CONTINUOUS COMBINATORICS of a LATTICE GRAPH in the CANTOR SPACE Edward Krohne

CONTINUOUS COMBINATORICS OF A LATTICE GRAPH IN THE CANTOR SPACE Edward Krohne

Dissertation Prepared for the Degree of DOCTOR OF PHILOSOPHY

UNIVERSITY OF NORTH TEXAS May 2016

APPROVED:

Su Gao, Major Professor and Chair of the Department of Mathematics Steve Jackson, Major Professor Charles Conley, Committee Member Krohne, Edward. Continuous Combinatorics of a Lattice Graph in the Cantor

Space. Doctor of Philosophy (Mathematics), May 2016, 85 pp., 28 figures, references, 4 titles.

We present a novel theorem of Borel Combinatorics that sheds light on the types of continuous functions that can be defined on the Cantor space. We specifically consider the part X=F(2 ) from the Cantor space, where the group G is the additive group of integer pairs ². That is, X is the set of aperiodic {0,1} labelings of the two- dimensional infinite lattice graph. We give X the Bernoulli shift action, and this action induces a graph on X in which each connected component is again a two-dimensional lattice graph. It is folklore that no continuous (indeed, Borel) function provides a two- coloring of the graph on X, despite the fact that any finite subgraph of X is bipartite.

Our main result offers a much more complete analysis of continuous functions on this space. We construct a countable collection of finite graphs, each consisting of twelve

"tiles", such that for any property P (such as "two-coloring") that is locally recognizable in the proper sense, a continuous function with property P exists on X if and only if a function with a corresponding property P' exists on one of the graphs in the collection. We present the theorem, and give several applications. Copyright 2016 by Edward Krohne

ii ACKNOWLEDGMENTS

I would like to thank Tamara Knox, a dear and old friend who has supported me through both a thesis and a dissertation, and from whom I have learned so much.

iii TABLE OF CONTENTS

Page

ACKNOWLEDGMENTS iii

LIST OF FIGURESv

CHAPTER 1 INTRODUCTION1 1.1. Basic Deﬁnitions and Notations3

1.2. The Borel and Continuous Chromatic Numbers of F (2Z) 14

CHAPTER 2 MAIN RESULT 19

2 2.1. Passing from F (2Z ) to Finite Graphs 22

2 Z 2.2. Passing from Γn,p,q to F (2 ) 30 2.3. Main Result in Summary 43

CHAPTER 3 EXAMPLES AND IMMEDIATE COROLLARIES 44

CHAPTER 4 UNDIRECTED GRAPH HOMOMORPHISMS 54

2 CHAPTER 5 SUBSETS OF F (2Z ) WITH CONTINUOUS GROUP ACTIONS 75

CHAPTER 6 CONCLUSIONS AND FUTURE WORK 83

BIBLIOGRAPHY 85

iv LIST OF FIGURES

Page Figure 1.1. An illustration of pullbacks.6

Figure 2.1. A simpliﬁed notation for a quotient Z2-graph of a Z2-graph consisting of rectangular grid graphs. 20

Figure 2.2. The ﬁrst four “tiles” of Γn,p,q. 21

Figure 2.3. The next four “tiles” of Γn,p,q. 22

Figure 2.4. The long horizontal “tiles” of Γn,p,q. 23

Figure 2.5. The long vertical “tiles” of Γn,p,q. 24 Figure 2.6. Constructing the portion of the hyper-aperiodic point that shows how to

0 deﬁne ϕ on Tca=ac. 26 Figure 2.7. Constructing the portion of the hyper-aperiodic point that shows how to

color Tdca=acd. 27 Figure 2.8. Edge boundaries for marker regions. 33

Figure 2.9. “Downward” propagation of an instance of Rd using tile Tda=ad through a

region mostly tiled with tile Tca=ac. 34

Figure 2.10. “Diagonal” propagation of an instance of Rd using tile Tdca=acd, through a

region mostly tiled with Tca=ac 35 Figure 2.11. An illustration of the construction of R 36

Figure 3.1. The four-coloring of Γ1,2,3. 44

Figure 3.2. The ﬁve-edge-coloring of Γ1,2,3. 48

Figure 3.3. A function of ϕ:Γ1,5,2 → 2 inducing a continuous function

2 2 Z ϕF (2Z ) : F (2 ) → 2 which has no inﬁnite monochromatic graph component. 49 Figure 3.4. Toast. 50

Figure 4.1. The complete graph of order 3, K3, with a weighting function. 61

v Figure 4.2. The Petersen graph ΓP . 62

Figure 4.3. Steve Jackson’s “Clamshell” graph ΓJ . 63

∗ Figure 4.4. A graph ΓK such that ΓK is a cell-complex homeomorphic to the Klein bottle. 64 Figure 4.5. An illustration of how to “connect” two copies of γ using a winding path. 68 Figure 4.6. A process for eliminating spurs in a cycle one at a time. 69 Figure 4.7. Construction of a hub. 70 Figure 4.8. The Chv´atalGraph. 71 Figure 4.9. The Gr¨otzsch Graph. 71

Figure 4.10. A rectangular grid graph Tr ⊆ Γ establishing the homotopy group relation

∗ r = 1 in π1(Γ ). 73

2 Figure 5.1. On a Z -pullback ϕZ2 of ϕΓ, with ϕZ2 ◦ σ = ϕΓ. 77 Figure 5.2. Almost lined-up marker regions. 81

vi CHAPTER 1

INTRODUCTION

We are interested in viewing familiar graph theoretic questions such as chromatic number and edge chromatic number under the lens of definability. For example, it is a simple fact that any finite (indeed, any well-ordered) acyclic graph has chromatic number at most two, i.e., each node in the graph can be assigned a color from a palette of two colors such that no two adjacent nodes have the same color. Of interest in this work is the fact that, if the nodes of the acyclic graph are drawn from a standard Borel space, even in a sensible way, the two-coloring function may not be reasonably definable. The two-coloring may be at least non-Borel, and it may not be clear that a two-coloring exists at all without the well-ordering principle.

This research grew out of the study of Borel equivalence relations on standard Borel spaces, e.g., [4]. One powerful result in this area is that of Feldman-Moore [1], which states that any countable Borel equivalence relation (i.e., any equivalence relation in which the equivalences are countable) on a standard Borel space X is the orbit equivalence relation of some Borel action of some countable group on X. That is, broadly speaking, countable Borel equivalence relations can be studied one group at a time.

We may begin this program of study with the space {0, 1}Z = 2Z (henceforth we denote k = {0, 1, . . . , k − 1}), under the product topology. This space is homeomorphic to

the Cantor space. Our action will be the action of Z acting on 2Z by the Bernoulli shift or 1 left-shift action “·”. That is, for x ∈ 2Z, g, h ∈ Z,

(g · x)(h) = x(g−1 · h).

Consider the orbits of this action on 2Z, where Z is considered a multiplicative group and generated by z. Each point x of 2Z is a map x: Z → 2. If x is periodic with period

1The left-shift action is so named because g appears on the left of h, not because it is a left action. Indeed, there is another action, called the right-shift action, which is nonetheless a left action as well; namely (g ·right x)(h) = x(h · g). All actions considered in this work will act on the left.

1 g ∈ Z, the orbit [x] of x is ﬁnite with cardinality equal to |k| with zk = g. If x is aperiodic, then Z acts freely on [x] and [x] is countably inﬁnite.

It is natural to consider the unit-shifts z±1 · x to be “adjacent” to x; this gives rise to

±1 a graph structure (Γ2Z , ∼) with x ∼ y iﬀ z · x = y, and we may study the properties of this

graph. Then each connected component of Γ2Z is simply an orbit in the action; finite orbits are cycles, and infinite orbits are infinite chains. Henceforth, we will abusively abbreviate

Z (Γ2Z , ∼) as 2 when such usage is unambiguous.

We are interested specifically in graph-theoretic behavior arising out of the requirement that all functions and constructions be definable; in our case either Borel or continuous. Therefore, we remove the finite graph components from consideration, and only consider the set F (2Z) of aperiodic points in 2Z. This leaves an acyclic graph, but it is folklore (the proof to be presented shortly) that there exists no Borel two-coloring of F (2Z); this justifies our decision to drop cycles from consideration as the acyclic components are quite interesting on their own. We also present a folklore construction of a three-coloring on F (2Z) which is not only Borel, it is continuous. Therefore both the Borel and continuous chromatic numbers of

F (2Z) are three.

As the case of F (2Z) is relatively well-understood, we focus our attention in this work

2 on the next most complicated group action, F (2Z ). Indeed, a continuous four-coloring of

2 F (2Z ) has long been known [2, Thm. 4.2], but the question of the existence of a Borel or continuous three-coloring remained open [2,4]; the present work gives a proof of nonexistence

2 of a continuous three-coloring of F (2Z ). In addressing this problem, we needed to consider intricate intermediate structures, called marker structures, which serve as a “scaﬀolding” for building Borel or continuous functions. The questions of chromatic number, existence of particular ﬂavors of marker structures, and existence of other combinatorial constructions on

2 F (2Z ) turn out to be tightly related, especially in the continuous case. Our main theorem

2 unifies these ideas for the continuous case of F (2Z ). Specifically, given a discrete space D and a property P that can be witnessed locally in the proper sense (e.g., the property of being a k-coloring) our theorem provides a necessary and sufficient condition for the existence of

2 2 a continuous map ϕ: F (2Z ) → D with property P .

We note this result also is useful for purely Borel questions. Indeed, when attempting

2 to build a Borel map ϕ: F (2Z ) → X for some space X with property P , in practice one will typically build a collection of intermediate continuous maps and take one or more limits.

2 Thus, characterizing the continuous maps with property P from F (2Z ) constrains the search space for a Borel map with property P , and oﬀers insights into the structural idiosyncrasies of such a map. We are also optimistic that the main theorem can be generalized to spaces

2 beyond F (2Z ); this would provide insight into the general structure of Borel and continuous constructions on standard Borel spaces.

Before proceeding to the main result in Chapter2, we will establish our notation and

2 basic terms, and then warm up to F (2Z ) by proving that the continuous chromatic number of F (2Z) is three. We will then give the main result, and in the remainder of the work explore its consequences.

1.1. Basic Deﬁnitions and Notations

2 This paper will concentrate chieﬂy on the topological space F (2Z ). Nonetheless, we hope to generalize these results to arbitrary countable groups G acting on spaces F (2G), so we present our deﬁnitions in full generality. Thus for consistency, we take any group G for which we intend to study the shift action on 2G as a multiplicative group, even when G is

Abelian. We will use 1G for the identity of multiplicative groups G and 0G for the identity of additive groups G; we sometimes abbreviate these to 1 and 0 when no confusion will arise.

We are interested only in deﬁnable constructions, so unless otherwise noted we will fastidiously avoid the Axiom of Choice (AC). We will, however, freely use the Axiom of Countable Choice (ACω).

2 We ﬁrst extend the notion of “graph” slightly to account for the fact that in F (2Z ), there is a natural notion of not only adjacency but also “above”, “below”, “left”, and “right”. In general, for F (2G), orientation is given in terms of the generators of G. The following deﬁnition captures this directionality.

3 Definition 1.1. Let G be a group, and ﬁx a generator set G0 ⊆ G. For simplicity, we will use antisymmetric generator sets unless stated otherwise. If Γ is a directed multigraph whose edges are labeled with generators from G0, we call Γ a G-graph. We call the labels of the G-graph orientation labels. The generator set G0 will typically be clear from the group

G, and we will often omit mention of G0. If there is an edge from u to v with label g in Γ, g g−1 we write u −→ v and v −→ u.

In this deﬁnition, the −→ relation need be neither transitive nor reﬂexive: we do not

1G g1g0 g0 g1 in general have that u −→ u or u −→ w when u −→ v and v −→ w. Indeed, neither 1G nor

g1g0 need be included in the generating set.

Note that we consider a subgraph of a G-graph to inherit orientation labels, therefore we write simply “subgraph” instead of the awkward “sub-G-graph”.

Definition 1.2. If Γ, ∆ are G-graphs and ϕ:Γ → ∆ is a graph homomorphism that g g preserves labels, (i.e., ϕ(u) −→ ϕ(v) whenever u −→ v), we call ϕ a homomorphism of G-graphs.

2 z0 In the case of a Z -graph (with generators z0, z1), we say v is above u if u −→ v, v is to the right of u if u −→z1 v, v is below u if u is above v, and v is to the left of u if u is to the right of v. When illustrating Z2-graphs, we will omit explicit orientation labels to reduce clutter. Rather, we will indicate directionality by relative location in the picture: a node u above v will be drawn immediately above v.

Observe that G-graphs generalize directed graphs in the following sense: for any directed graph, we may canonically produce a Z-graph simply by labeling all edges with the sole generator z of Z. For this reason, we will refer to directed graphs as Z-graphs and use the term “undirected graph” to refer to any graph that is not a G-graph.

Further observe that G-graphs also generalize left actions in the following sense: if X g is a set with a left G-action , there is a canonical G-graph ΓX on X given by x −→ y iﬀ

g x = y for g ∈ G0. Where usage is clear, we abuse terminology and use X to refer to ΓX .

4 Definition 1.3. For any group G, the Cayley graph of G is the canonical G-graph induced by the action of G on itself. Again we abusively refer to this graph as simply G.

We recall the deﬁnition of a k-coloring, and extend it to the Borel and continuous cases.

Definition 1.4. Given a graph Γ (G-graph or undirected), a k-coloring ϕ:Γ → k (where k = {0, 1, 2, . . . , k − 1}) is any function such that whenever u and v are adjacent, ϕ(u) 6= ϕ(v). The chromatic number χ(Γ) of Γ is the least k such that a k-coloring exists on Γ. The Borel chromatic number χB(Γ) and continuous chromatic number χc(Γ) are respectively the least k such that there exists a Borel or continuous k-coloring on Γ.

Past work on F (22m ) has made extensive use of restrictions of functions on F (22m ) to j × k “boxes”. We will need to speak at length of analogous structures on general G graphs, so we formalize the notion of box, then generalize the notion of restriction of a function to a “pullback”, given shortly.

Definition 1.5. For two integers j, k, a j × k grid graph (or more tersely, a j × k box) is any rectangular full subgraph of the Cayley graph of Z2 which has exactly j nodes on each side and exactly k nodes on each of the top and bottom. Such a graph thus has j − 1 edges on each side and k − 1 edges on each of the top and bottom.

Definition 1.6. Let G be a group, S be a set, Γ, ∆ be G-graphs, and ϕ:Γ → S be a function. A ∆-pullback of ϕ is a function ψ : ∆ → S such that there is a homomorphism of G-graphs σ : ∆ → Γ with ψ = ϕ ◦ σ. That is, the following diagram commutes: ϕ Γ S

σ ψ ∆

In the case of G = Z2, we will use the term k-pullback of ϕ to refer to a Γ-pullback where Γ

is an k × k grid graph. If ψ is a Γ-pullback of ϕ, then we write ψ ≤Γ ϕ or ψ ≤ ϕ. Similarly,

if ψ is an k-pullback of ϕ, we write ψ ≤k ϕ.

5 1 2 1

2 0 2 1 2 1 3 4 3 1 2 1 2 2 ψ : ∆ → 5 4 0 4 1 2 1 3 4 3 3 4 3 ϕ:Γ → 5 4 0 4

3 4 3 ψ0 : ∆ → 5

Figure 1.1. An illustration of pullbacks. The graph Γ shown at left is a Z2- graph, with orientation indicated by relative position on the page. The map ϕ admits exactly two ∆-pullbacks ψ and ψ0, which are shown at right.

Pullbacks generalize restrictions of functions in the following way. Suppose ∆ is a subgraph of some G-graph Γ, and ϕ:Γ → S is a G-graph homomorphism. Viewing ϕ as a set of pairs as in the formal deﬁnition of function, ϕ∆ ⊆ ϕ. But here ϕ∆ ≤ ϕ by the identity map id∆.

Example 1.7. Two pullbacks ψ, ψ0 of a function ϕ are presented in Figure 1.1. Here, the graphs ∆ and Γ are Z2-graphs.

Recall that according to our convention for orientation labels, the center node has two nodes to its left, each of which has a single node above it but no nodes to its own left, and so forth. Γ consists of two disjoint copies of ∆ modded out by their respective center

0 nodes. Correspondingly, there are exactly two ∆-pullbacks ψ, ψ ≤∆ ϕ. It is convenient to speak of pullbacks, because pullbacks witness various combinatorial properties. For example, in Figure 1.1, it is sufficient to examine the two ∆-pullbacks of ϕ to conclude that ϕ is indeed a five-coloring. Indeed, if some map θ :Γ → 5 were not a five-coloring, θ would have a ∆-pullback that was also not a five-coloring. This fact remains

6 true regardless of the structure of ∆ and Γ, so long as ∆ remains large enough. However, pullbacks may have behavior quite diﬀerent from restrictions.

Example 1.8. Consider the case of ϕ:Γ → S and ψ : ∆ → S with ψ ≤ ϕ, where Γ

2 Z 2 is the Cayley graph of ( 5Z) (i.e., a 5 × 5 torus graph) and ∆ is the Cayley graph of Z , both as Z2-graphs.

Here, unlike the case in Example 1.7, ∆ is much larger than Γ. Because of the Z2- graph structure, if ϕ:Γ → S is a map and ψ : ∆ → S is a pullback of Γ, then ψ is a

2 2 2 Z doubly periodic tiling of Z by ϕ. Indeed, if ψ = ϕ ◦ σ, where σ : Z →( 5Z) , then σ is determined entirely by σ(g) for any one g ∈ Z2, thus there are exactly 25 possible such σ. This phenomenon, where Z2-pullbacks of certain graphs are simply tilings, will be exploited later when building tilings. We return our attention now to graphs on Polish spaces. Let S be a set. Recall that in the usual product topology basis, a basic open set U of the space 2S is a set for which there is a ﬁnite list of factors Λ ⊆ S and an element x ∈ U such that

U = {x0 ∈ 2S : ∀λ ∈ Λ, x(λ) = x0(λ)}.

When S is countable, 2S is homeomorphic to the Cantor space 2ω and, in particular, is compact and Polish. For 2ω, the space of inﬁnite sequences of ones and zeroes, it is convenient instead to deﬁne basic open sets by initial strings. That is, a basic open set U is a set such that there is a word s: {0, 1, . . . , k − 1} → 2 of length k such that

U = {x ∈ 2S : s is an initial string of x}.

When S is a countable group G, we lack initial strings, so instead we typically use some convenient sequence of sets

1G ∈ Λ0 ⊆ Λ1 ⊆ Λ2 ⊆ · · ·

Sω m such that k=0 Λk = G. For Z , these sets are typically squares and cubes of increasing size centered at the origin.

7 To keep our main definitions consolidated in this section, we repeat our definitions of left-shift action, the free part F (2G), and our definition of the graph on 2G, this time for arbitrary G, while expanding upon the terminology. In particular, we recast many notions in terms of pullbacks for ease of use later.

