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Problems on well quasi-ordering and hereditary classes of relational structures.

Maurice POUZET

Institut Camille Jordan,Universit´ e Claude-Bernard, Lyon 1, France and University of Calgary, Canada. Well quasi-orders in Computer Science Dagstuhl seminar

January 19, 2016

January 19, 2016 1 / 44

Abstract

I present some problems on well-quasi-ordering and hereditary classes of relational structures. Some of these problems go back to the seventies. Most of the problems discussed here can be formulated in terms of graphs (undirected, with no loops). For several reasons, I consider structures which are more general.

January 19, 2016 2 / 44

Basic notions.

Relational structures

A relational structure is a realization of a language whose non-logical symbols are predicates. This is a pair R ∶= (V ,(ρi)i∈I) made of a set V and of a family of mi-ary relations ρi on V . The set V is the domain or base of R, we set V (R) for the base of R. The family µ ∶= (mi)i∈I is the signature of R. The substructure induced by R on a subset A of V , simply called the restriction of R to A, is the relational structure R↾A ∶= (A,(Ami ∩ ρi)i∈I). Notions of isomorphism and local isomorphism from a relational structure to an other one are defined in a natural way as well as the notion of isomorphic type.

January 19, 2016 3 / 44

Basic notions.

Embeddability

A relational structure R is embeddable into a relational structure R′ and we set R ≤ R′ if R is isomorphic to some restriction of R′. Embeddability is a quasi-order on the class of relational structures. In the late forties, Fra¨ ıss´ e, following the work of Cantor, Hausdorff and Sierpinski, pointed out the role of this quasi-order in the theory of relations. It turns

• ut that the basic notions about ordered sets (posets) have a direct

counterpart in terms of relational structures.

January 19, 2016 4 / 44

Basic notions.

Hereditary classes, ideals and ages

This is the case of initial segments, ideals, chains and antichains. For example, a class C of structures is hereditary if it contains every structure which can be embedded into some member of C. Clearly, hereditary classes are initial segments of the class of relational structures quasi-ordered by embeddability. If R is a relational structure, the age of R is the set Age(R) of finite restrictions of R considered up to isomorphy (a set introduced by R. Fra¨ ıss´ e). This is an ideal of the poset made of finite structures considered up to isomorphy and ordered via embeddability. As shown by Fra¨ ıss´ e, 1948, every countable ideal has this form. Well-quasi-ordering, one of the most important notion in the theory of

• rdered sets, has a fundamental role in the theory of relations, a

cornerstone being Laver’s theorem on scattered chains. Recent years have seen a renewed interest for the study of hereditary classes particularly those made of finite structures. In this talk, I will list some problems concerning the quasi-order of embeddability.

January 19, 2016 5 / 44

Basic notions.

Well foundation, Well-quasi-order, Height

A poset P is well founded if every non-empty subset has some minimal

• element. It is well-quasi-ordered , in brief w.q.o., if every non-empty

subset contains finitely many minimal elements(this number being non-zero). A final segment F of a poset P is finitely generated if for some finite subset K of P, F equals the set ↑ K ∶= {y ∈ P ∶ x ≤ y for some x ∈ K}. Let P be a well founded poset; the height h(x,P) of an element x ∈ P is an ordinal defined by induction by the formula: h(x,P) = Sup{h(y,P) + 1 ∶ y ∈ P,y < x}.

January 19, 2016 6 / 44

Basic notions.

Ordinal length

An important result on w.q.o. is de Jongh-Parikh theorem (1977).

Theorem

If a poset P is w.q.o. then all the linear extensions of P are well-ordered and there is one having the largest possible order type. This largest order type, denoted o(P), is the ordinal length of P. For example, if Q is a w.q.o. then o(Q) = h(Q,I(Q)) where I(Q) is the set of initial segments of Q (this is an equivalent formulation of de Jongh-Parikh’s theorem).

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The ordinal length of several posets have been computed. For example, the ordinal length of the direct sum, resp. product, of two posets is the Heissenberg sum, resp. product, of their ordinal lengh (Carruth (1946), de Jongh and Parikh (1977)); if A∗ is the set of words over un alphabet A made of k letters then o(A∗) = ωωk−1 (de Jongh and Parikh (1977)), this formula was extended to an arbitrary wqo (see Schmidt (1978)); the

• rdinal length of the collection of binary trees is the ordinal ǫ0 (Schmidt

(1978)); for more, see her habilitation and Rathjen and Weiermann (1993)).

