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+%
+% so2.tex -- Illustration of so(2) -> SO(2)
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+% Erstellt durch Roy Seitz
+%
+% !TeX spellcheck = de_CH
+\bgroup
+
+\begin{frame}[t]
+  \setlength{\abovedisplayskip}{5pt}
+  \setlength{\belowdisplayskip}{5pt}
+  \frametitle{Von der Lie-Gruppe zur -Algebra}
+  \vspace{-20pt}
+  \begin{columns}[t,onlytextwidth]
+    \begin{column}{0.48\textwidth}
+      \uncover<1->{
+        \begin{block}{Lie-Gruppe}
+          Darstellung von \gSO2:
+          \begin{align*}
+            \mathbb R 
+            &\to 
+            \gSO2
+            \\
+            t
+            &\mapsto
+            \begin{pmatrix}
+              \cos t &         -\sin t \\ 
+              \sin t & \phantom-\cos t
+            \end{pmatrix}
+          \end{align*}
+        \end{block}
+      }
+      \uncover<2->{
+        \begin{block}{Ableitung am neutralen Element}
+          \begin{align*}
+            \frac{d}{d t}
+            &
+            \left.
+            \begin{pmatrix}
+              \cos t &         -\sin t \\ 
+              \sin t & \phantom-\cos t
+            \end{pmatrix}
+            \right|_{ t = 0}
+            \\
+            =
+            & 
+            \begin{pmatrix} -\sin0 & -\cos0 \\ \phantom-\cos0 & -\sin0 \end{pmatrix}
+            = 
+            \begin{pmatrix} 0 & -1 \\ 1 &  \phantom-0 \end{pmatrix}
+          \end{align*}
+        \end{block}
+      }
+    \end{column}
+    \begin{column}{0.48\textwidth}
+      \uncover<3->{
+        \begin{block}{Lie-Algebra}
+          Darstellung von \aso2:
+          \begin{align*}
+            \mathbb R 
+            &\to 
+            \aso2
+            \\
+            t
+            &\mapsto
+            \begin{pmatrix}
+              0 &         -t \\ 
+              t & \phantom-0
+            \end{pmatrix}
+          \end{align*}
+        \end{block}
+      }
+    \end{column}
+  \end{columns}
+\end{frame}
+
+
+\begin{frame}[t]
+  \setlength{\abovedisplayskip}{5pt}
+  \setlength{\belowdisplayskip}{5pt}
+  \frametitle{Von der Lie-Algebra zur -Gruppe}
+  \vspace{-20pt}
+  \begin{columns}[t,onlytextwidth]
+    \begin{column}{0.48\textwidth}
+      \uncover<1->{
+      \begin{block}{Differentialgleichung}
+        Gegeben:
+        \[
+        J
+        =
+        \dot\gamma(0) = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix}
+        \]
+        Gesucht:
+        \[ \dot \gamma (t) = J \gamma(t) \qquad \gamma \in \gSO2 \]
+        \[ \Rightarrow \gamma(t) = \exp(Jt) \gamma(0) = \exp(Jt) \]
+      \end{block}
+    }
+    \end{column}
+    \begin{column}{0.48\textwidth}
+      \uncover<2->{
+      \begin{block}{Lie-Algebra}
+        Potenzen von $J$:
+        \begin{align*}
+          J^2 &= -I &
+          J^3 &= -J &
+          J^4 &=  I &
+          \ldots
+        \end{align*}
+      \end{block}
+    }
+    \end{column}
+  \end{columns}
+\uncover<3->{
+  Folglich:
+  \begin{align*}
+    \exp(Jt)
+    &= I + Jt 
+    + J^2\frac{t^2}{2!} 
+    + J^3\frac{t^3}{3!}
+    + J^4\frac{t^4}{4!}
+    + J^5\frac{t^5}{5!}
+    + \ldots \\
+    &= \begin{pmatrix}
+      \vspace*{3pt}
+      1 - \frac{t^2}{2} + \frac{t^4}{4!} - \ldots
+      &
+      -t + \frac{t^3}{3!} - \frac{t^5}{5!} + \ldots
+      \\ 
+      t - \frac{t^3}{3!} + \frac{t^5}{5!} - \ldots
+      &
+      1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \ldots
+    \end{pmatrix}
+    =
+    \begin{pmatrix}
+      \cos t &         -\sin t \\ 
+      \sin t & \phantom-\cos t
+    \end{pmatrix}
+  \end{align*}
+  }
+\end{frame}
+\egroup