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Signed (2d advisor, if applicable)------------------t;: l< i>l•'c i-l Fo riv'S .j-c ,Ana ly S(.s 'B d1ih li'( ct Thesis title ~c..111lly ot M~:il±;rl(~k _,,.,. 1..: P"rt,·tu;.r M,.ps-T~" 0, are positive integers, a0 . Z, and a0 is the integer part of x, a1 is the integer part of x-1 a0 , 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 the continued 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 de.nitions was taken directly from that document and appears throughout the introductory sections with the original author’s consent.
1.3. The Triangle Map
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 identi.cation 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 fractions. 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.
1.3 The Triangle Map
This section largely follows the outline set in [5] and [2].
Let us .rst 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
. = {(x, y):1 = x = y = 0}
into smaller triangles given by
Dk = {(x, y) .. :1 - x - ky = 0 > 1 - x - (k + 1)y}
for every integer k = 0. The partitioning is represented in the following diagram:
1.3. The Triangle Map
The triangle map
8
T : Dk .D k=0
is then given by
T (x, y) = y , 1 - x - ky
x x
if (x, y) . Dk.
To each (a, b) .D assign the sequence (a1,a2, ...) if for every k> 0,
T k(a, b) .Dak .
If for some k it happens that T k(a, b) .{(x, 0) : 0 = x = 1}, we stop the iterative process and terminate the sequence. We call this sequence the triangle sequence associated with (a, b).
We want to represent the triangle map using matrix notation. To do this, start by de.ning a cone D* such that
D * = {(b0,b1,b2): b0 = b1 = b2 = 0},
and construct a projection map p : R3 . R2 given by
b1 b2p(b0,b1,b2)=,.
b0 b0
1.3. The Triangle Map
Then p(D*)= D, our “base” triangle. Now de.ne the vectors
.. .. ..
111 v1 = . 0 .,v2 = . 1 . ,v3 = . 1 . 001
and note that p maps v1,v2, and v3 to the vertices of D.
This implies that
..
1 11
..
(v1,v2,v3)= 011 = B 001 is the R3 representation of our base triangle D; operations on this matrix will facilitate the desired partitioning. To subdivide D in a way identical to the triangle map constructed above, we de.ne
matrices A0 = . . 0 1 0 0 0 1 1 0 1 . ., A1 = . . 1 0 0 0 1 0 1 0 1 . .
and notice that
(v1,v2,v3)A0 =(v2,v3,v1 + v3)
and (v1,v2,v3)A1 =(v1,v2,v1 + v3).
This implies that the action of A0 and A1 on B produces a disjoint partition of D. Now apply A1 k times to B, and then apply A0 once; in this process, D’s vertices are sent to the vertices of
Dk = {(x, y) .D :1 - x - ky = 0 > 1 - x - (k + 1)y} .
This allows us the de.ne the triangle map as
8
T : Dk .D k=0
where
..T
A-kB-1
T (x, y)= p(1, x, y) BA-01 1
1.4. A Family of 216 Multidimensional Continued Fraction Algorithms: TRIP Maps
if (x, y) .Dk. Doing out the matrix multiplication, the above de.nition yields
y 1 - x - ky
T (x, y)= ,,
xx
which is identical to the map we de.ned above.
This enables us to assign the triangle sequence (a1,a2, ...) to a point (a, b) .D by letting ai equal k if T i(x, y) .Dk; again, if for any k we have T k(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 . R3 before we project the output into R2 , de.ned by
T (1, x, y) = (1, x, y) BA-01A-1 kB-1 T
which reduces to .. 00 1
..
T (1, x, y) = (1, x, y) 10 -1 . 01 -k If the original components of the point (x, y) .D,x and y, are both algebraic, each will be algebraic of degree no more than 3 if we can .nd matrices B and A, where both B and A can be written as products of matrices having the form
..
00 1
..
10 -1
01 -k
such that (1, x, y) is some eigenvector of the matrix AB-1; i.e.,
(1, x, y)AB-1 = .(1, x, y),
where . is the associated eigenvalue. Of course, we require AB-1 .
= I.
1.4 A Family of 216 Multidimensional Continued Fraction Algorithms: TRIP Maps
Again, this section relies on material presented in [2] and [5].
1.4. A Family of 216 Multidimensional Continued Fraction Algorithms: TRIP Maps
Dasaratha, et. al [2] conjecture that there exists no unique, single multidimensional continued 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 D we allow for three permutations of the vertices of its R3 representation.
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 s . S33 , by some t1 . S33 after applying A1, and by some t0 . S33 after applying A0.
This leads us to de.ne the matrices
F0 = sA0t0,F1 = sA1t1
for each (s, t0,t1) . S33 . In particular, we denote the permutation matrices as follows:
..
100 e = . 010 . , 001
. .
010
. .
(12)= 100 , 001
. .
001
. .
(13)= 010 , 100
. .
100
. .
(23)= 001 , 010
. .
010
. .
(123)= 001 , 100
. .
001
. .
(132)= 100 . 010
1.5. TRIP Sequences and TRIP Tree Sequences
The application of F0 and F1 to (the R3 representation of) any triangle in (the R3 representation of) D partitions it into two triangles. Hence, instead of using A0 and A1, we can subdivide D using F0 and F1. Using these F0 and F1, we can de.ne 216 triangle partition maps (TRIP maps, for short), each for one of the 216 permutation triplets in S33 .
Let us make this de.nition more rigorous. Let Dk be the triangle de.ned by the three points obtained from applying the projection map p to the columns of the matrix BF 1 kF0.
8
Then de.ne a TRIP map as Ts,t0,t1 : k=0 Dk .D where
Ts,t0,t1 (x, y)= p (1, x, y) BF 0 -1F1 -kB-1 T
if (x, y) .Dk. Of course, the original triangle map corresponds to Te,e,e. We have calculated the explicit forms of Ts,t0,t1 (x, y) and T -1 (x, y) for all (s, t0,t1) .
s,t0,t1
S33; they are discussed in Chapters 2 and 3, and are explicitly presented in Appendices A and B.
1.5 TRIP Sequences and TRIP Tree Sequences
This section relies on material presented in [2].
The above de.nition of Ts,t0,t1 (x, y) allows us to create the analogue of the triangle sequence for all 216 TRIP maps: given (x, y) .D, let ai equal k if T i (x, y) .Dk. Then
s,t0,t1
the sequence (a1,a2, ...) is the TRIP sequence induced by Ts,t0,t1 that is assigned to (x, y). The sequence assigned to (a, b) .D terminates if there exists a k such that
8
T k(a, b) .{(x, y):(x, y) ./Di}.
i=0
Of course, the triangle sequence is simply the TRIP sequence induced by Te,e,e that is assigned to (x, y) .D.
It is important to note that there exists another related way of obtaining a sequence from a point (x, y) .D. In particular, given (s, t0,t1) . S33 , de.ne the triangle
D(i1,i2..., in)= BFi1 Fi2 ··· Fin .
1.6. Transfer Operators
Then, given (x, y) .D, de.ne ij .{0, 1} such that (x, y) .D(i1,i2,...,ij,...,in). The sequence thus obtained, (i1,i2,...,in), is the TRIP tree sequence induced by Ts,t0,t1 that is assigned to (x, y).
It is easy to convert between the TRIP and TRIP tree sequences of a point (x, y) .D induced by a given Ts,t0,t1 : 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,... ).
1.6 Transfer Operators
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
8
0
11
Lf(x)= f ;
(n + x)2 n + x
n=1
under certain conditions on f(x), the largest eigenvalue of this transfer operator is 1, with
associated eigenfunction 1
h(x)=
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 de.nition). A transfer operator linearly maps functions de.ned on X in some vector space to functions de.ned on X in some vector space. For a more concrete de.nition, choose a function g : X . R. The transfer operator, call it LT , acting on f : X . R is given by
0
LT f(x)= g(y)f(y).
y:T (y)=x
1.8. Polynomial-and Non-Polynomial-Growth TRIP Maps
If T is di.erentiable, we usually choose g = 1 .
|Jac(T )|
Extending this de.nition to the case of TRIP maps, we de.ne our transfer operators to
be
0
1
LTs,t0,t1 f(x, y)= f(a, b).
|Jac(Ts,t0,t1 (a, b))|
(a,b):Ts,t0,t1 (a,b)=(x,y)
As an example,
8
0
11 x
f(x, y)= f,.
LTe,e,e
(1 + kx + y)3 1+ kx + y 1+ kx + y
k=0
We have calculated the explicit form of LTe,e,e f(x, y) for all (s, t0,t1) . S33; these are presented in Appendix E. A sample calculation of the explicit form of a transfer operator is presented in Chapter 6.
1.7 Interlude
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.
1.8 Polynomial-and Non-Polynomial-Growth TRIP Maps
We see that the transfer operator corresponding to Te,e,e has a particularly nice form, in that the denominator of the factor 1 is (non-trivially) polynomial in k. In general, we will
(1+kx+y)3 be concerned with the form of the transfer operator LTs,t0,t1 where
0
1
LTs,t0,t1 f(x, y)= f(a, b)
|Jac(Ts,t0,t1 (a, b))|
(a,b):Ts,t0,t1 (a,b)=(x,y)
If the denominator of 1 is (non-trivially) polynomial in k (also allowing for
Jac(Ts,t0,t1 (a,b))
factors of (-1)k) for a given Ts,t0,t1 , then that Ts,t0,t1 gives rise to a polynomial-growth
1.9. Combo TRIP Maps and Polynomial-Growth
transfer operator, and is itself polynomial-growth; otherwise, both LTs,t0,t1 and Ts,t0,t1 are non-polynomial-growth.
By direct calculation of the explicit form of the transfer operators, presented in Appendix E, we have shown that exactly half of the 216 TRIP maps are polynomial-growth. In addition, we identi.ed the origin of polynomial-growth by showing that that a TRIP map Ts,t0,t1 is polynomial-growth if and only if the eigenvalues of the associated F1 = sA1t1 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
8
representations of the partitions k=0 Dk induced on D by Ts,t0,t1 .
