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-rw-r--r--vorlesungen/slides/7/Makefile.inc4
-rw-r--r--vorlesungen/slides/7/bch.tex76
-rw-r--r--vorlesungen/slides/7/chapter.tex4
-rw-r--r--vorlesungen/slides/7/dg.tex4
-rw-r--r--vorlesungen/slides/7/einparameter.tex6
-rw-r--r--vorlesungen/slides/7/integration.tex66
-rw-r--r--vorlesungen/slides/7/liealgbeispiel.tex78
-rw-r--r--vorlesungen/slides/7/vektorlie.tex206
8 files changed, 439 insertions, 5 deletions
diff --git a/vorlesungen/slides/7/Makefile.inc b/vorlesungen/slides/7/Makefile.inc
index 7512612..52c37d8 100644
--- a/vorlesungen/slides/7/Makefile.inc
+++ b/vorlesungen/slides/7/Makefile.inc
@@ -16,7 +16,10 @@ chapter5 = \
../slides/7/einparameter.tex \
../slides/7/ableitung.tex \
../slides/7/liealgebra.tex \
+ ../slides/7/liealgbeispiel.tex \
+ ../slides/7/vektorlie.tex \
../slides/7/kommutator.tex \
+ ../slides/7/bch.tex \
../slides/7/dg.tex \
../slides/7/zusammenhang.tex \
../slides/7/quaternionen.tex \
@@ -24,5 +27,6 @@ chapter5 = \
../slides/7/ueberlagerung.tex \
../slides/7/hopf.tex \
../slides/7/haar.tex \
+ ../slides/7/integration.tex \
../slides/7/chapter.tex
diff --git a/vorlesungen/slides/7/bch.tex b/vorlesungen/slides/7/bch.tex
new file mode 100644
index 0000000..0148dc4
--- /dev/null
+++ b/vorlesungen/slides/7/bch.tex
@@ -0,0 +1,76 @@
+%
+% bch.tex -- slide template
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+%
+\bgroup
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Baker-Campbell-Hausdorff-Formel}
+$g(t),h(t)\in G
+\uncover<2->{\Rightarrow \exists A,B\in LG\text{ mit }
+g(t)=\exp At, h(t)=\exp Bt}$
+\uncover<3->{%
+\begin{align*}
+g(t)
+&=
+I + At + \frac{A^2t^2}{2!} + \frac{A^3t^3}{3!} + \dots,
+&
+h(t)
+&=
+I + Bt + \frac{B^2t^2}{2!} + \frac{B^3t^3}{3!} + \dots
+\end{align*}}
+\uncover<5->{%
+\begin{block}{Kommutator in G: $c(t) = g(t)h(t)g(t)^{-1}h(t)^{-1}$}
+\begin{align*}
+\uncover<6->{c(t)
+&=
+\biggl(
+ {\color<7,9-11,13-15,19-21>{red}I}
+ + {\color<8,16-19>{red}A}t
+ + \frac{{\color<12>{red}A^2}t^2}{2!}
+ + \dots
+\biggr)
+\biggl(
+ {\color<7,8,10-12,14-15,17-18,21>{red}I}
+ + {\color<9,16,19-20>{red}B}t
+ + \frac{{\color<13>{red}B^2}t^2}{2!}
+ + \dots
+\biggr)
+\exp(-{\color<10,14,17,19,21>{red}A}t)
+\exp(-{\color<11,15,18,20-21>{red}B}t)
+}
+\\
+&\uncover<7->{={\color<7>{red}I}}
+\uncover<8->{+t(
+ \uncover<8->{ {\color<8>{red}A}}
+ \uncover<9->{+ {\color<9>{red}B}}
+ \uncover<10->{- {\color<10>{red}A}}
+ \uncover<11->{- {\color<11>{red}B}}
+)}
+\uncover<12->{+\frac{t^2}{2!}(
