# Is there a Fraisse limit whose automorphism group contains dense but not generic automorphisms?

It is well known that $\mathsf{Aut}(\mathbb{Q},<)$ has generic automorphisms (i.e., a comeagre conjugacy class under the diagonal action) but does not admit ample generics. The automorphism group $\mathsf{Aut}(\omega,E)$ of the structure $(\omega,E)$ where $nEm \iff n = m \mod{2}$ does not even have a dense conjugacy class. I am looking for an example 'in the midst' of the previous two, i.e., an automorphism group of a Fraisse limit that has a dense conjugacy class but does not have a generic automorphism.

I know that there are examples of Polish groups that have a dense conjugacy but do not have a generic automorphism. However, it is unknown to me or impossible to construct these examples as automorphism groups of Fraisse limits.

The first example that I know of is $\mathsf{Aut}(X,\mu)$ where $(X,\mu)$ is the standard probability space, i.e., a standard Borel space $X$ with a probability measure $\mu$. We cannot express $(X,\mu)$ as a Fraisse limit since this space is not countable. Another example of a Polish group that has dense automorphisms is $U(H)$, that is, the unitary group of the infinite dimensional Hilbert space. However, the last example is not even constructed as an automorphism group.

Is there an easy example of a Fraisse limit whose automorphism group has a dense conjugacy class but not a comeagre one?

• Could you add some details on the examples you mention (in the second paragraph)? – Primo Petri Jun 10 '16 at 18:05
• @PrimoPetri I just added some details on the examples. – namsap Jun 11 '16 at 9:08

I assume you're familiar with the paper Turbulence, amalgamation, and generic automorphisms of homogeneous structures by Kechris and Rosendal. If not, you should have a look, since it's about exactly these issues.

In the language of that paper, we take a Fraïssé class $K$ and expand it to a class $K_p$ consisting of pairs $(A,\varphi)$, where $A\in K$ and $\varphi$ is a partial automorphism of $A$ (i.e. an isomorphism between two substructures of $A$). Then an embedding $(A,\varphi)\to (B,\psi)$ is an embedding $A\to B$ such that $\psi$ extends the image of $\varphi$. If $M$ is the Fraïssé limit of $K$, Kechris and Rosendal show that $\mathrm{Aut}(M)$ has a dense conjugacy class if and only if $K_p$ has the joint embedding property (JEP), and $\mathrm{Aut}(M)$ has a comeager conjugacy class if and only if $K_p$ has the weak amalgamation property (WAP) and the JEP.

For a category of structures and embeddings, the WAP is the assertion that every $A$, there is an embedding $f\colon A\to B$, such that for any embeddings $g_1\colon B\to C_1$ and $g_2\colon B\to C_2$, there are embeddings $h_1\colon C_1\to D$ and $h_2\colon C_2\to D$, such that $h_1\circ g_1\circ f = h_2\circ g_2\circ f$.

So to answer your question, we just need an example of a Fraïssé class $K$ such that $K_p$ has the JEP but not the WAP. (Edit: I've realized that there's a much simpler example, so I've replaced my original example).

Let $K$ be the class of finite ordered graphs in the language $\{\leq,R\}$. Then $K$ is a Fraïssé class. Its Fraïssé limit is the ordered random graph.

$K_p$ has the JEP: Given $(A,\varphi)$ and $(B,\psi)$, embed them in $(C,\theta)$, where $C$ is the disjoint union of $A$ and $B$, $\theta$ is the union of $\varphi$ and $\psi$, all the elements of $B$ are greater than all the elements of $A$ in the order, and no new edge relations hold.

But $K_p$ does not have the WAP: Let $(A,\varphi)$ be any structure in $K_p$ such that there is some $a\in A$ such that $\varphi(a)$ is defined and $\varphi(a)> a$. Suppose $(A,\varphi)$ embeds in $(B,\psi)$. We will show that $(B,\psi)$ cannot witness the WAP for $(A,\varphi)$. Note that the sequence $\dots, \psi^{-1}(a), a, \psi(a), \psi^2(a), \dots$ is strictly increasing, so, since $B$ is finite, there is some largest $m\geq 0$ such that $b = \psi^{-m}(a)$ is defined, and there is some largest $n\geq 1$ such that $e = \psi^{n}(a)$ is defined. We will define two structures $(C_1,\theta_1)$ and $(C_2,\theta_2)$ into which $(B,\psi)$ embeds.

In each, we will add a single new element $c$ and extend $\psi$ to $\theta_i$ so that $\theta_i(e) = c$. For all $d$ such that $\psi(d)$ is defined, we set $\psi(d) < c$ if and only if $d < e$ and $\psi(d)Rc$ if and only if $dRe$. This is enough to ensure that the $\theta_i$ are partial automorphisms. Now in $C_1$ we set $bRc$, and in $C_2$ we set $\lnot bRc$. This is consistent with the assignments above, since $b$ does not have a preimage under $\psi$. We may decide the rest of the order and edge relations between $c$ and elements of $B$ arbitrarily (but consistently with the assignments above).

If $(C_1,\theta_1)$ and $(C_2,\theta_2)$ embed in some $(D,\rho)$ over $(A,\varphi)$, the images of $b = \theta_i^{-m}(a)$ and $c= \theta_i^{n+1}(a)$ must be equal, but $C_1$ and $C_2$ disagree about the truth of $bRc$, so we have a contradiction.

1. In fact, this argument shows that $K_p$ does not even have the local weak amalgamation property (i.e. there is no $A\in K_p$ such that the weak amalgamation property holds for the class of structures into which $A$ embeds), so the automorphism group of the ordered random graph does not even have any non-meager conjugacy classes.