1.9 Combo TRIP Maps and Polynomial-Growth
We obtain a larger class of maps by allowing compositions of TRIP maps [2]. As an example, we might perform the .rst division of D using the permutation triplet (s, t0,t1), the next using the permutation triplet (s2,t02,t12), 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 . ., with . some indexing set such that . .{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 . ., and exponential in all kl such that l .{1, 2,...,i}\ ..
We have also identi.ed the origin of polynomial-growth in such arbitrary .nite composi
-1 -1
F -k1 TT
B-1 F -k2 B-1
tions of TRIP maps. Let M1 = BF -1 , M2 = BF -1 , and
01 01 -1
so on until Mi = BF 0 -1F1 -ki B-1 T . Then the composition T1 .T2 .....Ti is polynomial-growth in kj,j .{1, 2,...,i} if and only if the eigenvalues of F1j all have magnitude 1. This result is presented in Chapter 9.
1.11. Ergodic Theory
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 .nding an appropriate Banach space V on which LT acts, 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, de.ned by the property that for any measurable set A .D, and for any function f . L1(µ)
LT,µf(x, y)dµ = f(x, y)dµ,
AT -1A
where µ is the measure de.ned by the eigenfunction of eigenvalue 1 of the corresponding 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.
1.11 Ergodic Theory
This section serves as a very brief introduction to ergodic theory, exactly following the outline provided in [7], which in turn uses [17] as a reference.
Consider a set X and a collection of subsets of X, S. S is said to be a s-algebra if it
1.12. Ergodicity of TRIP Maps
satis.es the following three properties: 1. X .S, 2. If A .S, then Ac .S, and 3. If A1,A2,A3, ··· . S, then .8 .S.
n=1An A measure on S is a function µ : S. [0, 8) where we require that 1. µ(Ř)=0, and 2.
8
If A1,A2,A3, ···.S are pairwise disjoint, then µ (.n8 =1An)=n=1 µ(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 (D, B,.), where B stands for Borel s-algebra and . 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 de.ne 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.
1.12 Ergodicity of TRIP Maps
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 convergence for combo TRIP maps and show in Chapter 15 that no TRIP sequence induced by 24 TRIP maps corresponds to a unique point.
1.14. Computational Methodology
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.
1.14 Computational Methodology
Mathematica was used to perform almost all of the calculations in this thesis; a discussion regarding the general research approach, as well as more details regarding the use of Mathematica, are found in Chapter 17.
Chapter 2 Explicit Form of TRIP Maps
The explicit form of Ts,t0,t1 (x, y) has been calculated for all (s, t0,t1) . S33 . 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.
2.1 Sample TRIP Map Calculation
We will go through a sample calculation of the from of Te,23,e(x, y). By de.nition, for (x, y) .D,
Te,23,e(x, y)= p((1, x, y) BF 0 -1F1 -kB-1 T ).
Here,
..
010
..
F0 =(e)A0(23)= 100 , 011
so that .. 0 10
F -1
0 = . 1 00 .; -101
further,
..
101
..
F1 =(e)A1(e)= 010 , 001
so that . .
1 0 -1
F -1 1 = . 0 1 0 .,
0 0 1
2.1. Sample TRIP Map Calculation
and .. 10 -k F1 -k = . 01 0 . . 00 1
We also have that .. 1 -10 B-1 = . 01 -1 .. 00 1
From this, it follows that
. .
0 1 0
BF -1 0 F -k 1 B-1 = . 0 0 1 . ,
-1 1 k + 1
so that
F -kB-1 T
(1, x, y) BF 0 -11 =(x, y, (k + 1)y + x - 1),
and hence
T y (k + 1)y + x - 1
F -kB-1
Te,23,e(x, y)= p((1, x, y) BF 0 -11 )= ,.
xx
Chapter 3
Explicit Form of TRIP Map Inverses
The explicit form of T -1 (x, y) has been calculated for all (s, t0,t1) . S33 . These explicit
s,t0,t1
forms are presented in Appendix B. These were calculated from the de.nition
T -1 F -kB-1 T -1 (x, y)= p((1, x, y) BF -1 ).
s,t0,t1 01
The explicit form associated with Te,e,e had already been calculated in [5], but here we present
all 216 explicit forms.
3.1 Sample TRIP Map Inverse Calculation
We will go through a sample calculation of the from of T -1 (x, y).
e,23,e
We have that
T -1 F -kB-1 T -1 e,23,e(x, y)= p((1, x, y) BF 0 -11 ).
From Chapter 2, we know that
..
01 0 BF 0 -1F1 -kB-1 = . 00 1 . , -11 k +1
so that .. 1 10
-1 BF 0 -1F1 -kB-1 T = . k +1 0 1 .,
-1 00
and hence
-1
T 1 x
F -kB-1
p((1, x, y) BF -1 )= ,.
01 (k + 1)x - y +1(k + 1)x - y +1
Chapter 4
Recurrence Relations for TRIP Map Orbits
Given a point (y1(ak),y2(ak)) .D and the TRIP sequence associated with that point, (ak+1,ak+2 ... ) induced by a TRIP map Ts,t0,t1 , it is possible to derive recurrence relations for (xn,yn) in terms of (y1(ak+1),y2(ak+1)) = Ts,t0,t1 (y1(ak),y2(ak)) and ak+1. These recurrence relations have been calculated for all polynomial-growth (and a select few non-polynomialgrowth) (s, t0,t1) . S33 and are presented in Appendix C.
Let us explain the process in detail. For any (y1,y2) .D, consider the set {T ak (y1,y2)}8
k=0, where a0 =0, and set (y1,y2)=(y1(a0),y2(a0)), so that, for k = 0,
(y1(ak+1),y2(ak+1)) = T (y1(ak),y2(ak));
note that
(y1(ak), y2(ak)) = x2(ak) x1(ak), x3(ak) x1(ak) ,
where
F -kB-1 T
(x1(ak+1),x2(ak+1),x3(ak+1)) = (x1(ak),x2(ak),x3(ak)) BF 0 -11 .
These relations allow us to solve for y1(ak) and y2(ak) in terms of y1(ak+1),y2(ak+1), and ak+1.
4.1. Sample Recurrence Relation Calculation
4.1 Sample Recurrence Relation Calculation
Let us walk through a sample calculation for (e, e, e). In this case, we have that
F -kB-1 T
(x1(ak+1),x2(ak+1),x3(ak+1)) = (x1(ak),x2(ak),x3(ak)) BF 0 -11
=(x2(ak),x3(ak),x1(ak) - x2(ak) - ak+1x3(ak))
so that (y1(ak),y2(ak))
x1(ak+1) x1(ak+1)y1(ak+1)
is equivalent to , ,
x1(ak+1)+ak+1x1(ak+1)y1(ak+1)+x1(ak+1)y2(ak+1) x1(ak+1)+ak+1x1(ak+1)y1(ak+1)+x1(ak+1)y2(ak+1)
which simpli.es to
1 y1(ak+1) ak+1y1(ak+1)+ y2(ak+1)+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
( a1 1 , a21 a3y1(a3)+y2(a3)+1
+ +1
a3y1(a3)+y2(a3)+1 + +1 a3y1(a3)+y2(a3)+ +a2+1
y2(a3) 1 y2(a3)
++a3 1
++a3y1(a3) y1(a3) y1(a3) y1(a3)
1
a21
+ +1 ).
a3y1(a3)+y2(a3)+1 y2(a3) 1
++a3
a2 1 y1(a3) y1(a3)
++ + a1 +1
y2(a3) 1 a3y1(a3)+y2(a3)+1
a3y1(a3)+y2(a3)+1
++a3 a3y1(a3)+y2(a3)+ +a2+1
y1(a3) y1(a3) y2(a3)+ 1 +a3y1(a3) y1(a3)
4.1. Sample Recurrence Relation Calculation
We have thus expressed the original input pair (y1(a0),y2(a0)) .D in terms of the .rst 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.
Chapter 5
Partition Diagrams and TRIP Diagrams
De.ne a partition diagram as a visual representation of {Dk}8 induced on D by a TRIP
k=0
map. Given a TRIP sequence (a1,a2,... ) induced by some TRIP map, de.ne a TRIP diagram as a visual representation of triangles of the form BF 1 a1 F0F1 a2 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 D. 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.
5.1 Sample Partition and TRIP Diagram Calculation
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 Dk 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 Dk. In
.. ..
112 123 particular, here we have that BF 0F0 = . 111 . , BF 1F0 = . 111 ., BF 2F0 =
111
011 011
.. .. ..
134 145 156 . 111 ., BF 13F0 = . 111 . , BF 14F0 = . 111 . ; projecting the columns
011 011 011
5.1. Sample Partition and TRIP Diagram Calculation
x2 x3
of each matrix using p(x1,x2,x3) . ,, we obtain the vertices of triangles Dk,k .
x1 x1
{0, 1, 2, 3, 4} shown in the diagram below:
Now let us show how to construct the TRIP diagram. Recall that this is a visual repre
sentation of triangles of the form BF 1 a1 F0F1 a2 F0 .... Remembering that we can easily convert
between the TRIP sequence and the TRIP tree sequence, we can begin to get a feel for the
way the diagram looks by considering all possible strings of 0s and 1s of length n (to get
all possible TRIP tree sequences of length n), replacing the 0s by F0s and the 1s by F1s,
applying the resulting product to B, and projecting to columns of each matrix to R2 using
x2 x3
p(x1,x2,x3) . ,. Let us consider n =2. The matrices involved are
x1 x1
..
123
..
1. BF0F0 = 112 ,
111
..
113
..
2. BF0F1 = 112 ,
011
..
123
..
3. BF1F0 = 111 ,
011
..
113
4. BF1F1 = . 011 . ;
001
5.1. Sample Partition and TRIP Diagram Calculation
x2 x3
projecting the columns of each matrix using p(x1,x2,x3) . ,, we obtain the vertices
x1 x1
of triangles (labeled by the .rst two terms in the associated TRIP tree sequence) shown in the TRIP diagram below:
We can continue in a similar fashion to get more detail; for n =6, we obtain the TRIP diagram shown below:
Chapter 6
Explicit Form of all Transfer Operators LTs,t0,t1
We have calculated the explicit form of
0
1
LTs,t0,t1 f(x, y)= f(a, b)
|Jac(Ts,t0,t1 (a, b))|
(a,b):Ts,t0,t1 (a,b)=(x,y)
for all (s, t0,t1) . S33 . The forms of |Jac(Ts,t0,t1 (a, b))| and LTs,t0,t1 f(x, y) are presented in Appendices D and E. The explicit forms associated with Te,e,e had already been calculated in [5], but here we present the explicit forms associated with all 216 TRIP maps.
