Generalizing the Minkowski Function
Using Triangle Partition Maps
by
Peter McDonald
Professor Thomas Garrity, Advisor
A thesis submitted in partial fulfillment
of the requirements for the
Degree of Bachelor of Arts with Honors
in Mathematics
Williams College
Williamstown, Massachusetts
May 20, 2016
Abstract
In this paper, we present two previous attempts at generalizing the Minkowski
Question Mark Function before presenting a framework for generalizing ?(x) to
a family of 216 multidimensional continued fraction algorithms known as triangle
partition (TRIP) maps. Furthermore, we place these 216 maps into 15 classes
whose associated generalization of the Question Mark Function is related by a
linear transformation and show for 7 of these classes that this function is singular.
Acknowledgments
First of all, a big thank to Professor Garrity for introducing me to such an interesting topic and providing me with endless humor, encouragement, and motivation,
and for signing me up for the first thesis defense so that I actually was able to
finish this. I also owe a lot to Professor Silva, both for being my second reader
and for giving me the insight that lead to Section 3.2.1. Shoutout to Liz Frank
and Manny Daring, the other members of the multidimensional continued fraction gang, for being a sounding board for ideas and helping me understand TRIP
maps. Thank you to the entire Williams math department for providing such an
amazing learning opportunity over these past four years, and especially to Professor Loepp for being such a mentor to me over the years and for giving me the
chance to experience research for the first time last summer in SMALL. Finally,
thank you to my family and friends for making my last year at Williams a great
one.
Contents
1 Preliminaries
1.1 Singular Functions . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 The Minkowski Question Mark Function . . . . . . . . . .
1.1.2 Connection to Continued Fractions . . . . . . . . . . . . .
1.2 Multidimensional Continued Fractions and TRIP Maps . . . . . .
1.2.1 The Triangle Map . . . . . . . . . . . . . . . . . . . . . . .
1.2.2 A Family Multidimensional Continued Fraction Algorithms:
TRIP Maps . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Previous Attempts . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.1 Beaver and Garrity . . . . . . . . . . . . . . . . . . . . . .
1.3.2 Panti . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
14
14
17
2 Defining Our Function
19
3 Results
3.1 A Special Case: Mönkemeyer-Type Maps
3.2 Remaining Maps . . . . . . . . . . . . .
3.2.1 Triangle Tree Sequence Approach
3.2.2 Limit Definition Approach . . . .
3.2.3 Finding limn→∞ snn . . . . . . . .
3.3 Degenerate Farey Maps . . . . . . . . . .
29
29
30
31
33
38
45
4 Conclusion and Future Directions
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47
Chapter 1
Preliminaries
Singular functions are of interest in measure theory because they are continuous
functions that send a set of positive measure to a set of measure zero. In fact, in
the one dimensional case they are even more interesting, as they are monotonically
increasing or decreasing, and their derivative is zero almost everywhere. While singular functions may sound like pathologies, we have an extremely natural example
of a singular function from [0, 1] to [0, 1] known as the Minkowski Question Mark
Function, denoted ?(x). As it turns out, the definition of ?(x) is intimately related
to the well-known continued fraction algorithm. Thus, in searching for natural
examples of singular functions from R2 to R2 , we will study multidimensional
continued fraction algorithms. We will focus on a family of 216 multidimensional
continued fraction algorithms known as triangle partition (TRIP) maps in looking
for generalizations of the Question Mark Function and present a general framework for showing when the function we obtain is singular. Furthermore, we break
these 216 maps into 15 classes whose associated generalization of ?(x) is related
by a linear transformation and show for 7 of these classes that this function is
singular.
1.1
Singular Functions
Definition 1.1. A function f : [0, 1] → [0, 1] is singular if:
(1) f (0) = 0 and f (1) = 1,
3
(2) f is continuous,
(3) f is strictly increasing,
(4) f 0 (x) = 0 almost everywhere.
At first glance, conditions (3) and (4) seem contradictory. We have a function
that is strictly increasing, but at most points the rate of change is 0. If we were to
take the Lebesgue integral of f 0 (x) from 0 to 1, rather than getting f (1)−f (0) = 1
we would get 0. It is this fourth condition is essential to singularity and the one
that we will be examining as we attempt to find analogous singular functions from
R2 to R2 .
Looking at the definition, singular functions seem like more of a pathology
than a natural object to study. However, it turns out that there is one particular
singular function that happens to appear extremely naturally as a result of the
study of continued fractions.
1.1.1
The Minkowski Question Mark Function
To define the Minkowski Question Mark Function, we first define the Farey addition of two rationals. Given pq , rs ∈ Q define the operation +̂ as follows:
p r
p+r
+̂ =
q s
r+s
Thinking of the fractions pq and rs as the vectors (q, p), (s, r) ∈ Z2 , we notice
that this is simply vector addition. When we take the Farey sum of two rational
numbers, we are really adding them as vectors in Z2 and projecting back into Q.
Then we define the nth Farey set recursively, beginning with F0 = { 10 , 11 }. Then
0 0 1 1
0 1 1
F1 =
, +̂ ,
=
, ,
1 1 1 1
1 2 1
Following this pattern, we define Fn by taking the union of Fn−1 and the Farey
4
sums of adjacent elements of Fn−1 and ordering them appropriately. Then
0 1
,
F0 =
1 1
0 1 1
, ,
F1 =
1 2 1
0 1 1 2 1
F2 =
, , , ,
1 3 2 3 1
0 1 1 2 1 3 2 3 1
F3 =
, , , , , , , ,
1 4 3 5 2 5 3 4 1
Similarly, we define the nth barycentric set, Bn , by starting with B0 = {0, 1} and
creating Bn by taking the union of Bn−1 and the averages of adjacent elements of
Bn−1 and ordering appropriately. We present the first few iterations of this:
0 1
B0 =
,
1 1
0 1 1
, ,
B1 =
1 2 1
0 1 1 3 1
B2 =
, , , ,
1 4 2 4 1
0 1 1 3 1 5 3 7 1
B3 =
, , , , , , , ,
1 8 4 8 2 8 4 8 1
Notice that F0 = B0 and F1 = B1 but that after this point they diverge. Now we
are ready to define ?(x) as a limit of functions we will call ?n (x).
Define ?0 (x) = x on [0, 1]. This corresponds to identifying the first element
of F0 with the first element of B0 and the second element of F0 with the second
element of B0 and connecting these points with a straight line. Then we can define
?n (x) by ?n (fi,n ) = bi,n , where fi,n is the ith element of Fn and bi,n is the ith element
of Bn , and filling in the rest by linearity. This means each ?n (x) will be a piecewise
linear function. The first few ?i are shown below:
5
?1 (x)
?2 (x)
?3 (x)
Then we define
lim ?n (x) =?(x).
n→∞
In [1] Salem show that the Question Mark Function is singular by showing
that everywhere the derivative exists and is finite, the derivative is zero. Since the
derivative must exist and be finite almost everywhere, we have that ?(x) is singular.
A further discussion of alternate definitions and explorations of the singularity of
?(x) can also be found in [2]. In looking for higher dimensional singular functions
we will distill Salem’s argument to its essential ingredients by using the limit
definition of ?(x) . Each ?n can be though of as a map from In,F , the unit interval
partitioned according to Fn , to In,B , the unit interval partitioned according to
Bn . Because ?n is piecewise linear, the derivative at any point α ∈ (fj,n , fj+1,n ) is
simply
bj+1,n − bj,n
.
fj+1,n − fj,n
6
Working this out for the first few iterations of ?n (x) we get the following:
?01 (x) = 1

