From 753ff60029e269239a60f64cf036626d7ab7377e Mon Sep 17 00:00:00 2001 From: Franciszek Malinka Date: Thu, 21 Apr 2022 01:14:27 +0200 Subject: Main proof maybe finished --- lic_malinka.pdf | Bin 381541 -> 386819 bytes lic_malinka.tex | 78 +++++++++++++++++++++++++++++++++++++++++++++++--------- 2 files changed, 66 insertions(+), 12 deletions(-) diff --git a/lic_malinka.pdf b/lic_malinka.pdf index b46df85..5c50fff 100644 Binary files a/lic_malinka.pdf and b/lic_malinka.pdf differ diff --git a/lic_malinka.tex b/lic_malinka.tex index 47dea88..85541cf 100644 --- a/lic_malinka.tex +++ b/lic_malinka.tex @@ -717,7 +717,8 @@ 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 random graph such that $(\FrGr, g) = \Flim{\cH}$, i.e. the Fraïssé limit of finite - graphs with automorphism. By the Fraïssé theorem \ref{theorem:fraisse_thm} + graphs with automorphism. Precisely, $\bigcap^{\infty}_{i=0}B_{g_i} = \{g\}$. + By the Fraïssé theorem \ref{theorem:fraisse_thm} it will follow that $(\FrGr, g)\cong (\FrAut, \sigma)$. By the proposition \ref{proposition:graph-aut-is-normal} we can assume without the loss of generatlity that $\FrAut = \FrGr$ as a plain graph. Hence, @@ -732,11 +733,18 @@ i.e. $(A_n, \alpha_n)\subseteq (B_n, \beta_n)$. Also, it may be that $A_n$ is an empty graph. We enumerate the verticies of the random graph $\FrGr = \{v_0, v_1, \ldots\}$. - - Just for sake of fixing a technical problem, let $g_{-1} = \emptyset$. - Suppose that player \textit{I} in the $n$-th move chose a finite partial - isomoprhism $f_n$. We will construct $g_n\supseteq f_n$ such that - following properties hold: + + 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 naturaly induces + a well ordering on $\bN\times\bN$. This will prove useful later, as the + main argument of the proof will be constructed as a bookkeeping argument. + + Just for sake of fixing a technical problem, let $g_{-1} = \emptyset$ and + $X_{-1} = \emptyset$. + Suppose that player \textit{I} in the $n$-th move chooses a finite partial + isomoprhism $f_n$. We will construct $g_n\supseteq f_n$ 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 subgraph $\FrGr_n$, \item $g_n(v_n)$ and $g_n^{-1}(v_n)$ are defined, @@ -747,14 +755,15 @@ $(A_{n,k}, \alpha_{n,k})\subseteq (B_{n,k}, \beta_{n,k})$, and $f_{n,k}$ is an embedding of $(A_{n,k}, \alpha_{n,k})$ in the $\FrGr_{n-1}$ (which is the graph induced by $g_{n-1}$). Let - $(i, j) = \min\{\{0, 1, \ldots\} \times \bN \setminus X_{n-1}\}$. Then + $(i, j) = \min\{(\{0, 1, \ldots\} \times \bN) \setminus X_{n-1}\}$ (with the + order induced by $\gamma$). Then $X_n = X_{n-1}\cup\{(i,j)\}$ and $(B_{n,k}, \beta_{n,k})$ embedds in $(\FrGr_n, g_n)$ so that this diagram commutes: \begin{center} \begin{tikzcd} - (A_{i,j}, \alpha_{i,j}) \arrow[d, "\subseteq"'] \arrow[r, "f_{i,j}"] & (\FrGr_n, g_n) \\ - (B_{i,j}, \beta_{i,j}) \arrow[ur, dashed, "\hat{f}_{i,j}"'] + (A_{i,j}, \alpha_{i,j}) \arrow[d, "\subseteq"'] \arrow[r, "f_{i,j}"] & (\FrGr_{n-1}, g_{n-1}) \arrow[d, "\subseteq"] \\ + (B_{i,j}, \beta_{i,j}) \arrow[r, dashed, "\hat{f}_{i,j}"'] & (\FrGr_n, g_n) \end{tikzcd} \end{center} \end{enumerate} @@ -762,10 +771,55 @@ First item makes sure that no inifite orbit will not be present in $g$. The second item together with the first one are necessary for $g$ to be an automorphism of $\FrGr$. The third item is the one that gives weak - ultrahomogeneity. + ultrahomogeneity. Now we will show that indeed such $g_n$ may be constructed. + + First, we will suffice the (iii) item. Namely, we will construct $\FrGr'_n, g'_n$ + such that $g_{n-1}\subseteq g'_n$ and $f_{i,j}$ extends to an embedding of + $(B_{i,j}, \beta_{i,j})$ to $(\FrGr'_n, g'_n)$. Without the loss of generality + we may assume that $f_{i,j}$ is an inclusion and that $A_{i,j} = B_{i,j}\cap\FrGr_{n-1}$. + Then let $\FrGr'_n = B_{i,j}\cup\FrGr_{n-1}$ and $g'_n = g_{n-1}\cup \beta_{i,j}$. + Then $(B_{i,j}, \beta_{i,j})$ simply embedds by inclusion in $(\FrGr'_n, g'_n)$, + i.e. this diagram commutes: + + \begin{center} + \begin{tikzcd} + (A_{i,j}, \alpha_{i,j}) \arrow[d, "\subseteq"'] \arrow[r, "\subseteq"] & (\FrGr_{n-1}, g_{n-1}) \arrow[d, "\subseteq"] \\ + (B_{i,j}, \beta_{i,j}) \arrow[r, dashed, "\subseteq"'] & (\FrGr'_n, g'_n) + \end{tikzcd} + \end{center} + + The argument that those are actually embeddings is almost the same as in + proof of the amalgamation property of $\cH$. + + It may be also assumed without the loss of generality that $\FrGr'_n\subseteq \FrGr$. + Of course by the recurisve assumption $\FrGr_{n-1}\subseteq\FrGr$. The + $\FrGr'_n \setminus\FrGr_{n-1} = B_{i,j}\setminus A_{i,j}$ can be found in + $\FrGr$ by the random graph property -- we can find verticies of the remaining + part of $B_{i,j}$ each at a time so that all edges are correct. + + Now, by the WHP of $\bK$ we can extend the graph $\FrGr'_n\cup\{v_n\}$ together + with its partial isomorphism $g'_n$ to a graph $\FrGr_n$ together with its + automorphism $g_n\supseteq g'_n$, where without the loss of generality we + may assume that $\FrGr_n\subseteq\FrGr$. 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 $\FrGr$ such that $(\FrGr, g)$ has the age $\cH$ and has weak ultrahomogeneity. + It is of course an automorphism of $\FrGr$ as it is defined for every $v\in\FrGr$ + and is a sum of increasing chain of finite automorphisms. + + Take any finite graph 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 $(B, \beta)$ embedds into $(\FrGr_n, g_n)$, hence it embedds + into $(\FrGr, g)$. With a similar argument we can see that $(\FrGr, g)$ is + weakly ultrahomogenous. + + By this we know that $g$ and $\sigma$ conjugate in $G$. As we stated in the + beginning of the proof, the set 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. \end{proof} - - \printbibliography \end{document} -- cgit v1.2.3