a two-dimensional minkowski ðxÞ function · notethat0¼ 0 1 and1¼ 1 1: now,giventhepartitioni k...

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Journal of Number Theory 107 (2004) 105–134

A two-dimensional Minkowski ?ðxÞ function

Olga R. Beaver and Thomas Garrity�

Department of Mathematics and Statistics, Williams College, Williamstown, MA 01267, USA

Received 7 April 2003

Communicated by A. Granville

Abstract

A function from a triangle to itself is defined that has both interesting number theoretic and

analytic properties. This function is shown to be a natural generalization of the classical

Minkowski ?ðxÞ function. It is shown there exists a natural class of pairs of cubic irrational

numbers in the same cubic number field that are mapped to pairs of rational numbers, in

analog to ?ðxÞ mapping quadratic irrationals on the unit interval to rational numbers on the

unit interval. It is also shown that this new function satisfies an analog to the fact that ?ðxÞ;while increasing and continuous, has derivative zero almost everywhere.

r 2004 Elsevier Inc. All rights reserved.

1. Introduction

Any real number with an eventually periodic continued fraction expansion mustbe a quadratic irrational. This property linking periodicity of a number’s continuedfraction expansion with its being quadratic led Minkowski to define his remarkablequestion-mark function

? : ½0; 1�-½0; 1�:

(See [27, Vol. II, p. 50]; see also [15, article 196, p. 754], which appears to beessentially a translation of all of Minkowski’s number theory papers.) The question-

mark function is increasing, continuous, maps each rational number pq

to a pure

dyadic number of the form k2n; maps each quadratic irrational to a rational number,

and has the property that the inverse image of the rational numbers is exactly the setof quadratic irrationals. In order to understand the number theoretic properties of

ARTICLE IN PRESS

�Corresponding author. Fax: +1-413-597-4061.

E-mail addresses: [email protected] (O.R. Beaver), [email protected] (T. Garrity).

0022-314X/$ - see front matter r 2004 Elsevier Inc. All rights reserved.

doi:10.1016/j.jnt.2004.01.008

quadratic irrationals, it is natural to look at the function theoretic properties of ?ðxÞ:In particular, the question-mark function is not only continuous and monotonicallyincreasing but has derivative zero almost everywhere. As such, it is a naturallyoccurring example of a singular function. Moreover, it is, in fact, the diophantineproperties of continued fractions that lead to its derivative being zero a.e. Thus, theanalytic property of ?ðxÞ being both increasing and having derivative zero almosteverywhere is actually number theoretic in origin.

In this paper, we will construct a function similar to Minkowski’s question-markfunction, and will use that function in order to understand the properties of cubicirrationals.

Denjoy [4,5] and independently Salem [36] were the first to realize that ?ðxÞ issingular, although earlier, Ryde [34] proved in essence that ?ðxÞ was singular.However, Ryde showed that ?ðxÞ was singular without realizing its connection withMinkowski’s function (see also [35]). Recent work on ?ðxÞ is the work of Kinney [16],Girgensohn [9], Ramharter [33], of Tichy and Uitz [41], and of Viader et al. [42],Paradis et al. [31]. (In fact, the idea for the inequality that we prove in Section 6.1and use in 6.2 was inspired by the work of Viader et al. [42].)

A natural question to ask is: do cubic irrationals and other higher order algebraicnumbers lend themselves to similar analysis? An even more basic question to ask ishow to generalize the relation between periodicity for continued fractions andquadratic irrationals to cubics. In 1848, in a letter to Jacobi, Hermite [14] asked forsuch a generalization. Specifically, the Hermite problem is:

Find methods for expressing real numbers as sequences of positive integers so that the

sequence is eventually periodic precisely when the initial number is a cubic irrational.

Over the years there has been much work in trying to solve the Hermite problem.For an overview, see Schweiger’s Multidimensional Continued Fractions [40].For work up to 1980, see Brentjes’ overview in [3]. Other work is in [1,2,6–8,11–13,17,19–25,28,29,32,37–39].

On the other hand, there has been little attempt to approach the Hermite problemby generalizing the Minkowski ?ðxÞ: The only such attempt that we have found is inthe thesis of Kollros [18]. Kollros generalizes ?ðxÞ to a map from the unit square toitself. However, while he sets up various methods for associating points in the unitsquare with sequences of integers, he does not concern himself with the function-theoretic properties of this function. It does not appear that Kollros has solved theHermite problem. In particular, he was not interested in the differentiabilityproperties of his analogue to ?ðxÞ:

In this paper we develop a different, more natural, analog to ?ðxÞ: In Section 2, areview of the Minkowski question-mark function is given. In Section 3, we constructa map from a two-dimensional simplex (a triangle) to itself, as an analog to the mapof ?ðxÞ from a one-dimensional simplex to itself. The map will be determined bypartitioning the triangle, first via a ‘‘Farey’’ partition, and then by a barycentric(triadic) partitioning, which we will frequently call the ‘‘bary’’ partitioning. Wedefine a function dðx; yÞ from the Farey triangle to the barycentric triangle. In

ARTICLE IN PRESSO.R. Beaver, T. Garrity / Journal of Number Theory 107 (2004) 105–134106

Section 4, we see that the Farey iteration can be viewed as a multidimensionalcontinued fraction. We show that periodicity of the Farey iteration corresponds(usually) to a class of cubic irrationals. In Section 5 we show, by contrast, thatperiodicity for the barycentric iterations corresponds to a class of rational points. Thisresults in that our function will map a natural class of cubic points to a natural class ofrational points. Finally, in Section 6, we prove an analog of singularity by showingthat, a.e., the area of image triangles in the barycentric partitioning approaches zerofar more quickly than the area of the domain triangles in the Farey partitioning.