3



4
?02 (x) = 32



3
4
?03 (x) =

1



2



3



2


15



8
5
4



15



8



3



2



1
2
0<x<
1
3
1
3
<x<
2
3
2
3
<x<1
0<x<
1
4
1
4
<x<
1
3
1
3
<x<
2
5
2
5
<x<
3
5
3
5
<x<
2
3
2
3
<x<
3
4
3
4
<x<1
Then as we increase n, we find that this ratio goes to zero almost everywhere.
It turns out that this understanding of ?(x) is intimately linked to continued
fractions, as we will see in the next section.
1.1.2
Connection to Continued Fractions
Given a number α ∈ [0, 1] we can define the continued fraction expansion of α by
α=
1
a1 +
1
a2 + a
1
3 +···
for ai ∈ Z>0 . We say that α = [0; a1 , a2 , a3 , . . . ].
It can be shown that every α ∈ [0, 1] has a continued fraction expansion that
encodes some interesting properties about the number. In analogy to a number’s
decimal expansion telling us that a number is rational if and only if it’s expansion
terminates or is eventually periodic, a number is rational if and only if its continued fraction expansion terminates, and is a quadratic irrational if and only if
its continued fraction expansion is eventually periodic ([3]). Thus, continued fraction expansions of numbers give us more information about a number’s algebraic
properties than its decimal expansion.
7
Traditionally, a number’s continued fraction expansion is linked to the Gauss
Map
1
1
G(x) = −
.
x
x
We can understand the Gauss map by partitioning the unit interval into Ik where
1
, k1 ]:
Ik = ( k+1
I4 I3
1
5
1
4
I2
1
3
I1
1
2
Note that for α = [0; a1 , a2 , a3 , . . . ] ∈ [0, 1) we have α ∈ Ia1 . This becomes clear
when you realize α = a11+b where b = [0; a2 , a3 , . . . ]. Then because a11+1 < a11+b ≤
1
, α ∈ Ia1 . In fact, using the Gauss map, we get:
a1
Proposition 1.2. α ∈ [0, 1] equals [0; a1 , a2 , a3 , . . . ] if and only if Gi−1 (α) ∈ Iai
for all i.
It turns out that this algorithm for finding a number’s continued fraction
expansion is intimately linked to the Farey partition of the unit interval that we
used to define ?(x). To see this, we first define the following matrices:
!
!
1 0
1 1
A0 =
A1 =
1 1
0 1
Let pq , rs be two adjacent elements of some Fn with
the interval [ pq , rs ] by the following matrix:
!
p r
M=
q s
p
q
< rs . Then we can denote
Then the intervals [ pq , pq +̂ rs ] and [ pq +̂ rs , rs ] are represented by M A1 and M A0 respectively:
!
!
!
p r
1 1
p p+r
M A1 =
=
q s
0 1
q q+s
M A0 =
p r
q s
!
1 0
1 1
8
!
=
p+r r
q+s s
!
Recall that in F0 we wrote 0 =
with matrix
0
1
and 1 = 11 . Then we represent the unit interval
V =
0 1
1 1
!
Then V A1 and V A0 represent the partition of the unit interval corresponding to
F1 . In general, the partition of the unit interval corresponding to Fn is represented
by the 2n matrices V Ai1 AI2 · · · Ain .
Recall our partition of I into Ik . We claim each Ik is represented by the matrix
V Ak−1
1 A0 . To see this, we notice that
!
1 k−1
k−1
A1 =
0
1
so
V Ak−1
1 A0 =
0 1
1 1
!
1 k−1
0
1
!
1 0
1 1
!
=
1
1
k+1 k
!
.
In general we get that for α = [0; a1 , a2 , . . . ], α will always be contained in the
interval represented by V Aa11 A0 Aa12 A0 · · · Aa1i A0 for all i.
Similarly, we can use matrices to define our barycentric partition of [0, 1].
Define
!
!
1/2 0
1 1/2
B0 =
B1 =
1/2 1
0 1/2
, k ] in In,B . Then
Consider the interval [ k−1
2n 2n
!
!
2k − 1 k
k−1 k
B0 =
2n
2n
2n+1 2n
!
!
k−1 k
k − 1 2k − 1
B1 =
2n
2n
2n
2n+1
which are both matrices representing the subintervals of [ k−1
, k ] in In+1,B . Then
2n 2n
In,B is given by all matrices of the form V Bi1 · · · Bin for (i, . . . , in ) ∈ {0, 1}n .
This partitioning of the unit interval corresponds to finding the binary expansion of α ∈ [0, 1]. Then we get that α = .i1 i2 . . . if and only if α is contained in
9
the interval represented by V Bi1 Bi2 · · · Bin for all n. With a bit more work, this
leads us to an alternate definition of ?(x) that Salem describes in [1]
?([0; a1 , a2 , . . . ]) =
1
2a1 −1
−
1
2a1 +a2 −1
+
1
2a1 +a2 +a3 −1
− ···
Now that we understand how the Minkowski function is linked to our continued
fraction algorithm, it makes sense that our attempts to generalize the Minkowski
function to R2 would center around what are known as multidimensional continued
fractions.
1.2
Multidimensional Continued Fractions and TRIP
Maps
Recall that the continued fraction expansion of a number is eventually periodic
if and only if that number is a quadratic irrational ([3]). Writing a number α
in the shorthand notation for continued fractions, α = [0; a1 , a2 , a3 , . . . ], we can
think of α being represented by the sequence (a1 , a2 , a3 , . . . ). Then the continued
fraction algorithm is a way of assigning a sequence of positive integers to each real
number such that that sequence is eventually periodic if and only if the number
is a quadratic irrational. Thinking of continued fractions in this way gave rise to
the following natural question, known as the Hermite problem:
Question 1.3. Is there a method of associating a sequence of integers to a real
number such that the sequence is eventually periodic if and only if the number is
a cubic irrational?
Because of the similarity to simple continued fractions, we call attempts to
generalize this method to cubic irrationals and beyond multidimensional continued fractions, even though the sequences themselves do not actually correspond
to fractions in the traditional sense. The multidimensional aspect comes from the
fact that instead of subdividing the unit interval to find a number’s simple continued fractions expansion, we will now be subdividing triangles to find a pair
of numbers’ two-dimensional continued fraction expansion. While many multidimensional continued fraction algorithms have been known for years, we will
be focusing on a family of 216 continued fraction algorithms known as triangle
10
partition (TRIP) maps which encompass many of the existing multidimensional
continued fraction algorithms. The notation and theory laid out in sections 1.2.1
and 1.2.2 were developed Dasaratha et al. in [7].
1.2.1
The Triangle Map
Consider the triangle shown below with vertices (0, 0), (1, 0) and (1, 1).
(1, 1)
(0, 0)
(1, 0)
We can represent this triangle, which we will henceforth refer to as 4, by
expressing its vertices in projective coordinates using the following matrix:


1 1 1


V = 0 1 1 
0 0 1
where the values in the second row divided by the values in the first row correspond
to the x coordinate of the vertex and the values in the third row divided by the
values in the first row correspond to the y value of the vertex. The next step is to
come up with a partition of 4. To do this, we define the following matrices:
11




0 0 1
1 0 1




A0 =  1 0 0  A1 =  0 1 0 
0 1 1
0 0 1
Note thatgiven a triangle with vertices represented by the 3 × 3 matrix
v1 v2 v3 we get the following:
v1 v2 v3
v1 v2 v3
A0 =
A1 =
v2 v3 v1 + v3
v1 v2 v1 + v3
This means that for any triangle, A0 and A1 give a disjoint bipartition of that
triangle. Applying this to 4, we consider what happens if we apply A1 a nonnegative number of times, say k, followed by applying A0 once. Then the vertices
of the original triangle are mapped to the triangle
4k = {(x, y) ∈ 4 : 1 − x − ky ≥ 0 > 1 − x − (k + 1)y}
(1, 1)
40
41
42
43
44
(0, 0)
(1, 0)
Letting π : R3 → R2 be the projection map π(z, x, y) = ( xz , yz ), Define the map
T : 4 → 4 ∪ {(x, 0) : 0 ≤ x ≤ 1}
12
by
−k −1 T
T (α, β) = π( 1 α β
V A−1
)
0 A1 V
where (α, β) ∈ 4k .
This yields the following definition:
β 1 − α − kβ
,
T (α, β) =
α
α
if the pair (α, β) ∈ 4k .
We are now ready to define the triangle sequence of a pair of numbers. We say
that a pair of numbers (α, β) ∈ 4 has triangle sequence (a1 , a2 , . . . ) if for each ai
(α, β) is contained in the triangle represented by the matrix V Aa11 A0 Aa12 A0 · · · Aa1i A0 .
This is equivalent to saying T i (α, β) ∈ 4ai+1 .
It can also be helpful to consider how we get to a point (x, y) through repeated
applications of A0 and A1 , thinking of it as a sort of binary tree where at each
node we choose either the triangle determined by A0 or the triangle determined
by A1 as our next triangle. Given a point (x, y) ∈ 4 we define the triangle
tree sequence to be the sequence (i0 , i1 , . . . ) with ik ∈ {0, 1} such that (x, y) is
contained in the triangle determined by V Ai0 Ai1 · · · Ain for all n ≥ 0.
1.2.2
A Family Multidimensional Continued Fraction Algorithms: TRIP Maps
The Triangle Map is not only way to partition our triangle that gives us interesting
properties. Using the Triangle Map as a jumping off point, we can actually define
215 more maps that can be used to partition 4 in various ways. This comes from
the fact that we can chose to permute the matrices before and after applying our
matrices to the given triangle, leading to a different pattern of cutting.
Where exactly do these 215 maps come from though (and why do we get
216 maps total)? Take our triangle with ordered vertices v1 , v2 , and v3 . Before
applying either A0 or A1 , what if we permuted the vertices? Then rather than
connecting v2 and v1 + v3 to get our new triangles, we might get two new triangles
by connecting v1 and v2 + v3 or v3 and v1 + v2 . Then, once we apply A0 and A1 ,
we can permute the vertices of each of the two new triangles depending on which
13
matrix it corresponds to. This leads us to the following:
Definition 1.4. For a given (σ, τ0 , τ1 ) ∈ S33 , define
F0 = σA0 τ0 and F1 = σA1 τ1
by thinking of σ, τ0 , and τ1 as column permutation vertices.
We denote the TRIP map described by the matrices F0 and F1 as T(σ,τ0 ,τ1 ) .
In this notation, the triangle map is T(e,e,e) . Similar to the triangle map, we can
define T(σ,τ0 ,τ1 ) as map on 4 by the following:
T(σ,τ0 ,τ1 ) = π((1, x, y)(V F0−1 F1−k V )T ) when (x, y) ∈ 4k
where 4k denotes the image of 4 under F1k F0 . Recall that in defining the original
Question Mark Function we had a notion of both Farey and Bary partitioning.
The partition of 4 determined by F0 and F1 will be our Farey partition. Later,
we will define a notion of Bary partitioning for each TRIP map, but before that
we will examine existing generalizations of the Minkowski function.
1.3
1.3.1
Previous Attempts
Beaver and Garrity
The first attempt at generalizing the Minkowski Question Mark Function was
done in [5] and follows a similar line of reasoning to explaining the singularity of
?(x). Our initial triangle will still be the 4 we defined in the previous section. We
then define the Farey sum of three vectors in Q2 . Consider the points
!
!
!
p1 /r1
p2 /r2
p3 /r3
v1 =
v2 =
v3 =
q1 /r1
q2 /r2
q3 /r3
To each vi we can associate a v̄i ∈ Z3 :