6.1 Sample Transfer Operator Calculation
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 LT23,23,23 f(x, y).
By direct calculation, we have that
x - y -1 + (2 + k)x +(-1 - k)y
T23,23,23 = ,
xx
and
11 - x
T -1
= ,.
23,23,23
1 + (1 + k)x - y 1 + (1 + k)x - y
6.1. Sample Transfer Operator Calculation
We are interested in the Jacobian of T23,23,23 with any x replaced by the .rst component of
T -1
23,23,23 and any y replaced by the second component of T -1 23,23,23. The Jacobian is calculated
as follows:
.x-y .x-y .x x .x xJac(T23,23,23) = det. -1+(2+k)x+(-1-k)y. -1+(2+k)x+(-1-k)y .x x .x x
y -1 = det x2 x
1+y+ky -(1+k) x2 x
=1/x3
1 )3
Substituting the .rst component of T -1 for x, the Jacobian becomes 1/(=
23,23,23 1+(1+k)x-y
(1 + (1 + k)x - y)3 , so that
11
= .
Jac(T23,23,23) (1+(1+ k)x - y)3
Hence, we have shown that
0
1
f(x, y)= f(a, b)
LT23,23,23
|Jac(T23,23,23(a, b))|
(a,b):T23,23,23(a,b)=(x,y)
8
0
1 11 - x
= f,.
(1 + (1 + k)x - y)3 1 + (1 + k)x - y 1 + (1 + k)x - y
k=0
Chapter 7
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):
CHAPTER 7. EIGENFUNCTIONS OF EIGENVALUE 1 FOR SELECT TRANSFER OPERATORS
7.1. Sample Eigenfunction Veri.cation
7.1 Sample Eigenfunction Veri.cation
Let us verify that f(x, y)= 1 is indeed an eigenfunction of eigenvalue 1 of LT23,23,23 . The
x(x-y+1)
veri.cations 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 .,
1
.(x +1) = .(x)+ .
x
Finding these eigenfunctions, even with the aid of Mathematica, required a lot of work. From the previous section, we know that
80
1 11 - x
f(x, y)= f,
LT23,23,23
(1 + (1 + k)x - y)3 1 + (1 + k)x - y 1 + (1 + k)x - y
k=0
Noting that we have to replace any x in f(x, y) with the .rst component of T -1
23,23,23 and any
y by the second component, we obtain
80
1 1+ x + kx - y
f(x, y)=
LT23,23,23
(1 + (1 + k)x - y)3
1 + (2 + k)x - y
k=0
80
1
=
(1 + x + kx - y)(1 + (2 + k)x - y)
k=0
80
11
= -
x(-1 - 2x - kx + y) x(-1 - x - kx + y)
k=0
1
== f(x, y),
x(x - y + 1)
as desired.
Chapter 8
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 .rst present a well-known result we will need later:
..
ab c
..
Theorem 1 Consider a vector v = (1, x, y) and a matrix M = def . Then Jac(p(vM)) = ghi
det(M) (vM(1,0,0)T )3 .
Proof
Note that
b + ex + hy c + fx + iy
p(vM)= , .
a + dx + gy a + dx + gy
By direct calculation,
. .
.b+ex+hy .b+ex+hy .x a+dx+gy .y a+dx+gy
. .
Jac(p(vM)) = det
.c+fx+iy .c+fx+iy .x a+dx+gy .y a+dx+gy
-ceg + bfg + cdh - afh - bdi + aei
=
(a + dx + gy)3 det(M)
= .
(vM(1, 0, 0)T )3
CHAPTER 8. ORIGIN OF POLYNOMIAL-GROWTH IN TRIP MAPS
We are concerned with the form of the transfer operator LTs,t0,t1 where
0
1
LTs,t0,t1 f(x, y)= f(a, b)
|Jac(Ts,t0,t1 (a, b))|
(a,b):Ts,t0,t1 (a,b)=(x,y)
If the denominator of 1 is (non-trivially) polynomial in k (also allowing for
Jac(Ts,t0,t1 (a,b)) factors of (-1)k) for a given Ts,t0,t1 , then that Ts,t0,t1 gives rise to a polynomial-growth transfer operator, and is itself polynomial-growth; otherwise, both LTs,t0,t1 and Ts,t0,t1 are non-polynomial-growth. Here we state and prove a theorem regarding the polynomial-growth 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 Ts,t0,t1 is polynomial-growth if and only if the eigenvalues of the associated F1 = sA1t1 all have magnitude 1; it is non-polynomial-growth otherwise.
Proof By the previous theorem and the de.nition of Ts,t0,t1 , it follows that 1 =
Jac(Ts,t0,t1 (a,b))
det(Ms,t0,t1 )
where
((1,x,y)Ms,t0,t1 (1,0,0)T )3
F -kB-1)T )-1
Ms,t0,t1 = ((BF 0 -11
and B is the 3-space representation of our base triangle. Since det (Ms,t0,t1 ) 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 Ts,t0,t1 is polynomial-growth if and only if the .rst column of Ms,t0,t1 contains terms polynomial in k.
For simplicity, write Ms,t0,t1 =(C1,C2,C3) , where Ci are the columns of Ms,t0,t1 . Note
that the inverse of a TRIP map, T -1 s,t0,t1 , can be written as
T -1 s,t0,t1 (x, y) = p (1, x, y) BF -1 0 F -k 1 B-1 T -1
= p ((1, x, y)Ms,t0,t1 )
(1, x, y) · C2 (1, x, y) · C3
= (1, x, y) · C1 , (1, x, y) · C1 .
CHAPTER 8. ORIGIN OF POLYNOMIAL-GROWTH IN TRIP MAPS
Note that for any (x, y) .D,T -1 (x, y) must be bounded as it must land back inside
s,t0,t1
D. Since the inverse of each TRIP map is bijective and since the choice of k depends on which Dk the original (x, y) lies in, the .rst column of Ms,t0,t1 must depend on k. Thus, to show that the .rst column of Ms,t0,t1 depends on k polynomially, it is su.cient to show that Ms,t0,t1 exhibits only polynomial dependence on k – for then, by the above argument, the .rst row of Ms,t0,t1 must necessarily exhibit only polynomial dependence on k.
Note that
F -kB-1)T )-1
Ms,t0,t1 = ((BF -1
01
=(B-1)T ((F1 -1)k)T (F0 -1)T BT -1
=(BT )-1((F0 -1)T )-1(((F1 -1)k)T )-1((B-1)T )-1
=(BT )-1F0 T (F1 T )kBT .
Let A =(F1)T . We can .nd the Jordan decomposition of A; i.e. we can write it as
A = PJP -1
where P is some invertible matrix of dimensions identical to those of A, and J is the Jordan normal form of A. Performing the Jordan decomposition for all 36 unique As we see that there exist only six unique Js, namely
..
110
..
J1 =0 1 0 001
..
-100
..
J2 =0 11 , 0 01
..
100
..
J3 =0 1 1 , 001
..
10 0
v
J4 = .0 21 (1 - 5) 0 .,
v
00 21 (1+ 5)
CHAPTER 8. ORIGIN OF POLYNOMIAL-GROWTH IN TRIP MAPS
J5 = D(roots(-1 - t2 + t3)),
and
J6 = D(roots(-1 - t + t3)).
where D(roots(-1 - t2 + t3)) corresponds to a three-by-three square matrix with diagonal
entries de.ned by the roots of -1 - t2 + t3 = 0; similarly for D(roots(-1 - t + t3)).
It can be shown that ..
1 k 0
(J1)k = .010.
001
. .
(-1)k 00 (J2)k = . 01 k., 0 01
..
100 (J3)k = .01 k., 001
. .
10 0
v
(J4)k = . 0(12 (1 - 5))k 0v . , 00(12 (1+ 5))k
(J5)k = D(roots(-1 - t2 + t3)k),
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 .rst column of Ms,t0,t1 depends polynomially on k, and hence the associated Ts,t0,t1 is polynomial-growth. Otherwise, the .rst column of Ms,t0,t1 depends exponentially on k, and hence the associated Ts,t0,t1 is non-polynomial-growth.
Assume T is polynomial-growth in k. Then the .rst column of M must depend strictly polynomially on k. From the above decomposition of M, we see that the only place where kdependance may enter the .rst column of M is through A.; hence, by the above argument the
CHAPTER 8. ORIGIN OF POLYNOMIAL-GROWTH IN TRIP MAPS
eigenvalues of F1 all have magnitude 1. Now assume the eigenvalues of F1 all have magnitude
1. Running above argument in reverse, we see that T must necessarily be polynomial-growth in k.
Corollary 3 A TRIP map Ts,t0,t1 is non-polynomial-growth if and only if the associated Ms,t0,t1 depends exponentially on k.
Proof This follows immediately from the form of J1 through J6 in Theorem 1. .
Corollary 4 A TRIP map Ts,t0,t1 is polynomial-growth if and only if the associated F1 = sA1t1 is not diagonalizable; it is non-polynomial-growth otherwise.
Proof This follows immediately from the form of J1 through J6 in Theorem 1.
Theorem 5 If a TRIP map Ts,t0,t1 is polynomial-growth, the only k dependence exhibited in the explicit form of Ts,t0,t1 , aside from factors of (-1)k , is polynomial in k. If a TRIP map Ts,t0,t1 is non-polynomial-growth, the only k dependence exhibited in the explicit form of Ts,t0,t1 , is exponential in k.
Proof
Note that in this theorem, by “dependance” we do not mean that each of the terms in the output pair is itself necessarily polynomial in k or exponential in k; we allow for each of the two terms in the output pair to be a fraction involving solely polynomial (aside from factors of (-1)k) k dependence in both the numerator and denominator, or a fraction involving solely exponential k dependence in both the numerator and denominator.
