Applications of the profinite splinter theorem

The first step is to show that a regular bounded profile, for every sufficiently small setX of vertices, points towards a component ofG−X. That is to say:

bounded profiles do not behave like the ultrafilter tangles of [11].

Lemma 5.2.9. LetX be a finite set of vertices andP a regular bounded profile of order at least |X|+ 1in G. Then there is a unique componentC of G−X with (V rC , C∪X)∈P.

Proof. Suppose thatP contains for every component C of G−X the separa-tion (C∪X , V rC), we are going to construct an extension ofP to a profile ofS_ℵ0.

To determine the appropriate orientation of a separation {A, B} ∈ S_ℵ0, consider the components ofG−Xand letC_Abe the union of all those components which are contained inArB. Likewise letC_B be the union of all components contained inBrAandC_R the union of the remaining components, i.e., those which meet bothAandB. Since each of these needs to meetA∩B there are only finitely many.

By our assumptionP contains for every componentC ofG−X the separ-ation (C∪X , V rC). SinceCR is a union of only finitely many components ofG−X and sinceP is a profile, we have (CR∪X , V rCR)∈P.

We now want to prove that one of (VrCB, CB∪N(CB)) and (VrCA, CA∪ N(CB)) lies inP. Indeed, if this is not the case then their respective inverses are in P, so the profile property gives us

(CB∪CA∪N(CA∪CB), V r(CA∪CB))∈P.

This however would imply that the supremum

(CB∪CA∪N(CA∪CB), V r(CA∪CB)) ∨(CR∪X , V rCR) = (V, X) lies inP by the profile property and the fact that this is a separation of order|X|. This contradicts the regularity ofP.

This proves that one of (V rC_B, C_B∪N(C_B)) and (V rC_A, C_A∪N(C_B)) lies in P, and by consistency we cannot have both. We may thus define an orientationP⁰ ofS_ℵ0 by declaring that (A, B) shall be inP⁰ if and only if (V r C_B, C_B∪N(C_B)) is inP. This orientation is consistent since P is consistent and (A, B)6(V rC_B, C_B∪N(C_B)). Note thatP ⊆P⁰ and it only remains to show thatP⁰ is a profile.

Given (A, B), (C, D)∈P⁰we have (VrCB, CB∪N(CB)) and (VrCD, CD∪ N(CD)) inP, by definition. The profile property ofP then gives us

(V rC_B, C_B∪N(C_B))∨ (V rC_D, C_D∪N(C_D))∈P ⊆P⁰, so by the consistency ofP⁰ we have ((A, B)∨(C, D))^∗∈/ P⁰.

Our next intermediate step is to show that bounded profiles cannot give rise to a decreasing sequence of components, each of which is the sole right side of a separation in that profile, and whose intersection is empty:

Lemma 5.2.10. Let P be a k-profile of G and (C_i | i ∈ I) an infinite se-quence of non-empty connected vertex sets withCi⊇Cj for alli6j and such that(V rCi, Ci∪N(Ci))∈P for each i∈I. If T

i∈ICi is empty thenP can be extended to anℵ0-profile ofG.

Proof. Let us define an orientation ˜PofS_ℵ0and show that ˜Pis a profile extending P. Let ˜Pconsist of all (A, B)∈S_ℵ

0for which there is aC_iwithC_i⊆BrA. Since the Ci form a decreasing sequence of connected vertex sets withT

i∈NCi =∅, andA∩B is finite, this orients each{A, B} ∈S_ℵ0. Moreover this orientation clearly is consistent.

For the profile property consider (A, B) and (A⁰, B⁰) in ˜P. By definition there are Ci and Cj withCi ⊆BrA and Cj ⊆B⁰rA⁰. By symmetry we may assume thati6j and thusCj ⊆Ci. But thenCj⊆(BrA)∩(B⁰rA⁰), showing that ˜P contains (A∪A⁰, B∩B⁰), as required.

To finish the proof it thus remains to verify thatP ⊆P˜. However any (A, B)∈ PwithCi⊆ArBfor someCiwould be inconsistent with (V rCi, Ci∪N(Ci))∈P. Therefore ˜P is an extension ofP.

