Ordered structures and large conjugacy classes

TL;DR

This paper investigates the properties of automorphism groups of ordered Fraisse limits, demonstrating the existence of groups with large conjugacy classes and establishing conditions for their absence, with implications for group dynamics.

Contribution

It provides new results on the conjugacy class structure of automorphism groups of ordered Fraisse limits, including the universal ordered boron tree and poset.

Findings

Automorphism groups of certain ordered Fraisse limits have a comeager conjugacy class.

These groups lack a comeager 2-dimensional diagonal conjugacy class.

General conditions are given for the non-existence of large conjugacy classes in such groups.

Abstract

This article is a contribution to the following problem: does there exist a Polish non-archimedean group (equivalently: automorphism group of a Fraisse limit) that is extremely amenable, and has ample generics. As Fraisse limits whose automorphism groups are extremely amenable must be ordered, i.e., equipped with a linear ordering, we focus on ordered Fraisse limits. We prove that automorphism groups of the universal ordered boron tree, and the universal ordered poset have a comeager conjugacy class but no comeager $2$-dimensional diagonal conjugacy class. We also provide general conditions implying that there is no comeager conjugacy class, comeager $2$-dimensional diagonal conjugacy class or non-meager $2$-dimensional topological similarity class in the automorphism group of an ordered Fraisse limit. We provide a number of applications of these results.

Equations

$$\{(g{g}_1{g}^{-1},\dots ,g{g}_n{g}^{-1})\in G^n:g\in G\},$$

$${\mathcal{F}}_n=\{(A,p_1,\dots ,p_n):A\in \mathcal{F},p_i\text{ is a partial automorphism of }A,i\le n\}.$$

$${\mathrm{qftp}}_C(y_0/\mathrm{rng}(q))={\mathrm{qftp}}_C(y_1/\mathrm{rng}(q))$$

$${\mathrm{qftp}}_C(y_0/(\mathrm{def}(q)))\ne {\mathrm{qftp}}_C(y_1/(\mathrm{def}(q))).$$

$$d(y,x)=d(y,a)+d(a,x),$$

$$d(a^{\prime },x)=d(a^{\prime },y)+d(y,x).$$

$$S^A(a,b,c)⟺\varphi (a),\varphi (b){<}_{lex}^n\varphi (c)\text{ and }\mathrm{ht}(\mathrm{meet}(\varphi (a),\varphi (b)))>\mathrm{ht}(\mathrm{meet}(\varphi (b),\varphi (c))).$$

$$X^{\ast }=X\cup \{z=\text{mbox}{Cone}_Y:z\text{ is maximal }\text{ in }(\{\text{mbox}{Cone}_Z⊊\text{mbox}{Cone}_X:Z\in {\mathcal{P}}^A\},\subseteq )\}$$

$$p_k=p^{\prime }\upharpoonright [a_{-k},a_{n+k}{]}_{lex}\cup \{(c,c):c\in A,a_n{<}_{lex}c{<}_{lex}{b}_n\}\cup p^{\prime }\upharpoonright [b_{-k},b_{n+k}{]}_{lex},$$

$$A_B=\{y\in B:(\exists x\in A,k\in \mathbb{Z})q^k(x)=y\},$$

$$X^{\ast }=X\cup \{z=\text{mbox}{Cone}_Y:z\text{ is maximal }\text{ in }(\{\text{mbox}{Cone}_Z⊊\text{mbox}{Cone}_X:Z\in {\hat{\mathcal{P}}}^B\},\subseteq )\},$$

$$t(O,x)=\{n\in \mathbb{Z}:p^n(x)\in O\text{ and }x\prec p^n(x)\}.$$

$$t(O,N,x,y)=\{n\in \mathbb{Z}:p^n(y)\in N\text{ and }x\prec p^n(y)\}.$$

$$t(O_{B_2},N_{B_2},x,y)=\{kn+i:k\in \mathbb{N},0\le i<n,x\prec q^i(y)\}\cap B_2,$$

$$B_0^D<C_0^D<a_1<B_1^D<C_1^D<a_2<\dots <a_k<B_k^D<C_k^D.$$

$$c{l}_p=\{x\in M:x\text{ is cone-locked by an orbit of }p\}.$$

$$s^{N_0}w(p,q)(x)>\mathrm{supp}(q)\cap E_i$$

$$x\le s^{N_0}w(p,q)(y)\text{mbox}ifandonlyifq(x)\le s^{N_0}w(p,q)(y)$$

$$q^{\prime }(s^{N_0+N_1+n}w(p^{\prime },q^{\prime })(x))=s^{N_0+N_1+n}w(p^{\prime },q^{\prime })(x)$$

$$v{s}^{N_0+N_1+|v|+N}w(p^{\prime },q^{\prime })(x)\ge w(p^{\prime },q^{\prime })(x)$$

$$x\le s^{N}w(p^{\prime },q^{\prime })(y)\text{mbox}ifandonlyifq(x)\le s^{N}w(p^{\prime },q^{\prime })(y)$$

$$q^{\prime }(s^{N}w(p^{\prime },q^{\prime })(x))=s^{N}w(p^{\prime },q^{\prime })(x)$$

Full text

The first named author was supported by Narodowe Centrum Nauki grant 2016/23/D/ST1/01097.

1. Introduction

This article is a contribution to the following question: does there exist a Polish non-archimedean group, i.e., a Polish group with a neighborhood basis at the identity consisting of open subgroups, that simultaneously satisfies two frequently studied properties: it is extremely amenable, and it has ample generics.

A Polish group $G$ is extremely amenable if every continuous action of $G$ on a compact space has a fixed point. The group $G$ has ample generics if, for every $n\geq 1$, there exists an $n$-dimensional diagonal conjugacy class in $G$, i.e., a set of the form

[TABLE]

for some $g_{1},\ldots,g_{n}\in G$, which is comeager in $G^{n}$. Such a group admits only one Polish group topology, and all of its (abstract) homomorphisms into separable groups are continuous (Kechris-Rosendal [9].) In particular, every action by homeomorphisms of an extremely amenable group with ample generics on a compact separable space has a fixed point.

It is known that there exist Polish groups sharing both of these features. Pestov-Schneider [16] proved that, for any Polish group $G$, the group $L_{0}(G)$, i.e., the group of measurable functions with values in $G$, is extremely amenable, provided that $G$ is amenable, and Kaïchouh-Le Maître [10] proved that $L_{0}(G)$ has ample generics whenever $G$ has. As $S_{\infty}$, i.e., the group of all permutations of natural numbers, is amenable, and has ample generics, $L_{0}(S_{\infty})$ is extremely amenable and it has ample generics. However, it is still an open problem whether there are such groups in the non-archimedean realm.

Let $M$ be a first-order countable structure. It is well known that every Polish non-archimedean group is isomorphic to the automorphism group $\mbox{Aut}(M)$ of a structure $M$ (i.e., a set equipped with relations and functions) equal to the Fraïssé limit of a Fraïssé class $\mathcal{F}$ of finite structures (see the next section for precise definitions of notions used in the introduction.) The group $\mbox{Aut}(M)$ naturally acts on the compact space of linear orderings of $M$, viewed as a subspace of $\{0,1\}^{M\times M}$. This implies that when ${\rm{Aut}}(M)$ is extremely amenable, then there is a linear ordering of $M$ preserved by $\mbox{Aut}(M)$, see also [12]. Therefore if $\mbox{Aut}(M)$ is extremely amenable, we can actually assume that $\mathcal{F}$ is an order class, i.e., that each structure in $\mathcal{F}$ is equipped with a linear ordering $<$ of its elements. Thus, we can pose a more general question: does there exist a Fraïssé limit $M$ of an order class $\mathcal{F}$ such that the automorphism group $\mbox{Aut}(M)$ has ample generics. This article gives some partial answers as to when such a situation cannot happen.

Curiously enough, there are no known examples of Polish groups that do not have ample generics but they have a comeager diagonal conjugacy class for some $n\geq 2$. Thus, our article can be also viewed as a study of the question whether comeager diagonal conjugacy classes resemble weak mixing in topological dynamics, which implies weak mixing of all orders (see [5].)

One of our main tools is a theorem of Kechris-Rosendal, connecting the structure of diagonal conjugacy classes in the automorphism group of the Fraïssé limit $M$ of a Fraïssé class $\mathcal{F}$ with the joint embedding property (JEP), and the weak amalgamation property (WAP) in classes $\mathcal{F}_{n}$ of $n$-tuples of partial automorphisms of elements of $\mathcal{F}$. They prove (see also [6]) that $\mbox{Aut}(M)$ has a comeager $n$-dimensional diagonal conjugacy class if and only if $\mathcal{F}_{n}$ has JEP and WAP. Thus, showing that $\mbox{Aut}(M)$ does not have a comeager $n$-dimensional diagonal conjugacy class reduces to verifying that $\mathcal{F}_{n}$ has no JEP or WAP.

First, we study the one-dimensional case. A class of structures $\mathcal{F}$ has the $1$-Hrushovski property if every partial automorphism of an $A\in\mathcal{F}$ can be extended to an automorphism of some $B\in\mathcal{F}$. Clearly, if $\mathcal{F}$ is an order class of finite structures, then $\mathcal{F}$ does not have the $1$-Hrushovski property because in this case non-trivial orbits are necessarily infinite. We introduce the notions of strong splitting and always strong splitting in a Fraïssé class, which capture the idea of ‘flexible’ amalgamation. Then we prove (Theorem 3.5) that $\mathcal{F}_{1}$ has no WAP, provided that one of following holds: $\mathcal{F}$ is a Fraïssé class that does not have the $1$-Hrushovski property, and it always strongly splits, or $\mathcal{F}$ is a full order expansion of $\mathcal{K}$ (i.e., $\mathcal{F}$ is the class of all linear orderings on elements of $\mathcal{K}$), where $\mathcal{K}$ is a Fraïssé class that strongly splits. On the other hand, we show (Theorem 3.10) that the class $\mathcal{SB}_{1}$ of partial automorphisms of ordered boron trees, and (Theorem 3.13) the class $\mathcal{P}_{1}$ of partial automorphisms of ordered partial orders, have CAP, so, in particular, they have WAP. It seems that these are, except for $\mbox{Aut}(\mathbbm{Q})$ (see Truss [18]), the only known order classes such that the automorphism group of the limit has a comeager conjugacy class. We also give (Theorem 3.8) a short and elementary proof of a result of Slutsky [17] who showed that the class $(\mathcal{QU}_{\prec})_{1}$ of partial automorphisms of ordered metric spaces with rational distances has no WAP.

Next, we turn to the two-dimensional case. For a Fraïssé class $\mathcal{F}$, we formulate a simple but efficient condition (Proposition 4.1) implying that $\mathcal{F}_{2}$ has no WAP, and we verify it for a number of cases such as precompact Ramsey expansions of ultrahomogeneous directed graphs, in particular for $\mathcal{P}_{2}$. Using a similar approach, we also show (Theorem 4.4) that $\mathcal{SB}_{2}$ does not have WAP. Then we investigate topological similarity classes. For a Polish group $G$, $n\geq 1$, and an $n$-tuple $(f_{1},\ldots,f_{n})$ in $G$, the $n$-dimensional topological similarity class of $(f_{1},\ldots,f_{n})$ is the family of all $n$-tuples $(g_{1},\ldots,g_{n})$ in $G$ such that the mapping $f_{i}\mapsto g_{i}$ (uniquely) extends to a topological group isomorphism. Clearly, this is a generalization of the notion of the diagonal conjugacy class, and it is still not known whether there exists a Polish group $G$ such that for some $n\geq 2$ there is a non-meager $n$-dimensional topological similarity class, but all $n$-dimensional diagonal conjugacy classes are meager. Generalizing methods and results of Slutsky [17], we show (Theorem 5.5) that if $M$ is the Fraïssé limit of a Fraïssé class $\mathcal{F}$ that is a full order expansion and that satisfies certain additional conditions, then all $2$-dimensional topological similarity classes in $\mbox{Aut}(M)$ are meager. In particular, this is true if $\mathcal{K}$ is a class with free amalgamation, or the class of ordered tournaments (Theorem 5.7.)

2. Definitions

A topological group is Polish if its topology is separable, and completely metrizable. A Polish group is non-archimedean if it has a neighborhood basis at the identity consisting of open subgroups, or, equivalently, it is topologically isomorphic to the automorphism group $\mbox{Aut}(M)$ of a countable structure, equipped with the product topology (i.e., $\mbox{Aut}(M)\subseteq M^{M}$, where $M$ is regarded as a discrete space.)

By a structure we always mean a relational structure (i.e., a set equipped with relations), and we consider only classes of finite structures. Let $A$ be a structure, and let $B,C\subseteq A$. By ${\rm{qftp}}_{A}(B/C)$, we denote the quantifier-free type of $B$ over $C$ in $A$. Let $p$ be a partial automorphism of $A$. We write ${\rm{def}}(p)={\rm{dom}}(p)\cup{\rm{rng}}(p)$, and ${\rm{supp}}(p)=\{x\in{\rm{def}}(p)\colon p(x)\neq x\}$. An orbit of $p$ is a maximal subset $\mathcal{O}\subseteq A$ that can be enumerated into $\{a_{0},\ldots,a_{m}\}$ so that $p(a_{i})=a_{i+1}$ for $i<m$. If $p(a_{m})=a_{0}$, we say that $\mathcal{O}$ is a cyclic orbit. An orbit is trivial if it consists of a single element.

Let $\mathcal{F}$ be a class of structures in a given signature. We say that $\mathcal{F}$ has JEP (the joint embedding property) if any two $A,B\in\mathcal{F}$ can be embedded in a single $C\in\mathcal{F}$. We say that $\mathcal{F}$ has AP (the amalgamation property) if for every $A,B,C\in\mathcal{F}$ and embeddings $\alpha\colon A\to B$ and $\beta\colon A\to C$ there is $D\in\mathcal{F}$ and embeddings $\gamma\colon B\to D$, $\delta\colon C\to D$ such that $\gamma\circ\alpha=\delta\circ\beta$. In that case, we say that $B$ and $C$ *amalgamate * over $A$. We say that $\mathcal{F}$ has SAP (the strong amalgamation property) if, additionally, $\gamma[B]\cap\delta[C]=\gamma\circ\alpha[A]$. We say that $\mathcal{F}$ has CAP (the cofinal amalgamation property) if there is a cofinal (with respect to inclusion) subclass of $\mathcal{F}$ with AP. We say that $\mathcal{F}$ has WAP (the weak amalgamation property) if for every $A\in\mathcal{F}$ there is $A^{\prime}\in\mathcal{F}$ and an embedding $\phi\colon A\to A^{\prime}$ such that for every $B,C\in\mathcal{F}$ and embeddings $\alpha\colon A^{\prime}\to B$, $\beta\colon A^{\prime}\to C$ there is $D\in\mathcal{F}$ and embeddings $\gamma\colon B\to D$, $\delta\colon C\to D$ such that $\gamma\circ\alpha\circ\phi=\delta\circ\beta\circ\phi$. Clearly, if $\mathcal{F}$ has CAP, then it has WAP. Actually, in the definition of AP (CAP and WAP), it suffices to consider $B,C$ such that $B\cap C=A$ ($B\cap C=A^{\prime}$), and only trivial embeddings, i.e., inclusions.

A class of finite structures $\mathcal{F}$ is a Fraïssé class, if it is countable (up to isomorphism), closed under isomorphism, closed under taking substructures, and has JEP and AP. A countable first-order structure $M$ is ultrahomogeneous if every isomorphism between finite substructures of $M$ can be extended to an automorphism of the whole $M$. Then $\rm{Age}(M)$ – the class of all finite substructures embeddable in $M$ – is a Fraïssé  class. Conversly, by the classical theorem due to Fraïssé, for every Fraïssé  class $\mathcal{F}$ of finite structures, there is a unique up to isomorphism countable ultrahomogeneous structure $M$ such that $\mathcal{F}={\rm{Age}}(M)$. We call this $M$ the Fraïssé limit of $\mathcal{F}$.

A Fraïssé class $\mathcal{F}$ is called an order class if its signature includes a binary relation defining a linear ordering on each element of $\mathcal{F}$. If $\mathcal{F}$ is an order class, $\mathcal{F}^{-}$ denotes the reduct of $\mathcal{F}$ obtained by removing the order relation $<$ from the signature of $\mathcal{F}$. We call $\mathcal{F}$ a full order expansion if $\mathcal{F}=\mathcal{F}^{-}*\mathcal{LO}$, i.e., it is a class of elements of the form $(A,<)$, where $A\in\mathcal{F}^{-}$, and $<$ is any linear ordering of $A$. We will frequently use the observation that if $M$ is the Fraïssé limit of a full order expansion with SAP, then for any finite $A\subseteq B\subset M$, and any $<$-interval $I$ in $M$, there exists $C\subseteq I$ that is isomorphic to $B$ via an isomorphism that pointwise fixes $A$.

