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Representation and coding of rational pairs on a Triangular tree and Diophantine approximation in 2\mathbb{R}^{2}

Claudio Bonanno Dipartimento di Matematica, Università di Pisa, Largo Bruno Pontecorvo 5, 56127 Pisa, Italy claudio.bonanno@unipi.it  and  Alessio Del Vigna Dipartimento di Matematica, Università di Pisa, Largo Bruno Pontecorvo 5, 56127 Pisa, Italy delvigna@mail.dm.unipi.it
Abstract.

In this paper we study the properties of the Triangular tree, a complete tree of rational pairs introduced in [7], in analogy with the main properties of the Farey tree (or Stern-Brocot tree). To our knowledge the Triangular tree is the first generalisation of the Farey tree constructed using the mediant operation. In particular we introduce a two-dimensional representation for the pairs in the tree, a coding which describes how to reach a pair by motions on the tree, and its description in terms of SL(3,)SL(3,\mathbb{Z}) matrices. The tree and the properties we study are then used to introduce rational approximations of non-rational pairs.

MSC2020: 11B57, 37A44, 37E30
C. Bonanno is partially supported by the research project PRIN 2017S35EHN_\_004 “Regular and stochastic behaviour in dynamical systems” of the Italian Ministry of Education and Research, and by the “Istituto Nazionale di Alta Matematica” and its division “Gruppo Nazionale di Fisica Matematica”. This research is part of the authors’ activity within the DinAmicI community, see www.dinamici.org.

1. Introduction

The theory of multidimensional continued fractions has received increasing attention in the last years from researchers both in number theory and in ergodic theory. The origin of multidimensional continued fraction expansions may be traced back to a letter that Hermite sent to Jacobi asking for a generalisation of Lagrange’s Theorem for quadratic irrationals to algebraic irrationals of higher degree. It was for this reason that Jacobi developed what is now called the Jacobi-Perron algorithm. Unfortunately, despite numerous attempts and the introduction of many different algorithms, Hermite’s question remains unanswered. We refer the reader to [5] for a geometric description of the theory of multidimensional continued fractions. On the other hand, it is well-known that (regular) continued fraction expansions are related to the theory of dynamical systems as the expansion of a real number can be obtained by the symbolic representation of its orbit under the action of the Gauss map on the unit interval. The dynamical systems approach has led to new proofs of the Gauss-Kuzmin Theorem, Khinchin’s weak law and other metric results first obtained by Khinchin and Lévy, also thanks to the modern results of ergodic theory (see for example [11, 13]). In recent years the methods of ergodic theory have been applied also to maps related to multidimensional continued fraction algorithms, we refer to [15] for the first results in this research area. It is also interesting to mention that, in the opposite direction, number theoretical properties of real numbers have led to new results in ergodic theory, as the Three Gaps Theorem for instance, which can be interpreted in terms of return times for irrational rotations on the circle.

In this paper we continue the work initiated by the authors with Sara Munday in [7], where we have studied the properties of a tree of rational pairs, here called the Triangular tree, which was introduced as a two-dimensional version of the well-known Farey tree (or Stern-Brocot tree). The aim of this paper is to show that all the structures of the Farey tree can be found also in the Triangular tree and to construct approximations of real pairs using the tree. We believe that the Triangular tree will be useful for the study of interesting phenomena related to two-parameter systems, as this is the case for the Farey tree and one-parameter systems. For recent investigations on higher-dimensional phenomena we refer to [3, 4, 10].

The paper is structured as follows. In Section 2 we recall the construction of the Farey tree and its main properties based on the continued fraction expansions of real numbers, in particular the {L,R}\{L,R\} coding of real numbers which is based on the relations of the Farey tree with the group SL(2,)SL(2,\mathbb{Z}). In Section 3 we recall the main results of [7] concerning the Triangular tree. In particular we recall that it can be constructed in a dynamical way, by using the Triangle map defined in [8] which acts on the triangle {(x,y)2:1xy>0}\triangle\coloneqq\{(x,y)\in\mathbb{R}^{2}:1\geq x\geq y>0\} and its slow version SS introduced in [7], and in a geometric way, by using the notion of mediant of two pairs of fractions. The existence of these two possible constructions is the first basic property of the Farey tree that has been proved for the Triangular tree in [7]. Afterwards, Section 4 contains some preparatory results on the triangle sequences, that is the symbolic representation of the orbits of real pairs for the Triangle map. These sequences are the analogous of the continued fraction expansions for real numbers and the Gauss map, and are thus the two-dimensional continued fraction expansions of real pairs generated by the Triangle map. It is known that there are cases in which different real pairs have the same triangle sequence, this is discussed in details in Section 4.

Sections 5 and 6 contain the main results of the paper. First we use the slow map SS and its local inverses to introduce in Definition 5.3 a unique two-dimensional representation for rational pairs in widebar\widebar{\triangle}, thus improving the expansions obtained by using the Triangle map. For non-rational pairs the two-dimensional representation is that obtained by the Triangle map. Then in Theorem 6.6 we introduce a coding for rational pairs in the Triangular tree in terms of possible motions in the tree, in analogy with the coding of the Farey tree. This coding can be described also in terms of SL(3,)SL(3,\mathbb{Z}) matrices defined in (6.1). This coding and its descriptions are useful also to introduce approximations of real pairs, and of non-rational pairs in particular, in terms of the rational pairs on the Triangular tree. The definition of the approximations together with some examples are described in Section 7. Finally, in Section 8 we study the speed of the approximations introduced in the sense of the simultaneous approximations of couples of real numbers. We give results only for two classes of non-rational pairs, those with finite triangle sequence and those corresponding to fixed points of the Triangle map, leaving further developments for future research.

2. The Farey coding

In this section we recall the construction of the Farey tree, the coding it induces for the rationals in (0,1)(0,1), and its connection with the continued fraction expansion of real numbers.

The Farey tree is a binary tree which contains all the rational numbers in the interval (0,1)(0,1) and can be generated in a dynamical way using the Farey map, and in an arithmetic way using the notion of mediant between two fractions.

The Farey map is the map F:[0,1][0,1]F:[0,1]\rightarrow[0,1] defined to be

F(x){x1x,if 0x121xx,if 12x1F(x)\coloneqq\begin{cases}\frac{x}{1-x}\,,&\text{if }0\leq x\leq\frac{1}{2}\\[5.69046pt] \frac{1-x}{x}\,,&\text{if }\frac{1}{2}\leq x\leq 1\\ \end{cases}

Denoting with F0:[0,12][0,1]F_{0}:[0,\frac{1}{2}]\rightarrow[0,1] and F1:[12,1][0,1]F_{1}:[\frac{1}{2},1]\rightarrow[0,1] its two branches, FF admits two local inverses, ψ0F01\psi_{0}\coloneqq F_{0}^{-1} and ψ1F11\psi_{1}\coloneqq F_{1}^{-1}, given by

(2.1) ψ0(x)=x1+xandψ1(x)=11+x.\psi_{0}(x)=\frac{x}{1+x}\quad\text{and}\quad\psi_{1}(x)=\frac{1}{1+x}.

The Farey tree is then generated using the Farey map by setting 0{12}\mathcal{L}_{0}\coloneqq\{\frac{1}{2}\} for the root of the tree, and setting recursively nFn(12)\mathcal{L}_{n}\coloneqq F^{-n}\left(\frac{1}{2}\right) for all n1n\geq 1. Each level of the tree is written by ascending order as shown in Figure 1. The connections between the fractions of different levels are explained below. It is known that the Farey tree contains all the rational numbers in the interval (0,1)(0,1), that is n=0+n=n=0+Fn(12)=(0,1)\bigcup_{n=0}^{+\infty}\mathcal{L}_{n}=\bigcup_{n=0}^{+\infty}F^{-n}\left(\frac{1}{2}\right)=\mathbb{Q}\cap(0,1), and each rational number appears in the tree exactly once.

0\mathcal{L}_{0}1\mathcal{L}_{1}2\mathcal{L}_{2}3\mathcal{L}_{3}12\frac{1}{2}13\frac{1}{3}14\frac{1}{4}15\frac{1}{5}27\frac{2}{7}25\frac{2}{5}38\frac{3}{8}37\frac{3}{7}23\frac{2}{3}35\frac{3}{5}47\frac{4}{7}58\frac{5}{8}34\frac{3}{4}57\frac{5}{7}45\frac{4}{5}
Figure 1. The first four levels of the Farey tree.

We now describe the second way to construct the levels of the Farey tree. Let us consider the Stern-Brocot sets (n)n1(\mathcal{F}_{n})_{n\geq-1}, with 1{01,11}\mathcal{F}_{-1}\coloneqq\left\{\frac{0}{1},\frac{1}{1}\right\}, and for all n0n\geq 0 let n\mathcal{F}_{n} be obtained from n1\mathcal{F}_{n-1} by inserting the mediant of each pair of neighbouring fractions. We recall that the mediant of two fractions pq\frac{p}{q} and rs\frac{r}{s} with pq<rs\frac{p}{q}<\frac{r}{s} is

pqrsp+rq+s\frac{p}{q}\oplus\frac{r}{s}\coloneqq\frac{p+r}{q+s}

and that pq<pqrs<rs\frac{p}{q}<\frac{p}{q}\oplus\frac{r}{s}<\frac{r}{s}. We say that pqrs\frac{p}{q}\oplus\frac{r}{s} is the child of pq\frac{p}{q} and rs\frac{r}{s}, which are its left and right parent, respectively. It can be easily shown that two fractions pq<rs\frac{p}{q}<\frac{r}{s} are neighbours in a Stern-Brocot set if and only if qrps=1qr-ps=1 (see, for instance, [9]). Moreover, since the ancestors 01\frac{0}{1} and 11\frac{1}{1} are in lowest terms, it follows that all the fractions obtained through the mediant operation appear in lowest terms. The first sets n\mathcal{F}_{n} are

0={01,12,11},1={01,13,12,23,11},2={01,14,13,25,12,35,23,34,11}.\mathcal{F}_{0}=\left\{\frac{0}{1},\frac{1}{2},\frac{1}{1}\right\},\quad\mathcal{F}_{1}=\left\{\frac{0}{1},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{1}{1}\right\},\quad\mathcal{F}_{2}=\left\{\frac{0}{1},\frac{1}{4},\frac{1}{3},\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac{2}{3},\frac{3}{4},\frac{1}{1}\right\}.

The levels of the Farey tree are then given by n=nn1\mathcal{L}_{n}=\mathcal{F}_{n}\setminus\mathcal{F}_{n-1} for all n0n\geq 0. Two fractions in Figure 1 are connected if one of the fractions is a parent of the other. Notice that a fraction in a level n\mathcal{L}_{n} has a parent in the level n1\mathcal{L}_{n-1} and the other parent in a level m\mathcal{L}_{m} with m<n1m<n-1.

For later use we recall the notion of rank of a rational number xx in the open unit interval: rank(x)=r\operatorname*{rank}(x)=r if and only if xrx\in\mathcal{L}_{r}. For more details on the Farey tree, we refer to [6].

2.1. The Farey coding

We now recall the {L,R}\{L,R\} coding of rational numbers (see also [12, 6]). Let pq(0,1)\frac{p}{q}\in\mathbb{Q}\cap(0,1) be a fraction reduced in lowest terms. Since pq\frac{p}{q} belongs to a unique level of the Farey tree, we can write in a unique way

pq=lmrs,\frac{p}{q}=\frac{l}{m}\oplus\frac{r}{s},

with rmls=1rm-ls=1. That is, pq\frac{p}{q} is generated by taking the mediant between lm\frac{l}{m} and rs\frac{r}{s}, which are thus its parents in the tree. We associate to the fraction pq\frac{p}{q} the matrix

𝔪(pq)(rlsm)SL(2,),\mathfrak{m}\left(\frac{p}{q}\right)\coloneqq\begin{pmatrix}r&l\\ s&m\end{pmatrix}\in SL(2,\mathbb{Z}),

Introducing the two SL(2,)SL(2,\mathbb{Z}) matrices

(2.2) L(1011)andR(1101),L\coloneqq\begin{pmatrix}1&0\\ 1&1\end{pmatrix}\quad\text{and}\quad R\coloneqq\begin{pmatrix}1&1\\ 0&1\end{pmatrix},

the left and right children of pq=l+rm+s\frac{p}{q}=\frac{l+r}{m+s} on the tree, that are lml+rm+s=2l+r2m+s\frac{l}{m}\oplus\frac{l+r}{m+s}=\frac{2l+r}{2m+s} and l+rm+srs=l+2rm+2s\frac{l+r}{m+s}\oplus\frac{r}{s}=\frac{l+2r}{m+2s} respectively, can be obtained by right multiplication with the matrices LL and RR respectively, since

𝔪(2l+r2m+s)=𝔪(pq)L=(l+rlm+sm)and𝔪(l+2rm+2s)=𝔪(pq)R=(rl+rsm+s)\mathfrak{m}\left(\frac{2l+r}{2m+s}\right)=\mathfrak{m}\left(\frac{p}{q}\right)L=\begin{pmatrix}l+r&l\\ m+s&m\end{pmatrix}\quad\text{and}\quad\mathfrak{m}\left(\frac{l+2r}{m+2s}\right)=\mathfrak{m}\left(\frac{p}{q}\right)R=\begin{pmatrix}r&l+r\\ s&m+s\end{pmatrix}

In other words, the action of LL and RR given by the right matrix multiplication corresponds to travelling downward along the Farey tree by moving to the left or to the right, respectively. We finally notice that L=𝔪(12)L=\mathfrak{m}\left(\frac{1}{2}\right) is the matrix of the root of the Farey tree.

Proposition 2.1 ([12]).

Let x(0,1)x\in\mathbb{Q}\cap(0,1) with rank(x)=r\operatorname*{rank}(x)=r, and let x=[a1,,an]x=[a_{1},\ldots,a_{n}] be its continued fraction expansion. Then

rank(x)=i=1nai2\operatorname*{rank}(x)=\sum_{i=1}^{n}a_{i}-2

and the matrix associated to xx is

(2.3) 𝔪(x)=𝔪(12)i=1rMi={LLa11Ra2Lan1Ran1if n is evenLLa11Ra2Ran1Lan1if n is odd\mathfrak{m}\left(x\right)=\mathfrak{m}\left(\frac{1}{2}\right)\prod_{i=1}^{r}M_{i}=\begin{cases}LL^{a_{1}-1}R^{a_{2}}\cdots L^{a_{n-1}}R^{a_{n}-1}&\text{if $n$ is even}\\ LL^{a_{1}-1}R^{a_{2}}\cdots R^{a_{n-1}}L^{a_{n}-1}&\text{if $n$ is odd}\end{cases}

where Mi=LM_{i}=L if the ii-th turn along the path joining 12\frac{1}{2} with xx on the Farey tree goes left, and Mi=RM_{i}=R if it goes right111The leftmost symbol LL denotes 12\frac{1}{2} and is not associated to a move on the Farey tree..

The digits of the continued fraction expansion also have a dynamical meaning in terms of the Gauss map, hence of the Farey map. In fact we recall that the Gauss map is a fast version of the Farey map, precisely the Gauss map is the jump transformation of FF on the interval (12,1](\frac{1}{2},1]. If x=[a1,,an]x=[a_{1},\ldots,a_{n}] we have

x=ψ0a11ψ1ψ0an1ψ1(0)x=\psi_{0}^{a_{1}-1}\psi_{1}\cdots\psi_{0}^{a_{n}-1}\psi_{1}(0)

and (ai)i=1,,n(a_{i})_{i=1,\,\ldots,\,n} is the sequence of return times to (12,1](\frac{1}{2},1] of the orbit of xx under FF. Since ψ0ψ1(0)=12\psi_{0}\psi_{1}(0)=\frac{1}{2}, we can also express explicitly every rational number x(0,1)x\in(0,1) as a backward image of 12\frac{1}{2} under the Farey map. If an>1a_{n}>1 we have x=ψ0a11ψ1ψ0an2(12)x=\psi_{0}^{a_{1}-1}\psi_{1}\cdots\psi_{0}^{a_{n}-2}\left(\frac{1}{2}\right), and if an=1a_{n}=1 we have x=ψ0a11ψ1ψ0an11(12)x=\psi_{0}^{a_{1}-1}\psi_{1}\cdots\psi_{0}^{a_{n-1}-1}\left(\frac{1}{2}\right).

Example.

Let x=712x=\frac{7}{12}. This rational number appears at the fourth level of the Farey tree, so x=7124x=\frac{7}{12}\in\mathcal{L}_{4} and rank(x)=4\operatorname*{rank}(x)=4. Starting from the root 12\frac{1}{2}, the path to reach xx on the Farey tree is RLLRRLLR, thus 𝔪(712)=LRLLR=LRL2R\mathfrak{m}\left(\frac{7}{12}\right)=LRLLR=LRL^{2}R. Indeed

LRL2R=(3457)LRL^{2}R=\begin{pmatrix}3&4\\ 5&7\end{pmatrix}

and 4735=712\frac{4}{7}\oplus\frac{3}{5}=\frac{7}{12}. Moreover we have 712=[1,1,2,2]\frac{7}{12}=[1,1,2,2], so that 712=ψ1ψ1ψ0ψ1(12)\frac{7}{12}=\psi_{1}\psi_{1}\psi_{0}\psi_{1}\left(\frac{1}{2}\right).

The coding for all the real numbers in the closed unit interval extends the above construction and is given by a map π:[0,1]{L,R}\pi:[0,1]\rightarrow\{L,R\}^{\mathbb{N}} which associate to each x[0,1]x\in[0,1] an infinite sequence π(x)\pi(x) over the alphabet {L,R}\{L,R\}. First we set

π(01)=Landπ(11)=R.\pi\left(\frac{0}{1}\right)=L^{\infty}\quad\text{and}\quad\pi\left(\frac{1}{1}\right)=R^{\infty}.

Let x(0,1)x\in\mathbb{Q}\cap(0,1), so that it has a finite continued fraction expansion, say x=[a1,,an]x=[a_{1},\ldots,a_{n}]. Then note that there exists two infinite paths on the Farey tree which agree down to the node of xx. Both starts with the finite sequence coding the path from the root 12\frac{1}{2} to reach xx, according to (2.3) and terminating with either RLRL^{\infty} or LRLR^{\infty}. We let the infinite sequence terminate with RLRL^{\infty} or LRLR^{\infty} according to whether the number of partial quotients of xx is even or odd. Thus for x=[a1,,an]x=[a_{1},\ldots,a_{n}] we set

π(x)={La1Ra2Lan1RanLif n is evenLa1Ra2Ran1LanRif n is odd\pi(x)=\begin{cases}L^{a_{1}}R^{a_{2}}\cdots L^{a_{n-1}}R^{a_{n}}L^{\infty}&\text{if $n$ is even}\\ L^{a_{1}}R^{a_{2}}\cdots R^{a_{n-1}}L^{a_{n}}R^{\infty}&\text{if $n$ is odd}\end{cases}

In case x()(0,1)x\in(\mathbb{R}\setminus\mathbb{Q})\cap(0,1) the continued fraction expansion is infinite, say x=[a1,a2,]x=[a_{1},a_{2},\ldots], thus in this case we simply define

π(x)=La1Ra2La3Ra4.\pi(x)=L^{a_{1}}R^{a_{2}}L^{a_{3}}R^{a_{4}}\cdots.

3. Triangle maps and the Triangular tree

3.1. The setting

The Triangle Map has been introduced in [8] to define a two-dimensional analogue of the continued fraction algorithm. Let us consider the triangle

{(x,y)2: 1xy>0},\triangle\coloneqq\left\{(x,y)\in\mathbb{R}^{2}\,:\,1\geq x\geq y>0\right\},

and the pairwise disjoint subtriangles k{(x,y): 1xky0>1x(k+1)y}\triangle_{k}\coloneqq\left\{(x,y)\in\triangle\,:\,1-x-ky\geq 0>1-x-(k+1)y\right\}, with k0k\geq 0, and the line segment Λ{(x,0): 0x1}\Lambda\coloneqq\left\{(x,0)\,:\,0\leq x\leq 1\right\}. Note that widebar=k0kΛ\widebar{\triangle}=\bigcup_{k\geq 0}\triangle_{k}\cup\Lambda (see Figure 2). We also introduce

Σwidebar{x=y}andΥwidebar{x=1},\Sigma\coloneqq\widebar{\triangle}\cap\{x=y\}\quad\text{and}\quad\Upsilon\coloneqq\widebar{\triangle}\cap\{x=1\},

the slanting and the vertical side of \triangle, respectively. The Triangle Map T:widebarT:\triangle\to\widebar{\triangle} is then defined to be

T(x,y)(yx,1xkyx)for (x,y)k.T(x,y)\coloneqq\left(\frac{y}{x},\frac{1-x-ky}{x}\right)\quad\text{for }(x,y)\in\triangle_{k}.