!
r1
r2 //p2


v̄1 =  p1 
v2 =
q2
q1
14
v3 =
r3 //p3
q3
!
Then we define v̄ = v̄1 + v̄2 + v̄3 . Then


r1 + r2 + r3


v̄ =  p1 + p2 + p3 
q 1 + q2 + q 3
Then we define the Farey sum of v1 , v2 and v3 by
v̂ = v1 +̂v1 +̂v3 =
p1 +p2 +p3
r1 +r2 +r3
q1 +q2 +q3
r1 +r2 +r3
!
Applying this to the vertices of 4 we get the point (2/3, 1/3) and by connecting
this to each of the other vertices we end up with three subtriangles. Calling this
first subdivision of 4 P1 we can define Pn by taking the Farey sum of the vertices
of each of the 3n−1 subtriangles in Pn−1 and using this to create the 3n subtriangles
in Pn .
Now we define a sequence of barycentric partitions of P̃n according to a similar pattern. Given three points ( 3an1 , 3bn1 ), ( 3an2 , 3bn2 ), and ( 3an3 , 3bn3 ) we say that their
2 +a3 b1 +b2 +b3
, 3n+1 ). Then defining P̃ to be 4 we can inbarycenter is the point ( a1 +a
3n+1
ductively define P̃n from P̃n−1 by finding the barycenters of each of the 3n−1
subtriangles and using these to create the 3n subtriangles of P̃n . We present the
first few iterations of both partitions below:
41,F
41,B
42,F
42,B
15
Using the notation that 4n,F represents 4 after the nth Farey paritioning and
4n,B represents 4 after the nth Bary partitioning, note that
P0 = P̃0
P1 = P̃1
and
41,F = 41,B
Then we can define δ0 , δ1 : 4F → 4B as the identity maps on the vertices of the
subtriangles of P0 and P1 . Then for any n we define δn by sending the vertex of any
subtriangle of Pn to the corresponding vertex of P̃n and extending by continuity.
We will say that a point (x, y) ∈ 4 is an interior point if it is not on the vertex
or edge of any subtriangle in any 4n,F . Now we are ready to define our singular
function.
Definition 1.5. Define the Farey-Bary map δ : 4F → 4B by setting
δ(v) = lim δn (v)
n→∞
when v is an interior point and
δ(v) = δn (v)
when v is a vertex or edge of one of the triangles in 4n,F for some n.
Because we get that the subtriangles of 4n,B converge to points as n → ∞,
we find that δ is continuous at all interior points of 4F . Recall that the notion of
singularity for the Minkowski function is captured by the fact that
lim inf
k→∞
length of interval in range
=0
length of interval in the domain
almost everywhere. Thus we say that δ is singular if:
lim inf
k→∞
area of subtriangle in range
=0
area of subtriangle in domain
almost everywhere. By showing that this is the case, we have found our singular
function.
This method will be useful in determining singularity of many of the functions
we will explore later in this paper. However, for certain algorithms, it turns out
that we can follow a different approach that deals more with the fundamentals of
measure theory and how singular functions affect the measure of a set. We explore
one such attempt in the next section.
16
1.3.2
Panti
A second, more recent attempt at generalizing the Minkowski Function is presented in [4]. Recognizing that the Question Mark Function conjugates the Farey
map

 x
if 0 ≤ x < 1/2
F (x) = 1−x
 1−x if 1/2 ≤ x ≤ 1
x
with the tent map
T (x) =

2x
if 0 ≤ x < 1/2
2 − 2x if 1/2 ≤ x ≤ 1
and that this property characterizes ?(x), Panti turns to the n-dimensional generalizations of these maps, known as the Mönkemeyer map (M ) and the tent map
(T ) to find his generalization of ?(x). At this point the reader should note that
Dasratha et al. showed in [7] that the Mönkemeyer map in the 2-dimensional case
corresponds to the TRIP map Te,132,23 . Defining 4 as the n-dimensional simplex
that reduces to our familiar triangle in 2-dimensions, he presents the following
Theorem 1.6. There exists a unique homeomorphism Φ : 4 → 4 such that
T = Φ ◦ M ◦ Φ−1 .
Proving that this homeomorphism exists and is unique is the brunt of his
paper and comes largely from the fact that both M and T give partitions of 4
that converge to a point. Because of this, each point of {0, 1}N corresponds to a
unique point in 4 under an appropriate ordering. Then define
ϕ : {0, 1}N → 4F
ν : {0, 1}N → 4B
by sending the sequence (i1 , i2 , . . . ) to appropriate point in 4F and 4B respectively. It turns out that ϕ and µ are continuous, surjective, and have the same
fibers, so we can define an equivalence relation on {0, 1}N by a ≡ b if and
only if ϕ(a) = ϕ(b) if and only if ν(a) = ν(b). Then we can define bijections
ϕ̄, ν̄ : 0, 1N / ≡→ 4 as the natural quotient mappings of ϕ and ν respectively.
Then we define Φ as follows:
17
Definition 1.7. We define Φ : 4 → 4 as the homeomorphism Φ = ν̄ ◦ ϕ̄−1 .
Equivalently, Φ(p) = ν(a) for any a such that ϕ(a) = p.
With this definition of Φ we get the following commutative diagram, where S
denotes the shift map on {0, 1}N (i.e. S(a0 a1 a2 . . . ) = (a1 a2 a3 . . . )):
4
M
ϕ
Φ
{0, 1}N
ϕ
S
ν
4
4
{0, 1}N
Φ
ν
T
4
Now we are ready to explore the singularity of Φ with respect to the Lebesgue
measure λ. We normalize λ so that λ(4) = 1. We can define a probability measure
µ on 4 that is induced by
h(x1 , . . . , xn ) =
1
x1 (x1 − x2 + 1)(x1 − x3 + 1) · · · (x1 − xn + 1)
and properly normalizing µ we have that the Mönkemeyer map M preserves µ
and is ergodic with respect to it. It can also be shown that T is ergodic with
respect to λ and thus that M is ergodic with respect to the measure Φ−1
∗ λ where
−1
−1
Φ∗ λ(A) = λ(Φ(A)). It turns out that µ and Φ∗ λ are different but both ergodic
with respect to the same transformation M . This means that they are mutually
singular and thus that there is some measurable set A ⊆ 4 such that µ(A) = 1
and λ(Φ(A)) = 0. Because h ≥ 1 on 4 we have that µ ≥ Cλ for some C > 0.
It follows that µ and λ are absolutely continuous with respect to each other and
thus they have the same sets of full measure, meaning for the aforementioned A,
λ(A) = 1. Then Φ is singular with respect to λ.
18
Chapter 2
Defining Our Function
In this chapter, we will find a natural candidate for our singular function for each
TRIP map, which we will call Φ(σ,τ0 ,τ1 ) , related to the partition of 4 determined
by F0 and F1 . Furthermore, we will show that we do not have to check all 216
functions by reducing the number of cases we need to check to 15. Then, in the
next chapter, we will show which of these functions is singular.
Recall that in each of the examples of singular functions explored in the introduction, Minkowski’s, Beaver and Garrity’s, and Panti’s, the function arose as
a consequence of two different partitions: a Farey one and a Bary one. We will
be using the TRIP maps as our Farey partitions, so in order to find a singular
function, we are going to need a corresponding Bary map for each TRIP map.
Starting with A0 and A1 as defined above, we define a matrix B0 by replacing the
1’s in the third column of A0 with 1/2 and define B1 similarly:




0 0 1/2
1 0 1/2




B0 =  1 0 0 
B1 =  0 1 0 
0 1 1/2
0 0 1/2
Then for a general TRIP Map given by (σ, τ0 , τ1 ) we simply define G0 = σB0 τ0
and G1 = σB1 τ1 to get the corresponding barycentric partition. For notation,
we say Γ(a1 , . . . , an ) corresponds to the triangle whose vertices are given by
V Ga11 G0 . . . Ga1n G0 . Now we are ready to define our function Φ(σ,τ0 ,τ1 ) .
Consider the TRIP map Tσ,τ0 ,τ1 . Let (i1 , . . . , in ) ∈ {0, 1}n . Define 4(i1 , . . . , in )
to be the triangle corresponding to V Fi1 . . . Fin and Γ(i1 , . . . , in ) similarly. Then
19
define
Φ(σ,τ0 ,τ1 )n : 4 → 4
by Φ(σ,τ0 ,τ1 )n (4(i1 , . . . , in )) = Γ(i1 , . . . , in ) for every (i1 , . . . , in ) ∈ {0, 1}n by mapping vertices to vertices and extending by linearity. Then
Φ(σ,τ0 ,τ1 ) = lim Φ(σ,τ0 ,τ1 )n
n→∞
is the function we’re interested in.
To see how this works, we show the first few iterations of the Farey and Bary
partitions determined by (e, e, e):
41,F
41,B
42,F
42,B
Because we only care about the partitioning of 4 and not the actual TRIP maps
themselves, we want to understand when two TRIP maps give the same partition
20
of 4. As it turns out, no map uniquely partitions 4. Instead, every TRIP map
has a "twin" that is just the same map with matrices switched. This will reduce
the number of TRIP maps we need to study by half.
Lemma 2.1. (σ, τ0 , τ1 ) and ((13)σ, τ1 (12), τ0 (12)) give the same partition of 4.
Proof. Recall our matrices A0 and A1 :


