Housekeeping.
This commit is contained in:
Bokuan Li
@@ -128,21 +128,114 @@
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\begin{definition}[Universal Minimal Flow]
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\label{definition:universal-minimal-flow}
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Let $G$ be a group, then there exists a flow $(M, \alpha)$ such that:
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Let $G$ be a group, then there exists a flow $L$ such that:
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\begin{enumerate}
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\item $(M, \alpha)$ is a minimal $G$-flow.
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\item[(U)] For any minimal $G$-flow $(X, \beta)$, $x_0 \in X$, and $m_0 \in M$, there exists a unique continuous map $\Phi \in C(M; X)$ such that $\Phi(m_0) = x_0$ and the following diagram commutes
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\item $G \curvearrowright L$ is a minimal $G$-flow.
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\item[(U)] For any minimal $G$-flow $X$, there exists a continuous map $\Phi \in C(M; X)$ such that $\Phi(m_0) = x_0$ and the following diagram commutes
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\[
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\xymatrix{
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M \ar@{->}[r]^{\Phi} \ar@{->}[d]_{\alpha(g)} & X \ar@{->}[d]^{\beta(g)} \\
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M \ar@{->}[r]_{\Phi} & X
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L \ar@{->}[r]^{\Phi} \ar@{->}[d]_{g} & X \ar@{->}[d]^{g} \\
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L \ar@{->}[r]_{\Phi} & X
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}
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\]
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for all $g \in G$.
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\end{enumerate}
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The flow $(M, \alpha)$ is the \textbf{universal minimal $G$-flow}.
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\end{definition}
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\begin{proof}
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(1): Using \autoref{corollary:stone-cech-minimal-ideal-existence}, let $L \subset \beta G$ be a minimal left ideal, then $G \curvearrowright L$ is a minimal $G$-flow.
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(U): Let $G \curvearrowright X$ be a minimal $G$-flow, $x_0 \in X$, and $m_0 \in M$. By (U) of the \hyperref[universal flow]{definition:stone-cech-flow}, there exists a unique continuous map $\phi \in C(\beta G; X)$ such that $\phi(1_G) = x_0$ and the following diagram commutes
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\[
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\xymatrix{
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\beta G \ar@{->}[r]^{\Phi} \ar@{->}[d]_{g} & X \ar@{->}[d]^{g} \\
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\beta G \ar@{->}[r]_{\Phi} & X
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}
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\]
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for all $g \in G$.
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Let $\Phi = \phi|_{L} \in C(L; X)$, then as $L$ is $G$-invariant, $\Phi(L) \supset G\phi(x_0)$. By \autoref{proposition:compact-extensions}, $\Phi(L) = \ol{G\phi(x_0)} = X$.
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\end{proof}
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\begin{definition}[Minimal Idempotent]
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\label{definition:tds-minimal-idempotent}
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Let $S$ be a compact Hausdorff right topological semigroup and $e \in S$ be an idempotent, then $e$ is \textbf{minimal} if there exists a minimal left ideal $I \subset G$ such that $e \in I$.
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\end{definition}
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\begin{theorem}[Ellis]
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\label{theorem:ellis-semigroup}
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Let $S$ be a non-empty compact Hausdorff right topological semigroup and $I \subset S$ be a closed left ideal, then there exists a minimal left ideal $I \subset S$ and a minimal idempotent $e \in I$.
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\end{theorem}
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\begin{proof}
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By Zorn's Lemma, $I$ admits a minimal non-empty closed right topological subsemigroup $K \subset S$. For each $a \in K$, $Ka$ is closed by continuity of right multiplication and \autoref{proposition:compact-extensions}. By minimality of $K$, $K = Ka$, so there exists $e \in K$ such that $a = ea$. Let $E = \bracsn{b \in K|ba = a}$, then $E$ is also a closed and right topological subsemigroup. By minimality of $K$, $E = K$, and in particular $a \in E$.
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\end{proof}
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\begin{corollary}
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\label{corollary:stone-cech-minimal-idempotent}
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Let $G$ be an infinite group, then $\beta G$ admits a minimal idempotent in $\beta G \setminus G$.
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\end{corollary}
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\begin{proof}
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Let $L \subset (\beta G \setminus G)$ be a minimal left ideal, then $L$ is a non-empty compact Hausdorff right topological semigroup. By \hyperref[Ellis' Theorem]{theorem:ellis-semigroup}, there exists $e \in L$ with $e^2 = e$.