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Ordinal length of hereditary classes

Hereditary classes which are w.q.o. abund. A hereditary class C of finite structure is w.q.o. iff its antichains are finite. Hence if C is wqo it is countable and thus o(C) is a countable ordinal.

Problem

Is every countable ordinal attained by some hereditary class of finite structures with a finite signature? Is there a largest countable ordinal depending upon the signature? What about hereditary classes of graphs? If we allow unbounded signature, the answer to the problem above is negative: every ordinal below ω1 can be attained.

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Ideals, ages and ordinal length

An ideal of a poset P is any non empty initial segment of P which is up-directed. The poset P is wqo iff it is a finite union of ideals which are wqo. There is a relationship between the ordinal length of P and the height of the maximal ideals of P. For an example, if P is wqo and up-directed then

• (P) = h(P,I(P)) ≤ ωH(P) where H(P) ∶= h(P,J(P) and J(P) is the set
• f ideals of P.

A much more precise formula holds for some ages. An ideal of finite structures is wqo iff its antichains are finite; hence if it is wqo it is countable an thus this is the age of some relational structure. With Sobrani, we proved:

Theorem

Let A be an age. If J(A) is w.q.o then o(A) = ωα ⋅ q where α is such that ω ⋅ α ≤ H(A) < ω ⋅ (α + 1) and q is the number of ages included into A whose height is between ω ⋅ α and ω ⋅ (α + 1).

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Hereditary classes and beter quasi-ordering

We needed the condition that J(A) is w.q.o. This is related to the notion

• f better quasi-ordering and a question of Nash-Williams.

A poset P is better quasi-ordered (b.q.o.) if the class P<ω1 of countable

• rdinal sequences is w.q.o under embeddability of sequences (alternatively,

the transfinite iterates Iα(P), for ordinals α, of the set of initial segments

• f P are all w.q.o. Nash-Williams, who invented this notion and proved

the fundamental results, conjectured that natural classes of structures which are w.q.o are in fact b.q.o.

Problem

Is a hereditary class of finite structures with finite signature is b.q.o. whenever it is w.q.o.?

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Construction of counterexamples

The answer is negative if the signature is infinite and the arity is

• unbounded. There are w.q.o. ages which are not b.q.o.

Let V be an infinite set and F ⊆ [V ]<ω ∖ {∅}. Let MF ∶= (V ,(KF)F∈F) be the relational structure, where KF is the ∣F∣-ary relation on E defined by KF(x1,...,x∣F∣) = + if and only if {x1,...,x∣F∣} = F; let UF ∶= (V ,(UF)F∈F) where UF(x1,...,x∣F∣) = − everywhere. Given A ⊆ V , set

○

A∶= ⋃{F ∈ F ∶ F ⊆ A}. Define an equivalence on [V ]<ω, two elements A,B ∈ [V ]<ω being equivalent if ∣A∖

○

A∣ = ∣B∖

○

B∣ and

○

A=

○

B. The collection [E]<ω/F of equivalence classes is ordered via the inclusion relation: if U and V are two equivalence classes, U ≤ V means that U is the class of some A, V is the class of some B with A ⊆ B. Let F<ω (resp., F∪) be the collection of finite (resp., arbitrary) unions of members of F. Clearly F∪ is a complete lattice, the least element and the largest element being the empty set and ⋃F, respectively.

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Proposition

The ages Age(MF) and Age(UF) satisfy the following properties: 1) Age(MF)) is order-isomorphic to [V ]<ω/F. 2) The set D(A(UF)) of ages included into A(UF) is totally ordered and if V ∖ ⋃F is infinite or if ϕ[V ]<ω/F takes only finite values then D(A(UF),A(MF)) the set of ages beween A(UF) and A(MF) is

• rder-isomorphic to the complete lattice F∪.

Theorem

Let P be a poset. If P embeds into [ω]<ω then there are two ages A and B with A totally ordered and A ⊆ B such that the set D(A,B) of ages between A and B is isomorphic to I(P).