The map TT generates an expansion associated to each point of \triangle, the so-called triangle sequence. In particular, to a point (x,y)(x,y)\in\triangle we associate the sequence of non-negative integers [α0,α1,α2,][\alpha_{0},\alpha_{1},\alpha_{2},\ldots] if and only if Tk(x,y)αkT^{k}(x,y)\in\triangle_{\alpha_{k}} for all k0k\geq 0. In case Tk(x,y)ΛT^{k_{\ast}}(x,y)\in\Lambda for some k>0k_{\ast}>0 then we say that the triangle sequence terminates. An important result of [8] is that pairs of rational numbers have a finite triangle sequence. However, the converse is not true: also non-rational points can have a finite triangle sequence and actually there are entire line segments with every point having the same triangle sequence. As it is clear from the definition, note that if (x,y)(x,y) has triangle sequence [α0,α1,α2,][\alpha_{0},\alpha_{1},\alpha_{2},\ldots] then T(x,y)T(x,y) has triangle sequence [α1,α2,][\alpha_{1},\alpha_{2},\ldots]. In other words, the Triangle Map acts on triangle sequences as the left shift, exactly as the Gauss map does for the continued fraction expansions.

(0,0)(0,0)(1,0)(1,0)(1,1)(1,1)0\triangle_{0}1\triangle_{1}2\triangle_{2}3\triangle_{3}4\triangle_{4}
Figure 2. Partition of \triangle into {k}k0\{\triangle_{k}\}_{k\geq 0}.

The analogous of the Farey map in this two-dimensional setting has been introduced in [7]. Let {Γ0,Γ1}\{\Gamma_{0},\Gamma_{1}\} be the partition of widebar\widebar{\triangle} such that Γ00\Gamma_{0}\coloneqq\triangle_{0} and Γ1widebarΓ0\Gamma_{1}\coloneqq\widebar{\triangle}\setminus\Gamma_{0}. The map S:widebarwidebarS:\widebar{\triangle}\rightarrow\widebar{\triangle} is then defined to be

(3.1) S(x,y){(yx,1xx),(x,y)Γ0(x1y,y1y),(x,y)Γ1.S(x,y)\coloneqq\begin{cases}\left(\frac{y}{x},\,\frac{1-x}{x}\right)\,,&(x,y)\in\Gamma_{0}\\[5.69046pt] \left(\frac{x}{1-y},\,\frac{y}{1-y}\right)\,,&(x,y)\in\Gamma_{1}\end{cases}.

Notice that SS maps k\triangle_{k} onto k1\triangle_{k-1} for all k1k\geq 1 and that S(Γ0)Σ=S(Γ1)=widebarS(\Gamma_{0})\cup\Sigma=S(\Gamma_{1})=\widebar{\triangle}. As a consequence, if (x,y)k(x,y)\in\triangle_{k} for some k0k\geq 0 then

(3.2) T(x,y)=Sk+1(x,y)=S|Γ0S|Γ1k(x,y).T(x,y)=S^{k+1}(x,y)=S|_{\Gamma_{0}}\circ S|_{\Gamma_{1}}^{k}(x,y).

In other words, the Triangle Map TT is the jump transformation of SS on the set Γ0\Gamma_{0}, as the Gauss map is the jump transformation of the Farey map on (12,1](\frac{1}{2},1]. Thus the map SS can be thought of as a “slow version” of the Triangle Map TT.

The map SS also induces a coding for the points of the triangle \triangle. In particular, if the triangle sequence of (x,y)(x,y) is [α0,α1,][\alpha_{0},\alpha_{1},\ldots] then the itinerary under SS of a point (x,y)(x,y)\in\triangle with respect to the partition {Γ0,Γ1}\{\Gamma_{0},\Gamma_{1}\} is

(1,, 1α0, 0,1,, 1α1, 0,).(\underbrace{1,\,\ldots,\,1}_{\alpha_{0}},\,0,\,\underbrace{1,\,\ldots,\,1}_{\alpha_{1}},\,0,\,\ldots).

Many properties of the map SS have been proved in [7]: SS is ergodic with respect to the Lebesgue measure, it preserves the infinite Lebesgue-absolutely continuous measure with density 1xy\frac{1}{xy}, and it is pointwise dual ergodic. Finally, the role of the map SS as a two-dimensional version of the Farey map is confirmed by the construction of a complete tree of rational pairs, the Triangular tree, by using the inverse branches of SS, in the same way as the Farey tree is generated by the Farey map, and then, equivalently, by a generalised mediant operation. In Section 3.2 we recall the main steps of this construction.

3.2. Construction of the Triangular tree

We now briefly recap the construction of the Triangular tree and its main properties, following [7, Section 5]. The two inverse branches of the map SS are

(3.3) ϕ0(S|Γ0)1:widebarΣΓ0,ϕ0(x,y)=(11+y,x1+y)\phi_{0}\coloneqq(S|_{\Gamma_{0}})^{-1}:\widebar{\triangle}\setminus\Sigma\rightarrow\Gamma_{0},\quad\phi_{0}(x,y)=\left(\frac{1}{1+y},\ \frac{x}{1+y}\right)

and

(3.4) ϕ1(S|Γ1)1:widebarΓ1,ϕ1(x,y)=(x1+y,y1+y).\phi_{1}\coloneqq(S|_{\Gamma_{1}})^{-1}:\widebar{\triangle}\rightarrow\Gamma_{1},\quad\phi_{1}(x,y)=\left(\frac{x}{1+y},\ \frac{y}{1+y}\right).

We then introduce the map

(3.5) ϕ2:ΣΛ,ϕ2(x,x)(x,0)\phi_{2}:\Sigma\to\Lambda,\quad\phi_{2}(x,x)\coloneqq(x,0)

and restrict ϕ1\phi_{1} to the set widebarΛ=\widebar{\triangle}\setminus\Lambda=\triangle. The maps ϕ0\phi_{0} and ϕ1\phi_{1} so modified, and the map ϕ2\phi_{2} form all together the set of local inverses of a map S~:widebarwidebar\tilde{S}:\widebar{\triangle}\rightarrow\widebar{\triangle} which coincides with SS on \triangle, and satisfies S~(x,0)=(x,x)\tilde{S}(x,0)=(x,x) on Λ\Lambda. Thus the maps SS and S~\tilde{S} coincide up to a zero-measure set.

The levels of the Triangular tree will be denoted by (𝒯n)n1(\mathcal{T}_{n})_{n\geq-1}. We also use the notation n𝒯n\mathcal{B}_{n}\coloneqq\mathcal{T}_{n}\cap\partial\triangle and n𝒯n\mathcal{I}_{n}\coloneqq\mathcal{T}_{n}\cap\accentset{\circ}{\triangle} for the boundary points and the interior points of the nn-th level of the tree, respectively. We start by setting

𝒯1{(0,0),(1,0),(1,1)}and𝒯0{(12,0),(1,12),(12,12)}.\mathcal{T}_{-1}\coloneqq\left\{(0,0),\,(1,0),\,(1,1)\right\}\quad\text{and}\quad\mathcal{T}_{0}\coloneqq\left\{\left(\frac{1}{2},{0}\right),\,\left({1},\frac{1}{2}\right),\,\left(\frac{1}{2},\frac{1}{2}\right)\right\}.

We now describe precisely how the levels of the tree are generated, by showing all the possibilities for taking counterimages depending on the location of the point in widebar\widebar{\triangle} (see Figure 8 for reference).

  1. (R1)

    An interior point (pq,rq)n\left(\frac{p}{q},\frac{r}{q}\right)\in\mathcal{I}_{n} generates the two interior points (qr+q,pr+q)\left(\frac{q}{r+q},\frac{p}{r+q}\right) and (pr+q,rr+q)\left(\frac{p}{r+q},\frac{r}{r+q}\right) in n+1\mathcal{I}_{n+1}, through the application of ϕ0\phi_{0} and ϕ1\phi_{1}, respectively.

  2. (R2)

    A boundary point (pq,pq)n\left(\frac{p}{q},\frac{p}{q}\right)\in\mathcal{B}_{n} generates the point (pq,0)n\left(\frac{p}{q},{0}\right)\in\mathcal{B}_{n} through the application of ϕ2\phi_{2} and the boundary point (pp+q,pp+q)n+1\left(\frac{p}{p+q},\frac{p}{p+q}\right)\in\mathcal{B}_{n+1} through the application of ϕ1\phi_{1}.

  3. (R3)

    A boundary point (pq,0)n\left(\frac{p}{q},{0}\right)\in\mathcal{B}_{n} generates the point (1,pq)n\left({1},\frac{p}{q}\right)\in\mathcal{B}_{n} through the application of ϕ0\phi_{0}.

  4. (R4)

    A boundary point (1,pq)n\left({1},\frac{p}{q}\right)\in\mathcal{B}_{n} generates the boundary point (qp+q,qp+q)n+1\left(\frac{q}{p+q},\frac{q}{p+q}\right)\in\mathcal{B}_{n+1} and the interior point (qp+q,pp+q)n+1\left(\frac{q}{p+q},\frac{p}{p+q}\right)\in\mathcal{I}_{n+1}, through the application of ϕ0\phi_{0} and ϕ1\phi_{1}, respectively.

Note that, conversely to the Farey tree, taking a counterimage does not necessarily imply a change in the level of the tree.

We now describe the geometric way to obtain the same two-dimensional tree of rational pairs constructed above by counterimages (see Figure 3 for reference). We define the mediant of two couples of fractions (pq,rq)\left(\frac{p}{q},\frac{r}{q}\right) and (pq,rq)\left(\frac{p^{\prime}}{q^{\prime}},\frac{r^{\prime}}{q^{\prime}}\right) as

(pq,rq)(pq,rq)(p+pq+q,r+rq+q).\left(\frac{p}{q},\frac{r}{q}\right)\oplus\left(\frac{p^{\prime}}{q^{\prime}},\frac{r^{\prime}}{q^{\prime}}\right)\coloneqq\left(\frac{p+p^{\prime}}{q+q^{\prime}},\frac{r+r^{\prime}}{q+q^{\prime}}\right).

Note that we require that the two fractions of each couple have the same denominator, so that the mediant lies on the line segment joining the two points it is computed from. We further assume that the two fractions of each couple are reduced to their least common denominator.

\diamond : points of 𝒯1\mathcal{T}_{-1}\triangle : points of 𝒯0\mathcal{T}_{0}\star : points of 𝒯1\mathcal{T}_{1}\circ : points of 𝒯2\mathcal{T}_{2}\bullet : points of 𝒯3\mathcal{T}_{3}\diamond\diamond\diamond(0,0)(0,0)(1,0)(1,0)(1,1)(1,1)\triangle\triangle\triangle\star\star\star\star\star\star\star\circ\circ\circ\circ\circ\circ\circ\circ\circ\circ\circ\circ\circ\circ\circ\circ\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet
Figure 3. The first four levels of the Triangular tree as points in widebar\widebar{\triangle} with the sides of the triangles of the partitions.
Definition 3.1.

Consider a set ={𝔯i:i=1,,r}\mathfrak{R}=\{\mathfrak{r}_{i}\,:\,i=1,\,\ldots,\,r\} of rational points on a line segment, consisting of at least two points, and in ascending lexicographic order. The Farey sum of \mathfrak{R} is obtained by adding to \mathfrak{R} the mediant between each pair of neighbouring points, that is

{𝔯i𝔯i+1:i=1,,r1}.\mathfrak{R}^{\oplus}\coloneqq\{\mathfrak{r}_{i}\oplus\mathfrak{r}_{i+1}\,:\,i=1,\,\ldots,\,r-1\}\cup\mathfrak{R}.

To define the levels of the tree in this second, geometric way, we start from the set 𝒮1𝒯1\mathcal{S}_{-1}\coloneqq\mathcal{T}_{-1} of the vertices of \triangle and then we will define a sequence (𝒮n)n1(\mathcal{S}_{n})_{n\geq-1} of sets such that 𝒮1\mathcal{S}_{-1} are the three vertices of \triangle and 𝒮n𝒮n1\mathcal{S}_{n}\supseteq\mathcal{S}_{n-1} for all n0n\geq 0. In particular, we introduce a sequence (𝒫n)n0(\mathscr{P}_{n})_{n\geq 0} of measurable partitions of \triangle, each refining the previous one and such that the points of 𝒮n\mathcal{S}_{n} lie on the sides of the partition 𝒫n\mathscr{P}_{n}. Then the recursive construction is the following: given the set of points 𝒮n\mathcal{S}_{n} up to a certain level nn, we obtain 𝒮n+1\mathcal{S}_{n+1} by inserting the mediant between each pair of neighbouring points along each side of the triangles of the partition 𝒫n+1\mathscr{P}_{n+1}. More formally, let 𝒫0={widebar}\mathscr{P}_{0}=\{\widebar{\triangle}\} and let v0=(0,0)v_{0}=(0,0), v1=(1,0)v_{1}=(1,0), v2=(1,1)v_{2}=(1,1) be the three vertices of the triangle \triangle. We partition widebar\widebar{\triangle} into two subtriangles by the line segment joining v1v_{1} and v0v2=(12,12)v_{0}\oplus v_{2}=\left(\frac{1}{2},\frac{1}{2}\right). This determines the partition 𝒫1\mathscr{P}_{1}. Additionally, we label the vertices of the two subtriangles according to the rule shown in Figure 4. We now proceed inductively. Each triangle of 𝒫n\mathscr{P}_{n} is partitioned into two subtriangles by the line segment joining the vertex labelled “1” with the mediant of the vertex “0” and the vertex “2” and this gives us the next partition 𝒫n+1\mathscr{P}_{n+1}. Then, for n1n\geq-1,

𝒮n+1𝔖𝒮n+1(𝔖𝒮n),\mathcal{S}_{n+1}\coloneqq\bigcup_{\mathfrak{S}\in\mathscr{S}_{n+1}}(\mathfrak{S}\cap\mathcal{S}_{n})^{\oplus},

where 𝒮n\mathscr{S}_{n}, n0n\geq 0, is the set of sides of the triangles of the partition 𝒫n\mathscr{P}_{n}. To better understand this construction, we recall the conclusions of [7, Lemma 5.8] (for simplicity of notation, we continuously extend the maps ϕ0\phi_{0} and ϕ1\phi_{1} defined in (3.3) and (3.4) to widebar\widebar{\triangle}): for any finite binary word ω{0,1}\omega\in\{0,1\}^{*} of length nn, let ϕω\phi_{\omega} denote the composition ϕω0ϕω1ϕωn1\phi_{\omega_{0}}\circ\phi_{\omega_{1}}\circ\dots\circ\phi_{\omega_{n-1}}, then

  1. (i)

    the triangles of 𝒫n\mathscr{P}_{n} are given by all the possible counterimages ωϕω(widebar)\triangle_{\omega}\coloneqq\phi_{\omega}(\widebar{\triangle}) with |ω|=n|\omega|=n;

  2. (ii)

    𝒮n=𝒮0{ϕω()widebar:|ω|n1}\mathscr{S}_{n}=\mathscr{S}_{0}\cup\left\{\widebar{\phi_{\omega}(\ell)}\,:\,|\omega|\leq n-1\right\}, where \ell is the open line segment joining (1,0)(1,0) and (12,12)\left(\frac{1}{2},\frac{1}{2}\right).

011220112201122
Figure 4. Partition of a triangle of 𝒫n\mathscr{P}_{n} into two subtriangles and relabelling of the vertices.

The main properties concerning the triangular tree are contained in Theorem 5.4 and Theorem 5.8 of [7]. The first result states that the tree is complete, that is

n1𝒯n=2widebar,\bigcup_{n\geq-1}\mathcal{T}_{n}=\mathbb{Q}^{2}\cap\widebar{\triangle},

with every rational pair appearing exactly once in the tree. The second result establishes the level-by-level equivalence between the counterimages tree and the geometric tree defined above, that is 𝒯n=𝒮n𝒮n1\mathcal{T}_{n}=\mathcal{S}_{n}\setminus\mathcal{S}_{n-1} for all n0n\geq 0.

4. Triangle sequences: convergence and non-convergence

It is known that a triangle sequence does not necessarily represent a unique pair of real numbers, but could correspond to an entire line segment. If the triangle sequence terminates, we do not have uniqueness and Lemma 5.1 characterises the points having a given finite triangle sequence. Uniqueness is not guaranteed even when the triangle sequence is infinite: [8] gives a sufficient condition to have uniqueness and a criterion, equivalent to uniqueness is proved in the later work [1]. In this section we discuss this problem.

We start by introducing some notation. Let X=(qpr)X=\begin{pmatrix}q\\ p\\ r\end{pmatrix} be a three-dimensional vector with integer components and q0q\neq 0. We then define the correspondent rational pair

X^(pq,rq)\hat{X}\coloneqq\left(\frac{p}{q},\frac{r}{q}\right)

for which both components have the same denominator. For instance, the vertices v0v_{0}, v1v_{1} and v2v_{2} of \triangle are represented by (100)\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}, (110)\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}, and (111)\begin{pmatrix}1\\ 1\\ 1\end{pmatrix}, respectively. Note that the sum of two three-dimensional vectors corresponds to the mediant between the two correspondent two-dimensional vectors, that is

X+Y^=X^Y^.\widehat{X+Y}=\hat{X}\oplus\hat{Y}.

For a sequence (α0,α1,)(\alpha_{0},\alpha_{1},\ldots) of non-negative integers and for an integer k0k\geq 0 we define

(α0,,αk){(x,y):Tj(x,y)αj for all j=0,,k}.\triangle(\alpha_{0},\ldots,\alpha_{k})\coloneqq\left\{(x,y)\in\triangle\,:\,T^{j}(x,y)\in\triangle_{\alpha_{j}}\text{ for all }j=0,\,\ldots,\,k\right\}.

The set (α0,,αk)\triangle(\alpha_{0},\ldots,\alpha_{k}) is a triangle and consists of all those points whose first k+1k+1 triangle sequence digits are precisely α0,,αk\alpha_{0},\ldots,\alpha_{k} and thus these triangles are nested, that is

(α0)(α0,α1).\triangle\supset\triangle(\alpha_{0})\supset\triangle(\alpha_{0},\alpha_{1})\supset\cdots.

Let (Xk)k3(X_{k})_{k\geq-3} be the sequence of three-dimensional vectors defined as follows:

(4.1) X3=(001),X2=(100),X1=(110),Xk=Xk3+Xk1+αkXk2 for all k0,X_{-3}=\begin{pmatrix}0\\ 0\\ 1\end{pmatrix},\ X_{-2}=\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\ X_{-1}=\begin{pmatrix}1\\ 1\\ 0\end{pmatrix},\quad X_{k}=X_{k-3}+X_{k-1}+\alpha_{k}X_{k-2}\text{ for all $k\geq 0$},

then the vertices of (α0,,αk)\triangle(\alpha_{0},\ldots,\alpha_{k}) are X^k1\hat{X}_{k-1}, X^k\hat{X}_{k} and X^k2X^k\hat{X}_{k-2}\oplus\hat{X}_{k} (see [1, Theorem 3]). Figure 5 shows the recursive construction of the triangles (α0,,αk)\triangle(\alpha_{0},\ldots,\alpha_{k}).

When the triangle sequence is infinite, the infinite intersection k0(α0,,αk)\bigcap_{k\geq 0}\triangle(\alpha_{0},\ldots,\alpha_{k}) can be either a point or a line segment.

  1. (1)

    In the first case the nested triangles shrink to a point, which means that the triangle sequence (αk)k0(\alpha_{k})_{k\geq 0} denotes a unique pair of real numbers (α,β)(\alpha,\beta). As a consequence diam(α0,,αk)0\operatorname*{diam}\triangle(\alpha_{0},\ldots,\alpha_{k})\rightarrow 0 and the sequence (X^k)k3(\hat{X}_{k})_{k\geq-3} converges to (α,β)(\alpha,\beta). We will refer to this case as the convergent case.

  2. (2)

    In the second case the triangle sequence does not uniquely describe a point but instead identifies a line segment 𝔏\mathfrak{L} of length l>0l>0, such that all the points of 𝔏\mathfrak{L} have the same triangle sequence [α0,α1,][\alpha_{0},\alpha_{1},\ldots]. In this case diam(α0,,αk)l\operatorname*{diam}\triangle(\alpha_{0},\ldots,\alpha_{k})\rightarrow l and the sequence (X^k)k3(\hat{X}_{k})_{k\geq-3} does not admit a limit. More precisely, we have that the odd and even terms of (X^k)k3(\hat{X}_{k})_{k\geq-3} converge to the two endpoints of 𝔏\mathfrak{L} [1, Theorem 6]. In particular, d(X^k1,X^k)ld(\hat{X}_{k-1},\hat{X}_{k})\rightarrow l and it also holds that d(X^k,X^k1X^k)0d(\hat{X}_{k},\hat{X}_{k-1}\oplus\hat{X}_{k})\rightarrow 0, where d(,)d(\cdot,\cdot) is the Euclidean distance in 2\mathbb{R}^{2}. We will refer to this case as the non-convergent case.

