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Vorlesung vom 23.01.2014 geTeXt
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% Mitschrieb vom 09.01.2014 %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\chapter{Euklidische und nichteuklidische Geometrie}
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\begin{definition}
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Das Tripel $(X, d, G)$ heißt genau dann eine \textbf{Geometrie}\xindex{Geometrie},
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wenn $(X, d)$ ein metrischer Raum und $\emptyset \neq G \subseteq \powerset{X}$
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die Menge aller \textbf{Geraden}\xindex{Gerade} ist.
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\end{definition}
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\section{Axiome für die euklidische Ebene}
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Axiome\xindex{Axiom} bilden die Grundbausteine jeder mathematischen Theorie. Eine
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Sammlung aus Axiomen nennt man Axiomensystem\xindex{Axiomensystem}.
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@ -758,8 +765,8 @@ $\overset{\text{Strahlensatz}}{\Rightarrow} \frac{a}{h_c} = \frac{c}{h_a} \right
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die obere Halbebene bzw. Poincaré-Halbebene und $G = G_1 \cup G_2$
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mit
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\begin{align*}
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G_1 &= \Set{g_1 \subseteq \mdh | \exists m \in \mdr, r \in \mdr_{>0}: g_1 \in \mdc: g_1 = \Set{|z-m|=r}}\\
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G_2 &= \Set{g_2 \subseteq \mdh | \exists x \in \mdr: g_2 = \Set{z \in \mdc: \Re{z} = x} \cap \mdh}
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G_1 &= \Set{g_1 \subseteq \mdh | \exists m \in \mdr, r \in \mdr_{>0}: g_1 = \Set{z \in \mdc : |z-m|=r}}\\
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G_2 &= \Set{g_2 \subseteq \mdh | \exists x \in \mdr: g_2 = \Set{z \in \mdc: \Re(z) = x} \cap \mdh}
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\end{align*}
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Die Elemente von $\mdh$ heißen \textbf{hyperbolische Geraden}\xindex{Gerade!hyperbolische}
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@ -781,7 +788,21 @@ $\overset{\text{Strahlensatz}}{\Rightarrow} \frac{a}{h_c} = \frac{c}{h_a} \right
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Gegeben $z_1, z_2 \in \mdh$\\
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\textbf{Existenz:} $\Re(z_1) = \Re(z_2)$
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$\Rightarrow z_1$ und $z_2$ liegen auf
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\[g = \Set{z \in \mdc | \Re(z) = \Re{z_1} \land \mdh}\]
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\[g = \Set{z \in \mdc | \Re(z) = \Re(z_1) \land \mdh}\]
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\begin{figure}[ht]
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\centering
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\subfloat[Fall 1]{
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\input{figures/hyperbolische-geometrie-axiom-1-1.tex}
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\label{fig:hyperbolische-geometrie-axiom-1-1}
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}%
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\subfloat[Fall1]{
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\input{figures/hyperbolische-geometrie-axiom-1-2.tex}
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\label{fig:hyperbolische-geometrie-axiom-1-2}
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}%
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\label{fig:hyperbolische-geometrie-axiom-1-0}
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\caption{Zwei Punkte liegen in der hyperbolischen Geometrie immer auf genau einer Geraden}
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\end{figure}
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\item TODO
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\item Siehe \cref{fig:hyperbolische-halbebene-axiom-5}.
