title = From a Packing Problem to Quantitative Recurrence in $[0,1]$ and the
Lagrange Spectrum of Interval Exchanges, author = M. Boshernitzan, and V.
Delecroix, plaintextauthor = Michael Boshernitzan, and Vincent Delecroix,
plaintexttitle = From a Packing Problem to Quantitative Recurrence in [0,1]
and Lagrange spectrum of Interval Exchanges, runningtitle = From a Packing
Problem to Quantitative Recurrence, copyrightauthor = M. Boshernitzan, and V.
Delecroix, keywords = dynamical systems, diophantine approximation,
recurrence, interval exchange transformation, year=2017, number=10, received=2
December 2016, published=2 June 2017, doi=10.19086/da.1749,

[classification=text]

###### Abstract

In this work, we use use a solution to a packing problem in the plane to study
recurrence of maps on the interval [0,1]. First of all, we prove that
$1/\sqrt{5}$ is the optiaml recurrence rate of measurable applications of the
interval. Secondly, we analyze the bottom of the Lagrange spectrum of interval
exchange transformations.

##  1 Introduction

###  1.1 Recurrence rate and Lagrange constants

We study recurrence of maps of the unit interval that preserve Lebesgue
measure. Let $T\colon[0,1)\to[0,1)$ be a measurable map of the unit interval
into itself. The recurrence rate of $T$ at $x\in[0,1)$ is defined as

$r(T,x)=\liminf_{n\to\infty}\ n\,|T^{n}x-x|\in[0,\infty].$

Denote by $R_{\alpha}:[0,1)\to[0,1)$ the rotation by the angle
$\alpha\in\mathbb{R}$, that is $R_{\alpha}(x)=x+\alpha\pmod{1}$. Since the
group of such rotations acts transitively on $[0,1)$ by local isometries
(apart from at $x=0$), we note that
$r(R_{\alpha},x)=r(R_{\alpha},\frac{1}{2})$, for all $\alpha$ and $x\not=0$.
Set $r(R_{\alpha}):=r(R_{\alpha},\frac{1}{2})$. One verifies that, for every
$\alpha\in\mathbb{R}$,

$r(R_{\alpha})=\liminf_{q\to+\infty\atop q\in\mathbb{Z}}\ q\,\langle\\!\langle
q\alpha\rangle\\!\rangle,$

where $\langle\\!\langle t\rangle\\!\rangle=\min_{n\in\mathbb{Z}}\limits|t-n|$
stands for the distance of $t\in\mathbb{R}$ to the nearest integer.

We will provide two generalizations of the following well known result.

######  Theorem 1 (Hurwitz [Hu1891]).

For any real number $\alpha$ we have $r(R_{\alpha})\leq\frac{1}{\sqrt{5}}$.
Moreover, $r(R_{\alpha})=\frac{1}{\sqrt{5}}$ if and only if the continued
fraction expansion of $\alpha$ is eventually constant and equal to $1$. (In
particular, $r(R_{\beta})=\frac{1}{\sqrt{5}}$ for
$\beta=\frac{\sqrt{5}-1}{2}=[0;1,1,1,...]$). On the other hand, if
$r(R_{\alpha})\not=\frac{1}{\sqrt{5}}$ then
$r(R_{\alpha})\leq\frac{1}{2\sqrt{2}}$.

In other words, the inequality $r(R_{\alpha},x)\leq\frac{1}{\sqrt{5}}$ holds
for all rotations $R_{\alpha}\colon[0,1)\to[0,1)$ and $x\neq 0$. Note that
Hurwitz [Hu1891] also proved that the inequality $q\,\langle\\!\langle
q\alpha\rangle\\!\rangle<\frac{1}{\sqrt{5}}$ has infinitely many solutions in
integers $q\geq 1$.

_The Lagrange constant of_ $\alpha\in\mathbb{R}$ is defined as follows:

$L(\alpha)=\frac{1}{r(R_{\alpha})}=\limsup_{q\to\infty}\
\frac{1}{q\,\langle\\!\langle
q\alpha\rangle\\!\rangle}\in\big{[}\sqrt{5},\infty\big{]},$ (1)

and _the (classical) Lagrange spectrum_ is defined as the set of possible
finite Lagrange constants

$\mathcal{L}_{1}:=\big{\\{}L(\alpha):\,\alpha\in[0,1)\big{\\}}\setminus\\{+\infty\\}.$

An equivalent way of stating Hurwitz’s Theorem (Theorem 1) is

$\mathcal{L}_{1}\cap\big{[}0,2\sqrt{2}\big{)}=\big{\\{}\sqrt{5}\,\big{\\}}.$
(2)

###  1.2 Two generalizations of Hurwitz’s Theorem

The first proposed generalization is that the inequality
$r(T,x)\leq\frac{1}{\sqrt{5}}$ in Hurwitz’s Theorem actually holds almost
surely for all Lebesgue measure preserving transformations
$T\colon[0,1)\rightarrow[0,1)$ (and not just for the rotations
$T=R_{\alpha}$).

######  Theorem 2.

Let $T\colon[0,1)\rightarrow[0,1)$ be a measurable map of the unit interval
which preserves the Lebesgue measure $\mu$. Then,
$r(T,x)\leq\frac{1}{\sqrt{5}}$, for $\mu$-almost every $x\in[0,1)$.

The above theorem provides the optimal constant in the quantitative recurrence
result in [Bo93, Theorem 2.1] where, under the conditions of Theorem 2, a
weaker inequality was established (see Theorem 4 below).

Note that quantitative Poincaré recurrence results are possible in the more
general settings of transformations of arbitrary metric spaces having finite
Hausdorff dimension: see [Bo93] and the discussion following Theorem 4 below.

Our second generalization of Theorem 1 is related to interval exchange
transformations. An _interval exchange transformation_ (or _i.e.t._ for short)
is a bijection from an interval to itself that is a piecewise translation on
finitely many intervals. More precisely, given a permutation $\pi\in S_{k}$
and a vector $\lambda\in\mathbb{R}_{+}^{k}$, we define a $k$-i.e.t.
$T_{\pi,\lambda}$ as

$T_{\pi,\lambda}(x)=x-\sum_{j<i}\lambda_{j}+\sum_{\pi(j)<\pi(i)}\lambda_{j},\quad\text{
if \
$x\in[\lambda_{1}+\ldots+\lambda_{i-1},\lambda_{1}+\ldots+\lambda_{i})$}.$

Note that rotations are exactly the 2-i.e.t.s with permutations $\pi=(2,1)$.

The _singularities_ of $T$ are the points $x_{i}=\sum_{j\leq i}\lambda_{j}$
for $i=1,\dots,k-1$. An i.e.t. $T_{\pi,\lambda}$ is said to be _without
connection_ if there is no pair of singularities $x$ and $y$ of $T$ such that
$T^{n}x=y$ for some $n\geq 1$. It was shown by Keane [Ke75] that this
condition implies the minimality of the transformation $T$.

Given an i.e.t. $T=T_{\pi,\lambda}$ that satisfies the Keane condition, its
$n$-th iterate $T^{n}$ is also an i.e.t. but on $(k-1)n+1$ intervals. Let
$\varepsilon_{n}(T)$ be the smallest length of any of the intervals of
$T^{n}$. In particular $\varepsilon_{0}(T)=1$ and
$\varepsilon_{1}(T)=\min\lambda_{i}$. Note that the number
$\varepsilon_{n}(T)$ can alternatively be defined as the minimum distance
between the $n$-th first preimages of the singularities together with $0$ and
$1$.

For $T=T_{\pi,\lambda}$ we define

$\mathcal{E}(T):=\liminf_{n\to\infty}\,\frac{n\varepsilon_{n}(T)}{|\lambda|}\quad\text{where
}\ |\lambda|=\lambda_{1}+\ldots+\lambda_{k}.$

The value $L(T):=\mathcal{E}(T)^{-1}$ is called the _Lagrange constant_ of
$T$. It generalizes the Lagrange constant for the rotations $R_{\alpha}$. We
also recall that if $\pi\in S_{k}$ is _irreducible_ (or _indecomposable_) then
for Lebesgue-almost every $\lambda$ the Lagrange constant of $T_{\pi,\lambda}$
is infinite.

The Lagrange spectrum of i.e.t.s was introduced by S. Ferenczi in [Fe12] under
the name _lower Boshernitzan-Lagrange spectrum_. It was then further studied
by P. Hubert, L. Marchese and C. Ulcigrai in [HMU15]. The _Lagrange spectrum_
of the $k$-i.e.t.s is the following set of values:

$\mathcal{L}_{k-1}=\\{L(T)\colon\ \text{$T$ is a $k$-i.e.t. satisfying the
Keane condition and $L(T)<\infty$}\\}.$

Because $T^{n}$ is made of $(k-1)n+1$ intervals, $\inf\mathcal{L}_{k-1}\geq
k-1$. In [HMU15], the better bound $\inf\mathcal{L}_{k}\geq\frac{\pi}{2}k$ is
established. We prove the following.

######  Theorem 3.

There exists a constant $\varepsilon_{0}>0$ such that for any $d\geq 1$

$\mathcal{L}_{d}\cap\left[0,\
d\sqrt{5}+\frac{\varepsilon_{0}}{d}\right]=\left\\{d\sqrt{5}\right\\}.$

Moreover, for any permutation $\pi\in S_{d+1}$ such that $\pi(1)=d+1$, the
length vector $\lambda=(\frac{\sqrt{5}+1}{2},1,1,\ldots,1)$ is such that
$T_{\pi,\lambda}$ satisfies the Keane condition and
$L(T_{\pi,\lambda})=d\sqrt{5}$.

The proof of Theorem 3 will be given in Section 4. We will actually completely
characterize the $(d+1)$-i.e.t.s $T$ such that $L(T)=d\sqrt{5}$. Note that the
case $d=1$ is given by Hurwitz’s theorem (see in particular (2)) and that the
case $d=2$ was proven in [Fe12, Theorem 4.10].

The common tool in the proofs of Theorems 2 and 3 is Theorem 5 concerning an
unconventional packing problem in $\mathbb{R}^{2}$ (see Section 2 for precise
setting), which determines the relevant optimal constant $\frac{1}{\sqrt{5}}$.
Section 3 is dedicated to the proof of Theorem 2. And in Section 4 we provide
the proof for Theorem 3.

###  1.3 Further Comments

#### The Classical Lagrange spectrum

There are many results about the Lagrange spectrum $\mathcal{L}_{1}$ (of
rotations), including the following.

  1. 1.

$\mathcal{L}_{1}$ starts with a discrete sequence $\sqrt{5}$, $2\sqrt{2}$,
$\sqrt{221}/5$, …with an accumulation point at $3$ [Ma1879],

  2. 2.