0 0 1


A0 =  1 0 0 
0 1 1
Then we calculate (13)A0 (12) and


0 0 1
0


(13)A0 (12) =  0 1 0   1
1 0 0
0


0 0 1
1


(13)A1 (12) =  0 1 0   0
0
1 0 0


1 0 1


A1 =  0 1 0 
0 0 1
(13)A1 (12):

0 1
0

0 0  1
1 1
0

0 1
0

1 0  1
0 1
0
 
1 0
 
0 0 =
0 1
 
1 0
 
0 0 =

1 0 1

0 1 0  = A1
0 0 1

0 0 1

1 0 0  = A0
0 1
0 1 1
Then (e, e, e) and ((13), (12), (12)) give the same partition of 4, so (σ, τ0 , τ1 ) will
give the same partition as ((13)σ, τ1 (12), τ0 (12)).
With this pairing, this reduces the number of cases we need to check to 108.
In fact, as it turns out, there are only 15 main cases to check. To see this, we first
must first develop our notion of singularity. Recall from Beaver and Garrity in [5]
that the singularity of ?(x) was linked to the ratio of the lengths of intervals in
the range divided by the lengths of intervals in the domain went to zero almost
everywhere. Generalizing this two dimensions, they used the following as their
notion for singularity:
Definition 2.2. Given (σ, τ0 , τ1 ) ∈ S33 , we say that Φ(σ,τ0 ,τ1 ) : 4 → 4 is singular
if it is continuous and
|Γ(a1 , . . . , an )|
=0
lim
n→∞ |4(a1 , . . . , an )|
almost everywhere.
21
This is different from the traditional definition of a singular function in higher
dimensions, which we present below:
Definition 2.3. A function f : Rn → Rn is singular with respect to Lebesgue
measure if there exists a set A in the domain of f such that λ(A) = 1 while
λ(f (A)) = 0.
Though these definitions appear different, we have reason to believe that they
are essentially equivalent. While we have not proven this, we will later discuss why
we believe this to be true.
Under both these notions of singularity, we can further reduce the number of
cases by grouping permutations into classes whose partitionings of 4 are related
by a linear transformation.
Theorem 2.4. If the permutation (σ, τ0 , τ1 ) gives rise to a singular function from
0
4 → 4 under either definition 2.2 or 2.3, then for
any triangle
4 with vertices
given in projective coordinates by the matrix V 0 = v1 v2 v3 this permutation
gives rise to a singular function from 40 → 40 under the same definition.
Proof. First, we define the linear transformation M : R3 → R3 with the following
matrix:
M = v1 v2 − v1 v3 − v2 .
Note that M V = V 0 , so M sends 4 as projected on the plane z = 1 to 40 as projected on the plane determined by the endpoints of v1 , v2 and v3 . Now we can think
of 4 as the base of the tetrahedron T formed by the vectors (1, 0, 0), (1, 1, 0) and
(1, 1, 1) in R3 . We know that the volume of a tetrahedron is 31 (area(base))(height),
so letting P be the plane containing 4, we have that:
area(4) = 3
volume(T )
.
d((0, 0, 0), P)
Letting P 0 be the plane containing 40 and T 0 be the tetrahedron formed by v1 , v2
and v3 we know that
λ(T 0 ) = | det(M )|λ(T )
which means
area(40 ) = 3
| det(M )|volume(T )
.
d((0, 0, 0), P 0 )
22
Let T (a1 , . . . , an ) be the tetrahedra with 4(a1 , . . . , an ) as its base that extends
from the origin and B(a1 , . . . , an ) be the tetrahedra with Γ(a1 , . . . , an ) as its base
that extends from the origin. Then because the height of these tetrahedra will be
the same height as T , so we get
volume(B(a1 , . . . , an ))
area(Γ(a1 , . . . , an ))
=
area(4(a1 , . . . , an ))
volume(T (a1 , . . . , an ))
Then
area(Γ0 (a1 , . . . , an ))
area(40 (a1 , . . . , an ))
equals
3| det(M )|volume(B(a1 , . . . , an ))
d((0, 0, 0), P 0 )
,
d((0, 0, 0), P 0 )
3| det(M )|volume(T (a1 , . . . , an ))
which is
volume(B(a1 , . . . , an ))
.
volume(T (a1 , . . . , an ))
Then, for a given triangle sequence (a1 , a2 , . . . ), if
area(Γ(a1 , . . . , an ))
=0
n→∞ area(4(a1 , . . . , an ))
lim
then
area(Γ0 (a1 , . . . , an ))
=0
lim
n→∞ area(40 (a1 , . . . , an ))
We know that this limit is 0 almost everywhere on 4, so we would like that
to be true on 40 . To show this, we will examine how M interacts with sets of
measure 0 in 4. Let A ⊂ 4 be measure 0. Then suppose M (A) was not measure
0. Then there would be some open ball contained in M (A). Take the cone C with
tip at the origin and base this ball. Then λ(C) = δ for some nonzero δ. Then
δ
λ(M −1 (C)) = det(M −1 )λ(C) = det(M
6= 0. This implies that area(M −1 (C) ∩ P) 6=
)
0. But M −1 (C) ∩ P ⊂ A which has a measure 0. Then M (A) must have measure
0. This means that M sends sets of measure 0 to sets of measure 0 and sets of full
measure on 4 to sets of full measure on 40 . Then if
area(Γ(a1 , . . . , an ))
=0
n→∞ area(4(a1 , . . . , an ))
lim
23
almost everywhere,
area(Γ0 (a1 , . . . , an ))
=0
n→∞ area(40 (a1 , . . . , an ))
lim
almost everywhere, preserving Definition 2.2. Furthermore, if because M send sets
of measure 0 to sets of measure 0 and sets of full measure to sets of full measure,
a function Φ will be singular if and only if M ◦ Φ ◦ M −1 is singular, preserving
Definition 2.3.
This gives us the following nice result.
Corollary 2.5. If the Φ associated with the permutation (σ, τ0 , τ1 ) is singular,
then so is the Φ associated with (σρ, ρ−1 τ0 , ρ−1 τ1 ).
Proof. Note that these two partitions only different in the ordering of the initial
vertices of 4. Then this result follows directly from Theorem 2.4.
By this corollary, we can greatly reduce the number of cases that we need to
check. We will informally define a class of permutations to be those related by
some combination of Lemma 2.1 and Corollary 2.5. By Corollary 2.5, we know
that there will always be a permutation of the form (e, τ0 , τ1 ) in each class, which
means there are at most 36 classes to check. By direct calculation, we find 21
classes, represented by the following permutations:
(e, e, e)
(e, e, 12)
(e, 12, 12)
(e, 12, 13)∗ (e, 12, 23)
(e, 13, 132) (e, 23, e)
(e, e, 13)
(e, e, 23)
(e, e, 123)∗ (e, e, 132)
(e, 12, 132) (e, 13, e)
(e, 13, 12)∗
(e, 12, e)
(e, 13, 23)
(e, 23, 23)∗ (e, 23, 132) (e, 123, e)∗ (e, 123, 132) (e, 132, 132)∗
The starred permutations represent classes with 6 maps, while the unstarred ones
represent classes with 12 maps. Our final reduction in number of cases from 21 to
15 comes from the following lemmas.
We first standardize notation. Given a triangle tree sequence (i1 , i2 , . . . , in ) we
say that
V Fi1 Fi2 · · · Fin = v1,n v2,n v3,n
Then we are ready to continue. We present the following lemma from [7]
24
Lemma 2.6. The TRIP maps given by (e, 23, 23), (e, 23, 132), (e, 132, 23), (e, 132, 132),
(13, 132, 132), (13, 23, 132), (13, 132, 23), and (13, 23, 23) all give the same partition of 4.
Proof. The permutation

0

F0 =  1
0
(e, 23, 23) gives matrices:



1 0
1 1 0



F1 =  0 0 1 
0 0 
1 1
0 1 0
The permutation (e, 23, 132) gives matrices:


0 1 0


F0 =  1 0 0 
0 1 1
The permutation (e, 132, 23) gives matrices:


0 1 0


F0 =  0 0 1 
1 1 0
The permutation (e, 132, 132) gives matrices:


0 1 0


F0 =  0 0 1 
1 1 0


0 1 1


F1 =  1 0 0 
0 1 0


1 1 0


F1 =  0 0 1 
0 1 0


0 1 1


F1 =  1 0 0 
0 1 0
For the base case, we note that the new vertex of 4(i1 ) is on the edge between
(1, 0, 0) and (1, 1, 0) for each map. For the inductive step vk−1,1 is the unique
vertex of 4(i1 , . . . , ik−1 ) which is not a vertex of 4(i1 , i2 , . . . , ik−2 ). Because the
new vertex of 4(i1 , i2 , . . . , ik ) is vk−1,2 + vk−1,3 for either choice of ik , and this is
the sum of two vertices that were not new at the previous step, each map gives
the same partition. Thus each of these four maps gives the same partition of 4.
The remaining four maps come from applying Lemma 2.1 to these four maps.
This means that proving singularity for any of the classes represented by
(e, 23, 23), (e, 23, 132) or (e, 132, 132) proves singularity for the remaining two
classes. From [7] we know that the Mönkemeyer map is the same as Te,132,23 in two
dimensions and it is the map used by Panti in [4] to derive his generalization of the
25
Minkowski function, so we will call the family of maps represented by (e, 23, 132)
"Mönkemeyer-type maps."
This next lemma deal with maps that partition 4 into subtriangles that all
contain the vertex (1, 0) of 4, so in essence what they do is provide a Farey
partitioning of the hypotenuse of 4. We will use the notation that 4n,F refers to
the set of all 4(i1 , i2 , . . . , in ) for a given TRIP map.
Lemma 2.7. The TRIP maps given by (e, 12, e), (e, 123, e), (e, 12, 13), (e, 123, 13),
(13, 12, e), (13, 12, 13), (13, 123, e), and (13, 123, 13) all give the same partition of
4.
Proof. The permutation