Proof.

Since P embeds into [ω]<ω then ∣ ↓x∣ < ω for all x ∈ P. Let V ∶= P ∪ X where X is some infinite set disjoint from P, and F ∶= {↓x ∶ x ∈ P}. Clearly F∪ = I(P). The proof of the result follows from Proposition 5.

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Let A be the age of a countable structure; Since A is countable, ages included ino A coincide with ideals. Let D(A) be the collection of ages included into A. The following implications (i) ⇒ (ii) ⇔ (iii) ⇒ (iv) ⇒ (v) ⇒ (vi) hold: (i) A is w.q.o.; (ii) D(A) is countable; (iii) D(A) is topologically scattered; (iv) Chains in D(A) are at most countable; (v) D(A) is well-founded; (vi) The poset [ω]<ω of finite subsets of ω does not embed into A.

Problem

Is the reverse implication hold when the signature is finite ? What about ages of graphs?, of ordered graphs?, of permutations?

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Is there a combinatorial characterization of hereditary classes?

Consider the simpler case of an age. The age of R on a set V is the set of isomorphic types of the substructures induced on finite subsets of V . Say simply that two finite subsets A, A′ of V are equivalent if there is an isomorphism of the structure induced by R on A onto the structure induced by R′ on A′. This gives a partition of the finite subsets of V then an ordering on the parts: A part τ is less than τ ′ if there are A ⊆ A′ such that τ is the class of A and τ ′ the class of A′ (this amounts to every A′ contains some A). We can see that as an ordered quotient of the powerset ℘(V ).

• Problem. Characterize combinatorially these quotients.

Clearly, the members of each class have the same cardinality, next they have the same frequency vector: given A′, the frequency vector χA′ will be made of a sequence of numbers, counting for each class, how many members of that class are included into A′. In general, this does not

• suffice. But all properties of classes of relational structures that I can

prove use only that. This appeared in a paper with Ivo Rosenberg on Sperner propery for

January 19, 2016 15 / 44

A problem of Bonnet on bqo

Robert Bonnet asked the following question

Question

Is a wqo a countable union of bqo’s? We may ask for more

Question

If a poset P has no infinite antichain, is its domain a countable union of sets An such that the order on each An is a strenghtening of a finite intersection of linear orders? If P is well founded then the An are wqo, hence P is a countable union of bqo’s. This would anwer positively Bonnet’s question. Try to use induction on the rank r(P) of P, where r(P) is the height of the empty set in A(P) the poset of antichains of P ordered by reverse of

• inclusion. Avraham has proved that if r(P) < ω2

1 then P is a countable

union of chains.

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These question point to the class P of posets P which are countable unions of strenghtening of finite dimensional posets. Why strengthening? A strengthening of ω1 × ω1 is not necessarily a countable union of finite dimensional posets (use Erdos-Tarski).

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Profile

The profile of a relational structure R is the function φR which counts for every integer n the number φR(n) of substructures of R induced on the n-element subsets, isomorphic substructures being identified. Clearly, this function only depends upon the age Age(R) of R.

Problem

Is the profile of a relational structure R bounded by some exponential whenever the age of R is well-quasi-ordered under embeddability?

January 19, 2016 18 / 44

Theorem

Let C be a hereditary class of finite structures with finite signature. The profile of C is bounded by a polynomial iff C is w.q.o. and o(C) < ωω (in fact, o(C) = ωn.k + ν avec ν < ωn iff the growth of the profile is polynomial with degree n − 1).

January 19, 2016 19 / 44

Use the fact that the age A of a relational structure R with finite kernel has height ω.n + p iff its profile grows like a polynomial of degree n − 1. Furthermore, A has finitely many bounds and the collection of subages of A is w.q.o. (in fact A is the age of an almost multichainable structure). Hence, we may apply the formula in Theorem 3 obtained with Sobrani.

Theorem

If the age of a relational structure R is inexhaustible and ϕR has polynomial growth of degree k then ϕR(n) ≤ (n + k k ) for every integer n. Ages of height ω2 are also w.q.o. They may have infinitely many bounds. In fact, there are 2ℵ0 ages of undirected graphs of height ω2. There are larger ordinals. The ordinal length of the class of finite forest is ǫ0. The

• rdinal length of the class of finite series-parallel poset is Γ0 (Sobrani, I,

2003).

