l21

inputs/lecture_21.tex

@@ -6,12 +6,13 @@
     Consider the projection $\pi\colon Y \times S^1 \to Y$.
     By minimality of $Y$, we have $\pi(Z) = Y$.
     Note that for every $\theta \in S^1$, $\theta \cdot  Z$ is minimal,
-    so either $\theta \cdot  Z = Z$ or $(\theta \cdot  Z)\cap Z = \emptyset$.
+    so either $\theta \cdot  Z = Z$ or $(\theta \cdot  Z)\cap Z = \emptyset$.%
+    \footnote{actually $(1,\ldots,1, \theta) \cdot Z$, we identify $S^1$ and $\{0\}^d \times S^1 \subseteq Y \times S^1$.}
 
     Let $H = \{\theta \in S^1 : \theta \cdot  Z = Z\}$.
     $H$ is a closed subgroup of $S^1$.
     % H is a rotation of Z containing 1 (?)
-    Therefore either $H = S^1$ (but in that case $Z = Y \times S^1$),
+    Therefore either $H = S^1$ (but in that case $Z = Y \times S^1$, so this cannot be the case),
     or there exists $m \in \Z$ such that $H = \{ \xi \in S^1 : \xi^m = 1 \}$
     by \yaref{fact:tau1minimal}.
 
@@ -36,19 +37,22 @@
 
     Consider $f \colon  (y,\xi) \mapsto (y, \xi^m)$.
     Since $(\beta^{(y)} \cdot  t_i)^m = (\beta^{(y)})^m$
-    we get a continuous\todo{Why is this continuous?}
+    we get a continuous
     function $\phi\colon Y \to S^1$
     such that
     \[
-    Z = \{(y,\xi) \in Y \times S^1 : \xi^m = \phi(y)\}.
+    Z = \{(y,\xi) \in Y \times S^1 : \xi^m = \phi(y)\},
     \]
-    % namely
+    namely
-    % \begin{IEEEeqnarray*}{rCl}
+    \begin{IEEEeqnarray*}{rCl}
-    %     \phi\colon Y &\longrightarrow & S^1 \\
+        \phi\colon Y &\longrightarrow & S^1 \\
-    %     y &\longmapsto & \beta^{(y)}.
+        y &\longmapsto & (\beta^{(y)})^m
-    % \end{IEEEeqnarray*}
+    \end{IEEEeqnarray*}
+    Z is isomorphic to $m$ copies of the graph of that function, hence
+    the graph is closed, so the function is continuous.
 
-    Note that $f(Z)$ is homeomorphic to $Y$.\todo{Why?}
+    Note that $f(Z)$ is homeomorphic to $Y$
+    (for every $y \in Y$, $\phi(y)$ is the unique element such that $(y,\phi(y)) \in f(Z)$).
 
     \begin{claim}
         $\phi(S(y)) = \phi(y) \cdot  (\sigma(y))^m$.