We will be mostly interested in classes of tuples of partial automorphisms of structures coming from a given class $\mathcal{F}$. Formally, for $n\geq 1$, denote

[TABLE]

Often, we will think of elements of $\mathcal{F}_{n}$ simply as tuples of partial automorphisms. Then $(p_{1},\ldots,p_{n})$ is identified with $(\bigcup_{i}{\rm{def}}(p_{i}),p_{1},\ldots,p_{n})$. A map $\phi\colon(A,p_{1},\ldots,p_{n})\to(B,q_{1},\ldots,q_{n})$ will be called an embedding if it is an embedding of $A$ into $B$, and $\phi\circ p_{i}=q_{i}\circ\phi$ for $i\leq n$. Using this notion of embedding, we can also define properties JEP, AP, CAP, and WAP for classes $\mathcal{F}_{n}$. Then we have:

Theorem 2.1** (Kechris-Rosendal [9]).**

Let $\mathcal{F}$ be a Fraïssé class, and let $M$ be the Fraïssé limit of $\mathcal{F}$.

1. (1)

There exists a dense diagonal $n$-conjugacy class in ${\rm{Aut}}(M)$ iff $\mathcal{F}_{n}$ has JEP, 2. (2)

there exists a comeager diagonal $n$-conjugacy class in ${\rm{Aut}}(M)$ iff $\mathcal{F}_{n}$ has JEP and WAP.

In particular, it follows that if $\mathcal{F}_{n}$ has JEP but no WAP, then $\mbox{Aut}(M)$ has meager $n$-dimensional diagonal conjugacy classes.

Let $F_{n}=F_{n}(s_{1},\ldots,s_{n})$ denote the free group on $n$ generators $s_{1},\ldots,s_{n}$. For a word $w\in F_{n}$, and an $n$-tuple $\bar{f}=(f_{1},\ldots,f_{n})$ in $G$, the evaluation $w(\bar{f})$ denotes the element of $G$ obtained from $w$ by substituting $f_{i}$ for $s_{i}$, and performing the group operations on the resulting sequence. By a word, we will always mean a reduced word.

3. The one-dimensional case. Conjugacy classes

3.1. A condition that implies the failure of WAP

Recall that a family $\mathcal{F}$ of finite structures in a given signature has the Hrushovski property if for every $n\in\mathbb{N}$, $A\in\mathcal{F}$ and a tuple $(f_{1},\ldots,f_{n})$ of partial automorphisms of $A$, there exists $B\in\mathcal{F}$ such that $A\subseteq B$, and every $f_{i}$ can be extended to an automorphism of $B$. We say that $\mathcal{F}$ has the $n$-Hrushovski property if the above holds for a given $n$.

In [11, Theorem 4.7], we proved the following trichotomy.

Theorem 3.1**.**

Let $M$ be a Fraïssé limit of a Fraïssé family $\mathcal{F}$ such that algebraic closures of finite subsets of $M$ are finite. Then one of the following holds:

1. (1)

$\mathcal{F}$* has the Hrushovski property,* 2. (2)

$\mathcal{F}$* does not have the $1$-Hrushovski property,* 3. (3)

there exists $n$ such that none of $n$-dimensional topological similarity classes in $\mbox{Aut}(M)$ is comeager. In particular, $\mbox{Aut}(M)$ does not have ample generics.

It is well known that if a Fraïssé class $\mathcal{F}$ has the Hrushovski property, and sufficiently free amalgamation, then the automorphism group $\mbox{Aut}(M)$ of its limit $M$ has ample generics. By the above trichotomy, if $\mathcal{F}$ does not have the Hrushovski property, ample generics may be present in $\mbox{Aut}(M)$ only if $\mathcal{F}$ does not even have the $1$-Hrushovski property – which is true, in particular, for order classes. In this section, we prove (in Theorem 3.5) that such situations always presuppose a very rigid form of amalgamation in $\mathcal{F}$. In order to specify what ‘rigid’ is about in this context, let us introduce two definitions that capture what ‘flexible’ amalgamation means for us.

In [15, Definition 2.4], Panagiotopolus studies extensions of automorphisms of generic substructures of a given structure. He introduces the notion of splitting in a Fraïssé class $\mathcal{F}$. An element $C\in\mathcal{F}$ splits $\mathcal{F}$ if for every $D\in\mathcal{F}$ with $C\subseteq D$ there exist $D_{1},D_{2}\in\mathcal{F}$ with $D\subsetneq D_{1},D_{2}$, and a bijection $f\colon D_{1}\rightarrow D_{2}$ such that

1. (1)

$f$ pointwise fixes $D$, 2. (2)

$f\restriction(D_{1}\setminus C)$ is an isomorphism between $D_{1}\setminus C$ and $D_{2}\setminus C$, 3. (3)

$f$ is not an isomorphism between $D_{1}$ and $D_{2}$.

Analogously, we will say that $C\in\mathcal{F}$ strongly splits $\mathcal{F}$ if for all $D,D_{1}\in\mathcal{F}$ with $C\subseteq D\subsetneq D_{1}$ there exists $D_{2}\in\mathcal{F}$ with $D\subsetneq D_{2}$, and a bijection $f\colon D_{1}\rightarrow D_{2}$ such that Conditions (1)-(3) above hold. We will say that $\mathcal{F}$ strongly splits if there exists $C\in\mathcal{F}$ that strongly splits $\mathcal{F}$, and that $\mathcal{F}$ always strongly splits if every $C\in\mathcal{F}$ strongly splits $\mathcal{F}$.

We can think of $C$ in the above definitions as one ‘ear’ of an amalgamation diagram $U\subsetneq V,W$, i.e., $C=V\setminus U$ and $D=V$. Then $C$ strongly splits if for any other ‘ear’ $W\setminus U$ (i.e., $D_{1}\setminus D$), there are at least two non-equivalent ways in which we can define relations involving elements from the ‘ears’ $V\setminus U$ and $W\setminus U$ to form an amalgam of $V$ and $W$ over $U$: one represented by $D_{1}$ (where $W=D_{1}\setminus C$), the other one by $D_{2}$ (where $D_{2}\setminus C$ is an isomorphic copy of $W$.)

In particular, the property of always strong splitting can be also expressed as a variant of the amalgamation property: $\mathcal{F}$ always strongly splits if amalgamation in $\mathcal{F}$ is not too rigid, that is, if there is always more than one way of amalgamating structures. To be more precise, fix $C$, $D$ and $D_{1}$ as above, and think of $D_{1}$ as an amalgam of $D$ and $D_{1}\setminus C$ over $D\setminus C$. Then any other non-isomorphic amalgam with the same underlying sets gives a required $D_{2}$. In other words, a class $\mathcal{F}$ always strongly splits if for any $A,U,V\in\mathcal{F}$ with $A<U,V$ there exist two non-isomorphic amalgams $W_{1},W_{2}$ of $U$ and $V$ over $A$, with $U\cup V$ as the underlying set, and such that $U$ and $V$ embed in both $W_{1}$ and $W_{2}$ by the identity mapping.

Proving the next two lemmas is straightforward, and left to the reader.

Lemma 3.2**.**

If $\mathcal{F}$ is a full order expansion of a class that (always) strongly splits, then $\mathcal{F}$ also (always) strongly splits.

Lemma 3.3**.**

Let $p$ be a partial automorphism of a structure $A$, and let $x\in{\rm{rng}}(p)\setminus{\rm{dom}}(p)$. Suppose that $y,y^{\prime}\in A$ are such that $p\cup(x,y)$ is a partial automorphism, and ${\rm{qftp}}_{A}(y/{\rm{rng}}(p))={\rm{qftp}}_{A}(y^{\prime}/{\rm{rng}}(p))$. Then $p\cup(x,y^{\prime})$ is also a partial automorphism of $A$.

Lemma 3.4**.**

Let $\mathcal{F}$ be a full order expansion with SAP. Then for every $p\in\mathcal{F}_{1}$, there exists $q\in\mathcal{F}_{1}$ extending $p$ such that for every $r\in\mathcal{F}_{1}$ extending $q$, distinct orbits of $p$ are contained in distinct orbits of $r$.

Proof.

Fix $p\in\mathcal{F}_{1}$, and let $\mathcal{O}_{0},\ldots,\mathcal{O}_{n}$ be orbits of $p$. Fix $i<j\leq n$, and suppose that there exists an extension $q^{\prime}$ of $p$ such that $\mathcal{O}_{i}$ and $\mathcal{O}_{j}$ are in the same orbit of $q^{\prime}$. Then there must exist a partial automorphism $q^{\prime\prime}$ extending $p$, and $x,y\in{\rm{rng}}(q^{\prime\prime})\triangle{\rm{dom}}(q^{\prime\prime})$, $x$ in the orbit of $q^{\prime\prime}$ determined by $\mathcal{O}_{i}$, $y$ in the orbit of $q^{\prime\prime}$ determined by $\mathcal{O}_{j}$, and we can extend $q^{\prime\prime}$ by putting $q^{\prime\prime}(x)=y$ or $(q^{\prime\prime})^{-1}(x)=y$. Without loss of generality, we can assume that $x<y$, and $q^{\prime\prime}\cup(x,y)$ extends $q^{\prime\prime}$. Since $\mathcal{F}$ is a full order expansion with SAP, there exists $C\in\mathcal{F}$ with ${\rm{def}}(q^{\prime\prime})\subseteq C$, and $y^{\prime}\in C\setminus{\rm{def}}(q^{\prime\prime})$ with $y^{\prime}>y$, and such that ${\rm{qftp}}_{C}(y/{\rm{rng}}(q^{\prime\prime}))={\rm{qftp}}_{C}(y^{\prime}/{\rm{rng}}(q^{\prime\prime}))$. But then, by Lemma 3.3, $q=q^{\prime\prime}\cup(x,y^{\prime})$ also extends $q^{\prime\prime}$, and $x<y<q(x)$. Clearly, $\mathcal{O}_{i}$ and $\mathcal{O}_{j}$ stay distinct in any extension of $q$. By iterating this construction, we can find $q$ that works for all $i<j\leq n$. ∎

Theorem 3.5**.**

Let $\mathcal{F}$ be a Fraïssé class. Suppose that

1. (1)

$\mathcal{F}$* does not have the $1$-Hrushovski property, and it always strongly splits, or* 2. (2)

$\mathcal{F}$* is a full order expansion with SAP, and it strongly splits.*

Then $\mathcal{F}_{1}$ has no WAP.

Proof.

Assume that Condition (1) holds. Fix $p\in\mathcal{F}_{1}$ witnessing that $\mathcal{F}$ does not have the $1$-Hrushovski property. We show that $p$ also witnesses that $\mathcal{F}_{1}$ does not have WAP.

Fix $q\in\mathcal{F}_{1}$ that extends $p$. Clearly, there must exist an orbit $\mathcal{O}$ of $q$ intersecting ${\rm{dom}}(p)$ that is non-cyclic – otherwise the union of such orbits would be a structure in $\mathcal{F}$ invariant under $q$. Let $\mathcal{O}=\{o_{0},\ldots,o_{n}\}$ with $q(o_{i})=o_{i+1}$ for $i<n$. As $\mathcal{O}$ is not cyclic, $q$ is not defined on $o_{n}$. Fix $y_{0}\not\in{\rm{def}}(q)$ such that $q_{0}=q\cup(o_{n},y_{0})$ is a partial automorphism of $D_{1}={\rm{def}}(q)\cup\{y_{0}\}$. Since $\mathcal{F}$ always strongly splits, by putting $C={\rm{dom}}(q)\setminus{\rm{rng}}(q)$ (note that $o_{0}\in C$, so $C\neq\emptyset$), $D={\rm{def}}(q)$ (and $D_{1}={\rm{def}}(q)\cup\{y_{0}\}$), we can find $y_{1}$ such that $D_{2}={\rm{def}}(q)\cup\{y_{1}\}$ witnesses that $C$ strongly splits. However, this means that

[TABLE]

but

[TABLE]

Then (1) together with Lemma 3.3 implies that $q_{1}=q\cup(o_{n},y_{1})$ is also a partial automorphism. On the other hand, for every $r\in\mathcal{F}_{1}$ such that $q_{0}$ and $q_{1}$ can be embedded into $r$ by embeddings $e_{0}$ and $e_{1}$, respectively, that agree on ${\rm{def}}(q)$, we must have $e_{0}(y_{0})=e_{1}(y_{1})$, and this is impossible because of (2). Moreover, the same argument can be applied to any extension of $p$, so, in fact, if $e_{0}$, $e_{1}$ agreed on ${\rm{def}}(p)$, they would agree on ${\rm{def}}(q)$ as well. Thus, $q_{0}$, $q_{1}$ cannot be amalgamated over $p$. As $q$ was arbitrary, $p$ witnesses that $\mathcal{F}_{1}$ does not have WAP.

Assume now that Condition (2) holds. Fix $A\in\mathcal{F}$ witnessing that $\mathcal{F}$ strongly splits. Fix $p\in\mathcal{F}_{1}$ such that $A={\rm{dom}}(p)\setminus{\rm{rng}}(p)$ (this can be easily done using the assumption that $\mathcal{F}$ is is a full order expansion with SAP), and extend $p$ to a partial automorphism $q$ as in Lemma 3.4. Then for every extension $r\in\mathcal{F}_{1}$ of $q$, there exists $A^{\prime}\subseteq{\rm{dom}}(r)\setminus{\rm{rng}}(r)$ that is isomorphic with $A$, and so $A^{\prime}$ also witnesses that $\mathcal{F}$ strongly splits. Now we proceed as in the proof of (1). Fix a (non-cyclic) orbit $\mathcal{O}=\{o_{0},\ldots,o_{n}\}$ of $r$ intersecting ${\rm{dom}}(p)$, find $y_{0}$ such that $r_{0}=r\cup\{(o_{n},y_{0})\}$ is a partial automorphism, and put $C=A^{\prime}$, $D={\rm{def}}(r)$, $D_{1}={\rm{def}}(r)\cup\{y_{0}\}$ to obtain $y_{1}$ such that $r_{0}$ and $r_{1}=r\cup(o_{n},y_{1})$ cannot be amalgamated over $p$. ∎

Proposition 3.6**.**

The classes of ordered graphs, and ordered tournaments always strongly split.

Proof.

Let $\mathcal{F}$ be either the class of ordered graphs or ordered tournaments. Fix $C,D,D_{1}\in\mathcal{F}$ with $C\subseteq D\subsetneq D_{1}$. Fix $c\in C$ and $d_{1}\in D_{1}\setminus D$. For graphs, define $D_{2}$ to be the graph that differs from $D_{1}$ only in that $\{c,d_{1}\}$ is an edge in $D_{2}$ if and only if $\{c,d_{1}\}$ is not an edge in $D_{1}$. Similarly, for ordered tournaments, define $D_{2}$ to be the ordered tournament that differs from $D_{1}$ only in that $(d_{1},c)$ is an arrow in $D_{2}$ if and only if $(c,d_{1})$ is an arrow in $D_{1}$.

∎

Corollary 3.7**.**

1. (1)

The class of partial automorphisms of finite ordered tournaments has no WAP. 2. (2)

(Slutsky) The class of partial automorphisms of finite ordered graphs has no WAP.

Proof.

Finite ordered graphs and finite ordered tournaments are full order expansions with SAP. By Proposition 3.6, both of them always strongly split, so, by Theorem 3.5, they have no WAP. ∎

On the other hand, it is easy to see that both of these classes have JEP, which means (by Theorem 2.1), that automorphism groups of the universal ordered tournament and the universal ordered graph (i.e., Fraïssé limits of the above classes) have meager conjugacy classes.

Slutsky [17] also proved that the automorphism group $\mbox{Aut}(\mathbbm{QU}_{\prec})$ of the ordered rational Urysohn space $\mathbbm{QU}_{\prec}$, i.e., the full order expansion of finite ordered metric spaces with rational distances, has meager conjugacy classes. One of the ingredients of his proof is a deep theorem of Solecki saying that the class of finite metric spaces has the Hrushovski property. Theorem 3.5 cannot be used to recover Slutsky’s result because the class of ordered finite metric spaces with rational distances does not strongly split. However, a similar approach gives rise to a more elementary argument. We sketch it below. Note that $(\mathcal{QU}_{\prec})_{1}$ has JEP, therefore it will suffice to prove that it has no WAP.