The main result of [1, Section 6] is a criterion of uniqueness, which we now state. Let

λkd(X^k1,X^k+1)d(X^k1,X^kX^k2),\lambda_{k}\coloneqq\frac{d(\hat{X}_{k-1},\hat{X}_{k+1})}{d(\hat{X}_{k-1},\hat{X}_{k}\oplus\hat{X}_{k-2})},

and refer again to Figure 5 for the geometric interpretation of this quantity. The triangle sequence [α0,α1,][\alpha_{0},\alpha_{1},\ldots] does not correspond to a unique pair of real numbers if and only if it contains only a finite number of zeroes and kN(1λk)>0\prod_{k\geq N}(1-\lambda_{k})>0, where NN is such that αk>0\alpha_{k}>0 for all kNk\geq N222Note that λk=1\lambda_{k}=1 if and only if αk=0\alpha_{k}=0.. The convergence of the infinite product to a non-zero number is equivalent to the convergence of k0λk\sum_{k\geq 0}\lambda_{k}, which is in turn equivalent to λk0\lambda_{k}\rightarrow 0 sufficiently fast: the geometric meaning of λk\lambda_{k} suggests that this condition is equivalent to a sufficiently fast growth of the triangle sequence digits αk\alpha_{k}.

X^k1\hat{X}_{k-1}X^k\hat{X}_{k}X^kX^k2\hat{X}_{k}\oplus\hat{X}_{k-2}X^k+1\hat{X}_{k+1}X^k1X^k+1\hat{X}_{k-1}\oplus\hat{X}_{k+1}(α0,,αk)\triangle(\alpha_{0},\,\ldots,\,\alpha_{k})(α0,,αk+1)\triangle(\alpha_{0},\,\ldots,\,\alpha_{k+1})
Figure 5. Construction of the triangle (α0,,αk,αk+1)\triangle(\alpha_{0},\,\ldots,\,\alpha_{k},\,\alpha_{k+1}) starting from (α0,,αk)\triangle(\alpha_{0},\,\ldots,\,\alpha_{k})·

We now prove some results of convergence for the points of the closed triangle widebar\widebar{\triangle} in the non-convergent case. For two rational pairs X^\hat{X} and Y^\hat{Y} and a non-negative integer ss we introduce the notation

Y^sX^Y^X^X^s times,\hat{Y}\oplus_{s}\hat{X}\coloneqq\hat{Y}\oplus\underbrace{\hat{X}\oplus\cdots\oplus\hat{X}}_{s\text{ times}},

and recall that the maps ϕ0\phi_{0} and ϕ1\phi_{1} commute with the mediant operation for rational pairs whose components have the same denominator, that is

ϕi(X^Y^)=ϕi(X^)ϕi(Y^)for i=0,1.\phi_{i}\left(\hat{X}\oplus\hat{Y}\right)=\phi_{i}(\hat{X})\oplus\phi_{i}(\hat{Y})\quad\text{for $i=0,1$}.

Also, in the following we denote by ϕ0\phi_{0} and ϕ1\phi_{1} the continuous extension to widebar\widebar{\triangle} of the maps defined in (3.3) and (3.4).

Lemma 4.1.

Let k0k\geq 0 and let (αj)j=0,,k(\alpha_{j})_{j=0,\,\ldots,\,k} be a sequence of non-negative integers. It holds that

(α0,,αk)=ϕ1α0ϕ0ϕ1αkϕ0(widebarΣ),\triangle(\alpha_{0},\ldots,\alpha_{k})=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k}}\phi_{0}(\widebar{\triangle}\setminus\Sigma),

and more precisely X^k=ϕ1α0ϕ0ϕ1αkϕ0(1,0)\hat{X}_{k}=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k}}\phi_{0}(1,0).

Proof.

Let dd be a non-negative integer. We have d=ϕ1dϕ0(widebarΣ)\triangle_{d}=\phi_{1}^{d}\phi_{0}(\widebar{\triangle}\setminus\Sigma), which can be verified by explicitly computing the vertices of the two triangles. This in turn proves the result in the case k=0k=0.
For k1k\geq 1, by definition we have (x,y)(α0,,αk)(x,y)\in\triangle(\alpha_{0},\ldots,\alpha_{k}) if and only if

Tj(x,y)αjfor all j=0,,k.T^{j}(x,y)\in\triangle_{\alpha_{j}}\quad\text{for all }j=0,\,\ldots,\,k.

We can thus write

Tk(x,y)=T|αk1T|α0(x,y)=(3.2)S|Γ0S|Γ1αk1S|Γ0S|Γ1α0(x,y)αkT^{k}(x,y)=T|_{\triangle_{\alpha_{k-1}}}\circ\cdots\circ T|_{\triangle_{\alpha_{0}}}(x,y)\overset{\eqref{eq:T-jumpS}}{=}S|_{\Gamma_{0}}S|_{\Gamma_{1}}^{\alpha_{k-1}}\circ\cdots\circ S|_{\Gamma_{0}}\circ S|_{\Gamma_{1}}^{\alpha_{0}}(x,y)\in\triangle_{\alpha_{k}}

which holds if and only if (x,y)ϕ1α0ϕ0ϕ1αk1ϕ0(αk)(x,y)\in\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k-1}}\phi_{0}(\triangle_{\alpha_{k}}). Since αk=ϕ1αkϕ0(widebarΣ)\triangle_{\alpha_{k}}=\phi_{1}^{\alpha_{k}}\phi_{0}(\widebar{\triangle}\setminus\Sigma), the first part of the result follows.
For the second part of the lemma we argue by strong induction on k0k\geq 0. The case k=0k=0 follows from the first part of this proof, by the explicit computation of the vertices of (α0)\triangle(\alpha_{0}). Now let k1k\geq 1 and consider the triangle (α0,,αk1)=ϕ1α0ϕ0ϕ1αk1ϕ0(widebarΣ)\triangle(\alpha_{0},\ldots,\alpha_{k-1})=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k-1}}\phi_{0}(\widebar{\triangle}\setminus\Sigma). By inductive hypothesis two of its vertices are X^k1=ϕ1α0ϕ0ϕ1αk1ϕ0(1,0)\hat{X}_{k-1}=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k-1}}\phi_{0}(1,0) and X^k2=ϕ1α0ϕ0ϕ1αk2ϕ0(1,0)\hat{X}_{k-2}=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k-2}}\phi_{0}(1,0). Since ϕ0(0,0)=(1,0)\phi_{0}(0,0)=(1,0) and ϕ1(1,0)=(1,0)\phi_{1}(1,0)=(1,0), we can also write X^k2=ϕ1α0ϕ0ϕ1αk1ϕ0(0,0)\hat{X}_{k-2}=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k-1}}\phi_{0}(0,0). Thus, from (α0,,αk1)=ϕ1α0ϕ0ϕ1αk1ϕ0(widebarΣ)\triangle(\alpha_{0},\ldots,\alpha_{k-1})=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k-1}}\phi_{0}(\widebar{\triangle}\setminus\Sigma), it follows that the other vertex of (α0,,αk1)\triangle(\alpha_{0},\ldots,\alpha_{k-1}) has to be the backward image of (1,1)(1,1), that is X^k3X^k1=ϕ1α0ϕ0ϕ1αk1ϕ0(1,1)\hat{X}_{k-3}\oplus\hat{X}_{k-1}=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k-1}}\phi_{0}(1,1). Since the maps ϕ0\phi_{0} and ϕ1\phi_{1} commute with the mediant, we have

X^k\displaystyle\hat{X}_{k} =(X^k3X^k1)αkX^k2=ϕ1α0ϕ0ϕ1αk1ϕ0(1,1)αkϕ1α0ϕ0ϕ1αk1ϕ0(0,0)=\displaystyle=(\hat{X}_{k-3}\oplus\hat{X}_{k-1})\oplus_{\alpha_{k}}\hat{X}_{k-2}=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k-1}}\phi_{0}(1,1)\oplus_{\alpha_{k}}\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k-1}}\phi_{0}(0,0)=
=ϕ1α0ϕ0ϕ1αk1ϕ0(1αk+1,1αk+1)=ϕ1α0ϕ0ϕ1αk1ϕ0ϕ1αkϕ0(1,0),\displaystyle=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k-1}}\phi_{0}\left(\frac{1}{\alpha_{k}+1},\frac{1}{\alpha_{k}+1}\right)=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k-1}}\phi_{0}\phi_{1}^{\alpha_{k}}\phi_{0}(1,0),

and this completes the proof. ∎

Lemma 4.2.

Let X^=(pq,rq)\hat{X}=\left(\frac{p}{q},\frac{r}{q}\right) and Y^=(pq,rq)\hat{Y}=\left(\frac{p^{\prime}}{q^{\prime}},\frac{r^{\prime}}{q^{\prime}}\right), and let ss be a non-negative integer. Then

d(Y^sX^,X^)=d(X^,Y^)qq+sqandd(Y^sX^,Y^)=d(X^,Y^)sqq+sq.d(\hat{Y}\oplus_{s}\hat{X},\hat{X})=d(\hat{X},\hat{Y})\frac{q^{\prime}}{q^{\prime}+sq}\quad\text{and}\quad d(\hat{Y}\oplus_{s}\hat{X},\hat{Y})=d(\hat{X},\hat{Y})\frac{sq}{q^{\prime}+sq}.
Proof.

It is a straightforward computation involving X^\hat{X}, Y^\hat{Y} and Y^sX^=(p+spq+sq,r+srq+sq)\hat{Y}\oplus_{s}\hat{X}=\left(\frac{p^{\prime}+sp}{q^{\prime}+sq},\frac{r^{\prime}+sr}{q^{\prime}+sq}\right). ∎

For k3k\geq-3 we introduce the notation

Xk=(qkpkrk),X_{k}=\begin{pmatrix}q_{k}\\ p_{k}\\ r_{k}\end{pmatrix},

so that X^k=(pkqk,rkqk)\hat{X}_{k}=\left(\frac{p_{k}}{q_{k}},\frac{r_{k}}{q_{k}}\right). Note that (4.1) implies that the denominators satisfy the recurrence

qk=qk3+qk1+αkqk2,q_{k}=q_{k-3}+q_{k-1}+\alpha_{k}q_{k-2},

where q3=0q_{-3}=0, q2=1q_{-2}=1 and q1=1q_{-1}=1. This remark and Lemma 4.2 imply that for all k0k\geq 0 we have

λk=d(X^k1,X^k+1)d(X^k1,X^kX^k2)=qk2+qkqk2+qk+αk+1qk1=qk2+qkqk+1.\lambda_{k}=\frac{d(\hat{X}_{k-1},\hat{X}_{k+1})}{d(\hat{X}_{k-1},\hat{X}_{k}\oplus\hat{X}_{k-2})}=\frac{q_{k-2}+q_{k}}{q_{k-2}+q_{k}+\alpha_{k+1}q_{k-1}}=\frac{q_{k-2}+q_{k}}{q_{k+1}}.

It also easily follows that

(4.2) 1λk=αk+1qk1qk+1.1-\lambda_{k}=\alpha_{k+1}\frac{q_{k-1}}{q_{k+1}}.
Lemma 4.3.

Let (αk)k0(\alpha_{k})_{k\geq 0} be a non-convergent triangle sequence describing a line segment of length l>0l>0 (as described in point (2) above).

  1. (i)

    The ratio of consecutive denominators of the XkX_{k} diverges, that is limk+qkqk1=+\lim_{k\rightarrow+\infty}\frac{q_{k}}{q_{k-1}}=+\infty.

  2. (ii)

    It holds

    limk+d(X^k2X^k,X^k1)=l.\lim_{k\rightarrow+\infty}d(\hat{X}_{k-2}\oplus\hat{X}_{k},\hat{X}_{k-1})=l.
  3. (iii)

    For any non-negative integer ss, it holds

    limk+d(X^ksX^k1,X^k)=0andlimk+d((X^k2X^k)sX^k1,X^k2X^k)=0.\lim_{k\rightarrow+\infty}d(\hat{X}_{k}\oplus_{s}\hat{X}_{k-1},\hat{X}_{k})=0\quad\text{and}\quad\lim_{k\rightarrow+\infty}d((\hat{X}_{k-2}\oplus\hat{X}_{k})\oplus_{s}\hat{X}_{k-1},\hat{X}_{k-2}\oplus\hat{X}_{k})=0.
Proof.

(i) From the recurrence for the denominators we have

qkqk11+αkqk2qk1=(4.2)1+(1λk1)qkqk1,\frac{q_{k}}{q_{k-1}}\geq 1+\alpha_{k}\frac{q_{k-2}}{q_{k-1}}\overset{\textrm{\eqref{eq:1-lambda}}}{=}1+(1-\lambda_{k-1})\frac{q_{k}}{q_{k-1}},

so that qkqk11λk\frac{q_{k}}{q_{k-1}}\geq\frac{1}{\lambda_{k}}. Since λk0+\lambda_{k}\rightarrow 0^{+} as k+k\rightarrow+\infty, the thesis follows.
(ii) For kk large enough we have

d(X^k1,X^k)d(X^k2X^k,X^k1)d(X^k2X^k,X^k)+d(X^k1,X^k),d(\hat{X}_{k-1},\hat{X}_{k})\leq d(\hat{X}_{k-2}\oplus\hat{X}_{k},\hat{X}_{k-1})\leq d(\hat{X}_{k-2}\oplus\hat{X}_{k},\hat{X}_{k})+d(\hat{X}_{k-1},\hat{X}_{k}),

where the first inequality is shown in [1, Theorem 6] and the second inequality is the triangle inequality applied to (α0,,αk)\triangle(\alpha_{0},\,\ldots,\,\alpha_{k}). This is enough to conclude because we already know that d(X^k1,X^k)ld(\hat{X}_{k-1},\hat{X}_{k})\rightarrow l and d(X^k2X^k,X^k)0d(\hat{X}_{k-2}\oplus\hat{X}_{k},\hat{X}_{k})\rightarrow 0 (see point (2) above).
(iii) From Lemma 4.2 we have

d(X^ksX^k1,X^k)=d(X^k,X^k1)ss+qkqk1d(\hat{X}_{k}\oplus_{s}\hat{X}_{k-1},\hat{X}_{k})=d(\hat{X}_{k},\hat{X}_{k-1})\frac{s}{s+\frac{q_{k}}{q_{k-1}}}

and

d((X^k2X^k)sX^k1,X^k2X^k)=d(X^k2X^k,X^k1)ss+qkqk1+qk2qk1.d((\hat{X}_{k-2}\oplus\hat{X}_{k})\oplus_{s}\hat{X}_{k-1},\hat{X}_{k-2}\oplus\hat{X}_{k})=d(\hat{X}_{k-2}\oplus\hat{X}_{k},\hat{X}_{k-1})\frac{s}{s+\frac{q_{k}}{q_{k-1}}+\frac{q_{k-2}}{q_{k-1}}}.

Using d(X^k1,X^k)ld(\hat{X}_{k-1},\hat{X}_{k})\rightarrow l, (i), and (ii), the result easily follows. ∎

Proposition 4.4.

Let (αk)k0(\alpha_{k})_{k\geq 0} be a non-convergent triangle sequence and denote by 𝔭𝔏\mathfrak{p}_{\mathfrak{L}} and 𝔮𝔏\mathfrak{q}_{\mathfrak{L}} the two endpoints of the line segment 𝔏=k0(α0,,αk)\mathfrak{L}=\bigcap_{k\geq 0}\triangle(\alpha_{0},\,\ldots,\,\alpha_{k}), such that (X^2k)k0(\hat{X}_{2k})_{k\geq 0} and (X^2k+1)k0(\hat{X}_{2k+1})_{k\geq 0} converge respectively to 𝔭𝔏\mathfrak{p}_{\mathfrak{L}} and 𝔮𝔏\mathfrak{q}_{\mathfrak{L}}. Then for all (x,y)𝑤𝑖𝑑𝑒𝑏𝑎𝑟{(0,0)}(x,y)\in\widebar{\triangle}\setminus\left\{(0,0)\right\} it holds

limk+ϕ1α0ϕ0ϕ1α2kϕ0(x,y)=𝔭𝔏,limk+ϕ1α0ϕ0ϕ1α2k+1ϕ0(x,y)=𝔮𝔏,\lim_{k\rightarrow+\infty}\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}(x,y)=\mathfrak{p}_{\mathfrak{L}},\quad\lim_{k\rightarrow+\infty}\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k+1}}\phi_{0}(x,y)=\mathfrak{q}_{\mathfrak{L}},

and

limk+ϕ1α0ϕ0ϕ1α2kϕ0(0,0)=𝔮𝔏,limk+ϕ1α0ϕ0ϕ1α2k+1ϕ0(0,0)=𝔭𝔏.\lim_{k\rightarrow+\infty}\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}(0,0)=\mathfrak{q}_{\mathfrak{L}},\quad\lim_{k\rightarrow+\infty}\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k+1}}\phi_{0}(0,0)=\mathfrak{p}_{\mathfrak{L}}.
Proof.

We give the proof just for the case of even indices, the odd case is analogous. By notation we have that X^2k=ϕ1α0ϕ0ϕ1α2kϕ0(1,0)\hat{X}_{2k}=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}(1,0) converges to 𝔭𝔏\mathfrak{p}_{\mathfrak{L}} and that X^2k+1=ϕ1α0ϕ0ϕ1α2k+1ϕ0(1,0)\hat{X}_{2k+1}=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k+1}}\phi_{0}(1,0) converges to 𝔮𝔏\mathfrak{q}_{\mathfrak{L}}. Using that (1,0)=ϕ1ϕ0(0,0)(1,0)=\phi_{1}\phi_{0}(0,0), we can write X^2k+1=ϕ1α0ϕ0ϕ1α2k+2ϕ0(0,0)\hat{X}_{2k+1}=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k+2}}\phi_{0}(0,0), so that limk+ϕ1α0ϕ0ϕ1α2kϕ0(0,0)=𝔮𝔏\lim_{k\rightarrow+\infty}\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}(0,0)=\mathfrak{q}_{\mathfrak{L}}.
Let now (x,y)widebar{(0,0)}(x,y)\in\widebar{\triangle}\setminus\left\{(0,0)\right\} and notice that if limk+ϕ1α0ϕ0ϕ1α2kϕ0(x,y)\lim_{k\rightarrow+\infty}\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}(x,y) exists then it must lie on the line segment 𝔏\mathfrak{L}. We start considering the case when (x,y)(x,y) is on the boundary of widebar{(0,0)}\widebar{\triangle}\setminus\{(0,0)\}. Suppose that (x,y)Λ{(0,0)}(x,y)\in\Lambda\setminus\left\{(0,0)\right\}, so that (x,y)=(ξ,0)(x,y)=(\xi,0) for a certain ξ(0,1]\xi\in(0,1], and let ss be a positive integer such that 1s<ξ\frac{1}{s}<\xi. Thus (1s,0)=(1,0)s1(0,0)(\frac{1}{s},0)=(1,0)\oplus_{s-1}(0,0) . Writing X^2k\hat{X}_{2k} and X^2k1\hat{X}_{2k-1} as above, we have

ϕ1α0ϕ0ϕ1α2kϕ0(1s,0)=ϕ1α0ϕ0ϕ1α2kϕ0(1,0)s1ϕ1α0ϕ0ϕ1α2kϕ0(0,0)=X^2ks1X^2k1.\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}\left(\frac{1}{s},0\right)=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}(1,0)\oplus_{s-1}\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}(0,0)=\hat{X}_{2k}\oplus_{s-1}\hat{X}_{2k-1}.

where we have used that ϕ0\phi_{0} and ϕ1\phi_{1} commute with the mediant operation. Moreover ϕ0\phi_{0} and ϕ1\phi_{1} are monotonic along line segments (with respect to the lexicographic order) and (1s,0)lex(ξ,0)lex(1,0)(\frac{1}{s},0)\leq_{lex}(\xi,0)\leq_{lex}(1,0), hence

0d(ϕ1α0ϕ0ϕ1α2kϕ0(ξ,0),X^2k)d(ϕ1α0ϕ0ϕ1α2kϕ0(1s,0),X^2k)=d(X^2ks1X^2k1,X^2k).0\leq d(\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}\left(\xi,0\right),\hat{X}_{2k})\leq d\left(\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}\left(\frac{1}{s},0\right),\hat{X}_{2k}\right)=d\left(\hat{X}_{2k}\oplus_{s-1}\hat{X}_{2k-1},\hat{X}_{2k}\right).

