\documentclass[../lic_malinka.tex]{subfiles} \begin{document} Let $M$ be a countable $L$-structure. Recall, we define a topology on the $G=\Aut(M)$: for any finite function $f\colon M\to M$ we have a basic open set $[f]_{G} = \{g\in G\mid f\subseteq g\}$. \subsection{Prototype: pure set} In this section, $M=(M,=)$ is an infinite countable set (with no structure beyond equality). \begin{remark} \label{remark:cojugate-classes} If $f_1,f_2\in \Aut(M)$, then $f_1$ and $f_2$ are conjugate if and only if for each $n\in \bN\cup \{\aleph_0\}$, $f_1$ and $f_2$ have the same number of orbits of size $n$. \end{remark} \begin{proof} It is easy to see. \end{proof} \begin{theorem} Let $\sigma\in \Aut(M)$ be an automorphism with no infinite orbit and with infinitely many orbits of size $n$ for every $n>0$. Then the conjugacy class of $\sigma$ is comeagre in $\Aut(M)$. \end{theorem} \begin{proof} We will show that the conjugacy class of $\sigma$ is an intersection of countably many comeagre sets. Let $A_n = \{\alpha\in \Aut(M)\mid \alpha\text{ has infinitely many orbits of size }n\}$. This set is comeagre for every $n>0$. Indeed, we can represent this set as an intersection of countable family of open dense sets. Let $B_{n,k}$ be the set of all finite functions $\beta\colon M\to M$ that consist of exactly $k$ distinct $n$-cycles. Then: \begin{align*} A_n &= \{\alpha\in\ \Aut(M) \mid \alpha\text{ has infinitely many orbits of size }n\} \\ &= \bigcap_{k=1}^{\infty} \{\alpha\in \Aut(M)\mid \alpha\text{ has at least }k\text{ orbits of size }n\} \\ &= \bigcap_{k=1}^{\infty} \bigcup_{\beta\in B_{n,k}} [\beta]_{\Aut(M)}, \end{align*} where indeed, $\bigcup_{\beta\in B_{n,k}} [\beta]_{\Aut(M)}$ is dense in $\Aut(M)$: take any finite $\gamma\colon M\to M$ such that $[\gamma]_{\Aut(M)}$ is nonempty. Then also $\bigcup_{\beta\in B_{n,k}} [\beta]_{\Aut(M)}\cap[\gamma]_{\Aut(M)}\neq\emptyset$, one can easily construct a permutation that extends $\gamma$ and has at least $k$ many $n$-cycles. Now we see that $A = \bigcap_{n=1}^{\infty} A_n$ is a comeagre set consisting of all functions that have infinitely many $n$-cycles for each $n$. The only thing left to show is that the set of functions with no infinite cycle is also comeagre. Indeed, for $m\in M$ let $B_m = \{\alpha\in\Aut(M)\mid m\text{ has finite orbit in }\alpha\}$. This is an open dense set. It is a union over basic open sets generated by finite permutations with $m$ in their domain. Denseness is also easy to see. Finally, by the Remark \ref{remark:cojugate-classes}, we can say that $$\sigma^{\Aut(M)}=\bigcap_{n=1}^\infty A_n \cap \bigcap_{m\in M} B_m,$$ which concludes the proof. \end{proof} \subsection{More general structures} \begin{fact} \label{fact:conjugacy} Suppose $M$ is an arbitrary structure and $f_1,f_2\in \Aut(M)$. Then $f_1$ and $f_2$ are conjugate if and only if $(M,f_1)\cong (M,f_2)$ as structures with one additional unary function that is an automorphism. \end{fact} \begin{proof} Suppose that $f_1 = g^{-1}f_2g$ for some $g\in \Aut(M)$. Then $g$ is the isomorphism between $(M,f_1)$ and $(M,f_2)$. On the other hand if $g\colon (M, f_1)\to (M, f_2)$ is an isomorphism, then $g\circ f_1 = f_2 \circ g$ which exactly means that $f_1, f_2$ conjugate. \end{proof} \begin{theorem} \label{theorem:generic_aut_general} Let $\cC$ be a Fraïssé class of finitely generated $L$-structures. Let $\cD$ be the class of structures from $\cC$ with additional unary function symbol interpreted as an automorphism of the structure. If $\cC$ has the weak Hrushovski property and $\cD$ is a Fraïssé class, then there is a comeagre conjugacy class in the automorphism group of the $\Flim(\cC)$. \end{theorem} Before we get to the proof, it is important to mention that an isomorphism between two finitely generated structures is uniquely given by a map from generators of one structure to the other. This allow us to treat a finite function as an isomorphism of finitely generated structures (if it yields one) and \textit{vice