Merge pull request #13 from NaoPross/master

Paper title & Presentation

17 files changed, 2552 insertions, 196 deletions

diff --git a/buch/papers/punktgruppen/Makefile.inc b/buch/papers/punktgruppen/Makefile.inc
index 7c6e70d..629abca 100644
--- a/buch/papers/punktgruppen/Makefile.inc
+++ b/buch/papers/punktgruppen/Makefile.inc
@@ -3,12 +3,8 @@
 #
 # (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
 #
-dependencies-punktgruppen =						\
-	papers/punktgruppen/packages.tex					\
-	papers/punktgruppen/main.tex					\
-	papers/punktgruppen/references.bib				\
-	papers/punktgruppen/teil0.tex					\
-	papers/punktgruppen/teil1.tex					\
-	papers/punktgruppen/teil2.tex					\
-	papers/punktgruppen/teil3.tex
+dependencies-punktgruppen =	\
+	papers/punktgruppen/packages.tex \
+	papers/punktgruppen/main.tex \
+	papers/punktgruppen/references.bib
 
diff --git a/buch/papers/punktgruppen/main.tex b/buch/papers/punktgruppen/main.tex
index fc91913..603f293 100644
--- a/buch/papers/punktgruppen/main.tex
+++ b/buch/papers/punktgruppen/main.tex
@@ -3,34 +3,38 @@
 %
 % (c) 2020 Hochschule Rapperswil
 %
-\chapter{Thema\label{chapter:punktgruppen}}
-\lhead{Thema}
+\chapter{Crystal M\rotatebox[origin=c]{180}{a}th\label{chapter:punktgruppen}}
+\lhead{Crystal M\rotatebox[origin=c]{180}{a}th}
 \begin{refsection}
-\chapterauthor{Hans Muster}
-
-Ein paar Hinweise für die korrekte Formatierung des Textes
-\begin{itemize}
-\item
-Absätze werden gebildet, indem man eine Leerzeile einfügt.
-Die Verwendung von \verb+\\+ ist nur in Tabellen und Arrays gestattet.
-\item
-Die explizite Platzierung von Bildern ist nicht erlaubt, entsprechende
-Optionen werden gelöscht. 
-Verwenden Sie Labels und Verweise, um auf Bilder hinzuweisen.
-\item
-Beginnen Sie jeden Satz auf einer neuen Zeile. 
-Damit ermöglichen Sie dem Versionsverwaltungssysteme, Änderungen
-in verschiedenen Sätzen von verschiedenen Autoren ohne Konflikt 
-anzuwenden.
-\item 
-Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren
-Übersicht wegen, aber auch um GIT die Arbeit zu erleichtern.
-\end{itemize}
-
-\input{papers/punktgruppen/teil0.tex}
-\input{papers/punktgruppen/teil1.tex}
-\input{papers/punktgruppen/teil2.tex}
-\input{papers/punktgruppen/teil3.tex}
+\chapterauthor{Tim T\"onz, Naoki Pross}
+
+%% TODO: remove
+%% Some ideas to motivate the topic:
+%%  - Physics in a crystal lattice structure
+%%    - Birifrencenge and scattering of light / Xray in Crystals
+%%    - Electron density function in a lattice
+%%    - Heat diffusion with lattice model
+%%    - Ising model for ferromagnetism (?? => H.D. Lang)
+%%
+%%  - Homomorphic encryption (or lattice based cryptography)
+%%    + Q: Is it possible to edit encrypted data without decrypting it first?
+
+%% TODO: translated and move into a file {{{
+
+\section{Motivation}
+% birifrengence
+
+\section{Math}
+% lattice group
+% symmetry
+% space group
+
+\section{Physics}
+\subsection{Electromagnetic Waves}
+\subsection{Crystal Lattice}
+
+
+%% }}}
 
 \printbibliography[heading=subbibliography]
 \end{refsection}
diff --git a/buch/papers/punktgruppen/packages.tex b/buch/papers/punktgruppen/packages.tex
index 971bcfe..9953339 100644
--- a/buch/papers/punktgruppen/packages.tex
+++ b/buch/papers/punktgruppen/packages.tex
@@ -4,7 +4,4 @@
 % (c) 2019 Prof Dr Andreas Müller, Hochschule Rapperswil
 %
 
-% if your paper needs special packages, add package commands as in the
-% following example
-%\usepackage{packagename}
-
+\usepackage{tikz-3dplot}
diff --git a/buch/papers/punktgruppen/teil0.tex b/buch/papers/punktgruppen/teil0.tex
deleted file mode 100644
index 5a8278e..0000000
--- a/buch/papers/punktgruppen/teil0.tex
+++ /dev/null
@@ -1,22 +0,0 @@
-%
-% einleitung.tex -- Beispiel-File für die Einleitung
-%
-% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
-%
-\section{Teil 0\label{punktgruppen:section:teil0}}
-\rhead{Teil 0}
-Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam
-nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam
-erat, sed diam voluptua \cite{punktgruppen:bibtex}.
-At vero eos et accusam et justo duo dolores et ea rebum.
-Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum
-dolor sit amet.
-
-Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam
-nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam
-erat, sed diam voluptua.
-At vero eos et accusam et justo duo dolores et ea rebum.  Stet clita
-kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit
-amet.
-
-
diff --git a/buch/papers/punktgruppen/teil1.tex b/buch/papers/punktgruppen/teil1.tex
deleted file mode 100644
index 228af33..0000000
--- a/buch/papers/punktgruppen/teil1.tex
+++ /dev/null
@@ -1,55 +0,0 @@
-%
-% teil1.tex -- Beispiel-File für das Paper
-%
-% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
-%
-\section{Teil 1
-\label{punktgruppen:section:teil1}}
-\rhead{Problemstellung}
-Sed ut perspiciatis unde omnis iste natus error sit voluptatem
-accusantium doloremque laudantium, totam rem aperiam, eaque ipsa
-quae ab illo inventore veritatis et quasi architecto beatae vitae
-dicta sunt explicabo.
-Nemo enim ipsam voluptatem quia voluptas sit aspernatur aut odit
-aut fugit, sed quia consequuntur magni dolores eos qui ratione
-voluptatem sequi nesciunt
-\begin{equation}
-\int_a^b x^2\, dx
-=
-\left[ \frac13 x^3 \right]_a^b
-=
-\frac{b^3-a^3}3.
-\label{punktgruppen:equation1}
-\end{equation}
-Neque porro quisquam est, qui dolorem ipsum quia dolor sit amet,
-consectetur, adipisci velit, sed quia non numquam eius modi tempora
-incidunt ut labore et dolore magnam aliquam quaerat voluptatem.
-
-Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis
-suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur?
-Quis autem vel eum iure reprehenderit qui in ea voluptate velit
-esse quam nihil molestiae consequatur, vel illum qui dolorem eum
-fugiat quo voluptas nulla pariatur?
-
-\subsection{De finibus bonorum et malorum
-\label{punktgruppen:subsection:finibus}}
-At vero eos et accusamus et iusto odio dignissimos ducimus qui
-blanditiis praesentium voluptatum deleniti atque corrupti quos
-dolores et quas molestias excepturi sint occaecati cupiditate non
-provident, similique sunt in culpa qui officia deserunt mollitia
-animi, id est laborum et dolorum fuga \eqref{000tempmlate:equation1}.
-
-Et harum quidem rerum facilis est et expedita distinctio
-\ref{punktgruppen:section:loesung}.
-Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil
-impedit quo minus id quod maxime placeat facere possimus, omnis
-voluptas assumenda est, omnis dolor repellendus
-\ref{punktgruppen:section:folgerung}.
-Temporibus autem quibusdam et aut officiis debitis aut rerum
-necessitatibus saepe eveniet ut et voluptates repudiandae sint et
-molestiae non recusandae.
-Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis
-voluptatibus maiores alias consequatur aut perferendis doloribus
-asperiores repellat.
-
-
diff --git a/buch/papers/punktgruppen/teil2.tex b/buch/papers/punktgruppen/teil2.tex
deleted file mode 100644
index b48e785..0000000
--- a/buch/papers/punktgruppen/teil2.tex
+++ /dev/null
@@ -1,40 +0,0 @@
-%
-% teil2.tex -- Beispiel-File für teil2 
-%
-% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
-%
-\section{Teil 2 
-\label{punktgruppen:section:teil2}}
-\rhead{Teil 2}
-Sed ut perspiciatis unde omnis iste natus error sit voluptatem
-accusantium doloremque laudantium, totam rem aperiam, eaque ipsa
-quae ab illo inventore veritatis et quasi architecto beatae vitae
-dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit
-aspernatur aut odit aut fugit, sed quia consequuntur magni dolores
-eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam
-est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci
-velit, sed quia non numquam eius modi tempora incidunt ut labore
-et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima
-veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam,
-nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure
-reprehenderit qui in ea voluptate velit esse quam nihil molestiae
-consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla
-pariatur?
-
-\subsection{De finibus bonorum et malorum
-\label{punktgruppen:subsection:bonorum}}
-At vero eos et accusamus et iusto odio dignissimos ducimus qui
-blanditiis praesentium voluptatum deleniti atque corrupti quos
-dolores et quas molestias excepturi sint occaecati cupiditate non
-provident, similique sunt in culpa qui officia deserunt mollitia
-animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis
-est et expedita distinctio. Nam libero tempore, cum soluta nobis
-est eligendi optio cumque nihil impedit quo minus id quod maxime
-placeat facere possimus, omnis voluptas assumenda est, omnis dolor
-repellendus. Temporibus autem quibusdam et aut officiis debitis aut
-rerum necessitatibus saepe eveniet ut et voluptates repudiandae
-sint et molestiae non recusandae. Itaque earum rerum hic tenetur a
-sapiente delectus, ut aut reiciendis voluptatibus maiores alias
-consequatur aut perferendis doloribus asperiores repellat.
-
-
diff --git a/buch/papers/punktgruppen/teil3.tex b/buch/papers/punktgruppen/teil3.tex
deleted file mode 100644
index 94abd74..0000000
--- a/buch/papers/punktgruppen/teil3.tex
+++ /dev/null
@@ -1,40 +0,0 @@
-%
-% teil3.tex -- Beispiel-File für Teil 3
-%
-% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
-%
-\section{Teil 3
-\label{punktgruppen:section:teil3}}
-\rhead{Teil 3}
-Sed ut perspiciatis unde omnis iste natus error sit voluptatem
-accusantium doloremque laudantium, totam rem aperiam, eaque ipsa
-quae ab illo inventore veritatis et quasi architecto beatae vitae
-dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit
-aspernatur aut odit aut fugit, sed quia consequuntur magni dolores
-eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam
-est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci
-velit, sed quia non numquam eius modi tempora incidunt ut labore
-et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima
-veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam,
-nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure
-reprehenderit qui in ea voluptate velit esse quam nihil molestiae
-consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla
-pariatur?