$\mathcal{L}_{1}$ contains the half line $[c,+\infty)$ where
$c=\frac{2221564096+283748\sqrt{462}}{491993569}\simeq 4.528$, and this half
line is maximal (i.e. $\mathcal{L}_{1}$ does not contains a half-line
$[c^{\prime},+\infty)$ with $c^{\prime}<c$) [Ha47], [Fr75],

  3. 3.

$\operatorname{Hdim}(\mathcal{L}_{1}\cap[0,t])=0$ if and only if $t\leq 3$ and
$\operatorname{Hdim}(\mathcal{L}_{1}\cap[0,2\sqrt{3}])=1$ [Mo].

The interval exchange Lagrange spectrum $\mathcal{L}_{d}$ contains
$m\mathcal{L}_{d^{\prime}}$ if $d=md^{\prime}$. In particular
$\mathcal{L}_{d}$ contains $d\mathcal{L}_{1}$, from which some properties
follow (such as the existence of a half line). But nothing as precise as the
three above items for $\mathcal{L}_{1}$ is known in general for
$\mathcal{L}_{d}$.

#### Lebesgue-preserving maps on subsets in $\mathbb{R}^{n}$ of finite volume

Recall that quantitative Poincaré recurrence (almost everywhere) results are
possible in more general settings of transformations of arbitrary metric
spaces having finite Hausdorff dimension, see [Bo93]. In particular, the
following result holds.

######  Theorem 4 ([Bo93], Theorem 1.5).

Let $(X,d)$ be a metric space and let $\mu$ be a probability measure on it
which coincides, for some $\alpha\in(0,\infty)$, with the $\alpha$-Hausdorff
measure on $(X,d)$. Then, for any transformation $T\colon X\to X$ which
preserves the measure $\mu$, we have

$\liminf_{n\to\infty}n^{1/\alpha}d(x,T^{n}x)\leq 1,\qquad\text{for
$\mu$-almost every $x\in X$.}$

Now, denote by $\mu_{d}(\cdot)$ the Lebesgue measure on $\mathbb{R}^{d}$ where
$d\geq 1$. Let also $\rho$ be a norm on $\mathbb{R}^{d}$ and $B_{\rho}$ be the
unit ball for this norm. Let $X\subset\mathbb{R}^{d}$ be a measurable set of
finite non-zero measure and let $T\colon X\to X$ be a transformation which
preserves $\mu_{d}$. Then Theorem 4 implies that for almost every $x\in X$ we
have

$\liminf_{n\to\infty}\ n^{1/d}\rho({x}-T^{n}{x})\leq
2\left(\frac{\mu_{d}(X)}{\mu_{d}(B_{\rho})}\right)^{1/d}.$

In the particular case when $X=[0,1)^{d}$ is the unit cube, the above
inequality takes the form

$\liminf_{n\to\infty}\ n^{1/d}\rho({\bf x}-T^{n}{\bf x})\leq 2\
\mu_{d}(B_{\rho})^{-1/d}.$

In the even more special case where $X=[0,1)$ and $\rho$ is the absolute
value, we obtain a weaker version of Theorem 2 with the constant $1$ instead
of $\frac{1}{\sqrt{5}}$. The technique we use in this article to determine the
optimal constant $\frac{1}{\sqrt{5}}$ for $X=[0,1)$ does not seem to extend to
higher dimensions; not even for the square $X=[0,1)^{2}$.

#### Singular vectors and the Dirichlet spectrum

We have seen one extension of the Lagrange constant of 1-dimensional rotation
to interval exchange transformations. It can also be defined for higher-
dimensional rotations as follows. Given $\alpha\in\mathbb{R}^{d}$ we define,
similarly to (1),

$L(\alpha)=\limsup_{q\to\infty}\ \frac{1}{q^{1/d}\,\langle\\!\langle
q\alpha\rangle\\!\rangle}.$

where $\langle\\!\langle x\rangle\\!\rangle$ denotes the Euclidean distance to
the nearest integer lattice point. In both contexts, the Lagrange constant
also has a natural $\liminf$ counterpart that we discuss next.

Given $\alpha\in\mathbb{R}^{d}$ we define its _Dirichlet constant_ $D(\alpha)$
as

$D(\alpha)=\liminf_{n\to\infty}\frac{1}{n^{1/d}\varepsilon_{n}(\alpha)}\qquad\text{where}\qquad\varepsilon_{n}(\alpha)=\min_{1\leq
k\leq n}\langle\\!\langle k\alpha\rangle\\!\rangle$

Recall that a vector $\alpha$ is called _singular_ if $D(\alpha)=+\infty$
(such vector only exists if $d\not=1$). The set of singular vectors is known
to be of zero measure in any dimension, and its Hausdorff dimension has
recently been computed by N. Chevallier and Y. Cheung [CC16]. Recall that for
the Lagrange constant, a vector $\alpha$ such that $L(\alpha)<+\infty$ is
called _badly approximable_. For a Lebesgue generic $\alpha$ we have
$D(\alpha)=c_{d}$ and $L(\alpha)=+\infty$, where $0<c_{d}<+\infty$ is a
constant that only depends on the dimension.

Similarly, if $T$ is an i.e.t. that satisfies the Keane condition we define
its _Dirichlet constant_ as

$D(T)=\liminf_{n\to\infty}\frac{1}{n\varepsilon_{n}(T)}.$

The _Dirichlet spectrum_ 111 In [Fe12] the Dirichlet spectrum is called the
_upper Boshernitzan-Lagrange spectrum_. The reason for this is that M.
Boshernitzan proved that for i.e.t. the condition $D(T)<+\infty$ implies
unique ergodicity [Bo92]. is the set of possible Dirichlet constants for a
given class of systems (i.e., a dimension for rotations, or a number of
intervals for i.e.t.s, are fixed).

For rotations (or 2-i.e.t.s), the Dirichlet spectrum has a structure similar
to the Lagrange spectrum: that is, it starts with a discrete sequence and
contains an interval (see the discussion and references in the introduction of
[AS13]). But the situation changes dramatically when one goes to higher
dimensional situations. For instance, for both the 2-dimensional rotations
[AS13, Theorem 1] and 3-i.e.t.s [Fe12, Theorem 4.14] the Dirichlet spectrum is
an interval. Nothing seems to be known about the structure of Dirichlet
spectrum for rotations in $\mathbb{R}^{3}$ or 4-i.e.t.s.

##  2 An unconventional packing problem in $\mathbb{R}^{2}$

Denote by $\mathbb{C}$ the set of complex numbers. Given $z=x+iy$ we define
$\operatorname{\bf N}(z)=\sqrt{|xy|}$ (and $\operatorname{\bf
N}^{2}(z)=(\operatorname{\bf N}(z))^{2}=|xy|$). The function
$\operatorname{\bf N}$ can be thought as a generalization of a norm whose unit
ball is the region delimited by the hyperbolas $xy=\pm 1$. Unlike with a
genuine norm, the unit ball of $\operatorname{\bf N}$, namely
$\\{z\colon\operatorname{\bf N}(z)\leq 1\\}$, is not convex. However, it is
still star shaped, and satisfies $\operatorname{\bf
N}(tz)=|t|\operatorname{\bf N}(z)$ for all real $t$.

Given a polygon $P$ with vertices $z_{1}$, $z_{2}$, …, $z_{n}$, we define its
$\operatorname{\bf N}$-perimeter as $\displaystyle
p(P)=\sum_{i=1,\ldots,n}\limits\operatorname{\bf N}(z_{i+1}-z_{i})$ (where
indices are taken modulo $n$). Our main tool in the present paper is given by
the following result.

######  Theorem 5 ([Sm52]).

Let $\Gamma$ be a finite set of points in $\mathbb{C}$ such that $N(x-y)\geq
1$ for every pair of distinct points $(x,y)$ of $\Gamma$. Let $C$ be its
convex hull. Let $A$ and $p$ be respectively the area and $\operatorname{\bf
N}$-perimeter of $C$. Then

$\\#\Gamma\leq\frac{1}{\sqrt{5}}A+\frac{p}{2}+1.$

Moreover, if equality holds, then the set $\Gamma$ is a subset of a golden
lattice.

In the case of norms (i.e., when the unit ball is convex) the above result was
a conjecture of H. Zassenhaus, which was proven in full generality by N. Oler
[Ol61].

######  Theorem 6 ([Ol61]).

Let $\rho$ be a norm in $\mathbb{R}^{2}$ and let $\Gamma$ be a finite set of
points such that $\rho(x-y)\geq 1$ for all pairs $(x,y)$ of points of
$\Gamma$. Let $C$ be the convex hull of $\Gamma$, $A$ the area of $C$ and $p$
the $\rho$-perimeter of $C$. Then

$\\#\Gamma\leq\Delta(\rho)A+\frac{p}{2}+1$

where $\Delta(\rho)$ is the critical determinant of the unit ball of $\rho$.

The _critical determinant_ $\Delta(C)$ of a centrally symmetric convex body
$C$ in $\mathbb{C}$ is defined as follows. A lattice
$\Lambda\subset\mathbb{C}$ is said to be $C$-admissible if
$C\cap\Lambda=\\{0\\}$. Then

$\Delta(C):=\min\\{\det(\Lambda):\ \text{$\Lambda$ is $C$-admissible}\\}.$

In Theorem 5 the constant $\frac{1}{\sqrt{5}}$ is also a critical determinant
(for a star but non-convex body). This constant is achieved exactly by the
golden lattices. These two facts are just a reformulation of Hurwitz Theorem
(Theorem 1).

The rest of this section is devoted to the proof of Theorem 5. A complete
proof is given in the PhD thesis of N. E. Smith (a student of H. Zassenhaus),
see [Sm52]. Our proof uses the same path except that a delicate induction is
avoided by using Delaunay triangulations.

######  Remark 7.

As pointed out in the review paper [Za61] a weaker version of Theorem 5 can be
derived from Theorem 6 as shown in the PhD thesis of Sr. M. R. von Wolff
[Wo61]. Namely, we always have $N(z)\leq\rho(z)$ where $\rho(z)=(|x|+|y|)/2$.
Hence if $N(z)\geq 1$ then a fortiori $\rho(z)\geq 1$ and Oler’s result
applies. Luckily the critical determinants are the same for $N$ and $\rho$
equal to $1/\sqrt{5}$, though in the error term the $\rho$-perimeter is
generally larger than the $N$-perimeter. Note that this weaker result would
have been enough for our applications but we prefer to include a self-
contained and short proof of Theorem 5.

######  Remark 8.

It would be tempting to conjecture that Oler’s result actually holds for
centrally symmetric bodies. But this is actually false. Sr. M. R. von Wolff
provided a counterexample in her PhD thesis [Wo61].