We note that similar, but not exactly the same, type of Farey partitioning, orFarey nets, to solve the Hermite problem has been considered by both Monkemeyer[30] and more recently by Grabiner [10]. Both papers are quite interesting; neitheruse Farey nets in the way that we do and, more importantly for this paper, neitherattempt to generalize the Minkowski ?ðxÞ function. In actuality, this analyticapproach would not have been a natural succession in either of these papers, asMonkemeyer’s and Grabiner’s goals were not function theoretic.

2. A review of the Minkowski question-mark function

All of the discussion in this section is well-known. We include it here for sake ofcompleteness.

Recall that given two rational numbers p1

q1and p2

q2; each in lowest terms, the Farey

sum, þþ; of the numbers is

p1

q1þþ p2

q2¼ p1 þ p2

q1 þ q2:

The ? function is then defined as follows. Suppose we know the value of ?ðp1

q1Þ and

?ðp2

q2Þ: We then set

?p1 þ p2

q1 þ q2

� �¼

?ðp1

q1Þ þ ?ðp2

q2Þ

2:

Specifying the initial values

?ð0Þ ¼ 0 and ?ð1Þ ¼ 1;

we now know the values of ?ðxÞ for any rational number x:By continuity arguments we can determine the values of ?ðxÞ for any real number

x in the unit interval. Since we will be generalizing this continuity argument in thenext section, we discuss this now in some detail.

We produce two sequences of partitions, Ik and *Ik; of the unit interval. For each

kX0; each partition will split the unit interval into 2k subintervals. Both start withjust the unit interval itself:

I0 ¼ *I0 ¼ ½0; 1�:

ARTICLE IN PRESSO.R. Beaver, T. Garrity / Journal of Number Theory 107 (2004) 105–134 107

Note that 0 ¼ 01 and 1 ¼ 1

1: Now, given the partition Ik�1; suppose that the endpoints

of each of its 2k�1 open subintervals are rational numbers. Form the next partitionIk by taking the Farey sum of the endpoints of the partition Ik�1: Thus theendpoints of Ik consist of the Farey fractions of order k:

The partition *Ik is even simpler. It is just the partition given by the subintervals

½l�12k ;

l2k�:

Then the function ?ðxÞ can be seen to map the endpoints of each Ik to the

corresponding endpoints of *Ik:Now, as is shown, for example, in [36], ?ðxÞ is singular and hence has derivative

zero almost everywhere. Using the partitions defined above, we can recast the factthat ?ðxÞ is a singular function into the language of lengths of intervals. Fix aA½0; 1�:For each k; let Ik and Ik be the subintervals of the respective partitions Ik and *Ik

that contain the point a: Then, as shown in [36, p. 437].

Theorem 1. For almost all aA½0; 1�;

lim infk-N

length of Ik

length of Ik

¼ 0:

It is this theorem that provides the most natural language for generalizing thefailure of differentiability for our analog of the question-mark function.

The proof involves the idea that the Diophantine approximation properties ofcontinued fractions make the above denominator approach zero more slowly thanthe numerator.


3. The Farey–Bary map: a generalization of the Minkowski question-mark function

Our goal is to define a map from a two-dimensional simplex (a triangle) to itselfthat generalizes the Minkowski question-mark function. This will involve twoseparate partitionings of the triangle. We would like to have periodicity in thedomain correspond to cubic irrationals while periodicity in the range to implyrationality. Both of these goals will only be achieved in part, as we will show thatperiodicity will usually imply cubic irrationality in the domain case and will alwaysimply rationality for the range. At the same time, we want our generalization to obeysome sort of singularity property.

Although in a sense it would be most natural to denote our generalization by thesymbol ?ðx; yÞ; we have found that it is both awkward to say and awkward to read.Thus we will denote our generalization by dðx; yÞ:

3.1. The Farey sum in the plane

We will often need to refer to a point in the plane of the form

v ¼p=r

q=r

� �:

Here, since the coordinates share the same denominator, we can associate to thispoint a unique vector in space, namely

%v ¼p

q

r

0B@

1CA:

Conversely, a vector %v ¼p

q

r

0@

1A can be associated uniquely to the point

v ¼p=r

q=r

� �

in the plane. In what follows, we will usually refer to both the point and itscorresponding vector as v:

Consider three points in the plane, each of whose entries are nonnegative integers,each ria0; and such that each vector’s entries share no common factors:

v1 ¼p1=r1

q1=r1

� �; v2 ¼

p2=r2

q2=r2

� �; v3 ¼

p3=r3

q3=r3

� �:

These points define a triangle in the plane and, as noted above, can also berepresented as the vectors,

v1 ¼p1

q1

r1

0B@

1CA; v2 ¼

p2

q2

r2

0B@

1CA and v ¼

p3

q3

r3

0B@

1CA:


Summing the three vectors, we get

v ¼ v1 þ v2 þ v3 ¼p1 þ p2 þ p3

q1 þ q2 þ q3

r1 þ r2 þ r3

0B@

1CA:

This vector sum can be converted into a point v in the plane, where

v ¼

p1 þ p2 þ p3

r1 þ r2 þ r3

q1 þ q2 þ q3

r1 þ r2 þ r3

0BBB@

1CCCA:

This correspondence between points in the plane and a vector representation allowsus to define the Farey sum.