Definition 1.9. For G a group, the left-shift or Bernoulli shift action of G on 2G is given by

(g · x)(h) = x(g−1h) for g, h ∈ G, x ∈ 2G.

For x ∈ 2G, we use [x] to refer to the orbit of x under G. The symbol “·” will refer speciﬁcally to the Bernoulli shift action; when multiplying two elements of a group G, we will consistently use juxtaposition instead.

2 When dealing with Z2 or 2Z we use the distance function ρ deﬁned as follows.

2 2 Definition 1.10. Let Z = hz0, z1i in the usual way. If u, v are each in Z or in some

2 Z j k common orbit of F (2 ), and u, v diﬀer by z0z1 , let ρ(u, v) = max{|j|, |k|}. That is, ρ is the

`∞ metric.

We will always use ρ when discussing the distance between points; in particular, we will have no use for the metric that generates the topology.

Definition 1.11. A set S ⊆ 2G is invariant if, for all g ∈ G, we have g ·S = S where g · S = {g · x : g ∈ G}.

Observe that the complement of an invariant set is invariant, and the intersection or union of a collection of invariant sets is invariant.

Definition 1.12. Let X be a space with an equivalence relation E, and let M ⊆ X. If, for every equivalence class e of E, there is a point m ∈ e ∩ M, then M is called a complete section.

8 Definition 1.13. Let G be a group. The free part of G, denoted F (2G), is the set of all x ∈ G such that, for each g ∈ G, if g · x = x, then g is the identity of G. In other words, F (2G) is the set of all aperiodic x ∈ 2G.

Consider a G-orbit [x] ⊆ F (2G). Because G acts freely on [x], there is a one-to-one correspondence between points in [x] and elements of G. Also, our G-graph structure on [x] and our G-graph structure on G itself are deﬁned based on G, so these graphs are G- graph isomorphic. F (2G) is just an uncountable union of orbits, so F (2G) shares this G-like structure locally.

It is rather easy to forget2 whether the domain of a particular function is G, as in some x ∈ 2G meaning x: G → 2, or whether the domain is F (2G), as in ϕ: F (2G) → 2. In the latter case, each point x in the domain of ϕ that is mapped to ϕ(x) ∈ 2 is itself a function in much the same way that a number r on the real line can be viewed as a sequence of digits. We will attempt to resolve some confusion by preferring the terms “element” for group elements of G, “point” for points of F (2G), and “vertex” for vertices of a G-graph, however these terms unavoidably overlap.

If G is countable, 2G is compact. However, F (2G) is certainly not compact. For example, consider the sequence of functions xk : Z → 2 (with Z = hzi) deﬁned by  1 if j < k, j  xk(z ) =  0 otherwise.

Each zk is not periodic, but lim(xk) is just the constant function of all ones, which is periodic. k→ω This observation is easily generalized to an arbitrary group G.

Nonetheless, we will be concerned with clopen sets and continuous maps, and compactness will be indispensable for these arguments. For this purpose, it helps to ﬁnd compact invariant sets within 2G. Such a set can be conveniently constructed as the orbit [x] of a point x ∈ 2G, where x has the following property.

2The present author has made this mistake repeatedly.

9 Definition 1.14. Let G be a group, and let x ∈ 2G. The point x is called hyper-

G aperiodic if Cl2G ([x]) ⊆ F (2 ).

G The fact that Cl2G ([x]) is a compact subset of F (2 ) follows from the fact that

G Cl2G ([x]) is a closed subset of the compact space 2 . Some authors, e.g., [3], use the term “two-coloring” to refer to hyper-aperiodic x ∈ 2G. The idea behind this terminology is that a hyper-aperiodic point x, being aperiodic in a strong sense, will have many g1, g2 ∈ G such that x(g1) 6= x(g2); this is reminiscent of the deﬁnition of “k-coloring” in the sense of chromatic number. However, since we will be discussing “k-coloring” in the chromatic sense quite frequently, we will use “k-coloring” exclusively to mean a function ϕ:Γ → k on some graph Γ taking adjacent nodes to diﬀerent values, and use “hyper-aperiodic point” to refer to a point x ∈ 2G giving rise to a compact subset of F (2G). In the remainder of this section we develop a convenient characterization of hyper- aperiodic points in the case that G is Abelian.

g Proposition 1.15. If G is Abelian and h −→ h0 for some generator g and some g h, h0 ∈ G, then th −→ th0 for any t ∈ G.

g Proof. Suppose h −→ h0 as above. Then h0 = gh and th0 = tgh. Because G is Abelian,

0 g 0 th = gth. Then th −→ th .

g Proposition 1.16. If G is Abelian and x −→ x0 for some generator g and some g x, x0 ∈ 2G, then t · x −→ t · x0 for any t ∈ G.

Proof. Fix t. The following are readily seen to be equivalent:

g x −→ x0,

g · x = x0,

∀h ∈ G x(g−1h) = x0(h),

∀h ∈ G x(g−1t−1h) = x0(t−1h),

10 ∀h ∈ G x((gt)−1h) = x0(t−1h) (by commutativity),

∀h ∈ G (gt · x)(h) = (t · x0)(h), and

gt · x = t · x0,

g 0 t · x −→ t · x .

Proposition 1.17. Suppose G is Abelian and x ∈ 2G. If ψ ≤ x is a pullback, then for any g ∈ G, ψ ≤ g · x.

Proof. Fix g ∈ G. Let Γ be the domain of ψ and let σ :Γ → G be a witnessing map such that x ◦ σ = ψ. We deﬁne σ0 :Γ → G as

σ0(v) = gσ(v)

for v ∈ Γ. We claim σ0 is a homomorphism of G-graphs. Indeed, suppose u −→h v in Γ. Because σ is a homomorphism of G-graphs, σ(u) −→h σ(v) in G. Because G is Abelian, gσ(u) −→h gσ(v). But this is just σ0(u) −→h σ0(v). We also have, for each u ∈ Γ,

(g · x) ◦ σ0(v) = x(g−1σ0(v))

= x(g−1gσ(v))

= x ◦ σ(v)

= ψ(h)

0 so σ witnesses that ψ ≤ g · x.

Proposition 1.18. If G is Abelian and U ⊆ 2G is open and invariant, then there

S G exists a collection of ﬁnite pullbacks Ψ such that U = ψ∈Ψ Uψ where Uψ = {x ∈ 2 : ψ ≤ x}.

Proof. Because U is open, there is a collection Ψ of ﬁnite partial functions ψ : G → 2 such

S 0 0 G that U = ψ∈Ψ Uψ where Uψ = {x ∈ 2 : ψ ⊆ x}. This diﬀers from the statement to be proven in that we have dropped the word “invariant” and replaced “pullbacks (≤)” with

11 “partial functions (⊆)”. That is, each ψ must appear in a speciﬁc place in x and not merely

G anywhere. By Proposition 1.17, for each ψ ∈ Ψ, the set Uψ = {x ∈ 2 : ψ ≤ x} is simply the

0 set of translates of elements in Uψ. By invariance of U, Uψ ⊆ U for each ψ ∈ Ψ, therefore S U = ψ∈Ψ Uψ.

Proposition 1.19. Suppose G is Abelian and F ⊆ 2G is a set. Let Ψ = {ψ : ∃x ∈ F, ψ ≤ x}. F is closed and invariant iﬀ, for each x ∈ 2G, whenever each ﬁnite pullback ψ ≤ x is a member of Ψ, x ∈ F .

Proof. Suppose first that F is closed and invariant and let x be such that each pullback ψ ≤ x is a member of Ψ. If x∈ / F , x is in the open invariant set 2G \ F , whence there is a pullback ψ ≤ x such that any x0 ∈ 2G with ψ ≤ x0 has x0 ∈/ F . But this is impossible because ψ ∈ Ψ, which means there is an x0 ∈ F with ψ ≤ x0. Now suppose that for each x ∈ 2G, whenever each finite pullback ψ ≤ x is a member of Ψ, x ∈ F . Let x∈ / F . There exists a finite ψ ≤ x such that ψ∈ / Ψ. Because ψ∈ / Ψ, for any x0 such that ψ ≤ x0, x0 ∈/ F . We have thus shown that any arbitrary point x∈ / F has an open invariant set Uψ 3 x with F ∩ Uψ = ∅, so F is closed and invariant.

We have thus characterized open and closed invariant sets for 2G where G is Abelian.

2G For what it’s worth, we have in fact characterized the quotient topology on the space G of G-orbits of 2G in terms of ﬁnite pullbacks in much the same way the usual topology on 2G is characterized in terms of ﬁnite restrictions. This characterization is unfortunately not

2G worth very much, as G is not even T0.

Proposition 1.20. If G is countable and Abelian, then x ∈ 2G is hyper-aperiodic iﬀ, for each non-identity g ∈ G, there is a ﬁnite G-graph Γg such that Γg admits a G-graph

homomorphism to G and such that any ψ ≤Γg x witnesses that x is not periodic with period g. By “ψ witnesses x is not periodic with period g,” we mean that for some decomposition

km−1 km−2 k0 of g into its generators as g = gm−1 gm−2 ··· g0 with each ki = ±1, there exists a sequence of nodes k k0 k1 k2 m−1 g0 g1 g2 gm−1 v0 −→ v1 −→ v2 −→ · · · −→ vm

12 through Γg such that ψ(v0) 6= ψ(vm).

Proof. Let Ψ be the collection of ﬁnite pullbacks of x. Then Ψ is also the collection of

ﬁnite pullbacks of F = Cl2G ([x]) by Proposition 1.19. Suppose x is not hyper-aperiodic.

0 That is, suppose F has a periodic point x with period g 6= 1G. But then none of the ﬁnite pullbacks of x0 witness that x0 is g-aperiodic because x0 actually is periodic.

Indeed, consider a ﬁnite G-graph Γg together with a G-graph homomorphism σ :Γg →

km−1 km−2 k0 G. Let ψ = x ◦ σ; thus ψ ≤Γg x. Fix an arbitrary decomposition g = gm−1 gm−2 ··· g0 with nodes v0, v1, . . . , vm ∈ Γg such that

k k0 k1 m−1 g0 g1 gm−1 v0 −→ v1 −→ · · · −→ vm, supposing such a decomposition exists. Then

k k0 k1 m−1 g0 g1 gm−1 σ(v0) −→ σ(v1) −→ · · · −→ σ(vm)

where each σ(vi) ∈ G. That is to say,

km−1 k0 gm−1 ··· g0 σ(v0) = gσ(v0) = σ(vm).

0 0 0 The point x has period g, so x (σ(v0)) = x (σ(vm)) and ψ(v0) = ψ(vm). This holds for each decomposition of g together with each corresponding v0, v1, . . . , vm, therefore ψ does not witness g-aperiodicity of x0. Because ψ is a ﬁnite pullback of x0, ψ is in turn a pullback of x. We conclude there is no ﬁnite Γg which admits a homomorphism to G such that every

ψ ≤Γg x witnesses g-aperiodicity. Now, suppose that x is hyper-aperiodic. Fix an increasing sequence of ﬁnite G-graphs

Γ0 ⊆ Γ1 ⊆ Γ2 ⊆ · · ·

Sω such that k=0 Γk = G; we can do this because G is countable. Fix a g 6= 1G, and consider 0 0 some x ∈ F . Let ψk = x Γk for each k; then

ψ0 ≤Γ0 ψ1 ≤Γ1 ψ2 ≤Γ2 ···

13 0 0 x is not periodic, so there is some kx which is least such that ψkx0 witnesses the fact. Here 0 0 0 0 kx is a continuous function of x , because kx is determined by the pullback ψkx0 ≤ x . F is a 0 closed subset of a compact space, so F is compact, and we may take k = max{kx0 : x ∈ F }.

The set {xΓk : x ∈ F } includes every Γk-pullback in Ψ, so any Γk pullback in Ψ witnesses g-aperiodicity.

Hyper-aperiodic points are not particularly hard to ﬁnd, and are ﬂexible enough to be found exhibiting a variety of properties [3]. Additionally, a novel construction of a

2 hyper-aperiodic point for F (2Z ) is given as a sublemma to Theorem 2.1.

1.2. The Borel and Continuous Chromatic Numbers of F (2Z)

In this section, we present the background folklore that

Z Z χB(F (2 )) = χc(F (2 )) = 3.

This example illustrates some of the main ideas in the area of deﬁnable combinatorics prior to our research, which motivated the search for the Borel and continuous chromatic numbers

2 m of F (2Z ) and F (2Z ) in general.

Proposition 1.21. There exists no Borel two-coloring ψ : F (2Z) → 2.

Proof. Suppose for a contradiction that ψ as described exists. By a standard fact of

Z topology, there is a comeager set Y ⊆ F (2 ) such that ψY is continuous with respect to the subspace topology on Y . Pick a point y ∈ Y , and without loss of generality suppose

ψ(y) = 0. By continuity of ψY on Y , there is an integer n and a ﬁnite partial function ψ ⊆ y and such that for any y0 extending ψ in Y , ψ(y0) = 0.

We can also translate ψ and Y in the following manner. Let z generate Z, and let k be odd and be much larger than the (usual) norm of any point of dom(ψ). Consider zk · Y = {zk · y : y ∈ Y }. Let zk · ψ be deﬁned on zk · dom(ψ) as (zk · ψ)(g) = ψ(z−kg). Now, for any point y0 ∈ F (2Z), we have

ψ ⊆ y0 ∈ Y

14 iﬀ zk · ψ ⊆ zk · y0 ∈ zk · Y.

k Let Y∩ = Y ∩ (z · Y ). Y∩ is the intersection of two comeagre sets, whence it is comeagre and in particular dense. Let U ⊆ F (2Z) be the open set consisting of extensions of both ψ and zk · ψ; U is nonempty because of our choice of large k. Let y0 be in both U

0 0 k 0 k −k 0 and Y∩. We have ψ ⊆ y ∈ Y , so ψ(y ) = 0. Also, z · ψ ⊆ y ∈ z · Y , so ψ ⊆ z · y ∈ Y , and then ψ(z−k · y) = 0. But we have just found two points, y and z−k · y, at the odd distance k from each other, that have the same color. This situation is impossible because

2 in a two-coloring on a graph component isomorphic to Z , colors must strictly alternate.

Z Z We now have χB(F (2 )) > 2, when intuitively, χ(F (2 )) = 2. Indeed, our intuition apparently relies on AC.

Proposition 1.22 (AC). The chromatic number of F (2Z) is 2.

Proof. Apply AC to ﬁnd a set M such that for each orbit [x] ⊆ F (2Z), M ∩[x] is a singleton set. Let ϕ: F (2Z) → 2 be deﬁned as follows. For x ∈ F (2Z), ϕ(x) is the distance to M, mod

This M is called a selector in the literature; a selector is a complete section that meets each equivalence exactly once. This concept is important because a deﬁnable selector for F (2G) is, in some respects, exactly what we cannot have. However, we can avoid the

necessity of obtaining a selector when constructing a three-coloring of F (2Z) by using a “marker set” provided by the following lemma.

Lemma 1.23 (Basic marker lemma). Let G be a countably inﬁnite group and let

S ⊆ G be a nonempty ﬁnite symmetric subset of G with 1G ∈/ S. There exists a clopen subset M ⊆ F (2G) such that S · M = F (2G) \ M.

Proof. Consider the set Ψ of of finite, partial functions ψ : G → 2 such that for each s ∈ S, s · ψ is incompatible with ψ. Any sufficiently large finite restriction ψ of an aperiodic point x: G → 2 suffices, so Ψ is infinite. Ψ is countable because G is countable, so enumerate all

15 ψ ∈ Ψ as (ψk)0≤k<ω. We will iteratively construct an increasing sequence of sets (Mk)0≤k<ω such that

∀k, S · Mk ∩ Mk = ∅

and ω [ G S · Mk ∪ Mk = F (2 ). k=0

G For each k, let Uk be the (open) set of all x ∈ F (2 ) such that ψk ⊆ x. Observe

S · Uk ∩ Uk = ∅ by construction of Ψ. Let M0 = ∅. Now suppose Mk has been constructed, and let Mk+1 = Mk ∪ (Uk \ (S · Mk)); that is, simply add in all elements of Uk that don’t cause a contradiction. We must argue Mk+1 ∩ S · Mk+1 = ∅. Suppose x ∈ Mk+1; we will show x∈ / S · Mk+1. Here, either x ∈ Mk or x ∈ Uk \ (S · Mk), so we will show x∈ / S · Mk

and x∈ / S · (Uk \ (S · Mk)).

Suppose x ∈ Mk. In this case, x∈ / S · Mk by the inductive hypothesis. If x ∈

−1 S · (Uk \ (S · Mk)), then x = s · u for some s ∈ S, u ∈ U \ (S · Mk). But u = s · x, so by

symmetry of S, u ∈ S · Mk, which is a contradiction. So x∈ / S · (Uk \ (S · Mk).

Now suppose x ∈ Uk \(S·Mk). Then x∈ / S·Mk is immediate, and x∈ / S·(Uk \(S·Mk))

because x ∈ Uk implies x∈ / S · Uk.

We now have S · Mk ∩ Mk for all k. It remains to be shown that this scheme captures the whole space. Indeed, suppose

ω ! G [ x ∈ F (2 ) \ Mk . k=0

Here x is aperiodic, so there is some ψk ⊆ x (i.e., x extends ψk) such that ψk ∈ Ψ, as

mentioned previously. x is in Uk but not in Mk+1, so

ω [ x ∈ S · Mk ⊆ S · Mk. k=0 Sω Sω Let M = k=0 Mk whence S · M = k=0 S · Mk. Each Mk and S · Mk is built by ﬁnite

unions and intersections of clopen sets, so each Mk and each S · Mk is clopen. Thus M and

S · M are both open; but these are complements of each other, whence clopen.

16 Lemma 1.23 is not the traditional presentation of the marker lemma, but it has the advantage of making clear exactly what groups G and sets S are acceptable; namely G

countable and S nonempty, ﬁnite, and symmetric with 1G ∈/ S. A more familiar presentation is as follows.

Corollary 1.24 (Basic marker lemma, traditional form). Let ρG be a left-invariant

metric on a countable group G. Suppose, for some d > 0, the set of all points with ρG norm at most d is ﬁnite. Then there exists a clopen subset M ⊆ F (2G) such that for any two

0 0 G points m 6= m ∈ M, ρG(m, m ) > d and for any point x ∈ F (2 ), ρG(x, M) ≤ d.

Proof. Take as S the punctured ball {g ∈ G : 1 ≤ ρG(1G, g) ≤ d}.

Proposition 1.25. There exists a three-coloring ϕ: F (2Z) → 3.

j k Proof. Let ρZ be the usual metric on Z, namely ρZ(z , z ) = |k − j|. Apply Corollary 1.24

Z to F (2 ) using ρZ as the metric and with d = 4 to obtain a marker set M. We have already observed that the connected components of F (2Z) are just inﬁnite chains; every node in the graph has degree 2. Points of M are interspersed inﬁnitely in both directions along each

chain; there is a point of M every ﬁve to nine nodes. Deﬁne our three-coloring ϕ: F (2Z) → G

as follows. Let ϕ(x) = 2 if x ∈ M. If x∈ / M, let mx be the next point in M to the left of x.

