3.6 Unravelling S k
4.1.6 Applications of the canonical splinter theorem
possibilities contradicts the choice of ai as an extremal element of S
k∈KAk. Thus the casei4j is impossible.
Let us now consider the case that i and j are incomparable. Again, by the choice of aj, none of the corner separations ofai andaj can lie inAj by Lemma 2.1.1. Therefore condition(2) of splintering hierarchically yields the existence of a corner separation of ai and aj in Ai for each side of ai; this, however, contradicts the extremality ofai inS
k∈KAk as before.
Therefore each of the sets A0j with j ∈ J is non-empty, and hence the collectionA0= (A0j|j ∈J) splinters hierarchically with respect to4. Since|J|<
|I|we may apply the induction hypothesis to this collection to obtain a canonical nested setN0=N(A0) meeting allA0j. NowN =N0∪E is a nested subset ofU which meets everyAifori∈I. It remains to show thatN is canonical.
To see thatN is canonical let ϕbe an isomorphism of separation systems between S
i∈IAi and a subset of some universe U0 such that ϕ(A) splinters hierarchically with respect to4inU0. Then ϕ(E) =E(ϕ(A)), i.e., the set of extremal elements ofS
i∈Iϕ(Ai) is exactlyϕ(E). Therefore ϕ(E) meetsϕ(Ai) if and only if E meets Ai. Consequently the restriction of ϕ toS
j∈JA0j is an isomorphism of separation systems between S
j∈JA0j and its image in U0 with the property that ϕ(A0) splinters hierarchically with respect to 4on J. Moreover, forj∈J, the imageϕ(A0j) ofA0j is exactly the set of those separations in ϕ(Aj) that are nested withϕ(E).
Thus we can apply the induction hypothesis to find that N(ϕ(A0)) = ϕ(N(A0)). Together with the above observation that ϕ(E(A)) = E(ϕ(A)) this gives
ϕ(N(A)) =ϕ(E(A))∪ϕ(N(A0)) =E(ϕ(A))∪N(ϕ(A0)) =N(ϕ(A)), concluding the proof.
Lemma 4.1.7. AP splinters hierarchically with respect to 4.
Proof. Let r ∈ AP,P0 and s ∈ AQ,Q0 be given. By switching their roles if necessary we may assume that|r|6|s|. ThenQandQ0 both orient r; we may assume without loss of generality thatr ∈Q. We will make a case distinction depending on the wayQ0 orientsr.
Let us first treat the case thatQandQ0 orientrdifferently, i.e., thatr∈Q0. Thenr distinguishesQandQ0 and hence|r|=|s| by the efficiency ofs. This implies that {P, P0}and{Q, Q0} are either the same pair or else incomparable in 4. We may assume further without loss of generality thats ∈Qands∈Q0. Consider now the two corner separationsr∨s andr∧s: if at least one of these two has order at most|s|, then this corner separation would distinguishQandQ0 by the profile property. The efficiency ofswould then imply that this corner separation has order exactly|s|and hence lies inAQ,Q0. The submodularity of the order function implies that this is the case for at least one, and therefore for both of these corner separations, yielding the existence of two corner separations ofrandsfrom different sides of sinAQ,Q0 and showing that (2)is satisfied.
Let us now consider the case thatQandQ0 orientrin the same way, i.e., thatr ∈Q0. We make a further split depending on whether|r|=|s|or|r|<|s|. Suppose first that|r| =|s|; then neither{P, P0} ≺ {Q, Q0} nor {Q, Q0} ≺ {P, P0}. We may assume thatP and P0 orient s in the same way: for if P andP0 orientsdifferently, we may switch the roles ofrandsas well as{P, P0} and {Q, Q0} and apply the above case. So suppose that both of P and P0 contain s, say. Then neither of the corner separationsr∨s nor r∨s can have order strictly less than|r|=|s|, as these corner separations would distinguishQ andQ0 or P andP0, respectively, and would therefore contradict the efficiency ofsor ofr, respectively. The submodularity of| |now implies that both of these corner separations have order exactly|r|=|s|and hence lie inAQ,Q0 andAP,P0, respectively, showing that(2)holds.
Finally, let us suppose that|r|<|s|; then{P, P0} ≺ {Q, Q0}. Consider the two corner separations r∨s andr∨s: if both ofr∨s andr∨s have order strictly greater than|s|, then by the submodularity of the order function both of the other two corner separationsr∨s andr∨s have order strictly smaller than |r|. By the robustness of P one of these two corner separations would distinguishP andP0, contradicting the efficiency of r.
Thus we may assume at least one ofr∨s andr∨s has order at most|s|. Then that corner separation distinguishesQandQ0. In fact, it does so efficiently and hence lies inAQ,Q0, showing that (1)holds and concluding the proof.
We are now ready to deduce the fullTheorem 20fromTheorem 19:
Theorem 20 (Canonical tangle-tree theorem for separation universes [17, The-orem 3.6]). LetU = (U ,6,∗,∨,∧,| |) be a submodular universe of separations.
