Explicit forms for and some functional analysis behind a family of multidimensional continued fractions triangle partition maps and their associated transfer operators

And Some Functional Analysis BehindA Family of Multidimensional Continued Fractions – Triangle Partition Maps – And Their Associated Transfer Operators Ilya Amburg ida1@williams.edu Prof

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-t;: l< i>l•'c i-l Fo riv'S j-c

,-Thesis title ~c 111lly ot M~:il±;rl(~k

And Some Functional Analysis Behind

A Family of Multidimensional Continued Fractions

– Triangle Partition Maps – And Their Associated Transfer Operators

Ilya Amburg ida1@williams.edu 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, MA

June 6, 2014

The family of 216 multidimensional continued fractions known as known as triangle partition maps (TRIP maps for short) has been used in attempts to solve the Hermite problem [3], and is hence important in its own right This thesis focuses on the functional analysis behind TRIP maps We begin by finding the explicit form of all 216 TRIP maps and the corresponding inverses We proceed to construct recurrence relations for certain classes of these maps; afterward, we present two ways of visualizing the action of each of the 216 maps.

We then consider transfer operators naturally arising from each of the TRIP maps, find their explicit form, and present eigenfunctions of eigenvalue 1 for select transfer operators We observe that the TRIP maps give rise to two classes of transfer operators, present theorems regarding the origin of these classes, and discuss the implications of these theorems; we also present related theorems on the form of transfer operators arising from compositions of TRIP maps We then proceed to prove that the transfer operators associated with select TRIP maps are nuclear of trace class zero and have spectral gaps We proceed to show that select TRIP maps are ergodic while also showing that certain TRIP maps never lead to convergence to unique points We finish by deriving Gauss-Kuzmin distributions associated with select TRIP maps.

1

I would like to sincerely thank my advisor, Professor Thomas Garrity, for introducing me to TRIP maps and providing invaluable insights during the course of my work I would also like to thank Professor Cesar Silva for his willingness to be my second reader and useful feedback.

2

1 Introduction 7

1.1 Continued Fractions and Periodicity in Real Number Representations 7

1.2 The Hermite Problem 8

1.3 The Triangle Map 8

1.4 A Family of 216 Multidimensional Continued Fraction Algorithms: TRIP Maps 11

1.5 TRIP Sequences and TRIP Tree Sequences 13

1.6 Transfer Operators 14

1.7 Interlude 15

1.8 Polynomial- and Non-Polynomial-Growth TRIP Maps 15

1.9 Combo TRIP Maps and Polynomial-Growth 16

1.10 Nuclearity and Spectral Gaps for Transfer Operators Associated with Select TRIP Maps 17

1.11 Ergodic Theory 17

1.12 Ergodicity of TRIP Maps 18

1.13 Gauss-Kuzmin Distributions for TRIP Sequences 19

1.14 Computational Methodology 19

2 Explicit Form of TRIP Maps 20 2.1 Sample TRIP Map Calculation 20

3

3 Explicit Form of TRIP Map Inverses 22 3.1 Sample TRIP Map Inverse Calculation 22

4.1 Sample Recurrence Relation Calculation 24

5.1 Sample Partition and TRIP Diagram Calculation 26

6.1 Sample Transfer Operator Calculation 29

7 Eigenfunctions of Eigenvalue 1 for Select Transfer Operators 31 7.1 Sample Eigenfunction Verification 33

8.1 A Permutation Triplet Mapping that Preserves

Polynomial-Growth 39

11 Functional Analysis Behind Transfer Operators: Banach Space Approach 53 11.1 Transfer Operators as Linear Maps on Appropriate Banach Spaces 53 11.2 Spectral Gap Results 58

12 Functional Analysis Behind Transfer Operators: Hilbert Space Approach 61 12.1 Nuclearity of Select Transfer Operators 63

13.1 Calculations Leading to Ergodicity 70

4

14 Infinitely Many Zeroes Almost Everywhere for Combo TRIP Maps 72

C.1 Polynomial-Growth Maps 220 C.2 Select Non-Polynomial-Growth Maps 227

D.1 Polynomial-Growth Maps 231 D.2 Non-Polynomial-Growth Maps 238

F.2 Non-Polynomial-Growth Maps 401

G.1 Polynomial-Growth Maps 411 G.2 Non-Polynomial-Growth Maps 421

6

Num-ber Representations

This section relies on content1 in [5].