###  2.1 Admissible triangles

Given a triangle $(p,q,r)$ in $\mathbb{C}$, it is always inscribed in a
smallest rectangle, namely the rectangle
$R(p,q,r)=[x^{-},x^{+}]\times[y^{-},y^{+}]$ defined by

$\begin{array}[]{l@{\qquad}l}x^{-}=\min(\operatorname{Re}(p),\operatorname{Re}(q),\operatorname{Re}(r))&x^{+}=\max(\operatorname{Re}(p),\operatorname{Re}(q),\operatorname{Re}(r))\\\
y^{-}=\min(\operatorname{Im}(p),\operatorname{Im}(q),\operatorname{Im}(r))&y^{+}=\max(\operatorname{Im}(p),\operatorname{Im}(q),\operatorname{Im}(r)).\end{array}$

######  Definition 9.

We call a triangle $(p,q,r)$ in $\mathbb{C}$ _admissible_ if the three points
$p,q,r$ are on the boundary of the minimal rectangle $R(p,q,r)$ and no two of
them are on the same side.

######  Remark 10.

One can alternatively define admissible triangles as triangles for which the
sign of the slopes of the sides are not all the same. This is the definition
proposed in [Sm52] on page 7 in which admissible triangles are called _type
(a)_.

Let $(p,q,r)$ be an admissible triangle. On the rectangle $R(p,q,r)$ exactly
one vertex is in a corner. By convention we always label the sides $a,b,c$ so
that the two sides $a,b$ are adjacent to that corner and $a$, $b$, $c$ are
taken in counter-clockwise order. (See Figure 1 above.)

Figure 1: Two admissible triangles and one non-admissible triangle.

Recall that $\operatorname{\bf N}\colon\mathbb{C}\to\mathbb{R}$ is the
function defined by $\operatorname{\bf N}(x+iy)=\sqrt{|xy|}$. The main
ingredient of the proof of Theorem 5 is the following lemma which establishes
the fact that the area of an admissible triangle is completely determined by
the $\operatorname{\bf N}$-length of its sides.

######  Lemma 11.

Let $a,b,c$ be three sides of an admissible triangle $\Delta$ and let

$\alpha:=\operatorname{\bf N}(a)^{2},\quad\beta:=\operatorname{\bf
N}(b)^{2},\quad\gamma:=\operatorname{\bf N}(c)^{2}.$

Then

$\operatorname{\bf
area}(\Delta)=\frac{\sqrt{\alpha^{2}+\beta^{2}+\gamma^{2}-2\alpha\beta+2\alpha\gamma+2\beta\gamma}}{2}.$
(3)

###### Proof.

Applying $g_{t}$ and an homothety of $1/\sqrt{\gamma}$ we can assume (up to
symmetry) that $c=(1,-1)$ as shown in the picture below.

Then we have $\alpha=x(y+1)$, $\beta=y(x+1)$, $\gamma=1$ while
$\operatorname{\bf area}(\Delta)=\frac{1+x+y}{2}$, and the validation of the
formula (3) becomes straightforward. ∎

######  Corollary 12.

Let $\Delta$ be an admissible triangle with $a$, $b$, $c$, $\alpha$, $\beta$,
$\gamma$ as in Lemma 11. Let $m=\min\\{\alpha,\beta,\gamma\\}$ and
$M=\max\\{\alpha,\beta,\gamma\\}$. Then

$\frac{\sqrt{5}}{2}\ m\leq\operatorname{\bf
area}(\Delta)\leq\frac{\sqrt{m^{2}+4M^{2}}}{2}\leq\frac{\sqrt{5}}{2}\ M.$

If moreover $\operatorname{\bf area}(\Delta)\leq\sqrt{2}\ m$ then

$\frac{\sqrt{M^{2}+4m^{2}}}{2}\leq\operatorname{\bf area}(\Delta).$

###### Proof.

Let us set $m=\min(\alpha,\beta,\gamma)$ and $M=\max(\alpha,\beta,\gamma)$. By
symmetry we can assume that $\alpha\geq\beta$. Let
$f(\alpha,\beta,\gamma)=\alpha^{2}+\beta^{2}+\gamma^{2}-2\alpha\beta+2\alpha\gamma+2\beta\gamma$.
From Lemma 11 we have $\operatorname{\bf
area}(\Delta)=\sqrt{f(\alpha,\beta,\gamma)}/2$.

As $\alpha^{2}+\beta^{2}\geq 2\alpha\beta$, we have

$f(\alpha,\beta,\gamma)\geq\gamma^{2}+2\alpha\gamma+2\beta\gamma\geq 5m^{2}$

which proves the lower bound. For the upper bound, if $M=\alpha$ then we can
use the fact that $-2\alpha\beta+2\beta\gamma\leq 0$. If $M=\gamma$ then we
use

$f(\alpha,\beta,\gamma)=(\alpha-\beta)^{2}+\gamma^{2}+2\alpha\gamma+2\beta\gamma\leq(\gamma-\beta)^{2}+\gamma^{2}+2\alpha\gamma+2\beta\gamma\leq
4M^{2}+m^{2}.$

To prove the last statement we analyze the function
$f(\alpha,\beta,\gamma)=\alpha^{2}+\beta^{2}+\gamma^{2}-2\alpha\beta+2\alpha\gamma+2\beta\gamma$.
For each possibility of maximum and minimum, we just analyze $f$ as a one-
variable function. The values of the extrema can be computed by elementary
calculus. We summarize this information in the following array.

$\begin{array}[]{l|cccc}\text{order on
$\alpha,\beta,\gamma$}&\operatorname{argmin}(f)&\min(f)&\operatorname{argmax}(f)&\max(f)\\\
\hline\cr m=\beta\ \text{and}\
M=\gamma&\alpha=m&4Mm+M^{2}&\alpha=M&m^{2}+4M^{2}\\\ \hline\cr m=\beta\
\text{and}\ M=\alpha&\gamma=m&M^{2}+4m^{2}&\gamma=M&4M^{2}+m^{2}\\\ \hline\cr
m=\gamma\ \text{and}\ M=\alpha&&&&\\\ \alpha/\gamma\leq
2&\beta=m&M^{2}+4m^{2}&\beta=M&m^{2}+4Mm\\\ 2\leq\alpha/\gamma\leq
3&\beta=M-m&4Mm&\beta=M&m^{2}+4Mm\\\
3\leq\alpha/\gamma&\beta=M-m&4Mm&\beta=m&M^{2}+4m^{2}\\\ \hline\cr\end{array}$

In the two first cases $m=\beta,M=\gamma$ or $m=\beta,M=\alpha$ we have the
lower bound $f(\alpha,\beta,\gamma)\geq 4m^{2}+M^{2}$. In the case
$m=\gamma,M=\alpha$, the condition $\operatorname{\bf
area}(\Delta)\leq\sqrt{2}\ m$ implies that $\alpha/\gamma\leq 2$. Indeed, if
we had $M/m>2$ then $\operatorname{\bf area}{\Delta}\geq\sqrt{4Mm}/2>\sqrt{2}\
m$. And in the case $\alpha/\gamma\leq 2$, the lower bound $M^{2}+4m^{2}$ is
valid. ∎

Note that the gap between $\sqrt{5}/2\simeq 1.118$ and $\sqrt{2}\simeq 1.4142$
is not large. But having this gap is essential as it will allow us to get
lower bounds from upper bounds in Section 4 via the following lemma.

######  Corollary 13.

Let $\Delta$ be an admissible triangle with $a$, $b$, $c$, $\alpha$, $\beta$,
$\gamma$ as in Lemma 11. Let $m=\min\\{\alpha,\beta,\gamma\\}$ and
$M=\max\\{\alpha,\beta,\gamma\\}$. Assume that $\operatorname{\bf
area}(\Delta)\leq\left(\frac{\sqrt{5}}{2}+\varepsilon\right)m$, for some
$\varepsilon$, $0<\varepsilon<\sqrt{2}-\frac{\sqrt{5}}{2}$. Then
$M\leq(1+(2\sqrt{2}+\sqrt{5})\varepsilon)m$.

###### Proof.

Because $\varepsilon<\sqrt{2}-\frac{\sqrt{5}}{2}$ the second half of Corollary
12 holds: $M^{2}+4m^{2}\leq 4\left(\operatorname{\bf
area}(\Delta)\right)^{2}$. Using the hypothesis, we get $M^{2}+4m^{2}\leq
4(\frac{\sqrt{5}}{2}+\varepsilon)^{2}m^{2}$ and hence
$M^{2}\leq(1+4\sqrt{5}\varepsilon+4\varepsilon^{2})m^{2}<(1+4(\sqrt{2}+\frac{\sqrt{5}}{2})\varepsilon)m^{2}$.
Taking square roots in this last inequality and applying the inequality
$\sqrt{1+x}\leq 1+x/2$, which is valid for all $x>0$, we get the result. ∎

###  2.2 $L^{\infty}$-Delaunay triangulations

For a general reference on Delaunay triangulations we refer the reader to
[OBSN2000].

Figure 2: An $L^{\infty}$-Delaunay triangulation with one of the maximal
squares $S$.

Let $\Gamma\subset\mathbb{C}$ be a finite set of points. A _triangulation_ of
$\Gamma$ is a set of triangles with disjoint interiors whose vertex set is
contained in $\Gamma$. Note that we have no maximality assumption here. The
$L^{\infty}$-_Delaunay triangulation_ of $\Gamma$ is defined as follows: a
triangle with vertices $p,q,r\in\Gamma$ belongs to that triangulation if and
only if there exists a square $S\subset\mathbb{C}$ with horizontal and
vertical sides such that $S\cap\Gamma=(\partial S)\cap\Gamma=\\{p,q,r\\}$. An
example of a Delaunay triangulation is provided in Figure 2.

In some cases, there might be more than three points on the boundary of a
square. We will implicitly exclude the case where two points $z$ and
$z^{\prime}$ of $\Gamma$ are on the same horizontal or vertical line, as these
correspond to $\operatorname{\bf N}(z-z^{\prime})=0$. Assuming that
$\displaystyle\min_{\stackrel{{\scriptstyle
z,z^{\prime}\in\Gamma}}{{z\not=z^{\prime}}}}\operatorname{\bf N}(\Gamma)>0$,
there are either three or four points on maximal squares. In the latter case,
there is an ambiguity as there are two different ways of making two triangles
out of these four points. We will abuse the terminology and still speak about
_the_ Delaunay triangulation for one of the triangulations obtained after
making a choice in each quadruple of points in a maximal square.

######  Lemma 14.

Let $\Gamma\subset\mathbb{C}$ be a finite set that contains at least three
points and is such that no pair of points are on the same horizontal or
vertical line. Let:

  * (a)

$C$ be the convex hull of $\Gamma$;

  * (b)

$T$ be the finite collection of closed triangles determined by the
$L^{\infty}$-Delaunay triangulation of $\Gamma$ (the interiors of these
triangles are disjoint);

  * (c)

$U=\bigcup_{\Delta\in T}\Delta$ be the union of all these triangles.

