Josia Pietsch
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114 lines
4.1 KiB
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114 lines
4.1 KiB
TeX
\tutorial{11}{2024-01-09}{}
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An equivalent definition of subflows can be given as follows:
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\begin{definition}
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Let $(X,T)$ be a flow with action $\alpha_x$.
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Let $Y \subseteq X$ be a compact subspace of $X$.
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If $Y$ is invariant under $\alpha_x$, we say that
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$(Y,T)$ (with action $\alpha_x\defon{T \times Y}$
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is a subflow of $(X,T)$.
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\end{definition}
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\begin{example}[Flows with a non-closed orbit]
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\begin{enumerate}[1.]
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\item Consider $(S^1, \Z)$
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with action given by $1 \cdot x = x + c$ for
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a fixed $c \in \R\setminus\Q$.\footnote{We identify $S^1$ and $\faktor{\R}{\Z}$.}
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Then the orbit of $0$, $\{nc : n \in \Z\}$ is dense but consists only of irrationals
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(except $0$),
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so it is not closed.
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\item Consider $(S^1, \Q)$ with action $qx \coloneqq x + q$.
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The orbit of $0$, $\faktor{\Q}{\Z} \subseteq S^1$,
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is dense but not closed.
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$(S^1,\Q)$ is minimal.
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\end{enumerate}
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\end{example}
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\begin{example}[\vocab{Left Bernoulli shift}]
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Consider $(\{0,1\}^{\Z}, T)$,
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where $T = \Z$ and the action is given by
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\begin{IEEEeqnarray*}{rCl}
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\Z \times \{0,1\}^{\Z}&\longrightarrow & \Z \\
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(m, (x_n)_{n \in \Z})&\longmapsto & (x_{n+m})_{n \in \Z}.
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\end{IEEEeqnarray*}
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The orbit of $z \coloneqq (0)_{n \in \Z}$ consist of only on point.
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In particular it is closed.
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Let $x \coloneqq ( [n = 0])_{n \in \Z}$.
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Then $Tx = \{([n = m])_{n \in \Z} | m \in \Z\}$.
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Clearly $z \not\in Tx$.
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\begin{claim}
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$z \in \overline{Tx}$
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\end{claim}
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\begin{proof}
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Consider a basic open $z \in U_I = \{y : y_i = 0, i \in I\}$
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where $I \subseteq \Z$ is finite.
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Then $U_I \cap Tx \neq \emptyset$
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as we can shift the $1$ out of $I$,
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i.e.~$(\max I + 1) x \in U_I$.
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\end{proof}
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\end{example}
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Flows are always on non-empty spaces $X$.
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\begin{fact}
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Consider a flow $(X,T)$.
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The following are equivalent:
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\begin{enumerate}[(i)]
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\item Every $T$-orbit is dense.
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\item There is no proper subflow,
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\end{enumerate}
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If these conditions hold, the flow is called \vocab{minimal}.
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\end{fact}
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\begin{proof}
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(i) $\implies$ (ii):
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Let $(Y,T)$ be a subflow of $(X,T)$.
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take $y \in Y$. Then $Ty$ is dense in $X$.
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But $Ty \subseteq Y$, so $Y$ is dense in $X$.
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Since $Y$ is closed, we get $Y = X$.
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(ii) $\implies$ (i):
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Take $x \in X$. Consider $Tx$.
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It suffices to show that $\overline{Tx}$ is a subflow.
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Clearly $\overline{Tx}$ is closed,
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so it suffices to show that it is $T$-invariant.
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Let $y \in \overline{Tx}$ and $t \in T$.
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Take $ty \in U \overset{\text{open}}{\subseteq} X$.
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Since $t^{-1}$ acts as a homeomorphism
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we have $y \in t^{-1} U \overset{\text{open}}{\subseteq} X$.
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We find some $t'x \in t^{-1}U$ since $y \in \overline{Tx}$.
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So $tt'x \in Tx \cap U$.
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\end{proof}
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\begin{fact}
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Every flow $(X,T)$ contains a minimal subflow.
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\end{fact}
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\begin{proof}
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We use Zorn's lemma:
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Let $S$ be the set of all subflows of $(X,T)$
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ordered by $Y \le Y' :\iff Y \supseteq Y'$.
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We need to show that for a chain $\langle Y_i : i \in I \rangle$,
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there exists a lower bound.
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Consider $\bigcap_{i \in I} Y_i$. This a subflow:
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\begin{itemize}
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\item It is closed as it is an intersection of closed sets.
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\item It is $T$-invariant, since each of the $Y_i$ is.
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\item It is non-empty by \yaref{tut10fact}.
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\end{itemize}
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\end{proof}
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\begin{fact}
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\label{tut10fact}
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Let $X$ be a topological space.
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Then $X$ is compact iff every family of closed sets with
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FIP\footnote{finite intersection property, i.e.~the intersection of every finite sub-family is non-empty}
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has non-empty intersection.
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\end{fact}
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\begin{proof}
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Note that families of
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closed sets correspond to families of open sets by taking complements.
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A family of open sets is a cover iff the corresponding family
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has empty intersection,
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and is admits a finite subcover iff the corresponding family
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has the FIP.
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\end{proof}
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