_{2}$ and $q_{3}'>q_{2}$, so both children of a centered
rational Markov triple are centered. If $q_{2}>\max\{q_{1},q_{3}\}>1$,
then $\min\{q_{1},q_{3}\}\leq q_{2}'<\max\{q_{1},q_{3}\}$, so exactly
one parent of a non-singular centered rational Markov triple is
centered. Finally, consider a singular but non-integer centered
rational Markov triple. It is of the form $(n, n+\frac{1}{2}, n+1)$,
and its parents are the integer triples $(n-1,n,n+1)$ and
$(n,n+1,n+2)$, which are both centered. Therefore, starting with any
non-integer centered rational Markov triple and traversing up the
parental line along centered triples, one will eventually reach a pair
of adjacent integer triples. The generating rule (<ref>) is
equivalent to the equations for children.

§.§.§ Proof of Theorem <ref> (i).
Every Markov fraction is the middle element of a unique centered

By definition, every Markov fraction $\frac{p}{q}$ is contained in some
rational Markov triple. If $q$ is not the largest denominator in that
triple, iteratively replace the triple with the parent or child in
which the fraction with largest denominator is removed until $q$ is the
largest denominator. If $\frac{p}{q}$ is not the middle element in
this triple then it is the middle element in one of the adjacent
triples (<ref>). This shows that there exists a centered
rational Markov triple $m$ with $\frac{p}{q}$ in the middle.

To see that $m$ is the unique centered rational Markov triple with
$\frac{p}{q}$ in the middle, consider first the tree of centered
rational Markov triples that contains $m$. All descendants of $m$ have
a larger denominator in the middle, and all ancestors have a smaller
denominator in the middle. So no other centered triple in the same
tree has $\frac{p}{q}$ in the middle. But each tree in the forest of
centered rational Markov triples contains only Markov fractions in a
unique integer interval $[n,n+1]$. So if a Markov fraction is
contained in two centered rational Markov triples of different trees,
then it is an integer, and it is the last element in one triple and
the first element in the other. This shows that no two centered
rational Markov triples have the same middle element.

§.§ Companions of Markov fractions

This section deals with the other type of worst approximable rational
numbers, the companions of Markov fractions (see
Theorems <ref> and <ref>).

The right companions and left companions of a Markov
fraction $\frac{p}{q}$ are the rational numbers
$\gamma^{+}_{k}(\frac{p}{q})$ and $\gamma^{-}_{k}(\frac{p}{q})$
for $k\geq 2$, defined by

\begin{equation}
\label{eq:companions}
\gamma^{\pm}_{k}\,\Big(\frac{p}{q}\Big)\;=\;
\frac{p}{q}\;\pm\;
\frac{u_{k-1}}{q\,u_{k}},
\end{equation}

where $(u_{k})$ is the integer sequence determined by the recursion

\begin{equation}
\label{eq:un}
u_{0}\;=\;0,\qquad u_{1}\;=\;1,\qquad
\end{equation}

We will often use the following lemma implicitly (e.g., in the
statement of Theorem <ref>).

The common denominator on the right hand side of
equation (<ref>) is indeed $q\,u_{k}$, so the fraction representation

\begin{equation*}
\gamma^{\pm}_{k}\,\Big(\frac{p}{q}\Big)\;=\;
\frac{p\,u_{k}\pm u_{k-1}}{q\,u_{k}}
\end{equation*}

is reduced.

See Corollary <ref>.

For $k=1$, equation (<ref>) says

\begin{equation}
\label{eq:gamma_1}
\gamma^{\pm}_{1}\,\Big(\frac{p}{q}\Big)
\;=\;\frac{p}{q}\,,
\end{equation}

But Definition <ref> explicitly requires $k\geq 2$, so
the Markov fraction $\frac{p}{q}$ is by definition not a companion
of itself. Consequently, the $k$-th left and right companions
are $\gamma^{+}_{k+1}(\frac{p}{q})$ and
$\gamma^{-}_{k+1}(\frac{p}{q})$, respectively. One could fix this
notational nuisance either by changing the definition and counting
the Markov fractions as their own companions, or by shifting the
index in the definition of $\gamma^{\pm}_{k}$. The first option is
bad because Markov fractions and their companions are different both
regarding Diophantine approximation (compare
Theorems <ref>
and <ref>) and from the geometric point
of view (compare Lemmas <ref>
and <ref>). The second option is bad because it
would create other nuisances. For example, the
symmetries (<ref>) and (<ref>) of the
following remark would become more complicated.

The the recursion (<ref>) can be used to define $u_{k}$ and
hence $\gamma^{\pm}_{k}$ not only for $k\geq 0$ but for all
$k\in\Z$. Then

\begin{equation}
\label{eq:u_symm}
\end{equation}

and this implies
$\gamma^{\pm}_{0}\big(\frac{p}{q}\big)=\sminfty$ and the following
symmetry relating $\gamma^{+}$ and $\gamma^{-}$:

\begin{equation}
\label{eq:gamma_symm}
\gamma^{+}_{-k}\,\Big(\frac{p}{q}\Big)
\;=\;
\gamma^{-}_{k}\,\Big(\frac{p}{q}+3\Big)
\end{equation}

The following properties of the sequence $(u_{k})_{k\in\Z}$ are easily verified
by induction.

For any Markov number $q$, the sequence $(u_{k})_{k\in\Z}$ defined
by (<ref>) is

* strictly increasing and

* satisfies the equation

\begin{equation}
\label{eq:u_conserved}
\end{equation}

for all $k\in\Z$.

(i) For $q=1$,

\begin{equation*}
u_{k}\;=\;F_{2k}\quad\text{and so}\quad
\gamma^{\pm}_{k}(p)\;=\;p\,+\,\frac{F_{2k-2}}{F_{2k}},
\end{equation*}

where $(F_{k})$ is the sequence of
Fibonacci numbers,

\begin{equation*}
\end{equation*}

(ii) For $q=2$ and odd $p$,

\begin{equation*}
\quad\text{and so}\quad
\gamma^{\pm}_{k}\,\Big(\frac{p}{2}\Big)
\;=\;
\frac{p}{2}
\,\pm\,
\frac{P_{2k-2}}{2\,P_{2k}}\,,
\end{equation*}

where $(P_{k})$ is the sequence of Pell numbers,

\begin{equation*}
P_{0}\;=\;0,\quad P_{1}\;=\;1,\quad P_{k+1}\;=\;2\,P_{k}\,+\,P_{k-1}.
\end{equation*}

The best approximants of a companion
$\gamma^{\pm}_{k}\big(\frac{p}{q}\big)$, $k\geq 2$, are precisely
$\frac{p}{q}$ and $\gamma^{\pm}_{k-1}\big(\frac{p}{q}\big)$, i.e.,

\begin{equation}
\label{eq:companion_equal_quality}
\left(
\gamma^{\pm}_{k}\Big(\frac{p}{q}\Big)
\right)
\;=\;
\left|\,
\gamma^{\pm}_{k}\Big(\frac{p}{q}\Big)\,-\,\frac{p}{q}\,
\right|
\;=\;
\left|\,
\gamma^{\pm}_{k}
\Big(\frac{p}{q}\Big)
\,-\,
\gamma^{\pm}_{k-1}
\Big(\frac{p}{q}\Big)\,
\right|
\end{equation}

\begin{equation*}
\left(
\gamma^{\pm}_{k}\Big(\frac{p}{q}\Big)
\right)
\;<\;
\left|\,
\gamma^{\pm}_{k}\Big(\frac{p}{q}\Big)\,-\,\frac{a}{b}\,
\right|
\end{equation*}

for every rational number $\frac{a}{b}$ except $\frac{p}{q}$,
$\gamma^{\pm}_{k-1}\big(\frac{p}{q}\big)$, and
$\gamma^{\pm}_{k}\big(\frac{p}{q}\big)$. Moreover,

\begin{equation}
\label{eq:Cgamma}
\gamma^{\pm}_{k}\,
\Big(\frac{p}{q}\Big)
\right)
\;=\;
\;\geq\;
\frac{1}{3}\,,
\end{equation}

where equality holds if and only if $k=2$, i.e., for first

The first companions $\gamma_{2}^{\pm}(\frac{p}{q})$ with
approximation constant $C=\frac{1}{3}$ are therefore the best of the
worst approximable rational numbers. They are also distinguished by
the property that the approximation constant is attained for one unique
best approximant, $\frac{p}{q}=\gamma_{1}^{\pm}\big(\frac{p}{q}\big)$,
instead of two.

(i) The first right and left companions of the Markov
fraction $\frac{1}{2}$ are $\gamma^{+}_{2}(\frac{1}{2})=\frac{7}{12}$
(compare Table <ref>) and
$\gamma^{-}_{2}(\frac{1}{2})=\frac{5}{12}$. They are best
approximated by their associated Markov fraction $\frac{1}{2}$ and
no other rational number. The approximation constant is

(ii) The second right and left companions of $\frac{1}{2}$
are $\gamma^{+}_{3}(\frac{1}{2})=\frac{41}{70}$ (compare
Table <ref>) and
$\gamma^{-}_{3}(\frac{1}{2})=\frac{29}{70}$. They are best
approximated by exactly two rational numbers: their associated
Markov fraction $\frac{1}{2}$ and its first right and left
companion, respectively. The approximation number is

§ THE GEOMETRIC POINT OF VIEW

§.§ The modular torus

Let $G=\langle [A],[B]\rangle\subset\PSLTZ$ be the subgroup of
orientation preserving hyperbolic isometries generated by

\begin{equation*}
\begin{bmatrix}
1 & -1 \\ -1 & 2
\end{bmatrix}
\quad\text{and}\quad
\begin{bmatrix}
1 & 1 \\ 1 & 2
\end{bmatrix}.
\end{equation*}

The modular torus is the quotient space $M\;=\;H^{2}/G$ with
canonical projection

\begin{equation*}
\pi\,:\,H^{2}\,\longrightarrow\,M,\quad \pi(z)\;=\;G\, z\,.
\end{equation*}

The ideal quadrilateral with vertices $-1$, $0$, $1$, and $\sminfty$
is a fundamental domain, from which one obtains the torus $M$ by
gluing the pairs of opposite sides via the isometries $[A]$
and $[B]$, respectively (see
Figure <ref>).

[anchor=south west, inner sep=0pt] (fig) at (0,0)

[below] at (0.77, 0)$-1$;
[below] at (2.64, 0)$0$;
[below] at (4.4, 0)$1$;
[above] at (3.5, 2)$[A]$;
[above] at (1.75, 2)$[B]$;

Fundamental domain of the modular torus with gluing

The Farey triangulation of the projective plane $H^{2}$ projects to
the most symmetric ideal triangulation of $M$, and the Ford circles
project to the boundary of the largest embedded cusp neighborhood
in $M$.

