Simplicity of the automorphism groups of some binary homogeneous structures determined by triangle constraints
aa r X i v : . [ m a t h . L O ] J un Simplicity of the automorphism groups of somebinary homogeneous structures determined bytriangle constraints
Yibei Li
Abstract
We study some amalgamation classes introduced by Cherlin in the appendixof [Che98] and prove the simplicity of the automorphism groups of theFra¨ıss´e limits of these classes. We employ the machinery of stationary inde-pendence relations used by Tent and Ziegler in [TZ13a].
Given a relational language L , a countable L -structure M is homogeneous if every partial isomorphism between finite substructures of M extends toan automorphism of M . Some examples include the random graph, thegeneric k -uniform hypergraph, the rationals with linear order etc. Fra¨ıss´e’sTheorem [Fra53] provides one way of constructing homogeneous structuresby establishing a one-to-one correspondence between such structures and amalgamation classes (see Definition 1.5). We call the homogeneous struc-ture the Fra¨ıss´e limit of the corresponding amalgamation class. For exam-ple, the random graph is the Fra¨ıss´e limit of the set of all finite graphs, theUrysohn space is the completion of the Fra¨ıss´e limit of the set of all finiterational-valued metric spaces.Given finite L -structures A, B, C where B ⊆ A, C , the free amalgam of A, C over B is the L -structure D on the disjoint union of A, C over B andfor each relation R ∈ L , R D = R A ∪ R C . An amalgamation class C is free if it is closed under taking free amalgams, i.e. for A, B, C ∈ C such that B can be embedded in A, C , the free amalgam of
A, C over B is also in C . Ahomogeneous structure is free if it is the Fra¨ıss´e limit of a free amalgamationclass. The following theorem about free homogeneous structures was provedby Macpherson and Tent [MT11] using ideas and results from model theoryand topological groups: 1 heorem 1.1. ([MT11]) Let M be a countable free homogeneous relationalstructure. Suppose Aut ( M ) = Sym ( M ) and Aut ( M ) is transitive on M .Then Aut ( M ) is simple.Tent and Ziegler [TZ13a] generalised the theorem to a weaker notionof free homogeneous structures, namely a homogeneous structure with astationary independence relation. They applied their method to the Urysohnspace, which is not free, but has a local stationary independence relation (seeDefinition 3.1). They also used their approach to show that the isometrygroup of the bounded Urysohn space is simple in [TZ13b]. In this paper, weapply their method to the Fra¨ıss´e limits of all amalgamation classes givenby Cherlin in the appendix of [Che98]. Cherlin’s classes satisfy a weakerproperty, called semi-free amalgamation , defined as follows: Definition 1.2.
Given a relational language L , let C be an amalgamationclass of finite L -structures. We say C is a semi-free amalgamation class ifthere exists L ′ ( L such that for any finite structures A, B, C ∈ C andembeddings f : B → A, f : B → C , there exist D ∈ C and embeddings g : A → D, g : C → D such that g f ( B ) = g f ( B ) = g ( A ) ∩ g ( C ) andfor any a ∈ g ( A ) \ g f ( B ) , c ∈ g ( C ) \ g f ( B ), if a, c are related by some R ∈ L , then R ∈ L ′ . We call L ′ the set of solutions .B BB AC B D ac R ∈ L ′ f f g g We can view a free amalgamation class as a special case of semi-freeamalgamation classes.In this paper, we will only consider a language L consisting of binary,symmetric and irreflexive relations and classes of complete L -structures. Wesay an L -structure A is complete if every two distinct elements a, b ∈ A arerelated by exactly one relation. We denote this relation by r ( a, b ).In the appendix of [Che98], Cherlin identified 28 semi-free amalgamationclasses of complete structures for languages consisting of three and fourrelations, specified by triangle constraints, which are defined as the following:2 efinition 1.3. An L -structure is a triangle if it is a complete structure onthree points. Let S be a set of triangles. We define F orb c ( S ) to be the setof all complete structures that do not embed any triangle from S . We call S the set of forbidden triangles of F orb c ( S ).We can think of the structures in F orb c ( S ) as complete edge-colouredgraphs that do not embed some coloured triangles by taking the elementsof the structures as vertices and the relations as colours. Throughout thispaper, S is assumed to be a set of forbidden triangles such that the corre-sponding F orb c ( S ) is a semi-free amalgamation class. Then we can take itsFra¨ıss´e limit and denote it by M S .Our main goal is to prove the simplicity of the automorphism groups of M S for S listed in the appendix of [Che98] as well as some general cases.Motivated by Cherlin’s examples, we will define a special semi-free amal-gamation class, called prioritised semi-free amalgamation class (see Defini-tion 2.2). We will prove, in Section 3, that if F orb c ( S ) forms a prioritisedsemi-free amalgamation class, then we can find a stationary independencerelation on M S . We then apply Tent and Ziegler’s method in Section 4 and5 to show the simplicity of Aut ( M S ) for S satisfying some extra conditions(Condition 4.1). The main result of this paper is stated in Corollary 5.4. InSection 6.1, we will define two conditions on S and show that if S satisfiesone of the conditions, F orb c ( S ) forms a prioritised semi-free amalgamationclass and satisfies Condition 4.1. We observe that 27 of the 28 examples inCherlin’s list satisfy one of the conditions. These conditions also apply tosome general cases where the language consists of more than four relations.Hence, we can apply the main result to Cherlin’s examples as well as somegeneral cases. The remaining case, Theorem 1.4.
With the above notation, let M S be one of the countablehomogeneous structures listed in the appendix of [Che98] (see Table 1) and M S as defined above. Then Aut ( M S ) is simple. In this section, we introduce some concepts in model theory for readers notfamilar with them.We first fix a first-order relational language L , which is specified by aset of relation symbols { R i : i ∈ I } and each R i has arity r i ∈ N . Thenan L -structure is a set A together with a subset R Ai ⊆ A r i for each i ∈ I representing the structure on A . In this paper, since the graphs we areworking on are all undirected and loopless, the relations are always binary,i.e. r i = 2 for all i , symmetric and irreflexive. An L -structure B is a3 ubstructure of A if B ⊆ A and R Ai ∩ B r i = R Bi for each i ∈ I . For example,if A is a graph, then a substructure of A is an induced subgraph.Let A, B be finite substructures of M , we use the notation AB to denotethe substructure of M on the underlying set A ∪ B . We also simplify thenotation { a } B to aB . Let G = Aut ( M ) and denote the pointwise stabiliserof B by G ( B ) and the setwise stabiliser of B by G { B } . For a homogeneousstructure M , the model theoretic notion of an n -type over B correspondsto a G ( B ) -orbit of an n -tuple. For a ∈ M n , the type of a over B , denotedby tp ( a/B ), is the type over B whose corresponding G ( B ) -orbit contains a .So, we may use tp ( a/B ) to denote its corresponding G ( B ) -orbit. Note that a, a ′ have the same type over B if they lie in the same G ( B ) -orbit, i.e. thereexists an automorphism of M that takes a to a ′ and fixes B pointwise. Wesay a realise some type p over B if it lies in the corresponding G ( B ) -orbit.We say a type is algebraic if its set of realisations is finite, and non-algebraic otherwise. In all of our examples, tp ( a/B ) is algebraic if and only if a ∈ B .In this paper, the structures we study are all homogeneous and are con-structed from amalgamation classes, which are defined as follows: Definition 1.5.