Definition 2. Let

v1 ¼p1=r1

q1=r1

� �; v2 ¼

p2=r2

q2=r2

� �; v3 ¼

p3=r3

q3=r3

� �;

where, for each i; the pi; qi and ri share no common factor. The Farey sum, v of the vi

is then

v ¼ v1þþv2þþv3 ¼

p1 þ p2 þ p3

r1 þ r2 þ r3

q1 þ q2 þ q3

r1 þ r2 þ r3

0BBB@

1CCCA:

Note that the point v is inside the triangle determined by the vertices v1; v2 and v3

and that v corresponds to the vectorp1 þ p2 þ p3

q1 þ q2 þ q3

r1 þ r2 þ r3

0@

1A:

3.2. Farey and barycentric partitions

In this section we will define two partitions of the triangle

W ¼ fðx; yÞ : 1XxXyX0g:

The first partition of W will yield the domain of our desired function d; while thesecond partition will yield the range.

3.2.1. The Farey partition

We will define a sequence of partitions fPng such that each Pn will consist of 3n

subtriangles of W and each Pn will be a refinement of the previous Pn�1:


Let P0 be the initial triangle W: The three vertices of W are

v1 ¼0

0

� �¼

0=1

0=1

� �;

v2 ¼1

0

� �¼

1=1

0=1

� �;

v3 ¼1

1

� �¼

1=1

1=1

� �:

Taking the Farey sum of these vertices, we have

v1þþv2þþv3 ¼2

1

3

0B@

1CA:

This Farey vector corresponds to the point2=31=3

� �: In particular, the

point2=31=3

� �is an interior point of the triangle W and, in a natural way,

partitions W into three subtriangles. We will refer to the resulting interior point asthe Farey-center.

This determines the partition P1: We now proceed inductively. Suppose we havethe partition Pn; that determines 3n triangles. We now partition each of thesetriangles into three subtriangles, as follows. Suppose one of the triangles in Pn has

vertices p1=r1q1=r1

; p2=r2

q2=r2

and p3=r3

q3=r3

: Computing the Farey sum of the three vertices of

the triangle gives a point v; the Farey-center, in the interior of the subtriangle. TheFarey-center, v; yields a partition of the subtriangle. Computing in this way thepartition of each subtriangle of W determined by Pn; gives us the desired nextpartition Pnþ1 of W:

We denote this full partitioning of W by WF and call it the Farey partitioning.


3.2.2. The Barycentric partition

Again we will define a sequence of partitions *Pn of W such that each*Pn will consist of 3n triangles of W and each *Pn will be a refinement of the

previous *Pn�1:

As with the Farey partitioning, the zeroth level partition *P0 is simply the initialtriangle W:

To compute *P1; we again start with the original three vertices of W and compute

the barycenter of W; namely 2=31=3

: This point happens, in this case, to be the same

point obtained by computing the Farey sum of the coordinates of the vertices of W;the Farey-center. This is just a coincidence.

Proceed inductively as follows. Assume we have a partition *Pn of W into 3n

subtriangles. Further, assume at the nth stage that the coordinates of the vertices ofany subtriangle can be expressed as rational numbers with 3n in the denominator.

Then, if a given subtriangle in *Pn has vertices a1=3n

b1=3n

; a2=3

n

b2=3n

; and a3=3

n

b3=3n

; we

compute the Farey sum of the three vertices. This again gives the barycenter of the

subtriangle, namely,a1þa2þa3

3nþ1

b1þb2þb3

3nþ1

!: Computing, in this way, the partition of each

subtriangle of *W determined by *Pn gives us the desired next partition *Pnþ1 of W:We call this full partitioning of W the barycentric, or Bary, partitioning and

denote it by WB:

3.3. The Farey–Bary map

We are now ready to define the extension of the Minkowski question-markfunction to d : WF-WB: We will proceed in stages. At first, we will define a functiondnðx; yÞ from the vertices of the nth partition of WF to the vertices of the nthpartition of WB and then extend linearly dn to all of WF: Then we will use thesefunctions fdng to define our desired function dðx; yÞ on WF:

We first need to introduce some notation. Each of the partitions Pn and *Pn

determines subtriangles of WF and WB; respectively. Let Wn;F and Wn;B denote WF

and WB after the nth partitioning, respectively. The expression /v1ðnÞ; v2ðnÞ; v3ðnÞSwill denote a general subtriangle of Wn;F with vertices v1ðnÞ; v2ðnÞ; and v3ðnÞ: When

we need to refer to the 3n specific subtriangles, we will use /vs1ðnÞ; vs

2ðnÞ; vs3ðnÞS;

where sAf1;y; 3ng: In a similar fashion, we will refer to the subtriangles of *Pn by/v1ðnÞ; v2ðnÞ; v3ðnÞS; in the general case, and /v s

1ðnÞ; v s2ðnÞ; v s

3ðnÞS in the

specific case.Note that it happens to be the case that

P0 ¼ *P0; P1 ¼ *P1 and W1;F ¼ W1;B:

Definition 3. Define d0; d1 : WF-WB to be the identity maps on the vertices of thesubtriangles determined by P0 and P1: For any n; define dn to send any vertex in the


nth partition Pn to the corresponding vertex in the partition *Pn: That is, define dn onany subtriangle /v1ðnÞ; v2ðnÞ; v3ðnÞS of the partition Pn by

dnðviðnÞÞ ¼ viðnÞ

for i ¼ 1; 2; 3: Finally, for any point ðx; yÞ in the subtriangle with vertices/v1ðnÞ; v2ðnÞ; v3ðnÞS; set

dnðx; yÞ ¼ av1ðnÞ þ bv2ðnÞ þ gv3ðnÞ;

where

ðx; yÞ ¼ av1ðnÞ þ bv2ðnÞ þ gv3ðnÞ:

Note that, since the point ðx; yÞ is in the interior of the triangle/v1ðnÞ; v2ðnÞ; v3ðnÞS; we have that

aþ bþ g ¼ 1

with

0pa; b; gp1:

As defined, d0 and d1 are both the identity map since W0;F ¼ W0;B and W1;F ¼W1;B: However, the mappings start to become more complicated with d2:

At this stage we have the Farey partition:


and the Bary partition:

The correspondence between the vertices becomes

d2

2=3

1=3

� �¼

2=3

1=3

� �;

d2

3=5

1=5

� �¼

5=9

1=9

� �;

d2

4=5

2=5

� �¼

8=9

4=9

� �;

d2

3=5

2=5

� �¼

5=9

4=9

� �:


Going a few stages further, we get for the Farey partition:

and for the Bary partition:

(We find it interesting that the diagram for the Farey partition is much moreaesthetically pleasing than the one for the barycentric partition.)


By definition, d3ðvið2ÞÞ ¼ d2ðvið2ÞÞ for any vertex in a subtriangle/v1ð2Þ; v2ð2Þ; v3ð2ÞS; of W2;F: Thus to describe d3; we need only specify what

happens on the new vertices obtained in W3;F and W3;B:This new correspondence becomes:

d3

11=27

4=27

� �¼

5=9

2=9

� �;

d3

11=27

7=27

� �¼

5=9

3=9

� �;

d3

14=27

1=27

� �¼

4=9

1=9

� �;

d3

14=27

13=27

� �¼

4=9

3=9

� �;

d3

20=27

4=27

� �¼

6=9

2=9

� �;

d3

20=27

16=27

� �¼

6=9

4=9

� �;

d3

23=27

7=27

� �¼

7=9

3=9

� �;

d3

23=27

16=27

� �¼

7=9

4=9

� �;

d3

26=27

13=27

� �¼

6=9

3=9

� �:

Definition 4. A point vAWF is an interior point of the Farey partitioning if v is not avertex or on an edge of any triangle of any of the Farey partitions.

Since there are only a countable number of triangles making up theFarey partitioning, we see that almost all points in WF are interior points.It is the interior points that we are interested in. It is these points forwhich periodicity of our eventual Farey sequences will give candidates for cubicirrationals.

We need to understand a bit better the nature of these interior points before wecan define our map d: Fix an interior point vAWF: Then for each n; v must be in the


interior of one of the 3n subtriangles of Pn: Label this triangle WnðvÞ: Then we havethat, for all m bigger than n;

WmðvÞCWnðvÞ:

Thus the interior point v is in a nested sequence of Farey triangles. Further, if *WnðvÞdenotes the corresponding triangle with respect to the barycentric partitioning, wehave another nested sequence of triangles.

Theorem 5. For each interior point vAWF; the sequence of dnðvÞ converges to a point

in WB:

Proof. Fix an interior point vAWF: For each n; label the vertices of the triangle*WnðvÞ by /v1ðnÞ; v2ðnÞ; v3ðnÞS: We know, for all m4n; that the image dmðWmðvÞÞ ¼*WmðvÞ is contained in *WnðvÞ: Thus the sequence dnðvÞ will converge if the vertices ofthe triangles dnðviðnÞÞ converge.

A calculation shows that the distance from the barycenter of a triangle to any ofthe vertices is at most two thirds the distance of the longest side of the triangle. If our

initial point v is interior, then we can see that at most one of the vertices in *Wn willremain constant for infinitely many n: (Usually none remain fixed.) This forces the

vertices of the triangles *Wn to converge, forcing in turn the sequence dnðvÞ toconverge. &

Definition 6. Define the Farey–Bary map d :WF-WB by setting

dðvÞ ¼ limn-N

dnðvÞ

when v is an interior point and

dðvÞ ¼ dnðvÞ

when v is a vertex or edge of one of the triangles in Wn;F; for some n:

While the Farey–Bary map is not continuous at all points, it is so at all interiorpoints, which is almost all points. Again, it is only these interior points that are ofinterest.

Theorem 7. Let fvkg be a sequence of points in WF that converge to an interior

point v: Then fdðvkÞg will converge to dðvÞ: Thus d is a continuous function at all

interior points.

Proof. Given any small ball about the point dðvÞ; there is an n such that the

triangle *WnðvÞ is contained in this ball. The inverse image of *WnðvÞ is thetriangle WnðvÞ; of which v is in the interior. Hence, eventually the fvkg must

be in WnðvÞ: Hence their images fdðvkÞg must eventually be in *WnðvÞ; and hence


must eventually be in our chosen small ball containing dðvÞ: Thus fdðvkÞg convergesto dðvÞ: &

4. Farey iteration in the domain as multidimensional continued fraction

Minkowski’s ?ðxÞ provides a link between algebraic properties of numbers and thefailure of differentiability, almost everywhere, for ?ðxÞ: Our goal is to find analogouslinks for the Farey–Bary map dðx; yÞ: The key algebraic property of the Minkowskiquestion mark function is that ?ðxÞ maps quadratic irrationals to rational numbers.The goal of this section is to show that dðx; yÞ maps a class of pairs of cubicirrationals to pairs of rationals. Unfortunately, we cannot make the claim that dmaps all pairs of cubics (even in the same number field) to pairs of rationals.