§.§ The geometric characterization of rational Markov triples and
Markov fractions

Rational Markov triples correspond to the fundamental domains of the
modular torus $M$ that are ideal quadrilaterals with one vertex
at $\sminfty$. This is illustrated in
Figure <ref> for the triple $(-1,0,1)$,
in Figure <ref>

[anchor=south west, inner sep=0pt] (fig) at (0,0)

at (0.5, 0.35)$\frac{0}{1}$;
at (8.2, 0.35)$\frac{2}{5}$;
at (10.15, 0.35)$\frac{1}{2}$;
at (9, 8.7)$d$;
at (8, 2.9)$d$;

[anchor=south west, inner sep=0pt] (fig) at (0,0)

[below] at (1.78, 0.02)$\frac{2}{5}$;
[below] at (2.75, 0.02)$\frac{12}{29}$;
[below] at (8.82, 0.02)$\frac{1}{2}$;
[above] at (1.25, 5.95)$d'$;
[xshift=1.5, yshift=3] at (3, 1)$d'$;

Fundamental domains corresponding to the centered rational
Markov triples $\big(\frac{0}{1}, \frac{2}{5}, \frac{1}{2}\big)$
and $\big(\frac{2}{5},\frac{12}{29},\frac{1}{2}\big)$ of
Example <ref>. The approximation constants of
$\frac{2}{5}$ and $\frac{12}{29}$ are
$e^{\frac{1}{2}d}=C\big(\frac{2}{5}\big)=\frac{2}{5}$ and
respectively (see Lemma <ref> and
Corollary <ref>). The signed distances $d$ and
$d'$ are negative because the vertical geodesics at $\frac{2}{5}$
and $\frac{12}{29}$ intersect the Ford circles at the best

for the triples $(\frac{0}{1},\frac{2}{5},\frac{1}{2})$ and
$(\frac{2}{5},\frac{12}{29},\frac{1}{2})$, and schematically in
Figure <ref>.

[anchor=south west, inner sep=0pt] (fig) at (0,0)

[below] at (1.99, 0) $\dfrac{p_{1}}{q_{1}}$;
[left]  at (1.99, 4.05) $q_{1}$;
[below] at (5.6, 0) $\dfrac{p_{2}}{q_{2}}$;
[left]  at (5.6, 3) $q_{2}$;
[below] at (7.37, 0) $\dfrac{p_{3}}{q_{3}}$;
[left]  at (7.37, 3.15) $q_{3}$;
[below] at (10.4, 0) $\dfrac{p_{1}}{q_{1}}+3$;
[left]  at (10.4, 4.05) $q_{1}$;

[above] at (4.7, 1.6) $q_{3}$;
[above] at (6.4, 0.9) $q_{1}$;
[above] at (8.3, 1.45) $q_{2}$;

[above] at (3.8, 4.6) $\dfrac{q_{3}}{q_{1}q_{2}}$;
[above] at (6.5, 4.6) $\dfrac{q_{1}}{q_{2}q_{3}}$;
[above] at (8.9, 4.6) $\dfrac{q_{2}}{q_{3}q_{1}}$;

$(\frac{p_{1}}{q_{1}},\frac{p_{2}}{q_{2}},\frac{p_{3}}{q_{3}})$ is
a rational Markov triple, then the ideal quadrilateral with
vertices $\frac{p_{1}}{q_{1}}$, $\frac{p_{2}}{q_{2}}$,
$\frac{p_{3}}{q_{3}}$, and $\sminfty$ is a fundamental domain of
the modular torus. Dashed: If we cut along the diagonal
$q_{2}$ and glue the left ideal triangle
$\frac{p_{1}}{q_{1}},\frac{p_{2}}{q_{2}},\sminfty$ back to the
right ideal triangle
$\frac{p_{2}}{q_{2}},\frac{p_{3}}{q_{3}},\sminfty$, but along edge
$q_{3}$, we obtain another fundamental domain, which corresponds
to the rational Markov triple
(compare equation (<ref>)).

For the sake of clearer pictures,
Figures <ref>,
<ref>, and <ref>
do not show the correct geometry of the modular torus $M$ but some
other hyperbolic tori with one cusp. As a result, the objects in
these figures are more evenly sized and intersect each other less
than would be the case in the modular torus.

To put the correspondence between rational Markov triples and
fundamental domains more precisely, we formulate the following lemma:

The following statements for real numbers
$x_{1}<x_{2}<x_{3}$ are equivalent:

* $(x_{1},x_{2},x_{3})$ is a rational Markov triple.

* The ideal quadrilateral with vertices
$x_{1},x_{2},x_{3},\sminfty$ is a fundamental domain of the
modular torus $M$.

* The ideal triangles with vertices $x_{1},x_{2},\sminfty$ and
$x_{2},x_{3},\sminfty$, respectively, project to the two
triangles of some ideal triangulation of the modular torus $M$.

(For a proof, see Section <ref>.)

Since every simple geodesic with both ends in the cusp is an edge of
some ideal triangulation of the modular torus $M$, we get the
following characterization of Markov fractions:

For a real number $x\in\R$, the following statements are equivalent:

* $x$ is a Markov fraction.

* The vertical geodesic $g_{x}$ joining $x$ and $\sminfty$ in $H^{2}$
projects to a simple geodesic $\pi(g_{x})$ in the modular torus
$M$ with both ends in the cusp.

§.§ Proof of Lemma <ref> (Geometric characterization of rational Markov triples)

The equivalence of statements (ii) and (iii) of the lemma is
obvious. It remains to show that (i) and (iii) are equivalent.

(iii)$\Rightarrow$(i). Assuming (iii) implies that the
vertices $x_{k}\in\R$ are rational,

\begin{equation*}
\end{equation*}

because they project to the ideal point of the modular torus.

The denominators $q_{k}$ are equal to the weights of the edges of the
triangulation <cit.> (see
Figure <ref>). Indeed, the euclidean
diameter of the Ford circle at $\frac{p_{k}}{q_{k}}$ is $q_{k}^{-2}$,
so the distance to the horocycle $\{\im z=1\}$ is
$d_{k}=2\log q_{k}$, and the weight is

By Penner's geometric interpretation of Markov's
equation [25, 26], the denominators form a Markov
triple $(q_{1},q_{2},q_{3})$. Finally, the relation between the edge
weights and the horocyclic arcs of a decorated ideal triangle (see
Figure <ref>)

[y=0.80pt, x=0.80pt, yscale=-, xscale=, inner sep=0pt, outer sep=0pt]
[shift=(268.39,341.718),draw=black,fill=black!10,line cap=rect,line
join=bevel,line width=0.500pt,miter limit=4.00]
(78.8672,0.9785) –
(78.8184,1.0605)arc(0.039:59.987:136.044590 and 136.045) –
(11.0527,118.7715)arc(240.094:299.884:136.038910 and 136.039) –
(146.9629,118.7461)arc(120.001:179.923:135.985530 and 135.986) – cycle;

[draw=black,line cap=butt,line join=bevel,line width=0.500pt,miter
limit=4.00] (347.2577,421.2515) ellipse (2.2170cm and 2.2170cm);

[color=black,draw=black,line cap=butt,line join=bevel,line
width=0.500pt,miter limit=4.00] (294.3342,451.7493) ellipse (0.4929cm and

[color=black,draw=black,line cap=butt,line join=bevel,line
width=0.500pt,miter limit=4.00] (347.3597,371.5814) ellipse (0.8152cm and

[color=black,draw=black,line cap=butt,line join=bevel,line
width=0.500pt,miter limit=4.00] (401.6195,452.6762) ellipse (0.4461cm and

[color=black,draw=black,line cap=butt,line join=bevel,line
width=0.800pt,miter limit=4.00] (279.2122,460.4881) ellipse (0.0135cm and

[color=black,draw=black,line cap=butt,line join=bevel,line
width=0.800pt,miter limit=4.00] (347.2577,342.6965) ellipse (0.0135cm and

[color=black,draw=black,line cap=butt,line join=bevel,line
width=0.800pt,miter limit=4.00] (415.3525,460.4635) ellipse (0.0135cm and

[color=black,draw=black,line cap=butt,line join=bevel,line
width=0.800pt,miter limit=4.00] (311.2088,447.2282) ellipse (0.0135cm and

[color=black,draw=black,line cap=butt,line join=bevel,line
width=0.800pt,miter limit=4.00] (306.7663,439.4821) ellipse (0.0135cm and

[color=black,draw=black,line cap=butt,line join=bevel,line
width=0.800pt,miter limit=4.00] (335.4945,397.9246) ellipse (0.0135cm and

[color=black,draw=black,line cap=butt,line join=bevel,line
width=0.800pt,miter limit=4.00] (359.1031,397.9793) ellipse (0.0135cm and

[color=black,draw=black,line cap=butt,line join=bevel,line
width=0.800pt,miter limit=4.00] (390.1848,441.7627) ellipse (0.0135cm and

[color=black,draw=black,line cap=butt,line join=bevel,line
width=0.800pt,miter limit=4.00] (386.4734,448.1397) ellipse (0.0135cm and

[draw=black,line cap=butt,line join=bevel,line width=1.000pt,miter
limit=4.00] (335.5023,397.9052)arc(23.949:45.289:136.044590 and 136.045)
node[midway, above left=3pt]$q_{1}$;

[color=black,draw=black,line cap=butt,line join=bevel,line
width=1.000pt,miter limit=4.00]
(311.3172,447.2017)arc(254.681:286.626:136.038910 and 136.039)
node[midway, below=4pt]$q_{3}$;

[color=black,draw=black,line cap=butt,line join=bevel,line
width=1.000pt,miter limit=4.00]
(390.1180,441.6953)arc(133.280:155.987:135.985530 and 135.986)
node[midway, above right=3pt]$q_{2}$;

[color=black,draw=black,line cap=butt,line join=bevel,line
width=1.000pt,miter limit=4.00]
(358.9509,398.0386)arc(66.341:114.644:28.884882 and 28.885)
node[midway, below=3pt]$\frac{q_{3}}{q_{1}q_{2}}$;

[anchor=south west, inner sep=0pt] (fig) at (0,0)

[below] (x1) at (1.32, 0.02) $\dfrac{p_{1}}{q_{1}}$;
[left] (q1) at (1.32, 3) $\smash{q_{1}}$;
[below] (x3p) at (3.05, 0.02) $\dfrac{p_{3}'}{q_{3}'}$;
[left] (q3p) at (3.05, 3) $\smash{q_{3}'}$;
[below] (x2) at (4.3, 0.02) $\dfrac{p_{2}}{q_{2}}$;
[right] (q2) at (4.3, 3) $\smash{q_{2}}$;

[xshift=-2, yshift=3] (q1b) at (3.35, 0.58) $\smash{q_{1}}$;
[xshift=3, yshift=1] (q2b) at (2.7, 0.75) $\smash{q_{2}}$;
[xshift=0, yshift=4] (q3b) at (2.85, 1.51) $\smash{q_{3}}$;

Ideal triangle decorated with horocycles. Relation between
edge weights $q_{k}$ and horocyclic arcs

Fundamental domain consisting of ideal triangles
$\frac{p_{1}}{q_{1}},\frac{p_{3}'}{q_{3}'},\sminfty$ and

implies equations (<ref>). Therefore,
is a rational Markov triple.

(i)$\Rightarrow$(iii). The implication is true for any
integer rational Markov triple, i.e., if
$(x_{1},x_{2},x_{3})=(n-1,n,n+1)$ for some $n\in\Z$. We will show: If
the implication is true for a rational Markov triple
then it also holds for the adjacent triples (<ref>) and the
children (<ref>). Since every centered rational Markov
triple is a descendant of an integer triple, and every non-centered
rational Markov triple is adjacent to a centered one, this proves the
implication (i)$\Rightarrow$(iii) for all rational Markov triples.

So let
be a rational Markov triple for which the implication
(i)$\Rightarrow$(iii) holds. To see that the same is true for the
adjacent triples (<ref>), note that $z\mapsto z+6$ is a deck
transformation of the modular torus, and $z\mapsto z+3$ is a lift of
the elliptic involution.