A set of finite L -structures C is an amalgamation class ifit satisfies the following conditions:(i) it is closed under isomorphism and there are countably many isomor-phism types.(ii) if A ∈ C and B is a substructure of A , then B ∈ C .(iii) (Joint embedding property) if A, B ∈ C , then there exist C ∈ C andembeddings f : A → C, f : B → C (iv) (Amalgamation property) if A, B, C ∈ C and f : B → A, f : B → C are embeddings, then there exist D ∈ C and embeddings g : A → D, g : C → D such that g f = g f In the amalgamation property, we can take the embeddings to be inclu-sion maps. Hence in this paper, when we have an amalgamation class C andsay that B is a substructure of A, C for
A, B, C ∈ C , the precise meaningis that there are embeddings from B to A, C as in the amalgamation prop-erty. Since L is relational, it is enough to check all A, B, C ∈ C such that | A \ B | = | C \ B | = 1 in verifying the amalgamation property.Fra¨ıss´e’s theorem gives a one-to-one correspondence between countablehomogeneous structures and amalgamation classes. The theorem states thatgiven an amalgamation class C , we can construct a countable homogeneousstructure M , whose class of isomorphism types of finite substructures is C .Conversely, the set of all isomorphism types of finite substructures of anhomogeneous structure M forms an amalgamation class, called the age of M . By the construction in the proof of Fra¨ıss´e’s theorem, the homogeneous4tructure satisfies the Extension Property : if A ⊆ M and f : A → B is anembedding where B ∈ C , then there is an embedding g : B → M such that g ( f ( a )) = a for all a ∈ A . We make some remarks about overlaps between this work and [Ara+17],which studies metrically homogeneous graphs. For any undirected graph,we can put a metric on the graph by defining the distance between anytwo vertices to be the length of the shortest path between them. A graphis metrically homogeneous if it is homogeneous as a metric space. Cherlin[Che11] produced a list of such graphs, which is conjectured to be complete.In [Che11], Cherlin noted that some of the graphs M S we study in thispaper can be regarded as metrically homogeneous graphs by interpreting therelations as distances. Moreover, some of the graphs can be interpreted asmetrically homogeneous graphs in different ways. More precisely, in Table1 in Section 2, cases Y = 4 , R = 3 , G = 2 , X = 1; cases , X = 4 , R = 3 , G = 2 , Y = 1. This interpretation also works for cases G = 4 , R = 3 , Y = 2 , X = 1.The full list can be found in [Che11]. [Che19] is a more recent work byCherlin that provides a more thorough study on these graphs.[Ara+17] showed that all metrically homogeneous graphs with someexceptions in Cherlin’s catalogue have the Ramsey property by finding a‘completion algorithm’, which also provides stationary independence rela-tions on those graphs. This work overlaps with some of the results inthis paper. The completion algorithm is somewhat similar to our priori-tised semi-free amalgamation process, defined in the next section. The pa-per [HKN17] generalised these the results of [Ara+17] to Conant’s gener-alised metric spaces [Con15]. There is also ongoing work of Evans, Hubiˇcka,Koneˇcn´y and the author of this paper (paper in preparation) to apply Tentand Ziegler’s method on these generalised metric spaces.The author has also generalised Tent and Ziegler’s method so that thesymmetry axiom in the stationary independence relation is no longer re-quired (paper in preparation). Hence we can also apply the method to thedirected graphs in [Che98], which is also refered to as 2-multi-tournamentsin [Che19]. Note that the proofs in Section 3-5 d not depend on symmetry(except the proof of symmetry of the stationary independence relation inTheorem 3.3). 5 Prioritised Semi-Free Amalgamation classes
In the appendix of [Che98], Cherlin produced a list of semi-free amalga-mation classes for languages consisting of three and four binary relations,excluding free amalgamation classes and those with a non-trivial equivalencerelation on the vertices. We provide Cherlin’s list below. Each entry in thelist is a set of forbidden triangles S and we study F orb c ( S ), the set of allcomplete structures that do not embed any triangle from S , as defined inDefinition 1.3. To avoid any confusion in the notation, we change some ofthe letters used in Cherlin’s original list:Language: { R, G, X } { R, G, X, Y } L ′ ⊂ L . Hence, F orb c ( S ) is a semi-free amalgamation class for each S in the list. It can be shown that we may take L ′ = { R, G } in all the casesexcept for L ′ consisting of two relations for each F orb c ( S ). However, thisis not possible for L ′ = { R, G, Y } for Example 2.1.
Taking L ′ = { R, G } , we want to show that any amalgamation can be completed byeither R or G , i.e. there does not exist an amalgamation aB , cB over somefinite set B such that ( a, c ) cannot be coloured by R or G . Suppose we havesuch an amalgamation. Then there exist b , b ∈ B such that ab c forbids r ( a, c ) = R and ab c forbids r ( a, c ) = G . Since the only forbidden trianglescontaining R, G are
RXX, GGX , we may assume without loss of generalitythat r ( a, b ) = r ( b , c ) = X and r ( a, b ) = G, r ( b , c ) = X , as shown in thegraph below. However, we cannot find a colour for b b without creating aforbidden triangle as b ab forbids r ( b , b ) = G because of the forbiddentriangle GGX and b cb forbids r ( b , b ) ∈ { R, Y, X } because of the forbid-den triangles RXX, Y XX, XXX . Therefore, such an amalgamation doesnot exist. Thus, we may take L ′ = { R, G } . Similarly, we can prove the samestatement for the other cases. a cb b Also note that given an amalgamation, it is possible to have differentcompletions. For example, if we amalgamate two R -coloured edges ab, bc over the vertex b in a, c ) by any relation as there isno forbidden triangle containing RR . However, in order to find a stationaryindependence relation on M S that Tent and Ziegler used in [TZ13a], wewant to find a ‘unique’ way of amalgamating. In order to do this, we put alinear ordering on L ′ . Definition 2.2.