Lemma 4.3-(iii) gives d(X^2ks1X^2k1,X^2k)0d(\hat{X}_{2k}\oplus_{s-1}\hat{X}_{2k-1},\hat{X}_{2k})\rightarrow 0 for k+k\rightarrow+\infty, thus d(ϕ1α0ϕ0ϕ1α2kϕ0(ξ,0),X^2k)0d(\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}(\xi,0),\hat{X}_{2k})\rightarrow 0, which means that limk+ϕ1α0ϕ0ϕ1α2kϕ0(ξ,0)=𝔭𝔏\lim_{k\rightarrow+\infty}\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}(\xi,0)=\mathfrak{p}_{\mathfrak{L}}. If (x,y)Υ(x,y)\in\Upsilon or (x,y)Σ{(0,0)}(x,y)\in\Sigma\setminus\{(0,0)\} we can argue as above to conclude that limk+ϕ1α0ϕ0ϕ1α2kϕ0(x,y)=𝔭𝔏\lim_{k\rightarrow+\infty}\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}(x,y)=\mathfrak{p}_{\mathfrak{L}}.
Finally, let (x,y)(x,y) be an interior point of the triangle \triangle and let AA be a triangle containing (x,y)(x,y) as interior point and having all the vertices along {(0,0)}\partial\triangle\setminus\{(0,0)\}. The maps ϕ0\phi_{0} and ϕ1\phi_{1} map triangles into triangles and the same does the composite map ϕ1α0ϕ0ϕ1α2kϕ0\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}. Thus for all k0k\geq 0 the image ϕ1α0ϕ0ϕ1α2kϕ0(A)\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}(A) is a triangle containing ϕ1α0ϕ0ϕ1α2kϕ0(x,y)\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}(x,y) as an interior point. The thesis follows by observing that ϕ1α0ϕ0ϕ1α2kϕ0(A)\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{2k}}\phi_{0}(A) shrinks to 𝔭𝔏\mathfrak{p}_{\mathfrak{L}} because its three vertices converge to 𝔭𝔏\mathfrak{p}_{\mathfrak{L}}. ∎

The last result will be important in Section 5 to better understand the coding of real pairs in the non-convergent case.

5. A two-dimensional representation

In this section we begin to use our construction of the Triangular tree and the properties of the maps ϕ0\phi_{0}, ϕ1\phi_{1} and ϕ2\phi_{2} defined in (3.3)-(3.5) to introduce a new representation of real pairs of numbers in widebar\widebar{\triangle} by combining triangle sequences and continued fraction expansions. We recall that for any finite binary word ω{0,1}\omega\in\{0,1\}^{*} of length nn, we let ϕωϕω0ϕω1ϕωn1\phi_{\omega}\coloneqq\phi_{\omega_{0}}\circ\phi_{\omega_{1}}\circ\dots\circ\phi_{\omega_{n-1}}.

Lemma 5.1.

Let (α,β)(\alpha,\beta) be a point of 𝑤𝑖𝑑𝑒𝑏𝑎𝑟\widebar{\triangle} with finite triangle sequence [α0,αk][\alpha_{0}\,\,\ldots,\,\alpha_{k}]. If (α,β)(\alpha,\beta) is an interior point, then:

  1. (i)

    (α,β)ϕωϕ2(Σ{(0,0),(1,1)})(\alpha,\beta)\in\phi_{\omega}\phi_{2}(\Sigma\setminus\{(0,0),(1,1)\}), where ω=1α001αk101αk0\omega=1^{\alpha_{0}}0\cdots 1^{\alpha_{k-1}}01^{\alpha_{k}}0, so that there exists a unique ξ(0,1)\xi\in(0,1) such that

    (5.1) (α,β)=ϕ1α0ϕ0ϕ1αkϕ0ϕ2(ξ,ξ);(\alpha,\beta)=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k}}\phi_{0}\phi_{2}(\xi,\xi);
  2. (ii)

    ξ\xi is rational if and only if (α,β)2(\alpha,\beta)\in\mathbb{Q}^{2}.

Proof.

(i) By definition of triangle sequences we have

Tj(α,β)αjfor all j=0,,kandTk+1(α,β)Λ.T^{j}(\alpha,\beta)\in\triangle_{\alpha_{j}}\quad\text{for all $j=0,\,\ldots,\,k$}\quad\text{and}\quad T^{k+1}(\alpha,\beta)\in\Lambda.

Thus Tk+1(α,β)=(ξ,0)=ϕ2(ξ,ξ)T^{k+1}(\alpha,\beta)=(\xi,0)=\phi_{2}(\xi,\xi) for some ξ(0,1)\xi\in(0,1). From (3.2) and arguing as in Lemma 4.1 the thesis follows.
(ii) From (i) we have that (α,β)(\alpha,\beta) can be obtained from (ξ,0)(\xi,0) with the application of a finite number of maps between ϕ0\phi_{0} and ϕ1\phi_{1}, which are linear fractional maps. ∎

Remark 5.2.

Let (α,β)(\alpha,\beta) be an interior point having finite triangle sequence [α0,,αk][\alpha_{0},\,\ldots,\,\alpha_{k}]. Note that αk>0\alpha_{k}>0 because the only points in widebar\widebar{\triangle} for which the triangle sequence is defined and ends with 0 are located on (ΣΥ)Λ(\Sigma\cup\Upsilon)\setminus\Lambda. In light of this simple remark we can rewrite (5.1) as

(α,β)=ϕ1α0ϕ0ϕ1αk1(11+ξ,ξ1+ξ),(\alpha,\beta)=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k}-1}\left(\frac{1}{1+\xi},\frac{\xi}{1+\xi}\right),

so that (α,β)ϕω()(\alpha,\beta)\in\phi_{\omega}(\ell) where ω=1α001αk1\omega=1^{\alpha_{0}}0\cdots 1^{\alpha_{k}-1}.

Definition 5.3.

Let (α,β)(\alpha,\beta) be an interior point of widebar\widebar{\triangle} with finite triangle sequence [α0,,αk][\alpha_{0},\,\ldots,\,\alpha_{k}], and let [a1,a2,][a_{1},\,a_{2},\,\ldots] be the continued fraction expansion of the unique number ξ\xi given by Lemma 5.1-(ii). We associate to (α,β)(\alpha,\beta) the representation given by the pair

([α0,,αk],[a1,]).\Big{(}[\alpha_{0},\ldots,\alpha_{k}],[a_{1},\ldots]\Big{)}.

As for the second component, we write [0][0] and [1][1] to denote the continued fraction expansion of 0 and 11, respectively.

Remark 5.4.

If (α,β)(\alpha,\beta) is a rational pair, then ξ\xi is rational (Lemma 5.1-(iii)) and thus its continued fraction expansion is finite, say [a1,,an][a_{1},\,\ldots,\,a_{n}]. In this case we further assume that an>1a_{n}>1 when n>1n>1.

We give further details for the coding of boundary rational points and, in particular, for the vertices of the triangle \triangle.

  1. (1)

    A point on Σ{(0,0),(1,1)}\Sigma\setminus\{(0,0),(1,1)\} is of the kind (ξ,ξ)(\xi,\xi), for some ξ\xi in the open unit interval. If [a1,][a_{1},\,\ldots] is the continued fraction expansion of ξ\xi, then 1a11<ξ1a1\frac{1}{a_{1}-1}<\xi\leq\frac{1}{a_{1}}, so that (ξ,ξ)a11(\xi,\xi)\in\triangle_{a_{1}-1}. In case ξ=1a1\xi=\frac{1}{a_{1}} we have T(ξ,ξ)=(1,0)ΛT(\xi,\xi)=(1,0)\in\Lambda, so that the triangle sequence of (ξ,ξ)(\xi,\xi) is [a11][a_{1}-1]. Otherwise, if 1a11<ξ<1a1\frac{1}{a_{1}-1}<\xi<\frac{1}{a_{1}}, we have T(ξ,ξ)Υ{(1,0)}T(\xi,\xi)\in\Upsilon\setminus\{(1,0)\}, so that the triangle sequence of (ξ,ξ)(\xi,\xi) is [a11,0][a_{1}-1,0]. Thus we set the representation of (ξ,ξ)(\xi,\xi) to be respectively

    ([a11],[a1])or([a11,0],[a1,]).\Big{(}[a_{1}-1],[a_{1}]\Big{)}\quad\text{or}\quad\Big{(}[a_{1}-1,0],[a_{1},\,\ldots]\Big{)}.
  2. (2)

    A point in Λ\Lambda is of the kind (ξ,0)(\xi,0), for some ξ\xi in the unit interval, and its triangle sequence is not defined and assumed to be empty. We thus represent (ξ,0)(\xi,0) with

    ([],[a1,]),\Big{(}[],[a_{1},\,\ldots]\Big{)},

    where [a1,][a_{1},\,\ldots] is the continued fraction expansion of ξ\xi. In particular, the representation of the vertices (0,0)(0,0) and (1,0)(1,0) are ([],[0])\left([],[0]\right) and ([],[1])\left([],[1]\right), respectively.

  3. (3)

    A point in Υ{(1,0)}\Upsilon\setminus\{(1,0)\} is of the kind (1,ξ)(1,\xi), for some ξ(0,1]\xi\in(0,1], and has triangle sequence [0][0]. The representation of (1,ξ)(1,\xi) is thus

    ([0],[a1,]),\Big{(}[0],[a_{1},\,\ldots]\Big{)},

    where [a1,][a_{1},\,\ldots] is the continued fraction expansion of ξ\xi. In particular, the representation of the vertex (1,1)(1,1) is ([0],[1])\left([0],[1]\right).

The representation of real pairs with infinite triangle sequences depends on the convergence of the sequence. We have seen in Section 4 that if (α,β)(\alpha,\beta) is a real pair in widebar\widebar{\triangle} with convergent infinite triangle sequence [α0,α1,][\alpha_{0},\alpha_{1},\dots], then

{(α,β)}=k0(α0,,αk)\{(\alpha,\beta)\}=\bigcap_{k\geq 0}\,\triangle(\alpha_{0},\,\ldots,\,\alpha_{k})

This shows that in this case it is enough to associate to (α,β)(\alpha,\beta) its triangle sequence, since

(5.2) limkϕ1α0ϕ0ϕ1αkϕ0(x,y)=(α,β)\lim_{k\to\infty}\,\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k}}\phi_{0}(x,y)=(\alpha,\beta)

for all (x,y)widebar(x,y)\in\widebar{\triangle}. In the definition below we choose (x,y)=(12,0)(x,y)=(\frac{1}{2},0), but other choices would work as well.

Definition 5.5.

Let (α,β)(\alpha,\beta) be a point of widebar\widebar{\triangle} with convergent infinite triangle sequence [α0,α1,][\alpha_{0},\alpha_{1},\ldots]. We associate to (α,β)(\alpha,\beta) the representation given by the pair

([α0,α1,],[2]).\Big{(}[\alpha_{0},\alpha_{1},\ldots],[2]\Big{)}.

Let us now consider the case of non-convergent infinite triangle sequences. In this case a line segment 𝔏\mathfrak{L} is associated to such a sequence, and we refer to Proposition 4.4 for the notation of its endpoints and more properties. Using these results we give the following definition.

Definition 5.6.

Let (α,β)(\alpha,\beta) be a point of widebar\widebar{\triangle} with non-convergent infinite triangle sequence [α0,α1,][\alpha_{0},\alpha_{1},\ldots]. Then (α,β)(\alpha,\beta) belongs to a line segment 𝔏\mathfrak{L} of real pairs having the same triangle sequence, with endpoints 𝔭𝔏\mathfrak{p}_{\mathfrak{L}} and 𝔮𝔏\mathfrak{q}_{\mathfrak{L}}. Then we consider

([α0,α1,],[2]).\Big{(}[\alpha_{0},\alpha_{1},\ldots],[2]\Big{)}.

to be the representation of the segment 𝔏\mathfrak{L}.

6. The Triangular coding

In this section we use the ideas exposed in Section 5 to introduce a coding for the rational pairs in the Triangular tree in an analogous way as for the Farey coding. As recalled in Section 2, the continued fraction expansion of a rational number in (0,1)(0,1) is related to the path on the Farey tree to reach it starting from the root 12\frac{1}{2}. This information is contained in the {L,R}\{L,R\} coding, which in turn can be seen as the action by right multiplication of two matrices L,RSL(2,)L,R\in SL(2,\mathbb{Z}) on the matrix representation of rational numbers. We now generalise this setting for the two-dimensional case by first defining a coding for the possible moves from parents to children along the Triangular tree in figure 3. Then we introduce a matrix representation for rational pairs and convert the action of the moves on the tree into the action by right multiplication of specific SL(3,)SL(3,\mathbb{Z}) matrices.

Definition 6.1.

Let 𝔖\mathfrak{S} be a line segment in widebar\widebar{\triangle} obtained as a counterimage of Σ{(0,0),(1,1)}\Sigma\setminus\left\{(0,0),(1,1)\right\} by a combination ϕω~\phi_{\tilde{\omega}} of the maps ϕ0\phi_{0}, ϕ1\phi_{1} and ϕ2\phi_{2}, where ω~\tilde{\omega} is a finite word which is either empty or is the concatenation (ω2)(\omega 2) with ω{0,1}\omega\in\{0,1\}^{*}. We consider on 𝔖\mathfrak{S} the orientation induced by the lexicographic ordering on Σ\Sigma by the map ϕω~\phi_{\tilde{\omega}}.
A rational pair (pq,rq)(\frac{p}{q},\frac{r}{q}) in 𝔖\mathfrak{S} is obtained in the Triangular tree as the mediant of the neighbouring pairs, its parents. Then we define two actions, LL and RR, on (pq,rq)(\frac{p}{q},\frac{r}{q}), as follows: (pq,rq)L(\frac{p}{q},\frac{r}{q})\,L is the rational pair obtained as the mediant of (pq,rq)(\frac{p}{q},\frac{r}{q}) with its left parent and (pq,rq)R(\frac{p}{q},\frac{r}{q})\,R is the rational pair obtained as the mediant of (pq,rq)(\frac{p}{q},\frac{r}{q}) with its right parent.

Lemma 6.2.

Let ab(0,1)\frac{a}{b}\in\mathbb{Q}\cap(0,1) and let [a1,,an][a_{1},\ldots,a_{n}] be its continued fraction expansion. We have

(ab,ab)={(12,12)La11Ra2Lan1Ran1,if n is even(12,12)La11Ra2Ran1Lan1if n is odd\left(\frac{a}{b},\frac{a}{b}\right)=\begin{cases}\left(\frac{1}{2},\frac{1}{2}\right)L^{a_{1}-1}R^{a_{2}}\cdots L^{a_{n-1}}R^{a_{n}-1}\,,&\text{if $n$ is even}\\[2.84544pt] \left(\frac{1}{2},\frac{1}{2}\right)L^{a_{1}-1}R^{a_{2}}\cdots R^{a_{n-1}}L^{a_{n}-1}\,&\text{if $n$ is odd}\end{cases}

The same combination of actions sends (12,0)\left(\frac{1}{2},{0}\right) to (ab,0)\left(\frac{a}{b},{0}\right) on Λ\Lambda and (1,12)\left({1},\frac{1}{2}\right) to (1,ab)\left({1},\frac{a}{b}\right) on Υ\Upsilon.

Proof.

According to (2.3), the path on the Farey tree from 12\frac{1}{2} to ab\frac{a}{b} is La11Ra2Lan1Ran1L^{a_{1}-1}R^{a_{2}}\cdots L^{a_{n-1}}R^{a_{n}-1} if nn is even and La11Ra2Ran1Lan1L^{a_{1}-1}R^{a_{2}}\cdots R^{a_{n-1}}L^{a_{n}-1} if nn is odd. By the definition of LL and RR in Definition 6.1, this is also the path to reach (ab,ab)\left(\frac{a}{b},\frac{a}{b}\right) from (12,12)\left(\frac{1}{2},\frac{1}{2}\right). The same holds on Λ\Lambda and Υ\Upsilon. ∎

Remark 6.3.

In the case n=1n=1 the above formula reads (ab,ab)=(12,12)La12\left(\frac{a}{b},\frac{a}{b}\right)=\left(\frac{1}{2},\frac{1}{2}\right)L^{a_{1}-2}. The same holds also in Theorem 6.6 below.

The previous lemma shows that the basic moves LL and RR are enough to reach every boundary rational pair starting from the midpoints of the sides of \triangle. We now describe how to reach interior rational pairs along the tree always starting from (12,12)\left(\frac{1}{2},\frac{1}{2}\right). Recall that we denote by \ell the open line segment joining (1,0)(1,0) and (12,12)\left(\frac{1}{2},\frac{1}{2}\right). Since interior rational pairs are located on backward images ϕω()\phi_{\omega}(\ell) with ω{0,1}\omega\in\{0,1\}^{*}, our strategy is divided into two steps: first we describe a path from (12,12)\left(\frac{1}{2},\frac{1}{2}\right) to the mediant between the endpoints of ϕω()\phi_{\omega}(\ell), and then we encode the sequence of moves along ϕω()\phi_{\omega}(\ell) to reach the considered rational pair.

Definition 6.4.

Let ω{0,1}\omega\in\{0,1\}^{*} with n=|ω|0n=|\omega|\geq 0 and consider the triangle ω=ϕω(widebar)\triangle_{\omega}=\phi_{\omega}(\widebar{\triangle}), partitioned by ϕω()\phi_{\omega}(\ell) into two subtriangles. The action of the symbol II on ϕω(12,12)\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right), which we denote as a right action by ϕω(12,12)I\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right)I, gives the mediant between ϕω(12,12)\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right) and ϕω(1,0)\phi_{\omega}(1,0). In other words,

ϕω(12,12)Iϕω(12,12)ϕω(1,0)=ϕω(23,13).\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right)I\coloneqq\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right)\oplus\phi_{\omega}(1,0)=\phi_{\omega}\left(\frac{2}{3},\frac{1}{3}\right).

We now give a geometric interpretation of this definition, also clarified by Figure 6. The point ϕω(12,12)\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right) is the right endpoint of ϕω()\phi_{\omega}(\ell) and the action of II is to step from ϕω(12,12)\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right) to the mediant between the two endpoints of ϕω()\phi_{\omega}(\ell), which is one of the children of ϕω(12,12)\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right).

ϕω(0,0)\phi_{\omega}(0,0)ϕω(1,0)\phi_{\omega}(1,0)ϕω(1,1)\phi_{\omega}(1,1)ϕω(12,12)\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right)\bulletϕω(12,12)I\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right)I
Figure 6. Right action of the symbol II.
Lemma 6.5.

Let n0n\geq 0 and let ω{0,1}\omega\in\{0,1\}^{*} with |ω|=n|\omega|=n. Then

ϕω(12,12)=(12,12)i=1nWi,\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right)=\left(\frac{1}{2},\frac{1}{2}\right)\prod_{i=1}^{n}W_{i},

where Wi=IW_{i}=I if ωi=0\omega_{i}=0 and Wi=LW_{i}=L if ωi=1\omega_{i}=1.

Proof.

We argue by induction on n0n\geq 0. The conclusion is trivial when n=0n=0. Suppose that the thesis holds for all the finite binary words of length nn and let ω\omega be a word with |ω|=n+1|\omega|=n+1. Thus ω=ω~0\omega=\tilde{\omega}0 or ω=ω~1\omega=\tilde{\omega}1 for some word ω~\tilde{\omega} of length nn. By definition of II we have ϕω~(12,12)I=ϕω~(23,13)=ϕω~0(12,12)\phi_{\tilde{\omega}}\left(\frac{1}{2},\frac{1}{2}\right)I=\phi_{\tilde{\omega}}\left(\frac{2}{3},\frac{1}{3}\right)=\phi_{\tilde{\omega}0}\left(\frac{1}{2},\frac{1}{2}\right), thus the thesis holds for ω~0\tilde{\omega}0. For ω~1\tilde{\omega}1 note that the left and right parents of the point ϕω(12,12)\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right) are ϕω(0,0)\phi_{\omega}(0,0) and ϕω(1,1)\phi_{\omega}(1,1), respectively. Hence

ϕω~(12,12)L=ϕω~(12,12)ϕω~(0,0)=ϕω~(13,13)=ϕω~1(12,12),\phi_{\tilde{\omega}}\left(\frac{1}{2},\frac{1}{2}\right)L=\phi_{\tilde{\omega}}\left(\frac{1}{2},\frac{1}{2}\right)\oplus\phi_{\tilde{\omega}}(0,0)=\phi_{\tilde{\omega}}\left(\frac{1}{3},\frac{1}{3}\right)=\phi_{\tilde{\omega}1}\left(\frac{1}{2},\frac{1}{2}\right),

so that the thesis is also true for ω~1\tilde{\omega}1. ∎

Theorem 6.6.

Let (pq,rq)(\frac{p}{q},\frac{r}{q}) be the interior rational pair with representation ([α0,,αk],[a1,,an])([\alpha_{0},\ldots,\alpha_{k}],[a_{1},\ldots,a_{n}]) (see Definition 5.3) and let ω=1α001αk101αk1\omega=1^{\alpha_{0}}0\cdots 1^{\alpha_{k-1}}01^{\alpha_{k}-1}, so that (pq,rq)ϕω()(\frac{p}{q},\frac{r}{q})\in\phi_{\omega}(\ell). Then:

  1. (i)

    ϕω(12,12)=(12,12)Lα0ILαk1ILαk1\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right)=\left(\frac{1}{2},\frac{1}{2}\right)L^{\alpha_{0}}I\cdots L^{\alpha_{k-1}}IL^{\alpha_{k}-1};

  2. (ii)

    (pq,rq)={ϕω(12,12)ILa11Ra2Lan1Ran1if n is evenϕω(12,12)ILa11Ra2Ran1Lan1,if n is odd\left(\frac{p}{q},\frac{r}{q}\right)=\begin{cases}\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right)IL^{a_{1}-1}R^{a_{2}}\cdots L^{a_{n-1}}R^{a_{n}-1}\,&\text{if $n$ is even}\\[2.84544pt] \phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right)IL^{a_{1}-1}R^{a_{2}}\cdots R^{a_{n-1}}L^{a_{n}-1}\,,&\text{if $n$ is odd}\end{cases}.