versa}. \begin{proof} Let $\Gamma = \Flim(\cC)$ and $(\Pi, \sigma) = \Flim(\cD)$. First, by the Theorem \ref{theorem:isomorphic_fr_lims}, we may assume without the loss of generality that $\Pi = \Gamma$. Let $G = \Aut(\Gamma)$, i.e. $G$ is the automorphism group of $\Gamma$. We will construct a winning strategy for the second player in the Banach-Mazur game (see \ref{definition:banach-mazur-game}) on the topological space $G$ with $A$ being $\sigma$'s conjugacy class. By the Banach-Mazur theorem (see \ref{theorem:banach_mazur_thm}) this will prove that this class is comeagre. Recall, $G$ has a basis consisting of sets $\{g\in G\mid g\upharpoonright_A = g_0\upharpoonright_A\}$ for some finite set $A\subseteq \Gamma$ and some automorphism $g_0\in G$. In other words, a basic open set is a set of all extensions of some partial automorphism $g_0$ of finitely generated substructures of $\Gamma$. By $B_{g}\subseteq G$ we denote a basic open subset given by a partial isomorphism $g$. Again, Note that $B_g$ is nonempty because of ultrahomogeneity of $\Gamma$. With the use of Corollary \ref{corollary:banach-mazur-basis} we can consider only games where both players choose partial isomorphisms. Namely, player \textit{I} picks functions $f_0, f_1,\ldots$ and player \textit{II} chooses $g_0, g_1,\ldots$ such that $f_0\subseteq g_0\subseteq f_1\subseteq g_1\subseteq\ldots$, which identify the corresponding basic open subsets $B_{f_0}\supseteq B_{g_0}\supseteq\ldots$. Our goal is to choose $g_i$ in such a manner that $\bigcap^{\infty}_{i=0}B_{g_i} = \{g\}$ and $(\Gamma, g)$ is ultrahomogeneous with age $\cD$. By the Fraïssé theorem (see \ref{theorem:fraisse_thm}) it will follow that $(\Gamma, \sigma)\cong (\Gamma, g)$, thus by the Fact \ref{fact:conjugacy} we have that $\sigma$ and $g$ conjugate. Fix a bijection $\gamma\colon\bN\times\bN\to\bN$ such that for any $n, m\in\bN$ we have $\gamma(n, m) \ge n$. This bijection naturally induces a well ordering on $\bN\times\bN$. This will prove useful later, as the main ingredient of the proof will be a bookkeeping argument. For technical reasons, let $g_{-1} = \emptyset$ and $X_{-1} = \emptyset$. Enumerate the elements of the Fraïssé limit $\Gamma = \{v_0, v_1, \ldots\}$. Suppose that player \textit{I} in the $n$-th move chooses a partial automorphism $f_n$. We will construct a partial automorphism $g_n\supseteq f_n$ together with a finitely generated substructure $\Gamma_n \subseteq \Gamma$ and a set $X_n\subseteq\bN^2$ such that the following properties hold: \begin{enumerate}[label=(\roman*)] \item $g_n$ is a partial automorphism of $\Gamma$ and an automorphism of finitely generated substructure $\Gamma_n$, \item $g_n(v_n)$ and $g_n^{-1}(v_n)$ are defined. \end{enumerate} Before we give the third point, suppose recursively that $g_{n-1}$ already satisfy all those properties. Let us enumerate $\{\langle (A_{n,k}, \alpha_{n, k}), (B_{n,k}, \beta_{n,k}), f_{n, k}\rangle\}_{k\in\bN}$ all pairs of finitely generated structures with automorphisms such that the first substructure embed into the second by inclusion, i.e. $(A_{n,k}, \alpha_{n,k})\subseteq (B_{n,k}, \beta_{n,k})$, and $f_{n,k}$ is an embedding of $(A_{n,k}, \alpha_{n,k})$ in the $(\FrGr_{n-1}, g_{n-1})$. We allow $A_{n,k}$ to be empty. Although $g_{n-1}$ is a finite function, we may treat it as a partial automorphism as we have said before. \begin{enumerate}[resume, label=(\roman*)] \item Let $(i, j) = \min\{(\{0, 1, \ldots, n\} \times \bN) \setminus X_{n-1}\}$ (with the order induced by $\gamma$). Then $X_n = X_{n-1}\cup\{(i,j)\}$ and $(B_{i,j}, \beta_{i,j})$ embeds in $(\FrGr_n, g_n)$ so that this diagram commutes: \begin{center} \begin{tikzcd} & (\Gamma_n, g_n) & \\ (B_{i,j}, \beta_{i,j}) \arrow[ur, dashed, "\hat{f}_{i,j}"] & & (\FrGr_{n-1}, g_{n-1}) \arrow[ul, dashed, "\subseteq"'] \\ & (A_{i,j}, \alpha_{i,j}) \arrow[ur, "f_{i,j}"'] \arrow[ul, "\subseteq"] \end{tikzcd} \end{center} \end{enumerate} First, we will satisfy the item (iii). Namely, we will