Let L be a language consisting of n binary, symmetric,irreflexive relations. Let L ′ ( L and suppose L ′ = { R , ..., R m } . Suppose L ′ is ordered as R > · · · > R m . For every A, B, C ∈ F orb c ( S ), where B ⊆ A, C , define the following way to amalgamate A and C over B : for7ach a ∈ A \ B, c ∈ C \ B , first check whether abc form a forbidden trianglefor any b ∈ B if ( a, c ) ∈ R . If B = ∅ or colouring ( a, c ) by R does not formany forbidden triangle, we let r ( a, c ) = R . Otherwise, we check the samething for ( a, c ) ∈ R and so on so forth. In other word, r ( a, c ) = R i where i is the smallest possible integer such that r ( a, b ) r ( b, c ) R i / ∈ S for any b ∈ B .Denote the resulting amalgamation by A ⊗ B C . If A ⊗ B C does not embedany forbidden triangle, i.e. A ⊗ B C ∈ F orb c ( S ), we call it the prioritisedsemi-free amalgamation of A, C over B . If for any A, B, C ∈ F orb c ( S )where B ⊆ A, C , A ⊗ B C ∈ F orb c ( S ), then we say F orb c ( S ) is a prioritisedsemi-free amalgamation class with respect to the given ordering on L ′ . Example 2.3.
In
R > G . Then r ( a, c ) = R in the graphbelow since otherwise we would have a triangle of RXX , so r ( a, c ) = G . a c X XRR G
It is possible that the resulting amalgamation is not in
F orb c ( S ). Forexample, in G > R , to complete the following amalgamation,we let both ( a , c ) , ( a , c ) ∈ G , but we would get a forbidden triangle GGX at a a c . a cba Note that in A ⊗ B C , a forbidden triangle can only appear in AC \ B aswe ensure by construction that there is no forbidden triangle with a vertexfrom B . Following [TZ13a], we consider a ternary relation among finite substructuresof a homogeneous structure, called a stationary independence relation . Wewill show in this section that the Fra¨ıss´e limit of a prioritised semi-freeamalgamation class has a stationary independence relation.8 efinition 3.1.
Let M be a homogeneous structure and suppose A | ⌣ B C is a ternary relation among finite substructure A, B, C of M . We say that | ⌣ is a stationary independence relation if the following axioms are statisfied:(i) Invariance: for any g ∈ Aut ( M ), if A | ⌣ B C , then gA | ⌣ gB gC (ii) Monotonicity: A | ⌣ B CD ⇒ A | ⌣ B C , A | ⌣ BC D (iii) Transitivity: A | ⌣ B C , A | ⌣ BC D ⇒ A | ⌣ B D (iv) Symmetry: A | ⌣ B C ⇒ C | ⌣ B A (v) Existence: If p is an n -type over B and C is a finite set, then p has arealisation a such that a | ⌣ B C .(vi) Stationarity: If a and a ′ are n -tuples that have the same type over B and are both independent from C over B , then a and a ′ have the sametype over BC .We say A is independent from C over B if A | ⌣ B C . Remark 3.2.
It can be shown from the above axioms that A | ⌣ B C ⇔ AB | ⌣ B C ⇔ A | ⌣ B BC Hence, we can assume without loss of generality that B ⊆ A, C whenever A | ⌣ B C for arbitrary A, B, C . Theorem 3.3.
Let S be a set of forbidden triangles such that F orb c ( S ) isa prioritised semi-free amalgamation class. Let M S be the Fra¨ıss´e limit of F orb c ( S ). For any finite substructure A, B, C of M S , let A | ⌣ B C if ABC = AB ⊗ B BC where AB ⊗ B BC is the prioritised semi-free amalgamationdefined in Definition 2.2. Then | ⌣ is a stationary independence relation on M S . Proof.
Invariance: Given B ⊆ A, C ⊆ M S , the prioritised semi-free amal-gamation of A, C over B is dependent only on the types of A, B, C andhomogeneity of M S .Monotonicity: Let B ⊆ A ⊆ M S and B ⊆ C ⊆ D ⊆ M S . Suppose A | ⌣ B D , i.e. AD = A ⊗ B D , we want to show A | ⌣ B C and AC | ⌣ C D . Byconstruction, we have A ⊗ B C ⊆ A ⊗ B D = AD , so A ⊗ B C = ABC , i.e. A | ⌣ B C .To show that AC | ⌣ C D , for any a ∈ A \ C, d ∈ D \ C , we want to showthat ( a, d ) is completed by the same relation in both A ⊗ B D and AC ⊗ C D .Suppose ( a, d ) is completed by some R i ∈ L ′ in A ⊗ B D and ( a, d ) / ∈ R i in AC ⊗ C D . Then there exists c ∈ C \ B that forbids ( a, d ) to be colouredb R i , i.e. R i r ( a, c ) r ( c, d ) forms a forbidden triangle. However, ( a, c ) , ( c, d )9ave the same colours in A ⊗ B D as in A ⊗ C D , i.e. it would also be aforbidden triangle in A ⊗ B D , a contradiction. b c da A C DTransitivity: Let B ⊆ A ⊆ M S , B ⊆ C ⊆ D ⊆ M S . Suppose A | ⌣ B C and A | ⌣ C D , i.e. AC = A ⊗ B C , AD = AC ⊗ C D . We want to show A | ⌣ B D , i.e. A ⊗ B D = AD = AC ⊗ C D . We already have A ⊗ B C = AC ⊂ AD = AC ⊗ C D , so we only need to show that, for any a ∈ A \ B, d ∈ D \ B ,( a, d ) is coloured by the same relation in both A ⊗ B D and AC ⊗ C D . Suppose( a, d ) is completed by some R i ∈ L ′ in A ⊗ B D and ( a, d ) / ∈ R i in AC ⊗ C D .This implies that there exists c ∈ C \ B that forbids ( a, d ) to be coloured by R i , i.e. R i r ( a, c ) r ( c, d ) is a forbidden triangle. This would be a forbiddentriangle in A ⊗ B D , a contradiction.Symmetry: The symmetry of the independence relation follows from thesymmetry of the relations in the language.Existence: Let p be a type over B and C any finite set. Let a be arealisation of p , we can embed BC into aB ⊗ B BC . Then by the extentionproperty, we may assume aB ⊗ B BC ⊆ M S and a | ⌣ B C .Stationarity: Suppose a, a ′ have the same type over B and are bothindependent from C over B . Then aB ⊗ B BC is isomorphic to a ′ B ⊗ B BC since the relations between a, a ′ and any c ∈ C depend only on therelations between aB, a ′ B and cB respectively. Hence, aBC = aB ⊗ B BC is isomorphic to a ′ B ⊗ B C = a ′ BC , which implies a, a ′ have the same typeover BC .Therefore, we have shown that if F orb c ( S ) forms a prioritised semi-freeamalgamation class, then there is a stationary independence relation on itsFra¨ıss´e limit M S . For Section 4 and Section 5, we will only consider S such that F orb c ( S )forms a prioritised semi-free amalgamation class as defined in Definition2.2. M in this section is always a relational homogeneous structure.10heorem 1.1 was proved in [TZ13a], which uses the result from [MT11]that a non-trivial automorphism of a free homogeneous structure does notfix the set of realisations of any non-algebraic type pointwise. In this section,we will prove the same statement for M S , where S satisfies the followingcondition. Condition 4.1.