A number is algebraic if it is the root of an irreducible polynomial in one variable having integer coefficients In particular, an algebraic number that is the root of an nth−order irreducible polynomial is referred to as having degree n There is no known way of telling whether a number is algebraic by looking at its base-ten expansion, unless, of course, the number is rational; the continued fraction expansion of a real number, however, provides a link between quadratic irrationality of that number and periodicity of its continued fraction expansion.

Consider any real number x Its continued fraction representation is

x = a0+ 1

a1+ 1

a2+a3+ 1where ai, i > 0, are positive integers, a0 ∈ Z, and a0 is the integer part of x, a1 is the integer part of x−a1

0, and so on In this way, any real number x may be expressed in the form [a0; a1, a2, a3, ] Lagrange proved that x ∈ R is algebraic of degree 2 if and only if thecontinued fraction representation of x eventually becomes periodic.

1A majority of the material in this section, as well as in the rest of the introduction, relies on [5] Some of the LaTeX code for formulas and definitions was taken directly from that document and appears throughout the introductory sections with the original author’s consent.

7

1.2 The Hermite Problem

This section also relies on content in [5].

Naturally, Lagrange’s theorem leads us to wonder whether there exist ways of writing real numbers to facilitate the identification of nth−degree algebraic numbers Indeed, this is the famous Hermite problem, which according to [5] was posed by Hermite to Jacobi in [6] Explicitly, quoting from [5], the Hermite problem asks for algorithms “ for writing a real number (or an n−tuple of reals) as sequences of integers so that periodicity of the sequence corresponds to the initial real (or the n−tuple of reals) being algebraic of a given degree.” Currently, the Hermite problem remains unsolved Attempts to solve it have relied on multidimensional continued f ractions For background on multidimensional continued fractions, see Schweiger’s Multidimensional Continued Fractions [16] A particular family of multidimensional continued fraction algorithms – TRIP maps – has been used to construct maps such that a number being a cubic irrational (real and algebraic of degree 3) corresponds

to a certain kind of periodicity under those maps [3] This thesis will explore the functional analysis behind this family of multidimensional continued fractions.

This section largely follows the outline set in [5] and [2].

Let us first examine the original TRIP map, the triangle map, introduced in [4], from which the whole family of 216 TRIP maps originated.

Subdivide the triangle given by

We want to represent the triangle map using matrix notation To do this, start by defining a cone 4∗ such that

4∗ = {(b0, b1, b2) : b0 ≥ b1 ≥ b2 ≥ 0}, and construct a projection map π : R3 → R2 given by

π(b0, b1, b2) =  b1

b0 ,

b2b0



9

Then π(4∗) = 4, our “base” triangle.

Now define the vectors

v1 =





1 0 0





and note that π maps v1, v2, and v3 to the vertices of 4.

This implies that

(v1, v2, v3)A1 = (v1, v2, v1+ v3).

This implies that the action of A0 and A1 on B produces a disjoint partition of 4.

Now apply A1 k times to B, and then apply A0 once; in this process, 4’s vertices are sent to the vertices of

4k = {(x, y) ∈ 4 : 1 − x − ky ≥ 0 > 1 − x − (k + 1)y} This allows us the define the triangle map as

T :

∞[

k=0

4k → 4 where

T (x, y) = π

 (1, x, y) BA−10 A−k1 B−1
T



10

if (x, y) ∈ 4k Doing out the matrix multiplication, the above definition yields

T (x, y) =  y

x ,

1 − x − ky x

 ,

which is identical to the map we defined above.