4.1. Preliminary notation

Let ða; bÞAWF be an interior point. The Farey partitions of WF yields a nestedsequence of triangles containing the point ða; bÞ: (Note that it is not necessarily thecase that the vertices of these triangles converge to v:) Suppose that at the nth stageof the Farey partitioning, the triangle that contains ða; bÞ is /v1ðnÞ; v2ðnÞ; v3ðnÞS: Wewill maintain the notation viðnÞ to mean either the cartesian version of the vertex orthe vector in space that corresponds to the vertex. That is, viðnÞ will refer to

piðnÞ=riðnÞqiðnÞ=riðnÞ

; as well as to

piðnÞqiðnÞriðnÞ

!: Furthermore, we will order the vertices so that for

all n;

r1ðnÞpr2ðnÞpr3ðnÞ:

We want to relate the vertices of the ðn � 1Þth subtriangle that contains ða; bÞ withthe vertices of the subtriangle at the next iteration. For that, suppose thatða; bÞA/v1ðn � 1Þ; v2ðn � 1Þ; v3ðn � 1ÞSDWn�1;F: Applying the next partition, Pn;to Wn�1;F; we decompose /v1ðn � 1Þ; v2ðn � 1Þ; v3ðn � 1ÞS into three new sub-

triangles. If we let /v1ðnÞ; v2ðnÞ; v3ðnÞS denote the subtriangle into which ða; bÞ falls,we see that there are three possibilities for the vertices of /v1ðnÞ; v2ðnÞ; v3ðnÞS:

In case I, the vertices of the newly partitioned triangle will be

v1ðnÞ ¼ v1ðn � 1Þ;


v3ðnÞ ¼ v1ðn � 1Þ þ þþv2ðn � 1Þþþv3ðn � 1Þ:


Similarly, the vertices in case II will be



v3ðnÞ ¼ v1ðn � 1Þþþv2ðn � 1Þþþv3ðn � 1Þ:


For case III we have



v3ðnÞ ¼ v1ðn � 1Þþþv2ðn � 1Þþþv3ðn � 1Þ:

In the next section, we will streamline this notation.

4.2. Fixing notation

For each interior point ða; bÞ in WF we now associate a sequence of positiveintegers that will uniquely determine the nested sequence of Farey subtrianglescontaining ða; bÞ:

To motivate the eventual notation, consider the following three possibilities. Startwith a triangle, with vertices v1; v2; and v3; still keeping the convention thatr1pr2pr3: Suppose we perform k type I operations in a row. The resulting newtriangle will have vertices in the following form:

v1; v2; kv1þþkv2þþv3:

If we perform a type II operation on the triangle, and then k � 1 type I operations,the new triangle will have vertices:

v2; v3; v1þþkv2þþkv3:


If we perform a type III operation on the triangle, and then k � 1 type Ioperations, the new triangle will have vertices:

v1; v3; kv1þþv2þþkv3:

This suggests the following notation.Define a sequence fa1ði1Þ; a2ði2Þ;yg to be such that each akðikÞ is a

positive integer and each ik represents either cases I, II or III. The valueof akðikÞ denotes the operation of first applying a type ik operation and thenakðikÞ � 1 type I operations. We use the further convention that for kX2; ik can onlybe of type II or III.

Note that, in the notation of the previous section, by the time we are at stepakðikÞ; we have performed n ¼ a1ði1Þ þ a2ði2Þ þ?þ akðikÞ Farey partitionsof WF: We associate to each interior point ða; bÞAWF the sequence thatyields the corresponding Farey partitions that converge to ða; bÞ: This sequencewill be unique. We will also use /v1ðkÞ; v2ðkÞ; v3ðkÞS to denote the subtriangle ofWn;F containing ða; bÞ after n ¼ a1ði1Þ þ a2ði2Þ þ?þ akðikÞ steps partitioning WF:Finally, if we know what case we are in, that is, if we know ik; we will simply write ak

instead of aðikÞ:

Example 8. The shaded region below corresponds to all points ð2ðIIIÞ; 1ðIIÞ; 1ðIÞÞ:Note that in the notation of last section n ¼ 4 but that in the notation of this sectionand for the rest of the paper k ¼ 3:


We now have the following recursion formulas for the vertices. For case I at thekth step we get

v1ðkÞ ¼ v1ðk � 1Þ;


v3ðkÞ ¼ akv1ðk � 1Þþþakv2ðk � 1Þþþv3ðk � 1Þ;

For case II we have



v3ðkÞ ¼ v1ðk � 1Þþþakv2ðk � 1Þþþakv3ðk � 1Þ:

Finally, for case III we get



v3ðkÞ ¼ akv1ðk � 1Þþþv2ðk � 1Þþþakv3ðk � 1Þ:

We can put these recursion relations naturally into a matrix language. At each stepk; define Mk to be the three-by-three matrix

Mk ¼ ðv1ðkÞ v2ðkÞ v3ðkÞÞ ¼p1ðkÞ p2ðkÞ p3ðkÞq1ðkÞ q2ðkÞ q3ðkÞr1ðkÞ r2ðkÞ r3ðkÞ

0B@

1CA:

If, from the ðk � 1Þth step to the kth step, we are in case I, then

Mk ¼ Mk�1

1 0 ak

0 1 ak

0 0 1

0B@

1CA;

for case II we have

Mk ¼ Mk�1

0 0 1

1 0 ak

0 1 ak

0B@

1CA;


and for case III,

Mk ¼ Mk�1

1 0 ak

0 0 1

0 1 ak

0B@

1CA:

Denote in each of these cases the matrix on the right by AkðIÞ;AkðIIÞ and AkðIIIÞ;respectively. Then we have that each Mk is the product of M0 with a sequence ofvarious Am:

Theorem 9. Each Mk is in the special linear group SL(3,Z).