Let S be a set of forbidden triangles. Assume that(i) S does not contain any triangle involving R R or R R .(ii) Let a, b, c ∈ M S and B ⊆ M S such that a | ⌣ bB c . If r ( a, b ) ∈ L ′ , wehave a | ⌣ B c .Note that satisfying Condition 4.1 does not guarantee that F orb c ( S )forms a prioritised semi-free amalgamation class.For readers familiar with model theory, the idea in this section comesfrom imaginary elements . Since the proof here does not directly involvethem, I refer interested readers to Section 16.4 and 16.5 in [Poi00]. Wewill show that our structure M S has the intersection property , defined inthe following. Note that for our structure M S , the intersection property isequivalent to having weak elimination of imaginaries by Theorem 16.17 in[Poi00] since the algebraic closure is trivial in M S . We will use this propertyof M S to prove that any non-trivial automorphism of M S does not fix theset of realisations of any non-algebraic type pointwise. Definition 4.2.
Let M be a homogeneous structure and G = Aut ( M ). Wesay that M has the intersection property , if for all finite subsets A, B ⊆ M , h G ( A ) , G ( B ) i = G ( A ∩ B ) . Note that if g ∈ h G ( A ) , G ( B ) i , then g fixes A ∩ B pointwise. So, we alwayshave h G ( A ) , G ( B ) i≤ G ( A ∩ B ) . Proposition 4.3.
Let M be a homogeneous structure and G = Aut ( M ).Suppose h G ( A ) , G ( B ) i = G ( A ∩ B ) (4.1)for all finite A, B ⊆ M such that | A \ B | = | B \ A | = 1. Then (4.1) holdsfor all finite subsets A, B ⊆ M . In other words, to prove the intersectionproperty for a relational homogeneous structure M , it is enough to show(4.1) for finite A, B ⊆ M such that | A \ B | = | B \ A | = 1. Proof.
We prove this by induction on n = | A \ B | + | B \ A | . The inductionbase is given by the assumption.Let n ≥ | A \ B | ≥
2. Pick a ∈ A \ B and let A ′ =( A ∩ B ) ∪ { a } and B ′ = B ∪ { a } . Then, h G ( A ) , G ( B ) i ≥ h G ( A ) , G ( B ′ ) , G ( B ) i .
11y inductive hypothesis, we have h G ( A ) , G ( B ′ ) i = G ( A ′ ) . Hence, h G ( A ) , G ( B ) i ≥ h G ( A ′ ) , G ( B ) i . Again, by inductive hypothesis, h G ( A ) , G ( B ) i ≥ h G ( A ′ ) , G ( B ) i = G ( A ∩ B ) . Since h G ( A ) , G ( B ) i ≤ G ( A ∩ B ) , we have h G ( A ) , G ( B ) i = G ( A ∩ B ) . Lemma 4.4.
Let M be a homogeneous structure with a stationary inde-pendence relation | ⌣ and G = Aut ( M ). If A | ⌣ A ∩ B B , then h G ( A ) , G ( B ) i = G ( A ∩ B ) . Proof.
Let g ∈ G ( A ∩ B ) . By the existence axiom of | ⌣ , there exists A ′ havingthe same type as A over B such that A ′ | ⌣ B B ∪ gB. Since M is homogeneous and tp ( A/B ) = tp ( A ′ /B ), there is k ∈ G ( B ) such that kA = A ′ . Hence G ( A ′ ) = kG ( A ) k − and h G ( A ) , G ( B ) i ≥ h G ( A ′ ) , G ( B ) i . By invariance, we have kA | ⌣ k ( A ∩ B ) kB , i.e. A ′ | ⌣ A ∩ B B. By transitivity on the above two independence relations, we have A ′ | ⌣ A ∩ B B ∪ gB. By monotonicity, we have A ′ | ⌣ A ∩ B B and A ′ | ⌣ A ∩ B gB . We also have tp ( B/A ∩ B ) = tp ( gB/A ∩ B ). So, by stationarity, tp ( B/A ′ ) = tp ( gB/A ′ ),i.e. there is h ∈ G ( A ′ ) such that hB = gB . Then g ∈ hG ( B ) and hence, g ∈ h G ( A ′ ) , G ( B ) i ≤ h G ( A ) , G ( B ) i . Therefore, G ( A ∩ B ) ≤ h G ( A ) , G ( B ) i .Since h G ( A ) , G ( B ) i ≤ G ( A ∩ B ) , we have the required result.Now let S be a set of forbidden triangles satisfying Condition 4.1 and F orb c ( S ) be the set of all finite L -structures that do not embed any trianglefrom S as defined in Definition 1.3. Let M S be the Fra¨ıss´e limit of F orb c ( S )and G = Aut ( M S ). For any A, B, C ⊆ M S , let A | ⌣ B C if ABC = AB ⊗ B BC where AB ⊗ B BC is the prioritised semi-free amalgamation defined inDefinition 2.2. The main result of this section is:12 heorem 4.5. Let S be a set of forbidden triangles satisfying Condition4.1. For any a, c ∈ M S and finite subset B ⊆ M S , h G ( aB ) , G ( cB ) i = G ( B ) . (4.2)Hence, by Proposition 4.3, M S has the intersection property, and thus weakelimination of imaginaries. Proof.
By existence, we can find c ′ realising tp ( c/aB ) such that c ′ | ⌣ aB c .Then there exists k ∈ G ( aB ) such that kc = c ′ and hence, G ( c ′ B ) = kG ( cB ) k − .So we have h G ( aB ) , G ( cB ) i ≥ h G ( c ′ B ) , G ( cB ) i . We can also find c ′′ realising tp ( c/c ′ B ) such that c ′′ | ⌣ c ′ B c . Then thereexists h ∈ G ( c ′ B ) such that hc = c ′′ and hence, G ( c ′′ B ) = hG ( cB ) h − . So wehave h G ( c ′ B ) , G ( cB ) i ≥ h G ( c ′′ B ) , G ( cB ) i . Since c ′ | ⌣ aB c , we have r ( c ′ , c ) ∈ L ′ . By part (ii) of Condition 4.1, wecan obtain c ′′ | ⌣ B c from c ′′ | ⌣ c ′ B c .Then by Lemma 4.4, h G ( c ′′ B ) , G ( cB ) i = G ( B ) . Thus, h G ( aB ) , G ( cB ) i ≥ G ( B ) . Since h G ( aB ) , G ( cB ) i ≤ G ( B ) , we have h G ( aB ) , G ( cB ) i = G ( B ) .We now prove that if g ∈ Aut ( M S ) fixes some non-algebraic type over B setwise, then it fixes B setwise. We say P is a definable set over B if it isa union of G ( B ) -orbits and B is a defining set for P . Note that for P ⊆ M S if B , B are defining sets for P , then h G ( B ) , G ( B ) i ≤ G { P } . So, by theprevious theorem, B ∩ B is a defining set for P and hence it follows that P has a minimal defining set. Corollary 4.6.