Then let ϕ(x) = 0 if the distance to mx is even, and ϕ(x) = 1 if the distance to mx is odd. This construction is continuous. Indeed, because we can determine the color of x by checking the membership in M of ﬁnitely many (at most ten) points of F (2Z), the preimage of any color is clopen.

Note that in the two-coloring case, we had the extremely rigid property that for any two nodes x, x0 ∈ F (2Z), ϕ(x) = ϕ(x0) iff x, x0 are at even distance from each other. In other words, we had a strict sense of “parity” on any orbit. This requirement turned out to be too strong for a Borel construction. For a three-coloring, we break parity using the third color. That is, only when x, x0 are in the same finite region between marker points do we have ϕ(x) = ϕ(x0) iff x, x0 are at even distance. Breaking a space up into finite regions using

17 marker points and working on each region individually is an overarching theme of every positive construction in this work.

18 CHAPTER 2

MAIN RESULT

2 Consider a continuous map ϕ: F (2Z ) → D for some discrete1 space D. We will show that if such a map exists, there exists ϕ0 :Γ → D with similar properties where Γ is a particular ﬁnite Z2-graph. Conversely, if such a map ϕ0 exists, a map with similar properties

2 ϕ00 : F (2Z ) → D exists. By “similar properties” we mean that ϕ must have most of the pullbacks of ϕ0, and ϕ0 must have all of the pullbacks of ϕ00. For example, a map ϕ with

2 domain Γ or F (2Z ) is a k-coloring iﬀ each of its 2-pullbacks is a k-coloring. By the main

2 2 theorems, 2-pullbacks pass between F (2Z ) and appropriate Γ, so F (2Z ) is continuously k-colorable iﬀ there is some k-colorable Γ. Using a ﬁnite combinatorial argument, we will show that there is no three-coloring of any of the Z2-graphs Γ, but there is a four-coloring

2 Z of a graph Γ, and hence χc F (2 ) = 4.

The ﬁnite Z2-graphs Γ used for this purpose will constitute a countable collection

Γn,p,q where n ≤ p, q < ω. Fix n ≤ p, q.Γn,p,q will consist of the disjoint union of twelve rectangular grid graphs, called “tiles” and notated by T , modded out by speciﬁc rectangular grid subgraphs, notated by R. The resulting quotient will thus be a single connected G- graph.

To explain this quotient, we introduce a type of diagram to notate it, an example of which is presented in Figure 2.1. These diagrams will consist of rectangular blocks arranged in patterns. Suppose Γ is such a graph. Each rectangular block R will represent a rectangular grid graph, and adjacent blocks represent grid graphs in Γ that are adjacent in that together they form a larger grid graph. We adopt the convention that adjacent blocks have disjoint nodes rather than overlapping boundaries, so a 2 × 5 block to the left of a 2 × 3 block would yield a 2 × 8 block. Some blocks will be labeled; labeling a block indicates that it is identiﬁed, node for node, with every other block with the same label. Unlabeled blocks are not identiﬁed with anything but themselves; these could equivalently be given unique labels.

1Non-discrete codomains remain as future work.

19 RA

Figure 2.1. A simpliﬁed notation for quotient Z2-graphs of Z2-graphs consisting of rectangular grid graphs. Two rectangular sheets are “glued” together

by giving the center nodes the shared label RA.

Consider Figure 2.1. Here, the Z2-graph on the left consists of two 6 × 6 rectangular grid graphs modded out by a 2 × 2 grid graph in the center. These modded nodes are represented by the blocks labeled RA, and the remaining nodes by the unlabeled blocks.

Deﬁne R×,Ra,Rb,Rc,Rd as rectangular grid graphs with the following dimensions:

R× : n × n

Ra :(p − n) × n

Rb :(q − n) × n

Rc : n × (p − n)

Rd : n × (q − n)

20 R× Rc R× R× Rc R×

Ra Ra Rb Rb

R× Rc R× R× Rc R×

Tca=ac Tcb=bc

R× Rd R× R× Rd R×

Ra Ra Rb Rb

R× Rd R× R× Rd R×

Tda=ad Tdb=bd

Figure 2.2. The ﬁrst four “tiles” of Γn,p,q. These are all isomorphic to torus graphs.

The twelve tiles of Γn,p,q, with subblocks labeled for modding out, are given in Fig- ures 2.2, 2.3, 2.4, and 2.5. The notation, e.g., Tca=ac, is inspired by the homotopy theory used in later proofs. Because we mod out by the blocks R× etc., each of these blocks is considered to appear exactly once in Γn,p,q. When we refer to a tile, we will mean simply the union of its blocks as a subgraph of Γ, whence, e.g., Tca=ac is isomorphic as an undirected

2 Z graph to the p × p torus graph ( pZ) . 2 The term ”tile” is justiﬁed by the fact that, e.g., a Z pullback of a function ϕ: Tca=ac → 2 S on the toroidal tile Tca=ac is a tiling of Z by Tca=ac. The presence of the additional 11 tiles allow more complex tilings. Indeed, observe that for any non-boundary vertex v ∈ Γn,p,q, i.e.,

v∈ / R× ∪ Ra ∪ Rb ∪ Rc ∪ Rd, if v is used in a pullback from a function on Γ we have that the whole tile containing v is used. This is because each of the non-boundary nodes has a unique neighbor in each of the four directions. Thus, the structure of x ∈ Γn,p,q enforces tilings in the sense that the only

21 R× Rc R×

R R R Ra × Rd × Rc × Rb R Ra Ra × R× R R R × Rc × Rd × Rb Ra

Tdca=acd R× Rc R×

Tcba=abc

R× Rc R×

R R R Ra × Rc × Rd × Rb R Ra Ra × R× R R R × Rd × Rc × Rb Ra

Tcda=adc R× Rc R×

Tcab=bac

Figure 2.3. The next four “tiles” of Γn,p,q. Tiles Tcab=bac and Tcba=abc com-

mute Ra with Rb and tiles Tcda=adc and Tdca=acd commute Rc with Rd.

2 freedom in pulling back to Z from a function on Γn,p,q is in which tiles T ∗ are adjacent to which.

2 2.1. Passing from F (2Z ) to Finite Graphs

2 Having established our Z graphs {Γn,p,q}n≤p,q, the following theorem quantiﬁes ex-

2 Z actly which pullbacks pass from F (2 ) to which graphs Γn,p,q.

2 Theorem 2.1. There exists a compact invariant set K ⊆ F (2Z ), which may be taken

2 to intersect any nonempty open set U ⊆ F (2Z ), such that the following holds. Suppose D

2 is a discrete space and ϕ: F (2Z ) → D is a continuous map. Then, for each n and each

22 q copies of Rc, q + 1 copies of R×

R× Rc R× Rc R× Rc R× Rc R× Rc R×

Ra ··· Ra

R× Rd R× Rd R× Rd R× Rd Rd R×

p copies of Rd, p + 1 copies of R×

Tcqa=adp

p copies of Rd, p + 1 copies of R×

R× Rd R× Rd R× Rd R× Rd Rd R×

Ra ··· Ra

R× Rc R× Rc R× Rc R× Rc R× Rc R×

q copies of Rc, q + 1 copies of R×

Tdpa=acq

Figure 2.4. The long horizontal “tiles” of Γn,p,q.

0 0 p, q n, there is a function ϕ :Γn,p,q → D such that ϕK has each n-pullback of ϕ and ϕ 0 0 has each pullback of ϕ to a ﬁnite grid graph. That is, for each ψn ≤n ϕ , we have

ψn ≤n ϕK

23 R× Rc R× R× Rc R×

Ra Ra Rb Rb

R× R× R× R× Ra Ra p q Rb Rb q oisof copies 1 + of copies 1 + oisof copies 1 + q q R× p R× oisof copies of copies oisof copies R R Ra × × Ra

R× R R R× R R

b R b a a b R R R , , , × × Ra × Ra R× R×

R× R× Rb Rb . . Rb Rb Ra Ra

R× Rc R× R× Rc R×

Tcbp=aqc Tcaq=bpc

Figure 2.5. The long vertical “tiles” of Γn,p,q.

0 and for each ψm ≤m ϕ with m < ω, we have

ψm ≤m ϕ.

Proof. We ﬁrst construct the compact invariant set K. That is, we construct a hyper-

2 aperiodic point x with the properties we need, and then let K = Cl2Z ([x]). Because K 0 is compact, for any continuous function ϕ, there will exist a w < ω such that ϕK (x ) for x0 ∈ K is determined by x0 : Z2 → 2 restricted to a w × w box centered at the origin of Z2. Thus, by Proposition 1.17, each w-pullback of x determines exactly one point of ϕ: F (2G) → D. Our strategy is to arrange w-pullbacks in x in such a way that we can “read

0 off” a consistent definition of ϕ :Γn,p,q → D for some Γn,p,q, one node at a time. Moreover, we will include material x that allow us to read off Γn,p,q for each w, each n > 2w, and each

24 p, q ≥ n; thus we diagonalize against the deﬁning window size w for any function ϕ and can always produce the required ϕ0. We build x with reference to a hyper-aperiodic point y, which will remain ﬁxed throughout our construction. The point y is provided by the following sublemma, whose proof follows the current proof.

2 Sublemma. There exists a hyper-aperiodic point y ∈ F (2Z ).

As we are diagonalizing against w, ﬁx an arbitrary value w ≥ 1. Further ﬁx n ≤ p, q.

Let ψ×, ψa, ψb, ψc, ψd be arbitrary pullbacks of y to grid graphs with the following dimensions:

ψ× :(n + 2w) × (n + 2w)

ψa :(p − n − 2w) × (n + 2w)

ψb :(q − n − 2w) × (n + 2w)

ψc :(n + 2w) × (p − n − 2w)

ψd :(n + 2w) × (q − n − 2w)

These dimensions correspond to the dimensions of R∗, but adjusted by ±2w.

Now, suppose part of x consists of the pullbacks ψ×, ψa, ψc arranged in the same manner as the boxes R×,Ra,Rc in tile Tca=ac. This is shown in Figure 2.6; the edges of the pullbacks ψ∗ are given as solid lines. In dotted lines, we place the corresponding boxes

0 intended to be used to read oﬀ the deﬁnition of ϕ for each R∗. That is, each element g ∈ G

−1 circumscribed by dotted lines corresponds to a point g · x ∈ [x] for which a node v ∈ Γn,p,q has ϕ0(v) = ϕ(g−1 · x). Thus, we must ensure that any two points x0, x00 ∈ [x] corresponding to the same node v are taken to the same value.

As we have noted, we have by diagonalization that each point of ϕK is determined by a w-pullback. A w-pullback centered at any point in any of the boxes corresponding to R× is contained completely in a copy of ψ×, therefore ϕ must be defined identically on each of the distinct portions of [x] corresponding to R×. Now, ψc is narrower than Rc, but everywhere ψc appears, ψc is flanked by copies of ψ×. Thus ϕ is also defined identically on

25 w w

ψ× ψc ψ×

ψa ψa

ψ× ψc ψ×

Figure 2.6. Constructing the portion of the hyper-aperiodic point that shows

0 how to deﬁne ϕ on Tca=ac. The solid lines represent the pullbacks ψ∗ ≤ y with y an arbitrary hyper-aperiodic point, and the dotted lines represent the

corresponding boxes R∗. The pullbacks ψ∗ are larger than the boxes R∗ by

2w, to ensure that ϕR∗ is read oﬀ consistently for each R∗. each box corresponding to Rc. A similar argument applies for Ra. Therefore, ϕ restricted to the area bounded by the boxes corresponding to R×,Ra,Rc conforms to all of the structure

0 of tile Tca=ac, so we can read oﬀ ϕ Tca=ac from ϕ. Similarly, we can arrange to be able to read the remaining 11 tiles from elsewhere in [x]. For example, the pullbacks ψ∗, and corresponding boxes, for tile Tdca=acd are given in Figure 2.7. Here again, each box corresponding to R×, Ra, or Rc must be deﬁned the same way not only as the other boxes corresponding to R×, Ra, and Rc respectively in the

Tdca=acd construction, but also as those of the Tca=ac construction. Each of the remaining pullbacks ψa, ψb, ψc, ψd, despite being narrowed or shortened, is always ﬂanked by instances of ψ×, so ϕ is deﬁned in a consistent way on each of these boxes regardless of in which tile’s construction it appears.

For each w ≥ 1, each n, p, q with p, q ≥ n + 2w, and each of the twelve tiles, we have given an arrangement of pullbacks that x must exhibit. Enumerate these arrangements as Ak with 1 ≤ k ≤ ω; one value of k for each tile and each valid collection of parameters w, n, p, q.

26 ψ× ψd ψ× ψc ψ×

ψa ψa

ψ× ψc ψ× ψd ψ×

Figure 2.7. Constructing the portion of the hyper-aperiodic point that shows

how to color Tdca=acd. The solid lines represent the pullbacks ψ∗ and the dotted

lines represent the corresponding boxes R∗.

Each arrangement contains an unlabeled interior portion; we consider the unlabeled interior portion to not be part of Ak. We must ensure K ∩ U 6= ∅. Because U is open and nonempty, there is a pullback

ψU such that if ψU ≤ x,[x] ∩ U 6= ∅, so let A0 = ψU .

Place these arrangements Ak in x in such a way that the distance between Ak and

0 Ak0 for k 6= k increases with k. Fill all remaining space, and all space surrounded by any arrangement Ak of pullbacks ψ∗ by y. This completely deﬁnes x; we now show x is hyper-aperiodic.

2 Fix an arbitrary element g ∈ Z . Because y is hyper-aperiodic, there exists an Ng < ω such that any Ng-pullback of y witnesses that g is not a period of y. The distance between arrangements Ak grows with k, so there exists a fixed distance Lg Ng about the origin of x outside of which any distinct Ak, Ak0 will be at distance much greater than Ng, and n − 2w for each such Ak will also be much greater than Ng. To show x is hyper-aperiodic with respect to g, it suffices to find an Mg such that any Mg-pullback of x centered at least

Lg from the origin of x witnesses g-aperiodicity. If so, any (2Lg + Mg)-pullback of x will witness g-aperiodicity of x, and hyper-aperiodicity will follow by Proposition 1.20.

0 Consider a shift x of x of at least distance Lg from x. Then consider the closest

27 0 0 arrangement Ak of pullbacks ψ∗ to the center of x (x may indeed be centered within Ak).

Let w, n, p, q be the parameters associated with Ak, and δ be the distance from the center

0 0 of x to the nearest copy of some ψ∗ in Ak. We may have δ > 0 if x is centered in interior,

0 non-boundary nodes of Ak. Suppose δ < Ng. Then a 4Ng-pullback of x centered at x

contains a copy of ψ∗, which is at least of size (n−2w)×(n−2w). We ensured n−2w Ng,

0 so ψ∗ shares a full Ng-pullback with y. Now suppose δ ≥ Ng. In this case, x is centered in

0 the interior unlabeled portion of Ak, or x is centered outside of Ak. Regardless of whether

0 0 x is interior to Ak, however, there is a full Ng-pullback of y centered at x . We thus let

Mg = 4Ng and any (2Lg + Mg)-pullback of x witnesses g-aperiodicity.

0 2 With x hyper-aperiodic, we obtain our K as Cl2Z ([x]), and ϕ can be read oﬀ from 2 Z ϕ[x] as described. It remains to be shown that ϕ: F (2 ) → D has the appropriate pullbacks 0 0 from ϕ :Γn,p,q → D. Namely, ϕK must have each n-pullback of ϕ , and ϕ must have each m-pullback of ϕ0 for any m < ω.

0 Consider an n-pullback ψn of ϕ :Γn,p,q → D. Because all tiles overlap on their

boundaries by at least n nodes, ψn is actually a pullback of a single tile. But each single tile

0 worth of ϕ appears in its entirety in ψ[x], so ψn ≤ ϕK . 0 Now ﬁx m < ω and consider an arbitrary m-pullback ψm of ϕ . Let σ witness that

0 0 ψm ≤ ϕ , that is, ψm = ϕ ◦ σ. As we have observed, σ is a tiling of an m × m grid graph by

2 tiles from Γn,p,q. We may create, on a portion Z , the corresponding tiling using patterns ψ∗ 2 along with the unlabeled interior ﬁllers bordered by ψ∗. We can extend this tiling on Z to

2 2 0 Z a full aperiodic {0, 1} labeling of Z , that is, a point x ∈ F (2 ). For each vertex v ∈ Γn,p,q, ϕ0(v) was determined by a unique w-pullback of x, and enough of these w-pullbacks have

0 been included in x to force the entirety of ψm to appear in ϕ[x0]. Then, ϕm ≤ ϕ.

0 Remark. Note that we do not have that any m-pullback of ϕ is a pullback of ϕK – this would actually grant K periodic points, which would defeat the point of K. Only

0 0 n-pullbacks (and smaller) of ϕ need be pullbacks of ϕK ; m pullbacks of ϕ with n < m < ω may only be found in ϕ. Indeed, the very purpose of n – the purpose of having tiles overlap

28 by distance n – is to enable delicate arguments that require considering ϕK speciﬁcally. For example, the proof of Theorem 5.2 includes such an argument. Observe that because n is universally quantiﬁed in Theorem 2.1, we can always dispense with n by taking n = 1.

2 We now return to the sublemma providing us the hyper-aperiodic point y ∈ F (2Z ).

Proof of the sublemma of Theorem 2.1. For each 0 ≤ k < ω, let β(k) be the number

2 of ones in the binary expansion of k, mod 2. Consider the point y ∈ 2Z (i.e., y : Z2 → 2) where

k0 k1 y(z0 z1 ) = β(|k0|) + β(|k1|) mod 2.

We will show y is hyper-aperiodic.

k0 k1 Fix z0 z1 6= 1Z2 . Consider ﬁrst the case k0 ≥ k1 ≥ 0. We have k0 and k1 are not

both zero, thus k0 > 0. Let `0 be the location of the ﬁrst “one” bit in the binary expansion

of k0, i.e., `0 = blog2k0c. Let `1 be the location of the ﬁrst “one” bit for k1, or zero if k1 = 0. Then, let   `0 5 · 2 , if β(k0) = 0, m0 =  0, if β(k0) = 1.   0, if β(k1) = 0, m1 =  `1 1 · 2 , if β(k1) = 1.

Consider the case where β(k0) = 0. Here, β(m0) = 0, and (mod 2) we have

` ` β(m0 + k0) = β(1102 · 2 + k0 − 2 )

= 0 + β(k0) − 1

= 1

and β(m0) 6= β(m0 + k0). Similarly, if β(k0) = 1, then β(m0) = 0 and β(m0 + k0) = 1. In

either case, β(m0) 6= β(m0 + k0). It also is easy to check β(m1) = β(m1 + k1), so

(∗) β(m0) + β(m1) 6= β(m0 + k0) + β(m1 + k1).