Then the following statements hold.

  * 1.

The $L^{\infty}$-Delaunay triangulation $T$ contains only admissible
triangles.

  * 2.

The set $U$ is simply connected.

  * 3.

The $\operatorname{\bf N}$-length of $\partial U$ is smaller than the
$\operatorname{\bf N}$-length of $\partial C$ (where $\partial U$ and
$\partial C$ stand for the boundaries of $U$ and $C$, respectively).

###### Proof.

The first statement is immediate from the definition. Indeed, each triangle of
$T$ is inscribed in a square with its three vertices on the sides (by
definition of the Delaunay triangulation) and since no pair of points of
$\Gamma$ are on the same horizontal or vertical line, they belong to different
sides.

The segments $[p,q]$ on the boundary $\partial U$ of $U$ are the ones such
that there exist arbitrary large squares $S$ with $S\cap\Gamma=\\{p,q\\}$.
Such a segment needs to be on the boundary.

For the third statement, let $\gamma=[p,q]$ be an edge of the convex hull $C$
of $\Gamma$ that is not an edge of a triangle in $T$. (Thus $\gamma\in\partial
C$ but $\gamma\notin\partial U$). The line through $\gamma$ separates the
plane into two regions, and one of them contains all points of $\Gamma$ except
$p$ and $q$. Without loss of generality we assume that
$\operatorname{Re}(p)<\operatorname{Re}(q)$,
$\operatorname{Im}(p)<\operatorname{Im}(q)$ and that the points of $\Gamma$
are above the line through $p$ and $q$ as in the following picture.

The segment $\gamma$ is not an edge of a triangle in $T$ if and only if there
are points in the square $S$ which admits $\gamma$ as a chord.

Let $\Gamma_{0}$ be the set of points in the interior of $S$ that are
different from $p$ and $q$. Let $p_{1}$ be the point in $\Gamma_{0}$ with
lowest imaginary part. Now we proceed inductively until $\Gamma_{n}$ by
defining

$\Gamma_{n}:=\\{x\in\Gamma_{n-1}\mid\operatorname{Re}(x)>\operatorname{Re}(p_{n})\quad\text{and}\quad\operatorname{Im}(x)>\operatorname{Im}(p_{n})\\}.$

and, if $\Gamma_{n}\neq\emptyset$, we continue with picking
$p_{n+1}\in\Gamma_{n}$ with the lowest imaginary part.

Let $p_{1},\ldots,p_{n}$ be the points selected in the above way when the
process stops (i.e., when $\Gamma_{n}=\emptyset$). By adding two more points
$p_{0}=p$ and $p_{n+1}=q$, we end up with the $(n+2)$ points
$p_{0}=p,p_{1},\ldots,p_{n},p_{n+1}=q$, with the edges $[p_{k},p_{k+1}]$,
$0\leq k\leq n$, forming the contour $\phi_{p,q}$ of $\partial U$ between $p$
and $q$.

Now, we claim that $\operatorname{\bf
N}(q-p)\geq\sum_{i=0}^{n}\operatorname{\bf N}(p_{i+1}-p_{i})$. This is to say
that the triangle inequality is actually reversed! It follows from the fact
that $\operatorname{\bf N}$ restricted to the positive quadrant is concave.

This completes the proof of Lemma 14. ∎

###  2.3 Proof of Theorem 5

Let $\Gamma\subset C$ be a finite set of cardinality $s$ and $C$ its convex
hull. Let $T$ be the $L^{\infty}$-Delaunay triangulation of $\Gamma$ and let
$U$ be the union of the closed triangles in $T$ (notations just as in Lemma
14). Next, we establish a lower bound (see (4)) on the number $n$ of triangles
in $T$.

The set $\Gamma$ can be partitioned into the three subsets
$\Gamma=\Gamma_{2}\cup\Gamma_{bad}\cup\Gamma_{good}$ as follows:

  1. 1.

The set $\Gamma_{2}$ of $2$ _special_ points that lie on the extreme left and
extreme right of $\Gamma$,

  2. 2.

The set $\Gamma_{bad}=(\Gamma\cap\partial U)\setminus\Gamma_{2}$ of $s_{bad}$
points that lie on $\partial U$ but not in $\Gamma_{2}$,

  3. 3.

The set $\Gamma_{good}=\Gamma\setminus(\Gamma_{2}\cup\Gamma_{bad})$ of
remaining $s_{good}=s-2-s_{bad}$ points that lie in the interior of $U$.

Since the $\operatorname{\bf N}$-distance between any two points of $\Gamma$
is at least 1 we have that $s_{bad}+2$ is smaller than the $\operatorname{\bf
N}$-perimeter of $U$. But from Lemma 14 we know that the $\operatorname{\bf
N}$-perimeter of $U$ is actually smaller than that of $C$. Hence
$s_{bad}+2\leq p$.

Next, with each triangle $\Delta$ of $T$, we associate a point in $\Gamma$ as
follows. There is exactly one vertex of $\Delta$ for which the vertical line
through that point intersects the interior of the triangle as in the following
pictures:

It is easy to see that

  1. 1.

for the (two) points in $\Gamma_{2}$, there are no associated triangles,

  2. 2.

for each of the $s_{bad}$ points in $\Gamma_{bad}$ there is exactly one
associated triangle,

  3. 3.

for each of the $s_{good}$ points in $\Gamma_{good}$ there are exactly two
associated triangles.

In other words, the number of triangles in $T$ is given by the formula
$n=2s_{good}+s_{bad}$. Now, substituting $s_{good}=s-2-s_{bad}$ and taking
into account the inequality $s_{bad}+2\leq p$ we obtain

$n=2s_{good}+s_{bad}=2s-s_{bad}-4\geq 2s-p-2$ (4)

By Corollary 12, each of the triangles in $T$ has area at least
$\frac{\sqrt{5}}{2}$. So

$A\geq\frac{\sqrt{5}}{2}\,n\geq\frac{\sqrt{5}}{2}\left(2s-p-2\right),$

and the inequality claimed in Theorem 5 follows.

###  2.4 The rectangular case

The special case where $C\subset\mathbb{R}^{2}=\mathbb{C}$ is a rectangle with
sides parallel to the coordinate axes is addressed by the following result. It
is in this form that our packing result will be used in Section 3.

######  Theorem 15.

Let $C=[x^{-},x^{+}]\times[y^{-},y^{+}]\subset\mathbb{R}^{2}$ be a rectangle
of area $A=(x^{+}-x^{-})(y^{+}-y^{-})$, and let $\Gamma\subset C$ be a finite
subset of cardinality $s\geq 2$. Set $\delta=\min_{\begin{subarray}{c}x,y\in
C\\\ x\neq y\end{subarray}}\limits\,\operatorname{\bf N}(x-y)$ and
$A^{\prime}=A/\delta^{2}$. Then

$s\leq\frac{A^{\prime}}{\sqrt{5}}+\sqrt{2A^{\prime}}+1.$ (5)

In particular, for a given $\varepsilon>0$, we have

$s\leq\left(\frac{1}{\sqrt{5}}+\varepsilon\right)\cdot A^{\prime}$ (6)

provided that either $A^{\prime}$ or $s$ are large enough. More precisely, for
$0<\varepsilon<1/2$, each of the following two conditions

$\text{either \quad{\bf(c1)}}\ A^{\prime}>\tfrac{4}{\varepsilon^{2}},\hskip
34.1433pt\text{or \quad{\bf(c2)}}\ s>(\tfrac{3}{\varepsilon}+1)^{2},$ (7)

suffices for the inequality (6) to hold.

###### Proof.

We show how to deduce Theorem 15 from Theorem 5. We adopt the notation used in
these two results.

By Theorem 5, we have $s\leq\frac{A}{\delta^{2}\sqrt{5}}+\frac{p}{2\delta}+1$
where $p$ is the $\operatorname{\bf N}$-perimeter of the convex hull of
$\Gamma$. As all quantities in the above inequality are invariant by the
linear action of the diagonal flow $x+iy\mapsto e^{t}x+ie^{-t}y$ we can assume
that $C$ is a square with side length $\sqrt{A}$. The $\operatorname{\bf
N}$-perimeter is always smaller than $\sqrt{2}/2$ times the euclidean
perimeter (since the euclidean ball of radius $\sqrt{2}$ is contained in the
$N$-ball of radius $1$). Moreover, if $C_{1}\subset C_{2}$ are two convexes,
it is well known that the euclidean perimeter of $C_{1}$ is smaller than the
one of $C_{2}$. Hence $p\leq 2\sqrt{2}\,\sqrt{A}$. The equation (5) follows.

It remains to check the inequality $\frac{\sqrt{2A}+1}{A}<\varepsilon$,
assuming that $0<\varepsilon<1$ and that at least one of the conditions in
(7), either (c1) or (c2), holds.

The condition (c1), $A>\frac{4}{\varepsilon^{2}}$, implies that
$\frac{1}{\sqrt{A}}<\frac{1}{2}$, and then
$\frac{\sqrt{2A}+1}{A}<\frac{2}{\sqrt{A}}<\varepsilon$.

The condition (c2), $s>(\frac{2}{\varepsilon}+1)^{2}$, implies that $s\geq
25$, and then (5) implies that $s\leq(\sqrt{A}+1)^{2}$. We obtain
$\sqrt{A}\geq\sqrt{s}-1>\frac{2}{\varepsilon}$, and then
$A>\frac{4}{\varepsilon^{2}}$. Thus (c2) implies (c1), the case already
established.

This completes the deduction of Theorem 15 from Theorem 5. ∎

##  3 Recurrence in the interval

This section is dedicated to the proof of Theorem 2. In the first part we
prove a technical step involving an estimation of the measure of points with a
lower bound on the rate of recurrence. This proposition uses the
unconventional packing result of Theorem 5 (in its form given in Theorem 15)
and basic ergodic theory. In a second part we derive Theorem 2 using the
Lebesgue density theorem.

###  3.1 An estimate for the measure of badly recurrent points

The following estimate is used in the proof of Theorem 2.

######  Proposition 16.

Let $T\colon[0,1)\rightarrow[0,1)$ be a measurable map which preserves the
Lebesgue measure $\mu$.
Let $V\subset[0,1)$ be a non-empty open subinterval. Set

$\rho(x,V)=\inf_{\begin{subarray}{c}n\geq 1\\\\[1.13809pt] T^{n}x\in
V\end{subarray}}\ n\cdot|T^{n}(x)-x|\quad\text{\em(for }\ x\in V),$

and for $r>0$ define the subsets $V_{r}=\\{x\in V\mid\rho(x,V)\geq r\\}\subset
V$. Then $\displaystyle\mu(V_{r})\leq\frac{\mu(V)}{r\sqrt{5}}$.