It remains to show that the implication (i)$\Rightarrow$(iii) also
holds for the children (<ref>). We will present the
argument for the left child
see Figure <ref>. The argument for the right
child is analogous.

First we choose a matrix $S\in\SLTR$ that satisfies

\begin{equation*}
\begin{pmatrix}
p_{1} \\ q_{1}
\end{pmatrix}
\;=\;
\pm
\begin{pmatrix}
p_{2} \\ q_{2}
\end{pmatrix}
\qquad\text{and}\qquad
\begin{pmatrix}
p_{2} \\ q_{2}
\end{pmatrix}
\;=\;
\pm
\begin{pmatrix}
p_{1} \\ q_{1}
\end{pmatrix},
\end{equation*}

which determines $S$ up to sign. Let us choose

\begin{equation*}
\begin{split}
\begin{pmatrix}
p_{2} & p_{1} \\ q_{2} & q_{1}
\end{pmatrix}
\cdot
\begin{pmatrix}
-p_{1} & p_{2} \\ -q_{1} & q_{2}
\end{pmatrix}^{-1}
\\
\frac{1}{p_{2}q_{1}-p_{1}q_{2}}\;
\begin{pmatrix}
p_{1}q_{1}+p_{2}q_{2} & -p_{1}^{2}-p_{2}^{2} \\
q_{1}^{2}+q_{2}^{2} & -p_{1}q_{1}-p_{2}q_{2}
\end{pmatrix}.
\end{split}
\end{equation*}

Then the hyperbolic isometry $[S]\in\PSLTR$ maps the Ford circle
$h(p_{1},q_{1})$ at $\frac{p_{1}}{q_{1}}$ to the Ford circle
$h(p_{2},q_{2})$ at $\frac{p_{2}}{q_{2}}$ and vice versa (see
Section <ref>). This implies that
$[S]\in\PSLTR$ is a lift of the elliptic involution of the modular
torus $M$, and therefore $S\in\SLTZ$.

Since the ideal triangle with vertices $\frac{p_{1}}{q_{1}}$,
$\frac{p_{2}}{q_{2}}$, $\sminfty$ is a lift of one triangle of an
ideal triangulation of the modular torus, the ideal triangle with

\begin{equation*}
\text{and}\qquad
\end{equation*}

is a lift of the other triangle of the triangulation. (Here
$\;\cdot\;$ denotes the isometric action of $\PSLTR$ on $H^{2}$. )
Applying an edge flip, we obtain a triangulation of the modular torus
whose triangles lift to the ideal triangles with vertices

\begin{equation*}
\frac{p_{1}}{q_{1}}\,,\quad
\sminfty
\qquad
\text{and}
\qquad
\frac{p_{2}}{q_{2}}\,,  \quad
\sminfty,
\end{equation*}

respectively.  Finally, using
equation (<ref>) and $(q_{1}^{2}+q_{2}^{2})/q_{3}=q_{3}'$, we

\begin{equation*}
\begin{pmatrix}
1 \\ 0
\end{pmatrix}
\;=\;
\begin{pmatrix}
\end{pmatrix},
\quad\text{so}\quad
\end{equation*}

and the implication (i)$\Rightarrow$(iii) holds for the child
too. This completes the proof of Lemma <ref>.

§.§ The geometric characterization of companions

For a Markov fraction $x$, Figure <ref> schematically
illustrates the geometric interpretation of the the first right
companion $\gamma^{+}_{2}(x)$.

[anchor=south west, inner sep=0pt] (image) at (0,0)

(eta) at (0.8, 1.8) $\eta$;
(gx) at (2.4, 3) $\pi(g_{x})$;
[right] (gamma) at (4, 1) $\pi(g_{\gamma_{2}^{+}(x)})$;
[-stealth] (4.1,1.1) – (3.3,1.7);
[left] (C1) at (1.2,-0.1) $\pi(R_{x})$;
[-stealth] (1.1,-0.0) – (1.5,0.4);
[left] (C2) at (0.5, 3.5) $\pi(L_{x})$;
[-stealth] (0.4,3.4) – (0.8, 3);

[-stealth] (3.1,3.075) – (3,2.995);

Topological sketch of the modular torus $M$ showing the
geodesic $\pi(g_{x})$ corresponding to a Markov fraction $x$, and
the simple closed geodesic $\eta$ that does not intersect it. The
geodesic $\pi(g_{\gamma_{2}^{+}(x)})$ corresponds to the first
right companion $\gamma_{2}^{+}(x)$. It is contained in the
cylinder $\pi(R_{x})$ bounded by $\pi(g_{x})$ and $\eta$ and has
one self-intersection.

Generally, the vertical geodesic $g_{x}$ projects to a simple
geodesic $\pi(g_{x})$ in the Modular torus $M$ with both ends in the
cusp (by Corollary <ref>).  Then there is a
unique simple closed geodesic $\eta$ in $M$ that does not
intersect $\pi(g_{x})$. The geodesics $\pi(g_{x})$ and $\eta$ separate
$M$ into two cylinders, which are labeled $\pi(R_{x})$ and
$\pi(L_{x})$ in Figure <ref> for reasons that will become
clear in the following Section <ref>. Be that
as it may, $\pi(R_{x})$ is the cylinder to the right of $\pi(g_{x})$
if $g_{x}$ is oriented in the direction from $x$ to $\sminfty$, and
$\pi(L_{x})$ is the cylinder to the left.

Now let $y$ be a right or left companion of $x$, i.e.,
$y=\gamma^{+}_{k}(x)$ or $y=\gamma^{-}_{k}(x)$ for some $k\geq 2$.
Then the vertical geodesic $g_{y}$ projects to a non-simple geodesic
$\pi(g_{y})$ in the cylinder $\pi(R_{x})$ or $\pi(L_{x})$,
respectively, with $k-1$ self-intersections.

Formulating the converse statement is slightly more involved. To
understand why, note first of all that the isometry $z\mapsto z+6$ of
$H^{2}$ is a deck transformation of $M$, so the companions of $x$
correspond to the same vertical geodesics in $M$ as the companions of
$x+6n$ for any $n\in\Z$. Secondly, the isometry $z\mapsto z+3$
of $H^{2}$ projects to the hyperelliptic involution of $M$, which
reverses the orientations of $\pi(g_{x})$ and $\eta$. As a
consequence, the cylinder $\pi(R_{x})$ contains not only the projected
vertical geodesics for the right companions $\gamma^{+}_{k}(x)$, but
also those for $\gamma^{-}_{k}(x-3)$, the left companions of $x-3$
(compare Remark <ref>).

Altogether, the following lemma characterizes the right and left
companions of a Markov fraction geometrically:

Let $x$ be a Markov fraction, so that $\pi(g_{x})$ is a simple
geodesic in the modular torus $M$ with both ends in the cusp. Let
$\eta$ be the unique simple closed geodesic in $M$ that does not
intersect $\pi(g_{x})$. Then for a number $y\in\R$ and the vertical
geodesic $g_{y}$, the statements (i$^+$) and (ii$^+$) are
equivalent, and the statements (i$^-$) and (ii$^-$) are equivalent:

* $y$ is a right companion of $x$, i.e., $y=\gamma^{+}_{k}(x)$ for some $k\geq 2$.

* $y\in[x,\, x+\frac{3}{2}]$ and $\pi(g_{y})$ is a geodesic with
both ends in the cusp that intersects neither $\pi(g_{x})$ nor

* $y$ is a left companion of $x$, i.e., $y=\gamma^{-}_{k}(x)$ for some $k\geq 2$.

* $y\in[x-\frac{3}{2}, \,x]$ and $\pi(g_{y})$ is a geodesic with
both ends in the cusp that intersects neither $\pi(g_{x})$ nor

The geodesics $\pi(g_{\gamma^{+}_{k}(x)})$ and
$\pi(g_{\gamma^{-}_{k}(x)})$ have $k-1$ self-intersections.

§.§ Proof of Lemma <ref> (Geometric
characterization of companions)

The geodesics in the $G$-orbit $G\cdot g_{x}$ of the vertical
geodesic $g_{x}$ together with $\pi^{-1}(\eta)$, the $G$-orbit of
lifts of the closed geodesic $\eta$, separate the hyperbolic
plane $H^{2}$ into simply connected domains. Each one of these domains
covers one of the two cylinders into which $\pi(g_{x})$ and $\eta$
separate $M$. Of these domains, let $R_{x}$ and $L_{x}$ be the ones to
the right and to the left of $g_{x}$, respectively, when $g_{x}$ is
oriented in the direction from $x$ to $\sminfty$ (see
Figure <ref>).

[anchor=south west, inner sep=0pt] at (0,0)

[left] at (3.78, 3) $\smash{g_{x}}$;

[below] at (3.78, 0.03) $x\vphantom{3\frac{3}{2}}$;
[below] at (7.955, 0.03) $x+\frac{3}{2}\vphantom{3}$;
[below] at (12.13, 0.03) $x+3\vphantom{\frac{3}{2}}$;

[below, xshift=3, yshift=2] at (1.82, 0.03) 0.7$\gamma_{2}^{-}(x)$;
[below, xshift=3, yshift=0.5] at (1.22, 0.03) 0.47$\gamma_{3}^{-}(x)$;
[below, xshift=0, yshift=-0.5] at (0.98, 0.03) 0.3$\gamma_{4}^{-}(x)$;
[below, xshift=0, yshift=-1] at (0.65, 0.03) 0.47$\cdots$;

[below, xshift=1, yshift=2] at (5.71, 0.03) 0.7$\gamma_{2}^{+}(x)$;
[below, xshift=0, yshift=0.5] at (6.3, 0.03) 0.47$\gamma_{3}^{+}(x)$;
[below, xshift=2, yshift=-0.5] at (6.6, 0.03) 0.3$\gamma_{4}^{+}(x)$;
[below, xshift=0, yshift=-1] at (7, 0.03) 0.47$\cdots$;

[below, xshift=3, yshift=2] at (10.2, 0.03) 0.7$\gamma_{2}^{-}(x+3)$;
[below, xshift=0, yshift=-1] at (9.5, 0.03) 0.47$\cdots$;

[right] at (5.71, 2.5) $\smash{g_{\gamma_{2}^{+}(x)}}$;
at (8, 4.5) $[T]^{-1}\cdot g_{\gamma_{2}^{+}(x)}$;
at (4.9, 1.55) $[T]\cdot g_{\gamma_{2}^{+}(x)}$;

[above] at (8, 1.1) $\varrho_{x}$;
[-stealth] (8.01, 1.125) – (7.95, 1.125);
at (0.7, 0.8) $\lambda_{x}$;

at (8, 2.5) $R_{x}$;
at (8, 0.6) $L_{x}'$;
[right] at (0, 2.5) $L_{x}$;

The right companions $\gamma^{+}_{2}(x),\,\ldots\,$ are
ideal vertices of region $R_{x}$ between $x$ and $x+\frac{3}{2}$.
The left companions $\gamma^{-}_{2}(x),\,\ldots\,$ are ideal
vertices of the region $L_{x}$ between $x-\frac{3}{2}$ and
$x$. The dotted lines are the geodesics
$[T]^{k}\cdot g_{\gamma^{+}_{2}(x)}$.