Then for every robust setP of profiles inU there is a nested set T =T(P)⊆U of separations such that:
(i) every two profiles in P are efficiently distinguished by some separation inT;
(ii) every separation inT efficiently distinguishes a pair of profiles inP; (iii) for every automorphism αofU we have T(Pα) =T(P)α; (canonicity)
(iv) if all the profiles inP are regular, thenT is a regular tree set.
Proof. By Lemma 4.1.7the family AP splinters hierarchically. Thus we can apply Theorem 19 to AP to obtain a nested set N = N(AP) which meets every AP,P0. Clearly,N satisfies (i), (ii) and (iv) ofTheorem 20.
To see that N satisfies (iii), let α be an automorphism of U. Then the restriction ofαtoS
{P,P0}∈IAP,P0is an isomorphism of separation systems onto its image inU. We therefore have, byTheorem 19, thatα(N(AP)) =N(α(AP)).
For every AP,P0 in AP we have that α(AP,P0) is precisely the set of those separations inU which distinguishPαandP0α efficiently; in other words, we have α(AP) =APα, showing that (iii) is satisfied.
Clique separations
Regarding the profiles of clique separations discussed inSection 4.1.3,Lemma 4.1.4 not only suffices to show that the setsAP,P0 splinters, but can be used to show that the collection of these AP,P0 even splinters hierarchically, allowing us to applyTheorem 19: for this we simply define the same partial order4on the set of pairs{P, P0}as in the previous section, that is,{P, P0} ≺ {Q, Q0}if and only if|r|<|s| for some (equivalently: for all)r∈AP,P0 ands∈AQ,Q0.
To see this, letP, P0 andQ, Q0 be distinguishable pairs of profiles of clique separations. Letr∈AP,P0 ands∈AQ,Q0, and suppose without loss of generality that|r|6|s|. Ifrandsare nested, thenrandsthemselves are corner separations ofrandsthat lie inAP,P0andAQ,Q0, respectively. However, ifrandscross, then byLemma 4.1.4there are orientations of randssuch that|r∧s|,|r∧s|6|r|
and|r∧s|,|r∧s|,|r∧s|6|s|. By switching their roles if necessary we may assume thatr ∈P andr ∈P0, and likewise thats ∈Qands∈Q0.
Since (r ∧s),(r ∧s) 6 s and s ∈ Q0, the profile Q0 contains both of these corner separations by consistency. On the other hand, by the assumption that|r|6|s|, the separationrgets oriented byQ, and consequently by the profile propertyQmust contain the inverse of one of those two corner separations. This corner separation then distinguishes QandQ0, and in fact it does so efficiently, since its order is at most|s|, meaning that this corner separation lies inAQ,Q0. Therefore, if|r|<|s|, condition(1)of splintering hierarchically is satisfied.
So suppose further that |r| =|s|, and let us check that(2) of splintering hierarchically is satisfied. Observe that, similarly as above, P orientss, andP0 contains both (r ∧s) and (r ∧s) by consistency with r ∈ P0, implying as before that one of (r∧s) and (r∧s) also efficiently distinguishes P andP0, i.e., is an element ofAP,P0. If this corner separation in AP,P0 and the corner separation inAQ,Q0 found above are from different sides of eitherrors, then(2) of splintering hierarchically would be satisfied. So suppose not; that is, suppose that (r ∧s) distinguishes both P and P0 as well as Qand Q0 efficiently. In particular|r∧s|=|r|=|s|, and hence by the last part ofLemma 4.1.4, all four corner separations of r ands have order at most|r|. Consequently, sinceP0 orientss, one of (r∧s) and (r∧s) distinguishesP andP0 efficiently, which one depending on whethers ∈P0 ors ∈P0. In either case we have found a corner separation ofrandsinAP,P0, which together with (r∧s)∈AQ,Q0 witnesses that (2)is fulfilled.
Therefore, byTheorem 19we get that we can choose the set inTheorem 21 canonically:
Theorem 23. For every setP of profiles of clique separations of a graphG, there is a nested set N=N(P)of separations which efficiently distinguishes all the distinguishable profiles inP and is canonical, that is, such thatN(Pα) =N(P)α for every automorphism αof the underlying graphG.
Proof. Every automorphism of Ginduces an automorphism of the separation system. Hence we can obtain the claimed nested set by applying Theorem 19 to the family of the sets AP,P0 of those clique separations which efficiently distinguish the pairP, P0 of distinguishable profiles inP.
Circle separations
Another special case of separation systems are those ofcircle separationsdiscussed in [16]: given a fixed cyclic order on a ground-setV, acircle separationofV is a bipartition (A, B) ofV into two disjoint intervals in the cyclic order. Observe that the set of all circle separations is not closed under joins and meets and hence not a sub-universe of the universe of all bipartitions ofV:
Example 4.1.8. Consider the natural cyclic order on the set V ={1,2,3,4}. The bipartitions ({1},{2,3,4}) and ({3},{4,1,2}) ofV are circle separations.
