Descriptive Chromatic Numbers of Locally Finite and Everywhere Two Ended Graphs
DDescriptive Chromatic Numbers of LocallyFinite and Everywhere Two Ended Graphs
Felix WeilacherApril 7, 2020
Abstract
We construct Borel graphs which settle several questions in descriptivegraph combinatorics. These include “Can the Baire measurable chro-matic number of a locally finite Borel graph exceed the usual chromaticnumber by more than one?” and “Can marked groups with isomor-phic Cayley graphs have Borel chromatic numbers for their shift graphswhich differ by more than one?” We also provide a new bound for Borelchromatic numbers of graphs whose connected components all have twoends. A graph on a set X is a symmetric irreflexive relation G ⊂ X × X . In thissituation, the elements of X are called the vertices of G . Vertices x and y arecalled adjacent if ( x, y ) ∈ G , and in this case the pair { x, y } is called an edge of G . The degree of a vertex is the number of other vertices adjacent to it. G is called locally finite if every vertex has finite degree, is said to have boundeddegree d if every vertex has degree at most d , and is called d -regular if everyvertex has degree exactly d , where d is some natural number. A connectedcomponent of G is an equivalence class of the equivalence relation generatedby G .A (proper) coloring of G is a function, say, c : X → Y to some set Y suchthat if x and y are adjacent, c ( x ) (cid:54) = c ( y ). In this situation, the elements of Y are called colors . The sets c − ( { y } ) for y ∈ Y are called color sets . If | Y | = k , c is called a k -coloring. The chromatic number of G , denoted χ ( G ), is theleast k such that G admits a k -coloring.1 a r X i v : . [ m a t h . L O ] A p r escriptive graph combinatorics studies these notions in the descriptivesetting: Let X now be a Polish space. A graph G on X is called Borel if G isBorel in the product space X × X . A coloring c : X → Y is called Borel if Y is also a Polish space and c is a Borel function. The Borel chromatic number of G , denoted χ B ( G ), is the least k such that G admits a Borel k -coloring.Similarly, c is called Baire measurable if it is a Baire measurable function, andthe
Baire measurable chromatic number of G , denoted χ BM ( G ), is the least k such that G admits a Baire measurable k -coloring. For a survey covering thisexciting emerging field, see [4].For a Borel graph G , we of course have χ ( G ) ≤ χ BM ( G ) ≤ χ B ( G ), butit is natural to ask just how large χ BM ( G ) and χ B ( G ) can be compared to χ ( G ). There are many known examples [4] where χ ( G ) = 2 while χ BM ( G )and χ B ( G ) are infinite. However, for graphs of bounded degree d , Kechris,Solecki, and Todorcevic [5] proved χ B ( G ) ≤ d + 1. We therefore restrict ourattention to bounded degree graphs for the remainder of the paper.In [6], Marks proved that this bound is sharp, even for acyclic G (so inparticular, χ ( G ) = 2). Thus, χ B ( G ) can be arbitrarily large compared to χ ( G ). On the other hand, for Baire measurable chromatic numbers, Conleyand Miller proved the following [2] (Theorem B): Theorem 1.
Let G be a locally finite Borel graph such that χ ( G ) < ℵ . Then χ BM ( G ) ≤ χ ( G ) − . The question “How close to this bound can we get?” still remains. Pre-viously, not much seems to have been known regarding this: In fact, Kechrisand Marks pose the following problem [4] (Problem 4.7):
Problem 1.
Is there a bounded degree Borel graph G for which χ BM ( G ) >χ ( G ) + 1 ? The graphs constructed by Marks in [6] are not hyperfinite (see Section 4for a definition). Furthermore, an analogue of Theorem 1 holds for measurechromatic numbers if the extra assumption of hyperfiniteness is added (seeTheorem 5). This lead to the question of whether the 2 χ ( G ) − is enough to get this bound. Using techniques similarto those in [2] and some results from [7], we prove in Section 2 the followinganalogue of Theorem 1: 2 heorem 2. Let G be a locally finite Borel graph such that χ ( G ) < ℵ , andsuch that every connected component of G has two ends. Then χ B ( G ) ≤ χ ( G ) − . See section 2 for a definition of two endedness. Also note that this conditionis indeed a strengthening of hyperfiniteness [7].Similarly little seems to have been known regarding the sharpness of thisbound. In fact, one of the goals of the project which led to this paper was toresolve the following:
Problem 2.
