aa r X i v : . [ m a t h . L O ] J a n Borel Vizing’s Theorem for 2-Ended Groups
Felix WeilacherFebruary 1, 2021
Abstract
We show that Vizing’s Theorem holds in the Borel context for graphsinduced by actions of 2-ended groups, and ask whether it holds moregenerally for everywhere two ended Borel graphs.
For a graph G on a set X , let χ ′ ( G ) denote the edge chromatic number of G . That is, the smallest cardinal k such that there exists a function assigningeach edge in G to an element of k such that any two edges incident on thesame vertex are assigned different elements. Such a function is called a k -edgecoloring of G , and the elements of k are called colors. If G is a Borel graph ona standard Borel space X , let χ ′ B ( G ) denote the Borel edge chromatic numberof G . That is, the smallest cardinal k as above, but where only Borel (asfunctions) colorings are allowed.A classical theorem of Vizing states that if G is a graph of maximum degree d ∈ ω , χ ′ ( G ) ≤ d + 1. Note the tirival lower bound χ ′ ( G ) ≥ d , so that Vizing’stheorem implies χ ′ ( G ) ∈ { d, d + 1 } . We are interested in generalizations ofthis theorem to the Borel context. Marks has shown [1] that the direct gen-eralization fails, but on the other hand, Grebik and Pikhurko have shown [2]that the generalization holds if ‘Borel’ is weakened to ‘ µ -measurable’ for someBorel probability measure µ on X which is G -invariant.Recently, Weilacher has shown [4] that some combinatorial bounds whichhold in the measurable context but not generally in the Borel context can stillbe salvaged in the Borel context with an additional assumption: that everyconnected component of G has two ends. In the spirit of this, we show in thisnote the following: 1 heorem 1. Let Γ be a marked group with two ends, say with d generators.Let G be the shift graph of Γ , so that G is d regular. Then χ ′ B ( G ) ≤ d + 1 . One purpose of this note is to pose the question of whether the assumptionin Theorem 1 that G be generated by a group action is necessary, which seemsto be open. Problem 1.
Let G be a Borel graph of maximum degree d such that everyconnected component of G has two ends. Is χ ′ B ( G ) ≤ d + 1 ? In this section we prove Theorem 1. The proof is very simple and intuitive,but it takes some time to write down all of the details
Proof.
Fix a two ended marked group Γ with symmetric generating set S ofsize d . It is well known that since Γ has two ends, there is a finite normalsubgroup ∆ ≤ Γ such that Γ / ∆ ∼ = Z or D ∞ = h a, b | a = b = id i . Let usstart with the former case for ease of notation. The latter case can be handledin the same way and will be addressed at the end of this section.Partition S as S = F n ∈ Z S n , where S n = { γ ∈ S | γ = n } , where γ denotes the image of γ in the quotient Γ / ∆ identified with Z . Note that S − n = S − n = { γ − | γ ∈ S n } for each n . Let d n = | S n | for each n , so that P n d n = d .Let G be the shift graph of Γ with vertex set X . Let Y be the standardBorel space of ∆-orbits of X . The action of Γ on X descends to an actionof Z on Y . Let H denote the Borel multigraph on Y defined by placing anedge between y and n · y for each γ ∈ S n for n > y ∈ Y . In otherwords, the number of edges between y and n · y in H is always d n for n = 0.Let k = P n> d n , so that H is 2 k -regular. We claim that H admits a Borel2 k + 1-edge coloring. Since H is generated by an action of Z , it suffices to justprove our main Theorem when Γ = Z and multiplicity for the generators isallowed: Lemma 1.
Theorem 1 holds when
Γ = Z , even when multiplicity for genera-tors is allowed. It should be noted that Vizing’s Theorem fails in general for multigraphs,(although there is a generalization which still holds) so this lemma is somewhatsurprising. We now prove it: 2 roof.