Theorem 3.8** (Slutsky).**

The class $(\mathcal{QU}_{\prec})_{1}$ has no WAP.

Proof.

First fix $A\in\mathcal{QU}$, and $x\in A$. Without loss of generality, we can assume that $A\subseteq\mathbbm{QU}$. Suppose that $y\in\mathbbm{QU}$ is such that the type ${\rm{qftp}}_{A}(y/(A\setminus\{x\}))$ determines $d(x,y)$. By the triangle inequality, this is possible only when there are $a,a^{\prime}\in A$ such that

[TABLE]

[TABLE]

But then, in particular, $d(x,y)\leq{\rm{diam}}(A)$. Thus, if for some partial automorphism $p$ there was an automorphism $q$ extending $p$ such that any two extensions $r_{0}$, $r_{1}$ of $q$ could be amalgamated over $p$, then for every automorphism $\phi$ of $\mathbbm{QU}_{\prec}$ extending $q$, orbits of $\phi$ determined by $p$ would be bounded by ${\rm{diam}}({\rm{def}}(q))$. But it is well known (see, e.g., Section 3.1 in [4]) that every partial isometry $q$ of $\mathbbm{QU}$ with no cyclic orbits can be extended to an isometry of $\mathbbm{QU}$ with unbounded orbits. And since $\mathcal{QU}_{\prec}$ is a full order expansion of $\mathcal{QU}$, every partial automorphism of a finite subspace of $\mathbbm{UQ}_{\prec}$ also can be extended to an automorphism of $\mathbbm{QU}_{\prec}$ with unbounded orbits.∎

Remark 3.9**.**

Using a similar approach, and a construction as in the proof of Lemma 5.6, one can also prove that the class of partial automorphisms of ordered $K_{n}$-free graphs does not have WAP, for every $n\geq 3$. It is not hard to see that this class does not strongly split.

It is known that the class of partial automorphisms of finite linear orderings has JEP and WAP. In the next two sections, we present two others such order classes: ordered boron trees, and ordered posets.

3.2. Ordered boron trees - a comeager conjugacy class

In this section, we prove that the automorphism group of the universal ordered boron tree has a comeager conjugacy class. The class of boron trees was introduced by Cameron [2].

Let $\mathcal{T}$ denote the class of all graph-theoretic trees such that the valency of each vertex is equal to 1 or 3. If $T\in\mathcal{T}$ and $a,b,c,d\in T$, we let $ab|cd$ iff arcs $ab$ and $cd$ do not intersect. To each $T\in\mathcal{T}$ we assign a structure $(B(T),R^{B(T)})$ such that $B(T)$ is the set of endpoints of $T$, and $R^{B(T)}(a,b,c,d)$ iff $a,b,c,d$ are pairwise different and $ab|cd$. Structures $(B(T),R^{B(T)})$, together with the one point structure, we call boron trees, and we denote the class of all these structures by $\mathcal{B}$. The universal boron tree is the Fraïssé limit of $\mathcal{B}$.

Let $2^{<n}$ denote the set of binary sequences of the length $<n$, including the empty sequence. Let $T^{\prime}_{n}$ denote the binary tree, that is, a graph with the set of vertices equal to $2^{<n}$ and edges exactly between vertices $s$ and $si$, $i=0,1$, $s\in 2^{<n}$. Let $T_{n}$ be the graph obtained by removing the vertex $\emptyset$ from $T^{\prime}_{n}$ and replacing edges $[\emptyset,0]$ and $[\emptyset,1]$ by the edge $[0,1]$, and denote $B_{n}=B(T_{n})$. Let $\leq^{n}$ be the lexicographical order on $B_{n}$, i.e. we let $s\leq^{n}t$ iff $s=t$ or $s(i)<t(i)$, where $i$ is the least such that $s(i)\neq t(i)$. For $s\in 2^{<n}$ we define the *height * as the length of $s$, i.e. ${\rm{ht}}(s)=|s|$. For a fixed $n$ and $s,t\in T^{\prime}_{n}$, we let $s<t$ if $s$ is an initial segment of $t$, and let for $s,t\in T^{\prime}_{n}$ the meet of $s$ and $t$, denoted ${\rm{meet}}(s,t)$, be the least upper bound of $s$ and $t$ with respect to the partial order $<$. Let $(A,R^{A})\in\mathcal{B}$ and let $\phi\colon(A,R^{A})\to(B_{n},R^{B_{n}})$ be an embedding (with respect to $R$, this is not necessarily a graph embedding). We let $\leq_{lex}^{A}$ to be the order inherited from $\leq^{n}$ and define a ternary relation $S^{A}$ on $A$ as follows:

[TABLE]

There are multiple ways to expand an $(A,R^{A})\in\mathcal{B}$ by adding relations $S^{A}$ and $<^{A}_{lex}$. Intuitively, the relation $S^{A}$ adds a root at an edge of the tree $T$ such that $B(T)=A$, viewed as a graph. We illustrate the construction of an expansion in Figure 1. Structures $(A,R^{A},S^{A},\leq_{lex}^{A})$, we call ordered boron trees, and we denote the class of all these structures by $\mathcal{SB}$. The universal ordered boron tree is the Fraïssé limit of $\mathcal{SB}$. From the work of Jasiński [7], it follows that the automorphism group of the universal ordered boron tree is extremely amenable.

Note that for $a<_{lex}b<_{lex}c<_{lex}d$ we have $R(a,b,c,d)$ iff $S(a,b,c)$ or ($\neg S(a,b,c)$ and $\neg S(b,c,d)$). Therefore, we can recover $R$ from $S$ and $<_{lex}$.

Theorem 3.10**.**

The family $\mathcal{SB}_{1}$ has CAP.

For $A\in\mathcal{SB}$, denote by $T_{A}$ the binary tree such that $A=B(T_{A})$. For every $A$ there exists unique such a $T_{A}$. The root of $T_{A}$, denoted by $\rho_{A}$, is the $<$-least element of $T_{A}$. By $<_{A}$ we denote the usual tree partial ordering of being an initial segment on elements of $T_{A}$. In the sequel, the structure $A$ in symbols $\leq_{lex}^{A}$, $<^{A}$, $R^{A}$, and $S^{A}$ will be always clear from the context, so, in order to simplify notation, we will simply write $\leq_{lex}$, $<$, $R$, and $S$, respectively.

Let $(A,p)\in\mathcal{SB}_{1}$. We say that a non-trivial orbit $O=\{a_{0},\ldots,a_{n}\}$ of $p$ is increasing if $a_{0}<_{lex}\ldots<_{lex}a_{n}$; analogously, we define a decreasing orbit. Clearly, every orbit is either increasing, decreasing, or trivial. Note that, setting $t_{i}={\rm{meet}}(a_{i},a_{i+1})$, we either have $t_{0}<t_{1}<t_{2}<\ldots<t_{n-1}$ or $t_{n-1}<\ldots<t_{2}<t_{1}$. In the first case, we say that $O$ is meet-increasing, and in the second that it is meet-decreasing. If $(B,p)\in\mathcal{SB}_{1}$ extends $(A,p)$, then we will denote by $O_{B}$ the extension of $O$ in $B$.

Let $A=(A,p)\in\mathcal{SB}_{1}$. We will call two orbits $O=\{a_{0},\ldots,a_{m}\}$ and $P=\{b_{0},\ldots,b_{n}\}$ of $p$ intertwining if the $\leq_{lex}$-intervals $(\min\{a_{0},a_{m}\},\max\{a_{0},a_{m}\})_{lex}$ and $(\min\{b_{0},b_{n}\},\max\{b_{0},b_{n}\})_{lex}$ intersect. Say that $O$ is $\leq_{lex}$-contained in $P$ if $(\min\{a_{0},a_{m}\},\max\{a_{0},a_{m}\})_{lex}$ is contained in $(\min\{b_{0},b_{n}\},\max\{b_{0},b_{n}\})_{lex}$. A point $x\in A$ is meet-locked by $O$ if for every extension $(B,q)$ of $(A,p)$ such that $a_{-1}=q^{-1}(a_{0})$ and $a_{m+1}=q(a_{m})$ are defined, denoting $t_{i}={\rm{meet}}(a_{i},a_{i+1})$, we have the following. (1) If $O$ is increasing and meet-increasing, then $a_{m+1}<_{lex}x$ and $t_{-1}<{\rm{meet}}(x,a_{m+1})<t_{m}$, (2) if $O$ is decreasing and meet-increasing, then $x<_{lex}a_{m+1}$ and $t_{-1}<{\rm{meet}}(x,a_{m+1})<t_{m}$, (3) if $O$ is increasing and meet-decreasing, then $x<_{lex}a_{-1}$ and $t_{-1}<{\rm{meet}}(x,a_{-1})<t_{m}$, (4) if $O$ is decreasing and meet-decreasing, then $a_{-1}<_{lex}x$ and $t_{-1}<{\rm{meet}}(x,a_{-1})<t_{m}$. Two orbits $O$ and $P$ are meet-intertwining if there is $x\in O$ meet-locked by $P$ or there is $x\in P$ meet-locked by $O$. Note that if $O$ and $P$ are meet-intertwining then one of them is increasing and the other one is decreasing. Moreover, either both are meet-increasing or both are meet-decreasing.

We call a cone any set ${\rm Cone}_{t}=\{s\in A\colon t\leq s\}$, for some $t\in T_{A}$. The root of the orbit $O$ is the meet $t_{O}\in T_{A}$ of all points $a_{0},\ldots,a_{n}$ (which, in fact is the meet of two first elements in the orbit, if the orbit is meet-increasing, or the last two, if it is meet-decreasing), and the cone ${\rm Cone}_{O}$ of $O$ is defined as ${\rm Cone}_{t_{O}}$. Note that any two cones are either disjoint or one is contained in the other. Denote by ${\rm Cone}(p)$ the collection of all cones of orbits of $p$.

For $A\in\mathcal{SB}$, by a segment we mean an ordered pair $(x,y)$ with $x,y\in T_{A}$ such that $x<y$ and there is no $z\in T_{A}$ satisfying $x<z<y$. For $A,E\in\mathcal{SB}$ and a segment $(x,y)$ in $A$, let $K=A(x,y,E,\emptyset)\in\mathcal{SB}$ be the result of attaching $E$ to $A$ on $(x,y)$ on the left. Specifically, think that elements of each of $T_{A}$ and $T_{E}$ are binary sequences, in particular, if $x=s$ and $y=t$, then $t=s0$ or $t=s1$. We let $T_{K}$ to consist of the following binary sequences. If $r\in T_{A}$ does not extend properly $x=s$, we let $r\in T_{K}$. We let $t\in T_{K}$. If $tr\in T_{A}$ for some $r$, we let $t1r\in T_{K}$ and if $r\in T_{E}$, let $t0r\in T_{K}$. This defines $K\in\mathcal{SB}$. Analogously, define $K=A(x,y,\emptyset,F)\in\mathcal{SB}$ as the result of attaching $E$ to $A$ on $(x,y)$ on the right. More generally, for $A\in\mathcal{SB},\mathcal{E}=(E_{1},\ldots,E_{m})$ and $\mathcal{F}=(F_{1},\ldots,F_{n})$ and a segment $(x,y)$ in $A$, we define $K=A(x,y,\mathcal{E},\mathcal{F})\in\mathcal{SB}$ as the result of attaching $E_{1},\ldots,E_{m}$ to the segment $(x,y)$ on the left in a way that $E_{1}<_{lex}\ldots<_{lex}E_{m}$ and attaching $F_{1},\ldots,F_{n}$ to the segment $(x,y)$ on the right in a way that $F_{1}<_{lex}\ldots<_{lex}F_{n}$ and the root of each $E_{i}$ is below the root of each $F_{j}$. In that case, we may also write $(x,y,\mathcal{E},\mathcal{F})$ for $\{z\in T_{K}\colon x<{\rm{meet}}(y,z)<y\}$.

We say that $(A,p)\in\mathcal{SB}_{1}$ is in a simple normal form if (1) there are orbits $P=\{a_{0},\ldots,a_{n}\}<_{lex}Q=\{b_{0},\ldots,b_{n}\}$, $n\geq 2$, of $p$, such that for every $i=0,\ldots,n-1$, ${\rm{meet}}(a_{i},a_{i+1})<{\rm{meet}}(b_{i},b_{i+1})$ and for every $i=0,\ldots,n-2$, ${\rm{meet}}(b_{i},b_{i+1})<{\rm{meet}}(a_{i+1},a_{i+2})$, (2) any non-trivial orbit $O=\{c_{0},\ldots,c_{l}\}$ in $A$ is $\leq_{lex}$-contained in $P$ or in $Q$ and it holds $l=n-1$, (3) for any $x$ with $p(x)=x$ it holds $\max\{a_{0},a_{n}\}<_{lex}x<_{lex}\min\{b_{0},b_{n}\}$, where $\min$ and $\max$ are taken with respect to the $<_{lex}$ order.

We say that $(A,p)$ is in a normal form if (1) $A={\rm{def}}(p)$, and any non-constant orbit of $p$ has at least 3 elements, (2) $p$ cannot be extended to a partial automorphism $q$ such that some two orbits that did not intertwine (or meet-intertwine) in $p$ now they intertwine (or meet-intertwine, respectively) or they form one orbit in $q$, and (3) there is a partition $\mathcal{P}^{A}$ of $A$ into singletons $\{x\}$ and closed $\leq_{lex}$-intervals that will be grouped into pairs $([a,b],[c,d])$ so that if $\{x\}\in\mathcal{P}^{A}$, then $p(x)=x$, and if $([a,b],[c,d])\in\mathcal{P}^{A}$, then the structure $p\restriction([a,b]\cup[c,d])$ is in a simple normal form without a non-trivial orbit, witnessed by some $P=\{a_{0},\ldots,a_{n}\}$ and $Q=\{b_{0},\ldots,b_{n}\}$ with $a=a_{0}$, $b=a_{n}$, $c=b_{n}$, $d=b_{0}$ (or $b=a_{0}$, $a=a_{n}$, $d=b_{n}$, $c=b_{0}$). We will sometimes identify $([a,b],[c,d])\in\mathcal{P}^{A}$ with the set $[a,b]\cup[c,d]$.

Lemma 3.11**.**

Any $(A,p)\in\mathcal{SB}_{1}$ can be extended to $(A^{\prime},p^{\prime})\in\mathcal{SB}_{1}$, which is in a normal form.

Proof.

Conditions (1) and (2) can be easily satisfied. To have (3), consider the equivalence relation on orbits: $O$ and $P$ are equivalent iff there is a sequence of orbits $O=O_{1},\ldots,P=O_{n}$ such that for each $i$, $O_{i}$ and $O_{i+1}$ intertwine or meet intertwine. An equivalence class $E$ either is a singleton containing a constant orbit, or it does not contain a constant orbit. In the second case, after extending $A$ if necessary, the class $E$ contains two meet-intertwining orbits $P_{0}<_{lex}Q_{0}$ (there are usually many such choices). Extend $P_{0}$ to $P$ and $Q_{0}$ to $Q$ so that every orbit in $E$ is contained in $P$ or in $Q$. Finally, extend the remaining orbits in $E$ so that (2) in the definition of simple normal form is satisfied. We do the induction on the number of equivalence classes $E$. ∎

Let $(A,p)\in\mathcal{SB}_{1}$ be in a normal form and let $X\in\mathcal{P}^{A}$. Then $(X,p_{X})\in\mathcal{SB}_{1}$, where $p_{X}=p\restriction X$, is in a simple normal form. The tree $T_{X}$ can be naturally identified with a subtree of $T_{A}$ (in fact, $T_{X}$ is the closure of $X$ in $T_{A}$ under taking the meet), let $\rho_{X}$ be the root of $T_{X}$, and let $\mbox{Cone}_{X}=\mbox{Cone}_{\rho_{X}}$. To $X$ as above we associate $X^{*}\in\mathcal{SB}$, and $(X^{*},p_{X}^{*})\in\mathcal{SB}_{1}$ in a simple normal form as follows. If $X=\{x\}$, let $p^{*}(x)=x$. Otherwise, if $X=([a,b],[c,d])$, we let

[TABLE]

and we let $p_{X}^{*}$ to be the extension of $p_{X}$ that is equal to the identity on $X^{*}\setminus X$. The set of all $X^{*}$ obtained in this way we denote by $(\mathcal{P}^{A})^{*}$. Definitions introduced above are illustrated in Example 3.12.