The above proposition is trivial for $r\leq\frac{1}{\sqrt{5}}$ (because then
it follows immediately from the inclusion $V_{r}\subset V$). Note also that
one can recover the recurrence rate $r(T,x)$ from $\rho(x,V)$ as

$r(T,x)=\liminf_{\varepsilon_{1},\varepsilon_{2}\to
0}\rho(x,(x-\varepsilon_{1},x+\varepsilon_{2})).$

###### Proof.

Let $S\colon V\to V$ be the first return map on $V$ induced by $T$. Thus

$S(j)=T^{F(x)}(x)\quad\text{(for a.a. $x\in V$)},$

where $F(x)$ is the minimal integer $n\geq 1$ such that $T^{n}(x)\in V$. By
Kac’s lemma, the function $F\colon V\to\mathbb{N}$ is defined almost
everywhere and $\int_{V}F(x)\,dx\leq 1$. For $n\geq 0$, we have

$S^{n}(x)=T^{F_{n}(x)}(x),\quad\text{where }\
F_{n}(x)=\sum_{k=0}^{n-1}F(S^{k}x)\quad\text{(for a.a. $x\in V$}).$

By the Birkhoff Ergodic Theorem, the pointwise convergence

$\lim_{n\to\infty}\frac{F_{n}(x)}{n}=G(x)\quad\text{(for a.a. $x\in V$)},$ (8)

to some integrable function $G(x)\geq 1$ takes place, and

$\int_{V}G(x)\,dx=\int_{V}F(x)\,dx\leq 1.$ (9)

Denote by $V^{\prime}$ the subset of those $x\in V$ for which all the values
$S^{n}(x)$, $F_{n}(x)$ for all $n\geq 0$ and $G(x)$ are defined and (8) holds.
Clearly, $\mu(V^{\prime})=\mu(V)$.

Fix $x\in V^{\prime}$ and let $\\{z_{n}\\}_{n\geq 0}=\\{x_{n}+iy_{n}\\}_{n\geq
0}$ be the sequence in $\mathbb{C}$ defined by the formula

$\displaystyle x_{n}$
$\displaystyle=\operatorname{Re}(z_{n})=F_{n}(x)\in\mathbb{N}\cup\\{0\\},$
$\displaystyle y_{n}$
$\displaystyle=\operatorname{Im}(z_{n})=T^{x_{n}}(x)=T^{F_{n}(x)}(x)=S^{n}(x)\in
V.$

(Thus $\\{x_{n}\\}$ is a strictly increasing sequence of integers, with
$x_{0}=0$, $z_{0}=it$). For an integer $q\geq 1$, let

$\displaystyle\Gamma_{q}=\Gamma_{q}(x)$ $\displaystyle=\\{z_{n}\mid
n\in[0,q-1]\\}=\\{z_{0},z_{1},\ldots,z_{q-1}\\},$
$\displaystyle\Gamma^{\prime}_{q}=\Gamma^{\prime}_{q}(x)$
$\displaystyle=\big{\\{}z_{n}\mid n\in[0,q-1],\ \text{\small and
}\operatorname{\bf N}^{2}(z_{n}-z_{m})\geq r,\ \forall
m\in[n+1,q-1]\\}\big{\\}}\subset\Gamma_{q},$
$\displaystyle\Gamma^{\prime\prime}_{q}=\Gamma^{\prime\prime}_{q}(x)$
$\displaystyle=\Gamma_{q}\setminus\Gamma^{\prime}_{q}=\\{z_{n}\mid
n\in[0,q-2],\ \text{\small and }\exists m\in[n+1,q-1]\ \text{\small\ such that
}\operatorname{\bf N}^{2}(z_{n}-z_{m})<r\\}.$

Let also $C_{q}(x)=[0,x_{q}]\times V$ and $A_{q}(x)=\operatorname{\bf
area}(C_{q}(x))=\mu(V)\,x_{q}$. Given $\Gamma\subset\mathbb{C}$ as the one
defined above, we set

$\delta(\Gamma)=\min_{z,z^{\prime}\in\Gamma}\operatorname{\bf
N}(z-z^{\prime}).$ (10)

Next we establish the following inequality:

$M_{x}\colon\\!=\limsup_{q\to\infty}\limits\frac{|\Gamma^{\prime}_{q}(x)|}{q}\leq\frac{\mu(V)\,G(x)}{r\sqrt{5}}\quad\
\text{(for $x\in V^{\prime}$).}$ (11)

Let us fix $x\in V^{\prime}$ and chose an increasing sequence of positive
integers $\\{q_{n}\\}_{n\geq 0}$ so that

$M_{x}=\lim_{n\to\infty}\limits\frac{|\Gamma^{\prime}_{q_{n}}|}{q_{n}}.$ (12)

Let us also fix $\varepsilon>0$. Then the inequality

$|\Gamma^{\prime}_{q_{n}}|\leq\frac{A_{q_{n}}}{\delta^{2}(\Gamma^{\prime}_{q_{n}})}\left(\frac{1}{\sqrt{5}}+\varepsilon\right)$

holds for all large $n$, in view of (6) in Theorem 15 and where $\delta$ is
defined by (10). (Note that $\Gamma^{\prime}_{q}\subset C_{q}$ and
$\lim_{n\to\infty}\limits|\Gamma^{\prime}_{q_{n}}|=\infty$ because otherwise
$M_{x}=0$ and (11) becomes trivial).

Since $\delta^{2}(\Gamma^{\prime}_{q})\geq r$ (in view of the definition of
$\Gamma^{\prime}_{q}$) and $\varepsilon>0$ is arbitrary, we get
$\limsup_{n\to\infty}\limits\
\frac{|\Gamma^{\prime}_{q_{n}}|}{A_{q_{n}}}\leq\frac{1}{r\sqrt{5}}$ and hence
$\limsup_{n\to\infty}\limits\frac{|\Gamma^{\prime}_{q_{n}}|}{x_{q_{n}}}\leq\frac{\mu(V)}{r\sqrt{5}}$
(as $A_{q}=\mu(V)x_{q}$). Taking in account (12) and that
$\lim_{q\to\infty}\limits\frac{x_{q}}{q}=\lim_{q\to\infty}\limits\frac{F_{q}(x)}{q}=G(x)$,
we obtain
$M_{x}=\limsup_{q\to\infty}\limits\frac{|\Gamma^{\prime}_{q_{n}}|}{q_{n}}\leq\frac{\mu(V)\,G(x)}{r\sqrt{5}},$
and the inequality (11) follows.

Next we observe the inclusions
$V_{r}\cap\Gamma_{q}(x)\subset\Gamma^{\prime}_{q}(x)$ (see the definition of
$V_{r}$ in Proposition 16) and conclude that

$\sum_{n=0}^{q-1}\limits
1_{V_{r}}(S^{n}(x))=|V_{r}\cap\Gamma_{q}(x)|\leq|\Gamma^{\prime}_{q}(x)|\quad\
\text{(for $q\geq 2$)}$ (13)

where $1_{V_{r}}$ stands for the characteristic function of the set $V_{r}$.
Since the map $S\colon V^{\prime}\to V^{\prime}$ is measure preserving,
integrating (13) results in the inequality

$q\mu(V_{r})=\int_{V^{\prime}}\Big{(}\sum_{n=0}^{q-1}1_{V_{r}}(S^{n}(x))\Big{)}\,dx\leq\int_{V^{\prime}}|\Gamma^{\prime}_{q}(x)|\,dx,$

whence

$\mu(V_{r})\leq\int_{V^{\prime}}\frac{|\Gamma^{\prime}_{q}(x)|}{q}\,dx\leq\int_{V^{\prime}}\Big{(}\sup_{p\geq
q}\frac{|\Gamma^{\prime}_{p}(x)|}{p}\Big{)}\,dx.$

Passing to the limit $q\to\infty$, we get

$\mu(V_{r})\leq\int_{V^{\prime}}\\!\Big{(}\limsup_{q\to\infty}\frac{|\Gamma^{\prime}_{q}(x)|}{q}\Big{)}\,dx\leq\int_{V^{\prime}}\frac{\mu(V)\,G(x)}{r\sqrt{5}}\,dx\leq\frac{\mu(V)}{r\sqrt{5}},$

in view of (9) and (11). This completes the proof of Proposition 16. ∎

###  3.2 Derivation of Theorem 2 (from Proposition 16)

###### Proof of Theorem 2.

Let $r>\frac{1}{\sqrt{5}}$ be given. Let $\Phi_{N}$ be the finite collections
of subintervals of $[0,1]$ defined as below

$\Phi_{N}=\big{\\{}[\tfrac{n}{N},\tfrac{n+1}{N}]\mid 0\leq n\leq
N-1\big{\\}}\quad(\text{for }N\geq 2).$

Each $\Phi_{N}$ partitions the unit interval $[0,1]$ into $N$ subintervals of
equal lengths $\tfrac{1}{N}$ (up to their boundaries).

Recall that for every subinterval $V\subset[0,1]$, the inequality
$\mu(V_{r})\leq\frac{\mu(V)}{r\sqrt{5}}$ holds (by Proposition 16). In
particular,

$\mu(V_{r})\leq\tfrac{1}{Nr\sqrt{5}},\quad\text{for any }V\in\Phi_{N}.$

Next we introduce two sequences of sets

$W_{N}=\\!\\!\bigcup_{V\in\Phi_{N}}\limits\\!\\!V_{r};\qquad
W^{\prime}_{N}=\bigcap_{n\geq N}\limits\\!W_{n}\qquad\text{(for }\ N\geq 2),$

and two additional sets

$\displaystyle W=$ $\displaystyle\bigcup_{N\geq
2}W^{\prime}_{N}=\liminf_{N\to\infty}\ W_{N}=\big{\\{}x\in[0,1]\ \big{|}\ x\in
W_{N},\ \text{\small for all $N$ large enough}\big{\\}},$ (14) $\displaystyle
U=$ $\displaystyle\ [0,1]\\!\setminus\\!W=\\{x\in[0,1]\mid x\notin W_{N},\
\text{\small for infinitely many $N$}\big{\\}}.$ (15)

Every subinterval $J\subset[0,1]$ can be covered by $[\mu(J)N]+2$ subintervals
from the collection $\Phi_{N}$, so

$\limsup_{N\to\infty}\ \mu(W_{N}\cap
J)\leq\limsup_{N\to\infty}\big{(}\tfrac{[\mu(J)N]+2}{Nr\sqrt{5}}\big{)}=\tfrac{1}{r\sqrt{5}}\cdot\mu(J),$

and hence, for every $N\geq 2$,

$\mu(W^{\prime}_{N}\cap J)\leq\liminf_{n\to\infty}\,\mu(W_{n}\cap
J)\leq\tfrac{1}{r\sqrt{5}}\cdot\mu(J).$

Since $J\subset[0,1]$ is arbitrary and $\tfrac{1}{d\sqrt{5}}<1$, the sets
$W^{\prime}_{N}$ cannot have Lebesgue density points, thus
$\mu(W^{\prime}_{N})=0$, and hence $\mu(W)=0$. It follows that
$\mu(U)=\mu([0,1]\\!\setminus\\!W)=1$.