Let P ⊆ M S be an infinite definable set and B is a minimaldefining set for P . If g ∈ Aut ( M S ) fixes P setwise, then g fixes B setwise. Proof.
Since g fixes P setwise, gB is also a defining set for P . Hence, byminimality of B , we have gB = B .Since the set of realisations of a non-algebraic type is also a definable set,we have that if g ∈ Aut ( M S ) fixes some non-algebraic type over B setwise,then it fixes B setwise. Proposition 4.7.
Let g ∈ Aut ( M S ). For any non-algebraic type p oversome finite set B , if g fixes its set of realisations pointwise, then g = 1.13 roof. Without loss of generality, we may assume B is a minimal definingset for p . Let c ∈ M S be such that c | ⌣ B . Then r ( c, b ) = R for every b ∈ B .Claim 1: if g fixes p pointwise, then g fixes tp ( c/B ) pointwise.For c , c ∈ tp ( c/B ) distinct, there exists a realising p such that r ( c , a ) = R , r ( c , a ) = R since r ( c , b ) = r ( c , b ) = R for every b ∈ B and theredoes not exist forbidden triangle containing R R or R R as required byCondition 4.1. Since ga = a , gc = c .Since tp ( c/B ) contains every point in M S that is in relation R from B , g fixes tp ( c/B ) setwise. Therefore, gc = c , i.e. g fixes tp ( c/B ) pointwise.Claim 2: if g fixes tp ( c/B ) pointwise, then g = 1.For x, y / ∈ B distinct, there exists c ′ realising tp ( c/B ) such that r ( x, c ′ ) = R , r ( y, c ′ ) = R since r ( b, c ′ ) = R for every b ∈ B and there does not existforbidden triangle containing R R or R R . As gc ′ = c ′ , gx = y . ByLemma 4.6, g fixes B setwise, gx / ∈ B . Therefore, gx = x for every x / ∈ B .Then for every b, b ′ ∈ B distinct, there exists d / ∈ B such that r ( d, b ) = R , r ( d, b ′ ) = R . Hence gb = b for every b ∈ B . R R R R R R x bb ′ yc ′ d Therefore, combining the above two claims, we have the required result.We have shown that for any non-trivial g ∈ Aut ( M S ) and any non-algebraic type p over some finite set B , g moves some realisation a of p . Wecan then define a new type q over aB which is defined the same over B as p .This is possible because of the amalgamation property. Since we can repeatthis process countably many times, g moves countably many realisations of p . In this section, with the help of automorphisms that move almost maximally ,defined below, we apply the following theorem from [TZ13a], which Tent andZiegler used to study the automorphism group of the Urysohn Space.14 efinition 5.1.
Let M be a homogeneous structure with a stationary in-dependence relation | ⌣ . We say that g ∈ Aut ( M ) moves almost maximally if every 1-type over a finite set X has a realisation a such that a | ⌣ X ga . Theorem 5.2. ([TZ13a], 5.4) Suppose that M is a countable homogeneousstructure with a stationary independence relation. If g ∈ Aut ( M ) movesalmost maximally, then any element of Aut ( M ) is the product of sixteenconjugates of g .In this section, we prove the following theorem for M S where S satisfiesCondition 4.1. This then implies the simplicity of Aut ( M S ) by Theorem5.2. We let [ h, g ] denote the commutator h − g − hg . Theorem 5.3.
Let S be a set of forbidden triangles satisfying Condition 4.1such that F orb c ( S ) forms a prioritised semi-free amalgamation class. Let M S be the Fra¨ıss´e limit of F orb c ( S ). Then for any non-trivial automorphism g ∈ Aut ( M S ), there exist k, h ∈ Aut ( M S ) such that [ k, [ h, g ]] moves almostmaximally.This immediately implies the following: Corollary 5.4.
Under the same assumptions as in the previous theorem,
Aut ( M S ) is simple. In particular, if 1 = g ∈ Aut ( M S ), then every elementof Aut ( M S ) can be written as a product of 64 conjugates of g ± .In order to prove Theorem 5.3, we show the following results. Lemma 5.5.
Let S be a set of forbidden triangles such that F orb c ( S ) is asemi-free amalgamation class. Let M S be its Fra¨ıss´e limit. Then given anynon-trivial g ∈ Aut ( M S ), we can construct h ∈ Aut ( M S ) such that for anynon-algebraic 1-type p over some finite set X , there exist infinitely manyrealisations a of p such that r ( a, [ h, g ] a ) ∈ L ′ . Proof.
List all 1-types over a finite set as p , p , ... . We start with the emptymap and use a back-and-forth construction to build h . Suppose at the somestage, we have a partial isomorphism ˜ h : A → B such that for any p j ∈{ p , ..., p i − } , there exists a realisation a of p j such that r ( a, [˜ h, g ] a ) ∈ L ′ .Let p := p i be a 1-type over X . We want to extend ˜ h such that p hasa realisation a such that r ( a, [˜ h, g ] a ) ∈ L ′ . We may assume X ⊆ A byextending ˜ h .Since p is non-algebraic, by Proposition 4.7, p has a realisation a suchthat a / ∈ A ∪ g − ( aA ) and ˜ h · tp ( a/A ) has a realisation b such that b / ∈ B ∪ g − ( bB ). Extend ˜ h by sending a to b . It is well-defined since ˜ h · tp ( a/A ) = tp ( b/B ).By the extension property, there exists a realisation c of ˜ h − · tp ( gb/bB )such that c and ga are semi-freely amalgamated over aA . Since b / ∈ g − ( bB )15nd ˜ h · tp ( c/aA ) = tp ( gb/bB ), we have c / ∈ aA . We also have ga / ∈ aA , hence( c, ga ) is coloured using relations from L ′ , i.e. r ( c, ga ) ∈ L ′ .Extend ˜ h by sending c to gb . Since ˜ h · tp ( c/aA ) = tp ( gb/bB ), ˜ h is awell-defined partal isomorphism. Then c = ˜ h − g ˜ ha and we have r ( a, [˜ h, g ] a ) = r ( a, ˜ h − g − ˜ hga ) = r (˜ h − g ˜ ha, ga ) = r ( c, ga ) ∈ L ′ . At every other step, we can make sure X ⊂ B by extending ˜ h . Let h be the union of all ˜ h over each step, it is an automorphism since it is welldefined and bijective as we made sure every finite subset of M is containedin both domain and image. Theorem 5.6.