Proof. All we need to show is that for all k;

detðMkÞ ¼ 71:

This follows immediately from observing that detðM0Þ ¼ 1 and that the determi-nants of each of the various AkðIÞ; AkðIIÞ and AkðIIIÞ are also plus orminus one. &

4.3. Areas of Farey subtriangles

Given a finite sequence fa1ði1Þ; a2ði2Þ;y; akðikÞg of positive integers, we define

Wk ¼ fðx; yÞ : fa1ði1Þ;y; akðikÞg are the 1st k terms in Farey sequenceg:

A major goal of this paper is showing that the areas of these triangles Wk cannotgo to zero too quickly. For these calculations, we will need an easy formula for theareas of the Wk:

Theorem 10. The area of a triangle with vertices ðp1=r1; q1=r1Þ; ðp2=r2; q2=r2Þ; and

ðp3=r3; q3=r3Þ is

Area of triangle ¼ 1

2

det

p1 p2 p3

q1 q2 q3

r1 r2 r3

0B@

1CA

r1r2r3:

This is just a calculation involving cross products.

Corollary 11. Given any finite sequence fa1ði1Þ; a2ði2Þ;y; akðikÞg of positive integers,

2AreaðWkÞ ¼1

r1ðkÞr2ðkÞr3ðkÞ:

This follows since detðMkÞ ¼ 71:


4.4. Farey periodicity implies cubic irrationals

We want to determine when our Farey sequence can be used to determinealgebraic properties of the corresponding interior point ða; bÞ: We want to show:

Theorem 12. Suppose that the interior point ða; bÞAWF has an eventually periodic

Farey sequence. Further suppose that its nested sequence of Farey triangles converges

to this point ða; bÞ: Then both a and b are algebraic numbers with

degðaÞp3; degðbÞp3 and

dimQQ½a; b�p3:

This is why the Farey partitioning can be viewed as a multi-dimensional continuedfraction algorithm.

Proof. We will be heavily using two facts. First, an eigenvector ð1; a; bÞ of a 3 � 3matrix with rational coefficients has the property that

dimQQ½a; b�p3;

as seen in a similar argument in [8] in Section 8. Second, if we multiply amatrix, which has a largest real eigenvalue, repeatedly by itself, in thelimit the columns of the matrix converge to the eigenvector corresponding to thelargest eigenvalue, unless one of the columns is itself an eigenvector for anothereigenvalue.

Suppose that ða; bÞAW has an eventually periodic Farey sequence. Byassumption, the vertices of the corresponding Farey partition triangles converge tothe point ða; bÞ: We have seen above that the vertices of the partition trianglescorrespond to the columns of matrices that are the products of various AkðIÞ; AkðIIÞand AkðIIIÞ: With the assumption of periodicity, denote the product of the initialnon-periodic matrices be B and the product of the periodic part be A: Then some ofthe Farey partition triangles about the point ða; bÞ are given by

B;BA;BA2;BA3;y :

The columns of the matrices A;A2;A3;y must converge to a multiple of

B�1ð1; a; bÞT : But the columns of the Ak must also converge to an eigenvector and

hence B�1ð1; a; bÞT is an eigenvector of the matrix A: This will give us that a and bmust have the desired properties. &

Note that the assumption that the vertices of the Farey partition trianglesconverging to ða; bÞ is needed. For example, if the Farey sequence to ða; bÞeventually consists solely of terms akðIIIÞ; then the vertices will converge to a linesegment, not a point. This follows from the corresponding matrices. Ifbeyond a certain point, we just have terms akðIIIÞ; then we are eventually


multiplying together matrices of the form

1 0 ak

0 0 1

0 1 ak

0B@

1CA:

The presence of the first column being a one, then two zeros, means that beyond acertain point, all of the nested Farey partition triangles contain the same vertex. Theother two vertices will of course converge to a point. On the other hand, it can bechecked that this is the only way for the Farey partition triangles to not converge toa single point.

5. Iteration in the barycentric range

We have defined Farey partitions, Pn; in WF and barycentric partitions, *Pn inWB: In WF; the Farey partitions yielded an association between each interior pointða; bÞ and a sequence obtained from the convergence of the subtriangles resultingfrom successive applications of the partitions, Pn: This association depended only onthe successive partitioning of each subtriangle into three more subtriangles and noton the relative positioning of the new subtriangles. We can follow the sameprocedure in WB: That is, if we let ða; bÞAWB be the image of an interior point, wecan again associate with ða; bÞ a sequence of positive integers fa1ði1Þ; a2ði2Þ;ygwhich come from a sequence of barycentric triangles converging to the point ða; bÞ:

Label the triangle corresponding to fa1ði1Þ; a2ði2Þ;y; akðikÞg by

*Wfa1ði1Þ; a2ði2Þ;y; akðikÞg:

Recall that

*WB ¼ /v1ð0Þ; v2ð0Þ; v3ð0ÞS;

where

v1ð0Þ ¼0

0

1

0B@

1CA; v2ð0Þ ¼

1

0

1

0B@

1CA and v3ð0Þ ¼

1

1

1

0B@

1CA:

Associated with the sequence fa1ði1Þ; a2ði2Þ;y; akðinÞg will be vertices v1ðkÞ; v2ðkÞand v3ðkÞ and corresponding vectors

v1ðkÞ ¼��

3a1þ?þak

0B@

1CA; v2ðkÞ ¼

��

3a1þ?þak

0B@

1CA; v3ðkÞ ¼

��

3a1þ?þak

0B@

1CA;


where the other entries for the vectors are nonnegative integers. There are, of course,

matrices *Mn that map the vertices from a given level to the vertices of the next level,

in analogue to the matrices Mn: The *Mn are products of matrices of the form

3 0 1

0 3 1

0 0 1

0B@

1CA;

0 0 1

3 1 1

0 3 1

0B@

1CA;

3 0 1

0 0 1

0 3 1

0B@

1CA

depending if we are in cases I, II or III, respectively.Note that at each individual step of the barycentric partitioning, we are cutting the

area down by a factor of a 1=3: This leads to the following theorem.