29 There are many more values satisfying (∗). Indeed, any

0 `0+4 m0 ∈ { c · 2 + m0 : c ∈ Z, c ≥ 0} ∪

`0+4 {−c · 2 − m0 − k0 : c ∈ Z, c ≥ 0}, and

0 `0+4 m1 ∈ { c · 2 + m1 : c ∈ Z, c ≥ 0} ∪

`0+4 {−c · 2 − m1 − k1 : c ∈ Z, c ≥ 0}

also satisfy (∗). Then, any (2`0+6)-pullback of y witnesses two points

m0 m1 m0+k0 m1+k1 2 z0 z1 , z0 z1 ∈ Z with

m0 m1 m0+k0 m1+k1 y(z0 z1 ) 6= y(z0 z1 ), and this holds whenever k0 ≥ k1 ≥ 0.

The other cases follow by symmetry. When k1 ≥ k0, simply switch k0, k1 and the

0 corresponding values m0, m1. When k0 ≤ 0, run the above argument with −k0 and −m0.

Similarly when k1 ≤ 0.

2 Z 2.2. Passing from Γn,p,q to F (2 )

2 For the converse to Theorem 2.1, we will need to build a continuous map from F (2Z ) to Γn,p,q. Following almost all constructions of continuous functions on such spaces, we do this with the aid of a marker region structure. In our case, the structure is an equivalence relation on each orbit (i.e., a subequivalence of the orbit equivalence relation). When dealing with equivalence relations in the clopen setting (as opposed to merely Borel), care must be taken in our deﬁnitions. Observe, for example, that the orbit equivalence relation – the relation that holds iﬀ

2 2 there is a g ∈ G such that g · x = y – is not an open subset of F (2Z ) × F (2Z ). Indeed, even the equality relation itself is not open despite clearly being “deﬁnable” in any reasonable sense.

30 k 2 Thus, when we speak of a relation R ⊆ F (2Z ) (including functions), we will adopt a more forgiving version of “clopen” or “continuous”. To simplify notation in giving the deﬁnition, we provide this deﬁnition for a general X with an action (G, ).

Definition 2.2. Let X be a space with an action of a countable group G. Fix 1 ≤ k < ω. Let Ek be the k-ary generalization of the orbit equivalence relation:

k (x0, x1, . . . xk−1) ∈ E ⇐⇒ ∀i < k, [x0] = [xi]

k k A relation R ⊆ E ⊆ X is called open (closed, clopen) with respect to if there is an open (closed, clopen) subset U of X × Gk (G given the discrete topology) such that

(x0, x1, . . . , xk−1) ∈ R iﬀ there exist g1, g2, . . . , gk−1 ∈ G with gi x0 = xi for 1 ≤ i < k and

(x0, g1, g2, . . . , gk−1) ∈ U.

Remarks. (1) Naturally, we intend to use this deﬁnition in the case that is con-

tinuous (indeed, we intend = ·), but there is no apparent need to restrict itself to be continuous in the deﬁnition. (2) It is unnecessary to consider “Borel” with respect to an action, this would simply be “Borel” for any Borel action.

Here we have simply replaced all but one of the points in each tuple from R with corresponding group elements, drastically limiting the amount of information needed about a tuple needed to determine membership in R. We claim Deﬁnition 2.2 gives the natural notion of “clopen”, etc. when considering relations. For example, in Proposition 1.25, we

essentially decomposed each orbit of F (2Z) into finite marker regions. The relation R such that x R x0 iff x is in the same finite marker region as x0 is clopen with respect to the

Bernoulli shift ·, but not clopen as a subset of F (2Z). Indeed, the set S of x ∈ F (2Z) such that z · x is in the same region as x is a clopen set that is straightforward to determine from from the marker point set M. Conversely, given any some subequivalence R of the orbit equivalence which is clopen with respect to ·, the set of boundary nodes of R-equivalences is clopen.

31 0 Theorem 2.3. Suppose D is a discrete space and there is a function ϕ :Γn,p,q → D for some n ≤ p, q with p ⊥ q (i.e., p relatively prime to q). Then, there is a continuous

00 2 00 0 Z 2 function ϕ : F (2 ) → D such that ϕ ≤F (2Z ) ϕ .

We observe under the conditions of Theorem 2.3, any pullback of ϕ00 is a pullback of ϕ0 by transitivity.

2 Proof. We proceed by constructing a continuous Z2-graph homomorphism σ : F (2Z ) → 00 0 Γn,p,q; once completed we will simply let ϕ = ϕ ◦ σ. The map σ has the eﬀect of tiling

2 Z F (2 ) with the twelve tiles of Γn,p,q. The presence of the boxes R∗ on the boundaries of each

tile T∗ of Γn,p,q forces us to place tiles such that adjacent tiles overlap by n rows or columns.

Moreover, R× appears at the corners of each tile, so R× indicates where corners of adjacent

tiles may be placed. Similarly, Ra and Rb impose restrictions on horizontal adjacency and

Rc and Rd impose restrictions on vertical adjacency. We begin with a construction of marker regions with a special property. Namely, each orbit is partitioned into ﬁnite rectangular regoins such that the distance between any two not-immediately-adjacent corners of any two regions must be large – a minimum distance of p10q10 suﬃces. We use the following sublemma for this construction, whose proof follows the current proof.

Sublemma. For any integer d > 2 there is a subequivalence R ⊆ E2, relatively clopen with respect to the Bernoulli shift, such that each equivalence of R is a box and no two corners x and y of boxes have 1 < ρ(x, y) < d.

This lemma does not guarantee, and we do not require, that the regions be nearly square in any sense. Indeed, near-squareness, together with the corner conditions we just stipulated, would force the marker regions to line up into a well-deﬁned lattice pattern of their own, which would contradict Theorem 5.2. However, we do observe that the corner condition forces the height and width of each box to be at least p10q10.

We begin by tracing the boundaries between marker regions with instances of R×,

Ra, Rb, Rc, and Rd, thus isolating the remaining space to be ﬁlled in and simplifying further

32 Rb

Rd Rd Rd Rd Rd Rd

Figure 2.8. Edge boundaries for marker regions. R× alternates with R∗ with

R∗ ∈ Ra,Rb,Rc,Rd. Instances of Rb and Rd are labeled. This ﬁgure diﬀers

from the actual construction in that instances of Rb and Rd in the actual construction are much less common than depicted: we require that no two

5 5 instances of Rb or Rd be within distance p q of one other. construction. The boundary construction is illustrated in Figure 2.8, modulo a diﬃcult to illustrate requirement that instances of Rd and Rb be distributed very sparsely. First, centered at the lower-left corner of each marker region, we place an instance of R×. Now, following the vertical region boundaries between corners, we alternate between instances of R× and instances of either Ra or Rb. This alternation can be accomplished regardless of the distance between two corners, because

height(R×) + height(Ra) = p,

height(R×) + height(Rb) = q, and p is coprime to q.

We use Ra much more often than Rb; it suﬃces for the remainder of the proof to

5 5 ensure each instance of Rb is at least distance p q away from any other copy of Rb or any corner of a marker region. There is ample room for such separation because the distance between the lower-left corners of any two marker regions is at least p10q10 p5q5. As such

5 5 extravagant separation as p q is impractical to illustrate, instances of Rb are drawn much

33 Tca=ac Tca=ac Tda=ad Tca=ac Tca=ac

Tca=ac Tca=ac Tda=ad Tca=ac Tca=ac

Figure 2.9. “Downward” propagation of an instance of Rd using tile Tda=ad

through a region mostly tiled with tile Tca=ac.

more frequently in Figure 2.8 than in the actual construction. We ﬁll horizontal boundaries

with R×, Rc, and Rd analogously to the vertical boundaries, using Rc much more often than

Rd.

Having ﬁlled in the boundaries, we need only need consider each marker rectangle in isolation. Consider a marker region M. In the simplest case, if M has a boundary devoid

of instances of Rb or Rd, we simply ﬁll the entire marker region with overlapping copies of

tile Tca=ac. Indeed, as most of the boundaries consist of alternating copies of R×,Ra,Rc, we

ﬁll most of M with copies of Tca=ac, and use the other tiles to ﬁll in the gaps.

To see how, consider the situation in Figure 2.9. Using a line of copies of Tda=ad, we

propagate an instance of Rd downward from the top of the ﬁgure to the bottom. Similarly,

in Figure 2.10, we use tile Tdca=acd to propagate an instance of Rd diagonally downward and

to the right. With Tcda=adc we can similarly propagate Rd diagonally downward and to the

left. Using Tcb=bc, Tcba=abc, and Tcab=bac, we can propagate a copy of Rb left to right, again with optional diagonals.

With these propagation techniques in mind, we seek to connect each instance of Rd on the top of M to an instance of Rd on the bottom, and each instance of Rb on the left side of M to an instance of Rb on the right side. There are two obstacles to this approach.

34 Tda=ad

Tdca=acd

Tda=ad

Figure 2.10. “Diagonal” propagation of an instance of Rd using tile Tdca=acd,

through a region mostly tiled with Tca=ac. Labels omitted to reduce clutter.

First, a horizontal line of Rb may need to intersect a vertical line of Rd. For this situation,

we place tile Tdb=bd at the intersection. Second, there may not be a match for every instance

of Rb or Rd, or the match may be shifted too far to account for by diagonal propagations. The latter can occur quite easily because the marker regions are not guaranteed to be nearly

square. We overcome this obstacle using the remaining four “long” tiles Tcqa=adp , Tdpa=acq ,

Tcaq=bpc and Tcbp=aqc to gather and eliminate copies of Rd.

Consider a long, narrow box R whose top is ﬂush with the top of the line of R∗ on the top border of our marker region M. We let R have height p(p + 1) + n. Thus R is tall

enough to contain a vertical arrangement of p + 1 tiles of the height of Tca=ac, overlapping on their boundaries as usual. Rather, R is sufficiently tall for p + 1 “rows” of tiles. Let R be nearly, but not entirely, the width of M; specifically leave distance p3q3 > p(p + 1) + n between each side of R and the corresponding side of M. We size and place R horizontally so that each of its upper corners is flush with a single copy of R×, itself contained in R, from the boundary of M.

Observe R captures each copy of Rd on the top boundary of M because each copy

5 5 3 3 of Rd cannot be within distance p q > p q of a corner of M. Let j, k be the number of

35 j copies of Rc and k copies of Rd

j k k j k k k − p p copies of Rd j + q p copies of Rc

Figure 2.11. An illustration of the construction of R. The smallest tiles are

Tca=ac. The snaking paths of slightly larger tiles are Tda=ad and Tdca=acd as in

Figure 2.10. The widest tiles are Tcqa=adp and Tdpa=acq , with the former on the row above the latter. instanes of Rc and Rd respectively along the top of R. We will ﬁll R in such a way that its left and right sides consist of alternating copies of Ra and R×, and the bottom of R has exactly k mod p copies of Rd on the left followed by copies of Rc, alternating as usual with

R×. A possible such R, with solution, is illustrated in Figure 2.11. The box R gathers disparate copies of Rd, eliminating some multiple of p such instances, and arranges the remaining neatly at the side of the bottom of the box for easy propagation downward.

We ﬁll R entirely with Tca=ac, Tda=ad, Tdca=acd, Tcqa=adp , and Tdpa=acq . Each such tile is of the height of Tca=ac, namely p + n, so we have our well-deﬁned rows of tiles as we mentioned previously.

Measuring between left sides of consequtive boxes (i.e., including the width n of each copy alternating copy of R×), each copy of Rc requires length p and each copy of Rd requires length q. Neglecting a single instance of R× on the far right side, therefore, R has width jp+kq. Because p and q are coprime, in converting Rd on the top of R to Rc on the bottom, we must convert copies of Rd in multiples of p to Rc in multiples of q. Such conversions are exactly the eﬀect of tiles Tdpa=acq and Tcqa=adp . We ﬁll the bottom row of R as follows. On the far left, we place as few copies of

Tda=ad as possible, followed by as many copies of Tdpa=acq as possible, followed by as few

36 copies of Tca=ac as possible. Given the multiplicity requirement above, we will thus have k modp copies of Tda=ad and j modq copies of Tca=ac. Each copy of tile Tdpa=acq in the middle requires space pq, so by subtracting from the width of R, we place

j k k j j k jp + kq − q(k mod p) − p(j mod q) jp + kq − q k − p p − p j − q q = pq pq

j k k j j k qp p + pq q = pq k j = + p q

many copies of Tdpa=acq on the bottom row of R. Observe we have computed an integer

number of tiles, so it is possible to place copies of Tdpa=acq as speciﬁed.

Each of the k mod p tiles Tda=ad contributes a single Rd to the bottom of R. Because

j k k there are k copies of Rd on the top of R and k modp = k −p p copies of Rd on the bottom, j k k j k k we have must have converted p p copies of Rd to q p copies of Rc. That is, there are j k k p q j + q p copies of Rc on the bottom of R, each contained in a copy of Td a=ac or Tca=ac.

We now proceed to the second row up. The bottom row left us mostly copies of Rd;

the only copies of Rc on the bottom of the second row are the j mod q many from copies

of Tca=ac. Having mostly Rd on the left of the bottom of the second row is advantageous

because we can build most of the second row iteratively using tiles Tda=ad and Tcqa=adp , each

of which only has copies of Rd on its bottom. We ﬁll in the remainder of the second row

with Tca=ac as before.

Begin with a single copy of Tcqa=adp . Let ` be the right edge of the previous tile placed and `0 be the value of ` from the previous iteration, or the left edge of R on the ﬁrst iteration.

There is at most one copy of Rd on the top border whose left edge is (horizontally) between

0 ` and ` because copies of Rd are separated by distance pq. If there are zero such copies,

place another copy of Tcqa=adp . If there is one such copy, place tile Tda=ad ﬁrst, followed by

tile Tdpa=acq (if room). Proceed until there is no longer room for another copy of Tcqa=adp .

Finish, again, with j mod q copies of Rc.

37 The process for the second row uses as many copies of tile Tcqa=adp as possible to

j j k q p convert Rc above to Rd below. Each copy of Tc a=ad has q copies of Rc on its top, so q j j k q p copies of Tc a=ad are used, leaving j − q = j mod q copies of Rc for the Tca=ac on the right side of the row. This is row is consistent with the of R, because it has k many copies of Rd and j many copies of Rc on its top, just as R does. The second row is consistent with the bottom row because it uses as many copies of Rd as possible on the left, and then as few copies of Rc as possible on the right, just as the bottom row does. Now, as we have observed, our work on the second row exposes exactly one instance of Rd for each such instance on the top border of R. For each such instance of Rd, connect with a chain of copies of Tda=ad and Tdca=acd. We have p − 1 remaining rows to do this with, and will only need to shift diagonally leftward by p − 1 tiles, so there is room vertically to do this. By the sparsity condition on instances of Rd, none of the chains will interfere with each other. We complete R by ﬁlling in the remaining space with tile Tca=ac.

The construction of R is readily adapted to gather copies of Rd on the bottom as well, and gather copies of Rb on the left and right. In particular, the height of the top and bottom buffer zones must be (p + 1)p + n, which is also the width of the side buffer zones. This height/width is much less than p3q3, the minimum distance from R to the edge of the marker region M, so the buffer regions R cannot overlap.

We observe that the top and bottom buﬀer regions expose the same number of copies of Rd. Indeed let j and k be the number of instances of Rc and Rd on the top of the marker region M respectively, and j0 and k0 be the corresponding numbers on the bottom. We have

pj + qk = p0j + q0k

= p(j − j0) = q(k0 − k)

Because p ⊥ q, p|(k0 − k), thus k0 and k are congruent mod p. The buﬀer zones leave

0 (k mod p) = (k mod p) copies of Rd exposed, and leave them aligned horizontally. The

analogous statement is true of Rd. It is now straightforward to connect the copies of Rd

with Tda=ad and the copies of Rb with Tcb=bc, using Tdb=bd where the lines cross. We ﬁll the

38 remainder of M with Tca=ac. This completes the construction of our tiling σ, and we let ϕ00 = ϕ ◦ σ.

To see that σ, and thus ϕ00, is continuous, recall that the marker regions themselves are continuous with respect to the Bernoulli shift, and the boxes along the marker boundaries that we began with only require the (ﬁnite) knowledge of the marker regions sharing any particular boundary, so these are continuous. Then, ﬁlling in the marker regions only required

2 knowledge of the boundary of each region, so σ(x) for each x ∈ F (2Z ) is deﬁned using only

ﬁnitely much information from x.

Remark. It is not difficult to ensure that, when defining σ on the buffer regions R on

the top and bottom side of M, σ uses tile Tdca=acd (or Tcda=adc) exclusively over tile Tcda=adc

(or Tdca=acd), thus obviating the need for the latter. Similarly, either of tiles Tcba=abc or

Tcab=bac can be dispensed with. Thus, it is possible to get a simpler definition for Γn,p,q using ten tiles instead of twelve, and all results in this work apply to the simplified graphs. We are not aware of any further possible simplifications.

We now return to the sublemma used to generate the marker regions in the ﬁrst

2 place, namely, the decomposition of each orbit of F (2Z ) into boxes such that no two corners x and y of boxes have 1 < ρ(x, y) < d. This sublemma is an easy consequence of the big-marker-little-marker argument [2, Thm. 3.1], however, we give a direct proof2 here.

Proof of the sublemma of Theorem 2.3. Let D0 and D1 be integers with d D0

D1. Applying Corollary 1.24, the basic clopen marker lemma, to D0 and D1 respectively, we

get clopen sets S0 and S1. Let g0, g1, . . . , gk be an enumeration of the set

2 {g ∈ Z : ρ(1Z2 , g) ≤ D1},

with g0 = 1Z2 . Then k 2 Z [ F (2 ) = gi · S1. i=1

2This proof contributed by Su Gao.

39 We deﬁne a partition of S0, {A0,...,Ak}, inductively:

A0 = S0 ∩ S1 = S0 ∩ (g0 · S1), [ Ai = S0 ∩ (gi · S1) − Aj, for 1 ≤ i ≤ k. j

It is clear that {A0,...,Ak} is a partition of S0, and that for any i ≤ k and distinct x, y ∈ Ai,

ρ(x, y) > D1.

For the rest of the proof we will no longer need the set S1. Our next step is to associate to each point x in S0 a box H(x) so that the following requirements are met:

2 Z (a) for every y ∈ F (2 ), there is x ∈ S0 such that y ∈ H(x);

(b) there is > 0 such that d D0 and each H(x) has side lengths between (1−)4D0

and (1 + )4D0;

(c) for any distinct x, y ∈ S0 with ρ(x, y) ≤ 10D0, the horizontal sides of H(x) and H(y) are at least d + 2 apart vertically, and the vertical sides of H(x) and H(y) are at least d + 2 apart horizontally.