By de.nition, we have that
Ts,t0,t1 (x, y)= p (1, x, y) BF 0 -1F1 -kB-1 T = p (1, x, y)M-1
8.1. A Permutation Triplet Mapping that Preserves Polynomial-Growth
with M as de.ned in the proof of Theorem 2. Clearly, after doing out the matrix multiplication, we see that the k-dependence is found only in the term ((F1)T )-k . The Jordan-normal decomposition of ((F1)T )-1 will be the same as that of A-1 , with A as de.ned in the proof of Theorem 2. It follows that we can write ((F1)T )-1 = PJ-1P -1 , so that ((F1)T )-k = PJ-kP -1 . As per the proof of the theorem quoted above, there exist only 6 unique J matrices, namely, the matrices J1,...,J6. It is clear by direct computation that
J-k J-k J-k
1 ,J-k , and will contain polynomial k-dependence, while J-k,J-k , and 6 will con
23 45
tain exponential k-dependence. Since all polynomial-growth TRIP maps correspond to one of J1,J2, or J3 and all non-polynomial-growth TRIP maps correspond to one of J4,J5, or J6, the result follows immediately.
Theorem 6 A TRIP map Ts,t0,t1 is polynomial-growth if and only if, aside from factors of
1
(-1)k , the denominator of 1 3 , is linear in k.
Jac(Ts,t0,t1 (a,b))
Proof
This follows from direct examination; see Appendix D.
8.1 A Permutation Triplet Mapping that Preserves Polynomial-Growth
Theorem 7 Let s, t0,t1,. and . be three-by-three permutation matrices. If a TRIP map Ts,t0,t1 is polynomial-growth, then the TRIP map T.s,.,t1.-1 is also polynomial-growth; simi
larly, if Ts,t0,t1 is non-polynomial-growth, then T.s,.,t1.-1 is also non-polynomial-growth.
Proof For ease of notation let us set
(.s, ., t1.-1)= P.
8.1. A Permutation Triplet Mapping that Preserves Polynomial-Growth
The argument will rely on the forms of F0P and F1P ; naturally,
F0P = .sA0.
and F1P = .sA1t1.-1 = .F1.-1 . Again, we are concerned with the form of the transfer operator LTP where
0
1
f(x, y)= f(a, b).
LTP |Jac(TP (a, b))|
(a,b):TP (a,b)=(x,y) 1 det(MP )
We have already shown explicitly that = where
Jac(TP (a,b)) ((1,x,y)MP (1,0,0)T )3
F -kB-1)T )-1
MP =(BF -1
0P 1P
and B is the 3-space representation of our base triangle.
Note that
MP = ((BF -1 0P F -k 1P B-1)T )-1
= (B-1)T ((F -1 1P )k)T (F -1 0P )T BT -1
= (BT )-1((F -1 0P )T )-1(((F -1 1P )k)T )-1((B-1)T )-1
= (BT )-1F T 0P (F T 1P )kBT
= (BT )-1F T 0P (.F1.-1)kBT
= (BT )-1F T 0P .(F T 1 )k.-1BT .
Let A = F1 T . We can .nd the Jordan decomposition of A; i.e. we can write it as
A = NJN-1
where N is some invertible matrix of dimensions identical to those of A, and J is the Jordan normal form of A. These 36 A matrices are exactly the same as those addressed in the proof of the theorem regarding the origin of polynomial-growth; in that theorem it was shown that a TRIP map is polynomial-growth if and only if the diagonal entries of J (corresponding to
8.1. A Permutation Triplet Mapping that Preserves Polynomial-Growth
the eigenvalues of the associated A) all have magnitude 1; the TRIP map is non-polynomialgrowth otherwise. Since as demonstrated above the mapping
(s, t0,t1) . (.s, ., t1.-1)
does not change the form of the original A associated with Ts,t0,t1 ,T.s,.,t1.-1 will inherit (non-)polynomial-growth from Ts,t0,t1 .
Chapter 9
Origin of Polynomial-Growth in Combo TRIP Maps
We are concerned with the form of the transfer operator LT where
0
1
LT f(x, y)= f(a, b).
|Jac(T (a, b))|
(a,b):T (a,b)=(x,y)
and T is a .nite composition of n TRIP maps de.ned by
T = T1 . T2 .· ··. Ti .· ··. Tn
where, of course, each subscript is short for a permutation triplet. If the denominator of 1 is (non-trivially) polynomial in ki (also allowing for
Jac(T (a,b)) factors of (-1)ki ) then T gives rise to a transfer operator that is polynomial-growth in ki, and is itself polynomial-growth in ki; otherwise, T gives rise to a transfer operator that is non-polynomial-growth in ki, and is itself non-polynomial-growth. Here we state and prove a theorem regarding the polynomial-growth of combo TRIP maps, and present some corollaries. To avoid redundancy, we will refer to polynomial dependence on ki aside from factors of (-1)ki as polynomial dependence on ki or strictly polynomial dependence on ki.
Theorem 8 A combo TRIP map T = T1 . T2 .···. Ti .···. Tn is polynomial-growth in ki if and only if the eigenvalues of the associated F1i (the F1 matrix corresponding to Ti) all have magnitude 1; it is non-polynomial-growth in ki otherwise.
CHAPTER 9. ORIGIN OF POLYNOMIAL-GROWTH IN COMBO TRIP MAPS
1 det(M)
Proof We have already shown explicitly that = where
Jac(T (a,b)) ((1,x,y)M(1,0,0)T )3
M = M1M2 ...Mi ...Mn
and
F -kB-1)T )-1
Mi = ((BF 0 -11i ,
where B is the 3-space representation of our base triangle. Since det (M) is real and has magnitude 1, this implies that T is polynomial-growth in ki if and only if the .rst column of M contains terms polynomial in ki
For simplicity, let M =(C1,C2,C3) , where Ci are the columns of M. Recall that the inverse of the combo TRIP map, T -1 , can be written as
T -1(x, y) = p ((1, x, y)M)
(1, x, y) · C2 (1, x, y) · C3
= (1, x, y) · C1 , (1, x, y) · C1 .
Recall that for any (x, y) .D,T -1(x, y) must be bounded as it must land back inside D. Since the inverse of our combo TRIP map T is bijective and since the choice of ki depends on which Dki (with respect to TRIP map Ti found in composition) the original (x, y) lies in, the .rst column of M must depend on ki Thus, to show that the .rst column of M depends on ki polynomially, it is su.cient to show that M exhibits only polynomial dependence on ki – for then, by the above argument, the .rst row of M must necessarily exhibit only polynomial dependence on ki
Note that using a little linear algebra, we can simplify M by getting rid of the overall inverses associated with each of the Mi’s, allowing us to better see the dependence on ki. In
CHAPTER 9. ORIGIN OF POLYNOMIAL-GROWTH IN COMBO TRIP MAPS
particular, we can write
M =(BT )-1F0 T (F T )k1 BT (BT )-1F0 T (F T )k2 BT ...
11 12
(BT )-1F0 T (F T )ki BT
...
1i
(BT )-1F T (F T )kn BT
01n
=(BT )-1F T (F T )k1 F T (F T )k2 ...F T (F T )ki ...F T (F T )kn BT
011 012 01i 01n
Let Si =(F1i )T . We can .nd the Jordan decomposition of Si; i.e. we can write it as
Si = PJP -1
where P is some invertible matrix of dimensions identical to those of Si, and J is the Jordan
normal form of Si. Performing the Jordan decomposition for all 36 unique Si’s we see that
there exist only six unique J’s, namely
..
110
..
J1 =0 1 0 001
..
-100
..
J2 =0 11 , 0 01
..
100
..
J3 =0 1 1 , 001
..
10 0
v
= .0 1 (1 - 5) 0 .,
J42 v 00 21 (1+ 5)
J5 = D(roots(-1 - t2 + t3)),
and
J6 = D(roots(-1 - t + t3)).
where D(roots(-1 - t2 + t3)) corresponds to a three-by-three square matrix with diagonal entries de.ned by the roots of -1 - t2 + t3 = 0; similarly for D(roots(-1 - t + t3)).
CHAPTER 9. ORIGIN OF POLYNOMIAL-GROWTH IN COMBO TRIP MAPS
It can be shown that ..
1 k 0
(J1)k = .010.
001
. .
(-1)k 00 (J2)k = . 01 k., 0 01
..
100 (J3)k = .01 k., 001
. .
10 0
v
(J4)k = . 0(12 (1 - 5))k 0v . , 00(12 (1+ 5))k
(J5)k = D(roots(-1 - t2 + t3)k),
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 Si 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.
Assume T is polynomial-growth in ki. Then the .rst column of M must depend strictly polynomially on ki. From the above decomposition of M, we see that the only place where ki dependance may enter the .rst column of M is through Si; hence, by the above argument the eigenvalues of F1i all have magnitude 1. Now assume the eigenvalues of F1i all have magnitude
1. Running above argument in reverse, we see that T must necessarily be polynomial-growth in ki.
Corollary 9 A combo TRIP map T as de.ned in the above theorem is non-polynomialgrowth in ki if and only if the associated M depends exponentially on ki.
Proof This follows immediately from the form of J1 through J6 in Theorem 1.
CHAPTER 9. ORIGIN OF POLYNOMIAL-GROWTH IN COMBO TRIP MAPS
Corollary 10 A combo TRIP map T as de.ned in the above theorem is polynomial-growth in ki if and only if the associated F1i is not diagonalizable; it is non-polynomial-growth otherwise.
Proof This follows immediately from the form of J1 through J6 in Theorem 1.
Corollary 11 Given a composition of n TRIP maps T = T1.T2.· · ·.Ti.· · ·.Tn, assume that {Tj} where j . ., with . .{1, 2,...,n} are polynomial-growth, and the rest non-polynomialgrowth. Then T is polynomial-growth in all kj such that j . .; it is non-polynomial-growth in all other kh with h .{1, 2,...,n}\ ..
Proof
In the above theorem, we showed that T is polynomial-growth in ki if and only if the the associated M, as de.ned in that theorem, is (aside from factors of (-1)k) polynomial in ki. Now simply apply this result for all kj,j . ..