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\begin{figure}[htp]
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@ -806,9 +827,9 @@ $\overset{\text{Strahlensatz}}{\Rightarrow} \frac{a}{h_c} = \frac{c}{h_a} \right
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$\sigma \in \PSL_2(\mdr)$ mit $\sigma(x_0) = 0$,
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$\sigma(x_1) = 1$, $\sigma(x_\infty) = \infty$
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\item \label{prop:15.2d} $\SL_2(\mdr)$ wird von den Matrizen
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\[\begin{pmatrix}\lambda & 0\\ 0 & \lambda^-1\end{pmatrix}, \lambda \in \mdr\]
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\[\begin{pmatrix}1 & a\\ 0 & 1\end{pmatrix}, a \in \mdr\]
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\[\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}\]
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\[\begin{pmatrix}\lambda & 0\\ 0 & \lambda^-1\end{pmatrix}, \lambda \in \mdr \;\;\;
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\begin{pmatrix}1 & a\\ 0 & 1\end{pmatrix}, a \in \mdr\;\;\;
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\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}\]
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erzeugt
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\item $\PSL_2(\mdr)$ operiert auf $G$
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\end{enumerate}
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@ -816,23 +837,53 @@ $\overset{\text{Strahlensatz}}{\Rightarrow} \frac{a}{h_c} = \frac{c}{h_a} \right
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\begin{beweis}\leavevmode
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\begin{enumerate}[label=\alph*)]
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\item Sei $z = x + iy \in \mdh$, d.~h. $y>0$ und
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\item Sei $z = x + \iu y \in \mdh$, d.~h. $y>0$ und
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$\sigma=\begin{pmatrix}a&b\\c&d\end{pmatrix} \in \SL_2(\mdr)$
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\todo{Hier stimmt was nicht}
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\begin{align}
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\Rightarrow \sigma(z) &= \frac{ax + aiy + b}{cx + ciy +d}\\
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&= \frac{ax + aiy + b}{cx + ciy +d} \cdot \frac{cx+d-iy}{cx+d-iy}\\
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&= \frac{\Re(...) + i(aycx + ayd - ciyax - cyb)}{(cx+d)^2 + (cy)^2}\\
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&= \frac{\Re(...) + i(ad-bc)y}{(cx+d)^2 + (cy)^2}
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&\overset{\mathclap{\SL_2(\mdr)}}{=} \frac{\Re(...) + iy}{(cx+d)^2 + (cy)^2}
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&= \frac{ax + aiy + b}{cx + c \iu y +d} \cdot \frac{cx+d-\iu y}{cx+d-\iu y}\\
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&= \frac{\Re(...) + \iu (aycx + ayd - axy - yb)}{(cx+d)^2 + (cy)^2}\\
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&= \frac{\Re(...) + \iu (ad-bc)y}{(cx+d)^2 + (cy)^2}\\
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&\overset{\mathclap{\SL_2(\mdr)}}{=} \frac{\Re(...) + \iu y}{(cx+d)^2 + (cy)^2}
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\end{align}
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$\Rightarrow \Im(\sigma(z)) = \frac{y}{(cx+d)^2 + (cy)^2} > 0$
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\item TODO b)
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\item TODO c)
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\item Ansatz: $\sigma = \begin{pmatrix}a & b\\c & d\end{pmatrix}$
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$\sigma(x_0) = \frac{ax_0 + b}{c x_0 + d} \overset{!}{=} 0$
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$\Rightarrow a x_0 + b = 0 \Rightarrow b = -a x_0$\\
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$\sigma(x_\infty) = \infty \Rightarrow c x_\infty + d = 0 \Rightarrow d = - x_\infty$\\
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$\sigma(x_1) = 1 \Rightarrow a x_1 + b = c x_1 + d$\\
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$a (x_1 - x_0) = c (x_1 - x_\infty) \Rightarrow c = a \frac{x_1 - x_0}{x_1 - x_\infty}$\\
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$\Rightarrow - a^2 \cdot x_\infty \frac{x_1 - x_0}{x_1 - x_\infty} + a^2 x_0 \frac{x_1 - x_0}{x_1 - x_\infty} = 1$\\
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$\Rightarrow a^2 \frac{x_1 - x_0}{x_1 - x_\infty} (x_0 - x_\infty) = 1$
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$\Rightarrow a^2 = \frac{x_1 - x_\infty}{(x_1 - x_\infty) (x_1 - x_0)}$
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\item TODO d)
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\item Es genügt die Aussage für Matrizen aus \cref{prop:15.2d}
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zu zeigen.
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\begin{itemize}
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\item $\sigma = \begin{pmatrix}\lambda & 0\\ 0 & \lambda^{-1}\end{pmatrix}$, also $\sigma(z) = \lambda^2 z$
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\begin{figure}[ht]
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\centering
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\subfloat[Fall 1]{
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\input{figures/hyberbolische-geometrie-1.tex}
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\label{fig:prop15.2.e.fall1.1}
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}%
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\subfloat[Fall 2 (Strahlensatz)]{
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\input{figures/hyberbolische-geometrie-2.tex}
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\label{fig:prop15.2.e.fall1.1}
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}%
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\label{fig:prop15.2.e.fall1.0}
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\caption{TODO}
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\end{figure}
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\item Offensichtlich gilt die Aussage für $\sigma = \begin{pmatrix}1 & a\\0 & 1\end{pmatrix}$
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\item Sei nun $\sigma = \begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}$, also $\sigma(z) = - \frac{1}{z}$
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\begin{figure}[htp]
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\centering
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\input{figures/inversion-am-kreis.tex}
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\caption{Inversion am Kreis}
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\label{fig:inversion-am-kreis}
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\end{figure}
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\end{itemize}
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\end{enumerate}
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\end{beweis}
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