Proof of Theorem 3.10.

Let $(A,p)\in\mathcal{SB}_{1}$ and consider extensions $(B,q),(C,r)\in\mathcal{SB}_{1}$ of $(A,p)$. By $\phi$, $\psi$, we denote the identity embeddings of $(A,p)$ into $(B,q)$, $(C,r)$, respectively. We show that if $(A,p)$ is in a normal form, then we can amalgamate $(B,q)$ and $(C,r)$ over $(A,p)$. Since the family of all elements in a normal form is cofinal in $\mathcal{SB}_{1}$, this will finish the proof.

(1) Suppose that $(A,p)$ is in a simple normal form.

Let $P=\{a_{0},\ldots,a_{n}\}$ and $Q=\{b_{0},\ldots,b_{n}\}$ be as in the definition of the simple normal form. Without loss of generality, $P$ is increasing, and hence $Q$ is decreasing. Set $t_{i}={\rm{meet}}(a_{i},a_{i+1})$ and let $s_{i}={\rm{meet}}(b_{i},b_{i+1})$. Note that all trees $T_{X_{i}}$ with $X_{i}=[a_{i},a_{i+1}]_{lex}$ are isomorphic, and all trees $T_{Y_{i}}$ with $Y_{i}=[b_{i+1},b_{i}]_{lex}$ are isomorphic as well.

Pick some $N$ such that each of $q^{-N}(a_{0})$, $q^{N}(a_{n})$, $q^{-N}(b_{0})$, $q^{N}(b_{n})$, $r^{-N}(a_{0})$, $r^{N}(a_{n})$, $r^{-N}(b_{0})$, $r^{N}(b_{n})$ is undefined. Let $\overline{\phi}\colon T_{A}\to T_{B}$ and $\overline{\psi}\colon T_{A}\to T_{C}$ be the unique meet-preserving extensions of $\phi$ and $\psi$.

For every $k$, let $(A_{k},p_{k})$ be defined as follows. Take an extension $p^{\prime}$ of $p$ such that for every $x\in[a_{0},a_{1})_{lex}\cup(b_{1},b_{0}]_{lex}$, the values $q^{n+k}(x)$ and $q^{-k}(x)$ are defined and every orbit in $p^{\prime}$ extends an orbit in $p$. Then let

[TABLE]

where $a_{i}=q^{i}(a_{0})$ and $b_{i}=q^{i}(b_{0})$, and let $A_{k}={\rm{def}}(p_{k})$. Note that $p=p_{0}$ and that each $p_{k}$ is in a simple normal form as witnessed by $P_{k}=\{a_{-k},\ldots,a_{n+k}\}$ and $Q_{k}=\{b_{-k},\ldots,b_{n+k}\}$. Consider $D_{0}=A_{N}$, $s_{0}=p_{N}$ for $N$ as above.

Let

[TABLE]

and define $A_{C}$ similarly. Let $\alpha\colon A_{B}\to D_{0}$ and $\beta\colon A_{C}\to D_{0}$ be the unique embeddings that agree on $A$. Denote by $\bar{\alpha}\colon T_{A_{B}}\to T_{D_{0}}$ and $\bar{\beta}\colon T_{A_{C}}\to T_{D_{0}}$ the tree embeddings corresponding to $\alpha$ and $\beta$. We clearly have $\rho_{B}\leq\overline{\phi}(\rho_{A})$ or $\rho_{C}\leq\overline{\psi}(\rho_{A})$, where $\rho_{A},\rho_{B},\rho_{C}$ are roots of $T_{A},T_{B},T_{C}$, and both inequalities can be strict. Then, to obtain the required amalgam $(D,s)$, first consider $\widehat{T}_{D_{0}}$ obtained from $T_{D_{0}}$ by adding a new point $v$, which satisfies $v<\rho_{D_{0}}$, where $\rho_{D_{0}}$ is the root of $T_{D_{0}}$. Next, fix a segment $(x,y)$ in $\widehat{T}_{D_{0}}$, and let $(x^{B},y^{B})=(\bar{\alpha}^{-1}(x),\bar{\alpha}^{-1}(y))$, if defined, and $(x^{C},y^{C})=(\bar{\beta}^{-1}(x),\bar{\beta}^{-1}(y))$, if defined. Suppose that $\mathcal{E}^{B}$, $\mathcal{E}^{C}$, $\mathcal{F}^{B}$, $\mathcal{F}^{C}$ are such that $\{z\in T_{B}\colon x^{B}<{\rm{meet}}(y^{B},z)<y^{B}\}$ can be identified with $(x^{B},y^{B},\mathcal{E}^{B},\mathcal{F}^{B})$ and $\{z\in T_{C}\colon x^{C}<{\rm{meet}}(y^{C},z)<y^{C}\}$ can be identified with $(x^{C},y^{C},\mathcal{E}^{C},\mathcal{F}^{C})$. Then replace $(x,y)$ in $\widehat{T}_{D_{0}}$ by $(x,y,(\mathcal{E}^{B}\mathcal{E}^{C}),(\mathcal{F}^{C}\mathcal{F}^{B}))$, where $(\mathcal{E}^{B}\mathcal{E}^{C})$ and $(\mathcal{F}^{C}\mathcal{F}^{B})$ are concatenations of sequences $\mathcal{E}^{B}$ with $\mathcal{E}^{C}$ and $\mathcal{F}^{C}$ with $\mathcal{F}^{B}$, respectively. We additionally require that the root of every $T_{E^{C}_{i}}$ is above the root of every $T_{F^{B}_{j}}$ and the root of every $T_{F^{C}_{i}}$ is above the root of every $T_{E^{B}_{j}}$, where $E^{C}_{i}\in\mathcal{E}^{C}$, etc. The obtained tree $T$ defines $D\in\mathcal{SB}$ such that $T_{D}=T$, and it defines embeddings $\alpha\colon B\to D$ and $\beta\colon C\to D$ of structures in $\mathcal{SB}$. We let $s(a)=b$ iff $\alpha^{-1}(a),\alpha^{-1}(b)$ are defined and $q(\alpha^{-1}(a))=\alpha^{-1}(b)$, or $\beta^{-1}(a),\beta^{-1}(b)$ are defined and $r(\beta^{-1}(a))=\beta^{-1}(b)$, or $p^{N}(a)=b$. For every segment $(x,y)$ in $\widehat{T}_{D_{0}}$ and $z\in T_{D}$, $x<z<y$, we will call the subtree $\mbox{Cone}_{z}$ of $T_{D}$ a triangle (coming from $B$ or from $C$).

We have to show that $s$ is a partial automorphism of $D$. Clearly $<_{lex}$ is preserved. Let $\bar{s}_{0}$ be the meet-preserving extension of $s_{0}$ to $\widehat{T}_{D_{0}}$. Then, clearly, a triangle attached to a segment $(x,y)$ is mapped to a triangle attached to a segment $(\bar{s}_{0}(x),\bar{s}_{0}(y))$. The key observation is that for every $k$, and in particular for $k=N$, and every $X\subseteq{\rm{dom}}(p_{k})$, the tree $T_{X}$ are isomorphic with the tree $T_{s_{0}(X)}$ via the tree isomorphism extending the bijection $x\mapsto s_{0}(x)$, $x\in X$. Moreover, for $a,b\in D$ that lie in different triangles, ${\rm{meet}}(a,b)$ is equal to the meet of the roots of the triangles to which $a$ and $b$ belong. Note also that if $\rho\leq\rho^{\prime}$ are roots of two triangles then $\bar{s}(\rho)\leq\bar{s}(\rho^{\prime})$. Therefore, if $x,y,z\in D$ lie in different triangles, we have $S(x,y,z)$ iff $S(s(x),s(y),s(z))$. Clearly, if all $x,y,z$ lie in the same triangle, then the conclusion holds. If $x\leq_{lex}y\leq_{lex}z$ and $x$ and $y$ lie in a triangle $T$ and $z$ lies in a different triangle $S$, if $\rho_{T}$ and $\rho_{S}$ denote roots of $S$ and $T$, then $S(x,y,z)$ holds iff $\rho_{S}<\rho_{T}$ and $S(s(x),s(y),s(z))$ holds iff $\bar{s}_{0}(\rho_{S})<\bar{s}_{0}(\rho_{T})$, hence $S(x,y,z)$ iff $S(s(x),s(y),s(z))$. The case when $y$ and $z$ lies in the same triangle and $x$ lies in a different one is analogous.

(2) Suppose that $(A,p)$ is in a normal form.

Without loss of generality, $(B,q)$ and $(C,r)$ are also in the normal form. Let $\widehat{\mathcal{P}}^{B}$ be the partition of $B$ into points and pairs of closed $\leq_{lex}$-intervals, which is a coarsening of $\mathcal{P}^{B}$, and has the following property: for every $([a,b],[c,d])\in\widehat{\mathcal{P}}^{B}$ there is exactly one $([a_{0},b_{0}],[c_{0},d_{0}])\in\mathcal{P}^{A}$ such that $[a_{0},b_{0}]\subseteq[a,b]$ and $[c_{0},d_{0}]\subseteq[c,d]$. We define $(\mathcal{P}^{A})^{*}$ out of $\mathcal{P}^{A}$, and the corresponding partial automorphisms $p_{X}^{*}$, in a way explained earlier. We define $(\widehat{\mathcal{P}}^{B})^{*}$ and $(\widehat{\mathcal{P}}^{C})^{*}$ similarly, but with respect to $\widehat{\mathcal{P}}^{B}$ and $\widehat{\mathcal{P}}^{C}$. Let

[TABLE]

and we let $p_{X}^{*}$ to be the extension of $p_{X}$ that is equal to the identity on $X^{*}\setminus X$. The set of all $X^{*}$ obtained in this way we denote by $(\widehat{\mathcal{P}}^{B})^{*}$. We similarly define $(\widehat{\mathcal{P}}^{C})^{*}$. For a given $X\in\mathcal{P}^{A}$, let $X_{B}\in\widehat{\mathcal{P}}^{B}$ and $X_{C}\in\widehat{\mathcal{P}}^{C}$ be the unique sets containing $X$. Amalgamate $(X^{*}_{B},q^{*}_{X_{B}})$ and $(X^{*}_{C},r^{*}_{X_{C}})$ over $(X^{*},p^{*}_{X})$, as we did in (1). We obtain $D_{X}\in\mathcal{SB}$, a partial automorphism $s_{X}$ of $D_{X}$, and a pair of embeddings $\alpha_{X}\colon X^{*}_{B}\to(D_{X},s_{X})$ and $\beta_{X}\colon X^{*}_{C}\to(D_{X},s_{X})$ such that $\alpha_{X}\restriction X^{*}=\beta_{X}\restriction X^{*}$. Let $\rho_{X}$ be the root of $T_{X}$, and let $\rho_{1}^{X},\ldots,\rho_{l_{X}}^{X}$ be an enumeration of all $z=\mbox{Cone}_{Y}$ from the definition of $X^{*}$. To the tuple $(X^{*},\rho_{X},\rho_{1}^{X},\ldots,\rho_{l_{X}}^{X})$ we associate the tuple $(D_{X},\rho_{D_{X}},\rho_{1}^{D_{X}},\ldots,\rho_{l_{X}}^{D_{X}})$, where $\rho_{D_{X}}$ is the root of $T_{D_{X}}$, and $\rho_{i}^{D_{X}}=\alpha_{X}(\rho_{i}^{X})=\beta_{X}(\rho_{i}^{X})$ for each $i$. By the definition of $(\mathcal{P}^{A})^{*}$, the tree $T_{A}$ is the disjoint union of $\{T_{X^{*}}\colon X\in\mathcal{P}^{A}\}$ after the identification of each $\rho_{i}^{X}$ with a $\rho_{Y}$ for some $Y\in(\mathcal{P}^{A})$. Finally, let $T_{D}$ be obtained by taking the disjoint union of $\{T_{D_{X}}\colon X\in\mathcal{P}^{A}\}$, and then identifying a $\rho_{i}^{D_{X}}$ with a $\rho_{D_{Y}}$ if and only if $\rho_{i}^{X}$ was identified with a $\rho_{Y}$, $X,Y\in\mathcal{P}^{A}$. This $T_{D}$ defines $D\in\mathcal{SB}$ and let $s=\bigcup_{X\in\mathcal{P}^{A}}s_{X}\restriction X$. Then $(D,s)\in\mathcal{SB}_{1}$. Clearly, $s$ preserves $<_{lex}$. To show that it preserves $S$, observe first the following.

(i) If $x\in D_{X}$ and $y\in D_{Y}$ with $X\neq Y$, and $\rho_{D_{X}}$ and $\rho_{D_{Y}}$ are incomparable in $(T_{D},\leq)$, then ${\rm{meet}}(x,y)={\rm{meet}}(\rho_{D_{X}},\rho_{D_{Y}})={\rm{meet}}(s(x),s(y))$.

(ii) If $x\in D_{X}$ and $y\in D_{Y}$ with $X\neq Y$ and $\rho_{D_{X}}\leq\rho_{D_{Y}}$, then ${\rm{meet}}(x,y)={\rm{meet}}(x,\rho_{D_{Y}})$ and ${\rm{meet}}(s(x),s(y))={\rm{meet}}(s(x),\rho_{D_{Y}})$. Moreover, in that case, if $y^{\prime}\in D_{Y}$, then ${\rm{meet}}(x,y),{\rm{meet}}(x,y^{\prime})<{\rm{meet}}(y,y^{\prime})$.

Let $(x,y,z)$ be a $<_{lex}$ ordered triple of points in $D$. There are several cases to consider. Clearly, if there is some $X\in\mathcal{P}^{A}$ such that $x,y,z\in D_{X}$, then $S(x,y,x)$ iff $S(s(x),s(y),s(z))$. If $x,y\in D_{X}$, $z\in D_{Y}$ with $X\neq Y$ and $\rho_{D_{X}}$ and $\rho_{D_{Y}}$ are incomparable, then both $S(x,y,z)$ and $S(s(x),s(y),s(z))$ hold. Similarly, if $x\in D_{X}$ and $y,z\in D_{Y}$ with $X\neq Y$ and $\rho_{D_{X}}$ and $\rho_{D_{Y}}$ are incomparable then none of $S(x,y,z)$, $S(s(x),s(y),s(z))$ holds. In the case when $\rho_{D_{X}}\leq\rho_{D_{Y}}$, two of the $x,y,z$ belong to $D_{X}$ and the remaining point belongs to $D_{Y}$, $X\neq Y$, then using (ii) we get that $s$ preserves $S$ on $(x,y,z)$ because $p_{X}$ preserves $S$ (the point belonging to $D_{Y}$ we can replace with an appropriate $\rho^{D_{X}}_{i}$). If two of the $x,y,z$ belong to $D_{Y}$ and the remaining point belongs to $D_{X}$, use the second sentence of (ii) to get the conclusion. Finally, suppose that $x\in D_{X}$, $y\in D_{Y}$, $z\in D_{Z}$ with $X,Y,Z$ pairwise different. There are a few cases to consider, $\rho_{D_{X}}$ and $\rho_{D_{Y}}$ can be comparable or not, and the same for the other two pairs. Each time, reasoning similarly as above and using (i) and (ii), we get the required conclusion.

∎

Example 3.12**.**

Let $(A,p)\in\mathcal{SB}_{1}$ be as in Figure 2 with $p(a_{0})=a_{1}$, $p(a_{1})=a_{2}$, $p(a^{\prime}_{0})=a^{\prime}_{1}$, $p(a^{\prime}_{1})=a^{\prime}_{2}$, etc. Then $\mathcal{P}^{A}=\{X=([a_{0},a_{2}],[a^{\prime}_{2},a^{\prime}_{0}]),Y=([b_{0},b_{3}],[b^{\prime}_{3},b^{\prime}_{0}]),Z=([d_{0},d_{2}],[d^{\prime}_{2},d^{\prime}_{0}])\}$. Figure 3 illustrates $X^{*},Y^{*},Z^{*}$, where $p^{*}_{Z}(x)=x$ and $p^{*}_{Z}(y)=y$.**

3.3. Ordered posets - a comeager conjugacy class

In this part, we will show that the automorphism group of the universal ordered poset has a comeager conjugacy class.