-
-\subsection{De finibus bonorum et malorum
-\label{punktgruppen:subsection:malorum}}
-At vero eos et accusamus et iusto odio dignissimos ducimus qui
-blanditiis praesentium voluptatum deleniti atque corrupti quos
-dolores et quas molestias excepturi sint occaecati cupiditate non
-provident, similique sunt in culpa qui officia deserunt mollitia
-animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis
-est et expedita distinctio. Nam libero tempore, cum soluta nobis
-est eligendi optio cumque nihil impedit quo minus id quod maxime
-placeat facere possimus, omnis voluptas assumenda est, omnis dolor
-repellendus. Temporibus autem quibusdam et aut officiis debitis aut
-rerum necessitatibus saepe eveniet ut et voluptates repudiandae
-sint et molestiae non recusandae. Itaque earum rerum hic tenetur a
-sapiente delectus, ut aut reiciendis voluptatibus maiores alias
-consequatur aut perferendis doloribus asperiores repellat.
-
-
diff --git a/vorlesungen/punktgruppen/.gitignore b/vorlesungen/punktgruppen/.gitignore
new file mode 100644
index 0000000..3633a3d
--- /dev/null
+++ b/vorlesungen/punktgruppen/.gitignore
@@ -0,0 +1,20 @@
+# directories
+__pycache__
+media/Tex
+
+media/images/crystal
+media/images/freezeframes
+
+media/videos
+media/audio
+
+media/Punktgruppen
+media/Punktgruppen.mp4
+
+build
+
+# files
+script.log
+slides.log
+slides.vrb
+missfont.log
diff --git a/vorlesungen/punktgruppen/Makefile b/vorlesungen/punktgruppen/Makefile
new file mode 100644
index 0000000..302e976
--- /dev/null
+++ b/vorlesungen/punktgruppen/Makefile
@@ -0,0 +1,18 @@
+TEX=xelatex
+TEXARGS=--output-directory=build --halt-on-error --shell-escape
+
+all: slides.pdf script.pdf media
+
+.PHONY: clean
+clean:
+	@rm -rfv build
+
+%.pdf: %.tex
+	mkdir -p build
+	$(TEX) $(TEXARGS) $<
+	$(TEX) $(TEXARGS) $<
+	cp build/$@ .
+
+media:
+	poetry install
+	poetry run manim -ql crystals.py
diff --git a/vorlesungen/punktgruppen/crystals.py b/vorlesungen/punktgruppen/crystals.py
new file mode 100644
index 0000000..4a9836a
--- /dev/null
+++ b/vorlesungen/punktgruppen/crystals.py
@@ -0,0 +1,611 @@
+from manim import *
+
+import math as m
+import numpy as np
+import itertools as it
+
+# configure style
+config.background_color = '#202020'
+config.tex_template.add_to_preamble(
+    r"\usepackage[p,osf]{scholax}"
+    r"\usepackage{amsmath}"
+    r"\usepackage[scaled=1.075,ncf,vvarbb]{newtxmath}"
+)
+
+# scenes
+class Geometric2DSymmetries(Scene):
+    def construct(self):
+        self.wait(5)
+
+        self.intro()
+        self.cyclic()
+        self.dihedral()
+        self.circle()
+
+    def intro(self):
+        # create square
+        square = Square()
+        square.set_fill(PINK, opacity=.5)
+        self.play(SpinInFromNothing(square))
+        self.wait()
+
+        # the action of doing nothing
+        action = MathTex(r"\mathbb{1}")
+        self.play(Write(action))
+        self.play(ApplyMethod(square.scale, 1.2))
+        self.play(ApplyMethod(square.scale, 1/1.2))
+        self.play(FadeOut(action))
+        self.wait()
+
+        # show some reflections
+        axis = DashedLine(2 * LEFT, 2 * RIGHT)
+        sigma = MathTex(r"\sigma")
+        sigma.next_to(axis, RIGHT)
+
+        self.play(Create(axis))
+        self.play(Write(sigma))
+        self.play(ApplyMethod(square.flip, RIGHT))
+
+        for d in [UP + RIGHT, UP]:
+            self.play(
+                Rotate(axis, PI/4),
+                Rotate(sigma, PI/4, about_point=ORIGIN))
+
+            self.play(Rotate(sigma, -PI/4), run_time=.5)
+            self.play(ApplyMethod(square.flip, d))
+
+        self.play(FadeOutAndShift(sigma), Uncreate(axis))
+
+        # show some rotations
+        dot = Dot(UP + RIGHT)
+        figure = VGroup(square, dot)
+
+        rot = MathTex(r"r")
+        self.play(Write(rot), Create(dot))
+
+        last = rot
+        for newrot in map(MathTex, [r"r", r"r^2", r"r^3"]):
+            self.play(
+                ReplacementTransform(last, newrot),
+                Rotate(figure, PI/2, about_point=ORIGIN))
+            self.wait(.5)
+            last = newrot
+
+        self.play(Uncreate(dot), FadeOut(square), FadeOut(last))
+
+
+    def cyclic(self):
+        # create symmetric figure
+        figure = VGroup()
+        prev = [1.5, 0, 0]
+        for i in range(1,6):
+            pos = [
+                1.5*m.cos(2 * PI/5 * i),
+                1.5*m.sin(2 * PI/5 * i),
+                0
+            ]
+
+            if prev:
+                line = Line(prev, pos)
+                figure.add(line)
+
+            dot = Dot(pos, radius=.1)
+            if i == 5:
+                dot.set_fill(RED)
+
+            prev = pos
+            figure.add(dot)
+
+        group = MathTex(r"G = \langle r \rangle")
+        self.play(Write(group), run_time = 2)
+        self.wait(3)
+
+        self.play(ApplyMethod(group.to_edge, UP))
+
+        actions = map(MathTex, [
+            r"\mathbb{1}", r"r", r"r^2",
+            r"r^3", r"r^4", r"\mathbb{1}", r"r"])
+
+        action = next(actions, MathTex(r"r"))
+
+        self.play(Create(figure))
+        self.play(Write(action))
+        self.wait()
+
+        for i in range(5):
+            newaction = next(actions, MathTex(r"r"))
+            self.play(
+                ReplacementTransform(action, newaction),
+                Rotate(figure, 2*PI/5, about_point=ORIGIN))
+            action = newaction
+
+        self.wait()
+        newaction = next(actions, MathTex(r"r"))
+        self.play(
+            ReplacementTransform(action, newaction),
+            Rotate(figure, 2*PI/5, about_point=ORIGIN))
+        action = newaction
+        self.wait(2)
+
+        self.play(Uncreate(figure), FadeOut(action))
+
+        whole_group = MathTex(
+            r"G = \langle r \rangle" 
+            r"= \left\{\mathbb{1}, r, r^2, r^3, r^4 \right\}")
+
+        self.play(ApplyMethod(group.move_to, ORIGIN))
+        self.play(ReplacementTransform(group, whole_group))
+        self.wait(5)
+
+        cyclic = MathTex(
+            r"C_n = \langle r \rangle" 
+            r"= \left\{\mathbb{1}, r, r^2, \dots, r^{n-1} \right\}")
+
+        cyclic_title = Tex(r"Zyklische Gruppe")
+        cyclic_title.next_to(cyclic, UP * 2)
+
+        cyclic.scale(1.2)
+        cyclic_title.scale(1.2)
+
+        self.play(ReplacementTransform(whole_group, cyclic))
+        self.play(FadeInFrom(cyclic_title, UP))
+
+        self.wait(5)
+        self.play(FadeOut(cyclic), FadeOut(cyclic_title))
+
+    def dihedral(self):
+        # create square
+        square = Square()
+        square.set_fill(PINK, opacity=.5)
+
+        # generator equation
+        group = MathTex(
+            r"G = \langle \sigma, r \,|\,",
+            r"\sigma^2 = \mathbb{1},",
+            r"r^4 = \mathbb{1},",
+            r"(\sigma r)^2 = \mathbb{1} \rangle")
+
+        self.play(Write(group), run_time = 2)
+        self.wait(5)
+
+        self.play(ApplyMethod(group.to_edge, UP))
+        self.play(FadeIn(square))
+        self.wait()
+
+        # flips
+        axis = DashedLine(2 * LEFT, 2 * RIGHT)
+        sigma = MathTex(r"\sigma^2 = \mathbb{1}")
+        sigma.next_to(axis, RIGHT)
+        self.play(Create(axis), Write(sigma))
+        self.play(ApplyMethod(square.flip, RIGHT))
+        self.play(ApplyMethod(square.flip, RIGHT))
+        self.play(Uncreate(axis), FadeOut(sigma))
+
+        # rotations
+        dot = Dot(UP + RIGHT)
+        rot = MathTex(r"r^4 = \mathbb{1}")
+        rot.next_to(square, DOWN * 3)
+
+        figure = VGroup(dot, square)
+
+        self.play(Write(rot), Create(dot))
+        for i in range(4):
+            self.play(Rotate(figure, PI/2))
+        self.play(FadeOut(rot), Uncreate(dot))
+
+        # rotation and flip
+        action = MathTex(r"(\sigma r)^2 = \mathbb{1}")
+        action.next_to(square, DOWN * 5)
+
+        dot = Dot(UP + RIGHT)
+        axis = DashedLine(2 * LEFT, 2 * RIGHT)
+        self.play(Create(dot), Create(axis), Write(action))
+
+        figure = VGroup(dot, square)
+
+        for i in range(2):
+            self.play(Rotate(figure, PI/2))
+            self.play(ApplyMethod(figure.flip, RIGHT))
+            self.wait()
+
+        self.play(Uncreate(dot), Uncreate(axis), FadeOut(action))
+        self.play(FadeOut(square))
+
+        # equation for the whole
+        whole_group = MathTex(
+            r"G &= \langle \sigma, r \,|\,"
+            r"\sigma^2 = r^4 = (\sigma r)^2 = \mathbb{1} \rangle \\"
+            r"&= \left\{"
+            r"\mathbb{1}, r, r^2, r^3, \sigma, \sigma r, \sigma r^2, \sigma r^3"
+            r"\right\}")
+
+        self.play(ApplyMethod(group.move_to, ORIGIN))
+        self.play(ReplacementTransform(group, whole_group))
+        self.wait(2)
+
+        dihedral = MathTex(
+            r"D_n &= \langle \sigma, r \,|\,"
+            r"\sigma^2 = r^n = (\sigma r)^2 = \mathbb{1} \rangle \\"
+            r"&= \left\{"
+            r"\mathbb{1}, r, r^2, \dots, \sigma, \sigma r, \sigma r^2, \dots"
+            r"\right\}")
+
+        dihedral_title = Tex(r"Diedergruppe: Symmetrien eines \(n\)-gons")
+        dihedral_title.next_to(dihedral, UP * 2)
+
+        dihedral.scale(1.2)
+        dihedral_title.scale(1.2)
+
+        self.play(ReplacementTransform(whole_group, dihedral))