Chapter 10 Origin of Partition Geometry
Here we explain why partition diagrams look the way they do and present a theorem regarding the rate of decrease of the area of Dk as k increases.
Let us call the union of all the Dk the partition of D induced by Ts,t0,t1 . The geometry of the partition of D induced by Ts,t0,t1 , up to a rotation of the original pattern, may be determined directly by observing the Jordan-decomposed form of the associated (F1)T = (sA1t1)T .
Let A =(F1)T . Finding the Jordan decomposition involves writing
A = PJP -1
where P is some invertible matrix of dimensions identical to those of A, and J is the Jordan normal form of A. In particular, it can be shown that there exist only 6 unique J’s, namely
..
-100
..
J1 =0 11 , 0 01
..
110
..
J2 =0 1 0 , 001
..
100
..
J3 =0 1 1 , 001
. .
10 0
v
J4 = . 0 12 (1 - 5) 0 v ., 00 21 (1+ 5)
CHAPTER 10. ORIGIN OF PARTITION GEOMETRY
J5 = D(roots(-1 - t2 + t3)),
and
J6 = D(roots(-1 - t + t3)).
where D(roots(-1 - t2 + t3)) corresponds to a three-by-three square matrix with diagonal entries de.ned by the roots of -1 - t2 + t3 = 0; similarly for D(roots(-1 - t + t3)).
We have proved that a TRIP map Ts,t0,t1 is polynomial-growth if and only if the diagonal elements of the associated J (which correspond to the set of eigenvalues of the associated F1) all have magnitude 1. J1 through J3 correspond to polynomial-growth TRIP maps while J4 through J6 correspond to non-polynomial-growth TRIP maps. This allows us to separate the partitions of D induced by TRIP maps into two subsets: those corresponding to polynomial-growth TRIP maps, and those corresponding to non-polynomial-growth TRIP maps. Among the polynomial-growth partitions, there exist three unique subdivision patterns; similarly among the non-polynomial-growth partitions. Further, the subtriangles Dk in the partitions corresponding to J5 and J6 spiral inwards because two eigenvalues for each of J5 and J6 are imaginary.
As stated above, these subdivision patterns (up to a rotation of the pattern) arise directly from the form of the Jordan decomposition of the associated (F1)T . While J2 through J6 each lead to a distinct (up to rotations) partitions independent of the associated P s from the Jordan decomposition of the (F1)T s, J1 leads to two partitions. J2 and J3 lead to the same partition since they correspond to permutation triplets of the form (e, t0,e), (12,t0, 12), (13,t0, 13), (23,t0, 23), (132,t0, 123), and (123,t0, 132). Note that the permutation matrices 12, 13, and 23 are their own inverses, while the inverse of 123 is 132
– this implies that in the computation of the inverse image of any (x, y) .D under the TRIP map corresponding to any of the above-mentioned permutation triplets will obtain its (non-trivially-polynomial) k-dependence (again, allowing for factors of (-1)k) solely through the A1 matrix (refer to the calculations in the proof of the origin of polynomial-growth to easily verify this); hence, we expect to see the same partitions for J2 and J3.
CHAPTER 10. ORIGIN OF PARTITION GEOMETRY
Theorem 12 The Dk associated with polynomial-growth TRIP maps for which the Jordan-normal decomposition of F1 -1 involves J5 or J6 spiral inwards (see Figures 9.5 and 9.6) for increasing k.
Proof Two of the eigenvalues of J5 and of J6 are complex; no other Js have complex eigenvalues – the two complex eigenvalues are directly responsible for the inward spiraling of Dk with increasing k. .
Theorem 13 The area of Dk decreases polynomially in k for polynomial-growth TRIP maps, and exponentially in k for non-polynomial-growth TRIP maps.
Proof We know that Dk is given by applying the projection map p to the vertices of the matrix K = F1 kF0.
As per Beaver, et al. [1], the area of Dk, call it Ak, will satisfy
| det(K)| 1
Ak == 2r1r2r3 2r1r2r3
where r1, r2, and r3 are the elements of the .rst row of K. From the preceding proofs regarding the origin of polynomial growth, it is evident that the entries of F1 k exhibit polynomial growth in k if and only if that F1 gives rise to polynomial-growth TRIP maps; similarly, the entries of F1 k exhibit exponential growth in k if and only if that F1 gives rise to non-polynomial-growth TRIP maps. It is further evident that Ak must depend on k, and since Ak = 1 , the k
2r1r2r3 dependence must enter the area through the ri – which will exhibit polynomial growth in k if and only if the associated F1 gives rise to polynomial-growth maps, and exponential growth in k if and only if the associated F1 gives rise to non-polynomial-growth maps. Hence, the area of Dk decreases polynomially in k for polynomial-growth TRIP maps, and exponentially in k for non-polynomial-growth TRIP maps. .
The unique partitions (up to rotations) are presented below, categorized by the patterns described above.
CHAPTER 10. ORIGIN OF PARTITION GEOMETRY
Figure 10.1: J1
Figure 10.2: J1
CHAPTER 10. ORIGIN OF PARTITION GEOMETRY
Figure 10.3: J2 and J3
Figure 10.4: J4
CHAPTER 10. ORIGIN OF PARTITION GEOMETRY
Figure 10.5: J5
Figure 10.6: J6
Chapter 11
Functional Analysis Behind Transfer
Operators: Banach Space Approach
11.1 Transfer Operators as Linear Maps on Appropriate Banach Spaces
Before we consider the Banach space approach, we must present some well-known de.nitions. As per [15], a vector space B is said to be a Banach space under some norm if it is complete. An operator U is said to have a spectral gap if the di.erence between the two largest moduli of its eigenvalues is nonzero [18].
The goal here is to .nd a Banach space for select LTs,t0,t1 such that 1 is the largest eigenvalue and has multiplicity 1. We proceed by de.ning the appropriate Banach spaces and proving the above result directly; the argument relies on the knowledge of the eigenfunction of eigenvalue 1 for LTs,t0,t1 , as well as on the ergodicity of Ts,t0,t1 . The arguments follow the outline set up by Garrity in [4] for the original triangle map Te,e,e.
Let
V = {f(x, y): D. R : .C . R : |g(x, y)f(x, y)| < C, .(x, y) . D}.
V is a Banach space using the norm
lf(x, y)l = sup |g(x, y)f(x, y)|,
(x,y).D
where we require that f(x, y) is a continuous function.
11.1. Transfer Operators as Linear Maps on Appropriate Banach Spaces
11.1. Transfer Operators as Linear Maps on Appropriate Banach Spaces
11.1. Transfer Operators as Linear Maps on Appropriate Banach Spaces
11.1. Transfer Operators as Linear Maps on Appropriate Banach Spaces
Proof
First, linearity is obvious for all the operators LTX .
Let f(x, y) . V and say ||g(x, y)|f(x, y)| 0 and by hypothesis 0 = f(x, y) = w(x, y), then
|Jac(Ts,t0,t1 (a,b))|
0 =Ln f(x, y) =Ln w(x, y).
Ts,t0,t1 Ts,t0,t1
Theorem 16 The greatest eigenvalue of LTs,t0,t1 : V . V for every (s, t0,t1) . S3 for which
3
we have established that LTs,t0,t1 is a linear map from V to V and found a corresponding eigenfunction of eigenvalue 1 is 1.
Proof
Let f(x, y) . V and suppose that h(x, y) is an eigenfunction of eigenvalue 1 for LTs,t0,t1 . Further suppose that there exist constants 0 0 a.e., then S is ergodic.
Clearly, for us S = Ts,t0,t1 ,P = LTs,t0,t1 , and f* is the associated eigenfunction of eigenvalue 1.
Corollary 19 The permutation triplets (s, t0,t1) . S3 for which LTs,t0,t1 is an ergodic map
3
from V to V whose eigenfunction of eigenvalue 1 is known are (e, e, e), (e, 23,e), (23, e, 23), and (132, 12, 123). By the above theorem, the eigenvalue of each of these transfer operators
11.2. Spectral Gap Results
that has value 1 has multiplicity 1. The original argument for (e, e, e) was worked out by Garrity in [4]. Note that ergodicity is di.cult to prove, and has been worked out for Te,e,e in [14], and the rest in [7] and later sections of this thesis.
Chapter 12
Functional Analysis Behind Transfer
Operators: Hilbert Space Approach
Below we prove that transfer operators corresponding to select TRIP maps are nuclear of trace class zero using a Hilbert space approach; this is a direct analogue of work by Garrity in Section 4.2 of [4] for the transfer operator corresponding to the original triangle map.
Before we get into the technical details, let us .rst present some well-known de.nitions. As per [19], a vector space H is said to be a Hilbert space if for f, h . H we can de.ne an inner product (f, h) such that H under the norm ((h, h))1/2 is a metric space in which every Cauchy sequence converges to an element of H. An operator U : H . H acting on a Hilbert space H is said to be nuclear of trace class zero1 if for every f . H, we can write
8
0
Uf = .kek (ek,f) k=0
where {.k}
8
k=1
are the eigenvalues of U, the set {ek}
8
k=0
consists of orthonormal functions in
H, and where we require that
8
k=0
|.k| < 8. It immediately follows that if an operator U
is nuclear of trace class zero, its spectrum is discrete.
To get at the Hilbert space approach, we consider a transfer operator related to LT , namely LT,µ, de.ned by the property that for any measurable set A .D, and for any function f . L1(µ)
LT,µf(x, y)dµ = f(x, y)dµ,
AT -1A
1The wording of this particular version of the de.nition was inspired by Professor Thomas Garrity.
CHAPTER 12. FUNCTIONAL ANALYSIS BEHIND TRANSFER OPERATORS: HILBERT SPACE APPROACH
where dµ is the invariant measure de.ned by the eigenfunction of eigenvalue 1 of the corresponding LTs,t0,t1 .
The transfer operator LTs,t0,t1 ,µ has been calculated for all (s, t0,t1) for which we have been able to .nd an eigenfunction of eigenvalue 1 for LTs,t0,t1 ; these are listed in the table below, along with the appropriate invariant measure dµ. The constant appearing in the
invariant measure is chosen so that we obtain D dµ =1.