A poset is a shortcut for a partially ordered set. By an ordered poset, we mean a structure of the form $(P,\prec,<)$, where $(P,\prec)$ is a finite poset, and $<$ is a linear ordering of $P$ extending $\prec$. We denote the class of all finite ordered posets by $\mathcal{P}$. Then the universal ordered poset is the Fraïssé limit of $\mathcal{P}$. Kuske-Truss [13] showed that the class of partial automorphisms of finite posets has CAP, and hence the corresponding automorphism group has a comeager conjugacy class. The same turns out to be true for the class $\mathcal{P}_{1}$.

We will see that the proof of Kuske-Truss generalizes to our context.

Theorem 3.13**.**

The class $\mathcal{P}_{1}$ has CAP.

Below, we will always use the symbol $\prec$ for the poset relation, and $<$ for the linear order. For $A=(A,p)\in\mathcal{P}_{1}$, and an orbit $O=\{a_{0},a_{1},\ldots,a_{n}\}$ of $p$, we say that $O$ is $<$-increasing if $a_{0}<\ldots<a_{n}$, otherwise, it is $<$-decreasing. As with boron trees, if $(B,q)\in\mathcal{P}_{1}$ extends $(A,p)$, the orbit of $q$ extending $O$ will be denoted by $O_{B}$.

We say that a pair of orbits $(O,N)$ is determined if for any extensions $(B,q),(C,r)$ of $(A,p)$, such that for every $k\in\mathbb{Z}$ and $x\in O\cup N$ we have that $q^{k}(x)$ is defined iff $r^{k}(x)$ is defined, the following holds: $(O_{B}\cup N_{B},q\restriction(O_{B}\cup N_{B}))$ and $(O_{C}\cup N_{C},r\restriction(O_{C}\cup N_{C}))$ are isomorphic via a mapping which is the identity on $O\cup N$. Clearly, $(O,N)$ is determined iff $(N,O)$ is determined. An orbit $O$ is determined if the pair $(O,O)$ is determined.

Let $(A,p)$ be given. Let $O$ be an orbit in $(A,p)$ and $x\in O$. Denote

[TABLE]

Note that if $O$ is $<$-increasing then $t(O,x)$ consists of non-negative integers and if $O$ is $<$-decreasing then $t(O,x)$ consists of non-positive integers. Moreover, if $n\in t(O,x)$ then by the transitivity of $\prec$, for every positive integer $k$, we have that if $p^{kn}(x)\in O$ then $kn\in t(O,x)$. More generally, if $(O,N)$ are orbits in $(A,p)$ and $x\in O$, $y\in N$, then let

[TABLE]

We say that $(O,N,x,y)$ is positive determined if orbits $O$ and $N$ are determined and for any extensions $(B,q),(C,r)$ of $(A,p)$, such that for every $k\in\mathbb{Z}$ we have that $q^{k}(x)$ is defined iff $r^{k}(x)$ is defined and $q^{k}(y)$ is defined iff $r^{k}(y)$ is defined, we have that $t(O_{B},N_{B},x,y)\cap\mathbb{N}=t(O_{C},N_{C},x,y)\cap\mathbb{N}$. We similarly define when $(O,N,x,y)$ is negative determined. We let $(O,N,x,y)$ to be determined iff $(O,N,x,y)$ is both positive and negative determined. The $(O,x)$ is determined if for any extensions $(B,q),(C,r)$ of $(A,p)$, such that for every $k\in\mathbb{Z}$ we have that $q^{k}(x)$ is defined iff $r^{k}(x)$ is defined, it holds $t(O_{B},x)=t(O_{C},x)$. Note that a pair of orbits $(O,N)$ is determined iff for every $x\in O$, $y\in N$, the $(O,N,x,y)$ and $(N,O,y,x)$ are determined iff for some $x\in O$, $y\in N$, the $(O,N,x,y)$ and $(N,O,y,x)$ are determined. Similarly, $O$ is determined iff for some/every $x\in O$, the $(O,x)$ is determined.

An orbit $O$ will be called an antichain if for every extension $(B,q)$ of $(A,p)$ and for every/some $x\in O$, we have $t(O_{B},x)=\emptyset$. An $X\subset\mathbb{Z}$ is called positive eventually periodic if there exist $N,k\geq 0$ such that $X\cap[N,\infty)=\{N+kn\colon n\geq 0\}$. The number $k$ we will call the positive period of $X$. We similarly define a negative eventually periodic set and the negative period. We call a set eventually periodic if it is both positive and negative periodic. We will call two orbits $O=\{a_{0},\ldots,a_{m}\}$ and $N=\{b_{0},\ldots,b_{n}\}$ of $p$ intertwining if the $<$-intervals $(\min\{a_{0},a_{m}\},\max\{a_{0},a_{m}\})_{<}$ and $(\min\{b_{0},b_{n}\},\max\{b_{0},b_{n}\})_{<}$ intersect.

Proof of Theorem 3.13.

The proof is similar to the proof of Kuske-Truss [13].

Step 1. Every $(A,p)\in\mathcal{P}_{1}$ can be extended to some $(B,q)\in\mathcal{P}_{1}$, in which every orbit is determined. Moreover, we can do it so that we do not add new orbits.

Take an orbit $O$ and $x\in O$, and without loss of generality suppose that $O$ is $<$-increasing. If $O$ is an antichain then it is already determined. Otherwise, let (perhaps after passing to an extension) $k\in t(O,x)$. Now for every positive integer $n$, we have that if $p^{kn}(x)\in O$ then $kn\in t(O,x)$. Using this remark, we obtain $(B,q)$, an extension of $(A,p)$, such that for every $i=1,\ldots,k-1$, if there is an extension $(B_{1},q_{1})$ of $(B,q)$ with $n_{i}k+i\in t(O_{B_{1}},x)$, for some $n_{i}\geq 0$, then, in fact, for every extension $(B_{2},q_{2})$ of $(B,q)$, it holds $n_{i}k+i\in t(O_{B_{2}},x)$, as long as $n_{i}k+i\in O_{B_{2}}$. To obtain such a $(B,q)$, we construct a sequence of extensions $(A_{1},p_{1}),\ldots,(A_{k-1},p_{k-1})$ of $(A,p)$ such that $(A_{i},p_{i})$ has the required property for $i$, and we take $(B,q)=(A_{k-1},p_{k-1})$. Then clearly $O_{B}$ is determined.

Step 2. Every $(B,q)\in\mathcal{P}_{1}$ can be extended to some $(D,s)\in\mathcal{P}_{1}$, in which every pair of orbits is determined. For this we find a $(D,s)$ such that for every pair of orbits $(O,N)$, $x\in O$, $y\in N$, there is an almost periodic set $X\subseteq\mathbb{Z}$ such that for every extension $(D_{1},s_{1})$ of $(D,s)$, if $m_{1},m_{2}$ are the least such that ${s_{1}}^{m_{1}}(y)$ and ${s_{1}}^{-m_{2}}(y)$ are undefined, we have $X\cap(-m_{1},m_{2})=t(O_{D},x)$.

Step 2a. The $(B,q)$ obtained in Step 1 can be extended to some $(C,r)\in\mathcal{P}_{1}$ such that all pairs of orbits in which both orbits are antichains, are determined. Moreover, we can do it in a way that for each such a pair we add four new orbits, none of which is an antichain.

We fix a pair $(O,N)$ of such orbits and let $x\in O$. As in Kuske-Truss, after possibly extending $N$, find $y\in N$ and $0<n$ with $q^{n}(y)\in{\rm{rng}}(q)\setminus{\rm{dom}}(q)$ such that for every $z\in{\rm{def}}(q)\setminus N$, it holds: $z\prec y$ iff $z\prec q^{n}(y)$, $y\prec z$ iff $q^{n}(y)\prec z$, $z$ is incomparable with $y$ iff $z$ is incomparable with $q^{n}(y)$. For this pick some $y_{0}\in N$ and, possibly extending $q$, using the pigeonhole principle, choose $k_{1}<k_{2}$ sufficiently large so that $q^{k_{2}}(y_{0})\in{\rm{rng}}(q)\setminus{\rm{dom}}(q)$, $z\prec q^{k_{1}}(y_{0})$ iff $z\prec q^{k_{2}}(y_{0})$, $q^{k_{1}}(y_{0})\prec z$ iff $q^{k_{2}}(y_{0})\prec z$, $z$ is incomparable with $q^{k_{1}}(y_{0})$ iff $z$ is incomparable with $q^{k_{2}}(y_{0})$. Take $y=q^{k_{1}}(y_{0})$ and $n=k_{2}-k_{1}$.

Let us proceed to the construction. Take $\{a_{i}\colon 0\leq i\leq n\}$ and $\{b_{i}\colon 0\leq i\leq n\}$ disjoint from each other and from ${\rm{def}}(q)$ and such that for all $0\leq i<j\leq n$, $z\in{\rm{def}}(q)\setminus N$ :

1. (1)

$a_{i}$ and $a_{j}$ are incomparable except for $a_{0}\prec a_{n}$, 2. (2)

$b_{i}$ and $b_{j}$ are incomparable except for $b_{n}\prec b_{0}$, 3. (3)

$a_{i}\prec q^{i}(y)\prec b_{i}$, 4. (4)

$z\prec b_{i}$ iff $z\prec q^{i}(y)$, 5. (5)

$a_{i}\prec z$ iff $q^{i}(y)\prec z$, 6. (6)

$a_{i}\prec b_{i}$, $a_{0}\prec b_{n}$, $a_{n}\prec b_{0}$, 7. (7)

$a_{0}<\ldots<a_{n}<{\rm{def}}(q)<b_{n}<\ldots<b_{0}$.

Denote the obtained structure by $(B_{1},\prec,<)$. Let $q_{1}$ extend $q$ by $q_{1}(a_{i})=a_{i+1}$ and $q_{1}(b_{i+1})=b_{i}$, $i=0,\ldots,n-1$. It is straightforward to see that $\prec$ is transitive, $<$ extends $\prec$, $p$ preserves $\prec$ and $<$. As in Kuske-Truss, we have that in any extension $(B_{2},q_{2})$ of $(B_{1},q_{1})$,

[TABLE]

hence $(O,N,x,y)$ is positive determined. Indeed, let $(B_{2},q_{2})$ be an extension of $(B_{1},q_{1})$ and denote $X=\{kn+i\colon k\in\mathbb{N},0\leq i<n,x\prec q^{i}(y)\}\cap B_{2}$. Since $a_{0}\prec a_{n}$ and $a_{i}\prec q^{i}(y)$, we get $t(O_{B_{2}},N_{B_{2}},x,y)\supseteq X$. Since $b_{n}\prec b_{0}$ and $q^{i}(y)\prec b_{i}$, using (4), we get $t(O_{B_{2}},N_{B_{2}},x,y)\subseteq X$.

By considering $(B_{1},q_{1}^{-1})$ and repeating the argument above, we further extend $(B_{1},q_{1})$ to obtain $(C,r)$ in which $(O,N)$ is determined.

Step 2b. The $(C,r)$ obtained in Step 2a can be extended to some $(D,s)\in\mathcal{P}_{1}$ such that all pairs of orbits such that at least one of them is not an antichain are determined. Moreover, we can do it in a way that we do not add new orbits.

We fix a pair $(O,N)$ of such orbits and let $x\in O$ and $y\in N$. If for every extension $(C_{1},r_{1})$ of $(C,r)$, $t(O,N,x,y)=\emptyset$ and $t(N,O,y,x)=\emptyset$, then $(O,N)$ is already determined. Therefore, by passing to an extension if necessary, we assume that $t(O,N,x,y)\neq\emptyset$ or $t(N,O,y,x)\neq\emptyset$. Note that at least one of the sets $t(O,N,x,y)$ and $t(N,O,y,x)$ is empty. Without loss of generality, let us assume that $x\prec y$ and $t(N,O,y,x)=\emptyset$.

If $k\in t(O,x)$ and $l\in t(N,y)$, then for every $n_{1},n_{2}\geq 0$, if $q^{kn_{1}+ln_{2}}(y)\in N$, then $kn_{1}+ln_{2}\in t(O,N,x,y)$. Hence if there is a positive number $k\in t(O,x)$ or a positive number $l\in t(N,y)$, reasoning as in Step 1, we can find an extension in which $(O,N,x,y)$ is positive determined. Similarly, if there is a negative number $k\in t(O,x)$ or a negative number $l\in t(N,y)$, we can find an extension in which $(O,N,x,y)$ is negative determined. Therefore, without loss of generality, what is left to be shown is the following. Suppose that $O$ and $N$ are $<$-increasing, at least one of them is not an antichain, and $(O,N,x,y)$ is positive determined. Then we can extend $(C,r)$ to $(C_{1},r_{1})$ so that $(O_{C_{1}},N_{C_{1}},x,y)$ is negative determined.

Without loss of generality, $O$ is not an antichain. Take some $k$ such that $x\prec r^{k}(x)$. Then, clearly, for every $n>0$, it holds $x\prec r^{nk}(x)$. Take an extension $(C_{1},r_{1})$ such that for every $0\leq i<k$ either for some $n_{i}$, we have ${r_{1}}^{-(n_{i}k+i)}(y)\in C_{1}$ and $x\prec{r_{1}}^{-(n_{i}k+i)}(y)$ does not hold, or, for every extension $(C_{2},r_{2})$ of $(C_{1},r_{1})$ and every $n>0$, $x\prec{r_{1}}^{-(nk+i)}(y)$ holds . Then $(O_{C_{1}},N_{C_{1}},x,y)$ is negative determined. Indeed, note that if $i$ and $n_{0}$ are such that $x\prec{r_{1}}^{-(n_{0}k+i)}(y)$ does not hold, then in every extension $(C_{2},r_{2})$ of $(C_{1},r_{1})$ and $n_{1}>n_{0}$, $x\prec{r_{1}}^{-(n_{1}k+i)}(y)$, does not hold either by the choice of $k$.

We apply this procedure to every pair of orbits such that at least one of them is not an antichain. The resulting extension, which we denote by $(D,s)$, is as required.

We will show that $\mathcal{P}_{1}$ has the CAP, i.e. we will show that for every $(A_{0},p_{0})\in\mathcal{P}_{1}$ there is $(A,p)\in\mathcal{P}_{1}$ extending $(A_{0},p_{0})$ such that for any $(B,q),(C,r)\in\mathcal{P}_{1}$ extending $(A,p)$ there exists $(D,s)\in\mathcal{P}_{1}$, which is an amalgam of $(B,q)$ and $(C,r)$ over $(A,p)$. For this fix $(A_{0},p_{0})\in\mathcal{P}_{1}$ and extend it to an $(A,p)\in\mathcal{P}_{1}$ such that any pair of orbits in $A$ is determined, and there is no extension $(A_{1},p_{1})$ of $(A,p)$ in which some two orbits in $A$ that did not intertwine, become one orbit or they intertwine in $A_{1}$. Fix $(B,q),(C,r)\in\mathcal{P}_{1}$ extending $(A,p)$. Without loss of generality, we have that $A\subseteq{\rm{dom}}(q)\cap{\rm{rng}}(q)$, and similarly, $A\subseteq{\rm{dom}}(r)\cap{\rm{rng}}(r)$, as well as that for every $a\in A$ and $n\in\mathbb{Z}$, $q^{n}(a)$ is defined iff $r^{n}(a)$ is defined.

Enumerate the set $\{b\in B\colon(\exists a\in A,n\in\mathbb{Z})\ b=r^{n}(a)\}$ into a $<$-increasing sequence $a^{B}_{1},a^{B}_{2},\ldots,a^{B}_{k}$, and similarly, enumerate the set $\{c\in C\colon(\exists a\in A,n\in\mathbb{Z})\ c=r^{n}(a)\}$ into $a^{C}_{1},a^{C}_{2},\ldots,a^{C}_{k}$ so that it is $<$-increasing. By the assumptions on $A$, for any $a,b\in A$ and $m,n\in\mathbb{Z}$, if $x_{B}=q^{m}(a)$, $y_{B}=q^{n}(b)$, $x_{C}=r^{m}(a)$, $y_{C}=q^{n}(b)$ are defined, then $x_{B}<y_{B}$ iff $x_{C}<y_{C}$. Denote $B_{i}=\{b\in B\colon a_{i}^{B}<b<a^{B}_{i+1}\}$, $i=1,\ldots,k-1$, $B_{0}=\{b\in B\colon b<a^{B}_{1}\}$, $B_{k}=\{b\in B\colon a_{k}^{B}<b\}$, and similarly define $C_{i}$’s.

