Commit c60da4f4 authored by Harry Fuchs's avatar Harry Fuchs

2019-01-17

parent aca1578f
Pipeline #3676 passed with stage
in 9 minutes and 41 seconds
...@@ -12,6 +12,8 @@ ...@@ -12,6 +12,8 @@
\newtheorem{Satz}[Def]{Satz}% \newtheorem{Satz}[Def]{Satz}%
\newtheorem{Kor}[Def]{Korollar} \newtheorem{Kor}[Def]{Korollar}
\theoremstyle{remark} \theoremstyle{remark}
\newtheorem{Notation}[Def]{Notation}%
\newtheorem{Idee}[Def]{Idee}%
\newtheorem{Frage}[Def]{Frage}% \newtheorem{Frage}[Def]{Frage}%
\newtheorem{Bem}[Def]{Bemerkung}% \newtheorem{Bem}[Def]{Bemerkung}%
\newtheorem{Ueb}[Def]{Übung}% \newtheorem{Ueb}[Def]{Übung}%
...@@ -54,6 +56,11 @@ ...@@ -54,6 +56,11 @@
\newcommand{ \N }{ \mathbb N } \newcommand{ \N }{ \mathbb N }
\newcommand{ \K }{ \mathbb K } \newcommand{ \K }{ \mathbb K }
\newcommand{ \M }{ \mathbb M } \newcommand{ \M }{ \mathbb M }
\newcommand{ \bigcupdot }{ \dot \bigcup }
\newcommand{ \cupdot }{ \dot \cup }
\newcommand{ \bw }{{\bigwedge}}
\newcommand{\miso}[4]{ \begin{tikzcd} #1 \arrow[r, "#3"', shift left=-0.25ex] \pgfmatrixnextcell #4 \arrow[l, "#2"', shift left=-0.75ex] \end{tikzcd} } \newcommand{\miso}[4]{ \begin{tikzcd} #1 \arrow[r, "#3"', shift left=-0.25ex] \pgfmatrixnextcell #4 \arrow[l, "#2"', shift left=-0.75ex] \end{tikzcd} }
\newcommand{\quoteunquote}[1]{ \begin{array}{rcl} && \text{} \\ & #1 & \\ \text{} && \end{array} } \newcommand{\quoteunquote}[1]{ \begin{array}{rcl} && \text{} \\ & #1 & \\ \text{} && \end{array} }
\newcommand{ \wlw }{\wedge\ldots\wedge}
...@@ -5,8 +5,6 @@ ...@@ -5,8 +5,6 @@
% ref % ref
%TODO Nummerierung %TODO Nummerierung
TODO 2020-01-16
TODO 2020-01-17
TODO 2020-01-23 TODO 2020-01-23
TODO 2020-01-24 TODO 2020-01-24
TODO 2020-01-31 TODO 2020-01-31
...@@ -3302,6 +3300,8 @@ $$ ...@@ -3302,6 +3300,8 @@ $$
$$ $$
$v\otimes w$ spannen $(V\otimes W)$ auf $\Rightarrow \bar\psi = \bar\varphi$ $v\otimes w$ spannen $(V\otimes W)$ auf $\Rightarrow \bar\psi = \bar\varphi$
%2020-01-16
%TODO having enumerations within Def environment %TODO having enumerations within Def environment
\begin{Def} \begin{Def}
\end{Def} \end{Def}
...@@ -3499,6 +3499,141 @@ $$ ...@@ -3499,6 +3499,141 @@ $$
$$ $$
mit mit
$$ $$
\sigma\colon V^*\otimes V^* \to V^* \otimes V^*\quad \text{linear fortgesetzt} \sigma\colon V^*\otimes V^* &\to& V^* \otimes V^*\quad \text{linear fortgesetzt}
\\ \alpha \otimes \beta \mapsto \beta\otimes \alpha \\ \alpha \otimes \beta &\mapsto& \beta\otimes \alpha
$$
%2020-01-17
\begin{Bem}
$\{ \alpha\colon V\times V \to K\mathrel|\alpha\mathrel{\text{symetrisch}} \} =: \operatorname{Sym}_{2,0}(V) \leqslant M_{2,0}(V)$
