aboutsummaryrefslogtreecommitdiff
path: root/sections/conj_classes.tex
blob: 96522f5b2d39f31d14d5a0e2ac0b63cc3d915217 (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
\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 have 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 finite $L$-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}

  \begin{proof}
	Let $\Gamma = \Flim(\cC)$ and $(\Pi, \sigma) = \Flim(\cD)$. Let $G = \Aut(\Gamma)$,
	i.e. $G$ is the automorphism group of $\Gamma$. First, by the Theorem
	\ref{theorem:isomorphic_fr_lims}, we may assume without the loss of generality
	that $\Pi = \Gamma$.
	We will construct a strategy for the second player in the Banach-Mazur game
	on the topological space $G$. This strategy will give us a subset 
	$A\subseteq G$ and as we will see a subset of the $\sigma$'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 open
	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 finite partial
	isomorphism $g_0$ of $\Gamma$. By $B_{g}\subseteq G$ we denote a basic 
	open subset given by a finite partial isomorphism $g$. Note that $B_g$
	is nonemty because of ultrahomogeneity of $\Gamma$.

	With the use of Corollary \ref{corollary:banach-mazur-basis} we can consider 
	only games where both players choose finite 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 the resulting function
	% $g = \bigcap^{\infty}_{i=0}g_i$ will be an automorphism of the Fraïssé limit 
	% $\Gamma$ such that $(\Gamma, g) = \Flim{\cD}$. 
	% Precisely, we will find $g_i$'s such that 
	% $\bigcap^{\infty}_{i=0}B_{g_i} = \{g\}$ and by
	% the Fraïssé theorem \ref{theorem:fraisse_thm}
	% it will follow that $(\Gamma, g)\cong (\Pi, \sigma)$. Hence,
	% by the fact	\ref{fact:conjugacy}, $g$ and $\sigma$ conjugate in $G$.
	
	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.

	% Once again, by the Fraïssé theorem and by Lemma 
	% \ref{lemma:weak_ultrahom} constructing $g_i$'s in a way such that
	% age of $(\Gamma, g)$ is exactly $\cD$ and so that it is weakly ultrahomogeneous
	% will produce expected result. 
	First, let us enumerate all pairs of structures
	$\{\langle(A_n, \alpha_n), (B_n, \beta_n)\rangle\}_{n\in\bN},\in\cD$
	such that the first element of the pair embeds by inclusion in the second,
	i.e. $(A_n, \alpha_n)\subseteq (B_n, \beta_n)$. Also, it may be that 
	$A_n$ is an empty. We enumerate the elements of the Fraïssé limit 
	$\Gamma = \{v_0, v_1, \ldots\}$.

	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
	$X_{-1} = \emptyset$. 
	Suppose that player \textit{I} in the $n$-th move chooses a finite partial
	isomorphism $f_n$. We will construct a finite partial isomorphism $g_n\supseteq f_n$ 
	and a set $X_n\subseteq\bN^2$
	such that following properties hold:

	\begin{enumerate}[label=(\roman*)]
	  \item $g_n$ is an automorphism of the induced substructure $\Gamma_n$,
	  \item $g_n(v_n)$ and $g_n^{-1}(v_n)$ are defined,
	  \item let 
		$\{\langle (A_{n,k}, \alpha_{n, k}), (B_{n,k}, \beta_{n,k}), f_{n, k}\rangle\}_{k\in\bN}$
		be the enumeration of all pairs of finite structures of $T$ with automorphism 
		such that the first is a substructure of the second, 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 $\FrGr_{n-1}$ (which
		is the substructure induced by $g_{n-1}$). Let 
		$(i, j) = \min\{(\{0, 1, \ldots\} \times \bN) \setminus X_{n-1}\}$ (with the
		order induced by $\gamma$). Then $X_n = X_{n-1}\cup\{(i,j)\}$ and
		$(B_{n,k}, \beta_{n,k})$ 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 item makes sure that no infinite orbit will be present in $g$. The 
	second item together with the first one are necessary for $g$ to be an 
	automorphism of $\Gamma$. The third item is the one that gives weak
	ultrahomogeneity. Now we will show that indeed such $g_n$ may be constructed.
	
	First, we will suffice the item (iii). Namely, we will construct $\Gamma'_n, g'_n$
	such that $g_{n-1}\subseteq g'_n$ 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. Moreover, without
	the loss of generality we can assume that all embeddings are inclusions.

    \begin{center}
      \begin{tikzcd}
                                  & (\Gamma'_n, g'_n) & \\
        (B_{i,j}, \beta_{i,j}) \arrow[ur, dashed, "\subseteq"] &   & (\Gamma_{n-1}, g_{n-1}) \arrow[ul, dashed, "\subseteq"'] \\
								  & (A_{i,j}, \alpha_{i,j}) \arrow[ur, "\subseteq"'] \arrow[ul, "\subseteq"]
      \end{tikzcd}
    \end{center}

	By the weak ultrahomogeneity we may assume that $\Gamma'_n\subseteq \Gamma$:

	\begin{center}
	  \begin{tikzcd}
		(B_{i,j}\cup\Gamma_{n-1}, \beta_{i,j}\cup g_{n-1}) \arrow[d, "\subseteq"'] \arrow[r, "\subseteq"] & \Gamma \\
		(\Gamma'_{n}, g'_n)\arrow[ur, dashed, "f"'] 
	  \end{tikzcd}
	\end{center}

	Now, by the WHP of $\cK$ we can extend the graph $\Gamma'_n\cup\{v_n\}$ together
	with its partial isomorphism $g'_n$ to a graph $\Gamma_n$ together with its
	automorphism $g_n\supseteq g'_n$ and 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 $\cH$ and is weakly ultrahomogeneous.
	It is of course an automorphism of $\Gamma$ as it is defined for every $v\in\Gamma$
	and is a sum of increasing chain of finite automorphisms.

	Take any finite structure of $T$ with automorphism $(B, \beta)$. 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 
	$(i, j) = \min\{\{0,1,\ldots\}\times\bN\setminus X_{n-1}\}$.
	This means that $(B, \beta)$ embeds into $(\Gamma_n, g_n)$, hence it embeds
	into $(\Gamma, g)$, thus it has age $\cH$. 
	With a similar argument we can see that $(\Gamma, g)$ is weakly ultrahomogeneous.

	By this we know that $g$ and $\sigma$ conjugate in $G$. As we stated in the
	beginning of the proof, the set $A$ of possible outcomes of the game (i.e.
	possible $g$'s we end up with) is comeagre in $G$, thus $\sigma^G$ is also
	comeagre and $\sigma$ is a generic automorphism, as it contains a comeagre
	set $A$.
  \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 strucutre.
  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}