In order to better understand the infinite Schreier graphsΓξof spinal groups, we will encode their structure in sequences of integers characterizing its marked unlabeled isomorphism class. Each sequence may be used to build a tree describing the shape ofΓξ. This purely combinatorial object serves as a way to visualize the infinite Schreier graphs. For spinal groups on the binary tree, by Proposition 3.5.1 we know that all their unlabeled Schreier graphs are isomorphic except for one orbit. Therefore, we assumed≥3for all the section unless stated otherwise.
Definition 3.6.1. Letξ∈XN. We define its associated infinite sequence{cn}n≥1byc1= 0 and, for everyn≥0,
cn+2 =
0 ifξ0. . . ξn= (d−1)n0
k+ 1 ifξk+1. . . ξn= (d−1)n−k−10andξk6=d−1 n+ 1 ifξn6= 0
.
The value ofcn+2depends only onξ0. . . ξn, the prefix of lengthn+ 1ofξ. Intuitively, cn+2is the minimal length of the prefix we have to change toξ0. . . ξnto obtain(d−1)n0.
Notice therefore thatcn< nfor everyn≥1.
Proposition 3.6.2. Let ξ, ξ0 ∈ XN, and let{cn}n≥1 and{c0n}n≥1 be, respectively, their associated sequences. Thenξ∼ξ0 if and only ifcn=c0nfor everyn≥1.
Proof. First, suppose thatξ andξ0are compatible. This means thatξn = 0if and only if ξn0 = 0, and hencecn+2 = c0n+2 =n+ 1for everyn≥ 0such thatξn, ξ0n 6= 0. Now let n≥0such thatξn, ξn0 = 0. We can assume for a contradiction thatcn+2 > c0n+2, in which case we have, for somei∈X,i6= 0, d−1,
. . . k k+ 1 . . . n−1 n . . . ξ . . . i d−1 . . . d−1 0 . . . ξ0 . . . d−1 d−1 . . . d−1 0 . . . .
We decomposeξ asw00w10. . . andξ0 asw˜00 ˜w10. . ., with|wk| =|w˜k|for everyk, and letrbe such thatξkbelongs towkand, respectively,ξk0 belongs tow˜k. We see from the table above thatwrandw˜rcannot belong to the sameYm, which is a contradiction with the fact thatξandξ0 are compatible.
Conversely, let us suppose thatcn+2 =c0n+2, for everyn≥0. Now letn≥0such that ξn= 0, socn+2 < n+ 1, and thusc0n+2 < n+ 1. This implies thatξn0 = 0, soξandξ0do have zeros at the same positions. To see that they are indeed compatible, assume that they are not, and thus that there exists somensuch thatξn =ξn0 = 0for which there is some k < nsatisfyingξk+1. . . ξn=ξk+10 . . . ξn0 = (d−1)n−k−10andξk6=ξk0 =d−1, without loss of generality. In that case, we are actually in the same situation as in the table above, which implies thatcn+2 > c0n+2, which contradicts the fact that the sequences are equal.
Henceξandξ0must be compatible.
Definition 3.6.3. Letξ∈XN, with associated sequence{cn}n≥1. We define the tree ofξ, denotedT(ξ), as the graph with vertex setN={0,1,2, . . .}and undirected edges between nandcnfor everyn≥1. The following properties are easy to check:
• Ascn< nfor everyn≥1, the graphT(ξ)is indeed a tree.
• Any vertexn≥1has exactly1neighborksuch thatk < n.
• The degree of any vertexn ≥ 1is between1 and3. The degree of the vertex0is between1and2.
• T(ξ)and{cn}n≥1contain precisely the same amount of information. We can construct T(ξ)knowing{cn}n≥1 and vice versa.
Remark 3.6.4. We can regardT(ξ)as a tree with an induced ordering in the set of vertices.
Whenever we say that two such trees are equal, we assume that this ordering in the vertices is preserved.
Proposition 3.6.5. Letξ, ξ0 ∈XN, and letT(ξ)andT(ξ0)be, respectively, their associated trees. Thenξ∼ξ0 if and only ifT(ξ) =T(ξ0).
Proof. By Proposition 3.6.2,ξandξ0are compatible if and only if their associated sequences are the same, which happens if and only ifT(ξ) =T(ξ0).
The notion of the associated tree of a pointξ∈XNcan be used to visualize a schema of the graphΓξ. In order to do that, we will introduce an alternative, purely geometric definition forT(ξ), which will make clear how the tree describes the shape of the graph.