We first amalgamate $B$ and $C$ over $A$ in $\mathcal{P}$. Let $D$ be the disjoint union of $B$ and $C$ with $a_{i}^{B}$ and $a_{i}^{C}$ identified. Set $a_{i}=a_{i}^{B}=a_{i}^{C}$, and let

[TABLE]

Denote by $\prec^{B}$ the partial ordering relation on $B$, by $\prec^{C}$ the partial ordering relation on $C$, and let $\prec^{D}$ be the transitive closure of $\prec^{B}\cup\prec^{C}$. Then $(D,\prec^{D})$ is a partial ordering such that the linear ordering $<^{D}$ extends $\prec^{D}$, and so $D$ is an amalgam of $B$ and $C$ over $A$.

We finally let $s(x)=y$ iff $x=b_{1},y=b_{2}$ for some $b_{1},b_{2}\in B$ and $q(b_{1})=b_{2}$, or $x=c_{1},y=c_{2}$ for some $c_{1},c_{2}\in C$ and $r(c_{1})=c_{2}$. This is a partial automorphism. In particular, if for some $b\in B$, $c\in C$, it holds $b\prec^{D}c$ and $s(b),s(c)$ are defined, then there is $a\in\{a_{1},\ldots,a_{k}\}$ such that $b\prec^{B}a$ and $a\prec^{C}c$, or $b\prec^{C}a$ and $a\prec^{B}c$. Without loss of generality, it holds $b\prec^{B}a$ and $a\prec^{C}c$. Note that $q(a)$ and $r(a)$ are defined and hence we have $q(b)\prec^{B}q(a)$ and $r(a)\prec^{C}r(c)$. This implies $s(b)\prec^{D}s(c)$. Hence $(D,s)$ is the required amalgam.

∎

4. The two-dimensional case. Conjugacy classes.

We provide a condition, which we will use to obtain many examples of ordered structures $M$ such that ${\rm Aut}(M)$ has no comeager $2$-dimensional diagonal conjugacy class.

Given a partial automorphism $p$ of a structure $A$, and $a\in A$, say that $a\in A$ is locked by $p$ if there are $x\leq a\leq y$, $x,y\in A$ such that $p(x)=y$ or $p(y)=x$.

Proposition 4.1**.**

Let $\mathcal{F}$ be an Fraïssé order class. Suppose that for every $(A,p)\in\mathcal{F}_{1}$ and $a\in A$ not locked by $p$, there are extensions $(B,r),(C,s)\in\mathcal{F}_{1}$ of $(A,p)$ such that $r(a)<a$ and $a<s(a)$. Then $\mathcal{F}_{2}$ has no WAP.

Proof.

It suffices to show that for a given $(A^{\prime},p^{\prime},q^{\prime})\in\mathcal{F}_{2}$ and $x\in A$ such that $x<p^{\prime}(x)$ there exists $(A^{\prime\prime},p^{\prime\prime},q^{\prime\prime})\in\mathcal{F}_{2}$ which is an extension of $(A^{\prime},p^{\prime},q^{\prime})$, and a word $w(s,t)\in F_{2}$, such that $w(p^{\prime\prime},q^{\prime\prime})(x)$ is defined and not locked by $p^{\prime\prime}$ or $q^{\prime\prime}$. Indeed, let $(A,p,q)\in\mathcal{F}_{2}$ and $x\in A$ be such that $x<p(x)$ (wlog there is such an $x$ in $A$), let $(A^{\prime},p^{\prime},q^{\prime})\in\mathcal{F}_{2}$ be an arbitrary extension of $(A,p,q)$, and let $(A^{\prime\prime},p^{\prime\prime},q^{\prime\prime})\in\mathcal{F}_{2}$ and $w(s,t)$ be as above. Then $y=w(p^{\prime\prime},q^{\prime\prime})(x)$ is not locked (wlog) by $p^{\prime\prime}$. Apply the assumptions of the proposition to $(A^{\prime\prime},p^{\prime\prime})$ and $y$ to find the corresponding $(B,r),(C,s)\in\mathcal{F}_{1}$. Then $(B,p^{\prime\prime},r),(C,p^{\prime\prime},s)$ cannot be amalgamated over $(A,p,q)$.

We construct the required $(A^{\prime\prime},p^{\prime\prime},q^{\prime\prime})$ and $w$ inductively. Let $(A^{\prime\prime}_{0},p^{\prime\prime}_{0},q^{\prime\prime}_{0})=(A^{\prime},p^{\prime},q^{\prime})$, and $w_{0}=1$. Suppose that we already constructed $(A^{\prime\prime}_{n},p^{\prime\prime}_{n},q^{\prime\prime}_{n})$ and $w_{n}(s,t)$, and suppose that $w_{n-1}(p^{\prime\prime}_{n-1},q^{\prime\prime}_{n-1})(x)=w_{n-1}(p^{\prime\prime}_{n},q^{\prime\prime}_{n})(x)<w_{n}(p^{\prime\prime}_{n},q^{\prime\prime}_{n})(x)$. Denote $y=w_{n}(p^{\prime\prime}_{n},q^{\prime\prime}_{n})(x)$ and, if $y$ is locked by $p^{\prime\prime}_{n}$ or $q^{\prime\prime}_{n}$, proceed as follows. Let $z_{1}\leq y\leq z_{2}$, and $f\in\{p^{\prime\prime}_{n},(p^{\prime\prime}_{n})^{-1},q^{\prime\prime}_{n},(q^{\prime\prime}_{n})^{-1}\}$ be such that $f(z_{1})=z_{2}$. Take $(A^{\prime\prime}_{n+1},p^{\prime\prime}_{n+1},q^{\prime\prime}_{n+1})$ to be an extension of $(A^{\prime\prime}_{n},p^{\prime\prime}_{n},q^{\prime\prime}_{n})$ such that $|A^{\prime\prime}_{n+1}\setminus A^{\prime\prime}_{n}|\leq 1$ and $f(y)$ is defined; clearly, $z_{2}<f(y)$. Let $u=s$ if $f=p^{\prime\prime}_{n}$, let $u=s^{-1}$ if $f=(p^{\prime\prime}_{n})^{-1}$, let $u=t$ if $f=q^{\prime\prime}_{n}$ , and let $u=q^{-1}$ if $f=(q^{\prime\prime}_{n})^{-1}$, and set $w_{n+1}(s,t)=uw_{n}(s,t)$. Clearly, $w_{n}(p^{\prime\prime}_{n},q^{\prime\prime}_{n})(x)=w_{n}(p^{\prime\prime}_{n+1},q^{\prime\prime}_{n+1})(x)<w_{n+1}(p^{\prime\prime}_{n+1},q^{\prime\prime}_{n+1})(x)$. Since $p^{\prime}$ and $q^{\prime}$ are finite and the sequence $(w_{n}(p^{\prime\prime}_{n},q^{\prime\prime}_{n})(x))_{n}$ is increasing, after finitely many steps we will obtain $N$ such that $w_{N}(p^{\prime\prime}_{N},q^{\prime\prime}_{N})(x)$ is not locked by $p^{\prime\prime}_{N}$ or is not locked by $q^{\prime\prime}_{N}$. Set $(A^{\prime\prime},p^{\prime\prime},q^{\prime\prime})=(A^{\prime\prime}_{N},p^{\prime\prime}_{N},q^{\prime\prime}_{N})$, $w=w_{N}$, and $y=w_{N}(p^{\prime\prime},q^{\prime\prime})(x)$. ∎

Corollary 4.2**.**

Suppose that $\mathcal{K}$ is a Fraïssé class, and let $\mathcal{F}$ be a full order expansion of $\mathcal{K}$. Then $\mathcal{F}_{2}$ has no WAP.

Corollary 4.3**.**

The class $\mathcal{P}_{2}$ has no WAP

Proof.

Let $P=(P,\prec^{P},<^{P})$ be an ordered poset, and let $p$ be a partial automorphism of $P$. Let $(Q,<^{Q})$ be an extension of $(P,<^{P})$ such that $|Q\setminus P|=1$. Let $x\in P$, $y\in Q\setminus P$, and suppose that $q=p\cup\{(x,y)\}$ is a partial automorphism of $(Q,<^{Q})$. Then there is $\prec^{Q}$ extending $\prec^{P}$ such that $(Q,\prec^{Q},<^{Q})$ is an ordered poset, and $q$ is a partial automorphism of $(Q,\prec^{Q},<^{Q})$. Indeed, define $\prec^{Q}$ as follows: for $a<^{Q}y$, $a\in P$, we set $a\prec^{Q}y$ iff there is $a\leq^{Q}b<^{Q}y$, $b\in{\rm{rng}}(p)$, such that (1) $a\prec^{P}b$ if $a\neq b$ and (2) $\alpha^{-1}(b)\prec^{P}x$.

Similarly, for $y<^{Q}d$, $d\in P$, we set $y\prec^{Q}d$ iff there is $y<^{Q}c\leq^{Q}d$, $c\in{\rm{rng}}(p)$, such that (1) $x\prec^{P}\alpha^{-1}(c)$ and (2) $c\prec^{P}d$ if $c\neq d$.

Thus, assumptions of Proposition 4.1 are satisfied. ∎

It is straightforward to verify that $\mathcal{P}_{2}$ has JEP. Therefore Corollary 4.3 implies that the automorphism group of the universal ordered poset has all $2$-dimensional diagonal conjugacy classes meager.

We show that there is no comeager $2$-dimensional diagonal conjugacy class in the automorphism group of the universal ordered boron tree. In fact, since $\mathcal{SB}_{2}$ has JEP, which is not hard to check, this will imply that all $2$-dimensional diagonal conjugacy classes of the group are meager.

Theorem 4.4**.**

The class $\mathcal{SB}_{2}$ has no WAP.

Let $(A,p)\in\mathcal{SB}_{1}$, let $x\in A$, and let $O=\{a_{0},\ldots,a_{n}\}$ be an orbit of $p$, $n\geq 2$. Suppose that $O$ is increasing and meet-increasing, other 3 cases being similar. We have therefore $a_{0}<_{lex}\ldots<_{lex}a_{n}$, and if $t_{i}={\rm{meet}}(a_{i},a_{i+1})$, then $t_{0}<t_{1}<\ldots<t_{n-1}$. Now the following two claims easily follow from the definition of the relation $S$. We will use them frequently.

Claim 1. If $a_{n}<_{lex}x$ with $x\in{\rm Cone}_{O}$ and $(B,q)\in\mathcal{SB}_{1}$ is an extension of $(A,p)$ such that $q(x)$ is defined, then $t_{i-1}<{\rm{meet}}(x,a_{n})<t_{i}$ implies $t_{i}<{\rm{meet}}(q(x),a_{n})<t_{i+1}$, for $i=1,\ldots,n-2$, $t_{n-2}<{\rm{meet}}(x,a_{n})<t_{n-1}$ implies $t_{n-1}<{\rm{meet}}(q(x),a_{n})$, and $t_{n-1}<{\rm{meet}}(x,a_{n})$ implies $t_{n-1}<{\rm{meet}}(q(x),a_{n})$.

Claim 2. If $x\in{\rm Cone}_{O}$ is such that $x<_{lex}a_{0}$ and $(B,q)\in\mathcal{SB}_{1}$ is an extension of $(A,p)$ such that $q^{-1}(x)$ is defined, then ${\rm{meet}}(q^{-1}(x),a_{0}))<t_{0}$ (in particular, $q^{-1}(x)\notin\mbox{Cone}_{O}$).

A point $x\in A$ is locked by $O$ if for every extension $(B,q)$ of $(A,p)$ such that $q^{-1}(a_{0})$ and $q(a_{m})$ are defined, $x$ belongs to the $\leq_{lex}$-interval with endpoints $q^{-1}(a_{0})$ and $q(a_{m})$. It is locked by $p$ if it is locked by some orbit of $p$. This definition of locked, which we will use only in this section to discuss $\mathcal{SB}$, is slightly different than the one we used earlier in this section. A point $x\in A$ is cone-locked by the cone $C_{O}$, if it is contained in $C_{O}$, and it is locked or meet-locked by $O_{A}$. Finally, say that a point $x\in A$ is cone-locked by $p$ if it is cone-locked by some $C\in\mbox{Cone}(p)$.

Proof of Theorem 4.4.

Let us start with some observations. Take $(A,p)\in\mathcal{SB}_{1}$ such that every orbit has at least 3 points and let $a\in A$. If $a$ is not cone-locked by a cone from $\mbox{Cone}(p)$, then there are extensions $(B,q),(C,r)\in\mathcal{SB}_{1}$ of $(A,p)$ such that $q(a)<_{lex}a$ and $a<_{lex}r(a)$. To see this, simply take $v$, the immediate predecessor of $a$ in $T_{A}$ with respect to $<$, add a new point $b$ to obtain $B$ such that $b$ is the immediate predecessor of $a$ in $B$ with respect to $<_{lex}$, $v<{\rm{meet}}(b,a)<a$, and $q(a)=b$. The claim below shows that this gives the required $(B,q)$. We will then similarly define $(C,r)$.

Claim. The map $q$ preserves $S$

Proof of the Claim..

Since for every $x\in A$, ${\rm{meet}}(b,x)={\rm{meet}}(a,x)$, equivalently, we have to show the following.

$(\triangle)$ For every $a,b\in A$, we have $S(a,b,c)$ iff $S(p(a),p(b),c)$, and similarly, $S(a,c,b)$ iff $S(p(a),c,p(b))$ and $S(c,a,b)$ iff $S(c,p(a),p(b))$.

We have to consider a number of cases. Denote $m={\rm{meet}}(a,p(a))$, $n={\rm{meet}}(b,p(b))$, $m_{1}={\rm{meet}}(p(a),p^{2}(a))$ and $n_{1}={\rm{meet}}(p(b),p^{2}(b))$ (if necessary, extend $(A,p)$ so that $p^{2}(a)$ and $p^{2}(b)$ are defined). Let $O_{a}$ and $O_{b}$ be orbits to which $a$ and $b$ belong, respectively. Without loss of generality, suppose that $O_{a}$ is increasing and meet-increasing. We will frequently use the following simple observations:

(i) If $x<_{lex}a$ and $m<{\rm{meet}}(x,a)$ or $a<_{lex}x<_{lex}p^{2}(a)$, then $x$ is $O_{a}$ locked.

(ii) If $p(a)<_{lex}x$ and $m<{\rm{meet}}(x,p(a))<m_{1}$, then $x$ is meet-locked by $O_{a}$.

(iii) If $c<{\rm{meet}}(m,n)$ then $(\triangle)$ holds.

We have to consider the following cases.

1. (1)

It holds $b<_{lex}a$ and ${\rm{meet}}(b,a)<m$. In that case, we can have (a) $p(b)<_{lex}b$, or (b) $b<_{lex}p(b)$, and ${\rm{meet}}(p(b),a)<m$, or (c) $b<_{lex}p(b)$, and $m<{\rm{meet}}(p(b),a)$, in which case, by (i), $O_{a}$ and $O_{b}$ intertwine. 2. (2)

It holds $b<_{lex}a$ and $m<{\rm{meet}}(b,a)$, or $a<_{lex}b<_{lex}p(a)$, in which case, by (i), $O_{a}$ and $O_{b}$ intertwine. 3. (3)

It holds $p(a)<_{lex}b$ and $m<{\rm{meet}}(p(a),b)$. In this case, $p^{2}(a)<_{lex}p(b)$. We can have that:

(a) It holds $p(a)<_{lex}b<_{lex}p^{2}(a)$, in which case $O_{a}$ and $O_{b}$ intertwine.

(b) It holds $p^{2}(a)<_{lex}b$ and $m_{1}<{\rm{meet}}(b,p^{2}(a))$, in which case $p^{2}(a)<_{lex}p(b)$ and $m<{\rm{meet}}(p(b),p^{2}(a))$. In fact, we have $m_{1}<{\rm{meet}}(p(b),p^{2}(a))$, as otherwise $O_{b}$ would be an increasing orbit meet intertwining with $O_{a}$, which is impossible.

(c) It holds $p^{2}(a)<_{lex}b$ and $m<{\rm{meet}}(b,p^{2}(a))<m_{1}$, in which case, by (ii), $O_{a}$ and $O_{b}$ meet intertwine and hence $m_{1}<{\rm{meet}}(p(b),p^{2}(a))$. 4. (4)

It holds $p^{2}(a)<_{lex}b$ and ${\rm{meet}}(b,p^{2}(a))<m$. Then $p^{2}(a)<_{lex}p(b)$ and either ${\rm{meet}}(p(b),p^{2}(a))<m$ (with $b<_{lex}p(b)$ or $p(b)<_{lex}b$) or $m<{\rm{meet}}(p(b),p^{2}(a))<m_{1}$, in which case $O_{a}$ and $O_{b}$ meet intertwine.