0

F0 =  0
1
(e, 12, e) gives matrices:



1 0 1
0 1



F1 =  0 1 0 
1 0 
0 0 1
0 1
The permutation (e, 123, e)

1 0

F0 =  0 1
1 0
gives matrices:

0

0 
1
The permutation (e, 12, 13)

0 0

F0 =  0 1
1 0
gives matrices:

1

0 
1
The permutation (e, 123, 13) gives matrices:


1 0 0


F0 =  0 1 0 
1 0 1


1 0 1


F1 =  0 1 0 
0 0 1


1 0 1


F1 =  0 1 0 
1 0 0


1 0 1


F1 =  0 1 0 
1 0 0
Note that all these maps fix the vertex (1, 1, 0) of 4. Then 41,F is the same for
all these maps. Now suppose that 4n,F is the same for all these maps. Then for
a given T ∈ 4n,F we have that {T (0), T (1)}, where T (i) is obtained to applying
Fi to T , is the same for all these maps so we get that 4n+1,F is the same for all
these maps. The remaining four maps come from applying Lemma 2.1 to these
26
four maps.
This means that proving singularity for any of the classes represented by
(e, 12, e), (e, 123, e) or (e, 12, 13) proves singularity for the remaining two classes.
We will call the family represented by (e, 12, e) "degenerate Farey maps" because
the maps in this family simply give a Farey partioning of one of the sides of 4.
Lemma 2.8. The TRIP maps given by (e, e, 12), (e, e, 123), (e, 13, 12), (e, 13, 123),
(13, e, 12), (13, 13, 12), (13, e, 123) and (13, 13, 123) give the same partition of 4.
Proof. The permutation

0

F0 =  1
0
(e, e, 12) gives the following matrices:



0 1
0 1 1



F1 =  1 0 0 
0 0 
1 1
0 0 1
The permutation (e, e, 123)

0 0

F0 =  1 0
0 1
gives the following matrices:



1 1 0
1



F1 =  0 0 1 
0 
1 0 0
1
The permutation (e, 13, 12)

1 0

F0 =  0 0
1 1
gives the following matrices:



0
0 1 1



F1 =  1 0 0 
1 
0
0 0 1
The permutation (e, 13, 123) gives the following matrices:




1 0 0
1 1 0




F0 =  0 0 1 
F1 =  0 0 1 
1 1 0
1 0 0
Notice that all three permutations give the same initial partition of 4 and that
v2 (1) is the same in both triangles. Now suppose that 4n,F is the same and that
for corresponding triangles v2 (n) is the same. Then because the new vertex is
v1 (n)+̂v3 (n) and the only difference between partitions is which vertex is v1 (n)
and which vertex is v3 (n), we get that 4n+1,F is the same for these four maps.
The remaining four maps come from applying 2.1 to these four maps.
27
Then we have the proving singularity for any one of the classes (e, e, 12),
(e, e, 123) or (e, 13, 12) proves singularity for the other two classes. Then with
these duplicities, we’ve reduced the number of classes from 21 to the following 15:
(e, e, e)
(e, e, 12)
(e, e, 13)
(e, e, 23)
(e, e, 132)
(e, 12, e)
(e, 12, 12)
(e, 12, 23)
(e, 12, 132)
(e, 13, e)
(e, 13, 23)
(e, 13, 132)
(e, 23, e)
(e, 23, 132)
(e, 123, 132)
28
Chapter 3
Results
In this chapter we explicitly show when Φ(σ,τ0 ,τ1 ) is singular. We are able to do
this for 7 of the 15 classes. Of these 7 classes, one is the Mönkemeyer-type maps.
The proof of the singularity of the associated Φ comes directly from Panti’s work
in [4] and uses special properties of these TRIP maps that are not true for TRIP
maps in general. 4 of the remaining classes are shown through direct computation
and a notion of what the "usual" triangle tree sequence of a point looks like. One
of these classes is the degenerate Farey maps, which are interesting because they
allow us to easily write an explicit function for Φ in terms of the original question
mark function. The final two classes involve the ergodic properties of the TRIP
maps in these classes.
3.1
A Special Case: Mönkemeyer-Type Maps
As we mentioned in the introduction, Panti’s generalization of the Question Mark
Function comes from his recognizing that ?(x) is completely characterized by the
fact that it is the unique homomorphism that conjugates the Farey map with the
Tent map. Thus, given that the n-dimensional Mönkemeyer map is the generalization of the Farey map, it’s only natural that his function is the unique homorphism
that conjugates the n-dimensional Mönkemeyer map with the n-dimensional Tent
map.
In [4], Panti explicitly shows that his function is singular under the traditional
measure theoretic definition of singularity. From [7] we know that the Mönkemeyer
29
map in 2-dimensions corresponds to the permutation triple (e, 23, 132). Then by
a combination of Corollary 2.5 and Lemma 2.6 we have that the following permutations give rise to a singular function:
(e, 23, 23)
(e, 23, 132)
(e, 132, 23)
(e, 132, 132)
(13, 23, 23)
(13, 23, 132)
(13, 132, 23)
(13, 132, 132)
(12, 13, 13)
(12, 13, 123)
(12, 123, 13)
(12, 123, 123)
(123, 13, 13)
(123, 13, 123)
(123, 123, 13)
(123, 123, 123)
(23, e, e)
(23, e, 12)
(23, 12, e)
(23, 12, 12)
(132, e, e)
(132, e, 12)
(132, 12, e)
(132, 12, 12)
What’s nice about the Mönkemyer-type maps is that every triangle sequence
corresponds to a unique point. In fact, they seem to be the only maps that do so,
which leads us to the following conjecture:
Conjecture 3.1. The TRIP maps given by the partitions directly above are the
only TRIP maps for which every triangle sequence corresponds to a single point.
This allows us to construct a bijection between the points of 4 and points
of {0, 1}N modulo an equivalence relation. This is what allows Panti to set up
the various conjugations that he uses to both define Φ and prove its singularity.
Unfortunately, this is not the case for all TRIP maps and so we are forced to
resort to a different approach.
3.2
Remaining Maps
In this section, we prove singularity for the associated Φ for 4 other classes of TRIP
maps. We will show that under both definitions of singularity the associated Φ
are singular. We begin by presenting a clean framework for showing when Φ(σ,τ0 ,τ1 )
is singular under the traditional definition by way of the triangle tree sequence.
Then, we will go through a more computational proof for the limit definition, as
it is interesting in light of work by Beaver and Garrity.
30
3.2.1
Triangle Tree Sequence Approach
Throughout both both this and the next section we will need the following lemma,
whose proof is a geometric fact:
Lemma 3.2. The area of a triangle T whose vertices are given in projective
coordinates by the 3 × 3 matrix M with projective coordinates x, y and z being the
entries in the top row of M is given by
|T | =
1 | det(M )|
2 xyz
We also need the following fact about our Bary partition:
Lemma 3.3. The area of Γ(i1 , . . . , in ) is half the area of Γ(i1 , . . . , in−1 ).
Proof. Consider the following matrix:


1 0 0


B= 0 1 0 
0 0 1/2
Then B0 = A0 B and B1 = A1 B so we have that G0 = σA0 Bτ0 and G1 = σA1 Bτ1 .
Then
1
det(Gi ) = det(σAi Bτi ) = ±1 det(B) = ±
2
Now because V has all ones in its top row, we know that V G0 and V G1 will
have all 1’s in their top row as well. Then by Lemma 3.2 the both Γ(1) and Γ(0)
have area 12 . Now consider T ∈ 4n,B represented by the matrix V Gi1 · · · Gin and
suppose this matrix has all 1’s in the top row. Then by Lemma 3.2 we know
1
area(T ) = | det(V Gi1 · · · Gin )|.
2
At the next stage of our partition, we get that V Gi1 · · · Gin G0 and V Gi1 · · · Gin G1
both have 1’s as entries in their top row. Furthermore, we know
1
| det(V Gi1 · · · Gin G0 )| = | det(V Gi1 · · · Gin )|| det(G0 )| = | det(V Gi1 · · · Gin )|
2
1
| det(V Gi1 · · · Gin G1 )| = | det(V Gi1 · · · Gin )|| det(G1 )| = | det(V Gi1 · · · Gin )|
2
Then our two new triangles are both half the area of T .
31
We are now ready to understand what a typical triangle tree sequence looks
like with respect to the Bary partition.
Proposition 3.4. Let (σ, τ0 , τ1 ) ∈ S33 . Consider a point (x, y) ∈ 4 with triangle tree sequence (i1 , i2 , . . . ) with respect to the barycentric partition, ie (x, y) ∈
Γ(i1 , . . . , in ) for all n. Then for almost all (x, y) ∈ 4 we have
1
#{ik = 1, 1 ≤ k ≤ n}
=
n→∞
n
2
That is to say, the set of points with normal triangle tree sequences with respect
to the barycentric partitioning has measure one under Lebesgue measure.
lim
Proof. By Lemma 3.3, Γ(i1 , . . . , in ) has half the area of Γ(i1 , . . . , in−1 ). Then given
a point (x, y) ∈ 4(i1 , . . . , in−1 ), the probability that in = 1 is 0.5. Then by the
central limit theorem, we have that
#{ik = 1, 1 ≤ k ≤ n}
1
=
n→∞
n
2
lim
almost everywhere in 4.
Then in order to show that a given Φ(σ,τ0 ,τ1 ) is singular, we simply need to
show that the set of points that have normal triangle tree sequences with respect
to the Farey partitioning has measure 0.
We want a way of getting at the proportion of 1’s and 0’s in the triangle tree
sequence in a random (x, y) ∈ 4F . It turns out that we can do this using the triangle sequence. Consider a point (x, y) ∈ 4F with triangle sequence (a1 , a2 , a3 , . . . ).
We will define
s n = a1 + · · · + an .
There is an easy way of translating back and forth between the triangle sequence
and the triangle tree sequence by replacing all the strings of ones with the number
representing their sum (length). Replace every string of zeros of length n with a
string of zeros of length n − 1. This will give us the triangle sequence.
Then we have that
1
#{ik = 1, 1 ≤ k ≤ n}
lim
=
n→∞
n
2
32
is equivalent to
lim
n→∞
sn
1
= .
sn + n
2
Inverting, this is the same as
lim
n→∞
sn + n
n
= lim 1 +
=2
n→∞
sn
sn
which is true if and only if
sn
= 1.
n
Then showing Φ(σ,τ0 ,τ1 ) is singular under the traditional definition is equivalent to
showing that almost everywhere in 4,
sn
6= 1.
lim
n→∞ n
Later, we will show four classes for which this is true. However, we will first
turn our attention to exploring an approach at finding which classes for which the
limit definition of singularity holds. As it turns out, for each definition the same
six classes are singular.
lim
n→∞
3.2.2
Limit Definition Approach
While the above argument reduces the singularity of Φ(σ,τ0 ,τ1 ) to the number theoretic properties of T(σ,τ0 ,τ1 ) , we are still interested in the limit notion of singularity
used in [5]. Recall that before showing ?0 (x) = 0 almost everywhere was akin
to showing limn→∞ |I|IBF || = 0 almost everywhere. Thus, it is natural to use the
following notion to capture the singularity of Φ(σ,τ0 ,τ1 ) :
Definition 3.5. We say that Φ : 4 → 4 is singular if it is continuous and
lim
n→∞
|Γ(a1 , . . . , an )|
=0
|4(a1 , . . . , an )|
almost everywhere.
With this we are ready to find the area of Γ(a1 , . . . , an ). First, we fix the
notation that sn = a1 + · · · + an for a triangle sequence (a1 , . . . , an ). Then we have
the following:
33
Lemma 3.6. For the TRIP map Tσ,τ0 ,τ1
area(Γ(a1 , . . . , an )) =
1 1
2 2sn +n
Proof. This follows from Lemma 3.3 and noting that V Ga11 G0 · · · Ga1n G0 contains
sn G1 and n G0 .
Finding the area of a generic 4(a1 , . . . , an ) turns out to be impractical, so
we instead find a lower bound on the area that holds in general. First we set
notation. We will say that for a given TRIP map Tσ,τ0 ,τ1 and a given triangle
sequence (a1 , . . . , an ) that
V F1a1 F0 . . . F1an F0 = Xn Yn Zn
and we will say the entries of the top row of this matrix are xn , yn and zn respectively. Then we know, since det(V ) = det(F1 ) = det(F0 ) = ±1, that the area
of 4(a1 , . . . , an ) is simply xn y1n zn . Thus we find need to find an upper bound on
xn yn zn .
To do this, we will use the following fact from [6]
Jac(Tσ,τ0 ,τ1 (x, y)) = ((1, x, y)Mσ,τ0 ,τ1 (1, 0, 0)T )3
where
Mσ,τ0 ,τ1 = (V F1k F0 V −1 )T .
We want to explore how we can connect the Jacobian of a TRIP map with the
area of 4(a1 , . . . , an ).
First, we consider the following matrix:
T
a b c (V −1 )T (V F1a1 F0 F1a2 . . . F1an F0 )T 1 0 0
Calculating (a, b, c)(V −1 )T we get

1

a b c  −1
0
the following:

0 0

1 0 = a−b b−c c
−1 1
Then we can rewrite the above expression as
T T
a−b b−c c
Xn Yn Zn
1 0 0
34
Which we can rewrite as
T
= (a − b)xn + (b − c)yn + czn
a−b b−c c
xn yn zn
Choosing the following three inputs for (a, b, c) we are able to isolate xn , yn and
zn :
(1, 0, 0) → xn
(1, 1, 0) → yn
(1, 1, 1) → zn
Now this is interesting, but how does this help us? Recall the following expression:
Jac(Tσ,τ0 ,τ1 (x, y)) = ((1, x, y)Mσ,τ0 ,τ1 (1, 0, 0)T )3
Rewriting this, including the explicit expression for Mσ,τ0 ,τ1 , we get the following:
(1, x, y)(V F1k F0 V −1 )T (1, 0, 0)T = (1, x, y)(V −1 )T (V F1k F0 )T (1, 0, 0)T
This is very similar to what we had above, and in fact plugging in the points
(1, 0, 0), (1, 1, 0) and (1, 1, 1) will get us x1 , y1 and z1 respectively. This means that
we can calculate the area of any 4k for a given TRIP map using its Jacobian.
The Jacobians of 108 TRIP maps are calculated in [6]. The maps for which he
calculates this are called polynomial maps and have the following property:
Theorem 3.7. (Theorem 7 of [6]) A triangle partition map Tσ,τ0 ,τ1 has polynomial
1
behavior if and only if, aside from factors of (−1)k , Jac(Tσ,τ0 ,τ1 (a, b)) 3 is linear in
k.
This is equivalent to saying
Jac(Tσ,τ0 ,τ1 (x, y)) = f (x, y, k)3
where f is linear in x, y and k aside from factors of (−1)k . Then we have that
f (x, y, k) = (1, x, y)Mσ,τ0 ,τ1 (1, 0, 0)T
35
Then from our work above, we have that for 4k
f (0, 0, k) = x1
f (1, 0, k) = y1
f (1, 1, k) = z1
To proceed, beyond x1 y1 z1 we need to find a way to incorporate projective coordinates into our equation. To do this, write x̄ = xz and ȳ = yz for x, y, z ∈ Z. Then
because f (a, b, k) is linear in a, b and k we can rewrite f (x̄, ȳ, k) as a function
g(x, y, z) where g is just f with all k and constant terms multiplied by z, aside
from (−1)k . Then
g(x, y, z) = (z, x, y)Mσ,τ0 ,τ1 (1, 0, 0)T
and we can now plug in a points projective coordinates which come directly from
our matrix notation.
Lemma 3.8. Given a polynomial behaving TRIP map Tσ,τ0 ,τ1 and triangle sequence (a1 , . . . , an ),
xn yn zn ≤
n
Y
(cai + d)3
i=1
for some positive constants c and d.
Proof. We will proceed by induction. By direct observation of [6] we have that
Jac(Tσ,τ0 ,τ1 )1/3 ≤ (p0 + p1 k)x + (q0 + q1 k)y + (r0 + r1 k)z
for some positive constants pi , qi and ri . Then, given 4a1 we have
x1 y1 z1 ≤ (r0 + r1 a1 )(p0 + r0 + (p1 + r1 )a1 )(p0 + q0 + r0 + (p1 + q1 + r1 )a1 )
≤ (p0 + q0 + r0 + (p1 + q1 + r1 )a1 )3
Now suppose for all k < n we have
k
Y
area(4(a1 , . . . , ak )) ≤
((p1 + q1 + r1 )ai + p0 + q0 + r0 )3
i=1
Consider the following triangle
V F1a2 F0 · · · F1an F0 =
36
Un Vn Wn
We will say u1n , v1n and w1n represent the projective coordinate of each matrix.
Then we know
(1, 0, 0)(V −1 )T (V F1a2 F0 · · · F1an F0 )T = UnT
(1, 1, 0)(V −1 )T (V F1a2 F0 · · · F1an F0 )T = VnT
(1, 1, 1)(V −1 )T (V F1a2 F0 · · · F1an F0 )T = WnT
Then
UnT (V −1 )T (V F1a1 F0 )(1, 0, 0)T = UnT Mσ,τ0 ,τ1 (1, 0, 0)T = xn
VnT (V −1 )T (V F1a1 F0 )(1, 0, 0)T = VnT Mσ,τ0 ,τ1 (1, 0, 0)T = yn
WnT (V −1 )T (V F1a1 F0 )(1, 0, 0)T = WnT Mσ,τ0 ,τ1 (1, 0, 0)T = zn
By our formula for the Jacobian we get
(p0 + p1 k)u2n + (q0 + q1 k)u3n + (r0 + r1 k)u1n = xn
(p0 + p1 k)v2n + (q0 + q1 k)v3n + (r0 + r1 k)v1n = yn
(p0 + p1 k)w2n + (q0 + q1 k)w3n + (r0 + r1 k)w1n = zn
Because in our triangle y ≤ x ≤ 1 we have that u1n ≥ u2n ≥ u3n , with the same
holding for the v’s and w’s. Then, letting c = p1 + q1 + r1 and d = p0 + q0 + r0 we
get
xn ≤ (ca1 + d)u1n
yn ≤ (ca1 + d)v1n
zn ≤ (ca1 + d)w1n
Invoking our inductive hypothesis, we get the following:
xn yn zn ≤ (ca1 + d)3 u1n v1n w1n
n
Y
3
(cai + d)3
≤ (ca1 + d)
i=2
=
n
Y
(cai + d)3
i=1
37
This gives us the following:
area(Γ(a1 , . . . , an ))
xn yn zn
= sn +n
area(4(a1 , . . . , an ))
2
Qn
(cai + d)3
≤ i=1 sn +n
2
(c snn + d)3n
≤
sn +n
n
2sn
(c n + d)3
=
sn
2 n +1
Then because the numerator of this fraction is polynomial in snn while the
denominator is exponential, we have that the limit as n → ∞ will be 0 as long as
limn→∞ snn = ∞. Then we have reduced the problem of showing singularity under
the limit definition to showing snn → ∞ almost everywhere, which also implies the
traditional definition of singularity.
3.2.3
Finding limn→∞ snn
In this section we show directly that four the four classes of maps represented
by (e, e, e), (e, e, 12), (e, 12, 12) and (e, 12, 12) that limn→∞ snn = ∞. We will also
show that ergodicity in the case of five other classes gives us that this limit is also
infinity, and cite a result from Jensen and Amburg in [8] which will give us the
classes (e, 23, e) and (e, 13, 23).
Given (σ, τ0 , τ1 ) ∈ S33 , consider the following set:
sn
M := {(x, y) ∈ 4 : lim
< ∞}
n→∞ n
Showing limn→∞ sn /n = ∞ almost everywhere is equivalent to showing λ(M ) = 0
where λ is the Lebesgue measure. Defining
sn
MN := {(x, y) ∈ 4 : lim
< N}
n→∞ n
we note that
∞
[
M=
MN .
N =1
Then if we can show that λ(MN ) = 0 for each N , because M is a countable union
38
of sets of measure 0, we will have that λ(M ) = 0. Calculating the area of MN is
difficult though so we define the following set
M̃N := {(x, y) ∈ 4 : ∀n, an < nN }
Then MN ⊂ M̃N . To get a bound on λ(M̃N ) we recursively define the family of
sets M̃N (k) by
M̃N (1) := {(x, y) ∈ 4 : a1 < N }
M̃N (k) := {(x, y) ∈ M̃N (k − 1) : ak < kN }
Then
M̃N =
∞
\
M̃N (k)
n=1
λ(M̃N (k − 1)) for some constant c(k)
Our goal is to show that λ(M̃N (k)) ≤ c(k)−1
c(k)
that is linear in kN . If this is the case, then λ(M̃N ) = 0.
Lemma 3.9. Suppose λ(M̃N (k)) ≤
stants c and d. Then λ(M̃N ) = 0.
akN +c−1
λ(M̃N (k
akN +c
− 1)) for some positive con-
Proof. Assuming the hypothesis and using the fact that M̃N =
have that
∞
Y
akN + c − 1
λ(M̃N ) ≤
akN + c
k=2
S∞
k=1
M̃N (k) we
Showing this product is 0 is equivalent to showing that its reciprocal
∞ ∞
Y
Y
akN + c
1
=
1+
=∞
akN
+
c
−
1
akN
+
c
−
1
k=1
k=2
Taking logarithms, this is the same as showing that the series
∞
X
1
log 1 +
=∞
(aN )k + c − 1
k=2
which follows by the integral test. Then we are done.
We will set ckN + c = c(k) and show how to calculate c(k) for a given TRIP
map, noting that this method does not hold for all maps. Consider M̃N (k − 1).
We know this will be made up of the subtriangles of the form V F1a1 F0 · · · F1ak −1 F0
39
where each ai < iN . Consider one of these subtriangles and denote it T . We define
Tk := {(x, y ∈ T : ak ≥ kN }
Then
[
M̃N (k) =
(T − Tk ).
T ∈M̃N (k−1)
Now given T , let x, y and z denote the projective coordinates of its vertices. It
turns out that we can find formulas for F1n for any triangle map with polynomial
behavior, and using these formulas, we can get an expression for the area of Tk in
terms of x, y, z and k. Ideally, we get some that that looks like the following:
1 1
c(k) xyz
|Tk | ≥
where c(k) is some constant with respect to k. Then we get that
|T − Tk | ≤
c(k) − 1
|T |
c(k)
which implies that
λ(M̃N (k)) ≤
c(k) − 1
λ(M̃N (k − 1)).
c(k)
Then, if c(k) is linear in k, we have shown that this TRIP map gives rise to a
singular function. We will now go through the process of computing c(k) for the
four classes mentioned above. The reader should note that these calculations all
follow the same structure, but we present them all for sake of completeness.
We begin with (e, e, e). We have