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\end{proof}
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\begin{proposition}
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\label{proposition:stone-cech-flow-minimal-idempotent}
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Let $G$ be an infinite group, $G \curvearrowright X$ be a minimal $G$-flow, then there exists a minimal idempotent $\fU \in \beta G \setminus G$ such that $\fU x = x$.
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\end{proposition}
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\begin{proof}
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Let $L \subset \beta G$ be a minimal left ideal, then the mapping $\fU \mapsto \fU x$ is a continuous surjection. As $x \in X = \bracsn{\fU x|x \in L}$, there exists $\fU \in L$ with $\fU x = x$. Let $F = \bracsn{\fU \in L|\fU x = x}$, then $F$ is a non-empty closed right topological subsemigroup of $L$. By \hyperref[Ellis' Theorem]{theorem:ellis-semigroup}, $F$ contains an idempotent.
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\end{proof}
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\begin{theorem}[Infinite Ramsey Theorem for Pairs]
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\label{theorem:ramsey-pairs}
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Let
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\[
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[\natp]^2 = \bracsn{\bracsn{m, n} \in \nat^2|m, n \in \natp, m \ne n}
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\]
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be the set of unordered pairs of distinct elements of $\natp$ and $c: [\natp]^2 \to \bracsn{1, \cdots, r}$, then there exists $A \subset \natp$ infinite such that for any $m, n, m', n' \in A$, $c(\bracsn{m, n}) = c(\bracsn{m', n'})$.
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\end{theorem}
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\begin{proof}
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Let $\fU \subset 2^{\natp}$ be a free ultrafilter. For each $n \in \natp$ and $1 \le i \le r$, let $A(n, i) = \bracsn{m > n| c(\bracsn{m,n}) = i}$, then $\natp = \bracsn{1, \cdots, n} \bigsqcup_{i = 1}^r A(n, i)$. As $\fU$ is a free ultrafilter, there exists a unique $1 \le i(n) \le r$ such that $A(n, i(n)) \in \fU$.
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For each $1 \le i \le r$, let $B_i = \bracsn{n \in \natp|i(n) = i}$, then $\natp = \bigsqcup_{i = 1}^r B_i$. As $\fU$ is an ultrafilter, there exists a unique $1 \le i \le r$ with $B_i \in \fU$.
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Let $n_1 \in B_i$, so $i(n_1) = i$. Assume inductively that $\bracsn{n_j}_1^N \subset \natp$ has been defined such that:
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\begin{enumerate}[label=(\roman*)]
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\item For each $1 \le j \le N - 1$, $n_j < n_{j+1}$.
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\item For each $1 \le j \le N$, $i(n_j) = i$.
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\item For each $1 \le j < k \le N$, $c(\bracsn{n_j, n_k}) = i$.
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\end{enumerate}
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Let $n_{N+1} \in B_i \cap \bigcap_{j = 1}^N A(n_j, i)$, then:
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\begin{enumerate}
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\item As $n_{N+1} \in A(n_N, i)$, $n_{N+1} > n_N$.
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\item Since $n_{N+1} \in B_i$, $i(n_{N+1}) = i$.
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\item For each $1 \le j \le N$, $n_{N+1} \in A(n_j, i) = A(n_j, i(n))$, so $c(\bracsn{n_j, n_{N+1}}) = i$.
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\end{enumerate}
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By induction, there exists $\seq{n_j} \subset \natp$ such that $c(\bracsn{n_j, n_k}) = i$ for all $j, k \in \natp$ with $j \ne k$.
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\end{proof}
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\begin{lemma}[Idempotent Refinement]
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\label{lemma:idempotent-refinement}
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Let $S$ be a semigroup and $f \in \beta S$ be idempotent, then there exists a minimal idempotent $e \in \beta S$ such that $ef = fe = e$.
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\end{lemma}
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\begin{proof}
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By \autoref{proposition:compact-extensions}, $(\beta S)f$ is a closed left ideal in $S$. By \hyperref[Ellis' Theorem]{theorem:ellis-semigroup}, $(\beta S)f$ contains a minimal left ideal $L$, and a minimal idempotent $r \in L$. As $L \subset (\beta S)f$, there exists $t \in \beta S$ such that $r = tf$.
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Let $e = fr \in L$, then $e^2 = frfr = fr^2 = fr = e$. Moreover,
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\[
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ef = frf = ftf^2 = ftf = fr = e = fr = ffr = fe
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\]
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\end{proof}
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