However, their supremum in the universe of all bipartitions ofV is ({1,3},{2,4}), which is not a circle separation.
LetV be a ground-set with a fixed cyclic order andU = (U ,6,∗,∨,∧,| |) the universe of all bipartitions ofV with a submodular order function| |. LetS⊆U be the set of all separations inU that are circle separations ofV. Consequently we denote bySk the set of all those circle separations inS whose order is< k. Given fixed integersm>1 andn >3, we call a consistent orientation ofSk
a k-tanglein S if it has no subset in F=Fmn :=
F ⊆2U T
(A,B)∈FB
< mand|F|< n
.
AtangleinSis then ak-tangle for somek, and amaximal tangleinS is a tangle not contained in any other tangle inS. As usual, two tangles aredistinguishable if neither of them is a subset of the other; a separationsdistinguishestwo tangles if they orient s differently, and s does so efficiently if it is of minimal order among all separations inS distinguishing that pair of tangles.
UsingTheorem 19we can show that there is a canonical nested set of circle separations which efficiently distinguishes all distinguishable tangles inS: Theorem 24. The setS of all circle separations ofV contains a tree setT = T(S)that efficiently distinguishes all distinguishable tangles ofS. Moreover, this tree setT can be chosen canonically, i.e., so that for every automorphismαofS we have T(Sα) =T(S)α.
In order to proveTheorem 24we need the following short lemma:
Lemma 4.1.9. Letrandsbe two circle separations ofV. If randscross then all four corner separations ofrands are again circle separations.
Proof. Let r = (A, B) and s = (C, D). Since r ands cross, the sets A∩C and B∩D are non-empty and moreover intervals in the cyclic order. Thus B∪Dis also an interval and thereforer∧s = (A∩C , B∪D) is indeed a circle separation.
Let us now proveTheorem 24.
Proof of Theorem 24. For every pairP, P0of distinguishable tangles inSletAP,P0 be the set of all circle separations that efficiently distinguish P and P0. We define a partial order 4on the set of all pairs of distinguishable tangles by let-ting{P, P0} ≺ {Q, Q0} for two distinct such pairs if and only if the separations in AP,P0 have strictly lower order than those in AQ,Q0.
Let us show that the collection of these setsAP,P0 splinters hierarchically;
the claim will then follow from Theorem 19.
For this letP, P0 andQ, Q0 be two distinguishable pairs of tangles inS and let r∈AP,P0 ands∈AQ,Q0. Ifrandsare nested, thenrandsthemselves are corner separations from different sides of randsthat lie in AP,P0 andAQ,Q0, respectively, in which case there is nothing to show.
So suppose thatrandscross. Then byLemma 4.1.9all corner separations ofr andsare circle separations. By switching their roles if necessary we may assume that|r|6|s|; we shall treat the cases of|r|<|s|and|r|=|s|separately.
Let us first consider the case that|r|<|s|. Then{P, P0} ≺ {Q, Q0}, so it suffices to show that(1)is satisfied, i.e., to find a corner separation ofrand s inAQ,Q0. SinceQandQ0 both orients, which is of higher order thanr, bothQ andQ0 also orientr. By|r|<|s|and the efficiency of s,rcannot distinguishQ andQ0. Thus some orientationr ofrlies in bothQand Q0.
By renaming them if necessary we may assume that r ∈ P and r ∈ P0. Suppose now that one of r∨s andr∨s has order at most|s|. ThenQandQ0 would both orient that corner separation, and they would do so differently by the definition of a tangle. Thus that corner separation would lie inAQ,Q0, as desired.
Hence we may assume that both ofr∨s andr∨s have order higher than|s|. Then, by submodularity, bothr∧s andr∧s have order less than|r|. Therefore both of these corner separations get oriented byP andP0, but neither of them can distinguishP andP0 by the efficiency of r. In fact by the consistency ofP andP0we must have (r∧s),(r∧s)∈P∩P0. However the set{r ,(r∧s),(r∧s)} lies inF, contradicting the assumption thatP andP0 are tangles in S.
It remains to deal with the case that|r|=|s|and show that(2)is satisfied.
For this we shall find corner separations from different sides ofror ofsthat lie inAP,P0 andAQ,Q0, respectively. By the submodularity of the order function, and by switching the roles ofrandsif necessary, we may assume that there are orientations ofrandssuch that bothr∨s andr∨s have order at most|r|. By possibly renamingsands we may further assume thatr∨sdistinguishesP and P0. Then, by the efficiency ofr, we must have|r∨s|=|r|, and hence|r∨s|6|s|
by submodularity. Recall that we assumed |r∨s|=|r|=|s|, so one of r∨s andr∨s must distinguishQandQ0. Again, that corner separation must in fact distinguish QandQ0 efficiently, i.e., lie inAQ,Q0. Now this corner separation together withr∨s witnesses that(2)holds.