Hence

(pq,rq)={(12,12)Lα0ILαk1ILαk1ILa11Ra2Lan1Ran1,if n is even(12,12)Lα0ILαk1ILαk1ILa11Ra2Ran1Lan1,if n is odd\left(\frac{p}{q},\frac{r}{q}\right)=\begin{cases}\left(\frac{1}{2},\frac{1}{2}\right)L^{\alpha_{0}}I\cdots L^{\alpha_{k-1}}IL^{\alpha_{k}-1}IL^{a_{1}-1}R^{a_{2}}\cdots L^{a_{n-1}}R^{a_{n}-1}\,,&\text{if $n$ is even}\\[2.84544pt] \left(\frac{1}{2},\frac{1}{2}\right)L^{\alpha_{0}}I\cdots L^{\alpha_{k-1}}IL^{\alpha_{k}-1}IL^{a_{1}-1}R^{a_{2}}\cdots R^{a_{n-1}}L^{a_{n}-1}\,,&\text{if $n$ is odd}\end{cases}
Proof.

(i) It is a straightforward consequence of Lemma 6.5.
(ii) Let ab\frac{a}{b} the unique rational number associated to (pq,rq)(\frac{p}{q},\frac{r}{q}) according to Lemma 5.1-(ii), so that by definition [a1,,an][a_{1},\ldots,a_{n}] is its continued fraction expansion. Thus (1,ab)\left({1},\frac{a}{b}\right) can be reached from (1,12)\left({1},\frac{1}{2}\right) with the sequence La11Ra2Lan1Ran1L^{a_{1}-1}R^{a_{2}}\cdots L^{a_{n-1}}R^{a_{n}-1} if nn is even or with La11Ra2Ran1Lan1L^{a_{1}-1}R^{a_{2}}\cdots R^{a_{n-1}}L^{a_{n}-1} if nn is odd, from Lemma 6.2. Since the maps ϕ0\phi_{0} and ϕ1\phi_{1} commute with the mediant operation, it is easy to prove that also the parent-child relationship is preserved. Thus by the above sequence of LL and RR moves one obtains (pq,rq)(\frac{p}{q},\frac{r}{q}) from ϕω1(1,12)=ϕω(23,13)=ϕω(12,12)I\phi_{\omega 1}\left({1},\frac{1}{2}\right)=\phi_{\omega}\left(\frac{2}{3},\frac{1}{3}\right)=\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right)I, which is the mediant between the endpoints of ϕω()\phi_{\omega}(\ell). ∎

To each interior rational pair we have thus associated a finite word over the alphabet {L,R,I}\{L,R,I\}, with the geometric meaning of telling how to move on the Triangular tree to reach the rational pair under consideration starting from the root (12,12)\left(\frac{1}{2},\frac{1}{2}\right). In particular, this finite word is the concatenation between a word over {L,I}\{L,I\} ending necessarily with the symbol II and a word over {L,R}\{L,R\}: the first word gives the path to reach ϕω(12,12)I\phi_{\omega}\left(\frac{1}{2},\frac{1}{2}\right)I, which is the rational point on ϕω()\phi_{\omega}(\ell) appearing on the level of the tree with smallest index and also the mediant between its endpoints; and the second word gives the moves along the line segment ϕω()\phi_{\omega}(\ell) to reach the given point from the mediant of its endpoints.

To continue the analogy with the one-dimensional case, we now convert the actions of the symbols LL, RR, and II, into the actions by right multiplication of three SL(3,)SL(3,\mathbb{Z}) matrices, using a matrix representation for the rational pairs in widebar\widebar{\triangle}. Towards this aim we recall the correspondence between rational pairs and three-dimensional vectors we have introduced in Section 4.

Definition 6.7.

Let (pq,rq)(\frac{p}{q},\frac{r}{q}) be a rational pair in widebar𝒯1\widebar{\triangle}\setminus\mathcal{T}_{-1}. We associate to (pq,rq)(\frac{p}{q},\frac{r}{q}) the 3×33\times 3 matrix 𝔪(pq,rq)\mathfrak{m}\left(\frac{p}{q},\frac{r}{q}\right) defined as follows. The first two columns are respectively the right and the left parents of (pq,rq)(\frac{p}{q},\frac{r}{q}), expressed as three-dimensional vectors. The third column depends on the location of the point:

  1. (i)

    if the pair is a boundary point, the third column is the vertex of widebar\widebar{\triangle} which is opposite to the side containing the pair;

  2. (ii)

    if (pq,rq)ϕω()(\frac{p}{q},\frac{r}{q})\in\phi_{\omega}(\ell) for ω{0,1}\omega\in\{0,1\}^{*}, the third column is the vertex “2” of ω\triangle_{\omega}, that is ϕω(1,1)\phi_{\omega}(1,1).

The three-dimensional vector associated to (pq,rq)(\frac{p}{q},\frac{r}{q}) is obtained as the sum of the first two columns of its matrix.

Since they will take on great importance, we explicitly show the matrices representing the midpoints of the three sides of \triangle:

𝔪(12,12)=(111101100),𝔪(12,0)=(111101001),𝔪(1,12)=(111110100).\mathfrak{m}\left(\frac{1}{2},\frac{1}{2}\right)=\begin{pmatrix}1&1&1\\ 1&0&1\\ 1&0&0\end{pmatrix},\quad\mathfrak{m}\left(\frac{1}{2},{0}\right)=\begin{pmatrix}1&1&1\\ 1&0&1\\ 0&0&1\end{pmatrix},\quad\mathfrak{m}\left({1},\frac{1}{2}\right)=\begin{pmatrix}1&1&1\\ 1&1&0\\ 1&0&0\end{pmatrix}.

Let us now define

(6.1) L(100110001),R(110010001),I(101100010).L\coloneqq\begin{pmatrix}1&0&0\\ 1&1&0\\ 0&0&1\end{pmatrix},\quad R\coloneqq\begin{pmatrix}1&1&0\\ 0&1&0\\ 0&0&1\end{pmatrix},\quad I\coloneqq\begin{pmatrix}1&0&1\\ 1&0&0\\ 0&1&0\end{pmatrix}.

Note that all the above matrices are in SL(3,)SL(3,\mathbb{Z}) and that LL and RR extend their 2×22\times 2 counterpart defined in (2.2). A straightforward computation shows that the action of each of the above matrices by right multiplication on the matrix representing a rational pair yields the matrix of the child obtained according to the move bearing the same name of the matrix. As a direct consequence of Lemma 6.2 and Theorem 6.6, we thus have the following properties.

  1. (1)

    Let ab(0,1)\frac{a}{b}\in\mathbb{Q}\cap(0,1) and let [a1,,an][a_{1},\,\ldots,\,a_{n}] be its continued fraction expansion. Then

    𝔪(ab,ab)={𝔪(12,12)La11Ra2Lan1Ran1,if n is even𝔪(12,12)La11Ra2Ran1Lan1,if n is odd\mathfrak{m}\left(\frac{a}{b},\frac{a}{b}\right)=\begin{cases}\mathfrak{m}\left(\frac{1}{2},\frac{1}{2}\right)L^{a_{1}-1}R^{a_{2}}\cdots L^{a_{n-1}}R^{a_{n}-1}\,,&\text{if $n$ is even}\\[2.84544pt] \mathfrak{m}\left(\frac{1}{2},\frac{1}{2}\right)L^{a_{1}-1}R^{a_{2}}\cdots R^{a_{n-1}}L^{a_{n}-1}\,,&\text{if $n$ is odd}\end{cases}

    The same right action yields 𝔪(ab,0)\mathfrak{m}\left(\frac{a}{b},{0}\right) from 𝔪(12,0)\mathfrak{m}\left(\frac{1}{2},{0}\right) and 𝔪(1,ab)\mathfrak{m}\left({1},\frac{a}{b}\right) from 𝔪(1,12)\mathfrak{m}\left({1},\frac{1}{2}\right).

  2. (2)

    Let (pq,rq)(\frac{p}{q},\frac{r}{q}) be the interior rational pair with representation ([α0,,αk],[a1,,an])([\alpha_{0},\ldots,\alpha_{k}],[a_{1},\ldots,a_{n}]). Then

    𝔪(pq,rq)={𝔪(12,12)Lα0ILαk1ILαk1ILa11Ra2Lan1Ran1,if n is even𝔪(12,12)Lα0ILαk1ILαk1ILa11Ra2Ran1Lan1,if n is odd\mathfrak{m}\left(\frac{p}{q},\frac{r}{q}\right)=\begin{cases}\mathfrak{m}\left(\frac{1}{2},\frac{1}{2}\right)L^{\alpha_{0}}I\cdots L^{\alpha_{k-1}}IL^{\alpha_{k}-1}IL^{a_{1}-1}R^{a_{2}}\cdots L^{a_{n-1}}R^{a_{n}-1}\,,&\text{if $n$ is even}\\[2.84544pt] \mathfrak{m}\left(\frac{1}{2},\frac{1}{2}\right)L^{\alpha_{0}}I\cdots L^{\alpha_{k-1}}IL^{\alpha_{k}-1}IL^{a_{1}-1}R^{a_{2}}\cdots R^{a_{n-1}}L^{a_{n}-1}\,,&\text{if $n$ is odd}\end{cases}
Example.

Consider the point in with representation ([2,0,1,1],[2,2])([2,0,1,1],[2,2]). Using =ϕ1ϕ0ϕ2(Σ{(0,0),(1,1)})\ell=\phi_{1}\phi_{0}\phi_{2}(\Sigma\setminus\left\{(0,0),(1,1)\right\}) and (5.1), our point is located along the open line segment ϕω()\phi_{\omega}(\ell) with ω=110010\omega=110010, whose left and right endpoints can be readily computed and are respectively:

ϕ110010(1,0)=(38,28)andϕ110010(12,12)=(515,415).\phi_{110010}(1,0)=\left(\frac{3}{8},\frac{2}{8}\right)\quad\text{and}\quad\phi_{110010}\left(\frac{1}{2},\frac{1}{2}\right)=\left(\frac{5}{15},\frac{4}{15}\right).

By Theorem 6.6 we have that the path along the Triangular tree to reach this point from the root (12,12)\left(\frac{1}{2},\frac{1}{2}\right) is

L2IILIILR,L^{2}IILIILR,

which is the concatenation of the words L2IILIIL^{2}IILII and LRLR. The word L2IILIIL^{2}IILII brings (12,12)\left(\frac{1}{2},\frac{1}{2}\right) first to (12,12)L2IILI=(515,415)\left(\frac{1}{2},\frac{1}{2}\right)L^{2}IILI=\left(\frac{5}{15},\frac{4}{15}\right), the right endpoint of ϕ110010()\phi_{110010}(\ell), and then to (12,12)L2IILII=(515,415)I=(823,623)\left(\frac{1}{2},\frac{1}{2}\right)L^{2}IILII=\left(\frac{5}{15},\frac{4}{15}\right)I=\left(\frac{8}{23},\frac{6}{23}\right), the mediant between the endpoints of ϕ110010()\phi_{110010}(\ell). The word LRLR shows how to move along ϕ110010()\phi_{110010}(\ell) starting from (823,623)\left(\frac{8}{23},\frac{6}{23}\right) to obtain our point:

(823,623)LR=(1131,831)R=(1954,1454).\left(\frac{8}{23},\frac{6}{23}\right)LR=\left(\frac{11}{31},\frac{8}{31}\right)R=\left(\frac{19}{54},\frac{14}{54}\right).

Figure 7 shows the the path in the triangle to reach ϕ110010()\phi_{110010}(\ell) and the moves along this line segment to get to our point. The corresponding matrix multiplication yields

𝔪(12,12)L2IILIILR=(2331118114683),\mathfrak{m}\left(\frac{1}{2},\frac{1}{2}\right)L^{2}IILIILR=\begin{pmatrix}23&31&11\\ 8&11&4\\ 6&8&3\end{pmatrix},

which is by definition the matrix representing (1954,1454)\left(\frac{19}{54},\frac{14}{54}\right).

(12,12)\left(\frac{1}{2},\frac{1}{2}\right)(0,0)(0,0)(1,0)(1,0)(13,13)\left(\frac{1}{3},\frac{1}{3}\right)(14,14)\left(\frac{1}{4},\frac{1}{4}\right)(25,15)\left(\frac{2}{5},\frac{1}{5}\right)(38,28)\left(\frac{3}{8},\frac{2}{8}\right)(411,311)\left(\frac{4}{11},\frac{3}{11}\right)(515,415)\left(\frac{5}{15},\frac{4}{15}\right)\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet(38,28)\left(\frac{3}{8},\frac{2}{8}\right)(515,415)\left(\frac{5}{15},\frac{4}{15}\right)(823,623)\left(\frac{8}{23},\frac{6}{23}\right)(1131,831)\left(\frac{11}{31},\frac{8}{31}\right)(1954,1454)\left(\frac{19}{54},\frac{14}{54}\right)\bullet\bullet\bulletϕ110010()\phi_{110010}(\ell)(12,12)\left(\frac{1}{2},\frac{1}{2}\right)(0,0)(0,0)(1,0)(1,0)(13,13)\left(\frac{1}{3},\frac{1}{3}\right)(14,14)\left(\frac{1}{4},\frac{1}{4}\right)(25,15)\left(\frac{2}{5},\frac{1}{5}\right)(38,28)\left(\frac{3}{8},\frac{2}{8}\right)(411,311)\left(\frac{4}{11},\frac{3}{11}\right)(515,415)\left(\frac{5}{15},\frac{4}{15}\right)\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet(38,28)\left(\frac{3}{8},\frac{2}{8}\right)(515,415)\left(\frac{5}{15},\frac{4}{15}\right)(823,623)\left(\frac{8}{23},\frac{6}{23}\right)(1131,831)\left(\frac{11}{31},\frac{8}{31}\right)(1954,1454)\left(\frac{19}{54},\frac{14}{54}\right)\bullet\bullet\bulletϕ110010()\phi_{110010}(\ell)
Figure 7. Path to reach the right endpoint (515,415)\left(\frac{5}{15},\frac{4}{15}\right) of the line segment ϕ110010()\phi_{110010}(\ell) with the moves L2IILIL^{2}IILI. The line segment ϕ110010()\phi_{110010}(\ell) is enhanced in blue and the remaining moves ILRILR are represented in the bottom-right figure.

The last step to complete the similarity with the one-dimensional case is a way to use the coding we have introduced to express a rational pair as a counterimage of (12,12)(\frac{1}{2},\frac{1}{2}) and vice versa. The key observation is that using the local inverses ϕ0\phi_{0}, ϕ1\phi_{1}, and ϕ2\phi_{2}, it is possible to mimic the behaviour of the one-dimensional Farey map along Σ\Sigma in terms of the inverses ψ0\psi_{0} and ψ1\psi_{1} defined in (2.1). In fact a straightforward computation shows that for 0x10\leq x\leq 1 it holds

(6.2) ϕ1(x,x)=(ψ0(x),ψ0(x))andϕ02ϕ2(x,x)=(ψ1(x),ψ1(x)).\phi_{1}(x,x)=(\psi_{0}(x),\psi_{0}(x))\quad\text{and}\quad\phi_{0}^{2}\phi_{2}(x,x)=(\psi_{1}(x),\psi_{1}(x)).
Proposition 6.8.
  1. (i)

    Let ab\frac{a}{b} be a rational number with continued fraction expansion [a1,,an][a_{1},\,\ldots,\,a_{n}]. Then

    (ab,ab)\displaystyle\left(\frac{a}{b},\frac{a}{b}\right) =ϕ1a11ϕ02ϕ2ϕ1an11ϕ02ϕ2ϕ1an2(12,12),\displaystyle=\phi_{1}^{a_{1}-1}\phi_{0}^{2}\phi_{2}\cdots\phi_{1}^{a_{n-1}-1}\phi_{0}^{2}\phi_{2}\phi_{1}^{a_{n}-2}\left(\frac{1}{2},\frac{1}{2}\right),
    (ab,0)\displaystyle\left(\frac{a}{b},{0}\right) =ϕ2ϕ1a11ϕ02ϕ2ϕ1an11ϕ02ϕ2ϕ1an2(12,12),\displaystyle=\phi_{2}\circ\phi_{1}^{a_{1}-1}\phi_{0}^{2}\phi_{2}\cdots\phi_{1}^{a_{n-1}-1}\phi_{0}^{2}\phi_{2}\phi_{1}^{a_{n}-2}\left(\frac{1}{2},\frac{1}{2}\right),
    (1,ab)\displaystyle\left({1},\frac{a}{b}\right) =ϕ0ϕ2ϕ1a11ϕ02ϕ2ϕ1an11ϕ02ϕ2ϕ1an2(12,12).\displaystyle=\phi_{0}\circ\phi_{2}\circ\phi_{1}^{a_{1}-1}\phi_{0}^{2}\phi_{2}\cdots\phi_{1}^{a_{n-1}-1}\phi_{0}^{2}\phi_{2}\phi_{1}^{a_{n}-2}\left(\frac{1}{2},\frac{1}{2}\right).
  2. (ii)

    Let (pq,rq)(\frac{p}{q},\frac{r}{q}) be the interior rational pair with representation ([α0,,αk],[a1,,an])([\alpha_{0},\ldots,\alpha_{k}],[a_{1},\ldots,a_{n}]). Then

    (pq,rq)=ϕ1α0ϕ0ϕ1αkϕ0ϕ2ϕ1a11ϕ02ϕ2ϕ1an11ϕ02ϕ2ϕ1an2(12,12).\left(\frac{p}{q},\frac{r}{q}\right)=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k}}\phi_{0}\circ\phi_{2}\circ\phi_{1}^{a_{1}-1}\phi_{0}^{2}\phi_{2}\cdots\phi_{1}^{a_{n-1}-1}\phi_{0}^{2}\phi_{2}\phi_{1}^{a_{n}-2}\left(\frac{1}{2},\frac{1}{2}\right).
Proof.

(i) From the properties of the Farey map we have

ab=ψ0a11ψ1ψ0an2(12).\frac{a}{b}=\psi_{0}^{a_{1}-1}\psi_{1}\cdots\psi_{0}^{a_{n}-2}\left(\frac{1}{2}\right).

By using the correspondence between ψ0\psi_{0} and ϕ1\phi_{1} and between ψ1\psi_{1} and ϕ02ϕ2\phi_{0}^{2}\phi_{2} we can conclude for (ab,ab)\left(\frac{a}{b},\frac{a}{b}\right). The expressions for (ab,0)\left(\frac{a}{b},{0}\right) and (1,ab)\left({1},\frac{a}{b}\right) are now a trivial consequence.
(ii) Let ab\frac{a}{b} the unique rational number associated to (pq,rq)(\frac{p}{q},\frac{r}{q}) according to Lemma 5.1-(i), so that by definition [a1,,an][a_{1},\ldots,a_{n}] is its continued fraction expansion. Equation (5.1) gives (pq,rq)=ϕ1α0ϕ0ϕ1αkϕ0ϕ2(ab,ab)(\frac{p}{q},\frac{r}{q})=\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k}}\phi_{0}\phi_{2}\left(\frac{a}{b},\frac{a}{b}\right), so that the conclusion holds directly from (i). ∎

Remark 6.9.

(i) In the statement of the above proposition we have explicitly used the composition sign to emphasise the separation between the first group of maps, the map ϕ2\phi_{2}, and the second group, because they correspond to the different parts of the representation.
(ii) Note that the map ϕ2\phi_{2} separating the two groups of maps is an actual separator, meaning that if we exhibit a point as a counterimage of (12,12)\left(\frac{1}{2},\frac{1}{2}\right), it can be unambiguously identified. In fact it is the only map ϕ2\phi_{2} which is preceded by a single occurrence of ϕ0\phi_{0}.

Also the definition of rank can be extended to our two-dimensional setting. We say that rank(pq,rq)=m\operatorname*{rank}(\frac{p}{q},\frac{r}{q})=m if and only if x𝒯m𝒯m1x\in\mathcal{T}_{m}\setminus\mathcal{T}_{m-1}.

Corollary 6.10.