This enables us to assign the triangle sequence (a1, a2, ) to a point (a, b) ∈ 4 by letting

ai equal k if Ti(x, y) ∈ 4k; again, if for any k we have Tk(a, b) ∈ {(x, 0) : 0 ≤ x ≤ 1}, we stop the iterative process and terminate the sequence In order to see that periodicity in the triangle sequence implies cubic irrationality, we consider the action of the map T : R3 → R3before we project the output into R2, defined by

If the original components of the point (x, y) ∈ 4, x and y, are both algebraic, each will

be algebraic of degree no more than 3 if we can find matrices B and A, where both B and

A can be written as products of matrices having the form

where λ is the associated eigenvalue Of course, we require AB−1 6= I.

Frac-tion Algorithms: TRIP Maps

Again, this section relies on material presented in [2] and [5].

11

Dasaratha, et al [2] conjecture that there exists no unique, single multidimensional tinued fraction algorithm capable of solving the Hermite problem This conjecture motivates looking at families of multidimensional continued fractions In particular, the family of 216 TRIP maps (short for triangle partition maps) arises if, at each step of the division of the base triangle 4 we allow for three permutations of the vertices of its R3 representation.

con-To construct the other 215 TRIP maps, it is important to note that there was nothing special about the ordering of the vertices (v1, v2, v3) Hence, we will allow permutations of the initial vertices by some σ ∈ S3

3, by some τ1 ∈ S3

3 after applying A1, and by some τ0 ∈ S3

3after applying A0.

This leads us to define the matrices

The application of F0 and F1 to (the R3 representation of) any triangle in (the R3representation of) 4 partitions it into two triangles Hence, instead of using A0 and A1, we can subdivide 4 using F0 and F1 Using these F0 and F1, we can define 216 triangle partition maps (TRIP maps, for short), each for one of the 216 permutation triplets in S33.

Let us make this definition more rigorous Let 4k be the triangle defined by the three points obtained from applying the projection map π to the columns of the matrix BF1kF0 Then define a TRIP map as Tσ,τ0,τ1 : S∞k=04k→ 4 where

Tσ,τ0,τ1(x, y) = π  (1, x, y) BF0−1F1−kB−1
T 

if (x, y) ∈ 4k Of course, the original triangle map corresponds to Te,e,e.

We have calculated the explicit forms of Tσ,τ0,τ1(x, y) and Tσ,τ−10,τ1(x, y) for all (σ, τ0, τ1) ∈

S33; they are discussed in Chapters 2 and 3, and are explicitly presented in Appendices A and B.

This section relies on material presented in [2].

The above definition of Tσ,τ0,τ1(x, y) allows us to create the analogue of the triangle sequence for all 216 TRIP maps: given (x, y) ∈ 4, let ai equal k if Ti

σ,τ0,τ1(x, y) ∈ 4k Then the sequence (a1, a2, ) is the TRIP sequence induced by Tσ,τ0,τ1 that is assigned to (x, y) The sequence assigned to (a, b) ∈ 4 terminates if there exists a k such that

Tk(a, b) ∈ {(x, y) : (x, y) / ∈

∞[

i=0

4i}.

Of course, the triangle sequence is simply the TRIP sequence induced by Te,e,ethat is assigned

to (x, y) ∈ 4.

It is important to note that there exists another related way of obtaining a sequence from

a point (x, y) ∈ 4 In particular, given (σ, τ0, τ1) ∈ S3

3, define the triangle

4(i1, i2 , in) = BFi1Fi2· · · Fin.

13

Then, given (x, y) ∈ 4, define ij ∈ {0, 1} such that (x, y) ∈ 4(i1, i2, , ij, , in) The sequence thus obtained, (i1, i2, , in), is the TRIP tree sequence induced by Tσ,τ0,τ1 that is assigned to (x, y).

It is easy to convert between the TRIP and TRIP tree sequences of a point (x, y) ∈ 4 induced by a given Tσ,τ0,τ1 : say the TRIP tree sequence is (1a1, 0, 1a2, 0, 1a3, ), where 1ai

stands for 1 appearing ai times consecutively Then the corresponding TRIP sequence is (a1, a2, a3, ).