+        self.play(FadeInFrom(dihedral_title, UP))
+
+        self.wait(5)
+        self.play(FadeOut(dihedral), FadeOut(dihedral_title))
+
+    def circle(self):
+        circle = Circle(radius=2)
+        dot = Dot()
+        dot.move_to(2 * RIGHT)
+
+        figure = VGroup(circle, dot)
+        group_name = MathTex(r"C_\infty")
+
+        # create circle
+        self.play(Create(circle))
+        self.play(Create(dot))
+
+        # move it around
+        self.play(Rotate(figure, PI/3))
+        self.play(Rotate(figure, PI/6))
+        self.play(Rotate(figure, -PI/3))
+
+        # show name
+        self.play(Rotate(figure, PI/4), Write(group_name))
+        self.wait()
+        self.play(Uncreate(figure))
+
+        nsphere = MathTex(r"C_\infty \cong S^1 = \left\{z \in \mathbb{C} : |z| = 1\right\}")
+        nsphere_title = Tex(r"Kreisgruppe")
+        nsphere_title.next_to(nsphere, 2 * UP)
+
+        nsphere.scale(1.2)
+        nsphere_title.scale(1.2)
+
+        self.play(ReplacementTransform(group_name, nsphere))
+        self.play(FadeInFrom(nsphere_title, UP))
+
+        self.wait(5)
+        self.play(FadeOut(nsphere_title), FadeOut(nsphere))
+        self.wait(2)
+
+
+class Geometric3DSymmetries(ThreeDScene):
+    def construct(self):
+        self.improper_rotation()
+        self.tetrahedron()
+
+    def improper_rotation(self):
+        # changes the source of the light and camera
+        self.renderer.camera.light_source.move_to(3*IN) 
+        self.set_camera_orientation(phi=0, theta=0)
+
+        # initial square
+        square = Square()
+        square.set_fill(PINK, opacity=.5)
+
+        self.play(SpinInFromNothing(square))
+        self.wait(2)
+
+        for i in range(4):
+            self.play(Rotate(square, PI/2))
+            self.wait(.5)
+
+        self.move_camera(phi= 75 * DEGREES, theta = -80 * DEGREES)
+
+        # create rotation axis
+        axis = Line3D(start=[0,0,-2.5], end=[0,0,2.5])
+
+        axis_name = MathTex(r"r \in C_4")
+        # move to yz plane
+        axis_name.rotate(PI/2, axis = RIGHT)
+        axis_name.next_to(axis, OUT)
+
+        self.play(Create(axis))
+        self.play(Write(axis_name))
+        self.wait()
+
+        # create sphere from slices
+        cyclic_slices = []
+        for i in range(4):
+            colors = [PINK, RED] if i % 2 == 0 else [BLUE_D, BLUE_E]
+            cyclic_slices.append(ParametricSurface(
+                    lambda u, v: np.array([
+                        np.sqrt(2) * np.cos(u) * np.cos(v),
+                        np.sqrt(2) * np.cos(u) * np.sin(v),
+                        np.sqrt(2) * np.sin(u)
+                    ]),
+                    v_min=PI/4 + PI/2 * i,
+                    v_max=PI/4 + PI/2 * (i + 1),
+                    u_min=-PI/2, u_max=PI/2,
+                    checkerboard_colors=colors, resolution=(10,5)))
+
+        self.play(FadeOut(square), *map(Create, cyclic_slices))
+
+        cyclic_sphere = VGroup(*cyclic_slices)
+        for i in range(4):
+            self.play(Rotate(cyclic_sphere, PI/2))
+            self.wait()
+
+        new_axis_name = MathTex(r"r \in D_4")
+        # move to yz plane
+        new_axis_name.rotate(PI/2, axis = RIGHT)
+        new_axis_name.next_to(axis, OUT)
+        self.play(ReplacementTransform(axis_name, new_axis_name))
+
+        # reflection plane
+        self.play(FadeOut(cyclic_sphere), FadeIn(square))
+        plane = ParametricSurface(
+                lambda u, v: np.array([u, 0, v]),
+                u_min = -2, u_max = 2,
+                v_min = -2, v_max = 2,
+                fill_opacity=.3, resolution=(1,1))
+
+        plane_name = MathTex(r"\sigma \in D_4")
+        # move to yz plane
+        plane_name.rotate(PI/2, axis = RIGHT)
+        plane_name.next_to(plane, OUT + RIGHT)
+
+        self.play(Create(plane))
+        self.play(Write(plane_name))
+        self.wait()
+
+        self.move_camera(phi = 25 * DEGREES, theta = -75 * DEGREES)
+        self.wait()
+
+        condition = MathTex(r"(\sigma r)^2 = \mathbb{1}")
+        condition.next_to(square, DOWN);
+
+        self.play(Write(condition))
+        self.play(Rotate(square, PI/2))
+        self.play(Rotate(square, PI, RIGHT))
+
+        self.play(Rotate(square, PI/2))
+        self.play(Rotate(square, PI, RIGHT))
+        self.play(FadeOut(condition))
+
+        self.move_camera(phi = 75 * DEGREES, theta = -80 * DEGREES)
+
+        # create sphere from slices
+        dihedral_slices = []
+        for i in range(4):
+            for j in range(2):
+                colors = [PINK, RED] if i % 2 == 0 else [BLUE_D, BLUE_E]
+                dihedral_slices.append(ParametricSurface(
+                        lambda u, v: np.array([
+                            np.sqrt(2) * np.cos(u) * np.cos(v),
+                            np.sqrt(2) * np.cos(u) * np.sin(v),
+                            np.sqrt(2) * np.sin(u)
+                        ]),
+                        v_min=PI/2 * j + PI/4 + PI/2 * i,
+                        v_max=PI/2 * j + PI/4 + PI/2 * (i + 1),
+                        u_min=-PI/2 if j == 0 else 0,
+                        u_max=0 if j == 0 else PI/2,
+                        checkerboard_colors=colors, resolution=(10,5)))
+
+        dihedral_sphere = VGroup(*dihedral_slices)
+
+        self.play(FadeOut(square), Create(dihedral_sphere))
+
+        for i in range(2):
+            self.play(Rotate(dihedral_sphere, PI/2))
+            self.play(Rotate(dihedral_sphere, PI, RIGHT))
+            self.wait()
+
+        self.wait(2)
+        self.play(*map(FadeOut, [dihedral_sphere, plane, plane_name, new_axis_name]), FadeIn(square))
+        self.wait(3)
+        self.play(*map(FadeOut, [square, axis]))
+        self.wait(3)
+
+    def tetrahedron(self):
+        tet = Tetrahedron(edge_length=2)
+        self.play(FadeIn(tet))
+        
+        self.move_camera(phi = 75 * DEGREES, theta = -100 * DEGREES)
+        self.begin_ambient_camera_rotation(rate=.1)
+
+        axes = []
+        for coord in tet.vertex_coords:
+            axes.append((-2 * coord, 2 * coord))
+
+        lines = [
+            Line3D(start=s, end=e) for s, e in axes
+        ]
+
+        self.play(*map(Create, lines))
+        self.wait()
+
+        for axis in axes:
+            self.play(Rotate(tet, 2*PI/3, axis=axis[1]))
+            self.play(Rotate(tet, 2*PI/3, axis=axis[1]))
+
+        self.wait(5)
+        self.stop_ambient_camera_rotation()
+        self.wait()
+        self.play(*map(Uncreate, lines))
+        self.play(FadeOut(tet))
+        self.wait(5)
+
+
+class AlgebraicSymmetries(Scene):
+    def construct(self):
+        self.wait(5)
+        self.cyclic()
+        # self.matrices()
+
+    def cyclic(self):
+        # show the i product
+        product = MathTex(
+            r"1", r"\cdot i &= i \\",
+            r"i \cdot i &= -1 \\",
+            r"-1 \cdot i &= -i \\",
+            r"-i \cdot i &= 1")
+        product.scale(1.5)
+
+        for part in product:
+            self.play(Write(part))
+            self.wait()
+
+        self.play(ApplyMethod(product.scale, 1/1.5))
+
+        # gather in group
+        group = MathTex(r"G = \left\{ 1, i, -1, -i \right\}")
+        self.play(ReplacementTransform(product, group))
+        self.wait(2)
+
+        # show C4
+        grouppow =  MathTex(
+            r"G &= \left\{ 1, i, i^2, i^3 \right\} \\",
+            r"C_4 &= \left\{ \mathbb{1}, r, r^2, r^3 \right\}")
+        self.play(ReplacementTransform(group, grouppow[0]))
+        self.wait(2)
+
+        self.play(Write(grouppow[1]))
+        self.wait(4)
+        
+        self.play(ApplyMethod(grouppow.to_edge, UP))
+
+        # define morphisms
+        morphism = MathTex(r"\phi: C_4 \to G \\")
+        morphism.shift(UP)
+        self.play(Write(morphism))
+        self.wait()
+
+        # show an example
+        mappings = MathTex(
+            r"\phi(\mathbb{1}) &= 1 \\",
+            r"\phi(r) &= i \\",
+            r"\phi(r^2) &= i^2 \\",
+            r"\phi(r^3) &= i^3 \\")
+        mappings.next_to(morphism, 2 * DOWN)
+
+        self.play(Write(mappings))
+        self.wait(3)
+        self.play(FadeOutAndShift(mappings, DOWN))
+
+        # more general definition
+        homomorphism = MathTex(
+            r"\phi(r\circ \mathbb{1}) &= \phi(r)\cdot\phi(\mathbb{1}) \\",
+            r"&= i\cdot 1")
+        homomorphism.next_to(morphism, DOWN).align_to(morphism, LEFT)
+        for part in homomorphism:
+            self.play(Write(part))
+            self.wait()
+
+        hom_bracegrp = VGroup(morphism, homomorphism)
+
+        self.play(
+            ApplyMethod(grouppow.shift, 3 * LEFT),
+            ApplyMethod(hom_bracegrp.shift, 3 * LEFT))
+
+        hom_brace = Brace(hom_bracegrp, direction=RIGHT)
+        hom_text = Tex("Homomorphismus").next_to(hom_brace.get_tip(), RIGHT)
+        hom_text_short = MathTex(r"\mathrm{Hom}(C_4, G)").next_to(hom_brace.get_tip(), RIGHT)
+
+        self.play(Create(hom_brace))
+        self.play(Write(hom_text))
+        self.wait()
+        self.play(ReplacementTransform(hom_text, hom_text_short))
+        self.wait()
+
+        # add the isomorphism part
+        isomorphism = Tex(r"\(\phi\) ist bijektiv")
+        isomorphism.next_to(homomorphism, DOWN).align_to(homomorphism, LEFT)