In short, $R_{x}$ and $L_{x}$ are the domains to the left and right
of $g_{x}$ such that the restrictions $\pi|_{R_{x}}$
and $\pi|_{L_{x}}$ are universal covers of the two cylinders in $M$.

The geometric characterization of companions follows in a
straightforward manner from an explicit analytic description of the
domains $R_{x}$ and $L_{x}$. We will perform the calculations only
for $R_{x}$ and the right companions $\gamma^{+}_{k}(x)$. The domain
$L_{x}$ and the left companions $\gamma^{-}_{k}(x)$ can be treated
analogously. Alternatively, one may use the fact that $z\mapsto z+3$
projects to the elliptic involution to see that $L_{x}=R_{x}-3$
(compare Remark <ref>).

Let the Markov fraction be $x=\frac{p}{q}$. Then the group of deck
transformations of the universal cover $R_{x}\rightarrow\pi(R_{x})$ is
generated by the hyperbolic isometry

\begin{equation*}
[T]\;\in\; G\;\subset\;\PSLTZ
\end{equation*}

that maps the Ford circle $h(1,0)$ at $\sminfty$ to the Ford circle
$h(p,q)$ at $x$, and the Ford circle $h(p+3q, q)$ at $x+3$ to the Ford
circle $h(1,0)$ at $\sminfty$. To see this, contemplate
Figure <ref> with

This determines the matrix $T\in\SLTZ$ up to sign. Choosing

\begin{equation*}
\begin{pmatrix}
1 & p+3\,q \\
0 & q
\end{pmatrix}
\;=\;
\begin{pmatrix}
-p & 1 \\
-q & 0
\end{pmatrix},
\end{equation*}

which works because

\begin{equation*}
\det\,
\begin{pmatrix}
1 & p+3\,q \\
0 & q
\end{pmatrix}
\;=\;q\;=\;
\det\,
\begin{pmatrix}
-p & 1 \\
-q & 0
\end{pmatrix}
\end{equation*}

we obtain

\begin{equation}
\label{eq:T}
\begin{pmatrix}
-p & 3\,p+\frac{p^{2}+1}{q} \\
-q & 3\,q+p
\end{pmatrix}
\;\in\;\SLTZ\,.
\end{equation}

In particular, this implies $\frac{p^{2}+1}{q}\in\Z$, which provides
a geometric interpretation of the number theoretic
Corollary <ref>. More directly, such a
geometric interpretation can be obtained by considering the
orientation preserving hyperbolic isometry $[S_{p,q}]\in\PSLTR$ that
interchanges the horocycles $h(p,q)$ and $h(1,0)$.
This determines

\begin{equation*}
\begin{pmatrix}
-p & \frac{p^{2}+1}{q} \\
-q & p
\end{pmatrix}
\;\in\;\SLTR
\end{equation*}

up to sign. Since $[S_{p,q}]$ projects to the hyperelliptic
involution on $M$, we have $S_{p,q}\in\SLTZ$. (The map $[T]$ is the
composition of $[S_{p,q}]$ followed by $z\mapsto z+3$.)

Let $\varrho_{x}$ be the axis of the hyperbolic translation $[T]$.  It
projects to the closed geodesic $\eta$ in $M$, i.e.,
$\pi(\varrho_{x})=\eta$. The geodesic $\varrho_{x}$ connects the
repelling and attracting fixed points of $[T]$, let us call them
$\xi_{-}$ and $\xi_{+}$. By a straightforward calculation we

\begin{equation}
\label{eq:T_fp}
\xi_{\pm}\;=\;
\,\mp\,\sqrt{\frac{9}{4}\,-\,\frac{1}{q^{2}}}\;.
\end{equation}

Since $\trace(T)=3q$, the eigenvalues of $T$ are
$\frac{1}{2}\,(3\,q\pm\sqrt{9\,q^{2}-4})$. This also shows the
well-known formulas for the lengths of closed geodesics in the
modular torus,

\begin{equation*}
\left(
\frac{\length(\eta)}{2}
\right)
\;=\;3\,q\,,
\qquad
\left(
\frac{\length(\eta)}{2}
\right)
\;=\;
\end{equation*}

but we will not need them.

The region $R_{x}$ is bounded by the geodesic $\varrho_{x}$ together
with the infinite ideal polygon with sides $[T]^{k}\cdot g_{x}$ for
$k\in\Z$, which connect the ideal vertices $[T]^{k}\cdot\infty$. The
endpoints $\xi_{\pm}$ of $\varrho_{x}$ are the irrational limit points
of the rational bi-infinite sequence of ideal vertices:

\begin{equation}
\label{eq:Tk_lim}
\;\longrightarrow\;
\xi_{\pm}
\quad\text{as}\quad
\end{equation}

Now the equivalence of the statements (i$^+$) and (ii$^+$) follows
from the fact that

\begin{equation}
\label{eq:companion_geometric}
[T]^{k}\cdot \sminfty\;=\;\gamma_{k}^{+}(x)
\end{equation}

(with $\gamma_{k}^{+}(x)$ defined by equation (<ref>)).
To see this, let $(r_{k})_{k\in \Z}$ and $(s_{k})_{k\in \Z}$ be
defined by

\begin{equation}
\label{eq:rs}
\begin{pmatrix}
r_{k} \\ s_{k}
\end{pmatrix}
\;=\;
\begin{pmatrix}
-1 \\ 0
\end{pmatrix}\,,
\end{equation}

so that

\begin{equation*}
[T]^{k}\cdot \sminfty\;=\;\frac{r_{k}}{s_{k}}\;.
\end{equation*}

Then it suffices to show that

\begin{equation}
\label{eq:rs_of_u}
\begin{pmatrix}
\end{pmatrix}
\;=\;
\begin{pmatrix}
\end{pmatrix}
\end{equation}

for the sequence $(u_{k})_{k\in \Z}$ defined by (<ref>).

First note that $q$ divides $s_{k}$ for all $k\in\Z$. (This is easy to
see by induction in both directions starting from $s_{0}=0$.) Hence we
can define integer sequences $(t_{k})_{k\in\Z}$ and
$(u_{k})_{k\in\Z}$ by

\begin{equation*}
u_{k}\;=\;\frac{s_{k}}{q} \qquad\text{and}\qquad
\end{equation*}

We have to show that $(u_{k})_{k\in\Z}$ solves the
recursion (<ref>). The initial conditions are easily
verified. Using the relation

\begin{equation*}
\begin{pmatrix}
\end{pmatrix}
\;=\;
\underbrace{
\begin{pmatrix}
1 & p\\
0 & q
\end{pmatrix}
\begin{pmatrix}
\end{pmatrix},
\end{equation*}

\begin{smallmatrix}
\end{smallmatrix}
\big)
\big(
\begin{smallmatrix}
\end{smallmatrix}
\big)$
we obtain the first order recursion formula

\begin{equation*}
\begin{pmatrix}
\end{pmatrix}
\;=\;
\underbrace{
\begin{pmatrix}
0  & 1 \\
-1 & 3q
\end{pmatrix}
\begin{pmatrix}
\end{pmatrix}
\end{equation*}

Eliminate $(t_{k})$ using $t_{k}=u_{k-1}$ to obtain the
second-order recursion (<ref>) for $(u_{k})_{k\in\Z}$ and
equation (<ref>) for $r_{k}$ and $s_{k}$.  This completes
the proof of equation (<ref>), showing that
(i$^{+}$) and (ii$^{+}$) are equivalent.

Finally, to see that the geodesic $\pi(g_{\gamma^{+}_{k}(x)})$ in the
cylinder $\pi(R_{x})\subset M$ has $k-1$ self-intersections, note that
the vertical geodesic $g_{\gamma^{+}_{k}(x)}$ in the universal cover
$R_{x}$ intersects the $2\,(k-1)$ geodesics

\begin{equation}
[T]^{\ell}\cdot g_{\gamma^{+}_{k}(x)}
\quad\text{for}\quad
\ell\;\in\;\{1,\ldots,k-1\}\;\cup\;\{-(k-1),\ldots,-1\}\,.
\end{equation}

but the intersection points

\begin{equation}
\zeta_{\ell}\;=\;g_{\gamma^{+}_{k}(x)}
\;\cap\;
\left(
[T]^{\ell}\cdot g_{\gamma^{+}_{k}(x)}
\right)
\end{equation}

project in pairs to the same point in $M$,

\begin{equation}
\pi(\zeta_{\ell})\;=\;\pi(\zeta_{-\ell}),
\end{equation}

\begin{equation}
\zeta_{\ell}\;=\;[T]^{\ell}\cdot\zeta_{-\ell}\,.
\end{equation}

This completes the proof of Lemma <ref>.

Because $T\in\SLTZ$, equation (<ref>) implies that $r_{k}$ and
$s_{k}$ are coprime for all $k\in\Z$. So the above proof (together
with the analogous argument for the left companions) yields the
following corollary:

For all $k\in\Z$, the integers $p\,u_{k}+ u_{k-1}$ and $q\,u_{k}$
are coprime, and so are the integers $p\,u_{k}-u_{k-1}$ and

§.§ Proof of Theorem <ref> (ii)
(Best approximants of Markov fractions)

Part (i) of Theorem <ref> was proved
without recourse to geometry in
Section <ref>. The purpose of this
section is to prove part (ii). The argument begins algebraically, but
geometry will come into play shortly.

\begin{equation*}
\end{equation*}

be a centered rational Markov triple. Then we have

\begin{equation*}
\left(\,
\frac{p_{2}}{q_{2}}\;-\;\frac{p_{1}}{q_{1}}\,
\right)
\;\;\;\overset{\makebox[0pt]{\scriptsize{\eqref{eq:p1p2}}}}{=}\;\;\;
\frac{q_{1}\,q_{3}}{q_{2}}
\;\;\;\overset{\makebox[0pt]{\scriptsize{\eqref{eq:p2p3}}}}{=}\;\;\;
\left(\,
\frac{p_{3}}{q_{3}}\;-\;\frac{p_{2}}{q_{2}}\,.
\right)\,,
\end{equation*}

\begin{equation*}
\frac{1}{3}\;<\;\frac{q_{1}\,q_{3}}{q_{2}}\;\leq\;1
\end{equation*}

from equation (<ref>), together with (<ref>)
for the upper bound. It remains to show that for every rational number

\begin{equation}
\label{eq:check_ab}
\frac{q_{1}\,q_{3}}{q_{2}}
\;<\;
\end{equation}

To see this, we will rely on the geometric characterization of Markov
fractions (see Sections <ref>
and <ref>) to narrow down the set of potential
best approximants so we can check (<ref>) for the
remaining candidates.

So let $x=\frac{p}{q}$ be a Markov fraction, and let
$x^{*}=\frac{p^{*}}{q^{*}}$ be a best approximant, i.e., suppose

\begin{equation}
\;=\;(q^{*})^{2}
\,\cdot\,
\left|\;\frac{p}{q}\;-\;\frac{p^{*}}{q^{*}}\;\right|\,.
\end{equation}

By Lemma <ref>, this means that the Ford circle
$h(p^{*},q^{*})$ has the smallest signed distance to the vertical
geodesic $g_{x}$ among all Ford circles except $h(p,q)$ and
$h(1,0)$. This leads to the following necessary geometric condition
for best approximants:

If $x$ is Markov fraction and $x^{*}$ is a best approximant of $x$,
then no geodesic in the $G$-orbit of $g_{x}$ separates $g_{x}$
and $x^{*}$.