UsingLemma 5.2.9and5.2.10we can take the next step towards showing that the setsAP,P⁰ are closed by showing that for every infinite chain inAP,P⁰

the sequence of the separators is eventually constant:

Lemma 5.2.11. Let P and P⁰ be distinguishable bounded profiles in G and (A₁, B₁)6(A₂, B₂)6. . . an infinite increasing sequence in AP,P⁰. Then the sequence (Ai∩Bi)i∈N is eventually constant.

Proof. By switching their roles if necessary we may assume thatP⁰ contains (B1, A1). Then, by consistency, (Bi, Ai)∈P⁰ and consequently (Ai, Bi)∈P for

every i∈N.

For i ∈N let us write Xi := Ai∩Bi. By Lemma 5.2.9 there is a unique componentCi of G−Xi with (V rCi, Ci∪Xi)∈P. Observe that, just like (Ai, Bi), the separation (V rCi, Ci∪Xi) distinguishes P and P⁰ efficiently.

This efficiency implies thatN(Ci) =Xi. Observe further that the separations (V rC_i, C_i∪X_i) form an increasing chain inAP,P⁰whose sequence of separators

is (A_i∩B_i)i>N. In fact C_i ⊇C_j for i 6j. It thus suffices to show that the

sequence of theC_i is eventually constant. So suppose for a contradiction that the sequence of the C_i is strictly decreasing.

To obtain a contradiction it suffices byLemma 5.2.10to show thatT

i∈NC_iis empty. So suppose that there is a vertexv∈T

i∈NCi. By applyingLemma 5.2.9 toP⁰and eachXi, we obtain componentsC_i⁰ ofG−Xi such that (VrC_i⁰, C_i⁰∪ Xi)∈P⁰ for alli∈N. ClearlyCi6=C_i⁰ andN(C_i⁰) =Xi for alli∈N. Fix any w∈C₁⁰. Thenw∈C_i⁰ for everyi∈N, and eachXi is a minimal v-w-separator inG. The latter contradictsLemma 5.2.2by the assumption that allCi, and hence allXi, are distinct. Thus T

i∈NCi is indeed empty, from which we can derive a contradiction to the boundedness ofP usingLemma 5.2.10.

Moreover, we can even show that the same statement also holds not only for chains of separations, but even for the entire setAP,P⁰:

Lemma 5.2.12. LetP and P⁰ be distinguishable regular bounded profiles in G.

Then the separations(A, B)∈ AP,P⁰ have only finitely many distinct separat-orsA∩B.

Proof. Suppose for a contradiction thatA_P,P⁰ contains an infinite sequence of separations (A₁, B₁), (A₂, B₂), . . . whose separatorsAi∩Biare pairwise distinct.

We may assume without loss of generality that (Ai, Bi) ∈P for every i∈ N.

By Lemma 5.2.1P contains all finite joins of these separations. For each i∈N

letX_ibe the separator of the supremum of (A₁, B₁) up to (A_i, B_i), and letC_i be the component ofG−X_i with (V rC_i, C_i∪X_i)∈P as given byLemma 5.2.9.

ByLemma 5.2.11the sequence of theXiis eventually constant, and therefore so is the sequence of theCi asP is a profile. LetC:=T

i∈NCi 6=∅.

Analogously forP⁰ letX_i⁰ be the separator of the supremum of (B1, A1) up to (Bi, Ai) and letC_i⁰ be the component ofG−X_i⁰ with (V rC_i⁰, C_i⁰∪X_i⁰)∈P⁰. As before letC⁰:=T

i∈NC_i⁰ 6=∅.

Since C1 and C₁⁰ are disjoint so are C and C⁰. Fix vertices v ∈ C and w∈C⁰. We claim that every separatorAi∩Bi is a minimalv-w-separator inG, contradicting the assertion of Lemma 5.2.2. To see this, consider Ai∩Bi for some i∈N. Let ˜Ci and ˜C_i⁰ be the components ofG−(Ai∩Bi) obtained by applying Lemma 5.2.9 withP andP⁰, respectively. Then C ⊆Ci ⊆ C˜i and likewise C⁰ ⊆C_i⁰ ⊆C˜_i⁰, givingv ∈ C˜_i andw∈ C˜_i⁰. Moreover both ˜C_i and ˜C_i⁰ have all ofA_i∩B_i as their neighbourhood as (A_i, B_i) efficiently distinguishesP andP⁰, and henceA_i∩B_i is indeed a minimalv-w-separator inG.