+        self.play(Write(isomorphism))
+
+        iso_bracegrp = VGroup(hom_bracegrp, isomorphism)
+
+        iso_brace = Brace(iso_bracegrp, RIGHT)
+        iso_text = Tex("Isomorphismus").next_to(iso_brace.get_tip(), RIGHT)
+        iso_text_short = MathTex("C_4 \cong G").next_to(iso_brace.get_tip(), RIGHT)
+
+        self.play(
+            ReplacementTransform(hom_brace, iso_brace),
+            ReplacementTransform(hom_text_short, iso_text))
+        self.wait()
+
+        self.play(ReplacementTransform(iso_text, iso_text_short))
+        self.wait()
+
+        # create a group for the whole
+        morphgrp = VGroup(iso_bracegrp, iso_brace, iso_text_short)
+
+        self.play(
+            ApplyMethod(grouppow.to_edge, LEFT),
+            ApplyMethod(morphgrp.to_edge, LEFT))
+
+        # draw a complex plane
+        plane = ComplexPlane(x_range = [-2.5, 2.5])
+        coordinates = plane.get_coordinate_labels(1, -1, 1j, -1j)
+
+        roots = list(map(lambda p: Dot(p, fill_color=PINK), (
+            [1, 0, 0], [0, 1, 0], [-1, 0, 0], [0, -1, 0]
+        )))
+
+        arrow = CurvedArrow(
+            1.5 * np.array([m.cos(10 * DEGREES), m.sin(10 * DEGREES), 0]),
+            1.5 * np.array([m.cos(80 * DEGREES), m.sin(80 * DEGREES), 0]))
+        arrowtext = MathTex("\cdot i")
+        arrowtext.move_to(2 / m.sqrt(2) * (UP + RIGHT))
+
+        square = Square().rotate(PI/4).scale(1/m.sqrt(2))
+        square.set_fill(PINK).set_opacity(.4)
+
+        figuregrp = VGroup(plane, square, arrow, arrowtext, *coordinates, *roots)
+        figuregrp.to_edge(RIGHT)
+
+        self.play(Create(plane))
+        self.play(
+            *map(Create, roots),
+            *map(Write, coordinates))
+        self.wait()
+        self.play(FadeIn(square), Create(arrow), Write(arrowtext))
+
+        for _ in range(4):
+            self.play(Rotate(square, PI/2))
+            self.wait(.5)
+
+        self.play(
+            *map(FadeOut, (square, arrow, arrowtext)),
+            *map(FadeOut, coordinates),
+            *map(FadeOut, roots))
+        self.play(Uncreate(plane))
+        self.play(
+            FadeOutAndShift(grouppow, RIGHT),
+            FadeOutAndShift(morphgrp, RIGHT))
+
+        modulo = MathTex(
+            r"\phi: C_4 &\to (\mathbb{Z}/4\mathbb{Z}, +) \\"
+            r"\phi(\mathbb{1} \circ r^2) &= 0 + 2 \pmod 4").scale(1.5)
+        self.play(Write(modulo))
+        self.wait(2)
+
+        self.play(FadeOut(modulo))
+        self.wait(3)
+
+    def matrices(self):
+        question = MathTex(
+            r"D_n &\cong \,? \\"
+            r"S_n &\cong \,? \\"
+            r"A_n &\cong \,?").scale(1.5)
+
+        answer = MathTex(
+            r"D_n &\cong \,?\\"
+            r"S_4 &\cong \mathrm{Aut}(Q_8) \\"
+            r"A_5 &\cong \mathrm{PSL}_2 (5)").scale(1.5)
+
+        self.play(Write(question))
+        self.wait()
+        self.play(ReplacementTransform(question, answer))
+
+        self.wait(3)
diff --git a/vorlesungen/punktgruppen/media/images/nosignal.jpg b/vorlesungen/punktgruppen/media/images/nosignal.jpg
new file mode 100644
index 0000000..2beeb8b
--- /dev/null
+++ b/vorlesungen/punktgruppen/media/images/nosignal.jpg
diff --git a/vorlesungen/punktgruppen/poetry.lock b/vorlesungen/punktgruppen/poetry.lock
new file mode 100644
index 0000000..069d270
--- /dev/null
+++ b/vorlesungen/punktgruppen/poetry.lock
@@ -0,0 +1,743 @@
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+
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+
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+
+[[package]]
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+
+[package.dependencies]
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+
+[[package]]
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+
+[package.dependencies]
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+[package.extras]
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+
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+[package.extras]
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diff --git a/vorlesungen/punktgruppen/pyproject.toml b/vorlesungen/punktgruppen/pyproject.toml
new file mode 100644
index 0000000..527eb57
--- /dev/null
+++ b/vorlesungen/punktgruppen/pyproject.toml
@@ -0,0 +1,15 @@
+[tool.poetry]
+name = "presentation"
+version = "0.1.0"
+description = ""
+authors = ["Nao Pross <np@0hm.ch>"]
+
+[tool.poetry.dependencies]
+python = "^3.7"
+manim = "^0.6.0"
+
+[tool.poetry.dev-dependencies]
+
+[build-system]
+requires = ["poetry-core>=1.0.0"]
+build-backend = "poetry.core.masonry.api"
diff --git a/vorlesungen/punktgruppen/script.pdf b/vorlesungen/punktgruppen/script.pdf
new file mode 100644
index 0000000..91993fb
--- /dev/null
+++ b/vorlesungen/punktgruppen/script.pdf
diff --git a/vorlesungen/punktgruppen/script.tex b/vorlesungen/punktgruppen/script.tex
new file mode 100644
index 0000000..bc50e21
--- /dev/null
+++ b/vorlesungen/punktgruppen/script.tex
@@ -0,0 +1,214 @@
+\documentclass[a4paper]{article}
+
+\usepackage{amsmath}
+\usepackage{amssymb}
+
+\usepackage[cm]{manuscript}
+\usepackage{xcolor}
+
+\newcommand{\scene}[1]{\par\noindent[ #1 ]\par}
+\newenvironment{totranslate}{\color{blue!70!black}}{}
+
+\begin{document}
+\section{Das sind wir}
+\scene{Camera}
+
+\section{Ablauf}
+Zuerst werden wir Symmetrien in 2 Dimensionen anschauen, dann \"uberlegen wir
+kurz was es heisst f\"ur eine Symmetrie ``algebraisch'' zu sein. Von da aus
+kommt die dritte Dimension hinzu, die man besser mit Matrizen verstehen kann.
+Mit der aufgebauten Theorie werden wir versuchen Kristalle zu klassifizieren.
+Und zum Schluss kommen wir zu Anwendungen, welche f\"ur Ingenieure von
+Interesse sind.
+
+\section{intro}
+\scene{Spontan}
+
+\section{2D Geometrie}
+\scene{Intro}
+Wir fangen mit den 2 dimensionalen Symmetrien an, da man sie sich am
+einfachsten vorstellen kann. Eine Symmetrie eines Objektes beschreibt eine
+Aktion, welche nachdem sie auf das Objekt wirkt, das Objekt wieder gleich
+aussehen l\"asst.
+
+\scene{Viereck}
+Die einfachste Aktion, ist das Viereck zu nehmen, und wieder hinzulegen.
+Eine andere Aktion k\"onnte sein, das Objekt um eine Achse zu spiegeln,
+oder eine Rotation um 90 Grad.
+
+\scene{Zyklische Gruppe}
+Fokussieren wir uns auf die einfachste Klassen von Symmetrien: diejenigen die
+von einer reinen Drehung generiert werden. Wir sammeln diese in einer Gruppe
+\(G\), und notieren das sie von eine Rotation \(r\) generiert worden sind, mit
+diesen spitzen Klammern.
+
+Nehmen wir als Beispiel dieses Pentagon. Wenn wir \(r\) 5-mal anwenden, ist es
+dasselbe als wenn wir nichts gemacht h\"atten. Wenn wir es noch ein 6. mal
+drehen, entspricht dies dasselbe wie \(r\) nur 1 mal zu nutzen.
+
+\scene{Notation}
+So, die Gruppe setzt sich zusammen aus dem neutralen Element, und den Potenzen
+1 bis 4 von \(r\). Oder im allgemein Gruppen mit dieser Struktur, in welcher die
+Aktion \(n-1\) mal angewendet werden kann, heissen ``Zyklische Gruppe''.
+
+\scene{Diedergruppe}
+Nehmen wir nun auch noch die Spiegeloperation \(\sigma\) dazu. Weil wir jetzt 2
+Operationen haben, m\"ussen wir auch im Generator schreiben wie sie
+zusammenh\"angen.  Schauen wir dann uns genauer diesen Ausdr\"uck an.  Zweimal
+Spielegeln ist \"aquivalent zum neutralen Element, sowie 4 mal um 90 Grad
+drehen und 2 Drehspiegelungen, welche man auch Inversion nennt.
+
+\scene{Notation}
+Daraus k\"onnen wir wieder die ganze Gruppe erzeugen, die im allgemeinen den
+Symmetrien eines \(n\)-gons entsprechen.
+
+\scene{Kreisgruppe}
+Bis jetzt hatten wir nur diskrete Symmetrien, was nicht zwingend der Fall sein
+muss. Ein Ring kann man kontinuierlich drehen, und sieht dabei immer gleich
+aus.
+
+Diese Symmetrie ist auch als Kreisgruppe bekannt, die man sch\"on mit dem
+komplexen Einheitskreis definieren kann.
+
+\section{Algebra}
+\scene{Produkt mit \(i\)}
+\"Uberlegen wir uns eine spezielle algebraische Operation: Multiplikation mit
+der imagin\"aren Einheit. \(1\) mal \(i\) ist gleich \(i\). Wieder mal \(i\)
+ist \(-1\), dann \(-i\) und schliesslich kommen wir z\"uruck auf \(1\).  Diese
+fassen wir in eine Gruppe \(G\) zusammen. Oder sch\"oner geschrieben:. Sieht das
+bekannt aus?
+
+\scene{Morphismen}
+Das Gefühl, dass es sich um dasselbe handelt, kann wie folgt formalisiert
+werden.  Sei \(\phi\) eine Funktion von \(C_4\) zu \(G\) und ordnen wir zu
+jeder Symmetrieoperation ein Element aus \(G\). Wenn man die Zuordnung richtig
+definiert, dann sieht man die folgende Eigenschaft: Eine Operation nach eine
+andere zu nutzen, und dann die Funktion des Resultats zu nehmen, ist gleich wie
+die Funktion der einzelnen Operazionen zu nehmen und die Resultate zu
+multiplizieren. Dieses Ergebnis ist so bemerkenswert, dass es in der Mathematik
+einen Namen bekommen hat: Homorphismus, von griechisch "homos" dasselbe und
+"morphe" Form.  Manchmal auch so geschrieben. Ausserdem, wenn \(\phi\) eins zu
+eins ist, heisst es \emph{Iso}morphismus: "iso" gleiche Form. Was man
+typischerweise mit diesem Symbol schreibt.
+
+\scene{Animation}
+Sie haben wahrscheinlich schon gesehen, worauf das hinausläuft.  Dass die
+zyklische Gruppe \(C_4\) und \(G\) isomorph sind ist nicht nur Fachjargon der
+mathematik, sondern sie haben wirklich die selbe Struktur.