Theorem 13. Twice the area of *Wfa1ði1Þ; a2ði2Þ;y; akðikÞg is 13a1þa2þ?þak ðik Þ

:

5.1. Ternary periodicity implies rationality

Suppose that we have a point ða; bÞAWB for which the barycentric partitioning iseventually periodic. We want to show that both a and b are rational numbers. Thatis, we want the following theorem.

Theorem 14. Let ða; bÞAWB be the image of an interior point. If ða; bÞ has an

eventually perdiodic Barycentric sequence, then both a and b are rational.

Proof. This proof is almost exactly the same as the corresponding proof for theFarey case, whose notation we adopt. There is one significant difference, namely thatthe matrices whose columns yield the vertices of the barycentric partitioning are allmultiples of a stochastic matrix. This means that each matrix is a multiple of amatrix whose columns add to one. If the columns add to one, then it can easily beshown that the limit of the products of such a matrix converges to a matrix whose

rows are multiples of ð1; 1; 1Þ (see Chapter 6 in [26]). Thus the matrices A;A2;A3;yconverge to a matrix whose rows are multiples of ð1; 1; 1Þ: Since everything in sight is

rational, we can show that B�1ð1; a; bÞT will converge to a triple of rational numbers.Since the entries of B are integers, this yields that a and b are rational numbers. &

6. The Farey–Bary analog of singularness

The original Minkowksi ?ðxÞ function is singular, meaning that even though it isincreasing and continuous, it has derivative zero almost everywhere. The key to theproof lies in showing that at almost all points

lim infk-N

length of interval in range

length of interval in domain¼ 0;


for appropriately defined intervals. We will show a direct analog of this, where thelengths of intervals are replaced by areas of triangles. Thus we will show

lim infk-N

area of subtriangle in range

area of subtriangle in domain¼ 0;

again for appropriately defined subtriangles.

6.1. Almost everywhere lim sup a1þ?þan

n¼ N

This is the most technically difficult section of the paper. The goal is to show thefollowing theorem, which will be critical in the next section. Recall that given anyinterior point ða; bÞAW; we have associated a sequence fa1; a2;yg of positiveintegers. (Also, we know that the noninterior points in W have measure zero; for thisreason we will ignore these noninterior points.) We want to show that this sequenceof positive integers must increase to infinity, in some sense, almost everywhere. Theprecise statement is:

Theorem 15. The set of ða; bÞAW for which

lim supn-N

a1 þ?þ an

noN

has measure zero.

As it will only be apparent in the next section why to we need this theorem, werecommend on the first reading of this paper to go to the next section first.

Before proving the theorem, we need a preliminary lemma. First, let

v1 ¼x1

y1

z1

0B@

1CA; v2 ¼

x2

y2

z2

0B@

1CA; v3 ¼

x3

y3

z3

0B@

1CA

and let

T ¼ /v1; v2; v3S

be the corresponding triangle in the plane, where the vi are now viewed as points inthe plane.

Suppose that det v1v2v3 ¼ 1: Then we know that

2area of /v1; v2; v3S ¼ 1

z1z2z3:

Given a positive number L41; define TLð1Þ to be the triangle with vertices v1; v2 and

Lv1 þ Lv2 þ v3; TLð2Þ the triangle with vertices v2; v1 þ Lv2 þ Lv3 and v3 and TLð3Þ


the triangle with vertices v1;Lv1þþv2þþLv3 and v3:

Define

TL ¼ T � TLð1Þ � TLð2Þ � TLð3Þ:

We now state and then prove a lemma that is the technical heart of the proof of thetheorem:

Lemma 16. For all LX1;

areaðT � TLÞpL � 1

LareaðTÞ:

Proof. We know that

2areaðTÞ ¼ 1

z1z2z3:

For ease of notation, we set z1 ¼ x; z2 ¼ y; z3 ¼ z: Then

2areaðT � TLÞ

¼ 1

xyz� 1

xðLx þ y þ LzÞz �1

xyðLx þ Ly þ zÞ �1

ðx þ Ly þ LzÞyz

¼ 1

xyz1 � y

ðLx þ y þ LzÞ �z

ðLx þ Ly þ zÞ �x

ðx þ Ly þ LzÞ

� �:

Thus we must show that

1 � y



ðx þ Ly þ LzÞ

� �p

L � 1

L:


Setting

a ¼ x2y þ x2z þ xy2 þ xz2 þ y2z þ yz2;

we have that

1 � y



ðx þ Ly þ LzÞ

� �

¼ aðL3 � LÞ þ 2xyzðL3 � 1ÞðLx þ Ly þ zÞðLx þ y þ LzÞðx þ Ly þ LzÞ:

After a series of calculations, we get that this is equal to

ðL � 1Þ LðL þ 1Þaþ 2xyzðL2 þ L þ 1ÞLðL2 þ L þ 1Þaþ L2ðx3 þ y3 þ z3Þ þ xyzð3L2 þ 2L3 þ 1Þ

� �:

Thus we must show that

LðL þ 1Þaþ 2xyzðL2 þ L þ 1ÞLðL2 þ L þ 1Þaþ L2ðx3 þ y3 þ z3Þ þ xyzð3L2 þ 2L3 þ 1Þ

� �p

1

L;

which is equivalent to showing that

L2ðL þ 1Þaþ 2xyzLðL2 þ L þ 1Þ

pLðL2 þ L þ 1Þaþ L2ðx3 þ y3 þ z3Þ þ xyzð2L3 þ 3L2 þ 1Þ;

which in turn, reduces to showing that

2Lxyzpaþ L2ðx3 þ y3 þ z3Þ þ L2xyz þ xyz:

This last inequality follows from the fact that LX1: Thus the proof of the lemma isdone.