We are now ready to prove Proposition 5.2.8, i.e. that the setsA_P,P⁰ are closed subsets ofU:

Proof of Proposition 5.2.8. Let two distinguishable regular bounded profilesP andP⁰ inGbe given. ByLemma 5.2.12only finitely many sets, sayX1, . . . , Xn, appear as separators of separations inAP,P⁰. For eachXi letCi andC_i⁰ be the two components ofG−Xi given by applyingLemma 5.2.9toXi forP andP⁰, respectively.

We are now able to give a complete description of the setAP,P⁰: it is easy to check that a separation (A, B)∈S_ℵ

0 distinguishesP andP⁰ efficiently if and only if A∩B =Xi for someiwith one ofCi andC_i⁰ being a subset of Aand the other a subset ofB.

For eachX_i, the set of all (A, B)∈U withA∩B =X_i as well asC_i⊆A and C_i⁰ ⊆B is closed byLemma 5.2.4. Likewise the set of all (A, B)∈U with separatorX_i as well asC_i⁰⊆AandC_i ⊆B is closed, too. ThereforeA_P,P⁰ is the union of finitely many closed subsets ofU and hence closed.

Having established that the setsAP,P⁰ are closed inU, it thus remains for us to verify that the family of the setsAP,P⁰ splinters in order to deduceTheorem 39 fromTheorem 38. Since we shall need a slightly stronger property than splintering at a later point inSection 5.2.5, we will prove this stronger assertion here. In particular we will not make use of the assumptions that the profiles inP are regular and bounded.

To show that the setsAP,P⁰ splinter, we need to show that for all (A, B)∈ AP,P⁰ and (C, D)∈ AQ,Q⁰, either some corner separation of (A, B) and (C, D) lies inAP,P⁰ or inAQ,Q⁰, or one of (A, B) and (C, D) lies in bothAP,P⁰andAQ,Q⁰. In fact we will show that the first option always occurs.

We will split our proof of this into two separate lemmas, distinguishing the cases of equal and of distinct order of (A, B) and (C, D).

Let us first deal with the case that (A, B) is of strictly lower order than (C, D).

In this case we can say precisely which ofA_P,P⁰ andA_Q,Q⁰ will contain a corner separation of (A, B) and (C, D):

Lemma 5.2.13. Let (A, B) ∈ AP,P⁰ and (C, D) ∈ AQ,Q⁰ with |(A, B)| <

|(C, D)|. Then some corner separation of (A, B)and(C, D)lies in AQ,Q⁰.

Proof. Since|(A, B)|<|(C, D)|it follows that bothQandQ⁰ orient{A, B}the

same, say (A, B)∈Q∩Q⁰. If|(A, B)∨(C, D)|6|(C, D)|or|(A, B)∨(D, C)|6|(C, D)|, it follows that this corner separation efficiently distinguishesQandQ⁰byLemma 2.1.1,

so suppose that this is not the case. Then submodularity implies that|(B, A)∨(C, D)|<|(A, B)| and|(B, A)∨(D, C)|<|(A, B)|, which in turn contradicts the efficiency of (A, B),

since one of (B, A)∨(C, D) and (B, A)∨(D, C) would also distinguish the two robust profilesP andP⁰.

For separationsr ands the corner separations given by r ∨s andr ∨s (as well as their underlying unoriented separations) are referred to as opposite

corner separations.

The second case is that (A, B) and (C, D) are of equal order. Here we can show that there are two opposite corner separations of (A, B) and (C, D) that lie in AP,P⁰ or inAQ,Q⁰:

Lemma 5.2.14. Let (A, B) ∈ A_P,P⁰ and (C, D) ∈ A_Q,Q⁰ with |(A, B)| =

|(C, D)|. Then there is either a pair of two opposite corner separations of(A, B) and(C, D)with one element inA_P,P⁰ and one inA_Q,Q⁰, or else there are two pairs of opposite corner separations of (A, B) and (C, D), the first with both elements in AP,P⁰ and the second with both elements inAQ,Q⁰.

Proof. From |(A, B)|= |(C, D)| it follows that P and P⁰ both orient {C, D}, and likewise that QandQ⁰ both orient{A, B}.

Let us first treat the case that one ofP andP⁰orients both{A, B}and{C, D}

in the same way as one ofQandQ⁰ does. So suppose that, say, bothP andQ contain (A, B) as well as (C, D).