+
+\scene{Modulo}
+Das Beispiel mit der komplexen Einheit, war wahrscheinlich nicht so
+\"uberraschend. Aber was merkw\"urdig ist, ist das Beziehungen zwischen
+Symmetrien und Algebra auch in Bereichen gefunden werden, welche auf den ersten
+Blick, nicht geomerisch erscheinen. Ein R\"atsel für die Neugierigen: die Summe
+in der Modulo-Arithmetik.  Als Hinweis: Um die Geometrie zu finden denken Sie
+an einer Uhr.
+
+\section{3D Geometrie}
+2 Dimensionen sind einfacher zu zeichnen, aber leider leben wir im 3
+dimensionalen Raum.
+
+\scene{Zyklische Gruppe}
+Wenn wir unser bekanntes Viereck mit seiner zyklischer Symmetrie in 3
+Dimensionen betrachten, k\"onnen wir seine Drehachse sehen.
+
+\scene{Diedergruppe}
+Um auch noch die andere Symmetrie des Rechteckes zu sehen, ben\"otigen wir eine
+Spiegelachse \(\sigma\), die hier eine Spiegelebene ist.
+
+\scene{Transition}
+Um die Punktsymmetrien zu klassifizieren orientiert man sich an einer Achse, um
+welche sich die meisten Symmetrien drehen. Das geht aber nicht immer, wie beim
+Tetraeder.
+
+\scene{Tetraedergruppe}
+Diese Geometrie hat 4 gleichwertige Symmetrieachsen, die eben eine
+Symmetriegruppe aufbauen, welche kreativer weise Tetraedergruppe genannt wird.
+Vielleicht fallen Ihnnen weitere Polygone ein mit dieser Eigenschaft, bevor wir
+zum n\"achsten Thema weitergehen.
+
+\section{Matrizen}
+\scene{Titelseite}
+Nun gehen wir kurz auf den Thema unseres Seminars ein: Matrizen.  Das man mit
+Matrizen Dinge darstellen kann, ist keine Neuigkeit mehr, nach einem
+Semester MatheSeminar.  Also überrascht es wohl auch keinen, das man alle
+punktsymmetrischen Operationen auch mit Matrizen Formulieren kann.
+
+\scene{Matrizen}
+
+Sei dann \(G\) unsere Symmetrie Gruppe, die unsere abstrakte Drehungen und
+Spiegelungen enth\"ahlt. Die Matrix Darstellung dieser Gruppe, ist eine
+Funktion gross \(\Phi\), von \(G\) zur orthogonalen Gruppe \(O(3)\), die zu
+jeder Symmetrie Operation klein \(g\) eine Matrix gross \(\Phi_g\) zuordnet.
+
+Zur Erinnerung, die Orthogonale Gruppe ist definiert als die Matrizen, deren
+transponierte auch die inverse ist. Da diese Volumen und Distanzen erhalten,
+natuerlich nur bis zu einer Vorzeichenumkehrung, macht es Sinn, dass diese
+Punksymmetrien genau beschreiben.
+
+Nehmen wir die folgende Operationen als Beispiele. Die Matrix der trivialen
+Operation, dass heisst nichts zu machen, ist die Einheitsmatrix. Eine
+Spiegelung ist dasselbe aber mit einem Minus, und Drehungen sind uns schon
+dank Herrn M\"uller bekannt.
+
+\section{Kristalle}
+\scene{Spontan}
+
+\section{Piezo}
+\scene{Spontan}
+
+\section{Licht}
+Als Finale, haben wir ein schwieriges Problem aus der Physik. Das Ziel dieser
+Folie ist nicht jedes Zeichen zu versehen, sondern zu zeigen wie man von hier
+weiter gehen kann. Wir mochten sehen wie sich Licht in einem Kristall verhaltet.
+Genauer, wir m\"ochten die Amplitude einer
+elektromagnetischer Welle in einem Kristall beschreiben.
+
+Das Beispiel richtet sich mehr an Elektrotechnik Studenten, aber die Theorie
+ist die gleiche bei mechanischen Wellen in Materialien mit einer
+Spannungstensor wie dem, den wir letzte Woche gesehen haben.
+% Ganz grob gesagt, ersetzt man E durch Xi und epsilon durch das Sigma.
+
+Um eine Welle zu beschreiben, verwenden wir die Helmholtz-Gleichung, die einige
+von uns bereits in anderen Kursen gel\"ost haben.  Schwierig wird aber dieses
+Problem, wenn der Term vor der Zeitableitung ein Tensor ist (f\"ur uns eine Matrix).
+
+Zur Vereinfachung werden wir eine ebene Welle verwenden. Setzt man dieses E in
+die Helmholtz-Gleichung ein, erhält man folgendes zurück: ein Eigenwertproblem.
+
+Physikalisch bedeutet dies, dass die Welle in diesem Material ihre Amplitude in
+Abhängigkeit von der Ausbreitungsrichtung ändert.  Und die Eigenwerte sagen
+aus, wie stark die Amplitude der Welle in jeder Richtung skaliert wird.
+
+Ich sagte, in jede Richtung skaliert, aber welche Richtungen genau?
+Physikalisch hängt das von der kristallinen Struktur des Materials ab, aber
+mathematisch können wir sagen: in Richtung der Eigenvektoren!  Aber diesen
+Eigenraum zu finden, in dem die Eigenvektoren wohnen, ist beliebig schwierig. 
+
+Hier kommt unsere Gruppentheorie zu Hilfe. Wir können die Symmetrien unseres
+Kristalls zur Hilfe nehmen. Zu jeder dieser Symmetrien lässt sich bekanntlich eine
+einfache Matrix finden, deren Eigenraum ebenfalls relativ leicht zu finden ist.
+Zum Beispiel ist der Eigenraum der Rotation \(r\), die Rotationsachse, für die
+Reflexion \(\sigma\) eine Ebene, und so weiter.
+
+Nun ist die Frage, ob man diese Eingenraume der Symmetrienoperationen
+kombinieren kann um den Eigenraum des physikalisches Problems zu finden.
+
+Aber leider ist meine Zeit abgelaufen in der Recherche, also müssen Sie mir 2
+Dingen einfach glauben, erstens dass es einen Weg gibt, und zweitens dass eher
+nicht so schlimm ist, wenn man die Notation einmal gelernt hat.
+
+Nachdem wir an, wir haben den Eigenraum U gefunden, dann können wir einen
+(Eigen)Vektor E daraus nehmen und in ihm direkt lambda ablesen. Das sagt uns,
+wie die Amplitude der Welle, in diese Richtung gedämpft wurde.
+
+Diese Methode ist nicht spezifisch für dieses Problem, im Gegenteil, ich habe
+gesehen, dass sie in vielen Bereichen eingesetzt wird, wie z.B.:
+Kristallographie, Festkörperphysik, Molekülschwingungen in der Quantenchemie
+und numerische Simulationen von Membranen.
+
+\section{Outro}
+\scene{Camera}
+
+\end{document}
+% vim:et ts=2 sw=2:
diff --git a/vorlesungen/punktgruppen/slides.pdf b/vorlesungen/punktgruppen/slides.pdf
new file mode 100644
index 0000000..e1769f8
--- /dev/null
+++ b/vorlesungen/punktgruppen/slides.pdf
diff --git a/vorlesungen/punktgruppen/slides.tex b/vorlesungen/punktgruppen/slides.tex
new file mode 100644
index 0000000..cd3d7d7
--- /dev/null
+++ b/vorlesungen/punktgruppen/slides.tex
@@ -0,0 +1,895 @@
+\documentclass[12pt, xcolor, aspectratio=169, handout]{beamer}
+
+% language
+\usepackage{polyglossia}
+\setmainlanguage{german}
+
+% pretty drawings
+\usepackage{tikz}
+\usepackage{tikz-3dplot}
+
+\usetikzlibrary{positioning}
+\usetikzlibrary{arrows.meta}
+\usetikzlibrary{shapes.misc}
+\usetikzlibrary{calc}
+
+\usetikzlibrary{external}
+\tikzexternalize[
+  mode = graphics if exists,
+  figure list = true,
+  prefix=build/
+]
+
+% Theme
+\beamertemplatenavigationsymbolsempty
+
+% set look
+\usetheme{default} 
+\usecolortheme{fly}
+\usefonttheme{serif}
+
+%% Set font
+\usepackage[p,osf]{scholax}
+\usepackage{amsmath}
+\usepackage[scaled=1.075,ncf,vvarbb]{newtxmath}
+
+% set colors
+\definecolor{background}{HTML}{202020}
+
+\setbeamercolor{normal text}{fg=white, bg=background}
+\setbeamercolor{structure}{fg=white}
+
+\setbeamercolor{item projected}{use=item,fg=background,bg=item.fg!35}
+
+\setbeamercolor*{palette primary}{use=structure,fg=white,bg=structure.fg}
+\setbeamercolor*{palette secondary}{use=structure,fg=white,bg=structure.fg!75}
+\setbeamercolor*{palette tertiary}{use=structure,fg=white,bg=structure.fg!50}
+\setbeamercolor*{palette quaternary}{fg=white,bg=background}
+
+\setbeamercolor*{block title}{parent=structure}
+\setbeamercolor*{block body}{fg=background, bg=}
+
+\setbeamercolor*{framesubtitle}{fg=white}
+
+\setbeamertemplate{section page}
+{
+  \begin{center}
+    \Huge
+    \insertsection
+  \end{center}
+}
+\AtBeginSection{\frame{\sectionpage}}
+
+% Macros
+\newcommand{\ten}[1]{#1}
+
+% Metadata
+\title{\LARGE \scshape Punktgruppen und Kristalle}
+\author[N. Pross, T. T\"onz]{Naoki Pross, Tim T\"onz}
+\institute{Hochschule f\"ur Technik OST, Rapperswil}
+\date{10. Mai 2021}
+
+% Slides
+\begin{document}
+\frame{
+  \titlepage
+  \vfill
+  \begin{center}
+    \small \color{gray}
+    Slides: \texttt{s.0hm.ch/ctBsD}
+  \end{center}
+}
+\frame{\tableofcontents}
+
+\frame{
+  \begin{itemize}
+    \item Was heisst \emph{Symmetrie} in der Mathematik? \pause
+    \item Wie kann ein Kristall modelliert werden? \pause
+    \item Aus der Physik: Licht, Piezoelektrizit\"at \pause
+  \end{itemize}
+  \begin{center}
+    \begin{tikzpicture}
+      \begin{scope}[
+          node distance = 0cm
+        ]
+        \node[
+          rectangle, fill = gray!40!background,
+          minimum width = 3cm, minimum height = 2cm,
+        ] (body) {\(\vec{E}_p = \vec{0}\)};
+
+        \node[