+ \uncover<12->{ {\color<12>{red}A^2}}
+ \uncover<13->{+ {\color<13>{red}B^2}}
+ \uncover<14->{+ {\color<14>{red}A^2}}
+ \uncover<15->{+ {\color<15>{red}B^2}}
+)}
+\\
+&\phantom{\mathstrut=I}
+\uncover<12->{+t^2(
+ \uncover<16->{ {\color<16>{red}AB}}
+ \uncover<17->{- {\color<17>{red}A^2}}
+ \uncover<18->{- {\color<18>{red}AB}}
+ \uncover<19->{- {\color<19>{red}BA}}
+ \uncover<20->{- {\color<20>{red}B^2}}
+ \uncover<21->{+ {\color<21>{red}AB}}
+)}
+\uncover<22->{+t^3(\dots)+\dots}
+\\
+&\uncover<23->{=
+I + \frac{t^2}{2}[A,B] + o(t^3)
+}
+\end{align*}}
+\end{block}
+\end{frame}
+\egroup
diff --git a/vorlesungen/slides/7/chapter.tex b/vorlesungen/slides/7/chapter.tex
index 1c78ccc..172b78a 100644
--- a/vorlesungen/slides/7/chapter.tex
+++ b/vorlesungen/slides/7/chapter.tex
@@ -15,7 +15,10 @@
\folie{7/einparameter.tex}
\folie{7/ableitung.tex}
\folie{7/liealgebra.tex}
+\folie{7/liealgbeispiel.tex}
+\folie{7/vektorlie.tex}
\folie{7/kommutator.tex}
+\folie{7/bch.tex}
\folie{7/dg.tex}
\folie{7/zusammenhang.tex}
\folie{7/quaternionen.tex}
@@ -23,3 +26,4 @@
\folie{7/ueberlagerung.tex}
\folie{7/hopf.tex}
\folie{7/haar.tex}
+\folie{7/integration.tex}
diff --git a/vorlesungen/slides/7/dg.tex b/vorlesungen/slides/7/dg.tex
index 4447bac..f9528a4 100644
--- a/vorlesungen/slides/7/dg.tex
+++ b/vorlesungen/slides/7/dg.tex
@@ -45,7 +45,7 @@ Ableitung von $\gamma(t)$ an der Stelle $t$:
\vspace{-10pt}
\uncover<7->{%
\begin{block}{Differentialgleichung}
-\vspace{-10pt}
+%\vspace{-10pt}
\[
\dot{\gamma}(t) = \gamma(t) A
\quad
@@ -66,7 +66,7 @@ Exponentialfunktion
\vspace{-5pt}
\uncover<9->{%
\begin{block}{Kontrolle: Tangentialvektor berechnen}
-\vspace{-10pt}
+%\vspace{-10pt}
\begin{align*}
\frac{d}{dt}e^{At}
&\uncover<10->{=
diff --git a/vorlesungen/slides/7/einparameter.tex b/vorlesungen/slides/7/einparameter.tex
index 5171085..a32affd 100644
--- a/vorlesungen/slides/7/einparameter.tex
+++ b/vorlesungen/slides/7/einparameter.tex
@@ -41,7 +41,7 @@ D_{x,t+s}
\begin{column}{0.48\textwidth}
\uncover<5->{%
\begin{block}{Scherungen in $\operatorname{SL}_2(\mathbb{R})$}
-\vspace{-12pt}
+%\vspace{-12pt}
\[
\begin{pmatrix}
1&s\\
@@ -61,7 +61,7 @@ D_{x,t+s}
\vspace{-12pt}
\uncover<6->{%
\begin{block}{Skalierungen in $\operatorname{SL}_2(\mathbb{R})$}
-\vspace{-12pt}
+%\vspace{-12pt}
\[
\begin{pmatrix}
e^s&0\\0&e^{-s}
@@ -78,7 +78,7 @@ e^{t+s}&0\\0&e^{-(t+s)}
\vspace{-12pt}
\uncover<7->{%
\begin{block}{Gemischt}
-\vspace{-12pt}
+%\vspace{-12pt}
\begin{gather*}