This section relies on material from [5].

Much work has been devoted to working out the statistics of the terms appearing in the continued fraction representations of real numbers (see, for example, [9]) As per [5], some important research inspired by this line of work (see [12]) relies on use of the transfer operator

Lf (x) =

∞X

n=1

1 (n + x)2f

 1

n + x



; under certain conditions on f (x), the largest eigenvalue of this transfer operator is 1, with associated eigenfunction

h(x) = 1

1 + x [12].

It is natural to inquire if there exist analogous results for multidimensional continued fractions, and in particular for TRIP maps First, we must develop the notion of a transfer operator Consider a dynamical system (X, S, µ, T ) (see Section 1.11 for the definition) A transfer operator linearly maps functions defined on X in some vector space to functions defined on X in some vector space For a more concrete definition, choose a function g :

X → R The transfer operator, call it LT , acting on f : X → R is given by

LT f (x) = X

y:T (y)=x

g(y)f (y).

14

If T is differentiable, we usually choose g = |Jac(T )|1 .

Extending this definition to the case of TRIP maps, we define our transfer operators to be

k=0

1 (1 + kx + y)3f

We have calculated the explicit form of LTe,e,ef (x, y) for all (σ, τ0, τ1) ∈ S33; these are presented in Appendix E A sample calculation of the explicit form of a transfer operator is presented in Chapter 6.

The Hermite problem was presented above to motivate the development of TRIP maps

by showing a potential application Having placed TRIP maps in this context, thereby establishing their importance in their own right, from now on we focus on the functional analysis behind these TRIP maps, discussing their explicit form and ergodic properties – as well as the form, spectrum, and nuclearity of the associated transfer operators.

We see that the transfer operator corresponding to Te,e,e has a particularly nice form, in that the denominator of the factor (1+kx+y)1 3 is (non-trivially) polynomial in k In general, we will

be concerned with the form of the transfer operator LTσ,τ0,τ1 where

transfer operator, and is itself polynomial-growth; otherwise, both LTσ,τ0,τ1 and Tσ,τ0,τ1 are non-polynomial-growth.

By direct calculation of the explicit form of the transfer operators, presented in pendix E, we have shown that exactly half of the 216 TRIP maps are polynomial-growth In addition, we identified the origin of polynomial-growth by showing that that a TRIP map Tσ,τ0,τ1 is polynomial-growth if and only if the eigenvalues of the associated F1 = σA1τ1 all have magnitude 1; theorems regarding polynomial-growth in TRIP maps are presented in Chapter 8 We use these results in Chapter 10 to classify patterns appearing in the visual representations of the partitions S∞k=04k induced on 4 by Tσ,τ0,τ1.

We obtain a larger class of maps by allowing compositions of TRIP maps [2] As an example,

we might perform the first division of 4 using the permutation triplet (σ, τ0, τ1), the next using the permutation triplet (σ2, τ02, τ12), etc We can represent such compositions as T1◦

T2 ◦ Tn, where, of course, each subscript is short for a permutation triplet.

Further, we can consider a composition of TRIP maps T1◦ T2◦ ◦ Ti, where {Tj} , j ∈ λ, with λ some indexing set such that λ ⊂ {1, 2, , i} , are polynomial-growth maps, and the rest are non-polynomial-growth In Chapter 9, we prove that the corresponding transfer operator will be polynomial in all kj such that j ∈ λ, and exponential in all kl such that

16

1.10 Nuclearity and Spectral Gaps for Transfer

Oper-ators Associated with Select TRIP Maps

In Chapters 11 and 12, we use analogues of the methods developed by Mayer, et al in [12], and applied to the original TRIP map in [4] (the “Banach space approach” and the “Hilbert space approach”) to show that the transfer operators corresponding to many additional TRIP maps are also nuclear of trace class zero or possess spectral gaps In particular, we have used the Banach space approach to show that several additional TRIP maps have spectral gaps.