Is there a bounded degree Borel graph G whose connected com-ponents all have two ends for which χ B ( G ) > χ ( G ) + 1 ? In this paper, we answer Problems 1 and 2 as strongly as possible, provingthe bounds in Theorems 1 and 2 are sharp:
Theorem 3.
Let k ≥ . There is a Borel k − -regular graph, say, G k ,such that all the connected components of G k have two ends, χ ( G k ) = k , and χ BM ( G k ) = χ B ( G k ) = 2 k − . The graphs G k will be arise in the following way: A marked group (in thispaper) is a pair (Γ , S ), where Γ is a (typically infinite) finitely generated groupand S is a finite symmetric set of generators for it not containing the identity.When there is no confusion, we will sometimes refer to a marked group by itsunderlying group. Consider the group action Γ (cid:121) Γ given by( g · x )( h ) = x ( g − h ) (1)for g, h ∈ Γ and x ∈ Γ . This is called the left shift action . When 2 Γ is giventhe product topology, this action is clearly continuous. Let F (2 Γ ) = { x ∈ Γ | ∀ g ∈ Γ − { id } , g · x (cid:54) = x } . (2)This is a G δ subspace of 2 Γ , hence a Polish space. We can therefore forma Borel graph on F (2 Γ ) by putting an edge between x and y exactly when s · x = y for some s ∈ S . This is called the shift graph of (Γ , S ). We willalways refer to the shift graph by its underlying set, F (2 Γ ). The graphs G k will all have the form F (2 Γ k ) for some marked group Γ k .Let Cay(Γ) be the Cayley graph of (Γ , S ). This is the graph on Γ givenby putting an edge between group elements g and h exactly when sg = h forsome s ∈ S . Clearly, as a (discrete) graph, F (2 Γ ) is isomorphic to a disjointunion of continuum many (if Γ is infinite) copies of 2 Γ . It is therefore naturalto expect to get some information on the descriptive combinatorics of F (2 Γ )from the graph Cay(Γ). However, in [8] (Theorem 1), Weilacher showed thatCay(Γ) is not enough to determine χ B ( F (2 Γ )) or χ BM ( F (2 Γ )):3 heorem 4. Let k ≥ . There are marked groups Γ and ∆ with isomorphicCayley graphs for which χ B ( F (2 ∆ )) = χ BM ( F (2 ∆ )) = k but χ B ( F (2 Γ )) = χ BM ( F (2 Γ )) = k + 1 . This led to the natural question:
Problem 3.
Are there marked groups Γ and ∆ with isomorphic Cayley graphsfor which χ B ( F (2 Γ )) − χ B ( F (2 ∆ )) > ? What about for Baire measurablechromatic numbers? We answer this as well by producing for each k a marked group ∆ k whoseCayley graph is isomorphic to that of Γ k , but for which χ B ( F (2 ∆ k )) = χ BM ( F (2 ∆ k )) = k + 1. Thus we get Corollary 1.