1 0 k


F1k =  0 1 0 
0 0 1


0 k k+1


F1k F0 =  1 0
0 
0 1
1
40
This gives us the following recurrence relations:
xn = yn−1
yn = an xn−1 + zn−1
zn = (an + 1)xn−1 + zn−1
Then zn ≥ yn ≥ xn for all n. Then
area(Tk ) =
For (e, e, 12) we have

1 0

2k
F1 =  0 1
0 0

0 k

2k
F1 F0 =  1 k
0 1
1
1
1
≥
xy(kN x + z)
kN + 1 xyz

k

k 
1


0 1 k+1


F12k+1 =  1 0
k 
0 0
1


1 k+1 k+1


F12k+1 F0 =  0
k
k+1 
0
1
1

k+1

k 
1
This gives us the following recurrence relations:
an = 2kn
an = 2kn + 1
xn = yn−1
xn = xn−1
yn = kn (xn−1 + yn−1 ) + zn−1
yn = kn (xn−1 + yn−1 ) + xn−1 + zn−1
zn = kn (xn−1 + yn−1 ) + xn−1 + zn−1
zn = (kn + 1)(xn−1 + yn−1 ) + zn−1
Then zn ≥ yn ≥ xn for all n. Then
area(Tk ) ≥
1
1
1
≥
xy((kN + 1)/2(x + y) + z)
kN + 2 xyz
For (e, 12, e) we have


1 0 k


F1k =  0 1 0 
0 0 1
41


k 0 k+1


F1k F0 =  0 1
0 
1 0
1
This gives us the following recurrence relations:
xn = an xn−1 + zn−1
yn = yn−1
zn = (an + 1)xn−1 + zn−1
Then zn ≥ xn ≥ yn for all n. Then
area(Tk ) =
1
1
1
≥
xy(kN x + z)
kN + 1 xyz
For (e, 12, 12) we have


1 0 k


F12k =  0 1 k 
0 0 1


k 0 k+1


F12k F0 =  k 1
k 
1 0
1

0 1

2k+1
F1
= 1 0
0 0

k+1

2k+1
F1 F0 =  k
1

k+1

k 
1

1 k+1

0 k+1 
0
1
This gives us the following recurrence relations:
an = 2kn
an = 2kn + 1
xn = kn (xn−1 + yn−1 ) + zn−1
xn = kn (xn−1 + yn−1 ) + xn−1 + zn−1
yn = yn−1
yn = xn−1
zn = kn (xn−1 + yn−1 ) + xn−1 + zn−1
zn = (kn + 1)(xn−1 + yn−1 ) + zn−1
Then zn ≥ xn ≥ yn for all n. Then
area(Tk ) ≥
1
1
1
≥
xy((kN + 1)/2(x + y) + z)
kN + 2 xyz
Thus for the classes represented by (e, e, e), (e, e, 12), (e, 12, e), and (e, 12, 12)
we have a singular function. We list out the maps in (e, e, e), (e, e, 12) and (e, 12, 12)
42
below, leaving the class (e, 12, e) to be treated in greater detail in the next section.
(e, e, e)
(12, 12, 12)
(13, 13, 13)
(23, 23, 23)
(13, 12, 12) (123, e, e)
(e, 123, 123)
(132, 132, 132) (12, 23, 23)
(13, 13, 13)
(e, e, 12)
(12, 12, e)
(13, 13, 123)
(23, 23, 132)
(123, 132, 23)
(132, 123, 13)
(13, e, 12)
(123, 12, e)
(e, 13, 123)
(132, 23, 132)
(12, 132, 23)
(13, 123, 13)
(e, e, 123)
(12, 12, 23)
(13, 13, 12)
(23, 23, 13)
(123, 132, e)
(132, 123, 132)
(e, 13, 12)
(12, 132, 12) (13, e, 123)
(23, 123, 132)
(123, 12, 23)
(132, 23, 13)
(e, 12, 12)
(12, e, e)
(13, 123, 123) (23, 132, 132)
(123, 23, 23)
(132, 13, 13)
(13, e, e)
(123, 12, 12) (e, 13, 13)
(12, 132, 132)
(13, 123, 123)
(132, 23, 23)
(123, 132, 132) (132, 123, 123)
We can go through similar calculations as above to attempt to calculate c(k)
for any of the 108 polynomial TRIP maps. Unfortunately, for the remaining maps
not in these classes we run into one of two problems. Either c(k) is quadratic, which
does not give us what we want, or we can’t actually calculate c(k). However, we
do have the following
Proposition 3.10. If a given TRIP map T is ergodic and the associated 4k
1
, then
satisfy that the area of 4k is (k+1)(k+2)
sn
=∞
n→∞ n
lim
almost everywhere.
Proof. First we define fk as the characteristic function of 4k :