construct $\Gamma'_n, g'_n$ such that $g_{n-1}\subseteq g'_n$, $\Gamma_{n-1}\subseteq\Gamma'_n$, $g'_n$ gives an automorphism of $\Gamma'_n$ and $f_{i,j}$ extends to an embedding of $(B_{i,j}, \beta_{i,j})$ to $(\Gamma'_n, g'_n)$. But this can be easily done by the fact, that $\cD$ has the amalgamation property. It is important to note that $g'_n$ should be a finite function and once again, as it is an automorphism of a finitely generated structure, we may think it is simply a map from one generators of $\Gamma'_n$ to the others. By the weak ultrahomogeneity of $\Gamma$, we may assume that $\Gamma'_n\subseteq \Gamma$. Now, by the WHP of $\cC$ we can extend $\langle\Gamma'_n\cup\{v_n\}\rangle$ together with its partial isomorphism $g'_n$ to a finitely generated structure $\Gamma_n$ together with its automorphism $g_n\supseteq g'_n$ and (again by weak ultrahomogeneity) without the loss of generality we may assume that $\Gamma_n\subseteq\Gamma$. This way we've constructed $g_n$ that has all desired properties. Now we need to see that $g = \bigcap^{\infty}_{n=0}g_n$ is indeed an automorphism of $\Gamma$ such that $(\Gamma, g)$ has the age $\cD$ and is weakly ultrahomogeneous. It is of course an automorphism of $\Gamma$ as it is defined for every $v\in\Gamma$ and is an union of an increasing chain of automorphisms of finitely generated substructures. Take any $(B, \beta)\in\cD$. Then, there are $i, j$ such that $(B, \beta) = (B_{i, j}, \beta_{i,j})$ and $A_{i,j}=\emptyset$. By the bookkeeping there was $n$ such that $(i, j) = \min\{\{0,1,\ldots n\}\times\bN\setminus X_{n}\}$. This means that $(B, \beta)$ embeds into $(\Gamma_n, g_n)$, hence it embeds into $(\Gamma, g)$. Thus, $\cD$ is a subclass of the age of $(\Gamma, g)$. The other inclusion is obvious. Hence, the age of $(\Gamma, g)$ is $\cH$. It is also weakly ultrahomogeneous. Having $(A,\alpha)\subseteq(B,\beta)$, and an embedding $f\colon(A,\alpha)\to(\Gamma,g)$, we may find $n\in\bN$ such that $(i,j) = \min\{\{0,1,\ldots n-1\}\times X_{n-1}\}$ and $(A,\alpha) = (A_{i,j},\alpha_{i,j}), (B,\beta)=(B_{i,j},\beta_{i,j})$ and $f = f_{i,j}$. This means that there is a compatible embedding of $(B,\beta)$ into $(\Gamma_n, g_n)$, which means we can also embed it into $(\Gamma, g)$. Hence, $(\Gamma,g)\cong(\Gamma,\sigma)$. By this we know that $g$ and $\sigma$ are conjugate in $G$, thus player \textit{II} have a winning strategy in the Banach-Mazur game with $A=\sigma^G$, thus $\sigma^G$ is comeagre in $G$ and $\sigma$ is a generic automorphism. \end{proof} \begin{theorem} \label{theorem:key-theorem} Let $\cC$ be a Fraïssé class of finitely generated $L$-structures with WHP and canonical amalgamation. Then $\Flim(\cC)$ has a generic automorphism. \end{theorem} \begin{proof} It follows trivially from Corollary \ref{corollary:whp+canonical-iso} and the above Theorem \ref{theorem:generic_aut_general}. \end{proof} \subsection{Properties of the generic automorphism} Let $\cC$ be a Fraïssé class of finitely generated $L$-structures with weak Hrushovski property and canonical amalgamation. Let $\cD$ be the Fraïssé class (by the Theorem \ref{theorem:key-theorem} of the structures of $\cC$ with additional automorphism of the structure. Let $\Gamma = \Flim(\cC)$. \begin{proposition} \label{proposition:fixed_points} Let $\sigma$ be the generic automorphism of $\Gamma$. Then the set of fixed points of $\sigma$ is isomorphic to $\Gamma$. \end{proposition} \begin{proof} Let $S = \{x\in \Gamma\mid \sigma(x) = x\}$. First we need to show that it is an infinite. By the theorem \ref{theorem:generic_aut_general} we know that $(\Gamma, \sigma)$ is the Fraïssé limit of $\cD$, thus we can embed finite $L$-structures of any size with identity as an automorphism of the structure into $(\Gamma, \sigma)$. Thus $S$ has to be infinite. Also, the same argument shows that the age of the structure is exactly $\cC$. It is weakly ultrahomogeneous, also by the fact that $(\Gamma, \sigma)$ is in $\cD$. \end{proof} \end{document}