In this last requirement, we spoke of the vertical distance of horizontal sides of rectangles (and horizontal distance of vertical sides of rectangles). To be completely explicit, if L1 and L2 are two horizontal sides of rectangles, their vertical distance is the shortest distance of any two points on the extensions of L1 and L2.

The construction of the boxes H(x) is by induction on 0 ≤ i ≤ k for x ∈ Ai. Fix

2 0 < < 1 such that D0 d. For any x ∈ A0 deﬁne H(x) = [−2D0, 2D0] · x. Note that conditions (b) and (c) are satisﬁed so far: for (b) the side lengths of our boxes are all 4D0 +1, and (c) is vacuously true since any distinct x, y ∈ A0 satisfy ρ(x, y) > D1 10D0.

Suppose 0 < i ≤ k and H(x) have been deﬁned for all x ∈ A0 ∪ · · · ∪ Ai−1. We deﬁne

H(x) for all x ∈ Ai. Fix a particular x ∈ Ai. Note that there is an absolute bound C for the number of points y ∈ S0 so that ρ(x, y) ≤ 10D0; this bound C represents an optimum packing density of ρ-balls, and therefore is a constant that does not depend on the distance

D0 and the set S0. Therefore C can be ﬁxed before we picked D0 and . We may assume

D0 > C(d + 2).

40 There are at most C many points y ∈ S0 with ρ(x, y) ≤ 10D0, and hence there are at most C many H(y) that are already deﬁned. It follows that there are at most 2C many horizontal sides of these H(y) that we have to avoid by staying at least vertical distance d away. The upper horizontal side of H(x) can therefore be chosen in any way so that its distance to x is in between (1 − )D0 and (1 + )D0 and so that its vertical distance to any of the other at most 2C many horizontal sides is at least d. Similarly, we can choose the lower horizontal side and the vertical sides of H(x).

Now if x, y are distinct elements of Ai, then ρ(x, y) > D1 10D0, and hence the above construction of H(x) and H(y) are independent of each other, and condition (c) is fulﬁlled. By construction, we also fulﬁlled (b). By the marker property (ii) for the set S0 and condition (b), we also obtain condition (a).

We now deﬁne a subequivalence relation P of the orbit equivalence by

(x, y) ∈ P ⇐⇒ ∀s ∈ S0 (x ∈ H(s) ↔ y ∈ H(s)).

For any P -equivalence N, choose an s ∈ S0 in some sensible way such that [x]P ⊆ H(s). For example, we could always pick the upper-left-most such s.

We finally define a subequivalence relation R of P satisfying the requirement of our proposition. R is defined by specifying a partition of each P -region into rectangles, as follows. Fix a P -region N, and let s ∈ S0 be the corresponding marker. Let L1,...,Lm be all horizontal sides of other boxes H(t) intersecting H(s). Extend L1,...,Lm so as to create maximal horizontal line segments J1,...,Jm within N. Note that each Ji extends either an upper side or a lower side of another box H(t). Since H(t) and H(s) intersect, it must be that ρ(s, t) < 2D0, and therefore by our construction the vertical distance between any two lines in J1,...,Jm is at least d + 2, and the vertical distance between each horizontal side of

H(s) with each of J1,...,Jm is also at least d + 2. Finally we deﬁne the R-regions of N as follows. Given distinct x, y ∈ N, deﬁne (x, y) ∈ R if

• there is no Li in between x and y, and

41 • any of the following:

◦ neither x nor y lies on any Li; or

◦ x and y lie on the same Li; or

◦ one of x and y (say x) lies on some Li, which is an upper side of another box

H(t), and the other (y) is below Li and does not lie on any Lj; or

◦ one of x and y (say x) lies on some Li, which is a lower side of another box

H(t), and the other (y) is above Li and does not lie on any Lj.

We claim that each R-region is a rectangle. It is clear that each R-region is bounded

above and below by some lines Li (possibly inclusive) or horizontal sides of H(s). If an R-region is not a rectangle, then there is some horizontal (possibly partial) side of the region

other than the upper and lower sides given by Li. Such new side must also come from some other box H(t) that intersects H(s), by the deﬁnition of P . This is a contradiction to our construction since in the deﬁnition of R we took into account all sides of boxes H(t) intersecting H(s). We next verify that the corners of R-regions satisfy the conclusion of the proposition. For this we verify inductively that at any stage of the construction we do not create corners x, y with 1 < ρ(x, y) < d. Assume toward a contradiction that x, y are two corners with

1 < ρ(x, y) < d. Let Lx be the horizontal side of some H(s) from which x resulted being a corner. This means that either x is on an extension of Lx or x is distance 1 from it. Similarly let Ly be the horizontal side that caused y to be a corner. Then ρ(x, s) ≤ 4D0(1 + ) and

ρ(y, t) ≤ 4D0(1 + ), and so ρ(s, t) ≤ 10D0. Thus the vertical distance between Lx and

Ly is either ≤ 1 (it is possible that s = t) or the vertical distance of Lx and Ly is at least d + 2. This implies that the vertical distance of x and y is either ≤ 1 or at least d. A similar argument shows that the horizontal distance of x and y is also either ≤ 1 or at least d.

Remarks. (1) The construction clearly gives a clopen rectangular decomposition since the R-region of any point is identiﬁed with only information of points within distance

10D0. (2) It is clear that our R-regions are of bounded size. But with more work we can

42 restrict the possible side lengths to a ﬁxed (for instance 12) number of possibilities. (3) With more work we can eliminate double corners altogether. But this seems unnecessary for our purpose and therefore we omit this part.

2.3. Main Result in Summary

Theorems 2.3 and 2.1 can be combined into a single theorem, which sacriﬁces a bit of strength for clarity. Fix a discrete space D. Let Ψ be a collection of partial functions

ψ : Z2 → D whose domains are rectangular grid graphs. Suppose further that Ψ contains functions with domains of unbounded size and is closed downward in the pullback partial order ≤. We write Ψ ≤ ϕ if, for every ψ ≤ ϕ deﬁned on a grid graph, ψ ∈ Ψ.

Corollary 2.4. Let D and Ψ be as above. The following are equivalent:

2 ∃ϕ: F (2Z ) → D, Ψ ≤ ϕ with ϕ continuous,

∃n∃p ⊥ q ≥ n, ∃ϕ:Γn,p,q → D, Ψ ≤ ϕ, and

∀n∀p, q n, ∃ϕ:Γn,p,q → D, Ψ ≤ ϕ.

Remarks. (1) In this way, Ψ encodes an arbitrary property that must pass between

2 Z functions on F (2 ) and functions on Γn,p,q. (2) Corollary 2.4 is weaker than Theorems 2.1 and 2.3 in that it does not address

ϕK , but we will not need to consider ϕK for most examples. (3) We have that the second condition implies the third by successive application of Theorems 2.1 and 2.3. Nonetheless, it is not diﬃcult to modify the proof of Theorem 2.3 to

2 0 0 produce a homomorphism of Z graphs σ :Γn,p,q → Γn0,p0,q0 so long as p ⊥ q and p , q p, q. We would then have a direct argument that the second condition implies the third.

43 CHAPTER 3

EXAMPLES AND IMMEDIATE COROLLARIES

2 We begin with the continuous chromatic number of F (2Z ). A continuous four- coloring has been provided in previous work [3], but we present a simpler version here to showcase our results. We also show that no continuous three-coloring can exist.

2 Example 3.1. The continuous chromatic number of F (2Z ) is four.

2 2 Z Z Proof. We ﬁrst show χc F (2 ) ≤ 4 by providing a continuous four-coloring of F (2 ); By

Theorem 2.3, it suffices to show that there is a chromatic four-coloring ϕ:Γ1,2,3 → 4. Because chromatic colorings are reflection and rotation invariant, it suffices to color tiles Tca=ac,

ϕRx ϕRa ϕRb ϕRc ϕRd

ϕTca=ac ϕTda=ad ϕTdb=bd

ϕTcqa=adp ϕTcda=adc

2 Z Figure 3.1. The four-coloring of Γ1,2,3, whence F (2 ) can also be four-

colored. Thick lines separate boxes R∗, which must be consistent across the tiles.

44 Tda=ad, Tdb=bd, Tcqa=adp and Tcda=adc. We give this coloring in Figure 3.1. The ﬁgure is obviously a four-coloring on its tiles. We must also observe that each tile is consistent with the others in that each tile has copies of R∗ in the correct places. We have presented ϕR∗ by themselves for reference, and used thick lines to delineate the diﬀerent boxes R∗ where they appear in T∗.

2 Z Conversely, using Theorem 2.1, we show χc F (2 ) > 3 by showing no chromatic three-coloring ϕ:Γ1,p,q → 3 exists for any p ⊥ q.

Fix p ⊥ q and suppose ϕ exists. Tile Tcqa=adp was the only tile that we did not show was three-colorable itself in Figure 3.1, so we focus our attention there.

Let w(α) be the winding number about the triangle of any cycle α in K3. Then we have that w(αβ) = w(α) + w(β) where αβ is the concatenation of cycles α, β in K3. Each cycle of length 4 in K3 is trivial in that it contains some vertex twice, so the winding number of any cycle of length 4 of K3 is zero. The map ϕ is just a graph homomorphism to to the complete graph K3, so ϕ preserves cycles and cycle concatenations.

Consider the boundary of Tcqa=adp . Ra ∪ R× is just a cycle of length p; call this cycle a. Rc ∪ R× is also a cycle of length p, and Rb ∪ R× and Rd ∪ R× are each cycles of length q; call these cycles c, b, d respectively. If we consider α−1 to be the reversal of the nodes of

q −p −1 q −p −1 cycle α, we have that the boundary of Tcqa=adp is c ad a . The cycle c ad a is ﬁlled in by a grid of cycles of length 4, therefore cqad−pa−1 maps to a cycle with winding number

q p zero. That is, w(ϕ(c a)) = w(ϕ(ad )), hence the name Tcqa=adp . We compute

(w ◦ ϕ)(cqa) = (w ◦ ϕ)(adp),

q(w ◦ ϕ)(c) + (w ◦ ϕ)(a) = (w ◦ ϕ)(a) + p(w ◦ ϕ)(d),

q(w ◦ ϕ)(c) = p(w ◦ ϕ)(d).

Here, p divides q(w ◦ ϕ)(c), so p divides (w ◦ ϕ)(c), and pq divides (w ◦ ϕ)(cq). The numbers p and q are both coprime, so p and q are not both even, and one of c or d is an odd length cycle. All odd-length cycles in K3 have nonzero winding number, so (w ◦ ϕ)(c) and

45 (w ◦ϕ)(d) are not both zero. Suppose (w ◦ϕ)(c) = 0 (the other case is handled analogously). Now

(w ◦ ϕ)(c) 6= 0, which, because pq divides (w ◦ ϕ)(c), means that

|(w ◦ ϕ)(c)| ≥ pq.

But the length of the cycle cq is only pq, and it requires at least three nodes for each “wind”

around K3 in the sense of winding number, so

|(w ◦ ϕ)(c)| ≤ pq/3.

In the proof of Theorem 2.1, we observed that any n-pullback of Γn,p,q is indeed an

n-pullback of some single tile T∗, because tiles overlap by n. It is also true that any (n + 1)-

pullback of Γn,p,q is a pullback of a single tile. Strictly speaking, Figure 3.1 presented not ϕ, but rather ϕ restricted to each box and tile. Because each (n + 1) pullback of ϕ is a pullback of an individual tile, it suﬃces to ensure that ϕ restricted to each individual tile is a four-coloring to obtain that ϕ is a four-coloring itself. In the case of the four-coloring this is a completely pedantic distinction, but it will become more relevant with more complicated proofs.

The proof that there is no continuous three-coloring was the original goal of our research. Originally, we proved nonexistence of the three-coloring using a crude precursor to

the the hyper-aperiodic point that allows us to read oﬀ ϕTcqa=adp and ϕTdpa=acq . Generalizing

the three-coloring result to apply to increasingly complicated graphs beyond K3 led to the

main theorem; indeed, the “twelve-tiles” graphs Γn,p,q themselves emerged in an attempt to ﬁnd a graph Γ such that the question of existence of a continuous graph homomorphism

2 ϕ: F (2Z ) → Γ was hard. We were successful in ﬁnding a hard class of graphs in that we

found the class of graphs Γn,p,q to be “complete” for the problem in the sense of Theorems 2.1 and 2.3. We present the theory for undirected graphs developed during this process in

46 Chapter4. Another outcome of our research is that there does exist a Borel three-coloring,

2 Z i.e., χB F (2 ) = 3; this result is to appear. Related to the question of chromatic number is the question of edge chromatic num-

2 ber. Recall the edge chromatic number of the edge graph of F (2Z ), where two edges are considered adjacent if they share a vertex.

2 Example 3.2. The continuous edge chromatic number of F (2Z ) is ﬁve.

Proof. We represent an edge-coloring by assigning, to each node, an encoding of the colors

of the edges above it, below it, and to its sides. For an n-edge coloring, there are nP4 such possibilities. This representation comes with it the requirement that adjacent nodes must to agree on the edges they share. A solution with ﬁve colors is given in Figure 3.2; each square is ﬁlled with four colors, representing that node’s four adjacent edge colors.

2 Conversely, we claim that an edge-coloring of F (2Z ) with four colors cannot exist.

Fix p ⊥ q, and suppose ϕ is an encoding of edge colors on Γ1,p,q as in Figure 3.2. At least one

of p, q must be odd, so let m be p or q and be odd. Tile Tca=ac or Tile Tdb=bd (corresponding to p and q respectively) is an m × m torus graph. That is, part of ϕ gives four-edge-coloring an m × m torus. Pick a color η, and observe that every node has exactly one edge of color η, and each η-colored edge connects two nodes. Thus, the η-colored edges form a pairing on the nodes of the m × m torus. But the m × m torus graph has an odd number of nodes, so

a pairing is impossible.

2 Observe that we have also shown that no continuous pairing on F (2Z ) can exist. The

2 question of the Borel edge chromatic number of F (2Z ) remains open, but we suspect it to be ﬁve as well.

In a recent unpublished result, Marks showed that Martin’s conjecture holds iﬀ every

F (2) Borel function ϕF (2F (2)) : F (2 ) → 2, where F (2) is the free group with two generators, has

2 an inﬁnitely large monochromatic graph component. In the case of F (2Z ), the situation is much simpler.

47 ϕRx ϕRa ϕRb ϕRc ϕRd

ϕTca=ac ϕTda=ad ϕTcb=bc ϕTdb=bd

ϕTcda=adc

ϕTcba=abc ϕTcab=bac

ϕTdca=acd

ϕTcqa=adp

ϕTcaq=bpc ϕTcbp=aqc ϕTdpa=acq

Figure 3.2. The ﬁve-edge-coloring of Γ1,2,3. Each node is assigned four colors representing the colors of its adjacent edges. Also observe that the boundary nodes of each tile are colored in a consistent way, e.g., Rd is colored the same way everywhere it appears in the ﬁgure. 48 ϕRx ϕRa ϕRb ϕRc ϕRd

ϕTca=ac ϕTcb=bc ϕTdb=bd

ϕTcqa=adp ϕTcda=adc

Figure 3.3. A function of ϕ:Γ1,5,2 → 2 inducing a continuous function

2 2 Z ϕF (2Z ) : F (2 ) → 2 which has no infinite monochromatic graph component. Five of the twelve tiles are given; the other seven are rotations and/or reflec- tions of these five.

2 2 Z Example 3.3. There exists a continuous function ϕF (2Z ) : F (2 ) → 2 such that each monochromatic component is not only ﬁnite, it has size at most 3.

2 Proof. We construct a function ϕ:Γ1,5,2 → 2 such that the corresponding ϕF (2Z ) has only bounded monochromatic components. Five of the twelve tiles of ϕ are given by Figure 3.3; each of the other seven is a rotation and/or reﬂection of one of the tiles shown. Here we exploit the fact, mentioned after Example 3.1 that any 2-pullback of ϕ is in fact a 2-pullback of a single tile, and is therefore directly visible as a colored 2 × 2 subbox

in Figure 3.3. Observe that in each tile, wherever a vertex u ∈ Ra and a vertex v∈ / Ra

are shown, ϕ(u) 6= ϕ(v). This statement remains true if we replace Ra with any other R∗.

2 Z Because 2-pullbacks pass from F (2 ) to Γ1,5,2 and ultimately to individual tiles T∗, the set

2 Z 2 of boundary nodes between tiles of F (2 ) induced by ϕF (2Z ) consists of a union of ﬁnite

49 Figure 3.4. Toast.

maximal monochromatic components. Indeed, these components have size at most two.

For each tile T∗ of Γ1,5,2 the non-boundary nodes of T are readily seen to be a disjoint union of maximal monochromatic components of size at most three, hence the non-boundary

2 nodes in each tile of ϕF (2Z ) must be the same.

It is not diﬃcult to show that monochromatic graph components cannot be reduced to two nodes, even in the Borel case. Indeed, consider two horizontally adjacent nodes x, x0

0 2 2 such that ϕF (2Z )(x) = ϕF (2Z )(x ). If we require all monochromatic components to have size at most 2 and consider the two nodes x00, x000 immediately above x, x0 respectively, we have

00 0 000 00 000 2 2 2 2 2 2 ϕF (2Z )(x) 6= ϕF (2Z )(x ) and ϕF (2Z )(x ) 6= ϕF (2Z )(x ). But then ϕF (2Z )(x ) = ϕF (2Z )(x ). This fact forces colors to strictly alternate moving upward along the vertical line given by x, and we can modify the construction in Proposition 1.21 to ﬁnd a contradiction. Therefore the lowest bound for monochromatic components, in both the Borel and continuous cases, is 3. While not an application of Theorems 2.1 or 2.3, related to the question of chromatic

50 2 number of F (2Z ) is the “toast” fractal1 depicted in Figure 3.4. A motivation for toast was

2 to partition F (2Z ) into ﬁnite connected components with ﬁnite boundaries from which a

2 three-coloring of F (2Z ) could be built relatively simply. Since a continuous three coloring does not exist then neither does continuous toast; nonetheless, rather than demonstrate the construction of a three-coloring via toast, we show that no continuous toast can exist directly.

Definition. A toast structure is a pairwise disjoint collection T of cycles in the

2 2 graph of F (2Z ), such that for any two points x, x0 from the same orbit of F (2Z ), x and x0 are together circumscribed by some cycle in T .

Observe that the toast in Figure 3.4 is not layered; the outermost cycle shown has children of various levels from 0 to 2 as well as children with undeﬁned level.

Example 3.4. Continuous toast cannot exist.

Proof. Suppose a toast fractal exists, and it is continuous with respect to the Bernoulli

2 shift. Deﬁne a function ϕ: F (2Z ) → ω that takes each x to the diameter of the region enclosed by the smallest toast cycle circumscribing x. The function ϕ is continuous because ϕ(x) is determined by a pullback large enough to witness a cycle circumscribing x, and graph cycles are necessarily ﬁnite.