This reduces checking to the following cases.

Case 1: $O_{a}$ and $O_{b}$ intertwine. Without loss of generality, $a<_{lex}b<_{lex}p(a)$ (meaning that, if instead $a<_{lex}p^{n}(b)<_{lex}p(a)$, for $n=1$ or $n=2$, then the reasoning will be essentially the same). This has two subcases: (1a) $m<{\rm{meet}}(a,b)$, in which case $m_{1}<{\rm{meet}}(p(a),p(b))$ and (1b) $m<{\rm{meet}}(b,p(a))$, in which case $m_{1}<{\rm{meet}}(p(b),p^{2}(a))$.

Taking into account (i), (ii) and (iii), all we have to do is to directly verify that $(\triangle)$ holds in (1a) and (1b) for a $c$ such that $p^{2}(a)<_{lex}c$ and $m_{1}<c$.

Case 2: $O_{a}$ and $O_{b}$ meet intertwine. Without loss of generality, $m<b<m_{1}$ (again, meaning that, if instead $m<p(b)<m_{1}$, then the reasoning will be essentially the same).

Taking into account (i), (ii) and (iii), all we have to do is to directly verify that $(\triangle)$ holds when $p^{2}(a)<_{lex}c<_{lex}p(b)$ and either $m_{1}<{\rm{meet}}(c,p^{2}(a))$ or $m_{1}<{\rm{meet}}(c,p(b))$.

Case 3: $b,p(b)<_{lex}a$ and ${\rm{meet}}(b,a),{\rm{meet}}(p(b),a)<m$.

Then possibilities on $c$ are: (3a) $b,p(b)<_{lex}c<_{lex}a$ and ${\rm{meet}}(c,a)<m$, (3b) $p(a)<_{lex}~{}c$ and $m<{\rm{meet}}(p(a),c)$, (3c) $p(a)<_{lex}c$ and ${\rm{meet}}(b,a),{\rm{meet}}(p(b),a)<{\rm{meet}}(p(a),c)<m$ (3d) $p(a)<_{lex}c$ and ${\rm{meet}}(p(a),c)$ is between ${\rm{meet}}(b,a)$ and ${\rm{meet}}(p(b),a)$ (this cannot happen though, otherwise $c$ would be meet-locked by $O_{b}$).

Case 4: $p(a)<_{lex}b,p(b)$ and ${\rm{meet}}(b,p(a)),{\rm{meet}}(p(b),p(a))<m$. This is very similar to Case 3.

Case 5: $p^{2}(a)<_{lex}b,p(b)$ and $m_{1}<{\rm{meet}}(b,p^{2}(a)),{\rm{meet}}(p(b),p^{2}(a))$. Let $p=\max\{{\rm{meet}}(b,p^{2}(a)),{\rm{meet}}(p(b),p^{2}(a))\}$. Then possibilities on $c$ are: (5a) $p^{2}(a)<_{lex}c<_{lex}b$ and $p<{\rm{meet}}(c,p^{2}(a))$, (5b) $p^{2}(a)<_{lex}c<_{lex}b$ and ${\rm{meet}}(c,p^{2}(a))\leq p$, (5c) $b,p(b)<_{lex}c$ and $m_{1}<{\rm{meet}}(c,p(b))<n$. ∎

Now we show that for a given $(A,p,q)\in\mathcal{SB}_{2}$ and $x\in A$ such that $x<p(x)$ there exists $(A^{\prime},p^{\prime},q^{\prime})\in\mathcal{SB}_{2}$, an extension of $(A,p,q)$, and a word $w(s,t)\in F_{2}$, such that $w(p^{\prime},q^{\prime})(x)$ is defined and not cone-locked by $p^{\prime}$ or $q^{\prime}$. Then an argument presented in the first paragraph of the proof of Proposition 4.1, will finish the proof. Without loss of generality, every non-trivial orbit of $p$ and $q$ consists of at least three points.

As for any $(A,p)\in\mathcal{SB}_{1}$, $A$ is a substructure of the Fraïssé limit $M$ of $\mathcal{SB}$, we consider

[TABLE]

Note that for every orbit $O$ of $p$, the set $\{x\in M\colon x\text{ is cone-locked by }O\}$ is the union of two $\leq_{lex}$-intervals, one of them constituted of points locked by $p$, and the other one of points meet-locked by $p$. This implies that $cl_{p}$ is the union of at most $2m_{p}$ disjoint $\leq_{lex}$-intervals, where $m_{p}$ is the number of orbits in $p$. Denote this collection of $\leq_{lex}$-intervals by $\mathcal{I}_{p}$, and its cardinality by $n_{p}$. Observe that the following hold:

$(*)$ For every $I\in\mathcal{I}_{p}$ and $x\in I$, there is an extension $(A^{\prime},p^{\prime})$ of $(A,p)$ so that $(p^{\prime})^{m}(x)<_{lex}I<_{lex}(p^{\prime})^{n}(x)$ for some $m,n\in\mathbb{Z}$.

$(**)$ For every $(A^{\prime},p^{\prime})\in\mathcal{SB}_{1}$ extending $(A,p)$ with $A^{\prime}\setminus A=\{(p^{\prime})^{\epsilon}(a),\ldots,(p^{\prime})^{\epsilon k}(a)\}$ for some $a\in A$, $\epsilon\in\{-1,1\}$, and $k\in\mathbb{N}$, and for every $I\in\mathcal{I}_{p^{\prime}}$, there is $J\in\mathcal{I}_{p}$ such that $J\subseteq I$. In particular, $n_{p^{\prime}}\leq n_{p}$.

We construct the required $(A^{\prime},p^{\prime},q^{\prime})$ and $w$ inductively. Let $(A^{\prime}_{0},p^{\prime}_{0},q^{\prime}_{0})=(A,p,q)$ and $w_{0}=1$. Suppose that we already constructed $(A^{\prime}_{n},p^{\prime}_{n},q^{\prime}_{n})$ and $w_{n}(s,t)$ and suppose that $w_{n-1}(p^{\prime}_{n-1},q^{\prime}_{n-1})(x)=w_{n-1}(p^{\prime}_{n},q^{\prime}_{n})(x)<_{lex}w_{n}(p^{\prime}_{n},q^{\prime}_{n})(x)$. Denote $y=w_{n}(p^{\prime}_{n},q^{\prime}_{n})(x)$ and if $y$ is cone-locked by $p^{\prime}_{n}$ and $q^{\prime}_{n}$, proceed as follows. Let $I_{p,y}\in\mathcal{I}_{p^{\prime}_{n}}$ be the $\leq_{lex}$-interval containing $y$, and similarly define $I_{q,y}$. If the right endpoint of $I_{p,y}$ is $<_{lex}$-greater or equal than the right endpoint of $I_{q,y}$, there must exist $k\in\mathbb{Z}$ such that $y<_{lex}(q^{\prime\prime})^{k}(y)\notin I_{p,y}$ in some extension $(A^{\prime\prime},p^{\prime\prime},q^{\prime\prime})$ of $(A^{\prime}_{n},p^{\prime}_{n},q^{\prime}_{n})$. Take the smallest such $k$, set $w_{n+1}(s,t)=t^{k}w_{n}(s,t)$ and let $(A^{\prime}_{n+1},p^{\prime}_{n+1},q^{\prime}_{n+1})$ be a minimal such extension. This can be done by observation $(*)$. Similarly, If the right endpoint of $I_{q,y}$ is $<_{lex}$-greater than the right endpoint of $I_{p,y}$, there must exist $k\in\mathbb{Z}$ such that, in some extension $(A^{\prime\prime},p^{\prime\prime},q^{\prime\prime})$, we have $y<_{lex}(q^{\prime\prime})^{k}(y)\notin I_{q,y}$. Take the smallest such $k$, set $w_{n+1}(s,t)=s^{k}w_{n}(s,t)$ and let $(A^{\prime}_{n+1},p^{\prime}_{n+1},q^{\prime}_{n+1})$ be a minimal such extension.

Observation $(**)$ implies that after at most $n_{p}+n_{q}$ many steps, this construction will stop, i.e., that for some $n\leq n_{p}+n_{q}$, we will have that $w_{n}(p^{\prime}_{n},q^{\prime}_{n})(x)$ is not cone-locked by $p^{\prime}_{n}$ or by $q^{\prime}_{n}$. ∎

Directed ultrahomogeneous graphs were classified by Cherlin [3] and their precompact Ramsey expansions were described by Jasiński-Laflamme-Nguyen van Thé-Woodrow [8]. See page 74 in [3] for the list of them, see also page 5 in [8]. By Kechris-Pestov-Todorcevic [12] and Nguyen Van Thé [14], automorphism groups of those expansions are extremely amenable. The notation we will use comes from [8].

Theorem 4.5**.**

Let $M$ be a precompact Ramsey expansion of a directed ultrahomogeneous graph, and let $\mathcal{F}={\rm{Age}}(M)$. Then $\mathcal{F}_{2}$ does not have WAP.

Sketch of the proof.

Proposition 4.1 applies to the age of each of these structures. A number of those structures are directly taken care of by Corollary 4.2. These are rational numbers and precompact Ramsey expansions of: the random tournament $T^{\omega}$, $\Gamma_{n}$– the random directed graph that does not embed the edgeless graph on $n$ vertices, $n\leq\omega$, $\mathcal{T}$ – the random directed graph that does not embed finite tournaments from some fixed set $\mathcal{T}$. Moreover, structures $S(2)^{*}$ and $S(3)^{*}$, that is, precompact Ramsey expansions of $S(2)$ and $S(3)$, are first-order simply bi-definable to structures $Q_{2}$ and $Q_{3}$, discussed by Nguyen van Thé [14], whose age is of the form as in Corollary 4.2.

Furthermore, proofs for the precompact Ramsey expansion of the structures of the form $T[I_{n}]$, $I_{n}[T]$, where $I_{n}$ is the edgeless graph on $n$ vertices, $n\leq\omega$, and $T$ is a homogeneous tournament, as well as of $\widehat{\mathbb{Q}}$, $\widehat{T^{\omega}}$ and of the complete $n$-partite random directed graph, $n\leq\omega$, are essentially the same as those for the structures taken care of by Corollary 4.2 (i.e. perhaps there are some additional unary predicates, which do not change the proof in an essential way). Let us discuss here one of these structures. We describe ${\rm Age}(T[I_{n}]^{*})$ of the expansion $T[I_{n}]^{*}$ of $T[I_{n}]$, where $T$ is a generic tournament and $I_{n}$ is the edgeless directed graph on $n<\omega$ vertices with the usual ordering (inherited from the natural numbers). Consider the language $L=\{E,<,L_{1},\ldots,L_{n}\}$, where $E,<$ are binary predicates and $L_{1},\ldots,L_{n}$ are unary predicates. We will use $E$ for the edge relation and $<$ for the linear order. We let the ${\rm Age}(T[I_{n}]^{*})$ to consist of substructures of structures whose universe is of the form $S\times I_{n}$, where $S$ is a linearly ordered tournament (the choice of a linear ordering is arbitrary). A pair $((x,i),(y,j))$ is an edge in $S\times I_{n}$ iff the pair $(x,y)$ is an edge in $S$. The ordering we put on $S\times I_{n}$ is lexicographic with respect to the order on $S$ and on $I_{n}$. Finally, we set $L_{i}(x,j)$ iff $i=j$. It is clear how to modify the proof from Corollary 4.2 to prove that assumptions of Proposition 4.1 are satisfied for ${\rm Age}(T[I_{n}]^{*})$ as well.

We have already discussed the universal ordered poset in Corollary 4.3. The precompact Ramsey expansion $\mathcal{P}(3)^{*}$ of the ‘twisted’ universal ordered poset $\mathcal{P}(3)$ is first-order simply bi-definable to the Fraïssé limit of the family $\mathcal{K}_{0}$ of ordered posets, additionally equipped with 3 subsets (described using unary predicates) forming a partition of the universe of the ordered poset, see the bottom of the page 21 in [8].

Proposition 4.1 also applies to the age of $\mathcal{S}^{*}$, the precompact Ramsey expansion of the semigeneric directed graph $\mathcal{S}$, which is rather straightforward to check. In fact, for given $(A,p)$ and $a\in A$ not locked by $p$, the required $q(a)$ and $r(a)$ (notation taken from the statement of Proposition 4.1) can be chosen in the same equivalence class with respect to the non-edge equivalence relation in which $a$ is.

∎

5. The two-dimensional case. Similarity classes

Slutsky [17] showed that every $2$-dimensional topological similarity class in $\mbox{Aut}(\mathbbm{Q})$ is meager. In this section, we extract from Slutsky’s arguments a general condition on a structure $M$ that implies that every $2$-dimensional topological similarity class in $\mbox{Aut}(M)$ is meager (Theorem 5.5).

Let $\mathcal{F}$ be a Fraïssé class. Generalizing the terminology introduced in [17], for $A,B\in\mathcal{F}$ with $B\subseteq A$, and a partial automorphism $q$ of $A$ such that ${\rm{def}}(q)\cap B=\emptyset$, we say that $B$ is free from $q$ if for every $n$, every relation symbol $R$ in the signature of $\mathcal{F}$ of arity $n$, for all $x_{1},\ldots,x_{n}\in B\cup{\rm{dom}}(q)$ we have $R(x_{1},\ldots,x_{n})$ iff $R(y_{1},\ldots,y_{n})$, where $y_{i}=x_{i}$ if $x_{i}\in B$, and $y_{i}=q(x_{i})$ if $x_{i}\in{\rm{dom}}(q)$. In other words, we can extend $q$ so that $q(x)=x$ for every $x\in B$.

We say that $\mathcal{F}$ has liberating automorphisms if for any partial automorphisms $p,q$ of $A\in\mathcal{F}$ with no cyclic orbits there exists $N\in\mathbb{N}$ such that, for every $N^{\prime}>N$, $p$ can be extended to a partial automorphism $p^{\prime}$ of an element of $\mathcal{F}$ so that $(p^{\prime})^{n}[A]$ is free from $q$ for all $n$ with $N\leq n\leq N^{\prime}$.

Let $\mathcal{F}$ be an Fraïssé order class with an order relation $<$, and let $M$ be the limit of $\mathcal{F}$. Let $p$ be a partial automorphism of $M$. For a convex $A\subseteq M$ (i.e., $x,y\in A$, and $x<z<y$ entails that $z\in A$), we say that $p$ is increasing on $A$ if for every $x\in A$, $p$ can be extended so that $p(x)>x$; it is decreasing on $A$ if for every $x\in A$, $p$ can be extended so that $p(x)<x$; it is monotone on $A$ if it is increasing or decreasing on $A$. We say that an extension $p^{\prime}$ of $p$ does not change monotonicity of $p$ if there are no new fixed points in $p^{\prime}$, and $p^{\prime}$ is increasing (decreasing) on $(a,b)$ iff $p$ is increasing (decreasing) on $(a,b)$, for $a,b\in{\rm{def}}(p)$. We say that $p$ is eventually increasing if there exist $x,y\in{\rm{supp}}(p)$ such that $z<p(z)$ for every $z\in{\rm{supp}}(p)$ such that either $z\leq x$ or $y\leq z$ (i.e., the first and the last orbits of $p$ are increasing.)

Let $(p,q)$ be a pair of partial automorphisms of $M$. We say that $x$ is in a final segment (or initial segment) of $(p,q)$ if there exists a common fixed point $y$ of $p$ and $q$ such that $p$ is monotone on $[x,y)$ (or $(y,x]$.) We say that $(p,q)$ is elementary if both $p$ and $q$ are eventually increasing, and the only common fixed points of $p$ and $q$ are the minimum $\min({\rm{dom}}(p))=\min({\rm{dom}}(q))$, and the maximum $\max({\rm{dom}}(p))=\max({\rm{dom}}(q))$ of their domains. A pair $(p,q)$ is piecewise elementary if we can find $E_{0}\leq\ldots\leq E_{n}$, called elementary components of $(p,q)$, such that $\bigcup_{i}E_{i}={\rm{def}}(p)\cup{\rm{def}}(q)$, and $(p,q)$ is elementary when restricted to each $E_{i}$.