Proof of Theorem. For each positive integer N; set

MN ¼ ða; bÞAW : for all nX1;a1 þ?þ an

npN

n o:

We will show that

measureðMNÞ ¼ 0:

Since the union of all of the MN is the set we want to show has measure zero, we willbe done.

Now, a1þ?þan

npN if and only if

a1 þ?þ anpnN:


Since each aiX1; this last inequality implies

n � 1 þ anpnN

or

anpnðN � 1Þ þ 1:

Set

MN ¼ fða; bÞAW : for all nX1; anpnðN � 1Þ þ 1g:

Since MNCMN ; if we can show that measureðMÞN ¼ 0; we will be done.

Set

MNð1Þ ¼ fða; bÞAW : a1pðN � 1Þ þ 1g

and in general

MNðkÞ ¼ fða; bÞAMNðk � 1Þ : akpkðN � 1Þ þ 1g:

Then we have a decreasing nested sequence of sets with

MN ¼\Nk¼1

MNðkÞ:

But this puts us into the language of the above lemma. Letting L ¼ kðN � 1Þ þ 1;we can conclude that

measureðMNðkÞÞ ¼kðN � 1Þ

kðN � 1Þ þ 1measureðMNðk � 1ÞÞ

and hence

measureðMNÞ ¼YNk¼2

kðN � 1ÞkðN � 1Þ þ 1

:

We must show this infinite product is zero, which is equivalent to showing that itsreciprocal

YNk¼2

kðN � 1Þ þ 1

kðN � 1Þ ¼YNk¼2

1 þ 1

kðN � 1Þ

� �¼ N:

Taking logarithms, this is the same as showing that the series

XNk¼2

log 1 þ 1

kðN � 1Þ

� �¼ N:


This in turn follows since, for large enough k; we have

log 1 þ 1

kðN � 1Þ

� �X

1

2kðN � 1Þ:

We are done.

6.2. Almost everywhere

lim infðareað *WnÞ=areaðWnÞÞ ¼ 0:

The goal of this section, and for the entire paper, is:

Theorem 17. For any point ða; bÞAW; off of a set of measure zero,

lim infn-N

areað *Wfa1ði1Þ; a2ði2Þ;y; anðinÞgÞareaðWfa1ði1Þ; a2ði2Þ;y; anðinÞgÞ

¼ 0:

This is capturing the intuition that the determinant of the Jacobian of the mapd :WF-WB is zero almost everywhere, which in turn is a direct generalization thatthe Minkowski question-mark function is singular. In fact, our proof is in spirit ageneralization of Viader, Paradis and Bibiloni’s work in [42].

Proof. We know that, letting sn ¼ a1 þ?þ an;

areað *Wfa1ði1Þ; a2ði2Þ;y; anðinÞgÞ ¼1

2 � 3sn

and that

areaðWfa1ði1Þ; a2ði2Þ;y; anðinÞgÞ ¼1

2 � r1ðnÞr2ðnÞr3ðnÞ:

Thus we want to show that, almost everywhere,

lim infn-N

r1ðnÞr2ðnÞr3ðnÞ3sn

¼ 0:

We know that r3ðnÞ is

anr1ðn � 1Þ þ anr2ðn � 1Þ þ r3ðn � 1Þ;

anr1ðn � 1Þ þ r2ðn � 1Þ þ anr3ðn � 1Þ

or

r1ðn � 1Þ þ anr2ðn � 1Þ þ anr3ðn � 1Þ:


Thus we have, by the convention of our notation,

r1ðnÞpr2ðnÞpr3ðnÞpð2an þ 1Þr3ðn � 1Þ:

By iterating this inequality, we have

r1ðnÞpr2ðnÞpr3ðnÞpYn

i¼1

ð2aj þ 1Þ:

Thus


pQn

i¼1 ð2aj þ 1Þ3

3sn:

By the arithmetic-geometric mean,

Yn

i¼1

ð1 þ biÞp 1 þ b1 þ?þ bn

n

� �n

:

Setting bj ¼ 2aj; we get


pð1 þ 2sn

nÞ3n

3sn

pð3sn

nÞ3n

3sn

p27 � ðsn

nÞ3

3sn=n

!n

:

From the previous section, we know that sn=n-N; almost everywhere. Since the

above denominator has a 3sn=n term while the numerator only has a ðsn=nÞ3 term, theentire ratio must approach zero, giving us our result.

7. Questions

There are a number of natural questions. First, all of this can almost certainly begeneralized to higher dimensions.

More importantly, how much does the function theory of d influence thediophantine properties of points in W?

There are many multi-dimensional continued fraction algorithms. For any of thesethat involve partitioning a given triangle into three new subtriangles, a mapanalogous to our d can of course be defined. What are the properties of these newmaps?


Underlying most work on multidimensional continued fractions, thoughfrequently hidden behind view, are Lie theoretic properties of the special lineargroup. Can this be made more explicit?

Finally, the initial Hermite problem remains open.

Acknowledgments

We thank Lori Pedersen for making all of the diagrams. We thank AndrewMarder for pointing out an error on an earlier version of this paper and would like tothank Keith Briggs for pointing out some errors in the bibliography in anotherearlier version.

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a two-dimensional minkowski ðxÞ function · notethat0¼ 0 1 and1¼ 1 1: now,giventhepartitioni k...

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