Next fix $x\in U$ and set $H_{x}=\\{n\geq 2\mid x\notin
W_{n}\\}\subset\mathbb{N}$. By definition of $U$ (see (15)), the set $H_{x}$
is infinite. Fix $N\in H_{x}$. Observe that $x\in V$, for some $V\in\Phi_{N}$
(as $\bigcup_{V\in\Phi_{N}}\limits\\!\\!V=[0,1]$), while $x\notin V_{r}$ (as
$x\notin W_{N}$). It follows that

$\rho(x,V)=\inf_{\begin{subarray}{c}n\geq 1\\\\[1.13809pt] T^{n}x\in
V\end{subarray}}n\cdot|T^{n}(x)-x|<r.$

We obtain (in view of the implication $\\{x,T^{n}x\\}\subset
V\in\Phi_{N}\\!\implies\\!|T^{n}x-x|\leq\frac{1}{N}$) that

$\inf_{\begin{subarray}{c}n\geq 1\\\\[1.13809pt]
|T^{n}x-x|\leq\frac{1}{N}\end{subarray}}\\!\\!n\cdot|T^{n}(x)-x|<r,\qquad\text{for
all }\ N\in H_{x}.$

Taking in account that $H_{x}$ is infinite, one concludes that
$r(T,x)=\liminf_{n\to\infty}\ n\,|T^{n}(x)-x|\leq r$. As the selection $x\in
U$ and $r>\frac{1}{\sqrt{5}}$ are arbitrary and $\mu(U)=1$, the proof of
Theorem 2 follows. ∎

##  4 Lagrange constants of interval exchange transformations

Our proof of Theorem 3 uses translation surfaces that can be thought as
suspension flows of interval exchange transformations. The Lagrange spectrum
of an interval exchange transformation will now be studied through the
$\operatorname{\bf N}$-norm of edges that are part of some specific
triangulation.

###  4.1 From interval exchange transformations to translation surfaces

First, we define translation surfaces. For more details, we invite the reader
to consult the survey by Masur and Tabachnilov [MT].

A _translation surface_ is a surface that is obtained from gluing $2d$
euclidean triangles by identifying their edges by translation. The simplest
example are tori $\mathbb{C}/\Lambda$ where $\Lambda$ is a lattice. The torus
$\mathbb{C}/\Lambda$ with $\Lambda=\mathbb{Z}u\oplus\mathbb{Z}v$ is obtained
by gluing together the two triangles with sides respectively $(u,-u+v,-v)$ and
$(v,-u,u-v)$. In that case we have $d=1$. (See Figure 3).

Figure 3: The fundamental domain of the torus seen as the gluing of two
euclidean triangles by translations.

Equivalently, a translation surface is a compact surface $X$ endowed with an
atlas defined on $X$ minus a finite (non-empty) set of points $\Sigma\subset
X$ with values in $\mathbb{C}$ such that coordinate changes are translations.
(We assume that the atlas is maximal for sake of uniqueness). It is easy to
see that a surface glued from euclidean triangles has such a geometric
structure. Conversely, on a translation surface there always exists a
triangulation whose edges are flat segments, and therefore the two definitions
are equivalent.

Two translation surfaces $(X,\Sigma)$ and $(X^{\prime},\Sigma^{\prime})$ are
isomorphic if there exists a homeomorphism $\phi:X\to X^{\prime}$ such that
$\phi(\Sigma)=\Sigma^{\prime}$ and for every chart $g:U\to\mathbb{C}$ of
$X^{\prime}$, $g\circ\phi$ is a chart of $f$. If the two surfaces are given by
a triangulation, the surfaces are isomorphic if and only if one can pass from
one to the other using edge flips (that is, we are allowed to paste two
triangles together and cut the resulting quadrilateral along the other
diagonal).

A point in the surface that is not a vertex of a triangle is called a _regular
point_. The (image of the) vertices in the surface are called the
_singularities_. As there are identifications, there can be fewer
singularities in the surface than vertices. Around each singularity, there is
a well defined angle that is a multiple of $2\pi$. If a translation surface
$X$ is made of $2d$ triangles then the set of conical angles of the
singularities $2k_{1}\pi,2k_{2}\pi,\ldots,2k_{n}\pi$ satisfies the relation
$d=k_{1}+k_{2}+\ldots+k_{n}$. In the torus of Figure 3, we have $d=1$, $n=1$
and $k_{1}=2\pi$.

A translation surface inherits a translation structure: that is, given a point
$p$ on the surface and a vector $v\in\mathbb{C}$ one can define $p+v$ on the
surface unless the segment $p+tv$ contains a vertex for some $t$ with $0\leq
t<1$. In general, we do not have $p+(v+w)=(p+v)+w$. The _(vertical)
translation flow_ on $X$ is the flow defined (almost everywhere) by
$\phi_{t}(p)=p+t\sqrt{-1}$.

A _saddle connection_ in a translation surface is a straight-line segment that
joins two singularities.

An horizontal segment $I$ in $X$ is _admissible_ if the orbits of its left and
right extremities under the translation flow have the following property:
either in the past or in the future, the orbit hits a singularity before
returning to the interval.

######  Proposition 17.

Let $X$ be a translation surface made of $2d$ triangles and $I\subset X$ an
admissible horizontal interval. The first return map $T$ of the flow on $I$ is
an interval exchange transformation. Furthermore, if $X$ has no vertical
saddle connection then $T$ satisfies the Keane condition and the number of
subintervals in $T$ is $d+1$.

In the case of the torus ($d=1$) the first return maps on admissible intervals
are rotations or 2-i.e.t. We will denote by $\mathcal{C}(d)$ the set of
translation surfaces obtained by gluing $2d$ triangles 222In Teichmüller
theory, $\mathcal{C}(d)$ is just the finite union of strata of translation
surfaces with given dimension. For example one has
$\mathcal{C}(2)=\mathcal{H}(0)$, $\mathcal{C}(3)=\mathcal{H}(0,0)$,
$\mathcal{C}(4)=\mathcal{H}(0,0,0)\cup\mathcal{H}(2)$ and
$\mathcal{C}(5)=\mathcal{H}(0,0,0,0)\cup\mathcal{H}(2,0)\cup\mathcal{H}(1,1)$..

###  4.2 Lagrange constants, best approximations and Delaunay triangulations

Now we explain how Lagrange constants of interval exchange transformation can
be computed from the $\operatorname{\bf N}$-norm of holonomies of saddle
connections.

Let $X$ be a translation surface. To a saddle connection can be associated a
vector in $\mathbb{C}$ that is called its _holonomy_. It corresponds to the
displacement induced on the translation structure while traveling along this
segment. Given a translation surface $X$ we denote by $V(X)\subset\mathbb{C}$
the set of holonomies of saddle connection in $X$.

######  Theorem 18 ([Vo96], [HMU15]).

Let $X$ be a translation surface. We define

$\operatorname{a}(X):=\liminf_{\stackrel{{\scriptstyle v\in
V(X)}}{{\operatorname{Im}(v)\to\infty}}}\frac{\operatorname{\bf
N}^{2}(v)}{\operatorname{\bf area}(X)},$

where $\operatorname{\bf N}(v):=\sqrt{|\operatorname{Re}(v)|\
|\operatorname{Im}(v)|}$, for vectors $v\in\mathbb{C}$.

Let $T$ be an interval exchange transformation that is the induced map of the
translation flow of $X$ on some admissible horizontal interval. Then
$\mathcal{E}(T)=\operatorname{a}(X)$.

In the above theorem, the minimum is taken over all saddle connections in
$V(X)$. We will show in Lemma 20 that it is enough to restrict the $\liminf$
to a subset of edges. Then we show that this subset of edges is exactly the
set of edges of the Delaunay triangulations of some deformations of $X$.

We first need to introduce ”quadrants” of saddle connections from [Ma12] and
[DU]. Let $X$ be a translation surface. Given two segments with the same
starting point they have a well defined angle. If the starting point has a
total angle $2k\pi$ then the angle between the segments is between $0$ and
$2k\pi$. Let us consider the vertical half lines that start from the
singularities in direction $\sqrt{-1}$. Up to a change of orientation, a
saddle connection can always be made to start with an angle in
$[-\pi/4,\pi/4]$ with respect to one of these outgoing vertical separatrices.
Let us fix a numbering of these half lines from $1$ to $d$. We associate to
each of them a pair $V^{i}_{\ell}(X)$ and $V^{i}_{r}(X)$ of subset of saddle
connections that are the one with angle respectively in $[-\pi/4,0]$ and
$[0,\pi/4]$ with the corresponding vertical.

A saddle connection is called a _best approximation_ if it is the diagonal of
an immersed rectangle in the surface whose boundary edges are horizontal and
vertical. Equivalently, $\gamma\in V^{i}_{s}(X)$ where
$i\in\\{1,2,\ldots,d\\}$ and $s\in\\{\ell,r\\}$ is not a best approximation if
there exists $\eta\in V^{i}_{s}(X)$ (the same quadrant) so that
$|\operatorname{Re}(\eta)|<|\operatorname{Re}(\gamma)|$ and
$|\operatorname{Im}(\eta)|<|\operatorname{Im}(\gamma)|$.

We will denote by $V_{ba}(X)$ the set of holonomies of best approximation on
$X$. From its definition it follows that

$\operatorname{a}(X)=\inf_{v\in V_{ba}(X)}\operatorname{\bf N}(v)=\inf_{v\in
V(X)}\operatorname{\bf N}(v)$

where $\operatorname{a}(X)$ is defined in Theorem 18.

We now show that best approximations can be seen as the edges of some Delaunay
triangulations. The group $\operatorname{SL}(2,\mathbb{R})$ acts on the set of
(equivalence classes) of translation surfaces through its linear action on
$\mathbb{R}^{2}$. The subaction of the diagonal subgroup $\displaystyle
g_{t}=\begin{pmatrix}e^{t}&0\\\ 0&e^{-t}\end{pmatrix}$ is called the
_Teichmüller flow_. Note that this action is also well defined on points and
segments (that is, given a pair $(X,p)$ (or $(X,\gamma)$) made of a
translation surface and a point, or a segment, and a matrix $m$, the image of
$p$ (or $\gamma$) in $m\cdot X$ is well defined). As in the case of the plane,
one can define the Delaunay triangulation of a translation surface $X$ by
considering maximal immersed squares. The following two lemmas are elementary.