Let S be a set of forbidden triangles satisfying Condition4.1 and F orb c ( S ) be a prioritised semi-free amalgamation class as definedin Definition 2.2. Let M S be its Fra¨ıss´e limit and g ∈ Aut ( M S ) be anautomorphism of M S such that for any non-algebraic 1-type p over somefinite set, there exist infinitely many realisations a of p with r ( a, ga ) ∈ L ′ .Then there exists k ∈ Aut ( M S ) such that [ k, g ] moves almost maximally. Proof.
We again use a back-and-forth construction as in the previous proof.Suppose at some stage, we have a partial isomorphism ˜ k : A → B . It sufficesto show that we can extend ˜ k so that [˜ k, g ] moves some given 1-type p almostmaximally. We may assume p is non-algebraic.We may also assume X ⊆ A and ˜ kX ⊆ g − B by extending ˜ k , then wehave ˜ k − g ˜ kX ⊆ A .By the existence axiom of | ⌣ and the assumption about g , there existsa realisation a of p such that r ( a, ga ) ∈ L ′ and a | ⌣ X A. (5.1)By existence and Lemma 4.7, there exists a realisation b of ˜ k · tp ( a/A )such that gb = b and b | ⌣ B g − B. (5.2)Extend ˜ k by sending a to b . It is a well-defined partial isomorphism since˜ k · tp ( a/A ) = tp ( b/B ). From (5.1), by invariance, we get b | ⌣ ˜ kX B. (5.3)Applying transitivity on (5.2) and (5.3), we have b | ⌣ ˜ kX g − B. (5.4)16gain by the existence axiom, there exists a realisation c of ˜ k − · tp ( gb/bB )such that c | ⌣ aA ga. Since r ( a, ga ) ∈ L ′ , by part (ii) of Condition 4.1, we have c | ⌣ A ga. (5.5)Extend ˜ k by sending c to gb . We have tp ( gb/bB ) = ˜ k · tp ( c/aA ) and gb = b , so ˜ k is a well-defined partial isomorphism.Acting by ˜ k − g on (5.4), we then get˜ k − gb | ⌣ ˜ k − g ˜ kX ˜ k − B, which can be simplified to c | ⌣ ˜ k − g ˜ kX A. (5.6)Since ˜ k − g ˜ kX ⊆ A , we can apply transitivity on (5.5) and (5.6) to obtain c | ⌣ ˜ k − g ˜ kX ga . Acting by ˜ k − g − ˜ k on it, we have the required result, i.e. a | ⌣ X [˜ k, g ] a. Proof of Theorem 5.3.
Given any automorphism g ∈ Aut ( M S ), we can firstfind h ∈ Aut ( M S ) such that for any non-algebraic 1-type p over some finiteset, there exist infinitely many realisations a of p such that r ( a, [ h, g ] a ) ∈ L ′ .Then by the previous theorem, there exists k ∈ Aut ( M S ) such that [ k, [ h, g ]]moves almost maximally. With notation and assumption as stated in Definition 2.2, we fix L ′ andan ordering on L ′ . Let S be a set of forbidden triangles. To make sure A ⊗ B C ∈ F orb c ( S ), we can impose one of the following two conditions on S : Condition 6.1.
Assume S does not contain any triangle of the form R i R j R ′ where R i , R j ∈ L ′ and R ′ ∈ L . 17 ondition 6.2. Assume L ′ = { R , R } with order R > R and for somesubset L ⋆ ⊆ L\L ′ and ˆ L ⊆ L\ ( L ′ ∪ L ⋆ ), S contains all triangles of the form R ′ R R , R R R , R ˆ RR , where R ′ ∈ L \ ( ˆ L ∪ { R } ) , R , R ∈ L ⋆ , ˆ R ∈ ˆ L and S contains no other triangle involving R or R R .In this section, we will show that if S satisfies either one of the above twoconditions, then F orb c ( S ) forms a prioritised semi-free amalgamation classand satisfies Condition 4.1. Therefore, by Corollary 5.4, the automorphismgroup of its Fra¨ıss´e limit M S is simple. The following lemma states that 27of 28 Cherlin’s examples satsify either one of the conditions can be checkedby quick computation. Hence, the automorphism groups of their Fra¨ıss´elimits are also simple. Lemma 6.3.
In Cherlin’s list, the only case where L consists of three rela-tions satisfies Condition 6.2 with L ′ = { R, G } , L ⋆ = { X } and order R > G .For L consisting of four relations, L ′ = { R, G } works for all 27 cases execptcase L ⋆ = { X } , ˆ L = ∅ andthe order R > G . Case L ⋆ = { X } , ˆ L = { Y } and the order R > G . Case L ⋆ = { X, Y } , ˆ L = ∅ and the order G > R .In the next two lemmas, we show that for a set of forbidden triangles S satisfying either Condition 6.1 or Condition 6.2, the corresponding F orb c ( S )forms a prioritised semi-free amalgamation class as defined in Definition 2.2.Hence, by Theorem 3.3, there is a stationary independence relation on thecorresponding M S . Lemma 6.4.
Let S be a set of forbidden triangles satisfying Condition6.1. For any A, B, C ∈ F orb c ( S ) such that B ⊆ A, C , let A ⊗ B C be theamalgamation defined in Definition 2.2. Then A ⊗ B C ∈ F orb c ( S ). Proof.
Suppose there is a forbidden triangle in A ⊗ B C . Then it is eitherof the form acc ′ for some a ∈ A \ B, c, c ′ ∈ C \ B or aa ′ c for some a, a ′ ∈ A \ B, c ∈ C \ B .Without loss of generality, we may assume aa ′ c is a forbidden trian-gle in A ⊗ B C . Since ( a, c ), ( a ′ , c ) are amalgamated by relations from L ′ , r ( a, c ) , r ( a ′ , c ) ∈ L ′ . Then r ( a, c ) r ( a ′ , c ) r ( a, a ′ ) is of the form R i R j R ′ ∈ S where R i , R j ∈ L ′ and R ′ ∈ L , which contradicts the condition 6.1. There-fore, we have A ⊗ B C ∈ F orb c ( S ). Lemma 6.5.