A_t = I \cosh t + \begin{pmatrix}1&a\\0&-1\end{pmatrix}\sinh t
\\
diff --git a/vorlesungen/slides/7/integration.tex b/vorlesungen/slides/7/integration.tex
new file mode 100644
index 0000000..525e6de
--- /dev/null
+++ b/vorlesungen/slides/7/integration.tex
@@ -0,0 +1,66 @@
+%
+% integration.tex -- slide template
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+%
+\bgroup
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Invariante Integration}
+\vspace{-20pt}
+\begin{columns}[t,onlytextwidth]
+\begin{column}{0.48\textwidth}
+\begin{block}{Koordinatenwechsel}
+Die Koordinatentransformation
+$f\colon\mathbb{R}^n\to\mathbb{R}^n:x\to y$
+hat die Ableitungsmatrix
+\[
+t_{ij}
+=
+\frac{\partial y_i}{\partial x_j}
+\]
+\uncover<2->{%
+$n$-faches Integral
+\begin{gather*}
+\int\dots\int
+h(f(x))
+\det
+\biggl(
+\frac{\partial y_i}{\partial x_j}
+\biggr)
+\,dx_1\,\dots dx_n
+\\
+=
+\int\dots\int
+h(y)
+\,dy_1\,\dots dy_n
+\end{gather*}}
+\end{block}
+\end{column}
+\begin{column}{0.48\textwidth}
+\uncover<3->{%
+\begin{block}{auf einer Lie-Gruppe}
+Koordinatenwechsel sind Multiplikationen mit einer
+Matrix $g\in G$
+\end{block}}
+\uncover<4->{%
+\begin{block}{Volumenelement in $I$}
+Man muss nur das Volumenelement in $I$ in einem beliebigen
+Koordinatensystem definieren:
+\[
+dV = dy_1\,\dots\,dy_n
+\]
+\end{block}}
+\uncover<5->{%
+\begin{block}{Volumenelement in $g$}
+\[
+\text{``\strut}g\cdot dV\text{\strut''}
+=
+\det(g) \, dy_1\,\dots\,dy_n
+\]
+\end{block}}
+\end{column}
+\end{columns}
+\end{frame}
+\egroup
diff --git a/vorlesungen/slides/7/liealgbeispiel.tex b/vorlesungen/slides/7/liealgbeispiel.tex
new file mode 100644
index 0000000..a17de40
--- /dev/null
+++ b/vorlesungen/slides/7/liealgbeispiel.tex
@@ -0,0 +1,78 @@
+%
+% liealgbeispiel.tex -- slide template
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+%
+\bgroup
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Lie-Algebra Beispiele}
+\vspace{-20pt}
+\begin{columns}[t,onlytextwidth]
+\begin{column}{0.48\textwidth}
+\begin{block}{$\operatorname{sl}_2(\mathbb{R})$}
+Spurlose Matrizen:
+\[
+\operatorname{sl}_2(\mathbb{R})
+=
+\{A\in M_n(\mathbb{R})\;|\; \operatorname{Spur}A=0\}
+\]
+\end{block}
+\begin{block}{Lie-Algebra?}
+Nachrechnen: $[A,B]\in \operatorname{sl}_2(\mathbb{R})$:
+\begin{align*}
+\operatorname{Spur}([A,B])
+&=
+\operatorname{Spur}(AB-BA)
+\\
+&=
+\operatorname{Spur}(AB)-\operatorname{Spur}(BA)
+\\
+&=
+\operatorname{Spur}(AB)-\operatorname{Spur}(AB)
+\\
+&=0
+\end{align*}