Let (pq,rq)(\frac{p}{q},\frac{r}{q}) be the rational pair with representation ([α0,,αk],[a1,,an])([\alpha_{0},\ldots,\alpha_{k}],[a_{1},\ldots,a_{n}]). Then

rank(pq,rq)={j=0kαj+i=1nai+k2,if the triangle sequence exists and αk>0i=1nai2,otherwise\operatorname*{rank}\left(\frac{p}{q},\frac{r}{q}\right)=\begin{cases}\sum_{j=0}^{k}\alpha_{j}+\sum_{i=1}^{n}a_{i}+k-2\,,&\text{if the triangle sequence exists and $\alpha_{k}>0$}\\[2.84544pt] \sum_{i=1}^{n}a_{i}-2\,,&\text{otherwise}\end{cases}
Proof.

By definition, rank(12,12)=0\operatorname*{rank}\left(\frac{1}{2},\frac{1}{2}\right)=0. Consider the expression of a rational pair as a backward image of (12,12)\left(\frac{1}{2},\frac{1}{2}\right) given in Proposition 6.8 and look separately to its (at most) three blocks of compositions, starting from the right to the left.

  1. (1)

    In the first group of maps, the one corresponding to the continued fraction expansion, each application of ϕ1\phi_{1} and each block ϕ02ϕ2\phi_{0}^{2}\phi_{2} increases the rank by 1 (ϕ02ϕ2\phi_{0}^{2}\phi_{2} does not increase the rank more because it is applied to points on Σ\Sigma). Thus the rank of ϕ1a11ϕ02ϕ2ϕ1an11ϕ02ϕ2ϕ1an2(12,12)\phi_{1}^{a_{1}-1}\phi_{0}^{2}\phi_{2}\cdots\phi_{1}^{a_{n-1}-1}\phi_{0}^{2}\phi_{2}\phi_{1}^{a_{n}-2}\left(\frac{1}{2},\frac{1}{2}\right) is i=1nai2\sum_{i=1}^{n}a_{i}-2.

  2. (2)

    The application of the separator ϕ2\phi_{2} appears if and only if the point is not along Σ\Sigma and anyway does not change the level.

  3. (3)

    The third block appears if and only if the point is along Υ\Upsilon or the point is interior. Moreover, when the third block appears, its right end is exactly one application of the map ϕ0\phi_{0}, but since it acts on a point on Λ\Lambda, it does not change the level. Each of the subsequent maps increases the level by 11, for a total increase of j=0kαj+k\sum_{j=0}^{k}\alpha_{j}+k. Thus note that the block related to the triangle sequence contributes to the rank if and only if the point is an interior rational pair.

Example.

As in Example Example, consider the point (1954,1454)\left(\frac{19}{54},\frac{14}{54}\right) whose representation is ([2,0,1,1],[2,2])([2,0,1,1],[2,2]). By Proposition 6.8, we have

(1954,1454)=ϕ12ϕ0ϕ0ϕ1ϕ0ϕ1ϕ0ϕ2ϕ1ϕ02ϕ2(12,12)=ϕ12ϕ0ϕ0ϕ1ϕ0ϕ1ϕ0ϕ2(25,25),\left(\frac{19}{54},\frac{14}{54}\right)=\phi_{1}^{2}\phi_{0}\phi_{0}\phi_{1}\phi_{0}\phi_{1}\phi_{0}\circ\phi_{2}\circ\phi_{1}\phi_{0}^{2}\phi_{2}\left(\frac{1}{2},\frac{1}{2}\right)=\phi_{1}^{2}\phi_{0}\phi_{0}\phi_{1}\phi_{0}\phi_{1}\phi_{0}\circ\phi_{2}\left(\frac{2}{5},\frac{2}{5}\right),

where we have emphasised that our point is a backward image of (25,25)\left(\frac{2}{5},\frac{2}{5}\right), which corresponds to the second component of the representation since 25=[2,2]\frac{2}{5}=[2,2].

7. Approximations for non-rational pairs

In this section we use the structure of the Triangular tree to define approximations of real pairs in widebar\widebar{\triangle} by rational pairs. This is particularly important for non-rational pairs, which have an infinite number of rational approximations.

Non-rational pairs have either an infinite triangle sequence or a finite triangle sequence with infinite associated continued fraction expansion. An analogous of Proposition 6.8 holds also for non-rational pairs, apart from the non-convergent infinite case, for which Proposition 4.4 basically implies that such a result is not possible.

7.1. Finite triangle sequence

Let us first consider the case of real pairs (α,β)(\alpha,\beta) with representation of the form ([α0,,αk],[a1,a2,])([\alpha_{0},\,\ldots,\,\alpha_{k}],[a_{1},\,a_{2},\,\ldots]) (the second component being finite or infinite). By extending the ideas of Section 6, we associate to a non-rational pair (α,β)(\alpha,\beta) the infinite word 𝒲\mathcal{W} over the alphabet {L,R,I}\{L,R,I\} defined to be

𝒲(α,β)=Lα0ILαk1ILαk1ILa11Ra2La3.\mathcal{W}(\alpha,\beta)=L^{\alpha_{0}}I\cdots L^{\alpha_{k-1}}IL^{\alpha_{k}-1}IL^{a_{1}-1}R^{a_{2}}L^{a_{3}}\cdots.

Moreover we define the approximations of (α,β)(\alpha,\beta) to be the rational pairs (mjsj,njsj)(\frac{m_{j}}{s_{j}},\frac{n_{j}}{s_{j}}) with coding 𝒲(j)\mathcal{W}_{(j)}, the prefix of 𝒲\mathcal{W} of length jj.

Example.

We consider the rational pair (1954,1454)\left(\frac{19}{54},\frac{14}{54}\right) with representation ([2,0,1,1],[2,2])([2,0,1,1],[2,2]). As shown in Example Example the word associated to this point is 𝒲=LLIILIILR\mathcal{W}=LLIILIILR. Thus its approximations are the following:

(12,12)𝒲(0)=ε([1],[2])(13,13)𝒲(1)=L([2],[3])(14,14)𝒲(2)=LL([3],[4])(25,15)𝒲(3)=LLI([3],[2])(38,28)𝒲(4)=LLII([2,1],[2])(411,311)𝒲(5)=LLIIL([2,1],[2])(515,415)𝒲(6)=LLIILI([2,0,2],[2])(823,623)𝒲(7)=LLIILII([2,0,1,1],[2])(1131,831)𝒲(8)=LLIILIIL([2,0,1,1],[3])(1954,1454)𝒲(9)=LLIILIILR([2,0,1,1],[2,2])\begin{array}[]{cll}\left(\frac{1}{2},\frac{1}{2}\right)&\mathcal{W}_{(0)}=\varepsilon&([1],[2])\\[5.69046pt] \left(\frac{1}{3},\frac{1}{3}\right)&\mathcal{W}_{(1)}=L&([2],[3])\\[5.69046pt] \left(\frac{1}{4},\frac{1}{4}\right)&\mathcal{W}_{(2)}=LL&([3],[4])\\[5.69046pt] \left(\frac{2}{5},\frac{1}{5}\right)&\mathcal{W}_{(3)}=LLI&([3],[2])\\[5.69046pt] \left(\frac{3}{8},\frac{2}{8}\right)&\mathcal{W}_{(4)}=LLII&([2,1],[2])\\[5.69046pt] \left(\frac{4}{11},\frac{3}{11}\right)&\mathcal{W}_{(5)}=LLIIL&([2,1],[2])\\[5.69046pt] \left(\frac{5}{15},\frac{4}{15}\right)&\mathcal{W}_{(6)}=LLIILI&([2,0,2],[2])\\[5.69046pt] \left(\frac{8}{23},\frac{6}{23}\right)&\mathcal{W}_{(7)}=LLIILII&([2,0,1,1],[2])\\[5.69046pt] \left(\frac{11}{31},\frac{8}{31}\right)&\mathcal{W}_{(8)}=LLIILIIL&([2,0,1,1],[3])\\[5.69046pt] \left(\frac{19}{54},\frac{14}{54}\right)&\mathcal{W}_{(9)}=LLIILIILR&([2,0,1,1],[2,2])\\[5.69046pt] \end{array}
Example.

We now consider the non-rational pair (α,β)=(12,21)(\alpha,\beta)=\left(\frac{1}{2},{\sqrt{2}-1}\right), with triangle sequence [1,1][1,1]. Since T2(α,β)=(212,0)T^{2}(\alpha,\beta)=\left(\frac{\sqrt{2}-1}{2},0\right) and 212=[4,1widebar]\frac{\sqrt{2}-1}{2}=[\widebar{4,1}], we have that the representation of (α,β)(\alpha,\beta) is ([1,1],[4,1widebar])\left([1,1],[\widebar{4,1}]\right) and thus the infinite word associated to (α,β)(\alpha,\beta) is

𝒲=LIIL3RL4widebar.\mathcal{W}=LIIL^{3}\widebar{RL^{4}}.

Thus its first ten approximations are the following:

(12,12)𝒲(0)=ε([1],[2])(13,13)𝒲(1)=L([2],[3])(24,14)𝒲(2)=LI([2],[2])(36,26)𝒲(3)=LII([1,1],[2])(48,38)𝒲(4)=LIIL([1,1],[3])(510,410)𝒲(5)=LIILL([1,1],[4])(612,512)𝒲(6)=LIILLL([1,1],[5])(1122,922)𝒲(7)=LIILLLR([1,1],[4,2])(1734,1434)𝒲(8)=LIILLLRL([1,1],[4,1,2])(2346,1946)𝒲(9)=LIILLLRLL([1,1],[4,1,3])\begin{array}[]{cll}\left(\frac{1}{2},\frac{1}{2}\right)&\mathcal{W}_{(0)}=\varepsilon&([1],[2])\\[5.69046pt] \left(\frac{1}{3},\frac{1}{3}\right)&\mathcal{W}_{(1)}=L&([2],[3])\\[5.69046pt] \left(\frac{2}{4},\frac{1}{4}\right)&\mathcal{W}_{(2)}=LI&([2],[2])\\[5.69046pt] \left(\frac{3}{6},\frac{2}{6}\right)&\mathcal{W}_{(3)}=LII&([1,1],[2])\\[5.69046pt] \left(\frac{4}{8},\frac{3}{8}\right)&\mathcal{W}_{(4)}=LIIL&([1,1],[3])\\[5.69046pt] \left(\frac{5}{10},\frac{4}{10}\right)&\mathcal{W}_{(5)}=LIILL&([1,1],[4])\\[5.69046pt] \left(\frac{6}{12},\frac{5}{12}\right)&\mathcal{W}_{(6)}=LIILLL&([1,1],[5])\\[5.69046pt] \left(\frac{11}{22},\frac{9}{22}\right)&\mathcal{W}_{(7)}=LIILLLR&([1,1],[4,2])\\[5.69046pt] \left(\frac{17}{34},\frac{14}{34}\right)&\mathcal{W}_{(8)}=LIILLLRL&([1,1],[4,1,2])\\[5.69046pt] \left(\frac{23}{46},\frac{19}{46}\right)&\mathcal{W}_{(9)}=LIILLLRLL&([1,1],[4,1,3])\\[5.69046pt] \vdots&&\end{array}

Since the triangle sequence is finite, if (α,β)(\alpha,\beta) is a non-rational pair, after a finite transient the approximations have the same triangle sequence as (α,β)(\alpha,\beta), and they are simply given by truncating the continued fraction expansion in the second component of the representation of (α,β)(\alpha,\beta). We are using the fact that

limnϕ1α0ϕ0ϕ1αkϕ0ϕ2ϕ1a11ϕ02ϕ2ϕ1an11ϕ02ϕ2ϕ1an2(12,12)=(α,β),\lim_{n\rightarrow\infty}\phi_{1}^{\alpha_{0}}\phi_{0}\cdots\phi_{1}^{\alpha_{k}}\phi_{0}\circ\phi_{2}\circ\phi_{1}^{a_{1}-1}\phi_{0}^{2}\phi_{2}\cdots\phi_{1}^{a_{n-1}-1}\phi_{0}^{2}\phi_{2}\phi_{1}^{a_{n}-2}\left(\frac{1}{2},\frac{1}{2}\right)=(\alpha,\beta)\,,

which follows by Proposition 6.8 and the properties of the continued fraction expansion translated into our framework with (6.2).

By learning from the previous examples we can give a general way of constructing the coding associated to the approximations of a real pair (α,β)(\alpha,\beta) with representation ([α0,,αk],[a1,a2,])([\alpha_{0},\,\ldots,\,\alpha_{k}],[a_{1},\,a_{2},\,\ldots]). The correspondence between the words and the representation of the related approximation can be recovered by the following properties, which easily follows from Theorem 6.6 and Proposition 6.8.

  1. (1)

    For jj a non-negative integer, the word LjL^{j} corresponds to the representation ([j+1],[j+2])([j+1],[j+2]).

  2. (2)

    For jj, rr, and d0,,drd_{0},\,\ldots,\,d_{r} non-negative integers, the word Ld0ILdrILjL^{d_{0}}I\cdots L^{d_{r}}IL^{j} corresponds to the representation ([d0,,dr1,dr+1],[j+2])([d_{0},\,\ldots,\,d_{r-1},d_{r}+1],[j+2]).

  3. (3)

    For rr and ss non-negative integers, and for d0,,drd_{0},\ldots,d_{r}, and c0,,csc_{0},\,\ldots,\,c_{s} non-negative integers, the word Ld0ILdrILc0Rc1LcsL^{d_{0}}I\cdots L^{d_{r}}IL^{c_{0}}R^{c_{1}}\cdots L^{c_{s}} if ss is even (or the word Ld0ILdrILc0Rc1RcsL^{d_{0}}I\cdots L^{d_{r}}IL^{c_{0}}R^{c_{1}}\cdots R^{c_{s}} if ss is odd) corresponds to the representation ([d0,,dr1,dr+1],[c0+1,c1,,cs1,cs+1])([d_{0},\,\ldots,\,d_{r-1},d_{r}+1],[c_{0}+1,\,c_{1},\,\ldots,\,c_{s-1},\,c_{s}+1]).

7.2. Convergent infinite triangle sequence

If the real pair (α,β)(\alpha,\beta) has a convergent infinite triangle sequence [α0,α1,α2,][\alpha_{0},\,\alpha_{1},\,\alpha_{2},\,\ldots], its representation is ([α0,α1,α2],[2])([\alpha_{0},\,\alpha_{1},\,\alpha_{2}\,\ldots],[2]) and (5.2) holds. Thus we associate to (α,β)(\alpha,\beta) the infinite word 𝒲\mathcal{W} over the alphabet {L,R,I}\{L,R,I\} defined by

𝒲(α,β)=Lα0ILα1ILα2I,\mathcal{W}(\alpha,\beta)=L^{\alpha_{0}}IL^{\alpha_{1}}IL^{\alpha_{2}}I\cdots\,,

and the approximations of (α,β)(\alpha,\beta) are again the rational pairs (mjsj,njsj)(\frac{m_{j}}{s_{j}},\frac{n_{j}}{s_{j}}) with coding 𝒲(j)\mathcal{W}_{(j)}, the prefix of 𝒲\mathcal{W} of length jj. Rules (1) and (2) above still hold and show how to construct the representations of the approximations. Notice that the choice of [2][2] as the second component of the representation is arbitrary, and [2][2] can be replaced by any continued fraction expansion. However this choice does not change the form of the word 𝒲\mathcal{W}.

Example.

Let α0=1\alpha_{0}=1 and for k1k\geq 1 let αk=pk\alpha_{k}=p_{k}, with pkp_{k} being the kk-th prime number, so that the triangle sequence we are considering is [1, 2, 3, 5, 7, 11,][1,\,2,\,3,\,5,\,7,\,11,\,\ldots]. From (4.2), for k0k\geq 0 we have

1λk=αk+1qk1qk+1=αk+1αk+1+qk2+qkqk1αk+11+αk+1,1-\lambda_{k}=\alpha_{k+1}\frac{q_{k-1}}{q_{k+1}}=\frac{\alpha_{k+1}}{\alpha_{k+1}+\frac{q_{k-2}+q_{k}}{q_{k-1}}}\leq\frac{\alpha_{k+1}}{1+\alpha_{k+1}},

where the last inequality holds since qk2+qkqk1qkqk11\frac{q_{k-2}+q_{k}}{q_{k-1}}\geq\frac{q_{k}}{q_{k-1}}\geq 1. Each triangle sequence digit is non-zero, thus λk<1\lambda_{k}<1 for all k0k\geq 0 and we thus have

k=0(1λk)=pp1+p,\prod_{k=0}^{\infty}(1-\lambda_{k})=\prod_{p}\frac{p}{1+p},

where the right-hand-side product extends over all the prime numbers. We thus have

pp1+p=p(11p+1)<p>2(11p)=0,\prod_{p}\frac{p}{1+p}=\prod_{p}\left(1-\frac{1}{p+1}\right)<\prod_{p>2}\left(1-\frac{1}{p}\right)=0,

and the last product diverges to 0 because the sum of the reciprocals of the primes diverges. As a consequence, the given triangle sequence represents a unique point of widebar\widebar{\triangle}. This point is represented by the word

𝒲=LIL2IL3IL5IL7IL11I\mathcal{W}=LIL^{2}IL^{3}IL^{5}IL^{7}IL^{11}I\cdots

and its first approximations are

(12,12)𝒲(0)=ε([1],[2])(13,13)𝒲(1)=L([2],[3])(24,14)𝒲(2)=LI([2],[2])(35,15)𝒲(3)=LIL([2],[3])(46,16)𝒲(4)=LILL([2],[4])(58,28)𝒲(5)=LILLI([1,3],[2])(610,310)𝒲(6)=LILLIL([1,3],[3])(712,412)𝒲(7)=LILLILL([1,3],[4])(814,514)𝒲(8)=LILLILLL([1,3],[5])(1119,619)𝒲(9)=LILLILLLI([1,2,4],[2])\begin{array}[]{cll}\left(\frac{1}{2},\frac{1}{2}\right)&\mathcal{W}_{(0)}=\varepsilon&([1],[2])\\[5.69046pt] \left(\frac{1}{3},\frac{1}{3}\right)&\mathcal{W}_{(1)}=L&([2],[3])\\[5.69046pt] \left(\frac{2}{4},\frac{1}{4}\right)&\mathcal{W}_{(2)}=LI&([2],[2])\\[5.69046pt] \left(\frac{3}{5},\frac{1}{5}\right)&\mathcal{W}_{(3)}=LIL&([2],[3])\\[5.69046pt] \left(\frac{4}{6},\frac{1}{6}\right)&\mathcal{W}_{(4)}=LILL&([2],[4])\\[5.69046pt] \left(\frac{5}{8},\frac{2}{8}\right)&\mathcal{W}_{(5)}=LILLI&([1,3],[2])\\[5.69046pt] \left(\frac{6}{10},\frac{3}{10}\right)&\mathcal{W}_{(6)}=LILLIL&([1,3],[3])\\[5.69046pt] \left(\frac{7}{12},\frac{4}{12}\right)&\mathcal{W}_{(7)}=LILLILL&([1,3],[4])\\[5.69046pt] \left(\frac{8}{14},\frac{5}{14}\right)&\mathcal{W}_{(8)}=LILLILLL&([1,3],[5])\\[5.69046pt] \left(\frac{11}{19},\frac{6}{19}\right)&\mathcal{W}_{(9)}=LILLILLLI&([1,2,4],[2])\\[5.69046pt] \vdots&&\end{array}

7.3. Non-convergent infinite case

Let (α,β)(\alpha,\beta) have a non-convergent infinite triangle sequence [α0,α1,][\alpha_{0},\,\alpha_{1},\,\dots]. It means that (α,β)(\alpha,\beta) is in the line segment 𝔏\mathfrak{L} of points sharing the same triangle sequence. In this case we have seen that the representation ([α0,α1,],[2])([\alpha_{0},\,\alpha_{1},\,\ldots],[2]) encodes the whole segment, and it is impossible to distinguish different points on 𝔏\mathfrak{L} by using it.

Here we propose a possible way to construct approximations of (α,β)(\alpha,\beta) following the methods used in the other cases. This proposal is certainly not the only meaningful and not the most “natural” in any sense. For all j0j\geq 0, the point (α,β)(\alpha,\beta) is in (α0,,αj)\triangle(\alpha_{0},\ldots,\alpha_{j}), so that

(ξj,ηj)Tj(α,β)(αj).(\xi_{j},\eta_{j})\coloneqq T^{j}(\alpha,\beta)\in\triangle(\alpha_{j}).