Let S be a set of forbidden triangles satisfying Condition6.2. For any A, B, C ∈ F orb c ( S ) such that B ⊆ A, C , let A ⊗ B C be theamalgamation defined in Definition 2.2. Then A ⊗ B C ∈ F orb c ( S ). Proof.
Suppose there exists a forbidden triangle in AC \ B . As in the pre-vious lemma, we may assume it is aa ′ c for some a, a ′ ∈ A \ B, c ∈ C \ B r ( a, c ) , r ( a ′ , c ) ∈ L ′ . Since the only forbidden triangle involving R i R j R where R i , R j ∈ L ′ and R ∈ L is R R R where R ∈ L ⋆ , we have r ( a, c ) = r ( a ′ , c ) = R and r ( a, a ′ ) ∈ L ⋆ . This implies that there exists b ∈ B thatforbids ( a, c ) to be completed by R . a cba ′ R R L ⋆ L ⋆ L ⋆ Since the only forbidden triangle containing R is of the form R R R where R , R ∈ L ⋆ , we have r ( a, b ) , r ( b, c ) ∈ L ⋆ .As S contains all triangles of the form R ′ R R where R , R ∈ L ⋆ , R ′ ∈L \ ( ˆ L ∪ { R } ), this forces r ( a ′ , b ) to be either R or in ˆ L as r ( a, a ′ ) , r ( a, b ) ∈ L ⋆ . If r ( a ′ , b ) ∈ ˆ L , a ′ bc forms a forbidden triangle of the form R ˆ RR ,where R ∈ L ⋆ , ˆ R ∈ ˆ L . If r ( a ′ , b ) = R , a ′ bc forms a forbidden triangle ofthe form R R R , where R ∈ L ⋆ . Hence, we cannot find a colour for a ′ b without creating a forbidden triangle. Thus, A ⊗ B C ∈ F orb c ( S ). Remark 6.6.
1. It can be seen in the proofs of the previous two lemmasthat when S satisfies Condition 6.1, the order on L ′ does not affectwhether A ⊗ B C ∈ F orb c ( S ). However for S satisfying Condition 6.2,whether A ⊗ B C ∈ F orb c ( S ) is dependent on R > R .2. For S satisfying Condition 6.1 or Condition 6.2, S does not containtriangle involving R R or R R , hence it satisfies part (i) of Condition4.1. We will show part (ii) of Condition 4.1 in the next lemma. Lemma 6.7.
For S satisfying Condition 6.1 or Condition 6.2, let a, b, c ∈M S and B ⊆ M S such that a | ⌣ bB c . If r ( a, b ) ∈ L ′ , we have a | ⌣ B c . Proof. If b ∈ B , the statement is trivial. Let b / ∈ B . Since a | ⌣ bB c , wehave abBc = a ⊗ bB c . To show that a | ⌣ B c , we want to show that ( a, c ) iscoloured by the same relation in both a ⊗ B c and a ⊗ bB c . Suppose on thecontrary, ( a, c ) ∈ R i in a ⊗ B c for some R i ∈ L ′ and ( a, c ) / ∈ R i in a ⊗ bB c ,this implies that R i r ( a, b ) r ( b, c ) forms a forbidden triangle.For S satisfying Condition 6.1, since r ( a, b ) , R i ∈ L ′ , R i r ( a, b ) r ( b, c ) is ofthe form R i R j R ′ where R i R j ∈ L ′ , R ′ ∈ L , contradicting Condition 6.1.For S satisfying Condition 6.2, since L ′ = { R , R } with R > R , theonly possibility of ( a, c ) / ∈ R i in a ⊗ bB c is when i = 1. Then R r ( a, b ) r ( b, c )is a forbidden triangle in S . Since r ( a, b ) ∈ L ′ , r ( a, b ) = R or R . However,19 does not contain any triangle involving R R or R R as noted in Remark6.6.Therefore, given the assumption, it can be deduced from a | ⌣ bB c that a | ⌣ B c .Therefore, applying Theorem 5.6, we obtain the following theorem: Theorem 6.8.
Let S be a set of forbidden triangles satisfying Condition6.1 or Condition 6.2 and M S be the Fra¨ıss´e limit of F orb c ( S ). Then for anynon-trivial automorphism g ∈ Aut ( M S ), there exist k, h ∈ Aut ( M S ) suchthat [ k, [ h, g ]] moves almost maximally. Hence, Aut ( M S ) is simple. The remaining case is the following: L = { R, G, X, Y } L ′ = { R, G, Y } , F orb c ( S )forms a prioritised semi-free amalgamation class. In fact, { R, G, Y } is theonly possible set of solutions for F orb c ( S ) to be a semi-free amalgamationclass. Hence, we can find a stationary independence relation on M S andfinally we will prove the simplicity of the automorphism group of M S witha similar approach, but with some extra conditions. Lemma 6.9.
Let S be as in G > R > Y on L ′ = { R, G, Y } . For any A, B, C ∈ F orb c ( S ) such that B ⊆ A, C , let A ⊗ B C be the prioritised semi-free amalgamation defined in Definition 2.2. Then A ⊗ B C ∈ F orb c ( S ). Proof.