We have also used the Hilbert space approach to show that all transfer operators for which

we have found corresponding eigenfunctions of eigenvalue 1 are nuclear of trace class zero The Banach space approach involves finding an appropriate Banach space V on which LTacts, showing that LT is a linear map from V to V, and showing that the largest eigenvalue

of LT is 1 and has multiplicity 1.

To get at the Hilbert space approach, we consider a transfer operator related to LT, namely LT,µ, defined by the property that for any measurable set A ⊂ 4, and for any function f ∈ L1(µ)

LT The fact that LT and LT ,µ can be related by a simple transformation shows that these transfer operators have identical spectra Hence, if we can show that LT ,µ is nuclear of trace class zero, analogous results immediately follow for LT [4] In order to show LT ,µ is nuclear

of trace class zero, we write it as a sum over special functions satisfying particular properties.

satisfies the following three properties: 1 X ∈ S, 2 If A ∈ S, then Ac ∈ S, and 3 If

A1, A2, A3, · · · ∈ S, then ∪∞n=1An∈ S.

A measure on S is a function µ : S → [0, ∞) where we require that 1 µ(∅) = 0, and 2.

If A1, A2, A3, · · · ∈ S are pairwise disjoint, then µ (∪∞n=1An) = P∞n=1µ(An).

A set A ∈ X is measurable if A ∈ S.

We refer to the triplet (X, S, µ) as a measure space; in our discussion of the ergodicity of TRIP maps, we will focus on the measure space (4, B, λ), where B stands for Borel σ-algebra and λ stands for the Lebesgue measure.

To get into ergodic theory, we must consider transformations induced on a measure space.

So let (X, S, µ) be a measure space The transformation T : X → X is measurable if it is such that for every A ∈ S, we also have that T−1(A) ∈ S We refer to (X, S, µ, T ) as a dynamical system A measurable T is nonsingular if for all A ∈ S, µ(T−1(A)) = 0 if and only if µ(T (A)) = 0.

Further, A ∈ X is strictly T-invariant if A = T−1(A) We are now at a point where

we can define ergodicity Let T be a nonsingular transformation T is ergodic if for every measurable, strictly T −invariant A ∈ X, µ(A) = 0 or µ(Ac) = 0.

Messaoudi, et al [14] have shown that the original triangle map Te,e,e is ergodic Jensen has extended these arguments to show that the maps Te,23,e, Te,23,23, Te,132,23 and Te,23,132 are ergodic [7] In Chapter 13 we show that 2 more TRIP maps are ergodic using the analogues

of arguments outlined by Jensen in [7].

In addition to these ergodicity results, in Chapter 14 we present results regarding vergence for combo TRIP maps and show in Chapter 15 that no TRIP sequence induced by

con-24 TRIP maps corresponds to a unique point.

18

1.13 Gauss-Kuzmin Distributions for TRIP Sequences

The probability distribution for terms appearing in the standard continued fraction expansion

of a number x ∈ R is called the Gauss-Kuzmin distribution [8] In Chapter 16, we derive analogues of the Gauss-Kuzmin distribution for TRIP sequences induced by select TRIP maps, relying on the fact that these maps have been shown to be ergodic.

Mathematica was used to perform almost all of the calculations in this thesis; a sion regarding the general research approach, as well as more details regarding the use of Mathematica, are found in Chapter 17.

discus-19

Explicit Form of TRIP Maps

The explicit form of Tσ,τ0,τ1(x, y) has been calculated for all (σ, τ0, τ1) ∈ S3

3 These explicit forms are presented in Appendix A Several explicit forms had already been calculated in [5] and [2], but here we present all 216 explicit forms.

We will go through a sample calculation of the from of Te,23,e(x, y).