2 Z 2 Now, consider a compact invariant set K ⊆ F (2 ), such as K = ClF (2Z )([y]) with y provided by the sublemma of Theorem 2.1. The image ϕ[K] is bounded, which is impossible because cycles (and thus the regions they circumscribe) must grow without bound on each orbit of K.

Borel toast does, however, exist; the construction is to appear. Curiously however, “layered toast” (deﬁned below) does not exist, even with a Borel speciﬁcation; the proof is also to appear.

1The fractal was accidentally named by Ben Miller when a chalkboard drawing resembled toast with butter.

51 Definition 3.5. A layered toast structure is toast in which each cycle can be assigned a natural number such that cycles at level 0 circumscribe no cycles, and any cycle of level n circumscribes only cycles of level (n − 1) exactly.

2 From past work [2], it has been noted that F (2Z ) can be partitioned into almost- square marker regions, which for the purpose of the next theorem are rectangular regions of dimensions d × d, d × (d + 1), (d + 1) × d, or (d + 1) × (d + 1) for some ﬁxed d.A technical question, yet to be fully answered, is whether it is possible to eliminate any of the

2 four shapes and still partition F (2Z ).

Example 3.6. Fix d > 1. It is impossible, a continuous deﬁnition continuous with

2 respect to the Bernoulli shift, to partition F (2Z ) into rectangular regions if the above list of allowable dimensions omits either d × d or (d + 1) × (d + 1), i.e., it omits either of the squares.

Proof. Suppose (d+1)×(d+1) squares are not allowed; the argument for d×d is analogous.

2 Represent the partition by a continuous function ϕ: F (2Z ) → 2, where ϕ(x) = 1 iﬀ x is a left or bottom boundary node of some rectangle in the partition. Fix arbitrary p ⊥ q, and

0 0 0 use Theorem 2.1 to produce a ϕ :Γ1,p,q → 2. ϕ Tca=ac and ϕ Tdb=bd thus encode partitions of the torus graphs Tca=ac and Tdb=bd into rectangles. Because (d + 1) × (d + 1) boxes have been forbidden, all remaining rectangles have a

2 number of nodes divisible by d. One of p and q is not divisible d so either Tca=ac (with p

2 many nodes) or Tdb=bd (with q many nodes) cannot be tiled by the remaining rectangles. But then ϕ0 does not exist. This was for an arbitrary p ⊥ q, so the partition itself cannot exist by Theorem 2.1.

Observe that this proof goes through if we replace d and d+1 by two relatively prime numbers d and d0.

Whether a Borel version of such a partition exists is an open question, as well as the question of whether one we omit either of the non-square tiles in the continuous setting.

52 2 Indeed, there is evidence that it may be possible to partition F (2Z ) into only the square tiles with a continuous construction.

53 CHAPTER 4

UNDIRECTED GRAPH HOMOMORPHISMS

We now turn our attention to the question of whether there exist continuous undi-

2 rected graph homomorphisms ϕ: F (2Z ) → Γ for various undirected graphs Γ. This research

2 arose as a natural extension of determining the continuous chromatic number of F (2Z ) in

that a three-coloring is merely a graph homomorphism to the complete graph K3 (without loop edges).

2 Our proof that there is no continuous three-coloring of F (2Z ) relied very heavily on the notion of winding number of the graph. Most of our theory generalizing these results involve generalizing the notion of winding number. We do this using homotopy groups and group homomorphic images of homotopy groups; this theory will constitute the present section. We now reﬁne and expand upon some of the basic graph-theoretic terms used in Chapter3. For the remainder of the section, all graphs will be undirected; we disregard any G-graph structure. Graphs will not have loop edges unless otherwise noted; this is because there trivially exists a homomorphism from any graph to a graph with a loop edge.

Definition 4.1. A cycle of length ` or `-cycle is a graph homomorphism from the Cayley graph of Z (i.e., the cycle graph of length `) for some positive integer `.A subcycle `Z

γ0 of a cycle γ is a cycle whose image (considering both vertices and edges) is a subgraph of the image of γ.A spur of γ is a subcycle of γ of length 2; spurs frequently represent degeneracies that we use various means to remove.

Cycles will be denoted with tuples, e.g., (u, v, w) – here it is understood that u is “ﬁrst” in the sense that it is the image of the identity element of Z , and w connects to u `Z in the cycle.

0 0 0 0 Definition 4.2. For two cycles γ = (u0, u1, . . . , uk) and γ = (u0, u1, . . . , uk0 ) where

0 0 a 0 0 0 0 u0 = u0, the concatenation of γ and γ is denoted γ γ and is deﬁned as (u0, u1, . . . , uk, u0, u1, . . . , uk0 ).

54 We occasionally suppress the symbol a to emphasize algebraic considerations, and write γn to indicate iterated concatenation of γ. The notation γ−1 is used for the reversal of γ, maintaining its ﬁrst element; i.e., For any `-cycle γ, γ−1(k + Z) = γ(` − k + Z).

−1 Here, for example, for example, (u0, u1, u2, u3) = (u0, u3, u2, u1).

2 We have already made use of the fact that four-cycles are quite prevalent in F (2Z ). To extend this fact, we deﬁne a sense in which four-cycles are nullhomotopic, and thus the image of any four-cycle is nullhomotopic. We may then use homotopy arguments to draw conclusions about the existence of graph homomorphisms. We recall the deﬁnition of cell-complex for this reason.

Definition 4.3. A cell-complex is a Hausdorﬀ space Γ partitioned into sets called cells such that for each cell κ, there is a continuous map f from a k-dimensional closed unit ball to Γ such that

• k is ﬁnite, • the image of f is the closure of κ in Γ, • f restricted to the open unit ball is a homeomorphism to κ, and • the boundary of κ is a ﬁnite union of cells of dimension less than k.

Additionally, Γ must have the coarsest topology with these conditions; namely, a set is open in Γ iﬀ it meets the closure of every cell in an open set. We call k-dimensional cells k-cells. We call 0-cells vertices, 1-cells edges, and 2-cells faces.

This deﬁnition recasts the notion of graph in terms of topology – a graph is just a cell- complex with cells of dimension at most 1. Further, a graph homomorphism becomes a continuous function between one-dimensional cell complexes that preserves cells. A cell-complex homomorphism is the arbitrary-dimensional generalization of a graph homomorphism. The topology on a cell complex has nothing at all to do with the totally disconnected topologies 2G that we have been using thus far. Indeed, a cell complex is made up of sets that are homeomorphic to Euclidean balls. Each ball of dimension at least one has uncountably many points, and each such ball represents a single cell such as an edge or face. We use a

55 cell complex as a convenient combinatorial object generalizing a graph; the topology tells us which faces, edges, vertices, and other cells Γ may have and gives us a convenient way to generalize the relationship between an edge and its vertices. Namely, a cell κ is a boundary cell of κ0 if κ is in the topological boundary of κ0. A useful concept when dealing with cell complexes is that of homeomorphism.

Definition 4.4. Two cell complexes are homeomorphic if they are homeomorphic as topologies.

Observe here the cell structure is irrelevant to homeomorphism. When some properties of interest are preserved by homeomorphism, we will sometimes simplify a cell complex by working on a homeomorphic version of our complex with fewer cells. Purely combinatorial definitions of cell complex do exist; these are much more analogous to the usual definition of a graph. Such combinatorial definitions provide functions from cells to their boundary cells, where each cell is considered an atomic object much like the usual notions of vertices and edges. Indeed, there is even a purely combinatorial notion of homeomorphism that is equivalent to the topological notion; this works by combining and splitting adjacent cells. These combinatorial definitions are more cumbersome to use than the topological notions, especially for homeomorphism arguments, so we avoid them. We now consider homotopy on cell complexes. Throughout the following, we will assume that cell complexes are always topologically connected.

Definition 4.5. Let Γ be a cell complex. Let I be the unit interval of the real line. Call a continuous function ψ : I → Γ a path, and call it closed if ψ(0) = ψ(1). Consider closed paths γ0 and γ1. If there exists a continuous map ζ : I × I → Γ such that ζ(t, 0) = γ0(t),

ζ(t, 1) = γ1(t), and ζ(0, t) = ζ(1, t) for all t ∈ I, then γ0 and γ1 are mutually homotopic. The class of all closed paths homotopic to a closed path γ is called the homotopy class of γ; this is an equivalence relation on the closed paths of Γ. A closed path ψ is called nullhomotopic if it is homotopic to a constant path; i.e., if ψ is homotopic to a path whose image is a single point.

56 Definition 4.6. Let Γ be a cell complex. Choose a distinguished point x (we show

momentarily that it does not matter which). Consider two closed paths ψ0, ψ1 that each

start and end at x. Let ψ0 a ψ1 : I → Γ be deﬁned as   ψ0(2t), if t ≤ 0.5, ψ0 a ψ1(t) =  ψ1(2t − 1), if t > 0.5.

It is straightforward to show that this places a group structure on homotopy equivalences

of paths starting and ending x where [ψ0][ψ1] = [ψ0 a ψ1]. In particular, the identity is the class of nullhomotopic paths, and [ψ]−1 can be achieved by reversing the direction of ψ. This

group is called the homotopy group of Γ, notated π1(Γ). The notation π1 may also be applied

to paths ψ; i.e., π1(ψ) = [ψ].

As noted, it is simple to show that π1(Γ) does not depend on the choice of distinguished point x. Indeed, to change the distinguished point from u to v, pick a path ψ from v to u and conjugate (via concatenation) all closed paths starting and ending with u with the path ψ, and take homotopy equivalences as before. The resulting group will be isomorphic to the original.

It is also important to observe that the homotopy group of a cell-complex Γ depends only on the topology of Γ and not on the partition into cells. Thus if we can show that two cell-complexes have homeomorphic topologies (which will in general be clear visually), we will have that they share a common homotopy group.

2 To exploit the ubiquity of four-cycles in F (2Z ), we deﬁne a special operation “∗” to promote graphs to 2-dimensional cell-complexes. Namely, Γ∗ is Γ together with, for each four-cycle γ, a face whose boundary1 is the cycle γ. Also, for any graph homomorphism

∗ ∗ ∗ ϕ:Γ0 → Γ1, we have a corresponding homomorphism of cell complexes ϕ :Γ0 → Γ1.

2 Z ∗ Now, on any orbit [x] ⊆ F (2 ), we have that π1([x] ) is just the trivial group. It is easy to check that cell-complex homomorphisms preserve homotopy classes, so we have

1Here, and henceforth, we abusively ignore the distinction between the cycle γ and the path through the topology Γ corresponding to γ.

57 2 2 that, for any graph homomorphism ϕ: F (2Z ) → Γ, the image of any cycle in F (2Z ) is nullhomotopic in Γ∗. This observation will be the basis for the main results of this section.

∗ ∼ 2 Example 4.7. Let p ⊥ q. Consider Γ1,p,q. We have π1(Γ1,p,q) = Z .

Proof. In this graph, because n = 1, we have that R× is a singleton set; call this vertex x.

As we observed in Example 3.1, the union Ra ∪ R× is merely a cycle, which we may call a.

Likewise, each of Rb,Rc,Rd deﬁnes a cycle in this manner, which we call, respectively, b, c, d.

∗ Then, the tile Tca=ac establishes the homotopy π1(ca) = π1(ac) in Γ1,p,q, again hence the

name Tca=ac. Indeed, each tile corresponds to a relation in a presentation of the homotopy

∗ group π1(Γ1,p,q).

The toroidal tiles Tca=ac,Tcb=bc,Tda=ad,Tdb=bd grant that each of a, b commutes with

each of c, d. There is a visually obvious automorphism of Γ1,p,q taking a, b to c, d and c, d

∗ ∗ to a, b, so each pair (a, b) and (c, d) behaves identically in π1(Γ1,p,q). Then, π1(Γ1,p,q) is a product of two isomorphic subgroups, ha, bi and hc, di.

Consider the subgroup ha, bi. Inspecting tile Tcab=bac, we observe ab = ba. Inspecting

q p q p tile Tcaq=bpc, we observe a = b . A group with presentation ha, b | a = b i is isomorphic to

∗ Z. We have enough four-cycles, then, to see that π1(Γ1,p,q) is some homomorphic image of 2 Z ; with more work, we could ensure we have found all four-cycles in Γ1,p,q and thus show ∗ ∼ 2 that π1(Γ1,p,q) = Z .

With still more work, we can show that the choice of n ≤ p ⊥ q is irrelevant to these

∗ ∼ 2 computations, so π1(Γn,p,q) = Z .

Example 4.8. For any graph Γ, π1(Γ) is free.

∗ Proof. As Γ has no faces of its own (unlike Γ ), its homotopy group has no relations.

We now introduce our ﬁrst homotopy-based result concerning the existence of continuous graph homomorphisms.

Theorem 4.9 (Homotopy-based negative condition for graph homomorphisms). Sup- pose there are inﬁnitely many pairs (j, k) ∈ ω × ω, with min{j, k} unbounded but j, k not

58 necessarily relatively prime, such that for each j-cycle γ of Γ, γk is not a j-th power in

2 ∗ Z π1(Γ ). Then, no continuous graph homomoprhism ϕ: F (2 ) → Γ exists.

2 Proof. By contrapositive. Suppose a continuous map ϕ: F (2Z ) → Γ exists.

Fix j, k 1. Applying Corollary 2.3 to ϕ, we obtain a continuous graph homomor-

0 0 phism ϕ :Γ1,j,k → Γ. Each 2-pullback of ϕ can be found in ϕ, and these pullbacks are

0 0 suﬃcient to witness that ϕ is also a graph homomorphism. Consider ϕ Tcqa=adp . By tracing the boundary of this function from the upper-left hand corner, moving clockwise, we ﬁnd a nullhomotopic cycle of the form ckad−ja−1. Then, in Γ∗,

k j −1 π1(c ) = π1(ad a ),

−1 j = π1((ada ) ),

−1 j = π1(ada ) .

whence ck is a j-th power. This holds arbitrary j, k, not necessarily coprime but each

suﬃciently large.

In many cases, it’s possible to reduce to the following simpler condition.

Theorem 4.10 (Weighting-based negative condition for graph homomorphisms). Let Γ be a graph. Assign directions (forward and backward) to its edges arbitrarily. Consider a weighting of the edges w : E(Γ) → G, G an additive Abelian group. For any cycle γ of Γ, let w(γ) be the sum of the weights encountered along a traversal through the cycle, counting reverse edges with negative weights. Suppose that w(γ) = 0g for any four-cycle γ, and there are inﬁnitely many p such that, for any p-cycle γ, w(γ) is not a p-th multiple of an element

2 in G. Then, no graph homomorphism ϕ: F (2Z ) → Γ exists.

∗ Proof. First observe that w = f ◦ π1 for some group homomorphism f : π1(Γ ) → G.

∗ Indeed, each element g ∈ π1(Γ ) is of the form π1(γ) for some cycle γ in Γ, so f is deﬁned everywhere on its domain. If π1(γ) = π1(β), then γ and β diﬀer by a product of conjugates of

59 four-cycles. But for each such conjugate of a four cycle η, w(η) = 0G, therefore w(γ) = w(β). So f is well-deﬁned. Finally, for any cycles γ, β in Γ,

−1 (f ◦ π1(γ))(f ◦ π1(β)) = w(γ) − w(β)

−1 = w(γ a β )

−1 = f ◦ π1(γβ ), so f is a homomorphism.

We now prove the theorem, again by contrapositive. Suppose ϕ exists. Then, by Theorem 4.9, for all (p, q) with min{p, q} large enough, including p ⊥ q, there is a p-cycle

q ∗ γ such that γ is a p-power in π1(Γ ). Consider a pair p, q and cycles γ, β of Γ such that

q p π1(γ) = π1(β) . Let m, ` be integers such that mq + `p = 1, that is, `p = 1 − mq. We have

q p π1(γ ) = π1(γ ),

mq mp π1(γ) = π1(β) ,

1−`p mp π1(γ) = π1(β) ,

mp `p π1(γ) = π1(β) π1(γ) ,

f ◦ π1(γ) = (mp)f ◦ π1(β) + (`p)f ◦ π1(γ),

w(γ) = (mp)w(β) + (`p)w(γ),

= p(mw(β) + `w(γ)),

and this holds for all but ﬁnitely many p.

Here, by passing from π1 to w, we are able to leverage the fact that G is Abelian to dispense with the parameters in the hypothesis of Theorem 4.9. Also, for many graphs,

using G = Z suﬃces to prove nonexistence of a graph homomorphism, thus reducing the task of ﬁnding weights to an integer programming problem.

2 As a beginning example, we revisit the chromatic number of F (2Z ).

60 1 0

Figure 4.1. The complete graph of order 3, K3, with a weighting function.

2 Z Example 4.11. There is no continuous graph homomorphism ϕ: F (2 ) → K3, where

K3 is the complete graph of size 3.

Proof. We use Theorem 4.10 with G = Z and the weighting function given in Figure 4.1.

The graph K3 has no four-cycles, so we need not worry that the sum of the weights of any four-cycle is 0G. Of the remaining cycles, each odd cycle γ has w(γ) 6= 0G; indeed, a cycle γ with w(γ) = 0G must be equal to the trivial cycle modulo spurs, and thus must have even length. Fix a γ of odd length p. We are attempting to show |w(γ)| is low compared to the length of γ, so we may assume without loss of generality that γ is without spurs. At most

1 one third of the weights of γ may be nonzero, so 0 < 3 |w(γ)| ≤ p and p - w(γ).

Here, of course, w(γ) is just the winding number as we used previously.

The nonexistence of a continuous graph homomorphism to K3 leads directly to the following.

2 Z Example 4.12. There is no continuous graph homomorphism ϕ: F (2 ) → ΓP , where

ΓP is the Petersen graph.

Proof. The Petersen graph is given in Figure 4.2, along with a three-coloring. Since no

Γn,p,q is three-colorable, no Γn,p,q admits a homomorphism to ΓP .

Our treatments of Petersen graph and K3 were both trivial corollaries of Theo-

2 rem 4.10. The “clamshell” shaped graph ΓJ in Figure 4.3 is a more robust example. This graph was created so as to have χ(ΓJ ) > 3 and to have several nontrivial four-cycles, and

2This graph is due to Steve Jackson.

61 0

1 1 1 2 0

2 0 0 2

Figure 4.2. The Petersen graph ΓP , with a three-coloring on its nodes.

2 thus to be resistant to “trivial” methods of determining whether F (2Z ) admits a continuous homomorphism to it.

2 Z Example 4.13. There is no continuous graph homomorphism ϕ: F (2 ) → ΓJ .

Proof. We let G = Z and assign 1 to each forward edge in Figure 4.3. Careful inspection of ΓJ reveals that the four-cycles are:

0 (x, uj, x , uk) ∀0 ≤ j 6= k ≤ 4,

0 (x, x , u0, v0),

0 (x, u4, v4, x ),

(x, u1, v1, v0), and

0 (x , u3, v3, v4).