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Preserving w.q.o. by adding a linear order

The concatenation of two relational structure R and S on the same domain is the relational structure, denoted by R.S, made of the relations

• f R followed by the relations of S.

Problem

1 Let C be a hereditary class of finite structures. If C quasi-ordered

under embeddability is w.q.o., is it true that for every R ∈ C there is some linear order LR on V (R), the domain of R, such that the class

• f R.LR is w.q.o.?

2 Let R be a relational structure. If Age(R) is w.q.o. is Age(R.L)

w.q.o. for some linear order on V (R)?

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(2) → (1). If C is w.q.o. then C is a finite union of ideals I1,...,In. Each Ii is countable (for each integer m, members of size m form an antichain, since Ii is wqo such an antichain must be finite) thus is the age of some relational structure Ri. If (2) 2 has a positive answer, there is some linear

• rder Li such that Age(Ri.Li) is wqo. Then, ⋃i=1,n Age(Ri.Li) is wqo.

No clear that (1) → (2): Let R and let C be the class of R′.L′ such that R′ ∈ Age(R) and L′ is a linear order on V (R′). We may find a wqo subclass C′ such that Age(R) ∶= {R′ ∶ R′.L′ ∈ C′}. If C′ is hereditary, compactness theorem will allow to find L on R such that Age(R.L) is wqo. But, if C′ is not hereditary?.

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Adding unary predicates

Let P be a poset. A labelling of a relational structure R by P is a map f from V (R), the domain of R, into P. Let C be a class of relational structures of a fixed signature µ. Denote by C.P the class of relational structures R ∈ C labelled by P, that is the class of pairs (R,f ) with f ∶ V (R) → P. We quasi order C.P by dominance : (R,f ) ≤ (R′,f ′) if there is an embedding h ∶ R → R′ such that f (x) ≤ f ′(h(x)) for every x ∈ V (R).

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Examples

Suppose that P = {0,1} with 0 incomparable to 1. Then C.P with this quasi-order is isomorphic to the collection C1 made of the collection of relational structures quasi-ordered by embeddability which is made of concatenats R.U where R ∈ C and U is a unary relation on V (R). More generally, let m be an integer and P be an antichain of size 2m then C.P with the quasi-order above is isomorphic to the collection Cm made of the collection of relational structures quasi-ordered by embeddability and made of concatenats R.U1.....Um where R ∈ C and U1,...,Um are m unary relation on V (R). Indeed, let us identify the 2m-element antichain P with ℘({1,...,m}), the power set of {1,...,m}, ordered by the equality relation. To f ∶ V → ℘({1,...,m}) associate U(f ) ∶= (U1,...Um) where Ui ∶= {x ∈ V (R) ∶ i ∈ f (x)}. Clearly (R,f ) ≤ (R′,f ′) iff R.U(f ) ≤ R′.U(f ′). More interestingly, order ℘({1,...,m}) with the inclusion order, in this case R.U1 ...Um ≤ R′.U′

1 ...U′ m mean that there is

some embedding h ∶ R → R′ such that x ∈ Ui implies h(x) ∈ U′

i for every

x ∈ V (R) and i ∶= 1,...,m.

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Let R be a relational structure and m be a non-negative integer. A labelling of R with m constants is the structure R with m distinguished elements a1,...,am of V (R) added. An embedding of (R,a1,...am) into (R′,a′

1,...a′ m) is an embedding f of R into R′ such that f (ai) = a′ i for

i = 1,...m. If C is a class of structures, we denote by C.m− the class of (R,a1,...am) where R ∈ C.

Problem

Let C be a hereditary class of finite structures. If C.2− is wqo, is C.P wqo for every w.q.o P? Let P be a w.q.o. A class C of structures is P-wqo if C.P is wqo. A class C of relational structures is very-well-quasi-ordered, in short v.w.q.o., if for every integer m the class Cm made of R ∈ C added of m unary relations is w.q.o for the embeddability relation. Equivalently, C is P-wqo for every finite poset P. We say that C is hereditarily w.q.o., resp. hereditarily b.q.o. if C.P is w.q.o., resp b.q.o., for every w.q.o., resp. b.q.o. P.