Definition 3.6.6. Letξ ∈ XN. Consider the sequence{Γnξ}n≥0 of finite subgraphs ofΓξ. Recall thatΓnξ =Xnσnξfor everyn≥0, see Definition 3.2.3. AsΓn+1ξ containsΓnξ for everyn≥0, we have an ascending sequence of finite subgraphs, such that S
n≥0
Γnξ = Γξ. In some sense, these subgraphs play the role of balls centered atξ, although they are much more convenient to describeΓξ.
In addition, let us define the notion of center of the subgraphs Γnξ. For n ≥ 2, we define thecenterofΓnξ as its subgraphΛn−2ξ . Recall thatΛn−2ξ = (d−1)n−20Xσnξ, see Definition 3.2.3. For convenience, we set the centersΛ−2ξ = Γ0ξ ={ξ}andΛ−1ξ = Γ1ξ = Xσξ.
Notice that, for everyn≥0,Γnξ contains all centersΛkξ fork=−2, . . . , n−2, and does not contain any centerΛkξ, fork≥n−1. Moreover,Γnξ intersects one or two other centers.
It intersectsΛn−1ξ at the vertex(d−1)n−10σnξifn≥1. It intersects another centerΛkξ, for k≥n, if and only if the vertex(d−1)nσnξis not fixed byB.
Definition 3.6.7. Let ξ ∈ XN. We define the graphT0(ξ) as the graph with vertex set N={0,1,2, . . .}and undirected edges betweenkandnfor eachk > n≥0satisfying one of these two conditions:
• Λn−2ξ andΛk−2ξ have non-empty intersection, and this intersection is not contained in any otherΛlξ, forl≤n−3.
• Λn−2ξ andΛk−2ξ do not intersect, but there exists a path inΓξbetween them satisfying:
– The endpoint inΛn−2ξ does not belong toΛlξ, for anyl≤n−3.
– The endpoint inΛk−2ξ does not belong toΛlξ, for anyl≤k−3.
– The inner vertices of the path do not belong toΛlξfor anyl≥ −2.
Notice that asΛ−2ξ ={ξ} ⊂Xσξ= Λ−1ξ , there is always an edge from1to0.
These conditions illustrate how the centers ofΓξ are connected. This is sufficient in order to know how the subgraphsΓnξ are connected, and hence to recover the shape ofΓξ itself. The following notion of blocking centers simplifies the evaluation of these conditions.
Definition 3.6.8. Letr, s≥1. We say that the centersΛnξ1−2, . . . ,Λnξr−2blockthe centers Λkξ1−2, . . . ,Λkξs−2 if any path from any ofΛkξ1−2, . . . ,Λkξs−2to any other centerΛl−2ξ , with l6=n1, . . . , nrandl6=k1, . . . , kr, contains an inner vertex in one ofΛnξ1−2, . . . ,Λnξr−2.
In that case, the vertices k1, . . . , ks ofT0(ξ)can only have edges among them or to n1, . . . , nr, but they cannot have edges to any otherl.
Proposition 3.6.9. For everyξ∈XN,T(ξ) =T0(ξ). Equivalently,T(ξ)∼=T0(ξ)and they have the same vertex ordering.
Proof. It follows from the definition that both graphs have an edge between1and0. Let us show that, for everyn≥2, the set of edges betweenn+ 2and{0, . . . , n+ 1}is the same forT(ξ)andT0(ξ). Notice that in the first, there is always exactly one such edge, which goes fromn+ 2tocn+2.
Letn≥0, and definev= (d−1)n0σn+1(ξ), the only vertex in the intersection ofΓn+1ξ andΛnξ. Let alsopbe the smallest index such thatv ∈Λp−2ξ . Ifpwas2, . . . , n+ 1, then vp−2= 0, which contradicts the definition ofv. So the only possibilities arep= 0,1, n+ 2.
Ifp = 0, thenv =ξ, socn+2 = 0. On the other hand, because every path fromΛnξ to any ofΛlξwithl≤n−1goes throughv, the only edge betweenn+ 2and{0, . . . , n+ 1}
joinsn+ 2and0. Ifp= 1,v= (d−1)σ(ξ), socn+2 = 1. Similarly, inT0(ξ), we have the edgen+ 2to1for the same reason as before. Now assumep=n+ 2for the rest of the proof. Λn+2is the only center containingv. We have three possibilities, according to the value ofξn.