Suppose there exists a forbidden triangle in A ⊗ B C . As a forbiddentriangle can only appear in AC \ B by construction, we may assume aa ′ c is a forbidden triangle in A ⊗ B C for some a, a ′ ∈ A \ B, c ∈ C \ B . Since r ( a, c ) , r ( a ′ , c ) ∈ L ′ , these are not X and hence, there are five possible casesfor r ( a, c ) r ( a ′ , c ) r ( a, a ′ ), as listed below. In each case, since r ( a, c ) = G ,there exists b ∈ B that forbids ( a, c ) to be coloured by G . ( a ′ , b ) is the edgewhere we cannot find any relation and hence we have a contradiction. a cba ′ r ( a, c ) r ( a ′ , c ) r ( a, a ′ ) = RRX .Since r ( a, c ) = R , either r ( b, a ) r ( b, c ) = Y X or r ( b, a ) r ( b, c ) = XY .However, in the first case, r ( a, a ′ ) r ( a, b ) = XY forbids r ( a ′ , b ) to be Y, G, X and r ( c, a ′ ) r ( c, b ) = RX forbids r ( a ′ , b ) to be R, X . Therefore,we cannot find a colour for ( a ′ , b ) without creating a forbidden triangle.In the second case, r ( a, a ′ ) r ( a, b ) = XX forbids r ( a ′ , b ) to be R, G, X and r ( c, a ′ ) r ( c, b ) = RY forbids r ( a ′ , b ) to be Y . Hence, in both case,there would exist a forbidden triangle.II. r ( a, c ) r ( a ′ , c ) r ( a, a ′ ) = Y Y R
Since r ( a, c ) = Y , r ( b, a ) r ( b, c ) = XX . However, r ( a, a ′ ) r ( a, b ) = RX forbids r ( a ′ , b ) to be R, X and r ( c, a ′ ) r ( c, b ) = Y X forbids r ( a ′ , b ) tobe Y, G, X , a contradiction.III. r ( a, c ) r ( a ′ , c ) r ( a, a ′ ) = Y Y X
Similarly as in the previous cases, r ( b, a ) r ( b, c ) = XX . However, r ( a, a ′ ) r ( a, b ) = XX forbids r ( a ′ , b ) to be G, R, X and r ( c, a ′ ) r ( c, b ) = Y X forbids r ( a ′ , b ) to be Y, G, X , a contradiction.IV. r ( a, c ) r ( a ′ , c ) r ( a, a ′ ) = Y RY
We have r ( b, a ) r ( b, c ) = XX . However, r ( a, a ′ ) r ( a, b ) = Y X forbids r ( a ′ , b ) to be G, Y and r ( c, a ′ ) r ( c, b ) = RX forbids r ( a ′ , b ) to be R, X , a contradiction.V. r ( a, c ) r ( a ′ , c ) r ( a, a ′ ) = Y GX
In this case, we have r ( a, b ) r ( b, c ) = XX . However, r ( a, a ′ ) r ( a, b ) = XX forbids r ( a ′ , b ) to be R, G, X and r ( a ′ , c ) r ( c, b ) = GX forbids r ( a ′ , b ) to be Y , a contradiction.As noted earlier, there appears to be a claim in [Che98] that a solutionwith 2 colours gives a semi-free amalgamation class for Lemma 6.10.
Let S be as in F orb c ( S ) be the set of all complete L -structures that do not embed any triangle in S . Then { R, G, Y } is theonly possible set of soltions for F orb c ( S ) to be a semi-free amalgamationclass. Proof.
Since in the previous lemma, we showed that with L ′ = { R, G, Y } , F orb c ( S ) forms a prioritised semi-free amalgamation class, it remains toshow that L ′ has to contain G, R, Y by finding amalgamations where eachof them is the only choice.In the following amalgamation, ( a, c ) has to be coloured by G as Y Y forbids it to be coloured R or X and GX forbids it to be coloured Y . Hence G ∈ L ′ . 21 c Y YRG X
Similarly, in the following amalgamation, ( a, c ) ∈ R as XY forbids it tobe coloured G or Y and RR forbids it to be coloured X . Hence R ∈ L ′ a c Y XGR R
The following amalgamation implies Y ∈ L ′ since XX forbids ( a, c ) tobe coloured R , G or X . Hence Y ∈ L ′ . a c X X
Therefore, we have shown that { R, G, Y } ⊆ L ′ . Remark 6.11.
With order
G > R > Y on L ′ = { G, R, Y } , part (i) ofCondition 4.1 holds, but part (ii) does not as in the following example, wehave a | ⌣ bb ′ c and r ( a, b ) ∈ L ′ , but a is not independent from c over b ′ . a cb b ′ RY XGR R
However, we can prove a similar version, namely, for a | ⌣ bB c where both r ( a, b ) , r ( b, c ) ∈ L ′ , we have a | ⌣ B c , proved in the following lemma. We willthen show that the proof of our main result, Corollary 5.4 follows similarly. Lemma 6.12.
Let S be as in a, b, c ∈ M S and B ⊆ M S suchthat r ( a, b ) , r ( b, c ) ∈ L ′ and a | ⌣ bB c . Then a | ⌣ B c .22 roof. If b ∈ B , the statement is trivial. Let b / ∈ B . As in Lemma 6.7,to show that a | ⌣ B c , we want to show that ( a, c ) is coloured by the samerelation in both aB ⊗ B cB and abB ⊗ bB cbB .If r ( a, c ) = G in aB ⊗ B cB and r ( a, c ) = G in aB ⊗ bB cB , then G r ( a, b ) r ( b, c ) forms a forbidden triangle where r ( a, b ) , r ( b, c ) ∈ L ′ , i.e. r ( a, b ) , r ( b, c ) = X , by assumption. However, all forbidden triangles in S involving G also contains X , a contradiction.If r ( a, c ) = R in aB ⊗ B cB , then there exists b ′ ∈ B such that b ′ forbids( a, c ) to be coloured by G . Since XX forbids both G and R , r ( a, b ′ ) r ( b ′ , c ) = XY or Y X . However, XY forbids ( a, c ) to be coloured by G or Y and hencemakes R the only choice for r ( a, c ) in aB ⊗ bB cB . The same argument worksfor when r ( a, c ) = Y since XX forbids both G, R .Therefore, given the assumption, it can be deduced from a | ⌣ bB c that a | ⌣ B c .Note that only Theorem 4.5 and 5.6 depend on part (ii) of Condition4.1. Theorem 4.5 holds for tp ( c/c ′ B ) = tp ( c ′′ /c ′ B )and then we have r ( c, c ′ ) = r ( c ′′ , c ′ ) ∈ L ′ . So, we again have a | ⌣ B c from a | ⌣ bB c by Lemma 6.12 and the rest of the proof follows.For Theorem 5.6, let g ∈ Aut ( M S ) be a non-trivial automorphism of M S . As Lemma 5.5 only depends on F orb c ( S ) being a semi-free amalgama-tion class, we can construct h ∈ Aut ( M S ) such that for any non-algebraic1-type p over some finite set, there exist infinitely many realisations a of p such that r ( a, [ h, g ] a ) ∈ L ′ . We want to show that there exists k ∈ Aut ( M S )such that [ k, [ h, g ]] moves almost maximally. The proof is mostly the sameas the proof of Theorem 5.6 except in Equation (5.2), we choose b also sat-isfying r ( b, gb ) ∈ L ′ . Then r ( a, c ) = r ( ka, kc ) = r ( b, gb ) ∈ L ′ . Thus, bythe previous lemma, we again have c | ⌣ A ga from c | ⌣ aA ga . The rest of theproof of Theorem 5.6 follows, which proves the following corollary: Corollary 6.13.
Let S be as in M S be the Fra¨ıss´e limit of F orb c ( S ). Then for any non-trivial automorphism g ∈ Aut ( M S ), thereexist k, h ∈ Aut ( M S ) such that [ k, [ h, g ]] moves almost maximally. Hence, Aut ( M S ) is simple.Combining the results in this section, we then have shown Theorem 1.4. References [Ara+17] A. Aranda, D. Bradley-Williams, J. Hubiˇcka, M. Karamanlis, M.Kompatscher, M. Koneˇcn´y, and M. Pawliuk, “Ramsey expan-23ions of metrically homogeneous graphs,”
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