+          draw, rectangle, thick, white, fill = red!50,
+          minimum width = 3cm, minimum height = 1mm,
+          above = of body
+        ] (pos) {};
+
+        \node[
+          draw, rectangle, thick, white, fill = blue!50,
+          minimum width = 3cm, minimum height = 1mm,
+          below = of body
+        ] (neg) {};
+
+        \draw[white, very thick, -Circle] (pos.east) to ++ (1,0) node (p) {};
+        \draw[white, very thick, -Circle] (neg.east) to ++ (1,0) node (n) {};
+
+        \draw[white, thick, ->] (p) to[out = -70, in = 70] node[midway, right] {\(U = 0\)} (n);
+      \end{scope}
+      \begin{scope}[
+          node distance = 0cm,
+          xshift = 7cm
+        ]
+        \node[
+          rectangle, fill = gray!40!background,
+          minimum width = 3cm, minimum height = 1.5cm,
+        ] (body) {\(\vec{E}_p = \vec{0}\)};
+
+        \node[
+          draw, rectangle, thick, white, fill = red!50,
+          minimum width = 3cm, minimum height = 1mm,
+          above = of body
+        ] (pos) {};
+
+        \node[
+          draw, rectangle, thick, white, fill = blue!50,
+          minimum width = 3cm, minimum height = 1mm,
+          below = of body
+        ] (neg) {};
+
+        \draw[orange, very thick, <-] (pos.north) to node[near end, right] {\(\vec{F}\)} ++(0,1);
+        \draw[orange, very thick, <-] (neg.south) to node[near end, right] {\(\vec{F}\)} ++(0,-1);
+
+        \draw[white, very thick, -Circle] (pos.east) to ++ (1,0) node (p) {};
+        \draw[white, very thick, -Circle] (neg.east) to ++ (1,0) node (n) {};
+
+        \draw[white, thick, ->] (p) to[out = -70, in = 70] node[midway, right] {\(U > 0\)} (n);
+      \end{scope}
+    \end{tikzpicture}
+  \end{center}
+}
+
+\section{2D Symmetrien}
+%% Made in video
+{
+  \usebackgroundtemplate{
+    \includegraphics[height=\paperheight]{media/images/nosignal}}
+  \frame{}
+}
+
+\section{Algebraische Symmetrien}
+%% Made in video
+\frame{
+  \begin{columns}[T]
+    \begin{column}{.5\textwidth}
+      Produkt mit \(i\)
+      \begin{align*}
+        1 \cdot i &= i \\
+        i \cdot i &= -1 \\
+        -1 \cdot i &= -i \\
+        -i \cdot i &= 1
+      \end{align*}
+      \pause
+      %
+      Gruppe
+      \begin{align*}
+        G &= \left\{
+          1, i, -1, -i
+        \right\} \\
+        &= \left\{
+          1, i, i^2, i^3
+        \right\} \\
+        C_4 &= \left\{
+          \mathbb{1}, r, r^2, r^3
+        \right\}
+      \end{align*}
+      \pause
+    \end{column}
+    \begin{column}{.5\textwidth}
+      Darstellung \(\phi : C_4 \to G\)
+      \begin{align*}
+        \phi(\mathbb{1}) &= 1 & \phi(r^2) &= i^2 \\
+        \phi(r) &= i & \phi(r^3) &= i^3
+      \end{align*}
+      \pause
+      %
+      Homomorphismus
+      \begin{align*}
+        \phi(r \circ \mathbb{1}) &= \phi(r) \cdot \phi(\mathbb{1}) \\
+        &= i \cdot 1
+      \end{align*}
+      \pause
+      %
+      \(\phi\) ist bijektiv \(\implies C_4 \cong G\)
+			\pause
+			%
+			\begin{align*}
+				\psi : C_4 &\to (\mathbb{Z}/4\mathbb{Z}, +) \\
+				\psi(\mathbb{1}\circ r^2) &= 0 + 2 \pmod{4}
+			\end{align*}
+    \end{column}
+  \end{columns}
+}
+
+\section{3D Symmetrien}
+%% Made in video
+{
+  \usebackgroundtemplate{
+    \includegraphics[height=\paperheight]{media/images/nosignal}}
+  \frame{}
+}
+
+\section{Matrizen}
+\frame{
+  \begin{columns}[T]
+    \begin{column}{.5\textwidth}
+      Symmetriegruppe
+      \[
+        G = \left\{\mathbb{1}, r, \sigma, \dots \right\}
+      \]
+      \pause
+      Matrixdarstellung
+      \begin{align*}
+        \Phi : G &\to O(3) \\
+        g &\mapsto \Phi_g
+      \end{align*}
+      \pause
+      Orthogonale Gruppe
+      \[
+        O(n) = \left\{ Q : QQ^t = Q^tQ = I \right\}
+      \]
+    \end{column}
+    \pause
+    \begin{column}{.5\textwidth}
+      \begin{align*}
+        \Phi_\mathbb{1} &= \begin{pmatrix}
+          1 & 0 & 0 \\
+          0 & 1 & 0 \\
+          0 & 0 & 1
+        \end{pmatrix} = I \\[1em]
+        \Phi_\sigma &= \begin{pmatrix}
+          1 & 0 & 0 \\
+          0 & -1 & 0 \\
+          0 & 0 & 1
+        \end{pmatrix} \\[1em]
+        \Phi_r &= \begin{pmatrix}
+          \cos \alpha & -\sin \alpha & 0 \\
+          \sin \alpha & \cos \alpha & 0 \\
+          0 & 0 & 1
+        \end{pmatrix}
+      \end{align*}
+    \end{column}
+  \end{columns}
+}
+
+\section{Kristalle}
+\begin{frame}[fragile]{}
+  \begin{columns}
+    \onslide<1->{
+      \begin{column}{.5\textwidth}
+        \begin{center}
+            \begin{tikzpicture}[
+                dot/.style = {
+                  draw, circle, thick, white, fill = gray!40!background,
+                  minimum size = 2mm,
+                  inner sep = 0pt,
+                  outer sep = 1mm,
+                },
+              ]
+
+              \begin{scope}
+                \clip (-2,-2) rectangle (3,4);
+                \foreach \y in {-7,-6,...,7} {
+                  \foreach \x in {-7,-6,...,7} {
+                    \node[dot, xshift=3mm*\y] (N\x\y) at (\x, \y) {};
+                  }
+                }
+              \end{scope}
+              \draw[white, thick] (-2, -2) rectangle (3,4);
+
+              \draw[red!80!background, thick, ->] 
+                (N00) to node[midway, below] {\(\vec{a}_1\)} (N10);
+              \draw[cyan!80!background, thick, ->]
+                (N00) to node[midway, left] {\(\vec{a}_2\)} (N01);
+            \end{tikzpicture}
+        \end{center}
+      \end{column}
+    }
+    \begin{column}{.5\textwidth}
+      \onslide<2->{
+        Kristallgitter:
+        \(n_i \in \mathbb{Z}\),
+      }
+      \onslide<3->{
+        \(\vec{a}_i \in \mathbb{R}^3\)
+      }
+      \onslide<2->{
+        \[
+          \vec{r} = n_1 \vec{a}_1 + n_2 \vec{a}_2 \onslide<3->{+ n_3 \vec{a}_3}
+        \]
+      }
+      \vspace{1cm}
+
+      \onslide<4->{
+        Invariant unter Translation
+        \[
+          Q_i(\vec{r}) = \vec{r} + \vec{a}_i
+        \]
+      }
+    \end{column}
+  \end{columns}
+\end{frame}
+
+\begin{frame}[fragile]{}
+  \begin{columns}[T]
+    \begin{column}{.5\textwidth}
+      \onslide<1->{
+        Wie kombiniert sich \(Q_i\) mit der anderen Symmetrien?
+      }
+      \begin{center}
+        \begin{tikzpicture}[
+            dot/.style = {
+              draw, circle, thick, white, fill = gray!40!background,
+              minimum size = 2mm,
+              inner sep = 0pt,
+              outer sep = 1mm,
+            },
+          ]
+
+          \onslide<2->{
+            \node[dot] (A1) at (0,0) {};
+            \node[below left] at (A1) {\(A\)};
+          }
+
+          \onslide<3->{
+            \node[dot] (A2) at (2.5,0) {};
+            \node[below right] at (A2) {\(A'\)};
+
+            \draw[red!80!background, thick, ->] 
+              (A1) to node[midway, below] {\(\vec{Q}\)} (A2);
+          }
+
+          \onslide<4->{
+            \node[dot] (B1) at (120:2.5) {};
+            \node[above left] at (B1) {\(B\)};
+
+            \draw[green!70!background, thick, ->]
+              (A1) ++(.5,0) arc (0:120:.5)
+              node[midway, above, xshift=1mm] {\(C_n\)};
+            \draw[red!80!background, dashed, thick, ->] (A1) to (B1);
+          }
+
+          \onslide<5->{
+            \node[dot] (B2) at ($(A2)+(60:2.5)$) {};
+            \node[above right] at (B2) {\(B'\)};
+
+            \draw[green!70!background, thick, dashed, ->]
+              (A2) ++(-.5,0) arc (180:60:.5);
+            \draw[red!80!background, dashed, thick, ->] (A2) to (B2);
+          }
+
+          \onslide<6->{
+            \draw[yellow!80!background, thick, ->]
+              (B1) to node[above, midway] {\(\vec{Q}'\)} (B2);
+          }
+
+          \onslide<10->{
+		    \draw[gray, dashed, thick] (A1) to (A1 |- B1) node (Xl) {};
+			\draw[gray, dashed, thick] (A2) to (A2 |- B2) node (Xr) {};
+            \node[above left, xshift=-2mm] at (Xl) {\(x\)};
+            \node[above right, xshift= 2mm] at (Xr) {\(x\)};
+          }
+        \end{tikzpicture}
+      \end{center}
+    \end{column}
+    \begin{column}{.5\textwidth}
+      \onslide<7->{
+        Sei \(q = |\vec{Q}|\), \(\alpha = 2\pi/n\) und \(n \in \mathbb{N}\)
+      }
+      \begin{align*}
+        \onslide<9->{q' = n q \onslide<10->{&= q + 2x \\}}