12.1. Nuclearity of Select Transfer Operators
We can consider two Hilbert spaces on the triangle of functions, L2(dxdy) and L2(dµ). It is clear that LTs,t0,t1 ,µ is a map from L2(dµ) to L2(dµ), while LTs,t0,t1 is a map from L2(dxdy) to L2(dxdy). Here we focus on examining LTs,t0,t1 ,µ : L2(dµ) . L2(dµ). The fact that LT and LT,µ can be related by a simple transformation shows that these transfer operators have identical spectra. Namely, let hs,t0,t1 (x, y) be the eigenfunction of eigenvalue 1 of LTs,t0,t1 . Then for all cases presented in the above table,
LTs,t0,t1 f(x, y)= hs,t0,t1 (x, y)LTs,t0,t1 ,µ(f(x, y)/hs,t0,t1 (x, y)).
This relation further implies, as can be checked by direct calculation, that the eigenfunction of eigenvalue 1 for LTs,t0,t1 ,µ is f(x, y)=1.
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.
12.1 Nuclearity of Select Transfer Operators
12.1.1 Statement of Nuclearity Theorem
First, de.ne a di.erential dm(s) by s
dm(s)= ds,
es - 1
and an inner product by
. 8
(f(x, s),g(x, s)) = f(x, s)g(x, s)dm(s).
0
Theorem 20 For all permutation triplets X listed in the table below for which we have found eigenfunctions of eigenvalue 1 for the associated transfer operator LTX , we show that the associated LTX ,µ can be written in the form
. 88
0 w
-s
LTX ,µf(x, y)= e(k+w) f(x, s)dm(s) 0 (k + w)2
k=0
8
0
= (f(x, s),.k(s))Ek(x, y) k=0
12.1. Nuclearity of Select Transfer Operators
where .k(s) and Ek(x, y) are suitably chosen functions. We further show that, for constant x,
.k(y),Ek(x, y) . L2(dm(y))
and
0
|.k|·|Ek| < 8 k=0
with respect to L2(dm(y)), thus proving, as per [4], that these transfer operators are nuclear of trace class zero in y.
Note that the proofs for each of the transfer operators are, excluding crucial technical details, almost identical. Below we outline a general framework for proving nuclearity of trace class 0. It is important to note that the argument below is a direct analogue and
8
generalization of work by Garrity in [4], which proves the result for the transfer operator corresponding to Te,e,e.
12.1.2 Proof of Nuclearity Theorem
Proof
Using functions j(x, v) and h(x, v), and l(x, y) as listed in the table at the end of this proof, de.ne the modi.ed Laplace transform
. 8
U(f(x, x - v)) = j(x, v) e -sh(x,v)f(x, s)dm(s)
s=0
and let w = l(x, y).
Suppose that f(x, y)= U(f(x, y)). Then by direct calculation,
. 8
8
0 w
-s
LTX ,µf(x, y)= e(k+w) f(x, s)dm(s) 0 (k + w)2
k=0
8
0
= (f(x, s),.k(s))Ek(x, y) k=0
64
12.1. Nuclearity of Select Transfer Operators
where
k-s
se
.k(s)= (k + 1)!
and
. 8
-t(w-1)Lk
Ek(x, y)= we 1(t)dm(t).
0
From [4], we know that
0
8(w + m - 1)k
Ek(x, y)= kw
(w + m)k+2
m=0
By direct calculation, since it has been checked that in all cases the sum converges, there exists a constant C1 such that
8
0 (w + m - 1)k
Ek(x, y)= kw .
.(C) 100
These results prove that the 2 additional TRIP maps mentioned above are ergodic.
13.1 Calculations Leading to Ergodicity
In [7], it is shown that a TRIP sequence converges under the projection p :(x1,x2,x3) . (x2/x1,x3/x1) if and only if the sequence does so under the projection p :(x1,x2,x3) .
x1 x2 x3
,, .
x1+x2+x3 x1+x2+x3 x1+x2+x3
Represent the vertices of the triangle D(a1,a2,...,an) by X1,X2, and X3. For j . {1, 2, 3}, let |Xj| represent the l1 norm of Xj. Further, let Xf,X2f, and Xfrepresent the
13
vertices of D(a1,a2,...,an,b), with b> 0, and let let X1ff,X2ff, and X3ffrepresent the vertices of D(a1,a2,...,an, b, 0).
We can measure the distances between triangle vertices under the p projection by de.ning
D(Xi,Xj)= dEucl(p(Xi) - p(Xj)),
where dEucl represents the Euclidean distance. Now we can de.ne the diameter dn of D(a1,a2,...,an) with respect to p to be
dn = max{D(X1,X2),D(X2,X3),D(X1,X3))}.
Let
C = D(a1,a2,...,an),
and let
8
R = D(a1,a2,...,an, 0) .D(a1,a2,...,an, b, 0). b=1
As per [14], quoted from [7], we have the following theorem (taken almost word-for-word):
Theorem 21 Let A be a nonnegative n×n matrix such that | det(A)| =1, and let A1,...,An be its columns. Then
v
n 2
.(LA(Dn- 1)) = ,
(n - 1)! |A1|···|An|
13.1. Calculations Leading to Ergodicity
where . refers to the Lebesgue measure, |.| refers to the l1-norm, LA is a map de.ned by
Ax
LA : x .Dn-1 . .Dn-1,
|Ax|
and
Dn-1 = {x . Rn : |x| =1}.
+
v
.(R) (3)
From this de.nition, we can compute the ratio , where .(C)= .
.(C)2|X1||X2||X3|
In Appendix H we present relations between (X1,X2,X3) and (X1f ,X2f ,X3f ), as well as relations between (X1,X2,X3) and (Xff,Xff,Xff) for all permutation triplets corresponding
123
to polynomial-growth TRIP maps; we also present analytical lower bounds for .(R) for all permutation triplets corresponding to polynomial-growth TRIP maps. Proving that limn.8 dn = 0 while simultaneously .nding a numerical lower bound for .(R)
(which is su.cient to show ergodicity) is much more involved and has been carried out
.(C)
for the 2 additional maps mentioned above. This has been done by proving certain relations between the norms of the vertices of the triangles and using these relations as assumptions for further calculations; these results are summarized in the table appearing below.
Theorem 22 The maps T23,e,23 and T132,12,123 are ergodic.
Proof As per the above table, we have shown that for each of the maps mentioned in the
.(R)1
theorem, dn+1 . By Section 5 in [7],
.(C) 100
it follows that these maps are ergodic. .
Chapter 14
In.nitely Many Zeroes Almost Everywhere for Combo TRIP Maps
Theorem 23 Let Ta = Ts,t0,t1 and Tb = Ts,t0,t* (same s and t0). Suppose Ta and Tb are
1
each such that almost every point in D corresponds to a TRIP sequence containing in.nitely many zeroes. Then almost every point in D corresponds to a TRIP sequence under Ta . Tb containing in.nitely many zeroes.
Proof Let n be odd and call the proportion of the area of D(a1,a2,...,an) taken up by subtriangles of the form D(a1,a2,...,an,b)
and D(a1,a2,...,an, b, 0)
.(Ra)
. Since
.(C) D(a1,a2,...,an,b)= XF 1baF0a
and D(a1,a2,...,an, b, 0) = XF 1baF0aF0a,
.(Ra)
is the same as in the proof that Ta induces a TRIP sequence containing in.nitely many
.(C) zeroes for almost any point in D; namely, it is nonzero. Similarly for Tb, and we are done.
Chapter 15
Non-Uniqueness for Select TRIP Maps
We prove that no TRIP sequence corresponds to a unique point for 24 TRIP maps. Note that except for crucial technical details, the proofs for each of the 24 maps are almost identical.
Theorem 24 No TRIP sequence induced by the map Te,12,e corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), note that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
ai 0 ßi F1 ai F0 = . 010 ..
101
By direct computation, this means that D(a0,a1,... ) is of the form
..
a 1 b D(a0,a1,... )= .c 1 d. . c 0 d This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (1, 0) while the others will be at some (x2,x2) and (x3,x3). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
Theorem 25 No TRIP sequence induced by the map Te,12,13 corresponds to a unique point. Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 .... By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
. .
ai 0 ßi
F ai 1 F0 = . 0 1 0 ..
.i 0 di
By direct computation, this means that D(a0,a1,... ) is of the form
..
a 1 b D(a0,a1,... )= .c 1 d. . c 0 d This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (1, 0) while the others will be at some (x2,x2) and (x3,x3). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 26 No TRIP sequence induced by the map Te,123,e corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
ai 0 ßi F1 ai F0 = . 010 ..
101
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
By direct computation, this means that D(a0,a1,... ) is of the form
. .
a 1 b
D(a0, a1, . . . ) = .c 1 d. .
c 0 d
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (1, 0) while the others will be at some (x2,x2) and (x3,x3). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 27 No TRIP sequence induced by the map Te,123,13 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
ai 0 ßi F ai ..
1 F0 =0 1 0 . .i 0 di
By direct computation, this means that D(a0,a1,... ) is of the form
..
a 1 b D(a0,a1,... )= .c 1 d. .
c 0 d
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (1, 0) while the others will be at some (x2,x2) and (x3,x3). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 28 No TRIP sequence induced by the map T12,e,12 corresponds to a unique point.
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 .... By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
. .
1 0 0
F ai 1 F0 = .0 ai ßi..
0 1 1
By direct computation, this means that D(a0,a1,... ) is of the form
..
1 ab D(a0,a1,... )= .0 ab. .
0 cd
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (0, 0) while the others will be at some (1,x2) and (1,x3). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 29 No TRIP sequence induced by the map T12,e,132 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
10 0
F ai ..
1 F0 =0 ai ßi . 0 .i di
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
By direct computation, this means that D(a0,a1,... ) is of the form
. .
1 a b
D(a0, a1, . . . ) = .0 a b. .
0 c d
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (0, 0) while the others will be at some (1,x2) and (1,x3). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 30 No TRIP sequence induced by the map T12,23,12 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
10 0
F ai ..
1 F0 =0 ai ßi . 01 1
By direct computation, this means that D(a0,a1,... ) is of the form
..