For each such four-cycle γ, we can verify by hand w(γ) = 0.

We claim that for inﬁnitely many p, namely odd p, each p-cycle γ has p - w(γ). As with K3, we will proceed by showing 0 < |w(γ)| < p for any cycle γ of odd length p. Fix p odd, and a cycle γ of length p. We have w(γ) 6= 0 because each edge has weight ±1, and there is no odd sum of ±1 equal to zero.

To show |w(γ)| < p it suﬃces to conﬁrm that there is no cycle in ΓJ consisting entirely of forward edges; thus the magnitude of the weight must always be less than the length. Suppose γ consists entirely of forward edges. The node u0 cannot appear in γ

62 u4

x v0 v1 v2 v3 v4 x0

Figure 4.3. Steve Jackson’s “Clamshell” graph ΓJ . We take a weighting

function assigning 1 ∈ Z to each forward edge. because u0 has no outgoing edges. But then, x cannot appear in γ either, because the only

0 0 outgoing edge from x is to u0. Then x cannot be in γ because the only outgoing edges of x are to u0 and x. We now have that none of u0, u1, u2, u3, and u4 can appear in γ. Finally,

γ cannot consist entirely of nodes v0, v1, v2, v3, and v4 while containing only forward edges, so γ must have both forward and backward edges.

We now have for any p-cycle with p odd, p - w(γ), and by Theorem 4.10, there is no

2 Z continuous homomorphism from F (2 ) to ΓJ .

We note here that while simpler to use, the “weighting” condition is strictly stronger than the “homotopy” condition. Consider the graph ΓK in Figure 4.4. To make it easier to visualize the graph in question, we label several of the nodes to indicate identiﬁcation. That

63 x c1 c2 c3 c4 x

a1 a4

a2 a3

a3 a2

a4 a1

x c1 c2 c3 c4 x

∗ Figure 4.4. A graph ΓK such that ΓK is a cell-complex homeomorphic to the Klein bottle. Each labeled node is identiﬁed with each other node with the same label. Unlabeled nodes are not identiﬁed with other nodes.

is, each labeled node is identiﬁed with each other node with the same label; unlabeled nodes

are not identiﬁed with other nodes. Notice that ΓK would be a 5 × 5 torus graph, i.e., the

2 Z Cayley graph of ( 5Z) , except that the progression of nodes ai is downward ascending on the left side of the graph and upward ascending on the right side of the graph.

2 Z Example 4.14. There is no continuous graph homomorphism ϕ: F (2 ) → ΓK .

Proof. It is our intent that the spurless four-cycles of ΓK ares just the (exactly) ﬁfteen “obvious” four-cycles that appear explicitly in Figure 4.4, i.e., the spurless four-cycles that appear in the 6 × 6 grid graph before identifying each node with a common label. Observe that by removing any one column of horizontal edges and any one row of vertical edges, we obtain a 5 × 5 grid graph. Each spurless four-cycle has exactly four edges, and is therefore contained entirely in such a 5 × 5 grid. The only such spurless four-cycles appear in the pre-identiﬁcation 6 × 6 graph.

Let γ be the cycle (x, c1, c2, c3, c4) and α be the cycle (x, a1, a2, a3, a4). Because the

∗ four-cycles in ΓK are the obvious ones, ΓK is a cell-complex homeomorphic to the Klein ∗ −1 bottle. Thus, π1(ΓK ) is the fundamental group of the Klein bottle, i.e., hg, h|gh = hg i.

For convenience, we take g = π1(α), h = π1(γ). Recall, for the fundamental group of the Klein bottle, we can uniquely write any

64 i j i j element in the normal form g h (equivalently, π1(α) π1(γ) ) simply by commuting instances of g leftward. That is, consider a word which is some product of h±1, g±1. Let Σ+ be the total power of g for each instance of g±1 that have an even number of instances of h±1 to its left. Deﬁne Σ− analogously; only replace “even” with “odd”. Let ∆ be the total power of all instances of h±1. Then, we can write our word as

gΣ+−Σ− h∆

Uniqueness of this normal form follows from the observation that both multiplication by a conjugate of ghgh−1 and introduction/elimination of an instance of gg−1 or hh−1 preserve the closed form values of Σ+, Σ− and ∆ thus computed. This normal form is also the

∗ i j shortest word for any element of π1(ΓK ). Indeed, beginning with a word g h and repeatedly permuting two letters gh ↔ hg−1 or introducing an instance gg−1 or hh−1 mid-word, we can produce any other word for the same group element; doing so either preserves or increases the length of the word

Consider two large numbers p, q with p, q, 2 pairwise coprime. Consider a p-length

i j i j cycle η of ΓK . η is homotopic to α γ for some i, j. Now, much as g h is the shortest word for its group element, αiγj is the shortest path in its homotopy class. Indeed, similarly to before, we can produce any path homotopic to αiγj by repeatedly either appending a spur, or replacing two consecutive edges on a four-cycle with the other two edges on that four-cycle. So the length of αiγj is at most the length of η, i.e., 5|i| + 5|j| ≤ p. Then |i| + |j| < p. We then have that if p divides i (or j), then i (or j) must be zero.

We now show, by contradiction, that ηq is not a p-th power. Suppose it is. Then,

q i0 j0 p i j q π1(η ) = (g h ) = (g h ) for some i0, j0. The exponent on h of (gi0 hj0 )p is j0p (regardless of the exponent on g), and the exponent on h of (gihj)q is jq, so we have j0p = jq by uniqueness of the normal form. Thus, j is a multiple of p, so j is zero. Then, i0p = iq, and i is also zero, i.e., η is nullhomotopic. But this is impossible, because η has odd length.

65 We have found inﬁnitely many p such that no p cycle η has ηq a p-th power, so there

2 Z is no continuous graph homomorphism ϕ : F (2 ) → ΓK .

Consider, on the other hand, the weighting-based negative condition applied to Ex- ample 4.14.

Example 4.15. The weighting-based negative condition fails for the Klein bottle graph

ΓK .

Proof. Consider w : edges(ΓK ) → G, G additive Abelian, such that the sum along any

∗ four-cycle is 0G. Let f : π1(ΓK ) → G be such that w = f ◦ π1. Here

−1 0g = f(ghgh )

= 2f(g) + f(h) + f(h−1)

= 2f(g).

Then, for any p odd, f(g) is a p-th multiple – namely, f(g) = pf(g). Obviously 0G is also a

j k p-th multiple. If η is a four-cycle starting at x, and j, k are integers, we have f ◦π1(γ a η ) ∈

j k {0G, f(g)}, and thus is a p-th multiple. We can adjust j, k so that the length of γ a η , namely 5j + 4k, is equal to any p high enough. Therefore there can be no inﬁnite collection of p for which each p-cycle is not a p-th multiple in G.

Homotopy theory also gives a positive condition for the existence of a continuous graph homomorphism.

Theorem 4.16. Suppose there exists an odd-length cycle γ of Γ such that |π1(γ)|= 6 0

∗ ∗ in π1(Γ ), where |π1(γ)| is the order of π1(γ) in π1(Γ ). Then, there is a continuous coloring

2 ϕ: F (2Z ) → Γ.

Proof. By Theorem 2.3, it suﬃces to ﬁnd a mapping ϕ:Γ1,p,q → Γ for some p ⊥ q. Let

` be the length of γ, and let m = |π1(γ)|. Let p m` be prime, and let q = p + m. Because p > m, p does not divide m, so p does not divide q. Thus, p ⊥ q. This gives us our parameters p, q.

66 Before we proceed to the coloring, we make the additional observation that q must

∗ be odd. Indeed, π1(Γ ) is just π1(Γ) modded out by conjugates of four-cycles in Γ, and each such conjugate is a cycle of even length. Thus, no odd length cycle is nullhomotopic. Since γ has odd length and γm (with length m`) is nullhomotopic, m must be even, whence q is odd. So we have that both p, q are odd length.

Let γ0 and γ1 be the ﬁrst two elements of γ. Extend γ with alternating instances of

γ0, γ1 until it is length p, and call this extension α – we can do this because p is odd and

γ is odd length. Similarly, extend γ with alternating instances of γ0, γ1 until it is length q,

and call this extension β. In our coloring, γ0 will correspond to R×, α to Ra,Rc, and β to

Rb,Rd.

Consider, for the moment, the boundary of the long tile Tcqa=adp . Moving counterclockwise from the top left corner and disregarding the nullhomotopic “ﬁller” between copies

of γ, we encounter p copies of γ corresponding to Rc, one copy of γ corresponding to Ra, q

backwards copies of γ corresponding to Rd, and one ﬁnal backwards copy of γ corresponding

p−q to Ra again. Thus, the boundary of this tile is homotopic to γ . But we have arranged that m = p − q, so the boundary is actually nullhomotopic. Similarly, all twelve tiles have nullhomotopic boundaries, so our scheme has, so far, avoided homotopy objections.

Having filled the boundaries of the tiles, we may fill each tile in isolation. The general strategy will be to connect pairs of copies of γ at the boundary of a tile with special paths, and then fill the remaining space with alternating γ0, γ1. This process is illustrated in Figure 4.5. This process is quite flexible; a path of width ` + 1 and almost arbitrary shape can connect two copies of γ, with the right orientations.

Connecting paths in this manner suﬃces for the ﬁrst eight tiles, because each has the same number of “forward” copies of γ as “backward” copies, proceeding counterclockwise

around the tile. But for the “long” tiles Tcqa=adp , Tdpa=acq , Tcaq=bpc, and Tcbp=aqc, there will always be m unmatched copies of γ. To concretely manifest the fact that γm is nullhomotopic, we construct a rectangular region, which we call the hub, whose boundary consists of γm

together with alternating γ0, γ1. We place a copy of the hub and connect the remaining

67 γ :

Figure 4.5. An illustration of how to “connect” two copies of γ using a winding path. copies of γ to it. We can chose q > p m` depending on γ, so we can ensure there is plenty of room to ﬁt such a hub as well as the necessary connections.

m Qk Because γ is nullhomotopic, there exists a ﬁnite product i=0 λi = λk, λk−1, . . . , λ0 of conjugates of four-cycles such that

k ! ! Y m π1 λi γ = eπ1(Γ). i=0

Qk m That is, the cycle i=0 λi γ diﬀers from a degenerate cycle only by spurs. We are able to leverage this fact by building patterns that “eliminate” spurs in one direction. Suppose a particular cycle η diﬀers from the identity cycle only by spurs, i.e., η

∗ is nullhomotopic in π1(Γ) as opposed to merely nullhomotopic in π1(Γ ). We can build a rectangular region whose boundary consists of η plus alternating η0, η1 by successively writing versions of η produced by eliminating a single spur. An example of this process is shown in Figure 4.6. Each spur eliminated shortens the cycle by one vertex, but because

68 Figure 4.6. A process for eliminating spurs in a cycle one at a time.

we only ever shorten by a single vertex at a time, we can always situate the shortened cycle above its predecessor.

Now, suppose η = λi for some i; here as well we can produce a box eliminating η. Because η is conjugate to a four-cycle, we can let

η = (η0, η1, . . . , ηj, θ0, θ1, θ2, θ3, θ0, ηj, . . . , η1)

where θ = (θ0, θ1, θ2, θ3) is the four-cycle that η is conjugate to. Now, if η appears horizontally in some grid, we can eliminate θ in this cycle moving upwards by laying two rows on top of it:

... ηj−2 ηj−1 ηj θ1 ηj ηj−1 ηj−2 ...

η0 ... ηj−1 ηj θ1 θ4 θ1 ηj ηj−1 ... η0

η0 η1 ... ηj θ1 θ2 θ3 θ4 θ1 ηj ... η1 η0

In the middle of this arrangement we exploit the fact that θ is a four-cycle, and literally include it as a four-cycle in our map. The top of this arrangement diﬀers from the identity by spurs, and can be handled b spur elimination as above.

We construct our hub on a large enough rectangle by starting with a line consisting of

69 γm

1G 1G 1G

m λk ··· λ2 λ1 γ

Figure 4.7. Construction of a hub. We use conjugates of the four-cycles

m Qk −m λ1, λ2, . . . λk to eliminate γ when i=0 λk = γ .

Qk m i=1 λi γ . Above each individual λi, taper off to nothing using the four-cycle elimination Qk m method just established. Below the whole product i=1 λi γ , taper off to nothing using the spur-removal method of Figure 4.6. This process is shown in Figure4. The construction so far leaves only γm exposed; create a path to the boundary of the hub, and fill the rest of the space in with a checkerboard pattern of γ0, γ1.

2 When there exists a homomorphism from F (2Z ) to a graph Γ, Theorem 4.16 can reduce finding such a homomorphism to a search problem suitable for a computer. For example, Theorem 4.16 applies to two well-studied and well-known graphs, the Chvátal graph and the Grötsch graph.

Example 4.17. Let ΓC be the Chv´atalgraph and ΓG the Gr¨otschgraph. There exist

2 2 Z Z continuous graph homomorphisms ϕC : F (2 ) → ΓC and ϕG : F (2 ) → ΓG.

Proof. We can establish that each graph has a cycle γ with π1(γ) = 2 by coloring a box graph with the graph, placing γ at the top of the box, γ−1 at the bottom, and leaving the

70 0 1 0 11 7 2 1 0 4 5 3 10 6 1 0 3 11 6 0 3 9 5 4 0 10 7 3 2 8 4 0 3 9 8 0 1 2 7 11 0 3 2

Figure 4.8. The Chv´atal Graph, given at left. The odd cycle γ = (0, 11, 7, 2, 1, 0) has order 2 in this graph, because there is a (necessarily nullhomotopic) graph homomorphism from a grid graph to the Chv´atalGraph with boundary homotopic to γ2, given at right.

1 0 1 2 3 9 0 6 0 2 6 2 1 7 10 6 5 7 10 0 1 5 10 6 0 9 8 6 0 4 8 2 6 4 3 0 9 3 2 1 0

Figure 4.9. The Grötzsch Graph. The odd cycle γ = (0, 1, 2, 3, 9, 0) has order 2 for similar reasons as with the Chvátalgraph. sides alternating between two colors. The alternation between two colors will contribute nothing to homotopy class of the boundary of the box, whence γ2 will be homotopic. The Chvátaland Grötsch graphs are given along with their solutions in Figures 4.8 and 4.9 respectively.

The following lemma, while not directly used in any proof, was nonetheless instru- mental in this research. It allows us to cook up graphs with arbitrary homotopy groups, and generalizes the construction of the Klein bottle graph in Figure 4.4.

Lemma 4.18. Consider a group G with presentation hG0 | Ri and a group homomor-

Z ∼ phism f : G → 2Z. There exists a graph Γ such that π1(Γ) = G and, for each cycle γ of Γ,

71 γ has odd length iﬀ f ◦ π1(γ) = 1 + Z.

Proof. We will construct Γ by creating a cycle γg for every g ∈ G0 and a mesh of four- cycles creating a face for every relator in R. We begin Γ with a vertex x that will be common to each cycle γg for g ∈ G0. Then, for each g ∈ G0, we introduce the cycle

(0) (1) (2) (3) (`g−1) (0) γg = γg , γg , γg , γg , . . . , γg where γg = x and

  100, if f(g) = 0 + Z, `g =  101, if f(g) = 1 + Z.

(0) In this scheme, each cycle γg has disjoint nodes, except γg , which is the same for

every g ∈ G0 and always equal to x. Observe that if we stopped the construction here, we ∗ ∼ ∼ would have that π1(Γ ) = π1(Γ) = F (G0), where F (G0) the free group with generators from

G0. It remains to guarantee the relators from R. F (G ) We have G = 0 G by deﬁnition. That is, R ⊆ F (G ). Fix r = g g g . . . g hRi 0 0 o 1 2 (`r−1) −1 with each gi ∈ G0 ∪ (G0) . Let

γ = γ a γ a γ a ... a γ . r g0 g1 g2 g(`r−1)

Let L be the length of γr, and enumerate

(0) (1) (L−1) γr = γr , γr , . . . , γr .

We are given r = 1G, therefore f(g0) + ··· + f(g`r−1) = f(r) = 0 + Z, and L must be is even.

Thus, we may place γr as the border of a grid graph of height 8, as shown in Figure 4.10

In analogy with the twelve tiles T∗ that make up Γn,p,q, we name this conﬁguration S Tr. We let Γ = r∈R Tr, with nodes identiﬁed as labeled. As with the Klein bottle graph of Example 4.14, we seek to ensure that by modding out by labeled nodes, we do not introduce any unintended four-cycles. To be perfectly explicit about what we mean by “unintended”, consider Γ as a disjoint union of grid graphs of height 8, modded out by labeled nodes. Modding out in this manner adds a few four-cycles that were not present in the original

72 L L 0 1 2 3 2 −8 2 −7

L L−1 2 −6

L L−2 2 −5

L−3 L −4 ··· 2 L L−4 2 −3

L L−5 2 −2

L L−6 2 −1

L L−7 L−8 L−9 L−10 L +1 2 2

Figure 4.10. A rectangular grid graph Tr ⊆ Γ establishing the homotopy

∗ group relation r = 1 in π1(Γ ). Each labeled node is identiﬁed with each other node with the same label. Unlabeled nodes are not identiﬁed with other nodes.

(k) To avoid clutter, for each k < L, γr is abbreviated simply k.

disjoint union; we will show that any such “new” four-cycle is already a product of existing four-cycles.

Consider a four-cycle γ in Γ which does not correspond directly to a four-cycle in one of the original grid graphs. Cycle γ must include both boundary (labeled) nodes and

non-boundary (unlabeled) nodes, so γ must be of the form (u0, b0, u1, b1) where b0, b1 ∈ γr

for some r and u0, u1 ∈/ γr for any r. If u0 has only one one boundary neighbor, b0 = b1 and

γ is nullhomotopic. If b0 6= b1, then u0 has two distinct boundary neighbors, so u0 is in the corner of a tile. Speciﬁcally, for some r ∈ R, {b0, b1} is one of

L −8 L −6 (1) (L−1) ( 2 ) ( 2 ) γr , γr , γr , γr ,

L −1 L +1 ( 2 ) ( 2 ) (L−8) (L−6) γr , γr , or γr , γr .

73 (0) Consider the ﬁrst case. We have γr = x, so

(1) (L−1) u0, γr , x, γr , and

(1) (L−1) u1, γr , x, γr are each cycles in Γ and each correspond to original four-cycles before modding out by labeled

(1) nodes. Taking Γr as the distinguished point relative to which homotopy is deﬁned, γ is just a product of these two cycles. That is, the “new” four-cycle γ is already nullhomotopic. For each of the other three cases, there is a unique node v in Γ, which may or may not be x, such that v is adjacent to both b0 and b1. Using v in place of x, we can repeat the preceding argument, so all “new” four-cycles are already nullhomotopic.