Keeping in line with the notation established so far, let H be the shiftgraph of Z with our generating set with vertex set Y . Let 0 < n ≤ n ≤· · · ≤ n k list the positive generators with multiplicity, so that H is 2 k -regular.Let G ′ be the graph on Y induced by the action of Z with usual generators ± A ⊂ Y be a Borel subset of our shiftgraph such that the induced subgraph G ′ ↾ A has connected components allof size at least 2, and such that A is recurrent . That is, for each x ∈ Y , thereare m, l > m · x ∈ A and ( − l ) · x ∈ A . Then there is a Borel3-edge coloring of G ′ , say using the colors 1,2, and 3, such that the color 3only appears on edges between points in A .First, by the recurrence of A , we can clearly find a Borel G ′ -independentrecurrent set B ⊂ Y such that if x, y ∈ B are distinct points in the same con-nected component of G ′ , the unique path between them in G ′ passes through A . Suppose x, y ∈ B such that y = N · x for some N > B between x and y in the graph G ′ .We need to color the edges between x and y . If N is even, we color theedge ( m · x, ( m + 1) · x ) with the color 1 for m even and the color 2 for m oddfor 0 ≤ m < N . If N is odd, let 0 < M < N − M · x ∈ A . This exists by definition of B , and then ( M + 1) · x ∈ A bydefinition of A . Accordingly, we color the edge ( m · x, ( m + 1) · x ) with thecolor 1 for m even and the color 2 for m odd for 0 ≤ m < M , then color theedge ( M · x, ( M + 1 · x )) with the color 3, then color the edge ( m · x, ( m + 1) · x )with the color 1 for m odd and the color 2 for m even for M < m < N . Notethat for each x ∈ B , ( x, · x ) has color 1 and ( − · x, x ) has color 2, so we doindeed end up with a coloring. Furthermore, the color 3 was clearly only everused for edges between points of A .Now, returning to our original goal, we begin by partitioning Y into k Borelrecurrent sets A , . . . , A k such that for each i , the connected components of G ′ ↾ A i each have size at least 2 n k . By a result from [3], we can start by findinga Borel maximal 2 n k -discrete set B . In particular B will be recurrent. Wecan then partition B into k many recurrent sets B , . . . , B k using, for example,that same result. Now, for each i and each x ∈ B i , there will be a smallest N > N · x ∈ B . We then include m · x ∈ A i for each 0 ≤ m < N .This clearly works.Now, fix one of our generators n i . Consider only the edges in H corre-sponding to this generator, and call the resulting (simple) graph H i . Abusinglanguage slightly, observe that since n i ≤ n k , our set A i will be recurrent forthe graph H i (more precisely, for each y ∈ Y , there are m, l > mn i · y and ( − l ) n i · y ∈ A i ), and all the connected components of H i ↾ A i will3ave size at least 2. It follows from the statement of paragraph 3 of this proofthat we can Borel edge color H i , say using the colors 2 i, i + 1, and 2 k + 1, suchthat the color 2 k + 1 is only used for edges between vertices in A i . Do so foreach i . Now, the sets of colors we used for each H i were disjoint, save for thecolor 2 k + 1. This was only used to color edges between points in A i , though,so since the A i ’s are pairwise disjoint, this will not cause any conflicts. Thusin the end, we have a Borel edge coloring using the colors 1 , , . . . , k + 1, asdesired.We now return to our proof of Theorem 1 in the case Γ / ∆ ∼ = Z . Fix aBorel 2 k + 1-edge coloring c of H , say using the colors 1 , , . . . , k + 1Let γ ∈ S n for some n >
0. For each ∆-orbit y , γ corresponds to an edgefrom y to n · y in H . Suppose c assigns the color l to that edge. Then letus give the edges ( x, γ · x ) in G the color l for each x ∈ y . Of course, since x = x ′ ⇒ γ · x = γ · x ′ , this does not cause any conflicts. Also, since c was acoloring of H , doing this for all γ ∈ S \ S does not cause any conflicts.It remains to color the edges corresponding to generators in S . These arethe edges within each ∆-orbit. For every such orbit y , the induced subgraph G ↾ y is d -regular, so by Vizing’s theorem it can be d + 1-edge colored, saywith the colors 2 k + 2 , . . . , k + d + 2 = d + 2. Since there are only finitelymany such colorings for each orbit, we may choose one of them for each orbitin a Borel fashion. We have now d + 2-edge colored our graph G in a Borelfashion.Finally, for each ∆-orbit y , there must be some color l ∈ { , . . . , k + 1 } which does not appear on any edges incident to y in our coloring c of H , since H is 2 k -regular. This means that, in our d + 2-edge coloring above, none of theedges incident to a point in y have the color l . Therefore, in the d + 1-coloringof y we have, we may replace the color d + 2 with the color l without causingany new conflicts. Doing so, we improve our coloring to a d + 1-edge coloring,and so are done.Finally, let us address the case Γ / ∆ ∼ = D ∞ . The argument which showedthe sufficiency of Lemma 1 was completely general, so here it suffices to show Lemma 2.
Theorem 1 holds when
Γ = D ∞ , even when multiplicity for gen-erators is allowed. This can be proved similarly to Lemma 1. If γ ∈ D ∞ is an order twoelement, no two of the edges it corrresponds to share an edge, So they canbe Borel colored with a single color. Else, γ has infinite order, so the edgescorresponding to γ and γ − can be Borel 3-colored. As in the proof of Lemma4, the third color here can be the same for every such γ , and used sparselyenough for each γ so that there is no conflict in the end.Thus, Theorem 1 is proved. References [1] A. Marks, A determinacy approach to Borel combinatorics.
J. Amer.Math. Soc.
29 (2016), 579-600.[2] J. Greb´ık, O. Pikhurko, Measurable Versions of Vizing’s Theorem.
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374 (2020).[3] A.S. Kechris, S. Solecki, and S. Todorcevic, Borel chromatic numbers,
Adv. Math. , 141 (1999), 1-44.[4] F. Weilacher, Descriptive chromatic numbers of locally finite and every-where two ended graphs.