Ifξn 6= 0, d−1, thencn+2 =n+ 1, andΓnξ is only connected toΓξ\Γnξ through the vertexu= (d−1)n−10σn(ξ). Because this vertexubelongs toΛn−1ξ , this means thatΛn−1ξ blocks the centersΛlξfor everyl≤n−2. Therefore, the only edge inT0(ξ)betweenn+ 2 and{0, . . . , n+ 1}is fromn+ 2ton+ 1.
Suppose thatξn=d−1, so againcn+2 =n+ 1. Letube the vertex(d−1)n−10σn(ξ) andwbe the vertex(d−1)nσn(ξ). Notice thatu ∈ Λn−1ξ ∩Γnξ andw ∈ Γnξ, andΓnξ is only connected toΓξ\Γnξ throughuand possiblyw. Ifwis fixed byB, then we are in the same situation as above:Λn−1ξ blocksΛlξfor everyl≤n−2. Otherwise,w∈Λrξ, for some r ≥n+ 1, and thenΛn−1ξ andΛrξ block the centersΛlξ for everyl ≤n−2. In any case, there is only one edge inT0(ξ)betweenn+ 2and{0, . . . , n+ 1}, joiningn+ 2andn+ 1.
Assume now for the rest of the proof ξn = 0, and let k be the smallest such that ξk+1. . . ξn= (d−1)n−k−10. Asv6=ξ, necessarilyk≥1, and by minimalityξk 6=d−1.
In this case, we havecn+2=k+ 1.
Letu= (d−1)k0σk+1(ξ)∈Λkξ, and observe thatΓk+1ξ is connected toΓn+1ξ \Γk+1ξ only through the vertexu. The former contains all centersΛlξfor everyl ≤k−1, while the latter contains all centersΛlξ with k ≤ l ≤ n−1. Any path from Λnξ to Λlξ with k≤l≤n−1must then contain the vertexu∈Λkξ, which prevents any edge inT0(ξ)from n+ 2to{k+ 2, . . . , n+ 1}.
NowΓkξ is connected toΓξ\Γkξonly through the verticesu0= (d−1)k−10σk(ξ)∈Λk−1ξ and possiblyw = (d−1)kσk(ξ). Ifwis fixed byB, thenΛk−1ξ blocks all centersΛlξ for l ≤ k−2. Otherwise, we haveξk = 0, and in factu = w. ThenΛk−1ξ and Λkξ block all centersΛlξfor l ≤ k−2. In any case, this prevents any edge inT0(ξ)from n+ 2to {0, . . . , k}. Hence, the only possible edge fromn+ 2to{0, . . . , n+ 1}is tok+ 1.
Finally, any path withinXk(d−1)σk+1(ξ)joiningvand(d−1)k−10(d−1)σk+1(ξ)∈
Λk−1ξ connects the centersΛnξ andΛk−1ξ without intersecting any other center. Hence, there is indeed an edge inT0(ξ)betweenn+ 2andk+ 1. In any of the cases we showed that T(ξ)andT0(ξ)have the same edges betweenn+ 2and{0, . . . , n+ 1}for everyn ≥0.
This suffices to conclude thatT(ξ)andT0(ξ)are the same tree.
Remark 3.6.10. Recall that, by Theorem 3.5.11, (Γξ, ξ) and (Γη, η) are isomorphic as marked unlabeled graphs if and only ifξ andηare compatible. We also proved in Propo-sition 3.6.2 thatξ andη are compatible if and only if their associated sequences{cn}n≥1 coincide, or, equivalently, by Proposition 3.6.5, if and only ifT(ξ)andT(η)are the same tree (with the induced ordering). Now Proposition 3.6.9 implies that we have two ways of building this tree. The first way is purely combinatorial, only using the sequence{cn}n≥1, which is obtained straightforward from the digits ofξ. The second is purely geometrical, as it is based in paths between the centers insideΓξ.
Remark 3.6.11. These two additional characterizations of compatibility, namely the asso-ciated sequence and tree, are useful in both directions. On the one hand, we can take any ξ ∈XN, find its associated sequence{cn}n≥1and from it build its treeT(ξ). We may now use the tree to recover the Schreier graphΓξas follows:
1. LetΓbe an unlabeled copy of the graphΓ1without loops at the verticesu1 = 0and u2 =d−1. Let alsozbe the vertex labeled byξ0. Call the verticesuandvopenand the restclosed.
2. For everyn≥0, do the following:
(i) LetΛnbe an unlabeled copy ofΛωn from Proposition 3.1.3. Call all its vertices open. Letv∈Λn.