Let k be a natural number. There are marked groups Γ and ∆ with isomorphic Cayley graphs but for which χ B ( F (2 Γ )) − χ B ( F (2 ∆ )) = χ BM ( F (2 Γ )) − χ BM ( F (2 ∆ )) = k . In Section 3 we define the marked groups ∆ k and Γ k and compute theirvarious chromatic numbers. In Section 4 we note that everything said inthis paper about Baire measurable chromatic numbers can also be said aboutmeasure chromatic numbers in the hyperfinite setting. In this section we prove Theorem 2. The proof uses little more than someresults of Miller from [7], but nevertheless the result seems to be new, andmay be of interest to some.Let G be a graph on a set X . If A ⊂ X , we denote by G (cid:22) A the graph G ∩ ( A × A ) on A . We call G connected if it has one connected component,and A connected if G (cid:22) A is connected.A path between vertices x and y is a finite sequence x = x , . . . , x n = y such that ( x i , x i +1 ) ∈ G for all i . In this situation n is called the length of thepath. Note that a graph is connected if and only if there is a path betweenany two of its vertices. The path distance between x and y is the smallest n such that there is a path of length n between x and y , or ∞ if there is no pathbetween x and y . The path distance between two sets of vertices A and B isthe smallest path distance between any pair of vertices x ∈ A and y ∈ B . Agraph is called acyclic if it admits no paths as above with x = x n , but noother repeats among the x i ’s. 4ow assume G is connected and locally finite. We say a subset F ⊂ X divides G into n parts if G (cid:22) ( X − F ) has n infinite connected components.We say G has n ends if there is a finite set F dividing G into n parts, but nosuch F dividing G into m parts for any m > n . Note that if G has n ends,we can finite a finite set F dividing it into n parts such that F is furthermoreconnected. It should be noted this definition is different in general from theone used in [7], but is equivalent in the locally finite case.Now let G be a localy finite Borel graph on a space X whose connectedcomponents all have two ends. Denote by [ G ] < ∞ the standard Borel space offinite connected subsets of X . Let Φ ⊂ [ G ] < ∞ be the set of sets which dividetheir connected component into two parts. Miller (Lemma 5.3) proves thatthere is a maximal Borel set Ψ (cid:48) ⊂ Φ whose members are pairwise disjoint. Aneasy modification of their proof shows that we can instead get a maximal Borelset Ψ ⊂ Φ such that the path distance between any two distinct members ofΨ is at least 4. Fix such a Ψ.Let T be the set of pairs ( S, T ) with
S, T ∈ Ψ such that S (cid:54) = T and thereis a path from S to T which avoids all other points of (cid:83) Ψ. Miller provesthat T is an acyclic graph on Ψ, that S and T are connected in this graph ifand only if they are subsets of the same connected component of G , and thatevery element of Ψ is T -adjacent to at most two other elements (Lemma 5.5).(Strictly speaking, they prove these things for Ψ (cid:48) , but the proofs clearly stillapply to Ψ.) Lemma 1.
Every S ∈ Ψ is T -adjacent to exactly two other elements.Proof. Suppose some S ∈ Ψ has fewer than two T -neighbors. Let C be theconnected component of S . Let C − and C + be the two infinite connectedcomponents of G (cid:22) ( C − S ). WLOG, C + must contain no sets in Ψ. Thisfollows from the fact that any T ∈ Ψ with T ⊂ C must be T -connected to S .Let N be the set of points in C + whose path distance from S is exactly 4. N is finite since G is locally finite. We claim N divides C into 2 parts: ByK¨onig’s Lemma, we can find an injective sequence { x n | n ∈ ω } of points in C + such that ( x n , x n +1 ) ∈ G for all n . Since G is locally finite, there mustbe some M for which for all n ≥ M , the path distance between x n and S isat least 5. Then the sequence { x n | n ≥ M } does not pass through N , so itis contained in an infinite connected component of G (cid:22) ( C − N ). Also, C − iscontained in an infinite connected component of G (cid:22) ( C − N ), so it sufficesto show there is no path from C − to x M avoiding N . This is clear, though,as any path from C − to x M must pass through S , say at the point y , since x M ∈ C + . Then since the path distance from S to x M is greater than 4, there5ust be some point in C + along our path from y to x M whose path distancefrom S is exactly 4.Let D be the infinite connected component of G (cid:22) ( C − N ) not containing S . Let N (cid:48) ⊂ N be the set of elements of N adjacent to a point in D . Then N (cid:48) still divides C into 2 parts. Furthermore, we can find a finite subset A ⊂ D such that N (cid:48) ∪ A is connected. Then N (cid:48) ∪ A ∈ Φ, and furthermore since everypoint in D has path distance at least 5 from S , the path distance between S and N (cid:48) ∪ A is 4. However, since we assumed C + contains no sets in Ψ, thiscontradicts the maximality of Ψ. Lemma 2.