21

Explicit Form of TRIP Map Inverses

The explicit form of Tσ,τ−10,τ1(x, y) has been calculated for all (σ, τ0, τ1) ∈ S3

3 These explicit forms are presented in Appendix B These were calculated from the definition

Tσ,τ−1

0,τ1(x, y) = π((1, x, y)  BF0−1F1−kB−1
T

−1 ).

The explicit form associated with Te,e,ehad already been calculated in [5], but here we present all 216 explicit forms.

We will go through a sample calculation of the from of Te,23,e−1 (x, y).

We have that

Te,23,e−1 (x, y) = π((1, x, y)  BF0−1F1−kB−1
T

−1 ).

From Chapter 2, we know that



1 (k + 1)x − y + 1 ,

x (k + 1)x − y + 1



22

Recurrence Relations for TRIP Map Orbits

Given a point (y1(ak), y2(ak)) ∈ 4 and the TRIP sequence associated with that point, (ak+1, ak+2 ) induced by a TRIP map Tσ,τ0,τ1, it is possible to derive recurrence relations for (xn, yn) in terms of (y1(ak+1), y2(ak+1)) = Tσ,τ0,τ1(y1(ak), y2(ak)) and ak+1 These recurrence relations have been calculated for all polynomial-growth (and a select few non-polynomial- growth) (σ, τ0, τ1) ∈ S3

3 and are presented in Appendix C.

Let us explain the process in detail For any (y1, y2) ∈ 4, consider the set {Tak(y1, y2)}∞k=0, where a0 = 0, and set (y1, y2) = (y1(a0), y2(a0)), so that, for k ≥ 0,

(x1(ak+1), x2(ak+1), x3(ak+1)) = (x1(ak), x2(ak), x3(ak)) BF0−1F1−kB−1
T

These relations allow us to solve for y1(ak) and y2(ak) in terms of y1(ak+1), y2(ak+1), and

ak+1.

23

4.1 Sample Recurrence Relation Calculation

Let us walk through a sample calculation for (e, e, e) In this case, we have that

(x1(ak+1), x2(ak+1), x3(ak+1)) = (x1(ak), x2(ak), x3(ak)) BF0−1F1−kB−1
T



1 ak+1y1(ak+1) + y2(ak+1) + 1 ,

y1(ak+1) ak+1y1 (ak+1) + y2(ak+1) + 1



It is possible to use these relations to write (y1(ak), y2(ak))) using an increasing number

of terms in the TRIP sequence and the image of (y1(ak), y2(ak)) after more applications of a

TRIP map; i.e., if that we know (y1(ak+n), y2(ak+n)) for n ≥ 1, we can write (y1(ak), y2(ak)) in

terms of ak, ak+1, , ak+n, and (y1(ak+n), y2(ak+n)) This is a direct analogue of the continued

fraction expansion of a real number To see how this works, note that, doing out the recursion

up to k = 3, we obtain

(y1(a0), y2(a0)) =

1 a2

1 y2(a3) y1(a3)+y1(a3)1 +a3

a3y1(a3)+y2(a3)+a3y1(a3)+y2(a3)+1

y2(a3) y1(a3)+

1 y1(a3)+a3

1 y1(a3)+a3

+

a2

1 y2(a3) y1(a3)+y1(a3)1 +a3

+1

a3y1(a3)+y2(a3)+a3y1(a3)+y2(a3)+1

y2(a3) y1(a3)+

1 y1(a3)+a3

+a2+1 + a1 + 1

).

24

We have thus expressed the original input pair (y1(a0), y2(a0)) ∈ 4 in terms of the first three terms of the TRIP sequence corresponding to (y1(a0), y2(a0)), and in terms of (y1(a3), y2(a3)), which is the third iterate of (y1(a0), y2(a0)) under Te,e,e We can continue this process until any k > 0, at each step putting in dependence on additional terms in the TRIP sequence (through ak), while pushing back the dependence on the original input pair

to its kth iterate.