1 x ∈ 4
k
fk (x) =
0 x ∈
/4
k
Then using the fact that every TRIP map has an intrinsic measure µ, by Birkhoff’s
Ergodic Theorem we have that
Z
n
1X
i
lim
fk (T (x)) =
fk (x)dµ = |4k |
n→∞ n
4
i=1
for almost all x ∈ 4.
Using the notation P (k) = µ(4k ), this says that for almost all x ∈ 4, ai = k
43
on average P (k) of the time. Then we have
∞
sn X
=
lim
kP (k)
n→∞ n
k=1
almost everywhere. Because the intrinsic measure of the TRIP map is absolutely
1
continuous with respect to the Lebesgue measure, we have that P (k) > C (k+1)(k+2)
for some constant C. Using the integral test, we have that
Z
Z
Cx
(x + 2)2
xP (x)dx ≥
dx = C[log
+ c0 ]
(x + 1)(x + 2)
x+1
which diverges on [1, ∞). Thus we get
sn
=∞
n→∞ n
lim
almost everywhere.
1
The condition that the area of 4k is (k+1)(k+2)
characterizes 2/3 of the polynomial TRIP maps, namely the ones without a (−1)k in their Jacobian. Checking
through the various classes, we see that the following class contain a map that
satisfy this condition on the area of the 4k :
(e, 13, e)
(e, 13, 23)
(e, 13, 132)
(e, 23, e)
(e, 123, 132)
In the case of (e, 13, e) and (e, 23, e) this map is the class representative itself. For
the class (e, 13, 23) this map is (132, 12, 123). For the class (e, 13, 132) this map
is (13, 23, 123). For the class (e, 123, 132) this map is (13, 23, 13). Then if we can
show these maps are ergodic we will get these five classes as well.
In [8], Amburg and Jensen show that (e, 23, e) and (132, 12, 123) are ergodic,
which gives us that the following 24 maps give rise to singular Φ:
(e, 23, e)
(12, 123, 12)
(13, 132, 13) (23, e, 23)
(13, 12, 132)
(123, e, 13)
(e, 123, 23)
(e, 13, 23)
(12, 123, 132) (13, e, 123)
(13, 132, 123) (123, 13, 23)
(e, 23, 12)
44
(123, 13, 132) (132, 12, 123)
(132, 132, 12) (12, 23, 123)
(23, 13, e)
(23, 132, e)
(123, 12, 13)
(132, 23, 12)
(132, 12, 13)
(12, 123, e)
(23, e, 132)
This leaves us with the following uncharacterized cases:
(e, e, 13)
3.3
(e, e, 23)
(e, e, 132)
(e, 12, 23)
(e, 12, 23)
(e, 12, 132)
Degenerate Farey Maps
While we showed the class of degenerate Farey maps give rise to singular functions
in the previous section through direct calculation, we can also understand these
functions’ singularity directly using the question mark function. Recall that the
degenerate Farey maps fix one of the original vertices of 4 and partition the
opposite side according to the same Farey division of the unit interval. We show
the three possible partitionings below:
To show that these maps give rise to singular functions under the limit definition
we will use the following geometric fact about triangles:
Lemma 3.11. Suppose we have a triangle T with vertices v1 , v2 and v3 . Now
consider the triangle T 0 with vertices v1 , w2 , w3 where w2 and w3 are on v2 v3 .
Then
λ(T 0 )
d(w2 , w3 )
=
λ(T )
d(v2 , v3 )
Label the original vertices of 4 as x, y and z. Let 4(i1 , . . . , in ) have vertices
f1 , f2 , and z, where f1 and f2 are adjacent Farey numbers on the side of the
triangle opposite z (so either in the form (α, 0), (0, α), or (α, α) where α is the
Farey number in question, and Γ(i1 , . . . , in ) have vertices b1 , b2 and z where b1 and
b2 are the corresponding bary numbers. Then
λ(Γ(i1 , . . . , in ))
λ(Γ(i1 , . . . , in ))/λ(4)
d(b1 , b2 )/d(x, y)
d(b1 , b2 )
=
=
=
λ(4(i1 , . . . , in ))
λ(4(i1 , . . . , in ))/λ(4)
d(f1 , f2 )/d(x, y)
d(f1 , f2 )
45
Then this will converge to 0 almost everywhere, so have that Φ will be singular
under the limit definition. In fact, we know exactly the form that Φ will take.
For the permutations that fix (0, 0), we have that Φ(1, y) = (1, ?(y)), Φ(x, 0) =
(x, 0), Φ(x, x) = (x, x), and that any point (x, y) on the interior of 4 on the line
with slope α will be sent to the point on the line with slope ?(α) that the same
proportion of the distance along the line.
For the permutations that fix (1, 0), we have that Φ(x, x) = (?(x), ?(x)), Φ(x, 0) =
(x, 0), Φ(1, y) = (1, y), and that any point (x, y) on the interior of 4 on the line
that passes through (α, α) will be sent to appropriate point on the line passing
through (?(α), ?(α)).
For the permutations that fix (1, 1), we have that Φ(x, 0) = (?(x), 0), Φ(x, x) =
(x, x), Φ(1, y) = (1, y), and that any point (x, y) on the interior of 4 on the line
that passes through (α, 0) will be sent to appropriate point on the line passing
through (?(α), 0).
We claim by the singularity of ?(x), we have that Φ(σ,τ0 ,τ1 ) for any (σ, τ0 , τ1 )
belonging to the degenerate Farey class will be singular under the traditional
definition as well. To see this, consider the permutation (e, 12, e) which gives a
Farey partitioning of the hypotenuse of 4. Then take the set of points (A) of
measure 1 on which ?(x) is singular, meaning the measure of ?(A) is 0. Then the
set of line segments connecting the points in the form (a, a) where a ∈ A on the
hypotenuse to the vertex (1, 0) will have full measure on 4. Denote this set as B.
The Φ(e,12,e) (B) will have measure 0 because ?(A) has measure 0. Then because
Φ(e, 12, e) is singular we have that the remaining maps in this class are singular,
which we list below:
(e, 12, e)
(e, 12, 13)
(e, 123, e)
(e, 123, 13)
(13, 12, e)
(13, 12, 13)
(13, 123, e)
(13, 123, 13)
(12, e, 12)
(12, e, 132)
(12, 23, 12)
(12, 23, 132)
(123, e, 12)
(123, e, 132)
(123, 23, 12)
(123, 23, 132)
(23, 13, 23)
(23, 13, 123)
(23, 132, 23)
(23, 132, 123)
(132, 13, 23)
(132, 13, 123)
(132, 132, 23)
(132, 132, 123)
46
Chapter 4
Conclusion and Future Directions
The goal of this paper is to find singular functions from R2 to R2 by generalizing
the Minkowski Question Mark Function. To do this, we used the 216 TRIP maps
to define both a Farey and Bary partition of 4 and defined a function Φ(σ,τ0 ,τ1 ) :
4 → 4 by mapping a point with triangle tree sequence with respect to the Farey
partition to the point with the same triangle tree sequence but with respect to the
Bary partition. Through a series of lemmas we were able to reduce the problem
from looking at 216 individual TRIP maps to looking at 15 classes of TRIP maps
where the singularity of the Φ associated with one of the TRIP maps in the class
implied the singularity of the Φ of associated with all the remaining TRIP maps.
We then, using two different notions of singularity, developed a framework for
showing when Φ(σ,τ0 ,τ1 ) is singular. The essence of showing this, we found, was
showing that the set of points with normal triangle tree sequences with respect
to the Farey partition has measure zero, while the set of points that have normal
triangle tree sequences with respect to the Bary partition has measure one. We
could also phrase this problem in terms of how the subtriangles on the Bary side
were shrinking with respect to the subtriangles on the Farey side. Ultimately, we
were able to show singularity for 7 of the 15 classes, corresponding to 120 total
TRIP maps. Furthermore, we showed that given ergodicity, we could show that
3 more classes would give rise to singular Φ, which would give us 36 more TRIP
maps.
Going forward, an obvious goal is to show singularity for the remaining 8
cases, especially for the 5 that Lemma 3.10 doesn’t cover. One possible way of
47
doing this would be understanding the probability distribution of the triangle tree
sequences of these various maps, as it directly gives us the set on which Φ will
be singular under the traditional definition. It’s also important to understand the
properties of these functions. While we understand the functions corresponding to
the Mönkemeyer-type and degenerate Farey maps very well, we know very little
about the continuity of our Φ. It’s even possible that some of our Φ will not be
well-defined if, for a given triangle sequence, we have convergence to a point on
the Farey side but not on the Bary side.
Furthermore, it’s very likely that we could use higher-dimensional continued
fraction algorithms to find singular functions from Rn to Rn . Panti does this with
the Mönkemeyer Map, but if we can find a full analogue of TRIP maps for higher
dimensions, we could really begin to understand singular functions in general.
48
References
[1] R. Salem, On some singular monotonic functions which are strictly increasing,
Transactions of the American Mathematical Society, 53, (1943), 427-439.
[2] P. Viader, J. Paradis and L. Bibiloni, A New Light on Minkowski’s ?(x) Function, Journal of Number Theory, 73, (1998), 212-227.
[3] Fritz Schweiger, Multidimensional Continued Fractions. Oxford Science
Publications. Oxford University, Oxford 2000.
[4] Giovanni Panti, Multidimensional Continued Fractions and a Minkowski Function, arXiv:0705.0584v2, to appear in Monatshefte fur Mathematik.
[5] Olga R. Beaver and Thomas Garrity, A Two-Dimensional Minkowski ?(x)
Function, Journal of Number Theory, 107(1):105-134, 2004.
[6] Ilya Amburg and Thomas Garrity, Functional Analysis Behind a Family of
Multidimensional Continued Fractions: Triangle Partition Maps, in preparation.
[7] Krishna Dasaratha, Laure Flapan, Thomas Garrity, Chansoo Lee, Cornelia
Mihaila, Nicholas Neumann-Chun, Sarah Peluse, Matthew Stroffregen, A
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[8] Ilya Amburg and Stephanie Jensen, Ergodicity for Select Triangle Partition
Maps, in preparation.
49