(A geodesic $g$ in the hyperbolic plane $H^{2}$ separates a
subset $A\in H^{2}$ and an ideal point $p\in\partial\!H^{2}$ if $A$
and $p$ are contained in different connected components of
$(H^{2}\cup\partial\!H^{2})\setminus(g\cup\partial\!g)$, where
$\partial\!g$ is the set of the two ideal endpoints of $g$.)

Suppose $\tau\cdot g_{x}$ separates $g_{x}$ and $x'$ for some
$x'=\frac{p'}{q'}\in\Q$ and $\tau\in G$. We have to show that $x'$
is not a best approximant of $x$.

Since the shortest geodesic ray from $x'$ to $g_{x}$ intersects
$\tau(g_{x})$ before reaching $g_{x}$,

\begin{equation*}
\;>\;d\big(\tau(g_{x}), \,h(\,p',q'\,)\big)
\;=\;d\big(g_{x},\,\tau^{-1}(\,h(\,p',q'\,)\,)\big).
\end{equation*}

Since the Ford circle $\tau^{-1}(h(p',q'))$ has a smaller signed
distance to $g_{x}$ than $h(p',q')$, the number $\tau^{-1}\cdot x'$
approximates $x$ better than $x'$.

Lemma <ref> implies that $x^{*}$ is a rational
ideal boundary point of one of the domains

\begin{equation*}
R_{x}\quad\text{or}\quad L'_{x}
\end{equation*}

that are bounded by the geodesic $\varrho_{x}$ and geodesics in the
$G$-orbit of $g_{x}$ (see Figure <ref>)

[anchor=south west, inner sep=0pt] (image) at (0,0)

[-stealth] (5.7, 3.99) – (5.6, 3.99);

[below] at (0.37, 0) $x$;
[below] at (1.78, 0) $\xi_{+}$;
[below] at (9.68, 0) $\xi_{-}$;
[below] at (10.92, 0) $x+3$;

[right] at (0.38, 2.5) $g_{x}$;

[above] at (6.2, 3.95) $\varrho_{x}$;
at (5.7,5) $R_{x}$;
at (5.7,2.75) ${L_{x}'}$;

The domains $R_{x}$ and $L'_{x}$ bounded by $\varrho_{x}$
and geodesics in the $G$-orbit of $g_{x}$

or one of the domains

\begin{equation*}
L_{x}=R_{x}-3 \quad\text{and}\quad R'_{x}=L'_{x}-3
\end{equation*}

bounded by $\lambda_{x}$ and geodesics in the $G$-orbit of $g_{x}$. In
other words, $L'_{x}$ and $R'_{x}$ are the lifts of the cylinders
$\pi(L_{x})$ and $\pi(R_{x})$ that are adjacent to the geodesics
$\varrho_{x}$ and $\lambda_{x}$, respectively, while $L_{x}$ and
$R_{x}$ are the lifts adjacent to the geodesic $g_{x}$. (Compare also
Figure <ref> but note that it does not
represent the geometry of the modular torus $M$ correctly whereas
Figure <ref> does, as explained in
Remark <ref>.)

We will now consider all these rational ideal boundary points that are
candidates for best approximants in turn. First, let us consider the
rational ideal boundary points of $R_{x}$, which are:

* the Markov fraction $x=\frac{p}{q}$ itself, which does not

* the right companions $\gamma^{+}_{k}(x)$, with $k\geq 2$,

* the left companions $\gamma^{-}_{k}(x+3)$, also with $k\geq 2$, and

* the Markov fraction $x+3$.

The left companions of $x+3$ are ruled out as best approximants
because $\gamma^{-}_{k}(x+3)$ has the same denominator as
$\gamma^{+}_{k}(x)$ but is further away from $x$.

For a right companion

\begin{equation*}
\frac{a}{b}\;=\;\gamma^{+}_{k}\,
\Big(\,\frac{p}{q}\,\Big)\,,
\end{equation*}

we get using Lemmas <ref> and <ref>:

\begin{equation*}
\begin{split}
\;\;&=\;\;
\left|\;
\frac{p}{q}\;-\;
\gamma^{+}_{k}\,
\Big(\,\frac{p}{q}\,\Big)\;
\right|\\
\tfrac{1}{3}\,(\,u_{k}^{2}\,+\,u_{k-1}\,-\,1\,)\\
3\,q^{2}\;\geq\; 3
\end{split}
\end{equation*}

This means that a right companion $\gamma^{+}_{k}(\frac{p}{q})$
approximates the Markov fraction $\frac{p}{q}$ worse than the Markov
fractions $\frac{p_{1}}{q_{1}}$ and $\frac{p_{3}}{q_{3}}$ of the
centered rational Markov triple
with $\frac{p_{2}}{q_{2}}=\frac{p}{q}$.

One ideal boundary point of $R_{x}$ is left to check: The Markov
fraction $x+3$, for which we get $1^{2}\big|x-(x+3)\big|=3$.

Thus we conclude that the ideal boundary points of $R_{x}$ are not best
approximants of $x$. Similarly, the ideal boundary points of $L_{x}$
are not best approximants of $x$.

Note that the first companion $\gamma^{+}_{2}(x)$ and the Markov
fraction $x+3$ approximate the Markov fraction $x$ exactly equally
badly. In general, the $(k-1)$st right companion $\gamma^{+}_{k}(x)$
and the $(k-2)$nd left companion $\gamma_{k-1}(x+3)$ approximate $x$
with the same approximation quality. This due to a geometric

Now let $y$ be a rational ideal boundary point of $L'_{x}$. Then the
vertical geodesic $g_{y}$ projects to a simple geodesic $\pi(g_{y})$
in $M$ with both ends in the cusp. Furthermore, $\pi(g_{y})$ does not
intersect of the geodesic $\pi(g_{x})$. (To see this, note that
$g_{y}$ does not intersect any of the geodesics $[T]^{k}\cdot g_{y}$
for $k\in\Z\setminus\{0\}$, nor any geodesics in the $G$-orbit of

This implies, first, that $y$ is also Markov fraction. Moreover, since
$\pi(g_{x})$ and $\pi(g_{y})$ do not intersect, there are exactly two
ideal triangulations of $M$ containing $\pi(g_{x})$ and $\pi(g_{y})$
as edges. Since $x<y<x+3$, we can deduce that there is a unique
rational Markov triple

\begin{equation*}
\left(\frac{p_{1}'}{q_{1}'},\frac{p_{2}}{q_{2}},\frac{p_{3}'}{q_{3}'}\right)
\qquad\text{with}\qquad
\qquad
\end{equation*}

Conversely, if $m'$ is a rational Markov triple with
$x=\frac{p_{2}}{q_{2}}$, then $\frac{p_{3}'}{q_{3}'}$ is an ideal
boundary point of $L_{x}'$.

For the ideal boundary point $y$ of $L_{x}'$, we get the approximation quality

\begin{equation*}
\;=\;
\frac{q_{3}'\,q_{1}'}{q_{2}}\,,
\end{equation*}

and if $m'$ is not the unique centered rational Markov triple $m$, then

\begin{equation*}
\frac{q_{3}'\,q_{1}'}{q_{2}}
\;>\;
\frac{q_{3}\,q_{1}}{q_{2}}\,.
\end{equation*}

This shows that among all ideal boundary points of $L_{x}'$, the third
element $\frac{p_{3}}{q_{3}}$ of the centered rational Markov triple
$m$ with $x=\frac{p_{2}}{q_{2}}$ approximates $x$ best, and all other
ideal boundary points of $L_{x}'$ are worse.

In the same way, one sees that among all ideal boundary points of $R_{x}'$, the first
element $\frac{p_{1}}{q_{1}}$ of the centered rational Markov triple
$m$ with $x=\frac{p_{2}}{q_{2}}$ approximates $x$ best, and all other
ideal boundary points of $R_{x}'$ are worse.

This completes the proof of
Theorem <ref> (ii).

While many constructions and arguments in this paper generalize to
arbitrary hyperbolic tori with one cusp,
Theorem <ref> (ii) really depends on
the geometry of the modular torus $M$. For example, in the torus
shown in Figure <ref>, the horocycles at
$\gamma^{\pm}_{2}(x)$ are closer to $g_{x}$ than the horocycles at
the ideal boundary points of $L_{x}'$ and $R_{x}'$.

§.§ Proof of Theorem <ref> (Best
approximants of companions)

From the definition (<ref>), we get immediately

\begin{equation*}
\left|\,
\gamma^{\pm}_{k}\Big(\frac{p}{q}\Big)\,-\,\frac{p}{q}\,
\right|
\;=\;
\end{equation*}

To see that also

\begin{equation*}
\left|\,
\gamma^{\pm}_{k}
\Big(\frac{p}{q}\Big)
\,-\,
\gamma^{\pm}_{k-1}
\Big(\frac{p}{q}\Big)\,
\right|
\;=\;
\end{equation*}

note that

\begin{equation}
\label{eq:companion_gap}
\left|\,
\gamma^{\pm}_{k}
\Big(\frac{p}{q}\Big)
\,-\,
\gamma^{\pm}_{k-1}
\Big(\frac{p}{q}\Big)\,
\right|
\;=\;
\frac{1}{q\,u_{k}\,u_{k-1}}
\end{equation}

\begin{equation*}
\frac{u_{k-1}}{q\,u_{k}}
\,-\,
\frac{u_{k-2}}{q\,u_{k-1}}
\;=\;
\frac{1}{q\,u_{k}\,u_{k-1}}\;
\big(\,u_{k-1}^{2}\,-\,u_{k}\,u_{k-2}\,\big)
\end{equation*}

\begin{equation*}
\begin{split}
u_{k-1}^{2}\,-\,u_{k}\,u_{k-2} \;\;
\;\;
\;\;
\;\;
\end{split}
\end{equation*}

This shows the right equality in (<ref>).

Alternatively, we could avoid these calculations here and infer from
a geometric symmetry that $\frac{p}{q}$ and
$\gamma^{\pm}_{k-1}\big(\frac{p}{q}\big)$ both approximate
$\gamma^{\pm}_{k}\big(\frac{p}{q}\big)$ with the same quality. But
we will need equation (<ref>) anyway for the proof
of the classification Theorem <ref> (see
Section <ref>).

To see the inequality in (<ref>), note that

\begin{equation*}
\;\;
\overset{\makebox[0pt]{\scriptsize\eqref{eq:un}}}{=}
\;\;
\frac{1}{3}\,(\,u_{k}\,+\,u_{k-2}\,)\,.
\end{equation*}

So for $k\geq 2$,

\begin{equation*}
\;=\;
\frac{1}{3}
\left(
\right)
\;\geq\;
\frac{1}{3}\;,
\end{equation*}

with equality only for $k=0$.

It remains to show that

\begin{equation}
\label{eq:companion_candidate_test}
\;<\;
\left|\,
\gamma^{\pm}_{k}\Big(\frac{p}{q}\Big)\,-\,\frac{a}{b}\,
\right|
\end{equation}

for every rational number $\frac{a}{b}$, except $\frac{p}{q}$,
$\gamma^{\pm}_{k-1}\big(\frac{p}{q}\big)$, and
$\gamma^{\pm}_{k}\big(\frac{p}{q}\big)$. The general strategy is the
same as in the previous section. We will use
Lemma <ref> to eliminate all but a manageable
subset of candidates $\frac{a}{b}$ that we have to check. However,
instead of $G$-orbits as in Lemma <ref>, we will
consider orbits with respect to a larger group $\widehat{G}$, of which
$G$ is a subgroup of index $2$:

Let $\widehat{G}\subset\PSLTZ$ be the group generated by the group
$G$ of deck transformations of $M$ and any lift of the elliptic
involution, e.g., $z\mapsto z+3$.