######  Lemma 19.

Let $X$ be a translation surface with neither a horizontal nor a vertical
saddle connection. Then for all $t\in\mathbb{R}$, $g_{t}X$ has a well defined
Delaunay triangulation except for a discrete set of times $t_{n}$ for which
some quadruple of singularities are on the boundary of an immersed square (in
which case there is no uniqueness of Delaunay triangulation).

######  Lemma 20.

Let $X$ be a translation surface. For a saddle connection $\gamma$ in $X$ the
following are equivalent:

  1. 1.

$\gamma$ is a best approximation,

  2. 2.

there exists $t\in\mathbb{R}$ such that the $L^{\infty}$-Delaunay
triangulation of $g_{t}X$ contains $g_{t}\gamma$ as an edge.

###### Proof.

We just need to remark that any rectangle immersed in $X$ can be turned into a
square using the $g_{t}$ action. ∎

###  4.3 At the bottom of the Lagrange spectrum are golden surfaces

Let us introduce the surfaces that will be shown to be exactly the ones that
minimize the quantity $\operatorname{a}(X)$ of Theorem 18. Let $\Lambda$ be
the lattice $\Lambda=\mathbb{Z}(-1,1)\oplus\mathbb{Z}(\phi-1,\phi)$ where
$\phi=(\sqrt{5}+1)/2$ is the golden ratio. We call any lattice in the family
$g_{t}\Lambda$ a _golden lattice_ and the associated quotients
$\mathbb{C}/g_{t}\Lambda$ a _golden torus_. We also call any parallelogram
generated by a basis of $\Lambda$ a _golden parallelogram_. A _golden surface_
in $\mathcal{C}(d)$ is a translation surface obtained by gluing together $d$
identical golden parallelograms (each obtained by gluing $2$ triangles). In
geometric terms, such surface is a ramified covering of degree $d$ of a golden
torus.

In this section we prove the following result

######  Theorem 21.

There exists $\varepsilon_{0}$ such that the following holds. If $X$ is a
translation surface in $\mathcal{C}(d)$ such that

$\inf_{v\in V(X)}\frac{\operatorname{\bf N}^{2}(v)}{\operatorname{\bf
area}(X)}\geq\frac{1}{d\sqrt{5}}-\frac{\varepsilon_{0}}{d^{2}}$

then $X$ is a golden surface.

Together with Theorem 18, Theorem 21 implies Theorem 3. In order to simplify
notation in the proof, we will from now on always deal with surfaces such that
$\operatorname{\bf area}(X)=d$. That way the mean area of a triangle is $1/2$
independently of $d$.

######  Lemma 22.

There exist constants $\varepsilon_{1}>0$ and $C_{1}>1$ such that the
following holds. For any area $d$ translation surface $X$ in $\mathcal{C}(d)$
such that for some $\varepsilon<\varepsilon_{1}/d$ we have

$\inf_{v\in V_{ba}(X)}\operatorname{\bf
N}^{2}(v)\geq\frac{1}{\sqrt{5}}-\varepsilon.$

Then

$\sup_{v\in V_{ba}(X)}\operatorname{\bf
N}^{2}(v)\leq\frac{1}{\sqrt{5}}+C_{1}d\varepsilon.$

###### Proof.

Let us consider a translation surface $(X,\omega)$ of area $d$ and let
$\varepsilon<\varepsilon_{1}/d=\frac{2\sqrt{2}-\sqrt{5}}{20d}\simeq\frac{0.03}{d}$.
Assume that

$\inf_{v\in V_{ba}(X)}\operatorname{\bf
N}^{2}(v)\geq\frac{1}{\sqrt{5}}-\varepsilon.$

As a consequence of Corollary 12, we have for any translation surface

$\inf\\{\operatorname{\bf area}(\Delta):\text{$\Delta$ admissible triangle in
$X$}\\}\geq\frac{\sqrt{5}}{2}\inf_{v\in V_{ba}(X)}\operatorname{\bf
N}^{2}(v)\geq\frac{\sqrt{5}}{2}\left(\frac{1}{\sqrt{5}}-\varepsilon\right).$
(16)

Let us consider the $L^{\infty}$-Delaunay triangulation $T$ of the surface $X$
and let $\Delta_{0}$ be one triangle in $T$. Using the fact that the sum of
the areas of the $2d$ triangles from $T$ is $\operatorname{\bf area}(X)=d$ we
have that

$\operatorname{\bf area}(\Delta_{0})\leq d-(2d-1)\ \min_{\Delta\in
T}\operatorname{\bf area}(\Delta).$

Now, using (16) we obtain that

$\displaystyle\operatorname{\bf area}(\Delta_{0})$ $\displaystyle\leq
d-\left(2d-1\right)\frac{\sqrt{5}}{2}\left(\frac{1}{\sqrt{5}}-\varepsilon\right)$
$\displaystyle\leq\left(\frac{1}{\sqrt{5}}-\varepsilon\right)\left(\frac{\sqrt{5}d}{1-\sqrt{5}\varepsilon}-d\sqrt{5}+\frac{\sqrt{5}}{2}\right)$
$\displaystyle\leq\left(\frac{1}{\sqrt{5}}-\varepsilon\right)\left(\sqrt{5}d(1+2\sqrt{5}\varepsilon)-d\sqrt{5}+\frac{\sqrt{5}}{2}\right)$
$\displaystyle\leq\left(\frac{1}{\sqrt{5}}-\varepsilon\right)\left(\frac{\sqrt{5}}{2}+10d\varepsilon\right)\leq\left(\frac{\sqrt{5}}{2}+10d\varepsilon\right)\inf_{v\in
V_{ba}(X)}\operatorname{\bf N}^{2}(v).$

Here we used the inequality $1/(1-\sqrt{5}\varepsilon)\leq
1+2\sqrt{5}\varepsilon$ which is valid since $\sqrt{5}\varepsilon<1/2$. From
our assumption, $10d\varepsilon<\sqrt{2}-\sqrt{5}/2$, and we can apply Lemma
13 and get that

$\sup_{v\in V_{ba}(X)}\operatorname{\bf
N}^{2}(v)\leq\left(1+\frac{\sqrt{5}+1}{2}10d\varepsilon\right)\left(\frac{1}{\sqrt{5}}-\varepsilon\right)\leq\frac{1}{\sqrt{5}}+C_{1}d\varepsilon$

where $C_{1}=10(2\sqrt{2}+\sqrt{5})+1/2\leq 52$. ∎

Let us define the following distance between vectors

$\operatorname{\bf
dist}((x_{1},y_{1}),(x_{2},y_{2}))=\left\\{\begin{array}[]{ll}1&\text{if
$\operatorname{sgn}(x_{1})\not=\operatorname{sgn}(x_{2})$ or
$\operatorname{sgn}(y_{1})\not=\operatorname{sgn}(y_{2})$}\\\
\max\left(\left|\log\left(\frac{x_{1}}{x_{2}}\right)\right|,\left|\log\left(\frac{y_{1}}{y_{2}}\right)\right|\right)&\text{otherwise}\end{array}\right.$

Note that $\operatorname{\bf dist}$ is invariant under the action of diagonal
invertible matrices: in other words, for any nonzero real numbers $\alpha$ and
$\beta$ we have

$\operatorname{\bf dist}((\alpha x_{1},\beta y_{1}),(\alpha x_{2},\beta
y_{2}))=\operatorname{\bf dist}((x_{1},y_{1}),(x_{2},y_{2})).$

Let also $\tau$ be the linear map defined by

$\tau(x,y)=(-x/\phi,\phi y),\qquad\text{where \ $\phi=\frac{1+\sqrt{5}}{2}$}.$

Note that the only lattices satisfying $\tau(\Lambda)=\Lambda$ are the golden
lattices.

######  Lemma 23.

There exist constants $\varepsilon_{2}>0$ and $C_{2}>1$ such that for any
$\varepsilon<\varepsilon_{2}$ and any
$\zeta_{\ell}=(x_{\ell},y_{\ell}),\zeta_{r}=(x_{r},y_{r}),\zeta=(x,y)\in\mathbb{C}$,
if the conditions

  * •

$x_{\ell}<0$ and $x_{r}>0$,

  * •

$0<y_{\ell}<y_{r}<y$,

  * •

all of $\operatorname{\bf N}^{2}(\zeta_{\ell})$, $\operatorname{\bf
N}^{2}(\zeta_{r})$, $\operatorname{\bf N}^{2}(\zeta)$, $\operatorname{\bf
N}^{2}(\zeta_{\ell}-\zeta_{r})$, $\operatorname{\bf
N}^{2}(\zeta-\zeta_{\ell})$ and $\operatorname{\bf N}^{2}(\zeta-\zeta_{r})$
are in between $1/\sqrt{5}-\varepsilon$ and $1/\sqrt{5}+\varepsilon$,

hold, then all of $\operatorname{\bf dist}(\zeta_{r},\,\zeta-\zeta_{\ell})$,
$\operatorname{\bf dist}(\zeta_{\ell},\,\zeta-\zeta_{r})$, $\operatorname{\bf
dist}(\tau(\zeta_{\ell}),\zeta_{r})$, $\operatorname{\bf
dist}(\tau(\zeta_{r}),\,\zeta-\zeta_{r})$, $\operatorname{\bf
dist}(\zeta_{\ell},\tau(\zeta-\zeta_{\ell}))$ are smaller than
$C_{2}\varepsilon$.

###### Proof.

Up to rescaling we can assume that $x_{\ell}=-1$. Once $x_{\ell}$ is fixed,
the maps

$(y_{\ell},x_{r},y_{r})\mapsto(\operatorname{\bf
N}^{2}(\zeta_{\ell}),\operatorname{\bf N}^{2}(\zeta_{r}),\operatorname{\bf
N}^{2}(\zeta_{r}-\zeta_{\ell}))\quad\text{and}\quad(y_{\ell},x,y)\mapsto(\operatorname{\bf
N}^{2}(\zeta_{\ell}),\operatorname{\bf N}^{2}(\zeta),\operatorname{\bf
N}^{2}(\zeta-\zeta_{\ell}))$

have invertible derivative at the point

$y_{\ell}=\frac{1}{\sqrt{5}},\ x_{r}=\phi-1,\ y_{r}=\frac{\phi}{\sqrt{5}},\
x=\phi-2,\ y=\frac{\phi^{2}}{\sqrt{5}}.$

Moreover, the above point is the unique solution of $\operatorname{\bf
N}^{2}(\zeta_{\ell})=\operatorname{\bf N}^{2}(\zeta_{r})=\operatorname{\bf
N}^{2}(\zeta)=\operatorname{\bf N}^{2}(\zeta-\zeta_{\ell})=\operatorname{\bf
N}^{2}(\zeta-\zeta_{r})=\frac{1}{\sqrt{5}}$. One concludes using the inverse
function theorem. ∎

Recall that $V_{ba}(X)$ denote the holonomies of best approximations. Given a
quadrant $i$ in $X$ we denote by $V_{ba}^{(i)}(X)$ the set of holonomies of
saddle connections restricted to the $i$-th quadrant.