IfP⁰contains (D, C), then (C, D)∈ AP,P⁰ andLemma 5.2.1gives (A, B)∨(C, D)∈ AP,P⁰

and (B, A)∨(D, C)∈ AP,P⁰. Thus byproperty Pwe also have (A, B)∨(C, D)∈ AQ,Q⁰, producing the desired pair of opposite corner separations. If Q⁰ con-tains (B, A) we argue analogously.

So suppose that (C, D)∈P⁰ and (A, B)∈Q⁰. Then (B, A)∨(C, D)∈P⁰ and (A, B)∨(D, C)∈Q⁰ by the profile property, since by submodularity and the efficiency of (A, B) and (C, D) both of these corner separations have order exactly |(A, B)|. These two separations, then, are opposite corner separations of (A, B) and (C, D) with the first lying inAP,P⁰ and the second lying inAQ,Q⁰.

The remaining case is that no two of the four profiles agree in their orientation of{A, B}and{C, D}. But then both of (A, B) and (C, D) lie inAP,P⁰ as well as inAQ,Q⁰, and the existence of two pairs of opposite corner separations, one with both elements inAP,P⁰ and one with both inAQ,Q⁰, follows fromLemma 5.2.1 and the disagreement of the four profiles on{A, B}and{C, D}.

We now have all the ingredients necessary for a proof ofTheorem 39:

Proof of Theorem 39. By Proposition 5.2.8, we can applyTheorem 38. Thus we only need to show that the collection of these setsAP,P⁰ splinters. However, this follows fromLemma 5.2.13andLemma 5.2.14.

We remark that even in locally finite graphs it is not generally possible to find atree-decomposition which efficiently distinguishes all the distinguishable robust regular bounded profiles, as witnessed by the following example:

K

⁸

K

¹⁶

K

³²

K

⁶⁴

K

¹²⁸

K

²⁵⁶

K

¹⁰

Figure 5.1: A locally finite graph where no tree-decomposition distinguishes all the finite-order tangles efficiently. The green separation is the only separation efficiently distinguishing the tangle induced by the K⁶⁴ and the tangle induced by theK¹²⁸.

Example 5.2.15. Consider the graph displayed in Fig. 5.1. This graph is constructed as follows: for everyn∈Npick a copy ofK²ⁿ⁺² together withn+ 3 verticeswⁿ₁, . . . , wⁿ_n+3. Pick 2ⁿ vertices of theK²ⁿ⁺² and call themuⁿ₁, . . . , uⁿ₂n. Additionally, pick 2ⁿ⁺¹ vertices from K²ⁿ⁺², disjoint from the set ofuⁿ_i, and call themvⁿ₁, . . . , vⁿ₂n+1. Now identifyuⁿ⁺¹_i withvⁿ_i and add edges between everyw_iⁿ and everyw_jⁿ⁺¹ as well as betweenw_iⁿ andv₁ⁿ=uⁿ⁺¹₁ .

Finally we pick one copy ofK¹⁰ and join one vertex v₁⁰ of thisK¹⁰ to u¹₁ andu²₁. Additionally we pick two verticesw₁⁰, w⁰₂which are distinct fromv₁⁰from thisK¹⁰ and add an edge between eachw⁰_i and eachw_j¹.

Now each of the chosenK²ⁿ⁺²induces a robust profilePnof order²₃·2ⁿ⁺¹. The only separation which efficiently distinguishes Pn andPn+1 is the separationsn

with separator{v₁ⁱ |i < n} ∪ {uⁿ⁺¹_j }.

Additionally, theK¹⁰induces a robust profileP₀of order 4. However the only separation that efficiently distinguishes P₀andP₁has the separator{v₁⁰, w⁰₁, w⁰₂}. But these separationss₁, s₂, ...,ands₀can be oriented such as to form a chain of order typeω+ 1. This chain witnesses that there cannot be a tree-decomposition which distinguishes all bounded profiles efficiently: the separations given by such a tree-decomposition would have to contain this chain of order typeω+ 1, and it was shown in [29] that a nested set of nontrivial separations corresponds to a tree-decomposition if and only if it does not contain a chain of order typeω+ 1.

Im Dokument Tangles and where to find them (Seite 117-122)