(ii) Ifcn+2 = 0, identifyvwithzinΓand call the identified vertexclosed.
Ifcn+2 = 1andu1is open, identifyvwithu1inΓand call the identified vertex closed. Otherwise identifyvwithu2inΓand call the identified vertex closed.
Ifcn+2=k+ 2fork≥0, letw∈Λkbe an open vertex. LetΓ0k+1be a copy of Γk+1without loops on the verticesv0= (d−1)k+1andw0 = (d−1)k0. Call all its vertices closed. InΓ, identifyvwithv0andwwithw0, and call both identified vertices closed.
3. Finally, for everyn≥0and every open vertexw∈Λn, letΓ0n+1 be a copy ofΓn+1 without loops on the vertexw0= (d−1)k0. Call all its vertices closed. InΓ, identify wwithw0, and call the identified vertex closed.
4. The resulting marked graph(Γ, z)is isomorphic to the Schreier graph(Γξ, ξ), with Λncorresponding to the centerΛnξ for everyn≥0.
On the other hand, we can work backwards if we start from a marked graph(Γ, z)and we want to know if there exists someξ ∈XNsuch that(Γ, z)∼= (Γξ, ξ). In this case, we can build the tree fromΓusing the second definition, and extract the sequence{cn}n≥1from it. Finally, we only have to use its definition to unravel a set of conditions on the digits ofξ, which will determine its compatibility class.
We may also wonder what happens if instead of prescribing a graph we prescribe a tree T(or, equivalently, a sequence{cn}n≥1) and we try to findξ ∈XNsuch thatT =T(ξ). As it turns out, not every tree or sequence can be realized as the associated of pointξ ∈XN. Let us characterize the sequences{cn}n≥1arising in such a way.
Proposition 3.6.12. Let{dn}n≥1⊂N.{dn}n≥1is the associated sequence of someξ∈XN if and only if
• dn≤n−1, ∀n≥1.
• dn+26=n+ 1 =⇒dr+2≥n+ 1, ∀r > n.
Proof. Every sequence defined as in Definition 3.6.1 satisfies these conditions. For the converse, let{dn}n≥1satisfy both conditions. LetN∗ =N\ {0}. We define a pointξ∈XN in the following way:
1. SetI ={n∈N∗ |dn+2 < n+ 1}andJ ={n∈N∗ |dn+2 =n+ 1}. By the first condition,I tJ =N∗.
2. Ifdn+2= 0for somen∈I, we defineξ0. . . ξn= (d−1)n0. By the second condition, there can only be one suchn, so this is well defined.
3. For eachksuch that0≤k≤n−1, ifdn+2=k+ 1for somen∈I, then we define ξk+1. . . ξn = (d−1)n−k−10. This is well defined because ifn0 ∈ I, withn0 < n, then by the second conditiondn+2≥n0+ 1, so they define disjoint sets of digits ofξ.
4. After this procedure, we set any undefined digit ofξto1.
Let{cn}n≥1 be the associated sequence of thisξ. Let us verify that, indeed,dn=cn for everyn≥1. First, we havec1 =d1 = 1, andcn+2 =n+ 1for everyn≥0such that ξn6= 0. By construction,ξn6= 0if and only ifn∈J, equivalently ifdn+2=n+ 1.
Now letn≥0such thatcn+2= 0. Then,ξ0. . . ξn= (d−1)n0. Again by construction ofξ,dn+2= 0too.
Finally, ifcn+2 =k+ 1with0≤k≤n−1, thenξk+1. . . ξn = (d−1)n−k−10, and ξk6=d−1. But then once againdn+2must also bek+ 1by construction ofξ.
In addition, we can use the tree description to illustrate the computation of the growth of the graphsΓξ, previously shown in [9].
Proposition 3.6.13. LetGωbe a spinal group withω∈Ωd,mwithd≥2andm≥1, and letξ∈XN. The growth of the Schreier graphΓξis polynomial of degreelog2(d).
Proof. Without loss of generality (see Remark 3.2.6), let us prove the statement for spinal groups Gd, defined by m = 1 and ω = πN, withπ mapping the generator of B to the generator ofA. Moreover, in [10] it is proved that the growth rate of the graphsΓξis the same for everyξ∈XN, so let us restrict to the pointξ= (d−1)N.