+        \onslide<11->{nq &= q + 2q\sin(\alpha - \pi/2) \\}
+        \onslide<12->{n &= 1 - 2\cos\alpha}
+      \end{align*}
+      \onslide<13->{
+        Somit muss
+        \begin{align*}
+          \alpha &= \cos^{-1}\left(\frac{1-n}{2}\right) \\[1em]
+          \alpha &\in \left\{ 0, 60^\circ, 90^\circ, 120^\circ, 180^\circ \right\} \\
+          n &\in \left\{ 1, 2, 3, 4, 6 \right\}
+        \end{align*}
+      }
+    \end{column}
+  \end{columns}
+\end{frame}
+
+\begin{frame}[fragile]{M\"ogliche Kristallstrukturen}
+  \begin{center}
+    \begin{tikzpicture}[]
+      \node[circle, dashed, draw = gray, 
+        thick, fill = background,
+        minimum size = 4cm] {};
+      \node[gray] at (.9,-1.2) {674};
+
+      \node[circle, draw = white, thick,
+        fill = orange!40!background,
+        xshift = -3mm, yshift = 2mm,
+        minimum size = 2.75cm,
+        outer sep = 1mm] (A) {};
+      \node[white, yshift = 2mm] at (A) {230};
+      \node[white, font=\large, above right = of A] (Al) {Raumgruppe};
+      \draw[white, thick, ->] (Al.west) to[out=180, in=60] (A);
+
+      \node[circle, draw = white, thick, 
+        fill = red!20!background,
+        xshift = -5mm, yshift = -5mm,
+        minimum size = 1cm,
+        outer sep = 1mm] (B) {32};
+      \node[white, font=\large, below left = of B, xshift=-4mm] (Bl) {Kristallklassen};
+      \draw[white, thick, ->] (Bl.east) to[out = 0, in = 180] (B);
+    \end{tikzpicture}
+  \end{center}
+\end{frame}
+
+{
+  \usebackgroundtemplate[fragile]{
+    \begin{tikzpicture}[
+        overlay,
+        xshift = .45\paperwidth,
+        yshift = .47\paperheight,
+        classcirc/.style = {
+          draw = gray, thick, circle,
+          minimum size = 12mm,
+          inner sep = 0pt, outer sep = 0pt,
+        },
+        classlabel/.style = {
+          below right = 5mm
+        },
+        round/.style = {
+          draw = yellow, thick, circle,
+          minimum size = 1mm,
+          inner sep = 0pt, outer sep = 0pt,
+        },
+        cross/.style = {
+          cross out, draw = magenta, thick,
+          minimum size = 1mm, 
+          inner sep = 0pt, outer sep = 0pt
+        },
+      ]
+      \matrix [row sep = 3mm, column sep = 0mm] {
+        \node[classcirc] (C1)  {} node[classlabel] {\(C_{1}\)};  & 
+        \node[classcirc] (C2)  {} node[classlabel] {\(C_{2}\)};  &
+        \node[classcirc] (C3)  {} node[classlabel] {\(C_{3}\)};  &
+        \node[classcirc] (Ci)  {} node[classlabel] {\(C_{i}\)};  &
+
+        \node[classcirc] (Cs)  {} node[classlabel] {\(C_{s}\)};  &
+        \node[classcirc] (C3i) {} node[classlabel] {\(C_{3i}\)}; &
+        \node[classcirc] (C2h) {} node[classlabel] {\(C_{2h}\)}; &
+        \node[classcirc] (D2)  {} node[classlabel] {\(D_{2}\)};  \\
+
+        \node[classcirc] (D3d) {} node[classlabel] {\(D_{3d}\)}; &
+        \node[classcirc] (C2v) {} node[classlabel] {\(C_{2v}\)}; & 
+        \node[classcirc] (D2h) {} node[classlabel] {\(D_{2h}\)}; & 
+        \node[classcirc] (D3)  {} node[classlabel] {\(D_{3}\)};  &
+
+        \node[classcirc] (C4)  {} node[classlabel] {\(C_{4}\)};  &
+        \node[classcirc] (C6)  {} node[classlabel] {\(C_{6}\)};  &
+        \node[classcirc] (D3dP) {} node[classlabel] {\(D_{3d}\)}; &
+        \node[classcirc] (S4)  {} node[classlabel] {\(S_{4}\)};  \\
+
+        \node[classcirc] (S3)  {} node[classlabel] {\(S_{3}\)};  &
+        \node[classcirc, dashed] (T)   {} node[classlabel] {\(T_{}\)};   &
+        \node[classcirc] (C4h) {} node[classlabel] {\(C_{4h}\)}; &
+        \node[classcirc] (C6h) {} node[classlabel] {\(C_{6h}\)}; &
+
+        \node[classcirc, dashed] (Th)  {} node[classlabel] {\(T_{h}\)};  &
+        \node[classcirc] (C4v) {} node[classlabel] {\(C_{4v}\)}; &
+        \node[classcirc] (C6v) {} node[classlabel] {\(C_{6v}\)}; &
+        \node[classcirc, dashed] (Td)  {} node[classlabel] {\(T_{d}\)};  \\
+        
+        \node[classcirc] (D2d) {} node[classlabel] {\(D_{2d}\)}; &
+        \node[classcirc] (D3h) {} node[classlabel] {\(D_{3h}\)}; &
+        \node[classcirc, dashed] (O)   {} node[classlabel] {\(O_{}\)};   &
+        \node[classcirc] (D4)  {} node[classlabel] {\(D_{4}\)};  &
+
+        \node[classcirc] (D6)  {} node[classlabel] {\(D_{6}\)};  &
+        \node[classcirc, dashed] (Oh)  {} node[classlabel] {\(O_{h}\)};  &
+        \node[classcirc] (D4h) {} node[classlabel] {\(D_{4h}\)}; &
+        \node[classcirc] (D6h) {} node[classlabel] {\(D_{6h}\)}; \\
+      };
+
+
+      \node[cross] at ($(C1)+(4mm,0)$) {};
+
+      
+      \node[cross] at ($(C2)+(4mm,0)$) {};
+      \node[cross] at ($(C2)-(4mm,0)$) {};
+
+
+      \node[cross] at ($(C3)+(  0:4mm)$) {};
+      \node[cross] at ($(C3)+(120:4mm)$) {};
+      \node[cross] at ($(C3)+(240:4mm)$) {};
+
+
+      \node[cross] at ($(Ci)+(4mm,0)$) {};
+      \node[round] at ($(Ci)-(4mm,0)$) {};
+
+
+      \node[cross] at ($(Cs)+(4mm,0)$) {};
+      \node[round] at ($(Cs)+(4mm,0)$) {};
+
+
+      \node[cross] at ($(C3i)+(  0:4mm)$) {};
+      \node[cross] at ($(C3i)+(120:4mm)$) {};
+      \node[cross] at ($(C3i)+(240:4mm)$) {};
+      \node[round] at ($(C3i)+( 60:4mm)$) {};
+      \node[round] at ($(C3i)+(180:4mm)$) {};
+      \node[round] at ($(C3i)+(300:4mm)$) {};
+
+
+      \node[cross] at ($(C2h)+(4mm,0)$) {};
+      \node[cross] at ($(C2h)-(4mm,0)$) {};
+      \node[round] at ($(C2h)+(4mm,0)$) {};
+      \node[round] at ($(C2h)-(4mm,0)$) {};
+
+
+      \node[cross] at ($(D2)+( 20:4mm)$) {};
+      \node[cross] at ($(D2)+(200:4mm)$) {};
+      \node[round] at ($(D2)+(160:4mm)$) {};
+      \node[round] at ($(D2)+(340:4mm)$) {};
+
+
+      \foreach \x in {0, 120, 240} {
+        \node[cross] at ($(D3d)+({\x+15}:4mm)$) {};
+        \node[cross] at ($(D3d)+({\x-15}:4mm)$) {};
+      }
+
+
+      \foreach \x in {0, 180} {
+        \node[cross] at ($(C2v)+({\x+15}:4mm)$) {};
+        \node[cross] at ($(C2v)+({\x-15}:4mm)$) {};
+      }
+
+
+      \foreach \x in {0, 180} {
+        \node[cross] at ($(D2h)+({\x+15}:4mm)$) {};
+        \node[cross] at ($(D2h)+({\x-15}:4mm)$) {};
+        \node[round] at ($(D2h)+({\x+15}:4mm)$) {};
+        \node[round] at ($(D2h)+({\x-15}:4mm)$) {};
+      }
+
+
+      \foreach \x in {0, 120, 240} {
+        \node[cross] at ($(D3)+({\x+15}:4mm)$) {};
+        \node[round] at ($(D3)+({\x-15}:4mm)$) {};
+      }
+
+
+      \foreach \x in {0, 90, 180, 270} {
+        \node[cross] at ($(C4)+(\x:4mm)$) {};
+      }
+
+
+      \foreach \x in {0, 60, 120, 180, 240, 300} {
+        \node[cross] at ($(C6)+(\x:4mm)$) {};
+      }
+
+
+      \foreach \x in {0, 120, 240} {
+        \node[cross] at ($(D3dP)+({\x+15}:4mm)$) {};
+        \node[cross] at ($(D3dP)+({\x-15}:4mm)$) {};
+        \node[round] at ($(D3dP)+({\x+15+60}:4mm)$) {};
+        \node[round] at ($(D3dP)+({\x-15+60}:4mm)$) {};
+      }
+
+
+      \node[cross] at ($(S4)+(4mm,0)$) {};
+      \node[cross] at ($(S4)-(4mm,0)$) {};
+      \node[round] at ($(S4)+(0,4mm)$) {};
+      \node[round] at ($(S4)-(0,4mm)$) {};
+
+
+      \foreach \x in {0, 120, 240} {
+        \node[cross] at ($(S3)+(\x:4mm)$) {};
+        \node[round] at ($(S3)+(\x:4mm)$) {};
+      }
+
+
+      %% TODO: T
+
+
+      \foreach \x in {0, 90, 180, 270} {
+        \node[cross] at ($(C4h)+(\x:4mm)$) {};
+        \node[round] at ($(C4h)+(\x:4mm)$) {};
+      }
+
+
+      \foreach \x in {0, 60, 120, 180, 240, 300} {
+        \node[cross] at ($(C6h)+(\x:4mm)$) {};
+        \node[round] at ($(C6h)+(\x:4mm)$) {};
+      }
+
+
+      %% TODO: Th
+
+
+      \foreach \x in {0, 90, 180, 270} {
+        \node[cross] at ($(C4v)+(\x+15:4mm)$) {};
+        \node[cross] at ($(C4v)+(\x-15:4mm)$) {};
+      }
+
+
+
+      \foreach \x in {0, 60, 120, 180, 240, 300} {
+        \node[cross] at ($(C6v)+(\x+10:4mm)$) {};
+        \node[cross] at ($(C6v)+(\x-10:4mm)$) {};
+      }
+
+
+      %% TODO: Td
+
+
+      \foreach \x in {0, 180} {
+        \node[cross] at ($(D2d)+({\x+15}:4mm)$) {};
+        \node[round] at ($(D2d)+({\x-15}:4mm)$) {};
+
+        \node[round] at ($(D2d)+({\x+15+90}:4mm)$) {};
+        \node[cross] at ($(D2d)+({\x-15+90}:4mm)$) {};
+      }
+
+
+      \foreach \x in {0, 120, 240} {
+        \node[cross] at ($(D3h)+({\x+15}:4mm)$) {};
+        \node[cross] at ($(D3h)+({\x-15}:4mm)$) {};
+        \node[round] at ($(D3h)+({\x+15}:4mm)$) {};