+$\Rightarrow$ $\operatorname{sl}_2(\mathbb{R})$ ist eine Lie-Algebra
+\end{block}
+\end{column}
+\begin{column}{0.48\textwidth}
+\begin{block}{$\operatorname{so}(n)$}
+Antisymmetrische Matrizen:
+\[
+\operatorname{so}(n)
+=
+\{A\in M_n(\mathbb{R})
+\;|\;
+A=-A^t
+\}
+\]
+\end{block}
+\begin{block}{Lie-Algebra?}
+Nachrechnen: $A,B\in \operatorname{so}(n)$
+\begin{align*}
+[A,B]^t
+&=
+(AB-BA)^t
+\\
+&=
+B^tA^t - A^tB^t
+\\
+&=
+(-B)(-A)-(-A)(-B)
+\\
+&=
+BA-AB
+=
+-(AB-BA)
+\\
+&=
+-[A,B]
+\end{align*}
+$\Rightarrow$ $\operatorname{so}(n)$ ist eine Lie-Algebra
+\end{block}
+\end{column}
+\end{columns}
+\end{frame}
+\egroup
diff --git a/vorlesungen/slides/7/vektorlie.tex b/vorlesungen/slides/7/vektorlie.tex
new file mode 100644
index 0000000..621a832
--- /dev/null
+++ b/vorlesungen/slides/7/vektorlie.tex
@@ -0,0 +1,206 @@
+%
+% viktorlie.tex -- slide template
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+%
+\bgroup
+\definecolor{darkgreen}{rgb}{0,0.6,0}
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Vektorprodukt als Lie-Algebra}
+%\vspace{-10pt}
+\centering
+\begin{tikzpicture}[>=latex,thick]
+\arraycolsep=2.4pt
+\def\Ax{0}
+\def\Ux{4.1}
+\def\Kx{7.2}
+\def\Rx{13.1}
+
+\def\Lx{2.2}
+\def\Ly{0}
+\def\Lz{-2.2}
+
+\fill[color=red!20] (\Ax,{\Lx-1.55}) rectangle ({\Ux-0.1},{\Lx+0.55});
+\fill[color=red!20] (\Ux,{\Lx-1.55}) rectangle ({\Kx-0.1},{\Lx+0.55});
+\fill[color=red!20] (\Kx,{\Lx-1.55}) rectangle ({\Rx},{\Lx+0.55});
+
+\fill[color=darkgreen!20] (\Ax,{\Ly-1.55}) rectangle ({\Ux-0.1},{\Ly+0.55});
+\fill[color=darkgreen!20] (\Ux,{\Ly-1.55}) rectangle ({\Kx-0.1},{\Ly+0.55});
+\fill[color=darkgreen!20] (\Kx,{\Ly-1.55}) rectangle ({\Rx},{\Ly+0.55});
+
+\fill[color=blue!20] (\Ax,{\Lz-1.55}) rectangle ({\Ux-0.1},{\Lz+0.55});
+\fill[color=blue!20] (\Ux,{\Lz-1.55}) rectangle ({\Kx-0.1},{\Lz+0.55});
+\fill[color=blue!20] (\Kx,{\Lz-1.55}) rectangle ({\Rx},{\Lz+0.55});
+
+\coordinate (A) at (\Ax,3.2);
+\coordinate (Ax) at (\Ax,\Lx);
+\coordinate (Ay) at (\Ax,\Ly);
+\coordinate (Az) at (\Ax,\Lz);
+
+\node at (A) [right]
+ {\usebeamercolor[fg]{title}Drehmatrix, $\operatorname{SO}(n)$\strut};
+
+\node at (Ax) [right] {$\displaystyle\tiny
+D_{x,\alpha}=\begin{pmatrix}
+1&0&0\\
+0&\cos\alpha&-\sin\alpha\\
+0&\sin\alpha&\cos\alpha
+\end{pmatrix}$};
+
+\node at (Ay) [right] {$\displaystyle\tiny
+D_{y,\alpha}=\begin{pmatrix}
+\cos\alpha&0&\sin\alpha\\
+0&1&0\\
+-\sin\alpha&0&\cos\alpha
+\end{pmatrix}$};
+
+\node at (Az) [right] {$\displaystyle\tiny
+D_{z,\alpha}=\begin{pmatrix}