Consider the treeT(ξ), which is a one-ended line with vertex labels inNand edges (n, n+ 1)for everyn≥0. Identifying each vertexnwith the centerΛn−2ξ , and each edge (n, n+ 1)with the finite subgraph connectingΛn−2ξ andΛn−1ξ inΓξ, we observe that, for anyn≥ 1, the subtree{0, . . . , n}represents the finite subgraphΓn−1ξ . This subgraph is, up to some loops, isomorphic to the finite Schreier graphΓn−1, and therefore has diameter 2n−1−1and containsdn−1vertices. Moreover, the ball of radius2n−1−1aroundξcoincides withΓn−1ξ .
Let r ≥ 0and defineksuch that2k−1 ≤ r ≤ 2k−1. The subgraph represented by {0, . . . , k}, of diameter2k−1−1, is then contained in the ball of radiusraroundξ, and this is contained in the subgraph represented by{0, . . . , k+ 1}, of diameter2k−1. The size of the ball is then bounded betweendk−1 anddk. Sincek−1 =dlog2(r)e, it is bounded betweenddlog2(r)eandddlog2(r)e+1, and is thus equivalent torlog2(d), as
dlog2(r) =dlog2(d) logd(r)=rlog2(d).
Let us now illustrate the computation of the sequences{cn}n≥1 and the treesT(ξ)with some examples.
Consider the Fabrykowski-Gupta group (d= 3,m= 1). We will compute the associated sequences{cn}n≥1 using Definition 3.6.1 and display the associated trees for the points 2N, 0N, (110)N and (210)N in XN. The first, third and fourth yield one-ended Schreier graphs (see Theorem 3.3.1), and the second a two-ended Schreier graph. For comparison, the last two are compatible points, so by Theorem 3.5.11 their unlabeled Schreier graphs are isomorphic.
In the illustrating Figures 3.10, 3.11, 3.12 and 3.13, vertex labels are to be concatenated with the appropriate shift of ξ. Subgraphs denotedΓn are copies withinΓξ of the finite Schreier graphsΓn, in the sense of Proposition 3.2.4. The associated treeT(ξ)is overlapped to show its relation withΓξ.
Γ1
Figure 3.10: Schema of the Schreier graphΓξfor the Fabrykowski-Gupta group withξ = 2N. The treeT(ξ)is overlapped in red.
Example 3.6.14. Letξ= 2N. Its sequence{cn}n≥1satisfiescn+2=n+ 1for everyn≥1, so it is(0,1,2,3,4,5,6, . . .). The associated treeT(ξ)is then a one-ended line labeled by the natural numbers in order. A schema ofΓξand the treeT(ξ)are displayed in Figure 3.10.
Γ1
Figure 3.11: Schema of the Schreier graphΓξfor the Fabrykowski-Gupta group withξ = 0N. The treeT(ξ)is overlapped in red.
Example 3.6.15. Letξ = 0N. Its sequence{cn}n≥1 satisfiescn+2 =nfor everyn ≥ 0, so it is(0,0,1,2,3,4,5, . . .). The associated tree T(ξ)is then a two-ended line labeled by increasing even numbers on one side of0and increasing odd numbers on the other. A schema ofΓξand the treeT(ξ)are displayed in Figure 3.11.
Example 3.6.16. Letξ= (210)N. Its sequence{cn}n≥1is given byc1= 0and, forn≥0, cn+2 =
( n ifn≡2 mod 3 n+ 1 ifn6≡2 mod 3 ,
so it is the sequence(0,1,2,2,4,5,5, . . .). The associated treeT(ξ)is the one-ended tree displayed in Figure 3.12 overlapped on a schema of the Schreier graphΓξ.
Γ1
Figure 3.12: Schema of the Schreier graph Γξ for the Fabrykowski-Gupta group with ξ = (210)N. The treeT(ξ)is overlapped in red.
Figure 3.13: Schema of the Schreier graph Γξ for the Fabrykowski-Gupta group with ξ = (110)N. The treeT(ξ)is overlapped in red.
Example 3.6.17. Letξ = (110)N. As it is compatible with (210)N, their associated se-quences coincide, hence{cn}n≥1is given byc1= 0and, forn≥0,
cn+2 =
( n ifn≡2 mod 3 n+ 1 ifn6≡2 mod 3 ,
so it is the sequence(0,1,2,2,4,5,5, . . .). A schema of the Schreier graphΓξis shown in Figure 3.13, together with the treeT(ξ). By comparing Figures 3.12 and 3.13, it becomes
evident that the unlabeled Schreier graphs are isomorphic and that both associated trees coincide.