In the previous section, it was sufficient to consider the
$G$-orbits of $g_{x}$, because the elliptic involution only reverses
the orientation of simple geodesics in $M$ with both ends in the
cups. Hence, the $G$-orbit and the $\widehat{G}$-orbit of $g_{x}$
are equal if we disregard the orientation of geodesics.

If $y$ is a companion of a Markov fraction and $y^{*}$ is a best
approximant of $y$, then no geodesic in the $\widehat{G}$-orbit of
$g_{y}$ separates $g_{y}$ and $y^{*}$.

Lemma <ref> can be proved in the same way as
the analogous Lemma <ref> for Markov
fractions. Next, we will use it to prove a second geometric lemma:

If $y$ is a companion of a Markov fraction $x$, and $y^{*}$ is a best
approximant of $y$, then no geodesic in the $G$-orbit of
$g_{x}$ separates $g_{y}$ and $y^{*}$.

Let us assume that $y$ is a right companion of $x$, i.e.,

\begin{equation}
\label{eq:y_gamma}
y\;=\;\gamma^{+}_{k}(x),\quad k\geq 2\,.
\end{equation}

The argument for left companions is analogous. The proof has two

Part 1.  We will first prove the following statement:
For all real numbers $y'<x$ there is a geodesic in the
$\widehat{G}$-orbit of $g_{y}$ that separates $g_{x}$ and $y'$.

To this end, we consider two cases separately:

* $y'\in\,(-\sminfty,\, y-3)$.  Let $\tau(z)=z-3$. Then
$\tau\in\widehat{G}$, and because $y-3<x$, the vertical geodesic
$\tau\cdot g_{y}$ separates $g_{x}$ and $y'$.

* $y'\in\,(\gamma^{+}_{k-1}(x)\,-\,3,\, x)$.  With $[T]\in G$ as in
Section <ref> and $\tau$ as above we have

\begin{equation*}
[T]^{-1}\cdot y \;=\; \gamma^{+}_{k-1}(x)
\qquad\text{and}\qquad
[T]^{-1}\cdot \sminfty \;=\; x+3.
\end{equation*}

The geodesic $\tau\circ[T]^{-1}\cdot g_{y}$ therefore has ideal
endpoints $\gamma^{+}_{k-1}(x)-3$ and $x$. So it separates $g_{y}$
and $y'$.

Since $\gamma^{+}_{k-1}(x)<y$, the two intervals cover
$(-\sminfty,\,x)$, and hence we have shown the statement for all $y'<x$.

Part 2.  Now to prove the lemma, suppose there is a $\sigma\in G$ such
that $\sigma\cdot g_{x}$ separates $g_{y}$ and $y'$. We have to show
that $y'$ is not a best approximant of $y$.

To see this, let $\hat{\sigma}\in\widehat{G}$ be the element with

\begin{equation*}
\hat{\sigma}\cdot g_{x}\;=\;\sigma\cdot g_{x}
\quad\text{and}\quad
y' \;\in\; \hat{\sigma}\cdot \R_{<x}\,.
\end{equation*}

(Either $\hat{\sigma}=\sigma$, or $\sigma^{-1}\hat{\sigma}$ is the
lift of the elliptic involution on $M$ that maps the simple geodesic
$g_{x}$ with both ends in the cusp to itself, but reversing its
orientation.) Then $\hat{\sigma}^{-1}\cdot y'<x$, so by (1) there is
a geodesic $g_{1}\in\widehat{G}\cdot g_{y}$ that
separates $g_{x}$ and $\hat{\sigma}^{-1}\cdot y'$. Hence
$\hat{\sigma}\cdot g_{1}$ separates $\hat{\sigma}\cdot g_{x}$ and
$y'$. Since $\hat{\sigma}\cdot g_{x}$ separates $g_{y}$ and $y'$ by
assumption, the geodesic $\hat{\sigma}\cdot g_{1}$ separates $g_{y}$
and $y'$. By Lemma <ref>, $y'$ is not a
best approximant of $y$.

To prove the theorem, let us consider a right companion $y$ as
in (<ref>). The arguments for a left companion are
completely analogous.

First, Lemma <ref> implies that the best
approximants of $y$ are among the rational ideal boundary points of
the domains $R_{x}$ and $L_{x}'$ (see Figure <ref>).

Let us first consider the rational ideal boundary points of $R_{x}$,
which are

\begin{equation*}
\gamma^{+}_{\ell}(x)\;=\;[T]^{\ell}\cdot \sminfty
\quad\text{for}\quad
\ell\;\in\;\Z
\end{equation*}

(see Section <ref>). Of these,
$y=\gamma^{+}_{k}(x)$ and $\sminfty=\gamma^{+}_{0}(x)$ are by
definition not best approximants of $y$.

Next, we will use Lemma <ref> to show that
$\gamma^{+}_{\ell}(x)$ is not a best approximant if $k<\ell$ or
$\ell<0$. To see this, note first that the ideal endpoints of the
geodesic $[T]^{m}\cdot g_{y}$ are

\begin{equation*}
\gamma^{+}_{m}(x)
\quad\text{and}\quad
\gamma^{+}_{m+k}(x).
\end{equation*}

Hence, if

\begin{equation*}
k\;\leq\; m\;<\;\ell\;<\;m+k
\end{equation*}

\begin{equation*}
\end{equation*}

then $[T]^{m}\cdot g_{y}$ separates $g_{y}$ and
$\gamma^{+}_{\ell}(x)$. Thus we have eliminated all rational ideal
boundary points of $R_{x}$ as best approximants except

\begin{equation}
\label{eq:Rx_candiates}
\gamma^{+}_{2}(x),\;
\ldots\quad
\gamma^{+}_{k-1}(x)\,.
\end{equation}

To eliminate all rational ideal boundary points of $L_{x}'$, let
$\iota\in\widehat{G}$ be the lift of the elliptic involution on $M$
that maps the geodesic $\varrho_{x}$ to itself, only reversing its
orientation. Then the rational ideal boundary points of $L_{x}'$ are
the numbers

\begin{equation*}
\iota\cdot\gamma^{+}_{\ell}(x)\;=\;\iota\circ [T]^{\ell}\cdot \sminfty
\quad\text{for}\quad
\ell\;\in\;\Z
\end{equation*}

Thus, if

\begin{equation*}
\end{equation*}

then the geodesic $\iota\circ [T]^{m}\circ g_{y}$ separates $g_{y}$
and $\iota\circ\gamma^{+}_{\ell}(x)$. By
Lemma <ref>, this excludes all rational ideal
boundary points of $L_{x}'$ as best companions of $y$.

Now the only candidates for best approximant of $y$ are the
numbers (<ref>), and we know that $x$ and
$\gamma^{+}_{k-1}(x)$ have the same approximation quality. To prove that
they are indeed the best approximants, it remains to
show (<ref>) for

\begin{equation*}
\frac{a}{b}\;=\;\gamma^{+}_{\ell}(x)
\quad\text{with}\quad
\end{equation*}

If $k=2$, there is nothing to show. For $k>2$, we will take the
geometric point of view to simplify the calculations.

We have to show that among the Ford circles at the the rational
numbers (<ref>), the minimal signed distance to the
vertical geodesic $g_{y}$ is attained precisely for the first and the
last (see Lemma <ref>). Note that the Ford circles at
$\gamma^{+}_{\ell}(x)$ with $\ell\in\Z$ are equally spaced translates
along the translation axis $\varrho_{x}$. In particular, all these
horocycles have the same signed distance to $\varrho_{x}$. Moreover,
$1\leq\ell\leq k-1$ if and only if $g_{y}$ separates $\varrho_{x}$ and
the center $\gamma^{+}_{\ell}(x)$.

To simplify the calculations, we apply a hyperbolic isometry that maps
the axis $\varrho_{x}$ to the upward oriented vertical geodesic
$g_{0}$, and the geodesic $g_{y}$ to the geodesic with ideal endpoints
$e^{-t_{0}}$ and $e^{t_{0}}$ for some positive real $t_{0}$ (see
Figure <ref>).

(-1,0) – (9,0);
(0,0) node[below] $0$ – (0,6);

(2,0) node[below] $e^{-t_{0}}$ arc (180:0:3) node[below] $e^{t_{0}}$;

[black] (0,0) circle (1pt);
[black] (2,0) circle (1pt);
[black] (2.828,0) node[below] $e^{t}$ circle (1pt);
[black] (4,0) node[below] $1$ circle (1pt);
[black] (8,0) circle (1pt);

[black] (0,4) node[left] $i$ circle (1pt);

[ultra thin] (2.818, 1.414) circle (1.414);
(1.8,3.1) node $\displaystyle h\big(e^{\frac{t}{2}},e^{\frac{-t}{2}}\big)$;

[ultra thin]  (2.828,0) arc (0:110.7:.9866);
[thick] ($(1.8418,0)+({.9866*cos(71.5)},{.9866*sin(71.5)})$) arc (71.5:110.7:.9866);
(1.8, 1.2) node $d$;

The signed distance $d$ between the geodesic with ideal
endpoints $e^{\pm t_{0}}$ and the horocycle

Then the images of the Ford circles at $\gamma^{+}_{\ell}(x)$ are
among the $1$-parameter family of translates of some horocycle
$h(a,a)$ along the axis $g_{0}$. To simplify the calculations further,
we will consider the translates of $h(1,1)$ instead. This scales the
horocycles evenly, so all distances change by the same additive
constant. That is, we consider the family of horocycles $h(p(t),q(t))$

\begin{equation*}
\begin{pmatrix}
\end{pmatrix}
\;=\;
\begin{pmatrix}
e^{\frac{t}{2}} & 0 \\
0 & e^{-\frac{t}{2}}
\end{pmatrix}
\,
\begin{pmatrix}
1 \\ 1
\end{pmatrix}
\;=\;
\begin{pmatrix}
e^{\frac{t}{2}} \\ e^{-\frac{t}{2}}
\end{pmatrix}
\end{equation*}

(see Section <ref> and <cit.>,
<cit.>). Among these, the scaled images of
the Ford circles at at $\gamma^{+}_{\ell}(x)$ are the horocycles
$h(e^{\frac{t}{2}},e^{-\frac{t}{2}})$ with

\begin{equation}
\label{eq:t_of_ell}
\end{equation}

(To see this, note that for integer values of $\ell$ between $0$ and
$k$, the $t$-values are evenly spaced reals between $-t_{0}$
and $t_{0}$.)