Every connected component of G (cid:22) ( X − (cid:83) Ψ) is finite.Proof. Let x ∈ ( X − (cid:83) Ψ). Let C be the connected component of x in the graph G , and let D be the connected component of x in the graph G (cid:22) ( C − (cid:83) Ψ).We want to show D is finite.By maximality there is some element of Ψ contained in C . Then, byLemma 1 along with the fact that T is acyclic, we can label the elements ofΨ contained in C as { S n | n ∈ Z } , where the indices are chosen such that( S n , S m ) ∈ T if and only if | n − m | = 1. By definition of Φ, for each n thegraph G (cid:22) ( C − S n ) has two infinite connected components, call them C n, − and C n, + . By definition of T , the sets S m for m > n must all lie in the sameconnected component of G (cid:22) ( C − S n ), and likewise for the sets S m for m < n .Therefore, by relabelling if necessary, we can assume S m ⊂ C n, + for all m > n and S m ⊂ C n, − for all m < n .Now, suppose D is infinite. Then, for each n , either D ⊂ C n, + or D ⊂ C n, − .Consider integers n , points y ∈ S n , and paths from x to y . Choose n, y , andsuch a path such that this path is of minimal length among all such choices.Then this path cannot pass through any sets S m for m (cid:54) = n . WLOG assume D ⊂ C n, + . We claim D ⊂ C n +1 , − . If not D ⊂ C n +1 , + , but then D and S n arein different connected components of G (cid:22) ( C − S n +1 ), so there can be no pathfrom x to S n avoiding S n +1 , a contradiction. Therefore D ⊂ C n, + ∩ C n +1 , − , sothis intersection is infinite. This implies, however, that the finite set S n ∪ S n +1 divides G (cid:22) C into at least three parts, a contradiction.We can now prove Theorem 2: Proof.
For each S ∈ Ψ, let S ∗ = S ∪ { x ∈ X | ∃ y ∈ S ( x, y ) ∈ G } . Since G is locally finite and each S is finite, each S ∗ is finite. Let B ∗ = (cid:83) S ∈ Ψ S ∗ . B ∗ is Borel since Ψ is Borel. Since distinct S ’s had path distances of atleast 4 between them, distinct S ∗ ’s have path distances of at least 2 betweenthem. Thus, every connected component of G (cid:22) B ∗ is a subset of some S ∗ .6n particular these connected components are all finite. Therefore, by theLusin-Novikov Uniformization Theorem (See [3], Lemma 18.12), there is aBorel χ ( G )-coloring, say c ∗ : B ∗ → { , , . . . , χ ( G ) } of G (cid:22) B ∗ . Let B = B ∗ − c ∗− ( { χ ( G ) } ) and c = c ∗ (cid:22) B . Then B is Borel and c is a Borel( χ ( G ) − G (cid:22) B .We claim that the connected components of G (cid:22) ( X − B ) are also allfinite. Suppose to the contrary that D ⊂ ( X − B ) is some infinite connectedcomponent. Let C be the connected component of G containing D . We firstclaim that D must contain infinitely many points not in B ∗ . If not, then D contains infinitely points from ( B ∗ − B ), and only finitely many not in B ∗ .By construction, though, B ∗ − B is independent, so since D is connected, forevery y ∈ ( B ∗ − B ) ∩ D , there must be some x ∈ D − B ∗ with ( x, y ) ∈ G . Thusthere is some x ∈ D − B ∗ connected to infinitely many such y ’s, contradictinglocal finiteness. Therefore, by Lemma 2, there are x, y ∈ D − B ∗ such that x and y are in different connected components of G (cid:22) ( C − (cid:83) Ψ). Let x = x , x , . . . , x n = y be a path from x to y consisting of points in D . Then theremust be some S ∈ Ψ and some 0 < i < n such that x i ∈ S . Then x i − , x i , and x i +1 are all in S ∗ . Since there are some edges between them, they can’t all beassigned the color χ ( G ) by c ∗ , but this means at least one of them is in B , acontradiction.Therefore, again by the Lusin-Novikov Uniformization Theorem, there isa Borel χ ( G ), coloring, say, c : ( X − B ) → { χ ( G ) , . . . , χ ( G ) − } , of G (cid:22) ( X − B ). Since c and c use disjoint sets of colors, c ∪ c is a Borel (2 χ ( G ) − G . Fix k ≥
3. In this section, we define the marked groups Γ k and ∆ k promisedin Section 1.We start with a finite marked group: Let Z k denote the cyclic group oforder k , which we will identify with the integers modulo k . Consider thegroup Z k × Z k with generating set S = { ( a, b ) | < a, b < n } . Let H be theCayley graph of this finite marked group. We’ll think of the vertices of H assitting on a k by k grid, with the horizontal axis corresponding to the firstcoordinate and the vertical to the second. Accordingly, by a row of H we meana set of the form { ( a, b ) | a ∈ Z k } for some fixed b ∈ Z k , and by a column of H we mean a set of the form { ( a, b ) | b ∈ Z k } for some fixed a ∈ Z k .An independent subset of a graph is a pairwise-non-adjacent set of vertices.Thus, a coloring is just a partition of the set of vertices into independent sets.7igure 1: A drawing of the graph H for k = 3. The edges shown are exactlythose meeting (0 , H of size greater than one must be eithercompletely contained in some row, or completely contained in some column,(and not both). Call such sets horizontal and vertical , respectively (See Figure1). Lemma 3.