+        \node[round] at ($(D3h)+({\x-15}:4mm)$) {};
+      }
+
+
+      %% TODO: O
+
+
+      \foreach \x in {0, 90, 180, 270} {
+        \node[cross] at ($(D4)+({\x+15}:4mm)$) {};
+        \node[round] at ($(D4)+({\x-15}:4mm)$) {};
+      }
+
+      \foreach \x in {0, 60, 120, 180, 240, 300} {
+        \node[cross] at ($(D6)+({\x+10}:4mm)$) {};
+        \node[round] at ($(D6)+({\x-10}:4mm)$) {};
+      }
+
+
+      % TODO Oh
+
+
+      \foreach \x in {0, 90, 180, 270} {
+        \node[cross] at ($(D4h)+(\x+15:4mm)$) {};
+        \node[cross] at ($(D4h)+(\x-15:4mm)$) {};
+        \node[round] at ($(D4h)+(\x+15:4mm)$) {};
+        \node[round] at ($(D4h)+(\x-15:4mm)$) {};
+      }
+
+
+      \foreach \x in {0, 60, 120, 180, 240, 300} {
+        \node[cross] at ($(D6h)+({\x+10}:4mm)$) {};
+        \node[cross] at ($(D6h)+({\x-10}:4mm)$) {};
+        \node[round] at ($(D6h)+({\x+10}:4mm)$) {};
+        \node[round] at ($(D6h)+({\x-10}:4mm)$) {};
+      }
+    \end{tikzpicture}
+  }
+  \begin{frame}[fragile]{}
+  \end{frame}
+}
+
+\section{Anwendungen}
+\begin{frame}[fragile]{}
+  \centering
+  \begin{tikzpicture}[
+      box/.style = {
+        rectangle, thick, draw = white, fill = darkgray!50!background,
+        minimum height = 1cm, outer sep = 2mm,
+      },
+    ]
+
+    \matrix [nodes = {box, align = center}, column sep = 1cm, row sep = 1.5cm] {
+      & \node (A) {32 Kristallklassen}; \\
+      \node (B) {11 Mit\\ Inversionszentrum}; & \node (C) {21 Ohne\\ Inversionszentrum}; \\
+      & \node[fill=red!20!background] (D) {20 Piezoelektrisch}; & \node (E) {1 Nicht\\ piezoelektrisch}; \\
+    };
+
+    \draw[thick, ->] (A.west) to[out=180, in=90] (B.north);
+    \draw[thick, ->] (A.south) to (C);
+    \draw[thick, ->] (C.south) to (D.north);
+    \draw[thick, ->] (C.east) to[out=0, in=90] (E.north);
+  \end{tikzpicture}
+\end{frame}
+
+\begin{frame}[fragile]{}
+  \begin{tikzpicture}[
+      overlay, xshift = 1.5cm, yshift = 1.5cm,
+      node distance = 2mm,
+      charge/.style = {
+        circle, draw = white, thick,
+        minimum size = 5mm
+      },
+      positive/.style = { fill = red!50 },
+      negative/.style = { fill = blue!50 },
+    ]
+
+    \node[font = {\large\bfseries}, align = center] (title) at (5.5,0) {Mit und Ohne\\ Symmetriezentrum};
+    \pause
+
+    \begin{scope}
+      \matrix[nodes = { charge }, row sep = 8mm, column sep = 8mm] {
+        \node[positive] {}; & \node[negative] (N) {}; & \node [positive] {}; \\
+        \node[negative] (W) {}; & \node[positive] {}; & \node [negative] (E) {}; \\
+        \node[positive] {}; & \node[negative] (S) {}; & \node [positive] {}; \\
+      };
+      \draw[gray, dashed] (W) to (N) to (E) to (S) to (W);
+    \end{scope}
+    \pause
+
+    \begin{scope}[xshift=11cm]
+      \foreach \x/\t [count=\i] in {60/positive, 120/negative, 180/positive, 240/negative, 300/positive, 360/negative} {
+        \node[charge, \t] (C\i) at (\x:1.5cm) {};
+      }
+
+      \draw[white] (C1) to (C2) to (C3) to (C4) to (C5) to (C6) to (C1);
+      \node[circle, draw=gray, fill=gray, outer sep = 0, inner sep = 0, minimum size = 3mm] {};
+      % \draw[gray, dashed] (C2) to (C4) to (C6) to (C2);
+    \end{scope}
+    \pause
+
+	%% 
+    \node[below = of title] {Polarisation Feld \(\vec{E}_p\)};
+
+	%% hex with vertical pressure
+    \begin{scope}[xshift=11cm, yshift=-4.5cm]
+      \node[charge, positive, yshift=-2.5mm] (C1) at ( 60:1.5cm) {};
+      \node[charge, negative, yshift=-2.5mm] (C2) at (120:1.5cm) {};
+      \node[charge, positive, xshift=-2.5mm] (C3) at (180:1.5cm) {};
+      \node[charge, negative, yshift= 2.5mm] (C4) at (240:1.5cm) {};
+      \node[charge, positive, yshift= 2.5mm] (C5) at (300:1.5cm) {};
+      \node[charge, negative, xshift= 2.5mm] (C6) at (360:1.5cm) {};
+
+      \draw[white] (C1) to (C2) to (C3) to (C4) to (C5) to (C6) to (C1);
+      % \draw[gray, dashed] (C2) to (C4) to (C6) to (C2);
+
+      \foreach \d in {C1, C2} {
+        \draw[orange, very thick, <-] (\d) to ++(0,.7);
+      }
+
+      \foreach \d in {C4, C5} {
+        \draw[orange, very thick, <-] (\d) to ++(0,-.7);
+      }
+
+      \node[white] (E) {\(\vec{E}_p\)};
+      \begin{scope}[node distance = .5mm]
+        \node[red!50, right = of E] {\(+\)};
+        \node[blue!50, left = of E] {\(-\)};
+      \end{scope}
+      % \draw[gray, thick, dotted] (E) to ++(0,2);
+      % \draw[gray, thick, dotted] (E) to ++(0,-2);
+    \end{scope}
+	\pause
+
+	%% square with vertical pressure 
+    \begin{scope}[yshift=-4.5cm]
+      \matrix[nodes = { charge }, row sep = 5mm, column sep = 1cm] {
+        \node[positive] (NW) {}; & \node[negative] (N) {}; & \node [positive] (NE) {}; \\
+        \node[negative] (W) {}; & \node[positive] {}; & \node [negative] (E) {}; \\
+        \node[positive] (SW) {}; & \node[negative] (S) {}; & \node [positive] (SE) {}; \\
+      };
+
+      \foreach \d in {NW, N, NE} {
+        \draw[orange, very thick, <-] (\d) to ++(0,.7);
+      }
+
+      \foreach \d in {SW, S, SE} {
+        \draw[orange, very thick, <-] (\d) to ++(0,-.7);
+      }
+
+      \draw[gray, dashed] (W) to (N) to (E) to (S) to (W);
+    \end{scope}
+    \pause
+
+	%% hex with horizontal pressure
+    \begin{scope}[xshift=5.5cm, yshift=-4.5cm]
+      \node[charge, positive, yshift= 2.5mm] (C1) at ( 60:1.5cm) {};
+      \node[charge, negative, yshift= 2.5mm] (C2) at (120:1.5cm) {};
+      \node[charge, positive, xshift= 2.5mm] (C3) at (180:1.5cm) {};
+      \node[charge, negative, yshift=-2.5mm] (C4) at (240:1.5cm) {};
+      \node[charge, positive, yshift=-2.5mm] (C5) at (300:1.5cm) {};
+      \node[charge, negative, xshift=-2.5mm] (C6) at (360:1.5cm) {};
+
+      \draw[white] (C1) to (C2) to (C3) to (C4) to (C5) to (C6) to (C1);
+      % \draw[gray, dashed] (C2) to (C4) to (C6) to (C2);
+
+      \draw[orange, very thick, <-] (C6) to ++(.7,0);
+      \draw[orange, very thick, <-] (C3) to ++(-.7,0);
+
+      \node[white] (E) {\(\vec{E}_p\)};
+      \begin{scope}[node distance = .5mm]
+        \node[blue!50, right = of E] {\(-\)};
+        \node[red!50, left = of E] {\(+\)};
+      \end{scope}
+      % \draw[gray, thick, dotted] (E) to ++(0,2);
+      % \draw[gray, thick, dotted] (E) to ++(0,-2);
+    \end{scope}
+    \pause
+	  
+
+  \end{tikzpicture}
+\end{frame}
+
+\frame{
+  \frametitle{Licht in Kristallen}
+  \begin{columns}[T]
+    \begin{column}{.45\textwidth}
+      \onslide<2->{
+        Helmholtz Wellengleichung
+        \[
+          \nabla^2 \vec{E} = \ten{\varepsilon}\mu
+          \frac{\partial^2}{\partial t^2} \vec{E}
+        \]
+      }
+      \onslide<3->{
+        Ebene Welle
+        \[
+          \vec{E} = \vec{E}_0 \exp\left[i
+            \left(\vec{k}\cdot\vec{r} - \omega t \right)\right]
+        \]
+      }
+      \onslide<4->{
+        Anisotropisch Dielektrikum
+        \[
+          (\ten{K}\ten{\varepsilon})\vec{E} 
+            = \frac{k^2}{\mu \omega^2} \vec{E}
+          \implies
+          \Phi \vec{E} = \lambda \vec{E}
+        \]
+      }
+    \end{column}
+    \begin{column}{.55\textwidth}
+      \onslide<5->{
+        Eingenraum
+        \begin{align*}
+          U_\lambda &= \left\{ v : \Phi v = \lambda v \right\}
+          = \mathrm{null}\left(\Phi - \lambda I\right)
+        \end{align*}
+      }\onslide<6->{
+        Symmetriegruppe und Darstellung
+        \begin{align*}
+          G &= \left\{\mathbb{1}, r, \sigma, \dots \right\} \\
+          &\Phi : G \to O(n)
+        \end{align*}
+      }\onslide<7->{
+        Kann man \(U_\lambda\) von \(G\) herauslesen?
+        \only<7>{
+          \[
+            U_\lambda \stackrel{?}{=} f\left(\bigoplus_{g \in G} \Phi_g\right)
+          \]
+        }\only<8>{
+          \begin{align*}
+            \mathrm{Tr}\left[\Phi_r(g)\right] 
+              &= \sum_i n_i \mathrm{Tr}\left[\Psi_i(g)\right] \\
+            |G| &= \sum_i\mathrm{Tr}\left[\Psi_i(\mathbb{1})\right]
+          \end{align*}
+        }
+      }
+    \end{column}
+  \end{columns}
+}
+
+% \begin{frame}[fragile]
+%   \centering
+%   \tdplotsetmaincoords{70}{110}
+%   \begin{tikzpicture}[scale=2, tdplot_main_coords]
+%     \node[draw=white, thick, minimum size = 3cm, circle] {};
+%     % \foreach \x in {0, 120, 240} {
+%     % }
+%   \end{tikzpicture}
+% \end{frame}
+
+
+\end{document}