######  Lemma 24.

There exist constants $\varepsilon_{3}>0$ and $C_{3}>0$ such that if $X$ is
any area $d$ surface in $\mathcal{C}(d)$ such that

$\forall v\in
V_{ba}(X),\quad\frac{1}{\sqrt{5}}-\varepsilon_{3}<\operatorname{\bf
N}^{2}(v)<\frac{1}{\sqrt{5}}+\varepsilon_{3}$

then in any quadrant $i$ of $X$ we have

$\forall u,v\in V^{(i)}_{ba}(X),u\not=v,\qquad\operatorname{\bf
dist}(u,v)>C_{3}.$

###### Proof.

Choose an $\varepsilon_{3}$ so that we can apply Lemma 23. In a given fixed
quadrant $i$ holonomies can be identified with saddle connections. Lemma 23
implies that best approximations are far apart. ∎

###### Proof of Theorem 21.

Let $d\geq 1$ and $\varepsilon>0$ be such that

$\varepsilon<\min\left(\frac{\varepsilon_{1}}{d},\frac{\varepsilon_{2}}{C_{1}d},\
\frac{\varepsilon_{3}}{C_{1}d},\ \frac{C_{3}}{C_{1}C_{2}d^{2}}\right).$

Let $X\in\mathcal{C}(d)$ be a translation surface so that

$\inf_{v\in V_{ba}(X)}\operatorname{\bf
N}^{2}(v)\geq\frac{1}{\sqrt{5}}-\varepsilon.$

We will show that $X$ is actually a golden surface. Because
$\varepsilon<\varepsilon_{1}/d$ we can apply Lemma 22. Denoting
$\varepsilon^{\prime}=C_{1}d\varepsilon$ we have that

$\forall v\in
V_{ba}(X),\quad\frac{1}{\sqrt{5}}-\varepsilon^{\prime}\leq\operatorname{\bf
N}^{2}(v)\leq\frac{1}{\sqrt{5}}+\varepsilon^{\prime}.$ (17)

Morever $\varepsilon^{\prime}$ satisfies

$\varepsilon^{\prime}<\min\left(\varepsilon_{2},\ \varepsilon_{3},\
\frac{C_{3}}{C_{2}d}\right).$

We now show that any best approximation is part of a triangulation close to a
golden triangulation. In a moment we will use a fixed point argument to show
that $X$ is itself a golden surface.

A quadrilateral $q$ in $X$ is called _admissible_ if there exists an immersed
rectangle $R$ with horizontal and vertical sides such that there is exactly
one vertex of $q$ on each side of $R$. It is easy to see that an admissible
quadrilateral can be decomposed into two admissible triangles in two ways by
adding either of the two diagonals of $q$. The two diagonals of an admissible
quadrilateral $q$ will be called left-right and top-bottom diagonals. Given an
admissible quadrilateral, we can identify each side by its position: bottom
right, bottom left, top right, top left. The slopes of the bottom right and
top left sides are positive and following [DU] we will say that they are
_right slanted_. Similarly the bottom left and top right sides are _left
slanted_.

(a) An admissible quadrilateral $q$. The sides $\zeta_{b\ell}$ and
$\zeta_{tr}$ are left slanted while $\zeta_{br}$ and $\zeta_{t\ell}$ are right
slanted.

(b) A left move followed by a right move on a golden parallelogram. The
initial quadrilateral has blue right slanted and green left slanted sides. The
top-bottom diagonals are in red. The sides of the new quadrilateral are
dashed.

Figure 4: Quadrangulations and diagonal changes.

Let $Q$ be a quadrangulation of $X$. We denote by $E_{\ell}(Q)$ and $E_{r}(Q)$
the left slanted and right slanted sides of $Q$. We measure how $Q$ is close
to a quadrangulation into golden parallelograms with the following function

$g(Q)=\max\left(\max_{\zeta\in E_{r},\zeta^{\prime}\in E_{r}}\operatorname{\bf
dist}(\zeta,\zeta^{\prime}),\,\max_{\zeta\in E_{\ell},\zeta^{\prime}\in
E_{\ell}}\operatorname{\bf dist}(\zeta,\zeta^{\prime}),\,\max_{\zeta\in
E_{r},\zeta^{\prime}\in E_{\ell}}\operatorname{\bf
dist}(\tau(\zeta),\zeta^{\prime}),\,\right).$

It is clear that $g(Q)=0$ if and only if $Q$ is a golden quadrangulation.

Let $\varepsilon^{\prime\prime}=dC_{2}\varepsilon^{\prime}$.

claim 1: Let $\zeta_{0}$ be a right slanted best approximation in $X$. Then
there exists a unique quadrangulation $Q_{0}$ of $X$ into admissible
quadrilaterals that admits $\zeta_{0}$ as one of its sides and so that
$g(Q_{0})<\varepsilon^{\prime\prime}$.

Since $\varepsilon^{\prime}<\varepsilon_{2}$, it follows from Lemma 23 that
$\zeta_{0}$ is the bottom right side of a unique quadrilateral so that its
bottom left side $\zeta_{b\ell}$, its top left side $\zeta_{t\ell}$ and its
top right side $\zeta_{tr}$ satisfy that all of $\operatorname{\bf
dist}(\zeta_{0},\zeta_{t\ell})$, $\operatorname{\bf
dist}(\zeta_{b\ell},\zeta_{tr})$ and $\operatorname{\bf
dist}(\tau(\zeta_{0}),\zeta_{b\ell})$ are smaller than
$C_{2}\varepsilon^{\prime}$. We denote this quadrilateral by $q(\zeta_{0})$.

Similarly, given a left slanted best approximation $\zeta$ one can also build
a unique quadrilateral so that its bottom left side $\zeta_{b\ell}$ is
$C_{2}\varepsilon^{\prime}$-close to $\tau^{-1}(\zeta)$.

Using these two rules we can build step by step a quadrangulation of $X$. More
precisely, starting from $q_{0}$ we consider its top sides and construct new
quadrilaterals from these two. Then we repeat the operations with the newly
created sides.

By construction, either the newly created top side will coincide with an
already constructed bottom side of another quadrilateral, or we will have some
non-trivial intersection between two constructed parallelograms. Let us show
that this second case cannot happen. Let us consider a chain
$q_{0},q_{1},\ldots,q_{k}$ of adjacent quadrilaterals such that the bottom
sides of $q_{0}$ and $q_{k}$ belong to the same bundle of $X$ and the bundles
of $q_{i}$ for $i=0,\ldots,k-1$ are disjoint. Necessarily $k\leq d$, and hence
the bottom sides of $q_{k}$ are $kC_{2}\varepsilon^{\prime}$-close to those of
$q_{0}$. As $kC_{2}\varepsilon^{\prime}\leq\varepsilon^{\prime\prime}\leq
C_{3}$, Lemma 24 implies that $q_{0}=q_{k}$. This finishes the proof of the
claim.

We will now use claim 1 to show that the surface $X$ is itself a golden
surface. In order to do so, we analyze how the different quadrangulations of
claim 1 are related to each other. The relation between the various
quadrangulations correspond to a particular case of the so called Ferenczi-
Zamboni induction [Fe15] (see also [DU] for a particular case related to
quadrangulations).

Given two admissible quadrangulations $Q$ and $Q^{\prime}$ of $X$ we say that
$Q^{\prime}$ is obtained from $Q$ by a _left move_ if the left slanted sides
of $Q$ and $Q^{\prime}$ are equal and the right slanted sides of $Q^{\prime}$
are the top-bottom diagonals of quadrilaterals of $Q$. We define right moves
similarly. See also Figure 4.

By the uniqueness in claim 1, there exists a biinfinite sequence of
quadrangulations of $X$ into admissible quadrilaterals …,
$Q_{-1},Q_{0},Q_{1},Q_{2},\ldots$ each of them satisfying the claim, and such
that $Q_{n+1}$ is obtained from $Q_{n}$ by a left move followed by a right
move.

Recall that the bundles in $X$ are numbered. Hence each quadrilateral also
inherits a number given by the bundle it belongs to. We say that two
quadrangulations $Q$ and $Q^{\prime}$ are _combinatorially equivalent_ if for
each $i\in\\{1,\ldots,d\\}$, the labels of the quadrilaterals on the top left
and top right of the quadrilateral labeled $i$ are the same for $Q$ and
$Q^{\prime}$. It is easy to see that there are finitely many possible
combinatorial types of quadrangulations. Moreover, if $Q^{\prime}$ is obtained
by a left or right move from $Q$ then the combinatorial type of $Q^{\prime}$
is only determined by the combinatorial type of $Q$. We refer the reader to
[DU] or [Fe15] for these two elementary facts. Combining these two facts, we
see that the sequence of combinatorial types of the quadrangulations
$(Q_{n})_{n\in\mathbb{Z}}$ is periodic for some period $p$. Since a left or
right move operates as a linear transformation on the holonomies of the sides,
there exists a $2d\times 2d$ matrix $A$ with non-negative integer coefficients
so that the holonomies of the sides of $Q_{p}$ are the images of the side of
$Q_{0}$ by $A$. Let $v_{0}$ be the holonomy of $\zeta_{0}$. Let
$Q^{\prime}_{0}$ be the quadrangulation with the same combinatorics of $Q_{0}$
but such that all right slanted sides have holonomy $v_{0}$ and all left
slanted sides have holonomy $\tau(v_{0})$. The quadrangulation
$Q^{\prime}_{0}$ is a golden quadrangulation of another translation surface.
By construction, the real and imaginary parts of holonomies of
$Q^{\prime}_{0}$ are eigenvectors of $A$ (namely real parts are multiplied by
$\phi^{-2p}$ and imaginary parts by $\phi^{2p}$). Applying the Perron-
Frobenius theorem to the real and imaginary parts of the holonomies we obtain
the uniqueness of the fixed point (up to scalar multiples). Hence,
$Q^{\prime}_{0}=Q_{0}$ and $X$ is a golden surface. ∎

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[mb] Micahel Boshernitzan
Rice University
Houston, Texas
http://math.rice.edu/ michael/ [vd] Vincent Delecroix
CNRS, Université de Bordeaux
Bordeaux, France
http://www.labri.fr/perso/vdelecro/