1 ab D(a0,a1,... )= .0 ab. .
0 cd
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (0, 0) while the others will be at some (1,x2) and (1,x3). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 31 No TRIP sequence induced by the map T12,23,132 corresponds to a unique point.
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 .... By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
. .
1 0 0
F ai 1 F0 = .0 ai ßi..
0 .i di
By direct computation, this means that D(a0,a1,... ) is of the form
..
1 ab
D(a0,a1,... )= . 0 ab. . 0 cd
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will
always be at (0, 0) while the others will be at some (1,x2) and (1,x3). From this it clearly
follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 32 No TRIP sequence induced by the map T13,12,e corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in
3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
ai 0 ßi F1 ai F0 = . 010 .. .i 0 di
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
By direct computation, this means that D(a0,a1,... ) is of the form
. .
a 1 b
D(a0, a1, . . . ) = .c 1 d. .
c 0 d
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (1, 0) while the others will be at some (x2,x2) and (x3,x3). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 33 No TRIP sequence induced by the map T13,12,13 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
101
F ai ..
1 F0 =0 1 0 . ai 0 ßi
By direct computation, this means that D(a0,a1,... ) is of the form
..
a 1 b D(a0,a1,... )= .c 1 d. .
c 0 d
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (1, 0) while the others will be at some (x2,x2) and (x3,x3). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 34 No TRIP sequence induced by the map T13,123,e corresponds to a unique point.
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 .... By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
. .
ai 0 ßi
F ai 1 F0 = . 0 1 0 ..
.i 0 di
By direct computation, this means that D(a0,a1,... ) is of the form
..
a 1 b
D(a0,a1,... )= . c 1 d. . c 0 d
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will
always be at (1, 0) while the others will be at some (x2,x2) and (x3,x3). From this it clearly
follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 35 No TRIP sequence induced by the map T13,123,13 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in
3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
101 F1 ai F0 = . 010 .. ai 0 ßi
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
By direct computation, this means that D(a0,a1,... ) is of the form
. .
a 1 b
D(a0, a1, . . . ) = .c 1 d. .
c 0 d
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (1, 0) while the others will be at some (x2,x2) and (x3,x3). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 36 No TRIP sequence induced by the map T23,13,23 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
ai ßi 0 F ai ..
1 F0 =1 10 . 0 01
By direct computation, this means that D(a0,a1,... ) is of the form
..
ab 1 D(a0,a1,... )= .cd 1. .
001
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (1, 1) while the others will be at some (x2, 0) and (x3, 0). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 37 No TRIP sequence induced by the map T23,13,123 corresponds to a unique point.
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 .... By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
. .
ai ßi 0
F ai 1 F0 = ..i di 0..
0 0 1
By direct computation, this means that D(a0,a1,... ) is of the form
..
ab 1
D(a0,a1,... )= . cd 1. . 001
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will
always be at (1, 1) while the others will be at some (x2, 0) and (x3, 0). From this it clearly
follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 38 No TRIP sequence induced by the map T23,132,23 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in
3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
ai ßi 0 F1 ai F0 = . 1 10.. 0 01
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
By direct computation, this means that D(a0,a1,... ) is of the form
. .
a b 1
D(a0, a1, . . . ) = .c d 1. .
0 0 1
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (1, 1) while the others will be at some (x2, 0) and (x3, 0). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 39 No TRIP sequence induced by the map T23,132,123 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
ai ßi 0 F ai ..
=0 .
1 F0 .i di
0 01 By direct computation, this means that D(a0,a1,... ) is of the form
..
ab 1 D(a0,a1,... )= .cd 1. . 001 This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (1, 1) while the others will be at some (x2, 0) and (x3, 0). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 40 No TRIP sequence induced by the map T123,e,12 corresponds to a unique point.
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 .... By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
. .
1 0 0
F ai 1 F0 = .0 ai ßi..
0 .i di
By direct computation, this means that D(a0,a1,... ) is of the form
..
1 ab
D(a0,a1,... )= . 0 ab. . 0 cd
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will
always be at (0, 0) while the others will be at some (1,x2) and (1,x3). From this it clearly
follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 41 No TRIP sequence induced by the map T123,e,132 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in
3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
10 0 F1 ai F0 = .01 1 .. 0 ai ßi
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
By direct computation, this means that D(a0,a1,... ) is of the form
. .
1 a b
D(a0, a1, . . . ) = .0 a b. .
0 c d
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (0, 0) while the others will be at some (1,x2) and (1,x3). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 42 No TRIP sequence induced by the map T123,23,12 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
10 0
F ai ..
1 F0 =0 ai ßi . 0 .i di
By direct computation, this means that D(a0,a1,... ) is of the form
..
1 ab D(a0,a1,... )= .0 ab. . 0 cd This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (0, 0) while the others will be at some (1,x2) and (1,x3). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 43 No TRIP sequence induced by the map T123,23,132 corresponds to a unique point.
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 .... By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
. .
1 0 0
F ai 1 F0 = .0 1 1 ..
0 ai ßi
By direct computation, this means that D(a0,a1,... ) is of the form
..
1 ab D(a0,a1,... )= .0 ab. .
0 cd
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will
always be at (0, 0) while the others will be at some (1,x2) and (1,x3). From this it clearly
follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 44 No TRIP sequence induced by the map T132,13,23 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in
3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
ai ßi 0 F ai ..
1 F0 = .i di 0 . 0 01
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
By direct computation, this means that D(a0,a1,... ) is of the form
. .
a b 1
D(a0, a1, . . . ) = .c d 1. .
0 0 1
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (1, 1) while the others will be at some (x2, 0) and (x3, 0). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 45 No TRIP sequence induced by the map T132,13,123 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
..
1 10
F ai ..
1 F0 = ai ßi 0 . 0 01
By direct computation, this means that D(a0,a1,... ) is of the form
..
ab 1 D(a0,a1,... )= .cd 1. .
001
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (1, 1) while the others will be at some (x2, 0) and (x3, 0). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
Theorem 46 No TRIP sequence induced by the map T132,132,23 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
. .
ai ßi 0
F ai 1 F0 = .di .i 0..
0 0 1
By direct computation, this means that D(a0,a1,... ) is of the form
..
ab 1 D(a0,a1,... )= .cd 1. .
001
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will always be at (1, 1) while the others will be at some (x2, 0) and (x3, 0). From this it clearly follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Theorem 47 No TRIP sequence induced by the map T132,132,123 corresponds to a unique point.
Proof
Given a TRIP sequence (a0,a1,... ), recall that the vertices of D(a0,a1,... ) are given (in 3-space representation) by the columns of the matrix
BF 1 a0 F0F1 a1 F0 ....
CHAPTER 15. NON-UNIQUENESS FOR SELECT TRIP MAPS
By direct calculation, we have that for every i .{0, 1, 2,... },F ai F0 is of the form
1
. .
1 1 0
F ai 1 F0 = .ai ßi 0..
0 0 1
By direct computation, this means that D(a0,a1,... ) is of the form
..
ab 1 D(a0,a1,... )= .cd 1. . 001
This implies that in the 2-space representation, one of the vertices of D(a0,a1,... ) will
always be at (1, 1) while the others will be at some (x2, 0) and (x3, 0). From this it clearly
follows that the vertices of D(a0,a1,... ) cannot converge to a unique point.
Chapter 16
Gauss-Kuzmin Distributions for TRIP Sequences
Let us .rst state the Birkho. ergodic theorem. As per [8] and [17], let (X, S, µ, T ) be a dynamical system. Suppose that µ(X) = 1 and let T be measure-preserving. If T is ergodic, then if f is an integrable function,
n-1.
0
lim 1 f(T i(x)) = f dµ
n.8 n
i=0
for almost every x . X.
For (x, y) .D, de.ne pn,k(x, y) as the proportion of integers ai in the TRIP sequence corresponding to (x, y) induced by Ts,t0,t1 that equal k for i .{1, 2,...,n}. Now de.ne p(k) = limn.8 pn,k(x, y), which clearly corresponds to the probability of seeing k as the ith term in the TRIP sequence.
Theorem 48 Let (x, y) .D and suppose Ts,t0,t1 is a TRIP map that is ergodic with respect
to the Lebesgue measure . and has associated invariant measure µ(A)= A h(x, y)dxdy for h(x, y) as shown in the table in Chapter 12. If µ and . are absolutely continuous, then
p(k)= Dk dµ for almost every (x, y) .D.
Proof
The argument below is an analogue of the one by Karpenkov for the Gauss map in [8]. Let (x, y) .D and suppose Ts,t0,t1 is a TRIP map that is ergodic with respect to the Lebesgue
CHAPTER 16. GAUSS-KUZMIN DISTRIBUTIONS FOR TRIP SEQUENCES
measure . and whose associated transfer operator LTs,t0,t1 has h(x, y) as its eigenfunction of eigenvalue 1. It is clear that
n-1
0
pn,k(x, y)= 1 (T i (x, y)).
IDk s,t0,t1
n
i=0
If Ts,t0,t1 is also known to be ergodic with respect to the measure µ(A)= A h(x, y)dxdy, by the Birkho. ergodic theorem
lim pn,k(x, y)= IDk dµ = dµ,
n.8 Dk
for almost every (x, y) .D. Thus Dk dµ would be the desired Gauss-Kuzmin distribution.
We will prove that if Ts,t0,t1 is ergodic with respect to ., it is also ergodic with respect to the measure µ de.ned above.
Let A .D be a measurable, strictly Ts,t0,t1 -invariant set. Then since Ts,t0,t1 is ergodic with respect to ., and since . and µ are absolutely continuous, it follows that .(A)=
0 or .(Ac)=0 .. IAd. = 0 or IAc d. =0 .. A dxdy = 0 or Ac dxdy =0 ..
A h dxdy = 0or Ac h dxdy =0 .. µ(A) or µ(Ac)=0 .. Ts,t0,t1 is ergodic with respect to µ.
-6Li2(1 )+p2-12 log2(2)
Corollary 49 For Te,e,e,p(0) = 4 p2 while for k> 0,p(k)=
611 1
p2 Li2 - Li2 + 4 log2(k + 1) - 2 log2 k+1 +1 - 2 log(k(k + 2)) log(k + 1) .