+\cos\alpha&-\sin\alpha&0\\
+\sin\alpha&\cos\alpha&0\\
+0&0&1
+\end{pmatrix}$};
+
+\coordinate (U) at (\Ux,3.2);
+\coordinate (Ux) at (\Ux,\Lx);
+\coordinate (Uy) at (\Ux,\Ly);
+\coordinate (Uz) at (\Ux,\Lz);
+\coordinate (Ex) at (\Ux,{\Lx-1});
+\coordinate (Ey) at (\Ux,{\Ly-1});
+\coordinate (Ez) at (\Ux,{\Lz-1});
+
+\uncover<2->{
+\node at (U) [right]
+ {\usebeamercolor[fg]{title}Ableitung, $\operatorname{so}(n)$\strut};
+
+\node at (Ux) [right] {$\displaystyle\tiny
+U_x=\begin{pmatrix*}[r]
+0&0&0\\
+0&0&-1\\
+0&1&0
+\end{pmatrix*}
+$};
+
+\node at (Uy) [right] {$\displaystyle\tiny
+U_y=\begin{pmatrix*}[r]
+0&0&1\\
+0&0&0\\
+-1&0&0
+\end{pmatrix*}
+$};
+
+\node at (Uz) [right] {$\displaystyle\tiny
+U_z=\begin{pmatrix*}[r]
+0&-1&0\\
+1&0&0\\
+0&0&0
+\end{pmatrix*}
+$};
+}
+
+\uncover<9->{
+\node at (Ex) [right] {$\displaystyle
+\, e_x = \tiny\begin{pmatrix}1\\0\\0\end{pmatrix}
+$};
+
+\node at (Ey) [right] {$\displaystyle
+\, e_y = \tiny\begin{pmatrix}0\\1\\0\end{pmatrix}
+$};
+
+\node at (Ez) [right] {$\displaystyle
+\, e_z = \tiny\begin{pmatrix}0\\0\\1\end{pmatrix}
+$};
+}
+
+\coordinate (K) at (\Kx,3.2);
+\coordinate (Kx) at (\Kx,\Lx);
+\coordinate (Ky) at (\Kx,\Ly);
+\coordinate (Kz) at (\Kx,\Lz);
+\coordinate (Vx) at (\Kx,{\Lx-1});
+\coordinate (Vy) at (\Kx,{\Ly-1});
+\coordinate (Vz) at (\Kx,{\Lz-1});
+
+\uncover<3->{
+\node at (K) [right]
+ {\usebeamercolor[fg]{title}Kommutator\strut};
+
+\node at (Kx) [right] {$\displaystyle
+\begin{aligned}
+[U_y,U_z] &\uncover<4->{=
+{\tiny
+\begin{pmatrix}
+0&0&0\\
+0&0&0\\
+0&1&0
+\end{pmatrix}}
+\uncover<5->{\mathstrut-
+\tiny
+\begin{pmatrix}
+0&0&0\\
+0&0&1\\
+0&0&0
+\end{pmatrix}}}
+\uncover<6->{=U_x}
+\end{aligned}
+$};
+}
+
+\uncover<7->{
+\node at (Ky) [right] {$\displaystyle
+\begin{aligned}
+[U_z,U_x] &=
+{\tiny
+\begin{pmatrix}
+0&0&1\\
+0&0&0\\
+0&0&0
+\end{pmatrix}
+-
+\begin{pmatrix}
+0&0&0\\
+0&0&0\\
+1&0&0
+\end{pmatrix}}
+=U_y
+\end{aligned}
+$};
+}
+
+\uncover<8->{
+\node at (Kz) [right] {$\displaystyle
+\begin{aligned}
+[U_x,U_y] &=
+{\tiny
+\begin{pmatrix}
+0&0&0\\
+1&0&0\\
+0&0&0
+\end{pmatrix}
+-
+\begin{pmatrix}
+0&1&0\\
+0&0&0\\
+0&0&0
+\end{pmatrix}}
+=U_z
+\end{aligned}
+$};
+}
+
+\uncover<10->{
+\node at (Vx) [right] {$\displaystyle \phantom{]}e_y\times e_z = e_x$};
+}
+
+\uncover<11->{
+\node at (Vy) [right] {$\displaystyle \phantom{]}e_z\times e_x = e_y$};
+}
+
+\uncover<12->{
+\node at (Vz) [right] {$\displaystyle \phantom{]}e_x\times e_y = e_z$};
+}
+
+\end{tikzpicture}
+\end{frame}
+\egroup