$$
\{ b\in V^*\otimes V^* \mathrel| \sigma (b) = b \} =: (V^*\otimes V^*)^{\sigma} \leqslant V^* \otimes V^*
$$
wobei
$$
\sigma : \begin{cases} V^*\otimes V^* &\to V^* \otimes V^* \\ \sum_{i}\alpha_i \otimes \beta_i &\mapsto \sum_i\beta_i\otimes \alpha_i \end{cases}
$$
\end{Bem}
* Äußere Algebra
\begin{Idee}
Koordinatenfreie Determinante
\end{Idee}
\begin{Def}
Sei $V$ ein $K$-Vektorraum. Die Tensoralgebra ist definiert als:
$$
T(V) := \bigoplus_{k=0}^\infty V^{\otimes k}, \quad V^{\otimes 0} := K
$$
mit Multiplikation $\otimes$
$$
(v_1\otimes \ldots \otimes v_n) \otimes (w_1\otimes \ldots \otimes w_m) := v_1\otimes \ldots \otimes v_n\otimes w_1\otimes \ldots\otimes w_m\in V^{\otimes (n+m)}
$$
\end{Def}
\begin{Def}
$$
V &=& \operatorname{Lin}(\{ e_1,\ldots, e_n \}) \Rightarrow T(V) \cong K\langle e_1,\ldots, e_n \rangle \quad \text{freie Algebra}
\\ I(V) &:=& \left( v\otimes v \mathrel| v\in V \right) \trianglelefteq T(V), \quad \bigwedge V := T(V) / I(V) \quad \text{äußere Algebra}
$$
\end{Def}
\begin{Notation}
$v_1 \wedge \ldots \wedge v_k := [v_1 \otimes \ldots \otimes v_k]_{I(V)}\in \bigwedge V$
\end{Notation}
Per Definition: $v \wedge v = 0 \in \bigwedge V, \quad\forall v\in V$.
$$
\Rightarrow 0 = (v+w)\wedge(v+w) =\underbrace{v\wedge v}_{=0} + v\wedge w ü w\wedge v + \underbrace{w\wedge w}_{=0} = v\wedge w + w\wedge v
\Rightarrow v\wedge w = -w \wedge v
$$
$$
{\bigwedge}^k V &:=& \left[ V^{\otimes k} \right]_{I(V)} \subseteq \bigwedge V
\\ &=& V^{\otimes k}/\left( V^{\otimes k} \cap I(V) \right), \quad V^{\otimes k} \cap I(V) =: I_k(V)
\\ {\bigwedge}^0 V &=& K/(K\cap I(V)) = K/\{ 0 \} = K
\\ {\bigwedge}^1 V &=& V/(V\cap I(V)) = V/\{0\} = V
\\ {\bigwedge}^k &=& \{0\},\quad \mathrel\text{falls} k> n := \dim V
$$
da $e_1\wedge\ldots\wedge e_n\wedge e_n = 0$ (ein Index immer doppelt)
\begin{Prop}
$(e_i)_{i=1,\ldots, n} \subsetneq V$ Basis, $\dim\left( {\bigwedge}^kV \right) = \dbinom{n}{k}, \quad (0\leqslant k\leqslant n)$ mit Basis $\{ e_{i_1}\wedge\ldots\wedge e_{i_k} \mathrel| i_1< \ldots < i_k \in \{ 1,\ldots, n\} \}$:
$$
\dim \left( \bigwedge V \right) = 2^n
$$
\end{Prop}
Beweis:
$$
\operatorname{Lin}\left( \{ e_{i_1} \wedge \ldots\wedge e_{i_k} \mathrel| i_1<\ldots<i_k\in \{ 1,\ldots, n \} \} \right) = {\bigwedge}^k V \quad \text{klar.}
$$
Lineare Unabhängig:
$$
\sum \lambda_{i_1,\ldots, i_k} e_{i_1}\wedge \ldots \wedge e_{i_k} = 0
$$
$$
\forall (i_1, \ldots, i_k) \exists (j_1, \ldots, j_{n-k}) : \{ i_1,\ldots, i_k\} \cupdot \{ j_{1k-1} j_{n-k} \} = \{ 1,\ldots, n \}