Let c : Z k × Z k → { , , . . . , k − } be a (2 k − -coloring of H .Exactly one of the following holds: • Every row contains a horizontal color set. • Every column contains a vertical color set.Proof.
Since there are k rows and k columns, for both to hold simultaneouslywould require 2 k colors. Therefore at most one holds.Suppose neither holds. Then there is some column C and some row R such that C does not contain a horizontal color set and R does not contain avertical color set. Then every point in R ∪ C must have a different color, but | R ∪ C | = 2 k −
1. Therefore at least one holds.We call c as in the lemma a horizontal coloring if the first condition holds,and a vertical coloring if the second holds.We can now define the marked group ∆ k : It will be the group ( Z k × Z k ) × Z ,with generating set S × {− , , } . Let G be the Cayley graph of ∆ k . It’s easyto see χ ( G ) = k : A k -coloring is given by sending the element (( a, b ) , n ) to a for all n ∈ Z and 0 ≤ a, b < k . Also note that G has two ends, as desired.8igure 2: A visual explanation of the proof of Lemma 4 in the case k = 3.The two squares enclose neighboring ( Z k × Z k )-orbits. The cirlces representcolor sets within each orbit. Most edges are omitted, but some are includedto show any horizontal color set from the first orbit must admit an edge toevery vertical color set from the second orbit. Others are included to showthat horizontal color sets in different rows of a single orbit always have edgesbetween them. The same is true for vertical color sets in different columns.For each n ∈ Z , the restriction of G to the ( Z k × Z k )-orbit ( Z k × Z k ) × { n } can be identified with H in the obvious way. Thus, if c : ( Z k × Z k ) × Z →{ , , . . . , k − } is a (2 k − G , the restriction of c to the orbit( Z k × Z k ) ×{ n } is, for each n , either a horizontal coloring or a vertical coloring.In the k -coloring defined in the previous paragraph, all these restrictions werehorizontal. The next lemma says that this was no accident: Lemma 4.
Let c : ( Z k × Z k ) × Z → { , , . . . , k − } be a (2 k − -coloringof G . Exactly one of the following holds: • The restriction of c to every ( Z k × Z k ) -orbit is horizontal. • The restriction of c to every ( Z k × Z k ) -orbit is vertical.Proof. By symmetry, it suffices to show that if the restriction of c to ( Z k × Z k ) × { n } is horizontal, then so is the restriction to ( Z k × Z k ) × { n + 1 } .Suppose instead that it is vertical. For 1 ≤ i ≤ k , let R i be a horizontal colorset contained in the i -th row of ( Z k × Z k ) × { n } , and let C i be a vertical color9et contained in the i -th column of ( Z k × Z k ) × { n + 1 } . Observe that forevery 1 ≤ i, j ≤ k , there is at least one edge between R i and C j (See Figure2). Furthermore, if i (cid:54) = j , there is at least one edge between R i and R j , as wellas between C i and C j (Again see Figure 2). Therefore each R i and C j musthave a distinct color, but this requires 2 k colors.This leads us to a natural definition of the marked group Γ k : Let ϕ ∈ Aut( Z k × Z k ) be the coordinate swapping map: ϕ ( a, b ) = ( b, a ). Γ k will be thesemi-direct product ( Z k × Z k ) (cid:111) (cid:55)→ ϕ Z , again with generating set S × {− , , } .Observe that the following gives an isomorphism between the Cayley graphsof Γ k and ∆ k :(( a, b ) , n ) (cid:55)→ (( a, b ) , n ) for n even, (( a, b ) , n ) (cid:55)→ (( b, a ) , n ) for n odd , (3)where a, b ∈ Z k and n ∈ Z . Thus we still have χ ( F (2 Γ k )) = χ (Cay(Γ k )) = k ,and this Cayley graph still has two ends as desired. We now compute theBorel and Baire measurable chromatic numbers of F (2 Γ k ), proving Theorem3: Proposition 1. χ B ( F (2 Γ k )) = χ BM ( F (2 Γ k )) = 2 k − .Proof. Theorem 2 gives us the upper bound χ B ( F (2 Γ k )) ≤ k −