The signed distance between the horocycle
$h(e^{\frac{t}{2}},e^{-\frac{t}{2}})$ and the geodesic with ideal
endpoints $e^{\pm t_{0}}$ is

\begin{equation*}
\left|\,
\right|\,,
\end{equation*}

where $f$ is a quadratic form on $\R^{2}$ with determinant $-1$ that is zero for
$\frac{p}{q}=e^{\pm t_{0}}$ (see
<cit.>). This determines $f$ up to sign and
we may choose

\begin{equation*}
\frac{1}{\sinh t_{0}}
\big(
p^{2}\,-\,2\,(\cosh t_{0})\,p\,q\,+\,q^{2})
\big)\,.
\end{equation*}

For $t\in (-t_{0},t_{0})$ we obtain

\begin{equation}
\label{eq:d_companions}
\frac{2}{\sinh t_{0}}\,
\big(\cosh t_{0}\,-\,\cosh t\big)\,.
\end{equation}

This is a concave even function of $t$ with limit zero as
$t\rightarrow \pm t_{0}$. Hence, for $t$ as in (<ref>)
with $\ell\in\Z$ and $1\leq\ell\leq k-1$, the signed distance $d$
attains its minimum precisely for $\ell=1$ and $\ell=k-1$. This
completes the proof of Theorem <ref>.

§.§ Empty intervals around Markov fractions

This section is about the largest intervals $I_{x}$ around each Markov
fraction $x$ that contain no other Markov fractions (see
Section <ref>). At the same time, $I_{x}$ is the
smallest closed interval that contains all companions of $x$. We will
prove Lemmas <ref> and <ref>, which will be used
in the following section to complete the proof of the classification
Theorem <ref>.

From the geometric point of view, Lemmas <ref>
and <ref> are well known, and even in greater
generality: for arbitrary hyperbolic tori with one cusp. This is the
origin of McShane's remarkable identity for the lengths of simple
closed geodesics [21] (see also Remark <ref>),
which has been reproved [5, 15, 29]
and generalized [2, 22, 23] in various
ways.  Nevertheless, it seems worthwhile to provide independent proofs
in the current context.

For every Markov fraction $x=\frac{p}{q}$, the interval $I_{x}$
defined by equation (<ref>) is the smallest closed interval
that contains all companions of $x$. It contains no Markov fractions
other than $x$, nor any of their companions. But the irrational
endpoints of $I_{x}$ are accumulation points of the set of Markov

By equations (<ref>) and (<ref>),
the right endpoint $\xi_{+}$ of $I_{x}$ is the limit of the
increasing sequence $(\gamma^{+}_{k}(x))_{k\geq 2}$ of right
companions of $x$. Analogously or by symmetry, the left endpoint is
the limit of the decreasing sequence of left companions. Thus no
smaller closed interval contains all companions.

The rational ideal boundary points of the domain $L_{x}'$ are Markov
numbers, and they accumulate at the right endpoint of $I_{x}$ (see
Section <ref> and
Figures <ref>
and <ref>). Similarly, the rational ideal boundary
points of $R_{x}'$ are Markov numbers accumulating at the left
endpoint of $I_{x}$. As accumulations points of Markov numbers, the
endpoints of $I_{x}$ are irrational (see
Section <ref>; in fact, it is well known that the
endpoints are Markov irrationals).

For every interior point $y$ of $I_{x}$ except $x$, the geodesic
$\pi(g_{y})$ in $M$ has self-intersections. To see this if $y>x$,
note that $g_{y}$ intersects $[T]\cdot g_{y}$ (see
Section <ref>). The same holds for $y<x$ by
an analogous argument or by symmetry. Since the endpoints of
$I_{x}$ are not even rational, this implies that $x$ is the only
Markov fraction in $I_{x}$.

In particular, this means that no other Markov fraction is between
$x$ and one of its companions. Since this is true for all Markov
fractions, and they accumulate at the endpoints of $I_{x}$, this
implies that $I_{x}$ does not contain any companions of other Markov
fractions than $x$.

For different Markov fractions $x\not=x'$, the interiors
$\mathring{I}_{x}$ and $\mathring{I}_{x'}$ are disjoint.

As $x$ ranges over the set of Markov fractions, the intervals
$I_{x}$ cover the set $\Q$ of rational numbers.

Since the intervals obviously cover the set of Markov fractions, let
us assume that $y\in\Q$ is not a Markov fraction. We have to find a
Markov fraction $x$ for which $y\in I_{x}$.

To this end, define the infinite sequence of centered Markov triples

\begin{equation*}
\left(\,
\frac{p_{k,1}}{q_{k,1}},\;
\frac{p_{k,2}}{q_{k,2}},\;
\frac{p_{k,3}}{q_{k,3}}\,
\right)
\end{equation*}

\begin{equation}
\label{eq:mk_nesting}
\frac{p_{k,1}}{q_{k,1}}
\;<\; y \;<\;
\frac{p_{k,3}}{q_{k,3}}
\end{equation}

recursively by

* $m_{0}
\;=\;
\left(\,
\lfloor y\rfloor,\;
\lfloor y\rfloor+\tfrac{1}{2},\;
\lfloor y\rfloor+1\,
\right)$, where $\lfloor y\rfloor$ denotes the largest integer not
larger than $y$, and

* for $k>0$, the triple $m_{k}$ is the child
satisfying (<ref>) among the two children of
$m_{k-1}$ in the tree of centered rational Markov triples (see
Lemma <ref>).

First consider the following two cases:

* The triple $m_{k}$ is the left child of $m_{k-1}$ for all but
finitely many $k$.

That is, the sequence of Markov fractions
$\frac{p_{k,1}}{q_{k,1}}$ eventually becomes constant, say

\begin{equation*}
\frac{p_{k,1}}{q_{k,1}}\;=\;x
\quad\text{for}\quad k\;>\;k_{0}\,,
\end{equation*}

while $\frac{p_{k,2}}{q_{k,2}}$ and $\frac{p_{k,3}}{q_{k,3}}$ are
consecutive ideal boundary points of $L_{x}'$ with

\begin{equation*}
\lim\,\frac{p_{k,2}}{q_{k,2}}
\;=\;
\lim\,\frac{p_{k,3}}{q_{k,3}}
\;=\;\xi_{+}
\end{equation*}

(see Section <ref> and
Figures <ref>
and <ref>). Hence $y\in I_{x}$

* The triple $m_{k}$ is the right child of $m_{k-1}$ for all but
finitely many $k$.

In this case,

\begin{equation*}
\frac{p_{k,3}}{q_{k,3}}\;=\;x
\quad\text{for}\quad k\;>\;k_{0}\,,
\end{equation*}

and like in case 1 we see that $y\in I_{x}$.

We will complete the proof of the lemma by showing that the third case
does not occur:

3. Infinitely many of the triples $m_{k}$ are left children
of $m_{k-1}$ and infinitely many are right children.

For such a sequence of rational Markov triples, we will show that

\begin{equation}
\label{eq:markov_squeeze}
\lim
\left(
\frac{p_{k,3}}{q_{k,3}}
\,-\,
\frac{p_{k,1}}{q_{k,1}}
\right)
\;=\;
\end{equation}

Then condition (<ref>) would imply

\begin{equation*}
\lim\,\frac{p_{k,1}}{q_{k,1}}
\;=\;
\;=\;
\lim\,\frac{p_{k,3}}{q_{k,3}}\,.
\end{equation*}

This would contradict $y$ being rational, because accumulation
points of the set of Markov fractions are irrational (see
Section <ref>).

It remains to show (<ref>) in case 3. To this end,
note that for any rational Markov triple
we have

\begin{equation*}
\begin{split}
\frac{p_{3}}{q_{3}}\,-\,\frac{p_{1}}{q_{1}}
\quad&=\quad
\left(
\frac{p_{3}}{q_{3}} \,-\, \frac{p_{2}}{q_{2}}
\right)
\,+\,
\left(
\frac{p_{2}}{q_{2}} \,-\, \frac{p_{1}}{q_{1}}
\right)\\
\frac{q_{1}}{q_{2}\,q_{3}}\,+\,\frac{q_{3}}{q_{1}\,q_{2}}\\
\frac{q_{2}}{q_{3}\,q_{1}}\,,
\end{split}
\end{equation*}

so it is enough to show

\begin{equation}
\label{eq:lim_q2_q3q1}
\lim\,\frac{q_{k,2}}{q_{k,3}\,q_{k,1}}\;=\;3.
\end{equation}

To see this, we will show that for any centered rational Markov
we have

\begin{equation}
\label{eq:q2_q3q1_ineq}
\frac{q_{2}}{q_{3}\,q_{1}}\;\geq\;
\frac{3}{
\,+\,
\frac{1}{q_{1}^{2}}
\,+\,
\frac{1}{q_{3}^{2}}
\end{equation}

Since both sequences $(q_{k,1})$ and $(q_{k,3})$ are unbounded in
case 3, equation (<ref>) follows.

Finally, to prove the estimate (<ref>) for a
centered rational Markov triple, note that (<ref>)

\begin{equation*}
\frac{q_{2}}{q_{3}\,q_{1}}
\;\geq\;
\max\,
\left\{\,
\frac{q_{1}}{q_{2}\,q_{3}}\,,\,
\frac{q_{3}}{q_{1}\,q_{2}}\,
\right\}\,,
\end{equation*}

and with equation (<ref>),

\begin{equation*}
\frac{q_{2}}{q_{3}\,q_{1}}\;\geq\;1\,,
\end{equation*}

Hence $\frac{q_{2}}{q_{1}}\geq q_{3}$ and therefore

\begin{equation*}
\frac{q_{1}}{q_{2}\,q_{3}}
\;=\;
\frac{q_{2}}{q_{3}\,q_{1}}
\cdot \frac{q_{1}^{2}}{q_{2}^{2}}
\;\leq\;
\frac{q_{2}}{q_{3}\,q_{1}}\cdot\frac{1}{q_{3}^{2}}\,.
\end{equation*}

In the same way we get $\frac{q_{2}}{q_{3}}\geq q_{1}$ and

\begin{equation*}
\frac{q_{3}}{q_{1}\,q_{2}}
\;\leq\;
\frac{q_{2}}{q_{3}\,q_{1}}\cdot\frac{1}{q_{1}^{2}}\,.
\end{equation*}

Now equation (<ref>) yields

\begin{equation*}
\frac{q_{2}}{q_{3}\,q_{1}}
\,
\left(
\frac{1}{q_{3}^{2}}
\,+\,1\,+\,
\frac{1}{q_{1}^{2}}
\right)
\end{equation*}

and hence (<ref>). This completes the proof of
Lemma <ref>.

McShane's identity [21] says that

\begin{equation}
\label{eq:mcshane}
\sum_{\gamma}\frac{1}{1\,+\,e^{|\gamma|}}
\;=\;
\frac{1}{2}\,,
\end{equation}

where the sum is taken over all simple closed geodesics in some
hyperbolic torus with one cusp, and $|\gamma|$ denotes the length of
the geodesic. For the modular torus $M$, this identity can be
derived as follows. The unoriented closed geodesics in $M$ are in
one-to-one correspondence with the
$(\operatorname{mod} 3)$-equivalence classes of Markov fractions.
The length of the closed geodesic $\eta_{x}$ and the
length of the interval $I_{x}$ are related by

\begin{equation}
\label{eq:eta_I_lengths}
\frac{1}{1\,+\,e^{|\gamma|}}
\;=\;
\tfrac{1}{6}\;|\,I_{x}\,|
\end{equation}

(see Remark <ref>). On the other hand,
Corollary <ref> and Lemma <ref> imply

\begin{equation}
\label{eq:mcshane_I}
\end{equation}

which yields McShane's identity.