25

Partition Diagrams and TRIP

Diagrams

Define a partition diagram as a visual representation of {4k}∞

k=0 induced on 4 by a TRIP map Given a TRIP sequence (a1, a2, ) induced by some TRIP map, define a TRIP diagram as a visual representation of triangles of the form BFa1

1 F0Fa2

1 F0 .

It is helpful to have both the partition and TRIP diagrams generated by each of the TRIP maps to help visualize the action of the maps on 4 The partition diagrams for all 216 maps are presented in Appendix F while the TRIP diagrams are presented in Appendix G Several similar diagrams had already been constructed in [5] and [2], but here we present diagrams for all 216 TRIP maps.

Let us explain how to arrive at the partition and TRIP diagrams generated by Te,e,e Let us begin with the partition diagram Recall that the vertices of 4k are found by applying the associated F1 = (e)A1(e) k times to the base triangle B, followed by applying

F1 = (e)A0(e) once; projecting the vertices into R2 gives 2−space representation of 4k In particular, here we have that BF0

of each matrix using π(x1, x2, x3) → x2

x1,x3

x1

 , we obtain the vertices of triangles 4k, k ∈ {0, 1, 2, 3, 4} shown in the diagram below:

projecting the columns of each matrix using π(x1, x2, x3) → x2

x1,x3

x1

 , we obtain the vertices

of triangles (labeled by the first two terms in the associated TRIP tree sequence) shown in the TRIP diagram below:

Explicit Form of all Transfer

in [5], but here we present the explicit forms associated with all 216 TRIP maps.

While the calculation of the explicit form of any non-polynomial-growth transfer operators

is incredibly involved, the calculation for certain polynomial-growth transfer operators is manageable We present one such polynomial-growth example below, calculating the explicit form of the transfer operator LT23,23,23f (x, y).

By direct calculation, we have that

29

We are interested in the Jacobian of T23,23,23with any x replaced by the first component of

T23,23,23−1 and any y replaced by the second component of T23,23,23−1 The Jacobian is calculated

∂x

x−yx

x2

−(1+k)x

1 (1 + (1 + k)x − y)3 Hence, we have shown that

k=0

1 (1 + (1 + k)x − y)3f

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Eigenfunctions of Eigenvalue 1 for

Select Transfer Operators

We have found the eigenfunction of eigenvalue 1 for transfer operators associated with 17 TRIP maps; these are presented below (note that the eigenfunction of eigenvalue 1 associated with Te,e,e was already known):

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HΣ,Τ0,Τ1L Eigenfunction of associated transfer operator

He, e, eL x Hy+1L 1

He, 23, eL x H1-yL 1

He, 132, eL x H1-yL 1 H12, 12, 12L Hy+1L H-x+y+1L 1 H12, 13, 12L H1-yL H-x+y+1L 1 H12, 123, 12L H1-yL H-x+y+1L 1 H13, 13, 13L Hx-2L H1-yL 1 H13, 23, 13L x H1-yL 1 H13, 132, 13L x H1-yL 1 H23, e, 23L x H-x+y+1L 1 H23, 12, 23L x H-x+y+1L 1 H23, 23, 23L x Hx-y+1L 1 H123, 13, 132L H1-yL H-x+y+1L 1 H123, 123, 132L H1-yL H-x+y+1L 1 H123, 132, 132L H1-yL Hx-y+1L 1 H132, e, 123L x H-x+y+1L 1 H132, 12, 123L x H-x+y+1L 1 H132, 123, 123L Hx-2L H-x+y+1L 1

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7.1 Sample Eigenfunction Verification

Let us verify that f (x, y) = x(x−y+1)1 is indeed an eigenfunction of eigenvalue 1 of LT23,23,23 The verifications for the some of the transfer operators presented in the above table are similar; however, many rely on more complicated techniques and properties of special functions In particular, we have exploited the property, stated in [13], that for the diagamma function ψ,

ψ(x + 1) = ψ(x) + 1

x . Finding these eigenfunctions, even with the aid of Mathematica, required a lot of work From the previous section, we know that

LT23,23,23f (x, y) =

∞X

k=0

1 (1 + (1 + k)x − y)3f

Noting that we have to replace any x in f (x, y) with the first component of T23,23,23−1 and any

y by the second component, we obtain

∞X

k=0

1 (1 + (1 + k)x − y)3

k=0

1 (1 + x + kx − y)(1 + (2 + k)x − y)

=

∞X

k=0

1 x(−1 − 2x − kx + y) − 1

Origin of Polynomial-Growth in TRIP Maps

While it is nice to have the explicit form of the transfer operators calculated, it is perhaps even more important to understand how these forms arise.