(k+1)2 (k+2)2
Proof
First, we know that Te,e,e is ergodic with respect to the Lebesgue measure. Since for Te,e,e, h(x, y)= p212 , it follows that µ and . are absolutely continuous. Hence, by Theorem 48,
x(1+y)
it follows that p(k)= Dk dµ. By direct calculation, it is evident that
. 1 x 1
.. 12 -6Li24 + p2 - 12 log2(2)
p(0) = dµ = dy dx = .
1
D02 1-x p2x(y + 1) p2
while for k> 0,
1-x 1 kk+1
.. 1 . .. x
12 12
p(k)= dµ = dy dx + dy dx
11-x 11-x
p2x(y + 1) p2x(y + 1)
Dkk+1 k+1 k+2 k+1
CHAPTER 16. GAUSS-KUZMIN DISTRIBUTIONS FOR TRIP SEQUENCES
611 1
= Li2 - Li2 + 4 log2(k + 1) - 2 log2 +1 - 2 log(k(k + 2)) log(k + 1) .
p2 (k+1)2 (k+2)2 k+1
Corollary 50 For Te,23,e,p(0) = 12 , while for k> 0,
1
k+1 6 k 6
.. x . 1 . 1-x
p(k)= 11-x dy dx + 11-x dy dx.
p2x(1-y) p2x(1-y)
k+2 k+1 k+1 k+1
Proof
First, we know that Te,23,e is ergodic with respect to the Lebesgue measure. Since for Te,23,e,h(x, y)= p26 , it follows that µ and . are absolutely continuous. Hence, by
x(1-y)
Theorem 48, it follows that p(k)= Dk dµ. By direct calculation, it is evident that
.. 1 . x
61
p(0) = dµ = dy dx = .
1 2
D0 1-x p2x(1 - y)2
1 k+1 6 k 6
... . 1-x
x . 1
while for k> 0,p(k)= dµ = 11-x p2dy dx + 11-x p2dy dx.
Dk x(1-y) x(1-y)
k+2 k+1 k+1 k+1
Chapter 17
Research Approach and Computational Methodology
The original goal of this thesis was to compute explicit forms of all TRIP maps and associated transfer operators, and to explore and determine the origin of polynomial-growth. This was accomplished via Mathematica programming. We have gone beyond this original goal, using the extra time to explore transfer operator functional analysis and TRIP map ergodicity, among other topics.
Essentially all calculations in this thesis have been carried out on Mathematica 8 or Mathematica 9. To ensure computational accuracy of the algorithms, select results were checked by hand or via deconstructed step-by-step versions of the original algorithms. Indeed, Mathematica has been an invaluable tool for applications such as .nding explicit forms of TRIP maps and associated transfer operators, and generating the partition and TRIP diagrams.
Further, most of the ergodicity and functional analysis results were obtained with the aid of Mathematica. For example, eigenfunctions for transfer operators were found via a Mathematica algorithm, as were appropriate Banach spaces for transfer operators. The appropriate changes of coordinates and forms of the Laplace transform in the Hilbert space argument were also found this way. An algorithm was developed for the very technical ergodicity argument that checked whether the argument could be applied to other permutation triplets. The non-uniqueness results for 24 TRIP maps were obtained by direct calculation
CHAPTER 17. RESEARCH APPROACH AND COMPUTATIONAL METHODOLOGY
in Mathematica.
Chapter 18
Conclusion
In this thesis, we have explored the functional analysis behind TRIP maps, examining their explicit forms – as well as convergence and ergodic properties – showing that no TRIP sequence corresponds to a unique point 24 maps two additional TRIP maps are ergodic, while also deriving Gauss-Kuzmin distributions associated with select TRIP maps. We have also examined the forms and spectral properties of transfer operators naturally arising from these TRIP maps, showing nuclearity of trace class zero and the existence of spectral gaps for select transfer operators. However, much room for exploration remains; potential avenues for future research include:
•
Find an alternative way to prove (non)-ergodicity for more TRIP maps – perhaps for whole classes of TRIP maps at once
•
Find eigenfunctions of eigenvalue 1 for (at least) all the polynomial-growth transfer operators. Given the complicated expressions for non-polynomial-growth transfer operators (see Appendix E), .nding eigenfunctions for them seems much harder
•
Prove/disprove the existence of spectral gaps for (at least) all the polynomial-growth transfer operators
•
Derive Gauss-Kuzmin distributions for TRIP sequences induced by more TRIP maps
•
Attempt to solve the Hermite problem using combo TRIP maps
Appendix A
Form of Ts,t0,t1(x, y)
A.1 Polynomial-Growth Maps
A.1. Polynomial-Growth Maps
A.1. Polynomial-Growth Maps
A.1. Polynomial-Growth Maps
A.1. Polynomial-Growth Maps
A.1. Polynomial-Growth Maps
A.1. Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2 Non-Polynomial-Growth Maps
Note that r1, r2, and r3 represent the roots of
x 3 - x - 1=0 while rs1, rs2, and rs3 represent the roots of
x 3 - x 2 - 1=0.
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
A.2. Non-Polynomial-Growth Maps
Appendix B
Form of T -1(x, y)
s,t0,t1
B.1 Polynomial-Growth Maps
B.1. Polynomial-Growth Maps
B.1. Polynomial-Growth Maps
B.1. Polynomial-Growth Maps
B.1. Polynomial-Growth Maps
B.1. Polynomial-Growth Maps
B.1. Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2 Non-Polynomial-Growth Maps
Note that r1, r2, and r3 represent the roots of
x 3 - x - 1=0 while rs1, rs2, and rs3 represent the roots of
x 3 - x 2 - 1=0.
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
B.2. Non-Polynomial-Growth Maps
Appendix C Recurrence Relations for
(y1(ak),y2(ak)) .D
C.1 Polynomial-Growth Maps
C.1. Polynomial-Growth Maps
C.1. Polynomial-Growth Maps
C.1. Polynomial-Growth Maps
C.1. Polynomial-Growth Maps
C.1. Polynomial-Growth Maps
C.1. Polynomial-Growth Maps
C.2. Select Non-Polynomial-Growth Maps
C.2 Select Non-Polynomial-Growth Maps
C.2. Select Non-Polynomial-Growth Maps
C.2. Select Non-Polynomial-Growth Maps
C.2. Select Non-Polynomial-Growth Maps
Appendix D
Form of |Jac(s, t0,t1)|
D.1 Polynomial-Growth Maps
D.1. Polynomial-Growth Maps
D.1. Polynomial-Growth Maps
D.1. Polynomial-Growth Maps
D.1. Polynomial-Growth Maps
D.1. Polynomial-Growth Maps
D.1. Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2 Non-Polynomial-Growth Maps
Note that r1, r2, and r3 represent the roots of
x 3 - x - 1=0 while rs1, rs2, and rs3 represent the roots of
x 3 - x 2 - 1=0.
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
D.2. Non-Polynomial-Growth Maps
Appendix E
Form of LTs,t0,t1 f(x, y)
E.1 Polynomial-Growth Maps
E.1. Polynomial-Growth Maps
E.1. Polynomial-Growth Maps
E.1. Polynomial-Growth Maps
E.1. Polynomial-Growth Maps
E.1. Polynomial-Growth Maps
E.1. Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2 Non-Polynomial-Growth Maps
Note that r1, r2, and r3 represent the roots of
x 3 - x - 1=0 while rs1, rs2, and rs3 represent the roots of
x 3 - x 2 - 1=0.
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
E.2. Non-Polynomial-Growth Maps
Appendix F Partition Diagrams
F.1 Polynomial-Growth Maps
F.1. Polynomial-Growth Maps
F.1. Polynomial-Growth Maps
F.1. Polynomial-Growth Maps
F.1. Polynomial-Growth Maps
F.1. Polynomial-Growth Maps
F.1. Polynomial-Growth Maps
F.1. Polynomial-Growth Maps
F.1. Polynomial-Growth Maps
F.1. Polynomial-Growth Maps
F.2. Non-Polynomial-Growth Maps
F.2 Non-Polynomial-Growth Maps
F.2. Non-Polynomial-Growth Maps
F.2. Non-Polynomial-Growth Maps
F.2. Non-Polynomial-Growth Maps
F.2. Non-Polynomial-Growth Maps
F.2. Non-Polynomial-Growth Maps
F.2. Non-Polynomial-Growth Maps
F.2. Non-Polynomial-Growth Maps
F.2. Non-Polynomial-Growth Maps
F.2. Non-Polynomial-Growth Maps
Appendix G TRIP Diagrams
G.1 Polynomial-Growth Maps
G.1. Polynomial-Growth Maps
G.1. Polynomial-Growth Maps
G.1. Polynomial-Growth Maps
G.1. Polynomial-Growth Maps
G.1. Polynomial-Growth Maps
G.1. Polynomial-Growth Maps
G.1. Polynomial-Growth Maps
G.1. Polynomial-Growth Maps
G.1. Polynomial-Growth Maps
G.2. Non-Polynomial-Growth Maps
G.2 Non-Polynomial-Growth Maps
G.2. Non-Polynomial-Growth Maps
G.2. Non-Polynomial-Growth Maps
G.2. Non-Polynomial-Growth Maps
G.2. Non-Polynomial-Growth Maps
G.2. Non-Polynomial-Growth Maps
G.2. Non-Polynomial-Growth Maps
G.2. Non-Polynomial-Growth Maps
G.2. Non-Polynomial-Growth Maps
G.2. Non-Polynomial-Growth Maps
Appendix H Calculations for Ergodicity Argument
APPENDIX H. CALCULATIONS FOR ERGODICITY ARGUMENT
APPENDIX H. CALCULATIONS FOR ERGODICITY ARGUMENT
APPENDIX H. CALCULATIONS FOR ERGODICITY ARGUMENT
APPENDIX H. CALCULATIONS FOR ERGODICITY ARGUMENT
APPENDIX H. CALCULATIONS FOR ERGODICITY ARGUMENT
APPENDIX H. CALCULATIONS FOR ERGODICITY ARGUMENT
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