§.§ Proof of Theorem <ref> (Classification)

Since we have already proved
Theorems <ref>
and <ref> in the two previous sections, we
only have to show that $C(y)<\frac{1}{3}$ if $y$ is a rational number
$y$ that is neither a Markov fraction nor a companion of a Markov

By Lemma <ref> there is a Markov fraction $x$ such that
$y\in I_{x}$. Since we assume that $y$ is not a Markov fraction or a
companion, Lemma <ref> implies that $y$ lies between two
companions of $x$ or between $x$ and one of its first companions. In
other words, there is an integer $k\geq 1$ such that

\begin{equation*}
\gamma^{-}_{k+1}(x)\;<\;y\;<\;\gamma^{-}_{k}(x)
\qquad\text{or}\qquad
\gamma^{+}_{k}(x)\;<\;y\;<\;\gamma^{+}_{k+1}(x)\,.
\end{equation*}

We will show that there is not enough space between the Ford circles
at $\gamma^{\pm}_{k}(x)$ and $\gamma^{\pm}_{k+1}(x)$ such that a
vertical geodesic could pass between them and stay at least a signed
distance of  $\log\frac{2}{3}$  away from both. With
Corollary <ref>, this shows that $C(y)<\frac{1}{3}$.

We already know that

\begin{equation}
\label{eq:companion_gap2}
\big|\,
\gamma^{\pm}_{k}(x)
\,-\,
\gamma^{\pm}_{k+1}(x)\,
\big|
\;=\;
\frac{1}{q\,u_{k}\,u_{k+1}}\,,
\end{equation}

and since the denominator a companion $\gamma^{\pm}_{k}(x)$ is
$qu_{k}$, the euclidean radius of its Ford circle is

\begin{equation*}
\frac{1}{2\,q^{2}\,u_{k}^{2}}
\end{equation*}

(see equation (<ref>),
Lemma <ref>, and
Section <ref>). A necessary condition for
$C(y)\geq\frac{1}{3}$ is that the vertical geodesic $g_{y}$ fits
between the Ford circles at $\gamma^{\pm}_{k}(x)$ and
$\gamma^{\pm}_{k+1}(x)$, scaled by $\frac{2}{3}$ in the euclidean
metric (see Corollary <ref> and
Figure <ref>). Therefore, to prove that $C(y)<\frac{1}{3}$, it
suffices to show that

\begin{equation}
\label{eq:gap_too_small}
\big|\,
\gamma^{\pm}_{k+1}(x)
\,-\,
\gamma^{\pm}_{k}(x)\,
\big|
\;<\;
\frac{1}{3\,q^{2}\,u_{k+1}^{2}}
\,+\,
\frac{1}{3\,q^{2}\,u_{k}^{2}}\,.
\end{equation}

And indeed we have

\begin{equation*}
\begin{split}
\frac{1}{3\,q^{2}\,u_{k+1}^{2}} \,+\,
\frac{1}{3\,q^{2}\,u_{k}^{2}} \;\;&=\;\;
\frac{u_{k}^{2}\,+\,u_{k+1}^{2}}{3\,q^{2}\,u_{k}^{2}\,u_{k+1}^{2}}\\
\frac{3\,q\,u_{k}\,u_{k+1}\,+\,1}{3\,q^{2}\,u_{k}^{2}\,u_{k+1}^{2}}\\
\frac{1}{q\,u_{k}\,u_{k+1}}
\,+\,\frac{1}{3\,q^{2}\,u_{k}^{2}\,u_{k+1}^{2}}\\
\big|\,
\gamma^{\pm}_{k}(x)
\,-\,
\gamma^{\pm}_{k+1}(x)\,
\big|\,.
\end{split}
\end{equation*}

§.§ Proof of Theorems <ref>
and <ref> (Triangle paths)

The Epstein–Penner convex hull construction [12]
provides the modular torus $M$ with a euclidean metric, with respect
to which the ideal hyperbolic triangles of the canonical triangulation
are equilateral euclidean triangles. We may assume that the developing
map $D:H^{2}\rightarrow\C$ maps the ideal triangle $\infty,0,1$ to the
euclidean triangle $0,1,\omega$ with $\omega$ as in
Theorem <ref>. The image of $D$ is then the
complement $\C\setminus\Gamma$ of the Eisenstein lattice
$\Gamma\;=\;\Z\,+\,\omega\Z$. The canonical projection $\pi$ factors
through $\C\setminus\Gamma$, giving rise to the projection

\begin{equation*}
\begin{tikzcd}
H^2 \arrow[r, "D"]\arrow[d, "\pi"] & \C\setminus\Gamma \arrow[dl, "\bar{\pi}"]\\
M &
\end{tikzcd}
\end{equation*}

If $m$ and $n$ are coprime integers, then the projection $\bar{\pi}$
maps the oriented line segment from $0$ to $z_{m,n}=m+n\omega$ to a
simple closed curve in $M$. There is a unique homotopic simple
geodesic in $M$ with both ends in the cusp (under homotopy with ends
in the cusp), and all simple geodesics with both ends in the cusp can
be obtained in this way. If $m$ and $n$ are both nonnegative, the
geodesic in $M$ lifts to a vertical geodesic $g_{x}$ in $H^{2}$,
oriented from $\sminfty$ to $0$, where $x$ is a Markov fraction
between $0$ and $1$. As the geodesic $g_{x}$ traverses finitely many
triangles of the Farey triangulation, the new vertices are obtained by
Farey addition. But the developing map $D$ maps the triangles of the
Farey triangulation that $g_{y}$ traverses to the regular euclidean
triangles that the line segment from $0$ to $z_{m,n}$
traverses. Hence, the Markov fraction $x$ can also be obtained by
Farey addition along the euclidean triangle strip. This proves
Theorem <ref>.

To prove Theorem <ref> in a similar fashion, note
that $\bar{\pi}$ projects the deformed straight line segments to
curves in $M$ with both ends in the cusp that cross neither the
respective simple geodesic with both ends in the cusp nor the
corresponding simple closed geodesic.

I would like to thank Gergely Harcos, who graciously made me aware of
the previous results by Flahive [14] and
Gurwood [17].

This research was supported by DFG SFB/TR
109 “Discretization in Geometry and Dynamics”.

[1]
M. Aigner.
Markov's theorem and 100 years of the uniqueness conjecture.
Springer, Cham, 2013.

[2]
H. Akiyoshi, H. Miyachi, and M. Sakuma.
A refinement of McShane's identity for quasifuchsian punctured
torus groups.
In In the tradition of Ahlfors and Bers, III, volume 355
of Contemp. Math., pages 21–40. Amer. Math. Soc., Providence, RI,

[3]
A. I. Bobenko, U. Pinkall, and B. A. Springborn.
Discrete conformal maps and ideal hyperbolic polyhedra.
Geom. Topol., 19(4):2155–2215, 2015.

[4]
F. Bonahon.
Low-dimensional geometry, volume 49 of Student
Mathematical Library.
Amer. Math. Soc., Providence, RI, 2009.

[5]
B. H. Bowditch.
A proof of McShane's identity via Markoff triples.
Bull. London Math. Soc., 28(1):73–78, 1996.

[6]
J. W. S. Cassels.
An introduction to Diophantine approximation.
Cambridge University Press, New York, 1957.

[7]
H. Cohn.
Approach to Markoff's minimal forms through modular functions.
Ann. of Math. (2), 61:1–12, 1955.

[8]
H. Cohn.
Representation of Markoff's binary quadratic forms by geodesics on
a perforated torus.
Acta Arith., 18:125–136, 1971.

[9]
J. H. Conway.
The sensual (quadratic) form, volume 26 of Carus
Mathematical Monographs.
Mathematical Association of America, Washington, DC, 1997.

[10]
T. W. Cusick and M. E. Flahive.
The Markoff and Lagrange spectra.
Amer. Math. Soc., Providence, RI, 1989.

[11]
M. Einsiedler and T. Ward.
Ergodic theory with a view towards number theory, volume 259 of
Graduate Texts in Mathematics.
Springer-Verlag London, Ltd., London, 2011.

[12]
D. B. A. Epstein and R. C. Penner.
Euclidean decompositions of noncompact hyperbolic manifolds.
J. Differential Geom., 27(1):67–80, 1988.

[13]
V. V. Fock and A. B. Goncharov.
Dual Teichmüller and lamination spaces.
In A. Papadopoulos, editor, Handbook of Teichmüller theory.
Vol. I, pages 647–684. Eur. Math. Soc., Zürich, 2007.

[14]
M. E. Gbur[Flahive] Gbur.
On the minimum of zero indefinite binary quadratic forms.
Mathematika, 25(1):94–106, 1978.

[15]
C. Goodman-Strauss and Y. Rieck.
Simple geodesics on a punctured surface.
Topology Appl., 154(1):155–165, 2007.

[16]
D. S. Gorshkov.
Geometry of Lobachevskii in connection with certain questions of
arithmetic (Russian).
Zap. Nauchn. Semin. Leningr. Otd. Mat. Inst. Steklova,
76:39–85, 1977.
English translation in J. Soviet Math. 16 (1981) 788–820.

[17]
C. Gurwood.
Diophantine approximation and the Markov chain.
PhD thesis, New York University, 1976.

[18]
A. Haas.
Diophantine approximation on hyperbolic Riemann surfaces.
Acta Math., 156(1-2):33–82, 1986.

[19]
G. H. Hardy and E. M. Wright.
An introduction to the theory of numbers.
Oxford University Press, Oxford, sixth edition, 2008.

[20]
A. Hatcher.
Topology of numbers.
Book preprint,
<https://www.math.cornell.edu/ hatcher/TN/TNpage.html> (accessed

[21]
G. McShane.
A remarkable identity for lengths of curves.
PhD thesis, University of Warwick, 1991.

[22]
G. McShane.
Simple geodesics and a series constant over Teichmuller space.
Invent. Math., 132(3):607–632, 1998.

[23]
G. McShane.
Weierstrass points and simple geodesics.
Bull. London Math. Soc., 36(2):181–187, 2004.

[24]
G. McShane and H. Parlier.
Multiplicities of simple closed geodesics and hypersurfaces in
Teichmüller space.
Geom. Topol., 12(4):1883–1919, 2008.

[25]
R. C. Penner.
The decorated Teichmüller space of punctured surfaces.
Comm. Math. Phys., 113(2):299–339, 1987.

[26]
R. C. Penner.
Decorated Teichmüller theory.
QGM Master Class Series. European Mathematical Society (EMS),
Zürich, 2012.

[27]
J. Propp.
The combinatorics of frieze patterns and Markoff numbers.
Integers, 20, 2020.

[28]
A. M. Rockett and P. Szüsz.
Continued fractions.
World Scientific Publishing Co., Inc., River Edge, NJ, 1992.

[29]
T. A. Schmidt and M. Sheingorn.
McShane's identity, using elliptic elements.
Geom. Dedicata, 134:75–90, 2008.

[30]
P. Schmutz.
Systoles of arithmetic surfaces and the Markoff spectrum.
Math. Ann., 305(1):191–203, 1996.

[31]
C. Series.
The geometry of Markoff numbers.
Math. Intelligencer, 7(3):20–29, 1985.

[32]
B. Springborn.
The hyperbolic geometry of Markov's theorem on Diophantine
approximation and quadratic forms.
Enseign. Math., 63(3-4):333–373, 2017.

[33]
D. Zagier.
On the number of Markoff numbers below a given bound.
Math. Comp., 39(160):709–723, 1982.

Boris Springborn<EMAIL_ADDRESS>
Technische Universität Berlin

Institut für Mathematik, MA 8-3

Str. des 17. Juni 136

10623 Berlin, Germany