∗ Having controlled the four-cycles, we now have that each Tr is homeomorphic, as a cell-complex, to a single face bordered by cycles corresponding to the decomposition of r into generators. This establishes that the homotopy group of Γ∗ has generators corresponding to

G0 and relations corresponding to R, and is thus isomorphic to G. We have also ensured, for each generator g ∈ G0, that f ◦ π1(γg) = 1 + Z iﬀ γg has odd length, and this passes to arbitrary elements of G.

Remark. Naturally, there is some flexibility in choosing `g so long as `g mod2 matches f(g). Smaller values of `g can be used and different values of `g can be used for different g; extreme cases may require manually checking that the resulting graph exhibits no unintended four-cycles. Furthermore, there is no reason the mesh Tr for each relator r need be a grid graph, so long as it is nullhomotopic and no unintended four-cycles arise.

74 CHAPTER 5

2 SUBSETS OF F (2Z ) WITH CONTINUOUS GROUP ACTIONS

Our next theorem concerns continuous free group actions on a clopen complete sec-

2 tion M of F (2Z ). For Theorem 5.2, informally speaking, we show that there cannot exist

2 any continuous construction that organizes parts F (2Z ) in such a way that we can (continuously) define a Z2 action on those parts. This result was proven as a generalization of our negative answer on the question of “almost-lined-up marker regions”, which we give in Example 5.4. We adjust Definition 2.2, the definition of a relation clopen relative to an action, to accommodate actions.

Definition 5.1. Let X be a space with a group action (G, ). Let Y ⊆ X. Let

(H,?) be a secondary action on Y . The action (H,?) is called continuous with respect to if there exists a continuous function θ : Y × H → G such that, for each y ∈ Y, h ∈ H,

h y = ϕ(y, h) ? y.

Here the function θ provides a G-translation to be applied to any y given an h; this G-translation is substituted for an actual value from Y in much the same manner as we substitute group elements for elements of X in Deﬁnition 2.2.

2 Theorem 5.2. Suppose M ⊆ F (2Z ) is a clopen complete section. Suppose further

there is deﬁned a secondary free Z2 action ? on M, which is continuous with respect to the left-shift action ·, and for any two points m, m0 ∈ M, m and m0 are equivalent in · iﬀ m and m0 are equivalent in ? (? being continuous relative to · only necessitates half of this

2 equivalence). Then M is trivial in the sense that M = F (2Z ).

2 Proof. Suppose for a contradiction that M 6= F (2Z ). Let K be as in Theorem 2.1, subject to the additional constraint that K \ M and K ∩ M are both nonempty. Because M is clopen and K \ M and K ∩ M are both nonempty, there is a distance n0 so that any n0 × n0 box of K contains a point of M and a point of K \ M. Then, picking n1 n0, we have that any n1 × n1 box of K contains many points of M and K \ M.

75 0 Any two points m, m in any n1 × n1 box are a bounded distance from each other

(i.e., n1). Thus, by compactness of K and continuity of ? relative ·, there exists an absolute

0 bound n2 such that any n2 ×n2 box centered at m or m witnesses a full path through (M,?),

2 using only the usual Z generators z0 and z1. That is, the n2 × n2 box fully witnesses the element g ∈ G such that g ? m = m0.

Enlarging slightly, we can ﬁnd n3 such that for any n1 × n1 box R, an n3 × n3 box

0 centered at the center of R witnesses, for any m, m in R ∩ M, a gm,m0 ∈ G such that

0 gm,m0 ? m = m . We will use n3 as our parameter n when applying Theorem 2.1.

2 To use Theorem 2.1, we will also need a continuous function ϕF (2Z ) encoding the

relevant local information of ?, namely, membership in M and the action of ?. Let z0, z1

2 be the usual generators of Z2, and let θ : F (2Z ) × G → G be the continuous function

2 witnessing that ? is continuous relative to · from Deﬁnition 5.1. Then we let ϕF (2Z ) encode

the characteristic function of M as well as the maps x 7→ θ(z0, x) and x 7→ θ(z1, x).

2 Apply Theorem 2.1 to ϕF (2Z ) with parameter n3 to produce a twelve-tiles graph

Γn3,p,q (henceforth abbreviated Γ) with p ⊥ q and a function ϕΓ :Γ → S such that every

2 n3-pullback of ϕΓ is a pullback of ϕF (2Z )K and every ﬁnite grid graph pullback of ϕΓ is a

2 pullback of ϕF (2Z ).

2 2 Consider an arbitrary Z -pullback ϕZ2 ≤ ϕΓ. As we have observed about Z pullbacks, 2 2 ϕZ2 is just a tiling of Z by the tiles of ϕΓ. The encoding ϕZ2 induces a set MZ2 ⊆ Z

corresponding to M, and what we will show is a well-deﬁned free action (MZ2 ,?Z2 ) on MZ2 .

Indeed, we must ﬁrst show that for a generator zi and a point zi ∈ MZ2 , zi ?Z2 m is itself

another member of MZ2 . Consider a ﬁnite grid pullback ϕm large enough to contain both m 2 2 2 Z 2 and zi ?Z m. ϕm ≤ ϕF (2Z ), so because ? is a closed operation on M ⊆ F (2 ), zi ?Z m must

be in MZ2 .

±1 ±1 To see that ?Z2 is well-deﬁned, consider a word w consisting of z0 , z1 such that w 0 is equivalent to eZ2 . Let m = w ?Z2 m, obtained by repeated actions of the generators. A ﬁnite pullback large enough to contain all intermediate members of M along the path will

0 2 be a pullback of ϕF (2Z ) and thus witness that m = m. By a similar argument, we have that

76 0 (R ) 2 (R ) 2 (R ) × Z c Z × Z2

RZ2

0 mZ2 m Z2

2 Figure 5.1. On a Z -pullback ϕZ2 of ϕΓ, with ϕZ2 ◦σ = ϕΓ. We seek to ﬁnd 0 c such that c ? 2 m 2 = m . R 2 has height n and thus witnesses a path of Z Z Z2 Z 1 0 nodes from m 2 to m – the black nodes. However, R 2 does not necessarily Z Z2 Z

witness the group elements for ?Z2 to bring each black node to the next along

the path. For the group elements, n3 pullbacks are needed, which reveal paths between the black nodes shown with dotted lines.

mZ2 is free.

Now that we can recover free actions from ϕΓ, we may reason about the structure of those actions. Let MΓ be the analog of M in Γ encoded by ϕΓ. Because R× ⊆ Γ is

2 an n3 × n3 box and every n0 × n0 box of ϕF (2Z ) contains a point in M, we can ﬁx a point mΓ ∈ MΓ ∩R× within an n0 ×n0 box centered in R×. Consider R× ∪Rc – this is a cylindrical

2 grid graph which is n3 nodes in height and p nodes in circumference. Z does not act on 2 MΓ, but nonetheless, we seek a group element c ∈ Z such that c brings mΓ to itself leftward 2 through R× ∪ Rc in the following sense. That is, consider a Z pullback ϕZ2 ≤ ϕΓ containing 0 arrangement shown in Figure 5.1. We have m 2 , m ∈ M 2 each corresponding to m Z Z2 Z Γ 0 respectively in boxes (R ) 2 , (R ) which correspond to R . These are separated by by × Z × Z2 × 0 (R ) 2 corresponding to R . We seek c such that for any such ϕ 2 , we have c ? 2 m 2 = m . c Z c Z Z Z Z2 0 Let R 2 be an n × (p + n ) box centered in (R ) 2 ∪ (R ) 2 ∪ (R ) . Because each Z 1 1 × Z c Z × Z2

2 2 n1 pullback of ϕΓ is a pullback of ϕF (2Z )K , and n1 × n1 boxes in ϕF (2Z )K contain many nodes of M, there is a sequence of nodes

0 m 2 = m , m , . . . , m = m 2 Z 0 1 k Z

through RZ2 separated by distances much less than n1 (the black nodes in the ﬁgure).

77 0 (R ) 2 ∪ (R ) 2 ∪ (R ) is a union of n × n boxes, and an n -pullback of ϕ 2 is a pullback × Z c Z × Z2 3 3 3 Z 2 2 2 of ϕF (2Z )K , thus a n3-pullbacks of ϕZ witness, for each i < k, a unique element gi ∈ Z such that gi ?Z2 mi = mi+1. We thus have c = gk−1, gk−2, . . . , g0, and this holds anywhere

R×,Rc,R× appear side-by-side in any pullback ϕZ2 ≤Z2 ϕΓ

Now, in Γ, Rd is similarly bracketed on the sides by R×, so we may apply the same

process to ﬁnd a corresponding value d taking mΓ to itself leftward through Rd. Ra and

Rb are bracketed by R× on the top and bottom, so we can let a and b bring mΓ to itself

downward through Ra and Rb respectively.

The structure of Γ will now permit us to reason about a, b, c, d. Consider tile Tcqa=adp .

In particular, consider a path around the boundary Tcqa=adp , hitting each mΓ every time it

passes through R×. If we keep a running product of the group elements a, b, c, d required to

bring mΓ to itself through each rectangle Ra,Rb,Rc,Rd encountered, we should obtain the

2 group identity. This is because a Z -pullback of ϕΓ, which may very well contain a pullback q p 2 of ϕΓTcqa=adp , must have a well-deﬁned ?Z action. We conclude c a = ad .

But Z2 is Abelian, so cq = dp. Because p, q are coprime, we can ﬁnd j, k such that jp + kq = 1. We have

(cjdk)p = cjpdkp

= cjpckq

= c

q p q p whence c is a p-th power. Similarly, using Tcaq=bpc, we have ca = b c whence a = b and a is also a p-th power.

We now claim a and c are independent, for which we will use Tca=ac. Observe that

2 2 Tca=ac, by itself, is a p × p torus Z -graph. Now, let σ : Z → Tca=ac be a periodic tiling of 2 2 Z by Tca=ac, and let ϕZ2 = ϕΓ ◦ σ, i.e., the Z pullback of ϕΓ witnessed by σ. Let mZ2 be j k such that σ(mZ2 ) = mΓ. Suppose that a and c are dependent; i.e., a c = eZ2 with j or k

78 p nonzero. Then, because a ?Z2 mZ2 = z0 mZ2 and likewise for c,

j k mZ2 = a c ?Z2 mZ2

jp kp = z0 z1 mZ2 which is clearly impossile. So a and c are independent. Z2 Because a, c are independent and multiples of p, the group ha, ci is nontrivial and has cardinality that is a multiple of p2; thus

2 Z 2 ≥ p = |Tca=ac|. ha, ci To complete the proof, we will further show

2 Z |MΓ ∩ Tca=ac| ≥ ha, ci

from which we will conclude Tca=ac ⊆ MΓ. A contradiction will soon follow, and we will be ﬁnished. To arrive at the necessary inequality we show that the periodicity of σ requires the

2 action ?Z2 : Z y MZ2 to induce, via σ, a free action

2 ? : Z (M ∩ T ) Tca=ac ha, ci y Γ ca=ac

(we henceforth abbreviate ?Tca=ac = ?T ) deﬁned in the obvious way

2 g ?T σ(xZ2 ) := σ(g ?Z2 xZ2 ) ∀g ∈ Z , xZ2 ∈ MZ2 .

2 We ﬁrst show that this action is well-deﬁned. Fix xZ2 . Let h ∈ Z be such that

h ?Z2 xZ2 = mZ2 , which, recall, was ﬁxed such that σ(mZ2 ) = mΓ. Then

a ?T σ(xZ2 ) = σ(a ?Z2 xZ2 )

−1 = σ(ah ?Z2 mZ2 )

−1 = h ?T σ(a ?Z2 mZ2 )

−1 p = h ?T σ(z0 mZ2 )

−1 = h ?T σ(mZ2 ) (by periodicity of σ)

79 −1 = σ(h ?Z2 mZ2 )

= σ(xZ2 ).

2 Z Similarly, c ?T σ(xZ2 ) = σ(xZ2 ), so the action of ha,ci is well-deﬁned on MΓ ∩ Tca=ac.

2 Now we show that ?T is free. Indeed, suppose for some xZ2 ∈ MZ2 and g ∈ Z ,

g ?T σ(xZ2 ) = σ(xZ2 ).

Then,

σ(g ?Z2 xZ2 ) = σ(xZ2 ).

By periodicity of σ, for some i, j,

ip jp g ?Z2 xZ2 = z0 z1 xZ2

i j = a c ?Z2 xZ2 .

2 i j Z Because ?Z2 is free, we have g = a c , so g is equivalent to the identity in ha,ci. Then ?T is free on MΓ ∩ Tca=ac. As promised, we now have

2 Z 2 |MΓ ∩ Tca=ac| ≥ ≥ p = |Tca=ac|. ha, ci

whence Tca=ac ⊆ MΓ. But now we are done, as R× ⊆ MΓ, but R× is large enough to contain

an n1-pullback of K, which must contain an element of K \ M.

Theorem 5.3. Theorem 5.2 fails for F (2Z).

Proof. Fix a distance d. Let M be a marker point set for distance d. We put the obvious action ? on M, namely, z ? m is the next element m0 ∈ M to the right of m. This action is continuous. Indeed, ﬁnite pullbacks of m witness m ∈ M, m0 ∈ M, no element between m

0 0 and m is in M, and thus witness z ? m = m .

80 Figure 5.2. Almost lined-up marker regions.

Theroem 5.2 is useful because Z2 actions sometimes come up when investigating

2 combinatorial structures on F (2Z ). For example, one longstanding open question was the existence of a clopen marker region structure such that not only is each marker region approximately square, but their boundaries line up as depicted in Figure 5.2. For this discussion, an approximately square marker structure will refer to a structure in which each marker region has dimensions (d + 1) × (d + 2) where 1, 2 are natural numbers with

1, 2 ≤ for ﬁxed d; this is easy to accomplish continuously by previous work. By almost lined up, we mean that for each horizontal boundary ` of a region, there is a another

0 0 horizontal boundary ` left of `, so that ` has a small vertical oﬀset ` ≤ from ` , and the analogous condition for vertical boundaries. Almost-square and almost-lined-up cannot be achieved together continuously; the Borel version of this question remains open.

Example 5.4. No clopen almost square, almost lined up marker regions structure

2 exists on F (2Z ).

Proof. We proceed by showing that such a structure would lead to a free Z2 action ?,

2 continuous relative to the Bernoulli shift ·, on a proper clopen subset M ⊆ F (2Z ); this would then be a contradiction by Theorem 5.2.

2 Let M be the set of lower-left corners of marker regions; since 0 < d, M ( F (2Z ).

For m ∈ M, we obtain z0 ? m by following the left boundary ` of m’s marker region upward,

81 and taking the point of M closest to the other end of `. There exists a point of M there by the almost lined up condition, and a trivial triangle inequality argument shows that z0 ? m is uniquely deﬁned. Deﬁne z1 ? m is analogously, moving rightward.

−1 −1 −1 −1 We need ? well-deﬁned; that is, z0z1z0 z1 ? m = m. But z0z1z0 z1 is exactly a trace around a square marker region, so ? is well-deﬁned. We also need ? free. Suppose, for

i j a contradiction, that z0z1 ? m = m with i, j not both zero. Either d|i| > j or d|j| > i; we i assume the former as the latter can be handled similarly. z0 ? m has horizontal distance at least id from m. Applications of z1 may reduce this horizontal distance by as much as per

i j application, but we just supposed d|i| < j, so the horizontal distance between z0z1 ? m and m is nonzero. Thus ? is free. We observe that the deﬁnitions of both M and ? used only local, ﬁnite information.

Indeed a pullback of size 3d suﬃces to witness membership in M as well as m 7→ z0 ?m, m 7→ z1 ? m, and ﬁnitely many such pullbacks can be applied iteratively to witness m 7→ g ? m for any g ∈ Z2. Thus M is clopen and ? is continuous with respect to ·, whence ? is ruled out by Theorem 5.2.

82 CHAPTER 6

CONCLUSIONS AND FUTURE WORK

2 Previous results in this area, particularly on F (2Z ), relied on marker region structures to construct “positive” results (something exists), and hyper-aperiodic points, often with compactness, to obtain “negative” results (something does not exist). A new marker region structure with a new construction was generally necessary for each new positive result, and likewise for negative results. In the case of continuous maps to discrete sets, the main result of this work hides the technical details of marker structures and hyper-aperiodic points. That is, at least in the case of discrete codomains, the hyper-aperiodic point x and the marker construction constituted by the tiling are the last such constructions that will be necessary.

Theorem 5.2 uses the main result heavily, and would likely have been infeasible without it. Prior proofs of nonexistence amounted to working on speciﬁc tiles of the twelve-tiles graph; for example, the proof of nonexistence of a continuous three-coloring used an early and crude version of the “long” tiles. Theorem 5.2, on the other hand, used several tiles simultaneously, and required care in passing pullbacks between diﬀerent spaces.

We wish to pursue this work further in two directions. First, it is not clear how best to extend these results to non-discrete codomains. For example, consider some map

2 ϕ: F (2Z ) → 2ω with certain properties of interest. A twelve-tiles graph analog of ϕ0 of ϕ can be constructed which gives ﬁnitely many bits of each real in 2ω mapped to by ϕ, but

to increase the number of bits, a larger Γn,p,q is needed. This gives rise to a sequence of

ω graphs Γn,pi,qi for i < ω giving increasingly many bits of each real in 2 . It is not clear what properties these sequences of graphs (and corresponding maps) might have, or what proofs might become possible.

Second, we hope to extend this work to other groups, beyond Z2. We have a correct but inelegant proof that a collection of “k-tiles” graphs exists for continuous maps from

m Z F (2 ) for any m ≥ 1. Indeed, if we deﬁne Γn,p,q as a Z-graph consisting of two cycles of length p,q respectively and sharing exactly n consecutive nodes, Theorems 2.1 and 2.3

83 become true of Z. Thus, we have a collection of “two-tiles” graphs for Z much in the same way as we had a collection of “twelve-tiles” graphs for Z2. We currently estimate that about 200 tiles will be needed for Z3. For G Abelian, such as Z3, obtaining negative results turns out to be relatively straightforward. For example, we have a proof that there is no continuous graph homomor-

3 Z 2 phism ϕ: F (2 ) → Γn,p,q where Γn,p,q is any suﬃciently large twelve-tiles graph from the Z case. This proof requires ﬁnding a single tile that could have been part of a k-tiles graph

3 for Z , and which cannot map to Γn,p,q; fittingly, the argumemnt uses the second homotopy ∗ group π2(Γn,p,q). Just as most of the proofs in this work chose a single tile out of the twelve- tiles graphs to work on, we have chosen a single tile out of a graph so large we haven’t fully mapped it, and carried out our argument there. The challenge, then, in the Abelian case, is produce a sufficient collection of tiles with which to generalize Theorem 2.3, hopefully with a sufficiently elegant construction that the resulting graphs are useable. For G non-Abelian, constructing hyper-aperiodic points with useful properties is sig- nificantly more difficult; we hope to modify constructions of hyper-aperiodic points for general groups to obtain negative results. The resulting “k-tiles” graphs will no doubt be vast and complex.

84 BIBLIOGRAPHY

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