1, so it remainsto show there is no Baire measurable (2 k − F (2 Γ k ).Suppose first that c : ( Z k × Z k ) (cid:111) (cid:55)→ ϕ Z → { , , . . . , k − } is a (2 k − k ). Note that the isomorphism (3) sends ( Z k × Z k )-orbitsto ( Z k × Z k )-orbits, but preserves the notions of “horizontal” and “vertical”for those with even Z -coordinate and flips those notions for those with odd Z -coordinate. Thus Lemma 4 has the following consequence for Γ k : If forsome n the restriction of c to ( Z k × Z k ) × { n } is horizontal, the restiction to( Z k × Z k ) × { n + 1 } must be vertical, and vice versa.Now suppose c : F (2 Γ k ) → { , , . . . , k − } is a Baire measurable (2 k − d : F (2 Γ k ) → { , } by sending a point x to 1 if therestriction of c to the ( Z k × Z k )-orbit of x is horizontal, and 2 if it is vertical.It is clear that d is Baire measurable since c was. By the previous paragraph, d ( x ) (cid:54) = d (((0 , , · x ) for all x .Now consider Z with generators {± } , and let g : F (2 Z ) → F (2 Γ k ) be themap given by g ( y )(( a, b ) , n ) = (cid:40) y ( n ) if ( a, b ) = (0 , g is continuous, and g (1 · y ) = ((0 , , · g ( y ) for all y . Therefore d ◦ g is a Baire measurable 2-coloring of the shift graph F (2 Z ). It was establishedin [5], though, that χ BM ( F (2 Z )) = 3. 10inally, we compute the Borel and Baire measurable chromatic numbers of F (2 ∆ k ), which gives Corollary 1 as promised: Proposition 2. χ B ( F (2 ∆ k )) = χ BM ( F (2 ∆ k )) = k + 1 .Proof. We first show there is no Baire measurable k -coloring c : F (2 ∆ k ) →{ , , . . . , k } . Suppose we had such a coloring. Observe that all k -coloringsof the Cayley graph of ∆ k look essentially like the one defined before Lemma4: Up to a relabeling of the colors, they either assign the color a to (( a, b ) , n )for all b and n , or the color b to (( a, b ) , n ) for all a and n . In particular, theelements g and ((0 , , · g always have the same color.Therefore, if we let C i = c − ( { i } ) for each i , each C i is sent to itself by theaction of the element ((0 , , C i is either meageror comeager. Since the C i ’s partition F (2 ∆ k ), at least one, say, C i , must bemeager. The sets (( a, b ) , · C i for a, b ∈ Z k cover F (2 ∆ k ), though, so this isa contradiction.It remains to construct a Borel ( k + 1)-coloring c : F (2 ∆ k ) → { , , . . . , k +1 } . A subset of x is called r -discrete , for r a natural number, if the pathdistance between any two points in A is greater than r . It is an easy corollaryof Proposition 4.2 in [5] that if G is a Borel graph of bounded degree, then X contains a Borel maximal r -discrete subset for every r .Applying this, let A ⊂ F (2 ∆ k ) be a Borel maximal 3 k -discrete set. Thenevery ( Z k × Z k )-orbit contains at most one element of A . For every x ∈ A ,color the ( Z k × Z k )-orbit of x by setting c ((( a, b ) , · x ) = a for 1 ≤ a, b ≤ k .We now color the ( Z k × Z k )-orbits between those meeting A . Let x ∈ A with ( Z k × Z k )-orbit E , and let N be the smallest positive number such that((0 , , N ) · E contains a point of A . Call that point y . Also note N > k bydefinition of A . There are elements 1 ≤ a , b ≤ k such that y = (( a , b ) , N ) · x .Also let E n denote the orbit ((0 , , n ) · E for n ∈ Z , so for example y ∈ A ∩ E N .We need to extend c by coloring all the E n ’s for 0 < n < N . We’ll proceedone n at a time:Given a ( Z k × Z k )-orbit E (cid:48) colored already by c and a positive integer n , let c n ( E (cid:48) ) denote the coloring on ((0 , , n ) · E (cid:48) given by c n ( E (cid:48) )( z ) = c (((0 , , − n ) · z ). We could try to extend our coloring c , by coloring E with c ( E ), then E with c ( E ), and so on. If we happened to have c (cid:22) E N = c N ( E ), this wouldwork out, but otherwise we will have a conflict. We can use our additionalcolor to fix