Since αj+\alpha_{j}\to+\infty, the second component ηj\eta_{j} is vanishing as jj increases. Let then pj(j)qj(j)=[a1(j),,aj(j)]\frac{p_{j}(j)}{q_{j}(j)}=[a_{1}(j),\,\ldots,\,a_{j}(j)] be the jj-th convergent of ξj\xi_{j}, obtained from the continued fraction expansion [a1(j),a2(j),a3(j),][a_{1}(j),\,a_{2}(j),\,a_{3}(j),\,\ldots] of ξj\xi_{j}. Then for ω=1α001αj10\omega=1^{\alpha_{0}}0\dots 1^{\alpha_{j-1}}0 it holds

(mjsj,njsj)ϕω(pj(j)qj(j),0qj(j))(α0,,αj)\left(\frac{m_{j}}{s_{j}},\frac{n_{j}}{s_{j}}\right)\coloneqq\phi_{\omega}\left(\frac{p_{j}(j)}{q_{j}(j)},\frac{0}{q_{j}(j)}\right)\in\triangle(\alpha_{0},\,\ldots,\,\alpha_{j})

and

limj|(α,β)(mjsj,njsj)|=0.\lim_{j\to\infty}\Big{|}(\alpha,\beta)-\left(\frac{m_{j}}{s_{j}},\frac{n_{j}}{s_{j}}\right)\Big{|}=0.

We then consider the approximations of (α,β)(\alpha,\beta) given by the rational pairs with representations

([α0,,αj],[a1(j),,aj(j)])([\alpha_{0},\,\ldots,\,\alpha_{j}],[a_{1}(j),\,\ldots,\,a_{j}(j)])

for j0j\geq 0, defined as above.

8. Speed of the approximations

In this final section we study the problem of the speed of the approximations introduced before. In particular given a non-rational pair (α,β)widebar(\alpha,\beta)\in\widebar{\triangle} we have defined a sequence (mjsj,njsj)(\frac{m_{j}}{s_{j}},\frac{n_{j}}{s_{j}}) of rational pairs in the Triangular tree which approximate (α,β)(\alpha,\beta). The speed of the approximations we consider concerns the supremum of the exponents η>0\eta>0 for which

(8.1) lim infjsjη|αmjsj||βnjsj|=0.\liminf_{j\to\infty}s_{j}^{\eta}\left|\alpha-\frac{m_{j}}{s_{j}}\right|\left|\beta-\frac{n_{j}}{s_{j}}\right|=0.

This problem is also known as the problem of simultaneous approximations of real numbers, and two famous problems in this research area are Dirichlet’s Theorem and Littlewood’s Conjecture (see [14] and [2] for more details).

We start with some notation. Let (α,β)(\alpha,\beta) be a point in widebar\widebar{\triangle} with α\alpha or β\beta irrational. Let ω{0,1}\omega\in\left\{0,1\right\}^{*} and ϕω\phi_{\omega} the correspondent concatenation of ϕ0\phi_{0} and ϕ1\phi_{1}, we use the notation

(8.2) ϕω(pq,rq)=(mω(p,r,q)sω(p,r,q),nω(p,r,q)sω(p,r,q)),\phi_{\omega}\left(\frac{p}{q},\frac{r}{q}\right)=\left(\frac{m_{\omega}(p,r,q)}{s_{\omega}(p,r,q)},\frac{n_{\omega}(p,r,q)}{s_{\omega}(p,r,q)}\right),

for the image of rational pairs (pq,rq)(\frac{p}{q},\frac{r}{q}) under ϕω\phi_{\omega}. In [7, Appendix A] we have introduced a matrix representation of the maps ϕ0\phi_{0} and ϕ1\phi_{1} by

(8.3) M0Mϕ0=(101100010)andM1Mϕ1=(101010001),M_{0}\coloneqq M_{\phi_{0}}=\begin{pmatrix}1&0&1\\ 1&0&0\\ 0&1&0\end{pmatrix}\qquad\text{and}\qquad M_{1}\coloneqq M_{\phi_{1}}=\begin{pmatrix}1&0&1\\ 0&1&0\\ 0&0&1\end{pmatrix},

from which we obtain a matrix representation with non-negative integers coefficients for any combination ϕωj\phi_{\omega^{j}} of ϕ0\phi_{0} and ϕ1\phi_{1}, which we denote by

Mωj(ρστρ1σ1τ1ρ2σ2τ2)M_{\omega^{j}}\coloneqq\begin{pmatrix}\rho&\sigma&\tau\\ \rho_{1}&\sigma_{1}&\tau_{1}\\ \rho_{2}&\sigma_{2}&\tau_{2}\end{pmatrix}

not including the dependence on jj in the notation for the coefficients of the matrix if not necessary. Using the matrix representation we write for all (x,y)2(x,y)\in\mathbb{R}^{2}

(8.4) ϕωj(x,y)=(ρ1+σ1x+τ1yρ+σx+τy,ρ2+σ2x+τ2yρ+σx+τy),\phi_{{}_{\omega^{j}}}(x,y)=\left(\frac{\rho_{1}+\sigma_{1}x+\tau_{1}y}{\rho+\sigma x+\tau y},\frac{\rho_{2}+\sigma_{2}x+\tau_{2}y}{\rho+\sigma x+\tau y}\right),

so that in particular it holds

(8.5) sωj(p,r,q)=ρq+σp+τrs_{\omega^{j}}(p,r,q)=\rho q+\sigma p+\tau r

in (8.2).

In this paper we begin to study the problem of the speed of the approximations by considering the first simple classes of real pairs of numbers with at least one irrational component: the pairs with finite triangle sequence and the pairs with periodic triangle sequence [d,d,d,][d,d,d,\,\dots] for d3d\geq 3.

8.1. Real pairs with finite triangle sequence

Let (α,β)widebar(\alpha,\beta)\in\widebar{\triangle}, with α\alpha or β\beta irrational, have finite triangle sequence [α0,,αk][\alpha_{0},\,\ldots,\,\alpha_{k}], then as proved in Lemma 5.1, in particular see (5.1), there exists ω{0,1}\omega\in\left\{0,1\right\}^{*} and ξ[0,1]\xi\in[0,1] such that (α,β)=ϕω(ξ,0)(\alpha,\beta)=\phi_{\omega}(\xi,0) (recall that (ξ,0)=ϕ2(ξ,ξ)(\xi,0)=\phi_{2}(\xi,\xi)). As remarked in Section 7, we can definitively consider the problem (8.1) for rational pairs (mjsj,njsj)(\frac{m_{j}}{s_{j}},\frac{n_{j}}{s_{j}}) with the same triangle sequence of (α,β)(\alpha,\beta). It follows that we can consider rational pairs as obtained in (8.2) with ϕω\phi_{\omega}. Hence we can write

|αmω(p,r,q)sω(p,r,q)|=|(ϕω(ξ,0))1(ϕω(pq,rq))1|\Big{|}\alpha-\frac{m_{\omega}(p,r,q)}{s_{\omega}(p,r,q)}\Big{|}=\Big{|}\left(\phi_{\omega}(\xi,0)\right)_{1}-\left(\phi_{\omega}\left(\frac{p}{q},\frac{r}{q}\right)\right)_{1}\Big{|}
|βnω(p,r,q)sω(p,r,q)|=|(ϕω(ξ,0))2(ϕω(pq,rq))2|\Big{|}\beta-\frac{n_{\omega}(p,r,q)}{s_{\omega}(p,r,q)}\Big{|}=\Big{|}\left(\phi_{\omega}(\xi,0)\right)_{2}-\left(\phi_{\omega}\left(\frac{p}{q},\frac{r}{q}\right)\right)_{2}\Big{|}

where the subscripts refer to the first and second component respectively, and consider (pq,rq)(\frac{p}{q},\frac{r}{q}) as approximations of (ξ,0)(\xi,0).

Using the matrix representation (8.4) for ϕω\phi_{\omega} we can write

αmω(p,r,q)sω(p,r,q)\displaystyle\alpha-\frac{m_{\omega}(p,r,q)}{s_{\omega}(p,r,q)} =ρ1+σ1ξρ+σξρ1q+σ1p+τ1rρq+σp+τr=\displaystyle=\frac{\rho_{1}+\sigma_{1}\xi}{\rho+\sigma\xi}-\frac{\rho_{1}q+\sigma_{1}p+\tau_{1}r}{\rho q+\sigma p+\tau r}=
=(σ1τστ1)ξr+(σ1ρσρ1)(ξqp)+(τρ1τ1ρ)r(ρ+σξ)(ρq+σp+τr)\displaystyle=\frac{(\sigma_{1}\tau-\sigma\tau_{1})\xi r+(\sigma_{1}\rho-\sigma\rho_{1})(\xi q-p)+(\tau\rho_{1}-\tau_{1}\rho)r}{(\rho+\sigma\xi)(\rho q+\sigma p+\tau r)}

and analogously

βnω(p,r,q)sω(p,r,q)=(σ2τστ2)ξr+(σ2ρσρ2)(ξqp)+(τρ2τ2ρ)r(ρ+σξ)(ρq+σp+τr)\beta-\frac{n_{\omega}(p,r,q)}{s_{\omega}(p,r,q)}=\frac{(\sigma_{2}\tau-\sigma\tau_{2})\xi r+(\sigma_{2}\rho-\sigma\rho_{2})(\xi q-p)+(\tau\rho_{2}-\tau_{2}\rho)r}{(\rho+\sigma\xi)(\rho q+\sigma p+\tau r)}

Using that for all ξ\xi\in\mathbb{R} there exist two sequences (pj)j(p_{j})_{j} of integers and (qj)j(q_{j})_{j} of positive integers with (pj,qj)=1(p_{j},q_{j})=1, such that |ξqjpj|1qj|\xi q_{j}-p_{j}|\leq\frac{1}{q_{j}} for all jj, and letting rj=0r_{j}=0 for all jj, so that it holds (pjqj,rjqj)widebar2\left(\frac{p_{j}}{q_{j}},\frac{r_{j}}{q_{j}}\right)\in\widebar{\triangle}\cap\mathbb{Q}^{2}, we obtain that there exist two sequences (pj)j(p_{j})_{j} of integers and (qj)j(q_{j})_{j} of positive integers such that

|αmω(pj,0,qj)sω(pj,0,qj)|c1qjsω(pj,0,qj)and|βnω(pj,0,qj)sω(pj,0,qj)|c1qjsω(pj,0,qj)\Big{|}\alpha-\frac{m_{\omega}(p_{j},0,q_{j})}{s_{\omega}(p_{j},0,q_{j})}\Big{|}\leq c\frac{1}{q_{j}s_{\omega}(p_{j},0,q_{j})}\quad\text{and}\quad\Big{|}\beta-\frac{n_{\omega}(p_{j},0,q_{j})}{s_{\omega}(p_{j},0,q_{j})}\Big{|}\leq c\frac{1}{q_{j}s_{\omega}(p_{j},0,q_{j})}

where the constant cc does not depend on pjp_{j} and qjq_{j}. Therefore choosing the two sequences (pj)j(p_{j})_{j} and (qj)j(q_{j})_{j} as before, we have that for all ε>0\varepsilon>0

lim infs\displaystyle\liminf_{s\to\infty} s(4ε)(infm|αms|)(infn|βns|)lim infjsj(4ε)|αmjsj||βnjsj|\displaystyle s^{(4-\varepsilon)}\Big{(}\inf_{m\in\mathbb{Z}}\Big{|}\alpha-\frac{m}{s}\Big{|}\Big{)}\Big{(}\inf_{n\in\mathbb{Z}}\Big{|}\beta-\frac{n}{s}\Big{|}\Big{)}\leq\liminf_{j\to\infty}s_{j}^{(4-\varepsilon)}\left|\alpha-\frac{m_{j}}{s_{j}}\right|\left|\beta-\frac{n_{j}}{s_{j}}\right|\leq
limjsω(pj,0,qj)(4ε)|αmω(pj,0,qj)sω(pj,0,qj)||βnω(pj,0,qj)sω(pj,0,qj)|\displaystyle\leq\lim_{j\to\infty}s_{\omega}(p_{j},0,q_{j})^{(4-\varepsilon)}\left|\alpha-\frac{m_{\omega}(p_{j},0,q_{j})}{s_{\omega}(p_{j},0,q_{j})}\right|\left|\beta-\frac{n_{\omega}(p_{j},0,q_{j})}{s_{\omega}(p_{j},0,q_{j})}\right|\leq
c2limjsω(pj,0,qj)(2ε)qj2=0,\displaystyle\leq c^{2}\lim_{j\to\infty}\frac{s_{\omega}(p_{j},0,q_{j})^{(2-\varepsilon)}}{q_{j}^{2}}=0,

since by (8.5) holds sω(pj,0,qj)cqjs_{\omega}(p_{j},0,q_{j})\leq c^{\prime}q_{j} for a suitable constant cc^{\prime}. This argument also implies that in this case (α,β)(\alpha,\beta) is is not a Bad (i1,i2)(i_{1},i_{2}) pair, for all admissible (i1,i2)(i_{1},i_{2}) (see [14] and [2]).

8.2. Real pairs with periodic triangle sequence [d,d,d,][d,d,d,\,\dots] for d3d\geq 3

Let (α,β)widebar(\alpha,\beta)\in\widebar{\triangle}, with α\alpha or β\beta irrational, have triangle sequence [d,d,d,][d,d,d,\,\dots] for d3d\geq 3. These pairs correspond to fixed points for the Triangle Map TT defined in Section 3 (all fixed points of TT are obtained by considering also the real pairs with triangle sequence [d,d,d,][d,d,d,\,\dots] for d=0,1,2d=0,1,2).

It is shown in [8] that if (α,β)(\alpha,\beta) has triangle sequence [d,d,d,][d,d,d,\,\dots] then β=α2\beta=\alpha^{2} and α(0,1)\alpha\in(0,1) is the largest root of the polynomial P(t)=t3+dt2+t1P(t)=t^{3}+dt^{2}+t-1. It follows that α\alpha is a cubic number, and α\alpha and β\beta are in the same cubic number field. We remark that the polynomial P(t)P(t) has three real roots for d3d\geq 3, and only one real root for d=0,1,2d=0,1,2. Hence here we only consider the simplest case for the roots of P(t)P(t).

In Section 7 we have constructed approximations of (α,β)(\alpha,\beta) by rational pairs in the Triangular tree with representation ([d,d,,d],[2])([d,d,\,\dots,d],[2]), where we recall that instead of [2][2] one could choose any fixed continued fraction expansion [a1,,an][a_{1},\dots,a_{n}]. Then using (8.3) we consider the matrix

MdM1dM0=(1d1100010)M_{d}\coloneqq M_{1}^{d}M_{0}=\begin{pmatrix}1&d&1\\ 1&0&0\\ 0&1&0\end{pmatrix}

which as in (8.4) represents the map

ϕω(x,y)=(11+dx+y,x1+dx+y)withω=(1110){0,1}d+1\phi_{\omega}(x,y)=\left(\frac{1}{1+dx+y},\frac{x}{1+dx+y}\right)\quad\text{with}\quad\omega=(11\cdots 10)\in\{0,1\}^{d+1}

and use approximations of (α,β)(\alpha,\beta) of the form

(8.6) (mksk,nksk)=ϕMdk(12,0)ϕω¯k(12,0)\left(\frac{m_{k}}{s_{k}},\frac{n_{k}}{s_{k}}\right)=\phi_{{}_{M_{d}^{k}}}\left(\frac{1}{2},0\right)\eqqcolon\phi_{\overline{\omega}^{k}}\left(\frac{1}{2},0\right)

where ω¯k{0,1}k(d+1)\overline{\omega}^{k}\in\{0,1\}^{k(d+1)} is the string obtained by concatenating kk copies of ω\omega. In matrix representation, (8.6) can be written as

(skmknk)=Mdk(210)\begin{pmatrix}s_{k}\\ m_{k}\\ n_{k}\end{pmatrix}=M_{d}^{k}\begin{pmatrix}2\\ 1\\ 0\end{pmatrix}

and we can then use linear algebra to study the properties of the approximations.

Remark 8.1.

As shown in (5.2) all points in widebar\widebar{\triangle} converge to (α,β)(\alpha,\beta) under repeated applications of ϕω\phi_{\omega}. Therefore different approximations of (α,β)(\alpha,\beta) can be constructed as in (8.6) by using a sequence (pkqk,rkqk)(\frac{p_{k}}{q_{k}},\frac{r_{k}}{q_{k}}) of possibly different rational pairs instead of the fixed pair (12,0)(\frac{1}{2},0).

The matrix MdM_{d} has characteristic polynomial pd(λ)=λ3λ2dλ1p_{d}(\lambda)=\lambda^{3}-\lambda^{2}-d\lambda-1 and, for d3d\geq 3, distinct eigenvalues λ1\lambda_{1}, λ2\lambda_{2}, and λ3\lambda_{3} satisfying

(8.7) λ1=1α2<λ2=1α1<0<1<λ3=1αwithλ3>|λ1|1>|λ2|\lambda_{1}=\frac{1}{\alpha_{2}}<\lambda_{2}=\frac{1}{\alpha_{1}}<0<1<\lambda_{3}=\frac{1}{\alpha}\quad\text{with}\quad\lambda_{3}>|\lambda_{1}|\geq 1>|\lambda_{2}|

where α1<α2<0<α<1\alpha_{1}<\alpha_{2}<0<\alpha<1 are the roots of P(t)P(t), and eigenvectors

v1=(1α2α22),v2=(1α1α12),v3=(1αβ).v_{1}=\begin{pmatrix}1\\[2.84544pt] \alpha_{2}\\[2.84544pt] \alpha_{2}^{2}\end{pmatrix},\quad v_{2}=\begin{pmatrix}1\\[2.84544pt] \alpha_{1}\\[2.84544pt] \alpha_{1}^{2}\end{pmatrix},\quad v_{3}=\begin{pmatrix}1\\[2.84544pt] \alpha\\[2.84544pt] \beta\end{pmatrix}.

We also recall that for d3d\geq 3 fixed, and given α\alpha the largest root of P(t)P(t), it holds

(8.8) α1=(d+α)(d+α)24α2andα2=(d+α)+(d+α)24α2.\alpha_{1}=\frac{-(d+\alpha)-\sqrt{(d+\alpha)^{2}-\frac{4}{\alpha}}}{2}\quad\text{and}\quad\alpha_{2}=\frac{-(d+\alpha)+\sqrt{(d+\alpha)^{2}-\frac{4}{\alpha}}}{2}.

For simplicity of notation in the following we let

h(d,α)(d+α)24αh(d,\alpha)\coloneqq\sqrt{(d+\alpha)^{2}-\frac{4}{\alpha}}

We are now ready to prove the following result.

Proposition 8.2.

Let (α,β)𝑤𝑖𝑑𝑒𝑏𝑎𝑟(\alpha,\beta)\in\widebar{\triangle}, with α\alpha or β\beta irrational, have triangle sequence [d,d,d,][d,d,d,\,\dots] for d3d\geq 3. There exist a constant c(α,d)c(\alpha,d), functions fα,di:3f^{i}_{\alpha,d}:\mathbb{Z}^{3}\to\mathbb{R} with i=1,2i=1,2 given by

fα,d1(q,p,r)\displaystyle f^{1}_{\alpha,d}(q,p,r) =h(d,α)(d+3α+h(d,α))(1+α2(d+2α))4α[q(3α+α2h(d,α))+\displaystyle=\frac{h(d,\alpha)\left(d+3\alpha+h(d,\alpha)\right)(1+\alpha^{2}(d+2\alpha))}{4\alpha}\Big{[}q\left(3-\alpha+\alpha^{2}h(d,\alpha)\right)+{}
+p(22dα2αh(d,α))+r(3αd+h(d,α))]\displaystyle\hskip 28.45274pt+p\left(-2-2d\alpha-2\alpha h(d,\alpha)\right)+r\left(-3\alpha-d+h(d,\alpha)\right)\Big{]}
fα,d2(q,p,r)\displaystyle f^{2}_{\alpha,d}(q,p,r) =h(d,α)(d+3αh(d,α))(1+α2(d+2α))4α[q(3αα2h(d,α))+\displaystyle=-\frac{h(d,\alpha)\left(d+3\alpha-h(d,\alpha)\right)(1+\alpha^{2}(d+2\alpha))}{4\alpha}\Big{[}q\left(3-\alpha-\alpha^{2}h(d,\alpha)\right)+{}
+p(22dα+2αh(d,α))+r(3αdh(d,α))]\displaystyle\hskip 28.45274pt+p\left(-2-2d\alpha+2\alpha h(d,\alpha)\right)+r\left(-3\alpha-d-h(d,\alpha)\right)\Big{]}

and functions gα,di:3g^{i}_{\alpha,d}:\mathbb{Z}^{3}\to\mathbb{R} with i=1,2,3i=1,2,3 given by

gα,d1(q,p,r)=d+3α+h(d,α)2[q(α)(d+α+h(d,α))2+pdα+h(d,α)2+r]g^{1}_{\alpha,d}(q,p,r)=\frac{d+3\alpha+h(d,\alpha)}{2}\left[q\frac{(-\alpha)\left(d+\alpha+h(d,\alpha)\right)}{2}+p\frac{d-\alpha+h(d,\alpha)}{2}+r\right]
gα,d2(q,p,r)=d3α+h(d,α)2[q(α)(d+αh(d,α))2+pdαh(d,α)2+r]g^{2}_{\alpha,d}(q,p,r)=\frac{-d-3\alpha+h(d,\alpha)}{2}\left[q\frac{(-\alpha)\left(d+\alpha-h(d,\alpha)\right)}{2}+p\frac{d-\alpha-h(d,\alpha)}{2}+r\right]
gα,d3(q,p,r)=h(d,α)[q1α+p(d+α)+r]g^{3}_{\alpha,d}(q,p,r)=-h(d,\alpha)\left[q\frac{1}{\alpha}+p(d+\alpha)+r\right]

such that for all k1k\geq 1 and all (pq,rq)𝑤𝑖𝑑𝑒𝑏𝑎𝑟(\frac{p}{q},\frac{r}{q})\in\widebar{\triangle} the rational pair (ms,ns)=ϕMdk(pq,rq)(\frac{m}{s},\frac{n}{s})=\phi_{M_{d}^{k}}(\frac{p}{q},\frac{r}{q}) satisfies

(8.9) αms=c(α,d)fα,d1(q,p,r)λ1k+fα,d2(q,p,r)λ2kgα,d1(q,p,r)λ1k+gα,d2(q,p,r)λ2k+gα,d3(q,p,r)λ3k\displaystyle\alpha-\frac{m}{s}=c(\alpha,d)\frac{f^{1}_{\alpha,d}(q,p,r)\lambda_{1}^{k}+f^{2}_{\alpha,d}(q,p,r)\lambda_{2}^{k}}{g^{1}_{\alpha,d}(q,p,r)\lambda_{1}^{k}+g^{2}_{\alpha,d}(q,p,r)\lambda_{2}^{k}+g^{3}_{\alpha,d}(q,p,r)\lambda_{3}^{k}}
βns=c(α,d)2(αd+h(d,α))fα,d1(q,p,r)λ1k+(αdh(d,α))fα,d2(q,p,r)λ2kgα,d1(q,p,r)λ1k+gα,d2(q,p,r)λ2k+gα,d3(q,p,r)λ3k\displaystyle\beta-\frac{n}{s}=\frac{c(\alpha,d)}{2}\frac{(\alpha-d+h(d,\alpha))f^{1}_{\alpha,d}(q,p,r)\lambda_{1}^{k}+(\alpha-d-h(d,\alpha))\ f^{2}_{\alpha,d}(q,p,r)\lambda_{2}^{k}}{g^{1}_{\alpha,d}(q,p,r)\lambda_{1}^{k}+g^{2}_{\alpha,d}(q,p,r)\lambda_{2}^{k}+g^{3}_{\alpha,d}(q,p,r)\lambda_{3}^{k}}

where λ1\lambda_{1}, λ2\lambda_{2}, and λ3\lambda_{3} are defined in (8.7).