January 19, 2016 25 / 44

I recall the following result.

Theorem

Let C be a class of finite relational structures, then:

1 C is v.w.q.o iff ↓ C, its downward closure, is v.w.q.o. 2 If ↓ C is v.w.q.o. then all the ages it contains are almost inexhaustible. 3 If ↓ C is v.w.q.o. and all of its members have the same finite signature

µ then it has only finitely many bounds,(I. 1972).

January 19, 2016 26 / 44

A problem of Atmitas and Lozin, Order, 2015

Let C be a hereditary class of finite graphs, undirected with no loops. If C is wqo and defined by finitely many obstructions, is C.P wqo for every wqo P? Does not hold for binary sructures (but the converse holds).

January 19, 2016 27 / 44

Extension of Laver’s theorem

Laver proved that the class of chains which are countable unions of scattered chains is b.q.o. hence w.q.o. under embeddability. Fra¨ ıss´ e asked how to extend this result to ages. There are bqo ages of graphs whose the collection of countable structures with this given age is not wqo (I found some examples in 1978, other were given in Sobrani thesis) . A relational structure is binary if it consists of relations which are unary or

• binary. If R ∶= (V ,(ρi)i∈I) is binary, a subset A of V is autonomous if

ρi(x,y) = ρi(x′,y) and ρi(y,x) = ρi(y,x′) for all i ∈ I, x,x′ ∈ A and y ∈ V ∖ A. A binary structure is indecomposable if all its autonomous subsets are trivial (that is are either ∅, singletons, or the whole set). Delhomm´ e showed:

Theorem

If a hereditary class C of finite binary structures of finite arity contains only finitely many indecomposable members then the class of countable structures whose age is included into C is b.q.o.

January 19, 2016 28 / 44

This result extends Thomass´ e’s result (1999) on the w.q.o. character of the class of countable series-parallel posets, which extends the famous Laver’s theorem (1971) on the w.q.o. character of the class of countable chains. With Oudrar, we conjecture:

Problem

If the set Ind(C) of indecomposable structures included into a hereditary class C of finite structures is hereditary b.q.o then the class of countable structures whose age is included into C is b.q.o.

January 19, 2016 29 / 44

Uniformly prehomogeneous structures

Homogeneous structures and their automorphism groups are subject to an intensive research. A related notion is this: A structure R is uniformly prehomogeneous (u.p.h.) if for every finite set F of the domain V of R there is a finite superset F ′ of F whose cardinality in bounded by some function θ of the cardinality of F such that every local isomorphism of R defined on F extends to an automorphism provided that it extends to F ′.

Problem

Let R be a uniformly prehomogeneous structure. If Age(R) w.q.o. is (R,L) u.p.h. and age(R,L) w.q.o. for some linear order L on V (R)?

January 19, 2016 30 / 44

A group G acting on a set V is oligomorphic if for each integer n the number of orbits of the action of G on n-element subsets of V is finite. A shown by Ryll-Nardzewski, the automorphism group of a countable structure R is oligomorphic iff R is ℵ0-categorical. Saracino and I obtained independently:

Theorem

A countable relational structure R such that Aut(R) is oligomorphic is equimorphic to a countable R′ which is uniformly prehomogeneous and furthermore Aut(R′) is oligomorphic.

Theorem

Let C be an ideal of finite structures. If C.m− is wqo for every non-negative integer m then there is a uniformly prehomogeneous structure R with age C.

January 19, 2016 31 / 44

• Hint. Fix an integer m. Let C(m) be the collection of (S,a1,...,am)

where S ∈ C. If C.m− is wqo then, Quasi-ordered by embeddability, C(m) is w.q.o. Hence it is a finite union of ideals. This is the test I gave in 1972 for the existence of a uniformly prehomogeneous structure R with age C.

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The most simple question

Suppose that a poset P is well founded with no infinite antichain. Is is possible to prove that the set of initial segments of P is well founded without the axiom of dependent choices?

January 19, 2016 33 / 44

Thank you for your attention.

January 19, 2016 34 / 44

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