Let us first present a well-known result we will need later:

Theorem 1 Consider a vector v = (1, x, y) and a matrix M =



∂

∂y

b+ex+hya+dx+gy



∂

∂x

c+f x+iya+dx+gy



∂

∂y

c+f x+iya+dx+gy

We are concerned with the form of the transfer operator LTσ,τ0,τ1 where

of TRIP maps, and present some corollaries To avoid redundancy, we will refer to polynomial dependence on k aside from factors of (−1)k as polynomial dependence on k or strictly polynomial dependence on k.

Theorem 2 A TRIP map Tσ,τ0,τ1 is polynomial-growth if and only if the eigenvalues of the associated F1 = σA1τ1 all have magnitude 1; it is non-polynomial-growth otherwise.

Proof By the previous theorem and the definition of Tσ,τ0,τ1, it follows that 1

Jac(Tσ,τ0,τ1(a,b)) =det(Mσ,τ0,τ1)

((1,x,y)Mσ,τ0,τ1(1,0,0)T)3 where

Mσ,τ0,τ1 = ((BF0−1F1−kB−1)T)−1and B is the 3-space representation of our base triangle Since det (Mσ,τ0,τ1) is real and has magnitude 1 because all matrices involved in its calculation also have real determinants of magnitude 1, this implies that a given Tσ,τ0,τ1 is polynomial-growth if and only if the first column of Mσ,τ0,τ1 contains terms polynomial in k.

For simplicity, write Mσ,τ0,τ1 = (C1, C2, C3) , where Ci are the columns of Mσ,τ0,τ1 Note that the inverse of a TRIP map, Tσ,τ−1

0,τ1, can be written as

Tσ,τ−10,τ1(x, y) = π

 (1, x, y)  BF0−1F1−kB−1
T

−1 

= π ((1, x, y)Mσ,τ0,τ1)

=  (1, x, y) · C2(1, x, y) · C1,

(1, x, y) · C3 (1, x, y) · C1



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Note that for any (x, y) ∈ 4, Tσ,τ−10,τ1(x, y) must be bounded as it must land back inside

4 Since the inverse of each TRIP map is bijective and since the choice of k depends on which 4k the original (x, y) lies in, the first column of Mσ,τ0,τ1 must depend on k Thus, to show that the first column of Mσ,τ0,τ1 depends on k polynomially, it is sufficient to show that

Mσ,τ0,τ1 exhibits only polynomial dependence on k – for then, by the above argument, the first row of Mσ,τ0,τ1 must necessarily exhibit only polynomial dependence on k.

J5 = D(roots(−1 − t2+ t3)), and

(J6)k = D(roots(−1 − t + t3)k).

It is well-known that the diagonal elements of each J are precisely the eigenvalues of the matrix A from which it originated; further, J4 through J6 are diagonal and each contain

at least one entry of magnitude greater than 1 on their respective diagonals From this,

it is clear that if (F1)T (and hence F1) has eigenvalues that all have magnitude 1, then the first column of Mσ,τ0,τ1 depends polynomially on k, and hence the associated Tσ,τ0,τ1 is polynomial-growth Otherwise, the first column of Mσ,τ0,τ1 depends exponentially on k, and hence the associated Tσ,τ0,τ1 is non-polynomial-growth.

Assume T is polynomial-growth in k Then the first column of M must depend strictly polynomially on k From the above decomposition of M, we see that the only place where k- dependance may enter the first column of M is through A.; hence, by the above argument the

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