Proof.

If (α,β)(\alpha,\beta) has triangle sequence [d,d,d,][d,d,d,\,\dots] then it is a fixed point of the Triangle Map TT with β=α2\beta=\alpha^{2} and, in particular using (3.2), (α,β)=Sd+1(α,β)=S|Γ0S|Γ1d(α,β)(\alpha,\beta)=S^{d+1}(\alpha,\beta)=S|_{\Gamma_{0}}\circ S|_{\Gamma_{1}}^{d}(\alpha,\beta) for the map SS defined in (3.1). It follows that (α,β)=ϕMdk(α,β)(\alpha,\beta)=\phi_{M_{d}^{k}}(\alpha,\beta) for all k1k\geq 1. We can thus write for fixed k1k\geq 1 and rational pair (pq,rq)widebar(\frac{p}{q},\frac{r}{q})\in\widebar{\triangle}

(8.10) αms\displaystyle\alpha-\frac{m}{s} =(ϕMdk(α,β))1(ϕMdk(pq,rq))1=\displaystyle=\left(\phi_{M_{d}^{k}}(\alpha,\beta)\right)_{1}-\left(\phi_{M_{d}^{k}}\left(\frac{p}{q},\frac{r}{q}\right)\right)_{1}=
=(ρσ1ρ1σ)(αqp)+(τσ1τ1σ)(αrβp)+(ρτ1ρ1τ)(βqr)(ρ+σα+τβ)(ρq+σp+τr)\displaystyle=\frac{(\rho\sigma_{1}-\rho_{1}\sigma)(\alpha q-p)+(\tau\sigma_{1}-\tau_{1}\sigma)(\alpha r-\beta p)+(\rho\tau_{1}-\rho_{1}\tau)(\beta q-r)}{(\rho+\sigma\alpha+\tau\beta)(\rho q+\sigma p+\tau r)}

using the matrix representation (8.4). Let Λ\Lambda be the diagonal matrix Λ=diag(λ1,λ2,λ3)\Lambda={\textrm{diag}}(\lambda_{1},\lambda_{2},\lambda_{3}), then Mdk=ΠΛkΠ1M_{d}^{k}=\Pi\Lambda^{k}\Pi^{-1} where Π\Pi is the matrix with columns the eigenvectors v1,v2v_{1},v_{2} and v3v_{3}. We use this representation to have an explicit form for the terms in (8.10), where the dependence on kk appears in the power of the eigenvalues and is of course hidden in the coefficients of MdkM_{d}^{k}. In what follows, all not explained steps are just straightforward computations. We have

Π=(111α2α1αα22α12β)andΠ1=1det(Π)(α1βα12αα12βαα1α22αα2ββα22α2αα12α2α1α22α22α12α1α2)\Pi=\begin{pmatrix}1&1&1\\[2.84544pt] \alpha_{2}&\alpha_{1}&\alpha\\[2.84544pt] \alpha_{2}^{2}&\alpha_{1}^{2}&\beta\end{pmatrix}\quad\text{and}\quad\Pi^{-1}=\frac{1}{\det(\Pi)}\begin{pmatrix}\alpha_{1}\beta-\alpha_{1}^{2}\alpha&\alpha_{1}^{2}-\beta&\alpha-\alpha_{1}\\[2.84544pt] \alpha_{2}^{2}\alpha-\alpha_{2}\beta&\beta-\alpha_{2}^{2}&\alpha_{2}-\alpha\\[2.84544pt] \alpha_{1}^{2}\alpha_{2}-\alpha_{1}\alpha_{2}^{2}&\alpha_{2}^{2}-\alpha_{1}^{2}&\alpha_{1}-\alpha_{2}\end{pmatrix}

with det(Π)=α1α(αα1)+α2α(α2α)+α1α2(α1α2)\det(\Pi)=\alpha_{1}\alpha(\alpha-\alpha_{1})+\alpha_{2}\alpha(\alpha_{2}-\alpha)+\alpha_{1}\alpha_{2}(\alpha_{1}-\alpha_{2}).
We first consider the denominator of (8.10). Let us write the first row of MdkM_{d}^{k} as a function of α\alpha and dd. It holds

(8.11) ρ\displaystyle\rho =1det(Π)[λ1kαα1(αα1)+λ2kαα2(α2α)+λ3kα1α2(α1α2)]\displaystyle=\frac{1}{\det(\Pi)}\Big{[}\lambda_{1}^{k}\alpha\alpha_{1}(\alpha-\alpha_{1})+\lambda_{2}^{k}\alpha\alpha_{2}(\alpha_{2}-\alpha)+\lambda_{3}^{k}\alpha_{1}\alpha_{2}(\alpha_{1}-\alpha_{2})\Big{]}
σ\displaystyle\sigma =1det(Π)[λ1k(α12α2)+λ2k(α2α22)+λ3k(α22α12)]\displaystyle=\frac{1}{\det(\Pi)}\Big{[}\lambda_{1}^{k}(\alpha_{1}^{2}-\alpha^{2})+\lambda_{2}^{k}(\alpha^{2}-\alpha_{2}^{2})+\lambda_{3}^{k}(\alpha_{2}^{2}-\alpha_{1}^{2})\Big{]}
τ\displaystyle\tau =1det(Π)[λ1k(αα1)+λ2k(α2α)+λ3k(α1α2)]\displaystyle=\frac{1}{\det(\Pi)}\Big{[}\lambda_{1}^{k}(\alpha-\alpha_{1})+\lambda_{2}^{k}(\alpha_{2}-\alpha)+\lambda_{3}^{k}(\alpha_{1}-\alpha_{2})\Big{]}

then we obtain

(8.12) ρ+σα+τβ=λ3k\rho+\sigma\alpha+\tau\beta=\lambda_{3}^{k}

and

rhoq+σp+τr=1det(Π)\displaystyle rhoq+\sigma p+\tau r=\frac{1}{\det(\Pi)} [λ1k(qα1α(αα1)+p(α12α2)+r(αα1))+\displaystyle\left[\lambda_{1}^{k}\Big{(}q\alpha_{1}\alpha(\alpha-\alpha_{1})+p(\alpha_{1}^{2}-\alpha^{2})+r(\alpha-\alpha_{1})\Big{)}\right.+
+λ2k(qα2α(α2α)+p(α2α22)+r(α2α))+\displaystyle\hskip 14.22636pt+\left.\lambda_{2}^{k}\Big{(}q\alpha_{2}\alpha(\alpha_{2}-\alpha)+p(\alpha^{2}-\alpha_{2}^{2})+r(\alpha_{2}-\alpha)\Big{)}+\right.
+λ3k(qα1α2(α1α2)+p(α22α12)+r(α1α2))]\displaystyle\hskip 28.45274pt+\left.\lambda_{3}^{k}\Big{(}q\alpha_{1}\alpha_{2}(\alpha_{1}-\alpha_{2})+p(\alpha_{2}^{2}-\alpha_{1}^{2})+r(\alpha_{1}-\alpha_{2})\Big{)}\right]

which using (8.8) gives

(8.13) ρq+σp+τr=1det(Π)(gα,d1(q,p,r)λ1k+gα,d2(q,p,r)λ2k+gα,d3(q,p,r)λ3k)\rho q+\sigma p+\tau r=\frac{1}{\det(\Pi)}\Big{(}g^{1}_{\alpha,d}(q,p,r)\lambda_{1}^{k}+g^{2}_{\alpha,d}(q,p,r)\lambda_{2}^{k}+g^{3}_{\alpha,d}(q,p,r)\lambda_{3}^{k}\Big{)}

with the functions gα,di(q,p,r)g^{i}_{\alpha,d}(q,p,r) defined in the statement. For the numerator of (8.10) let us write the second row of MdkM_{d}^{k} as a function of α\alpha and dd. It holds

ρ1\displaystyle\rho_{1} =1det(Π)[λ1k(αα1)+λ2k(α2α)+λ3k(α1α2)]\displaystyle=\frac{1}{\det(\Pi)}\Big{[}\lambda_{1}^{k}(\alpha-\alpha_{1})+\lambda_{2}^{k}(\alpha_{2}-\alpha)+\lambda_{3}^{k}(\alpha_{1}-\alpha_{2})\Big{]}
σ1\displaystyle\sigma_{1} =1det(Π)[λ1kα2(α12α2)+λ2kα1(α2α22)+λ3kα(α22α12)]\displaystyle=\frac{1}{\det(\Pi)}\Big{[}\lambda_{1}^{k}\alpha_{2}(\alpha_{1}^{2}-\alpha^{2})+\lambda_{2}^{k}\alpha_{1}(\alpha^{2}-\alpha_{2}^{2})+\lambda_{3}^{k}\alpha(\alpha_{2}^{2}-\alpha_{1}^{2})\Big{]}
τ1\displaystyle\tau_{1} =1det(Π)[λ1kα2(αα1)+λ2kα1(α2α)+λ3kα(α1α2)]\displaystyle=\frac{1}{\det(\Pi)}\Big{[}\lambda_{1}^{k}\alpha_{2}(\alpha-\alpha_{1})+\lambda_{2}^{k}\alpha_{1}(\alpha_{2}-\alpha)+\lambda_{3}^{k}\alpha(\alpha_{1}-\alpha_{2})\Big{]}

and then one can write the numerator of (8.10) as

1(det(Π))2(fα,d1(q,p,r)λ1kλ3k+fα,d2(q,p,r)λ2kλ3k+fα,d3(q,p,r)λ1kλ2k)\frac{1}{(\det(\Pi))^{2}}\Big{(}f^{1}_{\alpha,d}(q,p,r)\lambda_{1}^{k}\lambda_{3}^{k}+f^{2}_{\alpha,d}(q,p,r)\lambda_{2}^{k}\lambda_{3}^{k}+f^{3}_{\alpha,d}(q,p,r)\lambda_{1}^{k}\lambda_{2}^{k}\Big{)}

with fα,d3(q,p,r)0f^{3}_{\alpha,d}(q,p,r)\equiv 0, and fα,d1(q,p,r)f^{1}_{\alpha,d}(q,p,r) and fα,d2(q,p,r)f^{2}_{\alpha,d}(q,p,r) defined as in the statement. This gives the first of (8.9).

Now we can repeat the argument for the second term in (8.9) to get for fixed k1k\geq 1 and rational pair (pq,rq)widebar(\frac{p}{q},\frac{r}{q})\in\widebar{\triangle}

(8.14) βns\displaystyle\beta-\frac{n}{s} =(ϕMdk(α,β))2(ϕMdk(pq,rq))2=\displaystyle=\left(\phi_{M_{d}^{k}}(\alpha,\beta)\right)_{2}-\left(\phi_{M_{d}^{k}}\left(\frac{p}{q},\frac{r}{q}\right)\right)_{2}=
=(ρσ2ρ2σ)(αqp)+(τσ2τ2σ)(αrβp)+(ρτ2ρ2τ)(βqr)(ρ+σα+τβ)(ρq+σp+τr)\displaystyle=\frac{(\rho\sigma_{2}-\rho_{2}\sigma)(\alpha q-p)+(\tau\sigma_{2}-\tau_{2}\sigma)(\alpha r-\beta p)+(\rho\tau_{2}-\rho_{2}\tau)(\beta q-r)}{(\rho+\sigma\alpha+\tau\beta)(\rho q+\sigma p+\tau r)}

Equations (8.11), (8.12) and (8.13) give the denominator of (8.14). For the numerator we write the third row of MdkM_{d}^{k} as a function of α\alpha and dd, finding

1(det(Π))2(αd+h(d,α)2fα,d1(q,p,r)λ1kλ3k+αdh(d,α)2fα,d2(q,p,r)λ2kλ3k)\frac{1}{(\det(\Pi))^{2}}\Big{(}\frac{\alpha-d+h(d,\alpha)}{2}f^{1}_{\alpha,d}(q,p,r)\lambda_{1}^{k}\lambda_{3}^{k}+\frac{\alpha-d-h(d,\alpha)}{2}f^{2}_{\alpha,d}(q,p,r)\lambda_{2}^{k}\lambda_{3}^{k}\Big{)}

From this we obtain the second equation in (8.9). ∎

Corollary 8.3.

Let (α,β)𝑤𝑖𝑑𝑒𝑏𝑎𝑟(\alpha,\beta)\in\widebar{\triangle}, with α\alpha or β\beta irrational, have triangle sequence [d,d,d,][d,d,d,\,\dots] for d3d\geq 3, and consider the approximations (mksk,nksk)(\frac{m_{k}}{s_{k}},\frac{n_{k}}{s_{k}}) defined in (8.6). Then

limkskη|αmksk||βnksk|=0\lim_{k\to\infty}s_{k}^{\eta}\left|\alpha-\frac{m_{k}}{s_{k}}\right|\left|\beta-\frac{n_{k}}{s_{k}}\right|=0

for all η<2(1log|λ1|logλ3)\eta<2\big{(}1-\frac{\log|\lambda_{1}|}{\log\lambda_{3}}\big{)} if d4d\geq 4, where we are using (8.7), and for for all η<4\eta<4 if d=3d=3.

Proof.

We apply Proposition 8.2 with (q,p,r)=(2,1,0)(q,p,r)=(2,1,0) together with the relations (8.7). It follows

skη|αmksk||βnksk|c(α,d)(gα,d3(2,1,0)λ3k+o(λ3k))η2(fα,d1(2,1,0)λ1k+o(λ1k))2s_{k}^{\eta}\left|\alpha-\frac{m_{k}}{s_{k}}\right|\left|\beta-\frac{n_{k}}{s_{k}}\right|\leq c^{\prime}(\alpha,d)\Big{(}g^{3}_{\alpha,d}(2,1,0)\lambda_{3}^{k}+o(\lambda_{3}^{k})\Big{)}^{\eta-2}\Big{(}f^{1}_{\alpha,d}(2,1,0)\lambda_{1}^{k}+o(\lambda_{1}^{k})\Big{)}^{2}

where gα,d3(2,1,0)g^{3}_{\alpha,d}(2,1,0) and fα,d1(2,1,0)f^{1}_{\alpha,d}(2,1,0) do not vanish for d4d\geq 4. The result for this case immediately follows.
The case d=3d=3 is particular since α=1+2\alpha=-1+\sqrt{2} is a quadratic irrational. It also follows β=322\beta=3-2\sqrt{2}, h(3,α)=2h(3,\alpha)=\sqrt{2}, and

λ1=1,λ2=12,λ3=1+2.\lambda_{1}=-1,\quad\lambda_{2}=1-\sqrt{2},\quad\lambda_{3}=1+\sqrt{2}.

Moreover

fα,31(2,1,0)=c′′(α,3)(48α+2(α2α)h(3,α))=0f^{1}_{\alpha,3}(2,1,0)=c^{\prime\prime}(\alpha,3)\Big{(}4-8\alpha+2(\alpha^{2}-\alpha)h(3,\alpha)\Big{)}=0

and fα,32(2,1,0)f^{2}_{\alpha,3}(2,1,0) and gα,33(2,1,0)g^{3}_{\alpha,3}(2,1,0) do not vanish. Since λ2=λ31\lambda_{2}=-\lambda_{3}^{-1}, it follows that

skη|αmksk||βnksk|c′′′(α,3)(gα,33(2,1,0)λ3k+o(λ3k))η2λ32ks_{k}^{\eta}\left|\alpha-\frac{m_{k}}{s_{k}}\right|\left|\beta-\frac{n_{k}}{s_{k}}\right|\leq c^{\prime\prime\prime}(\alpha,3)\Big{(}g^{3}_{\alpha,3}(2,1,0)\lambda_{3}^{k}+o(\lambda_{3}^{k})\Big{)}^{\eta-2}\lambda_{3}^{-2k}

and the thesis follows. ∎

The result for d4d\geq 4 can be improved if we change the construction of the approximations as explained in Remark 8.1. For the moment we leave this problem and the study of the speed of the approximations for other real pairs to future research.

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𝒯0\mathcal{T}_{0}𝒯1\mathcal{T}_{1}𝒯2\mathcal{T}_{2}(12,12)\left(\frac{1}{2},\frac{1}{2}\right)(13,13)\left(\frac{1}{3},\frac{1}{3}\right)(14,14)\left(\frac{1}{4},\frac{1}{4}\right)(14,0)\left(\frac{1}{4},{0}\right)(1,14)\left({1},\frac{1}{4}\right)ϕ0\phi_{0}ϕ2\phi_{2}ϕ1\phi_{1}(13,0)\left(\frac{1}{3},{0}\right)(1,13)\left({1},\frac{1}{3}\right)(34,14)\left(\frac{3}{4},\frac{1}{4}\right)ϕ1\phi_{1}(34,34)\left(\frac{3}{4},\frac{3}{4}\right)(34,0)\left(\frac{3}{4},{0}\right)(1,34)\left({1},\frac{3}{4}\right)ϕ0\phi_{0}ϕ2\phi_{2}ϕ0\phi_{0}ϕ0\phi_{0}ϕ2\phi_{2}ϕ1\phi_{1}(12,0)\left(\frac{1}{2},{0}\right)(1,12)\left({1},\frac{1}{2}\right)(23,13)\left(\frac{2}{3},\frac{1}{3}\right)(34,24)\left(\frac{3}{4},\frac{2}{4}\right)ϕ0\phi_{0}(24,14)\left(\frac{2}{4},\frac{1}{4}\right)ϕ1\phi_{1}ϕ1\phi_{1}(23,23)\left(\frac{2}{3},\frac{2}{3}\right)(25,25)\left(\frac{2}{5},\frac{2}{5}\right)(25,0)\left(\frac{2}{5},{0}\right)(1,25)\left({1},\frac{2}{5}\right)ϕ0\phi_{0}ϕ2\phi_{2}ϕ1\phi_{1}(23,0)\left(\frac{2}{3},{0}\right)(1,23)\left({1},\frac{2}{3}\right)(35,25)\left(\frac{3}{5},\frac{2}{5}\right)ϕ1\phi_{1}(35,35)\left(\frac{3}{5},\frac{3}{5}\right)(35,0)\left(\frac{3}{5},{0}\right)(1,35)\left({1},\frac{3}{5}\right)ϕ0\phi_{0}ϕ2\phi_{2}ϕ0\phi_{0}ϕ0\phi_{0}ϕ2\phi_{2}ϕ0\phi_{0}ϕ0\phi_{0}ϕ2\phi_{2}
Figure 8. The levels 𝒯0\mathcal{T}_{0}, 𝒯1\mathcal{T}_{1} and 𝒯2\mathcal{T}_{2} of the Triangular tree generated through the rules (R1)-(R4).