$$
($j_1<\ldots < j_{n-k}$)
$$
0 &=& \left( \sum \lambda_{i_1,\ldots, i_k} e_{i_1} \wedge \ldots\wedge e_{i_k} \right) \wedge e_{j_1} \wedge \ldots \wedge e_{j_{n-k}}
\\&=& \pm \lambda_{i_1\cdots i_k} e_1\wedge\ldots\wedge e_n
$$
(fast alle Summanden fallen weg, da Vektor doppelt)
Zu zeigen: $e_1\wedge \ldots\wedge e_n \neq 0$
Trick: Suche $\omega\colon {\bigwedge}^n V \to K$, $\omega(e_1\wedge\ldots\wedge e_n)\neq 0$
\begin{center}
\begin{tikzcd}
\circ \\[-20pt]
\circ \\[-20pt]
\circ
\end{tikzcd}
\end{center}
\begin{Bem}
Sei $f\colon V\to V$ linear $\Rightarrow {\bigwedge}^n f \colon \begin{cases} {\bigwedge}^n V &\to {\bigwedge}^n V \\v_1\wedge\ldots\wedge v_n &\to f(v_1)\wedge \ldots\wedge f(v_n) \end{cases}$.
$$
\dim \left( {\bigwedge}^n V \right) = 1 &\Rightarrow& {\bigwedge}^nf\colon \begin{cases} {\bigwedge}^n V \to \bigwedge^n V \\ x\mapsto \alpha\cdot x \end{cases} \quad \mathrel{\text{mit}} \alpha = \det(f)
$$
$$
\bw^n (g\circ f) = \bw^ng\circ \bw^n f \Rightarrow \det(f\circ g) = \det(f)\cdot \det(g)
\\ f\mathrel{\text{invertierbar}} \Leftrightarrow \det(f)\in K^{\times}
$$
\end{Bem}
\begin{Prop}
$$
(1) \cong (2) \cong (3) :\Leftrightarrow
\\
\left( \bw^kV \right)^* \cong \bw^kV^* \cong \{ f\colon V^{\otimes k} \to K \mathrel{\text{multilinear, alternierend}} \}
$$
\end{Prop}
Beweis: $(1) \cong (3)$ per Definition:
$(1)\cong (2)$: Sei
$$
(\cdot, \cdot)\colon \begin{cases} \bw^k(V^*) \times \bw^k(V) &\to \mathbb R\\(v_1^* \wedge \ldots\wedge v_k^*, v_1\wedge\ldots\wedge v_k) &\mapsto \det \left( v_i^* (v_j) \right)_{i,j=1, \ldots, k} \end{cases}
$$ $$
In Basen
$\{ e_{i_1}^* \wedge\ldots\wedge e_{i_k}^* \mathrel| 1\leqslant i_1<\ldots<i_k \leqslant n \}\subseteq \bw^k(V^*)$
$\{e_{i_1}\wedge \ldots\wedge e_{i_k} \mathrel| 1\leqslant i_1 < \ldots < i_k \leqslant n\} \subseteq \bw^k(V)$
liefert Einheitsmatrix $E_{\binom nk}$:
$$
\left( (e_{i_1}^*\wedge\ldots\wedge e_{i_k}^*) , (e_{j_1}\wedge\ldots\wedge e_{j_k}) \right) &=& \det \left( e_{i_e}^* (e_{j_m}) \right)_{l,m=1, \ldots, k}
\\&=& \det (\delta_{i_e j_m})_{l,m=1,\ldots, k}
\\&=& \det \begin{cases} 1, & \mathrel{\text{falls}} (i_k,\ldots, i_k) = (j_1, \ldots, j_k) \\ 0, &\mathrel{\text{sonst}} \end{cases}
\\&=& \delta_{(i_1,\ldots, i_k)(j_1, \ldots, j_k)}
$$
\begin{Def}
$\dim V = n\Rightarrow \bw^n(V^*)$ ist Raum der \emph{Determinantenform} $w\in \bw^n(V^*)\setminus\{0\}$ heißt \emph{Volumenform} auf $V$
\end{Def}
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