Lemma 5.1**.**

Let $\mathcal{F}$ be a full order expansion with SAP. Let $(p,q)$ be an elementary pair. Then there exists an extension $(p^{\prime},q^{\prime})$ of $(p,q)$, and $w\in F(s,t)$, such that $p^{\prime}$ does not change monotonicity of $p$, and $w(p^{\prime},q^{\prime})[{\rm{def}}(q^{\prime})]$ is in the unique final segment of $(p^{\prime},q^{\prime})$.

Proof.

Without loss of generality, we can assume that $\min({\rm{supp}}(p))<{\rm{supp}}(q)$. Let $a_{0}<\ldots<a_{m}$ be the enumeration of ${\rm{def}}(p)\setminus\{\min({\rm{def}}(p)),\max({\rm{def}}(p))\}$. We can assume that $a_{0}$ is not a fixed point of $p$.

We construct $w_{i}\in F(s,t)$, $i\leq m$, and extensions $(p_{i},q_{i})$ of $(p_{i-1},q_{i-1})$ such that $w_{i}\ldots w_{0}(p_{i},q_{i})(a_{0})>a_{i}$. Moreover, we require that the only new element in ${\rm{def}}(p_{i})$ or ${\rm{def}}(q_{i})$ above $a_{i}$ is $w_{i}\ldots w_{0}(p_{i},q_{i})(a_{0})$.

Put $w_{-1}=\emptyset$, $p_{-1}=p$, $q_{-1}=q$. Fix $0\leq i\leq m$, and suppose that $w_{j},p_{j},q_{j}$ have been already constructed for $j<i$. Set $b=w_{i-1}\ldots w_{0}(p_{i-1},q_{i-1})(a_{0})$. If there is an extension $p_{i}$ of $p_{i-1}$ such that $(p_{i})^{\epsilon k}(b)>a_{i}$ for some $k\in\mathbb{N}$, and $\epsilon\in\{-1,1\}$, we take the least such $k$, and put $w_{i}=s^{\epsilon k}$, $q_{i}=q_{i-1}$.

Suppose otherwise. If $a_{i+1}$ is not a fixed point of $q$, put $c=a_{i}$, and $l=0$. Otherwise, as $(p_{i-1},q_{i-1})$ is elementary, it is not a fixed point of $p_{i-1}$, and so we can extend $p_{i-1}$ to some $p_{i}$ by adding only elements below $a_{i}$, so that, for some $l\in\mathbb{Z}$, $b<(p_{i})^{l}(a_{i})<a_{i}$, and $(b,(p_{i})^{l}(a_{i}))$ has empty intersection with both ${\rm{def}}(p_{i-1})$ and ${\rm{def}}(q_{i-1})$. Put $c=(p_{i})^{l}(a_{i})$.

Let $\epsilon\in\{-1,1\}$ be such that there exists an extension $q_{i}$ of $q_{i-1}$ with $(q_{i})^{\epsilon}(c)<c$. Because $\mathcal{F}$ is a full order expansion with SAP, there is an extension of $p_{i}$, which we will also denote by $p_{i}$, such that $p_{i}(b)\in((q_{i})^{\epsilon}(c),c)$. But then $(q_{i})^{-\epsilon}(p_{i}(b))>c$. Thus, for $w_{i}=s^{-l}t^{-\epsilon}s$, we have that $w_{i}\ldots w_{0}(p_{i},q_{i})(a_{0})>a_{i}$ and $w_{i}\ldots w_{0}(p_{i},q_{i})(a_{0})$ is the only new element in ${\rm{def}}(p_{i})$ or ${\rm{def}}(q_{i})$ above $a_{i}$.

∎

Lemma 5.2**.**

Let $\mathcal{F}$ be a full order expansion with SAP, and such that $\mathcal{F}^{-}$ has liberating automorphisms. Let $M$ be the limit of $\mathcal{F}$. Let $(p,q)$ be a piecewise elementary pair of partial automorphisms of $M$ such that, for some $w\in F(s,t)$, $w(p,q)(x)$ is in a final segment of $(p,q)$ for every $x\in{\rm{def}}(q)$. Then for any $v\in F(s,t)$, and $N\in\mathbb{N}$, there is $N^{\prime}\geq N$, and a pair $(p^{\prime},q^{\prime})$ extending $(p,q)$ such that $vs^{N^{\prime}}w(p^{\prime},q^{\prime})(x)$ is in a final segment of $(p,q)$ for $x\in{\rm{def}}(q^{\prime})$.

Proof.

Let $E_{0}\leq\ldots\leq E_{n}\subseteq M$ be elementary components of $(p,q)$. Because $w(p,q)(x)$ is in a final segment of $(p,q)$ for every $x\in{\rm{def}}(q)$, and $p$ is eventually increasing on each $E_{i}$, we can assume that there is $N_{0}\in\mathbb{N}$ such that

[TABLE]

for every $i\leq n$, and $x\in{\rm{supp}}(q)\cap E_{i}$. Then

[TABLE]

for every $x,y\in{\rm{def}}(q)$.

Because $\mathcal{F}^{-}$ has liberating automorphisms, and $\mathcal{F}$ is a full order expansion with SAP, we can find $N_{1}\in\mathbb{N}$, and an extension $p^{\prime}$ of $p$ such that $s^{N_{0}+N_{1}+n}w(p^{\prime},q)[{\rm{def}}(q)]$ is free from $q$ in $\mathcal{F}$ for $n\leq 2\left|v\right|+N$, and (3) still holds for ${\rm{def}}(q)$. But this means that $s^{N_{0}+N_{1}+n}w(p^{\prime},q)[{\rm{def}}(q)]$ is free from $q$ in $\mathcal{F}$, and we can extend $q$ to $q^{\prime}$ so that

[TABLE]

for $n\leq 2\left|v\right|+N$ and $x\in{\rm{def}}(q)$. It is easy to see that then

[TABLE]

for $n\leq N$, and so $vs^{N_{0}+N_{1}+\left|v\right|+N}w(p^{\prime},q^{\prime})(x)$ is in a final segment of $p^{\prime}$, for $x\in{\rm{def}}(q^{\prime})$. Therefore $N^{\prime}=N_{0}+N_{1}+\left|v\right|+N$ is as required. ∎

Lemma 5.3**.**

Let $\mathcal{F}$ be a full order expansion with SAP, and such that $\mathcal{F}^{-}$ has liberating automorphisms. Then for every piecewise elementary pair $(p,q)$ there exists a piecewise elementary pair $(p^{\prime},q^{\prime})$ extending $(p,q)$, and $w\in F(s,t)$ such that $w(p^{\prime},q^{\prime})(x)$ is in a final segment of $(p^{\prime},q^{\prime})$ for $x\in{\rm{def}}(q^{\prime})$.

Proof.

We prove the lemma by induction on the number $r$ of elementary components of $(p,q)$. For $r=1$, this follows from Lemma 5.1. Suppose that that the lemma is true for some $r$, and fix a piecewise elementary pair $(p,q)$ with $r+1$ elementary components.

Let us write $E=E_{0}\cup E_{1}$ so that $E_{0}\leq E_{1}$, $(p,q)$ is elementary when restricted to $E_{0}$, and there are $r$ elementary components in $(p,q)$ when restricted to $E_{1}$. Using Lemma 5.1, and the inductive assumption, we can fix an extension $(p^{\prime},q^{\prime})$ of $(p,q)$, so that, for $(p_{0},q_{0})$ denoting the restriction of $(p^{\prime},q^{\prime})$ to $E_{0}$, and $(p_{1},q_{1})$ denoting the restriction of $(p^{\prime},q^{\prime})$ to $E_{1}$, the following holds. The mapping $p_{0}$ does not change monotonicity of $p$ restricted to $E_{0}$, the pair $(p_{1},q_{1})$ is piecewise elementary, and there exist $w_{0},w_{1}\in F(s,t)$ such that $w_{0}(p_{0},q_{0})(x)$ is in the unique final segment of $(p_{0},q_{0})$ for $x\in{\rm{def}}(q_{0})$, and $w_{1}(p_{1},q_{1})(x)$ is in a final segment of $(p_{1},q_{1})$ for $x\in{\rm{def}}(q_{1})$.

Let $d_{0}=\min({\rm{supp}}(p_{0}))$. As $p_{0}$ does not change monotonicity of $p$ restricted to $E_{0}$, and so $p_{0}$ is increasing on the initial segment of $p_{0}$, in the case that $w_{1}(p_{0},q_{0})(d_{0})<d_{0}$, we can assume that there exists $N\in\mathbb{N}$ such that $s^{N}w_{1}(p_{0},q_{0})(d_{0})>d_{0}$. Then $w_{0}s^{N}w_{1}(p_{0},q_{0})(x)$ is in the final segment of $p_{0}$ for $x\in{\rm{def}}(q_{0})$.

Moreover, applying Lemma 5.2, we can assume that $N$ is large enough so that $w_{0}s^{N}w_{1}(p_{1},q_{1})(x)$ is in a final segment of $p_{1}$ for $x\in{\rm{def}}(q_{1})$. Thus, the pair $(p^{\prime},q^{\prime})$ extends $(p,q)$, and, $w(p^{\prime},q^{\prime})(x)$ is in a final segment of $p^{\prime}$ for $x\in{\rm{def}}(q^{\prime})$, if $w=w_{0}s^{N}w_{1}$.∎

Theorem 5.4**.**

Let $\mathcal{F}$ be a full order expansion with SAP, and such that $\mathcal{F}^{-}$ has liberating automorphisms. Let $M$ be the limit of $\mathcal{F}$. Then there are comeagerly many pairs in $\mbox{Aut}(M)^{2}$ generating a non-discrete group.

Proof.

Consider the following condition: for every pair $(p,q)$ of partial automorphisms of $M$ there exists an extension $(p^{\prime},q^{\prime})$ of $(p,q)$, and $w\in F_{2}$ such that $w(p^{\prime},q^{\prime})(x)=x$ for $x\in{\rm{def}}(q)$. It is easy to verify that if it holds, then the set of pairs $(f,g)\in\mbox{Aut}(M)^{2}$ generating a non-discrete group contains a dense $G_{\delta}$ subset of $\mbox{Aut}(M)^{2}$, that is, it is comeager in $\mbox{Aut}(M)^{2}$.

We verify this condition. Fix a pair $(p,q)$ of partial automorphisms of $M$. Without loss of generality, we can assume that it is piecewise elementary. By Lemma 5.3, we can also assume that there exists $w^{\prime}\in F(s,t)$ such that $w^{\prime}(p,q)(x)$ is in a final segment of $p$ for $x\in{\rm{def}}(q)$. But then, using our assumptions on $\mathcal{F}$, and the fact that $p$ is eventually increasing on each elementary component of $(p,q)$, we can find $N\in\mathbb{N}$, and an extension $(p^{\prime},q^{\prime})$ of $(p,q)$ such that

[TABLE]

for every $x,y\in{\rm{def}}(q^{\prime})$, and $s^{N}w^{\prime}(p^{\prime},q^{\prime})[{\rm{def}}(q^{\prime})]$ is free from $q^{\prime}$ in $\mathcal{F}$. Therefore we can put

[TABLE]

for $x\in{\rm{def}}(q^{\prime})$. Then for $w=(s^{N}w)^{-1}t(s^{N}w)$, we have $w(p^{\prime},q^{\prime})(x)=x$ for $x\in{\rm{def}}(q^{\prime})$.

∎

Theorem 5.5**.**

Let $\mathcal{F}$ be a full order expansion with SAP, and such that $\mathcal{F}^{-}$ has liberating automorphisms. Let $M$ be the limit of $\mathcal{F}$. Then every $2$-dimensional topological similarity class in $\mbox{Aut}(M)$ is meager.

Proof.

By the above theorem, there are comeagerly many pairs in $\mbox{Aut}(M)^{2}$ generating a non-discrete group. As automorphisms of order structures have only infinite non-trivial orbits, in fact, there are comeagerly many pairs in $\mbox{Aut}(M)^{2}$ generating a non-discrete and non-precompact group. By [11, Theorem 4.4], every $2$-dimensional topological similarity class in $\mbox{Aut}(M)$ is meager. ∎

Recall that a Fraïssé class $\mathcal{F}$ has free amalgamation if for every $A,B,C\in\mathcal{F}$ with $A\subseteq B,C$, the structure $D=B\cup C$ is an amalgam of $B$ and $C$ over $A$. In other words, no tuple in $D$ involving at the same time elements from $B\setminus A$ and from $C\setminus A$ is related in $D$. A typical example of a class with free amalgamation is the class of finite graphs.

Lemma 5.6**.**

If $\mathcal{F}$ is a Fraïssé class with free amalgamation or the class of finite tournaments, then $\mathcal{F}$ has liberating automorphisms.

Proof.

Suppose that $\mathcal{F}$ has free amalgamation. Let $(p,q)$ be a pair of partial automorphisms of $A\in\mathcal{F}$ with no cyclic orbits. Without loss of generality, we can assume that $A={\rm{def}}(p)$. Let $N\in\mathbb{N}$ be such that orbits in $p$ have size at most $N/2$. Set $C={\rm{dom}}(p)\setminus{\rm{rng}}(p)$, and fix $N^{\prime}>N$. We put $p_{0}=p$, and construct partial automorphisms $p_{i}$ and sets $D_{i}$, $0<i\leq N^{\prime}$ in the following manner. Assuming that $p_{i}$ is already constructed, with an aid of free amalgamation, we extend $p_{i}$ to a partial automorphism $p_{i+1}$ by defining it on $D^{\prime}={\rm{rng}}(p_{i})\setminus{\rm{dom}}(p_{i})$ in such a way that no relation involves at the same time elements from $C$ and $D=p_{i+1}[D^{\prime}]$ (where $D$ is disjoint from ${\rm{def}}(p_{i})$.) To be more precise, put $B={\rm{rng}}(p_{i})$. Then, for every relation $R$ of arity $n$, and every $n$-tuple $\bar{b}$ in $B\cup D$, whether $R(\bar{a})$ holds or not, is entirely determined by the requirement that $p_{i+1}$ is supposed to be a partial automorphism. Moreover, regardless of how we amalgamate $B\cup C$ and $B\cup D$ over $B$, to get a structure $E$ with underlying set $B\cup C\cup D$, $p_{i+1}$ will be a partial automorphism of $E$. Thus, $E$ obtained by freely amalgamating these structures works as ${\rm{def}}(p_{i+1})$. Finally, we put $D_{i+1}=D$.

Observe that no relation involves at the same time elements from $p_{N^{\prime}}^{n}[A]$ and $A$ for $N\leq n\leq N^{\prime}$, which means, because ${\rm{def}}(q)\subseteq A$, that each $p_{N^{\prime}}^{n}[A]$ is free from $q$. Indeed, by the construction of $p_{i}$, for $i>0$ and $x\in D_{i}$, no relation involves $x$ and elements from $C$. And then the same is true about any $p^{n}_{i+1}(x)$ and $p^{i}_{i+1}[C]$, $i\leq n\leq N^{\prime}$. As $p_{N^{\prime}}^{n}[A]\subseteq\bigcup_{N\leq i\leq N^{\prime}}D_{i}$, and $A\subseteq\bigcup_{i\leq N}p^{i}[C]$, this means that no relation involves at the same time elements from $p_{N^{\prime}}^{n}[A]$ and $A$, for $N\leq n\leq N^{\prime}$.

For finite tournaments, we proceed almost exactly as above. The only difference is that for every $x\in{\rm{dom}}(p_{i})\setminus{\rm{rng}}(p_{i})$, $y\in{\rm{rng}}(p_{i})\setminus{\rm{dom}}(p_{i})$, we choose $(x,y)$ as the arrow between $x$ and $y$. Then $(x,y)$ is an arrow for every $x\in{\rm{def}}(q)$ and $y\in p_{N^{\prime}}^{n}[{\rm{def}}(q)]$, $N\leq n\leq N^{\prime}$. ∎

Corollary 5.7**.**

Suppose that $\mathcal{F}$ is a full order expansion such that $\mathcal{F}^{-}$ has free amalgamation, or the class of finite ordered tournaments. Let $M$ be the Fraïssé limit of $\mathcal{F}$. Then every $2$-dimensional topological similarity class in $\mbox{Aut}(M)$ is meager.

Corollary 5.8**.**

Suppose that $\mathcal{F}$ is a full order expansion of a class with free amalgamation, and let $M$ be the Fraïssé limit of $\mathcal{F}$. Then every $2$-dimensional topological similarity class in $\mbox{Aut}(M)$ is meager.

Remark 5.9**.**

Compare the above corollary with Theorem 3.5 and Corollary 4.2.