this:Now, color E by using c ( E ), but then swapping the color k with the color k + 1. Since c (cid:22) E does not use the color k + 1, this is OK. Then c (cid:22) E does11ot use the color k , so we can color E by using c ( E ), but then swappingthe color a with the color k . Then c (cid:22) E does not use the color a , so we cancolor E by using c ( E ), but then swapping the color k + 1 with the color a .Now c (cid:22) E does not use the color k + 1, and furthermore it looks like c ( E ),but with the colors k and a swapped. Note that by performing this swap, wehave arragned that c N − ( E ) agrees with c (cid:22) E N on the a -th row.We can repeat this process k times, so that for each i ≤ k , E i will not usethe color k + 1 and c N − i ( E i ) will agree with c (cid:22) E N on i rows. In particular,we will have c N − k ( E k ) = c (cid:22) E N . Thus, we can color the remaining orbits E k + i for 0 < i < N − k using c i ( E k ). Thus we have a ( k + 1)-coloring c asdesired. Since A was Borel, it is clear that c is Borel, so we are done. In this section, we extend our results to the measurable setting.Let G be a Borel graph on a space X , now equipped with a Borel probabilitymeasure µ . Just as we defined Borel and Baire measurable colorings, wecan define µ - measurable colorings and the µ - measurable chromatic number ,denoted χ µ ( G ). The measure chromatic number of G , denoted χ M ( G ), is thesupremum of χ µ ( G ) over all Borel probability measures µ on X .An equivalence relation E on X is called Borel if it is Borel as a subset of X × X . E is called finite if its equivalence classes are all finite. E is called hyperfinite if it can be written as E = (cid:83) n ∈ ω E n for some increasing sequence E n of finite Borel equivalence relations. G is called hyperfinite if its connectedcomponent equivalence relation is hyperfinite.In [2] (Theorem A), Conley and Miller prove an analogue of Theorem 1 formeasure chromatic numbers with the added assumption of hyperfiniteness: Theorem 5.
Let G be a hyperfinite locally finite Borel graph such that χ ( G ) < ℵ . Then χ M ( G ) ≤ χ ( G ) − . As in the Baire measurable situation, the sharpness of this bound waspreviously unknown. All of the arguments we made in Section 3 in the Bairemeasurable setting still work in the measurable setting. Most crucially, we have χ M ( F (2 Z )) = 3 just as we did for the Baire measurable chromatic number,and the arguments regarding χ BM ( F (2 ∆ k )) in the proof of Proposition 2 stillgo through in the measure theoretic setting upon replacing “meager” and“comeager” with “measure 0” and “measure 1” respectively. Therefore, Proposition 3.
For all k ≥ , χ M ( F (2 Γ k )) = 2 k − and χ M ( F (2 ∆ k )) = k + 1 .
12s was noted in the introduction, these graphs are hyperfinite since theirconnected components all have two ends. Thus the bound in Theorem 5 isindeed sharp:
Theorem 6.
Let k ≥ . There is a Borel hyperfinite k − -regular graph,say, G k , for which χ ( G k ) = k but χ M ( G k ) = 2 k − . Similarly, alongside Theorem 4, Weilacher [8] proves that there are markedgroups with isomorphic Cayley graphs can have measure chromatic numberswhich differ by one, but notes that it is open whether or not these numberscan differ by more than one. By Proposition 3, we have resolved this as well:
Corollary 2.
Let k be a natural number. There are marked groups Γ and ∆ with isomorphic Cayley graphs but for which | χ M ( F (2 Γ )) − χ M ( F (2 ∆ )) | > k . Acknowledgements
We thank C. Conley for fruitful discussions and helpful comments on earlierdrafts of this paper.This work was partially supported by the ARCS foundation, Pittsburghchapter. 13 eferences [1] C.T. Conley, S. Jackson, A.S. Marks, B. Seward, and R. D. Tucker-Drob,Hyperfiniteness and Borel combinatorics,
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