Buchanfang Maßtheorie by Richard4321/ Die äußere Ableitung – „Mathe für Nicht-Freaks“

Wir können nun eine Ableitung einer k-Form zu einer (k+1)-Form definieren, indem wir einfach die Koeffizienten ableiten.

Die Ableitung ist mit dem Dachprodukt und gleichzeitig mit dem Vorschalten einer Funktion (Variablensubstitution) verträglich!

Zweifaches Anwenden der Ableitung ergibt Null.

Sei $g:\mathbb {R} ^{n}\rightarrow \mathbb {R}$ differenzierbar. Durch Multiplikation von $\nabla g$ mit einem Vektor $v$ erhält man eine $1$-Form, nämlich $dg(v)$

$\langle \nabla g,v\rangle =\sum _{i=1}^{n}{\frac {\partial g}{\partial x_{i}}}v_{i}=\sum _{i=1}^{n}{\frac {\partial g}{\partial x_{i}}}dx_{i}(v)=dg(v)$

Definition

Sei $w\in W^{k}(\mathbb {R} _{p}^{n})$. Die äußere Ableitung von w ist

${\begin{aligned}d:W^{0}(\mathbb {R} _{p}^{n})\rightarrow W^{1}(\mathbb {R} _{p}^{n}),g\mapsto &\sum _{i=1}^{n}{\frac {\partial g}{\partial x_{i}}}dx_{i}\\d:W^{k}(\mathbb {R} _{p}^{n})\rightarrow W^{k+1}(\mathbb {R} _{p}^{n}),\sum _{I}a_{I}dx_{I}\mapsto &\sum _{I}da_{I}\wedge dx_{I}\\&=\sum _{I}\sum _{j=1}^{n}{\frac {\partial a_{I}}{\partial x_{j}}}dx_{j}\wedge dx_{I}\end{aligned}}$

Beispiel

Für

$w=x_{1}x_{2}x_{3}dx_{1}+x_{2}x_{3}dx_{2}+(x_{1}+x_{3})dx_{3}$

gilt

${\begin{aligned}dw=&\left(x_{2}x_{3}\underbrace {\sum _{j=1}^{3}{\frac {\partial x_{1}}{\partial x_{j}}}dx_{j}} _{=dx_{1}}+x_{1}x_{3}\underbrace {\sum _{j=1}^{3}{\frac {\partial x_{2}}{\partial x_{j}}}dx_{j}} _{=dx_{2}}+x_{1}x_{2}\underbrace {\sum _{j=1}^{3}{\frac {\partial x_{3}}{\partial x_{j}}}dx_{j}} _{=dx_{3}}\right)\wedge dx_{1}\\&+\left(x_{2}\underbrace {\sum _{j=1}^{3}{\frac {\partial x_{3}}{\partial x_{j}}}dx_{j}} _{=dx_{3}}+x_{3}\underbrace {\sum _{j=1}^{3}{\frac {\partial x_{1}}{\partial x_{j}}}dx_{j}} _{=dx_{1}}\right)\wedge dx_{2}\\&+\left(\underbrace {\sum _{j=1}^{3}{\frac {\partial x_{1}}{\partial x_{j}}}dx_{j}} _{=dx_{1}}+\underbrace {\sum _{j=1}^{3}{\frac {\partial x_{3}}{\partial x_{j}}}dx_{j}} _{=dx_{3}}\right)\wedge dx_{3}\\=&(x_{3}-x_{1}x_{3})dx_{1}\wedge dx_{2}+(1-x_{1}x_{2})dx_{1}\wedge dx_{3}-x_{2}dx_{2}\wedge dx_{3}\end{aligned}}$

Satz

a) Für $a,b\in W^{0}(\mathbb {R} _{p}^{n}),c\in \mathbb {R}$ gilt: $d$ ist linear und lässt sich auf das Produkt anwenden

${\begin{aligned}d(a+b)=&da+db\\d(ab)=&bda+adb\end{aligned}}$

Seien $w_{1},w_{2}\in W^{k}(\mathbb {R} _{p}^{n}),c\in \mathbb {R}$ und $f:\mathbb {R} ^{n}\rightarrow \mathbb {R} ^{m}$ differenzierbar. Dann gilt $d$ ist linear, lässt sich auf das Dachprodukt anwenden, zweifache Anwendung von $d$ ergibt Null und $d$ vertauscht mit $f^{\ast }$.

${\begin{aligned}d(w_{1}+cw_{2})=&dw_{1}+cdw_{2}\\d(w\wedge r)=&dw\wedge r+(-1)^{k}w\wedge dr\\d(dw)=&d^{2}w=0\\d(f^{\ast }w)=&f^{\ast }(dw)\end{aligned}}$

Beweis

:

a)

${\begin{aligned}d(a+cb)=&\sum _{i=1}^{n}{\frac {\partial (a+cb)}{\partial x_{i}}}dx_{i}\\=&\sum _{i=1}^{n}{\frac {\partial a}{\partial x_{i}}}dx_{i}+c\sum _{i=1}^{n}{\frac {\partial b}{\partial x_{i}}}dx_{i}\\=&da+cdb\end{aligned}}$

${\begin{aligned}d(ab)=&\sum _{i=1}^{n}{\frac {\partial (ab)}{\partial x_{i}}}dx_{i}\\=&a\sum _{i=1}^{n}{\frac {\partial b}{\partial x_{i}}}dx_{i}+b\sum _{i=1}^{n}{\frac {\partial a}{\partial x_{i}}}dx_{i}\\=&adb+bda\end{aligned}}$

b)

${\begin{aligned}d(w_{1}+cw_{2})=&\sum _{I}d(a_{I}+cb_{I})\wedge dx_{I}\\{\stackrel {a)}{=}}&\sum _{I}(da_{I}+cdb_{I})\wedge dx_{I}\\=&\sum _{I}da_{I}\wedge dx_{I}+c\sum db_{I}\wedge dx_{I}\\=&dw_{1}+cdw_{2}\end{aligned}}$

c) Da $db$ an $dx_{1},\ldots ,dx_{k}$ vorbeigezogen wird, gilt

${\begin{aligned}d(w\wedge r)=&\sum _{I,J}d(a_{I}b_{J})\wedge dx_{I}\wedge dx_{J}\\=&\sum _{I,J}b_{J}da_{I}\wedge dx_{I}\wedge dx_{J}+\sum _{I,J}a_{I}db_{J}\wedge dx_{I}\wedge dx_{J}\\=&dw\wedge r+(-1)^{k}\sum _{I,J}a_{I}dx_{I}\wedge db_{J}\wedge dx_{J}\\=&dw\wedge r+(-1)^{k}\left(\sum _{I}a_{I}dx_{I}\right)\wedge \left(\sum _{J}db_{J}\wedge dx_{J}\right)\\=&dw\wedge r+(-1)^{k}w\wedge dr\end{aligned}}$

d) 1. Fall: Für eine 0-Form $w:\mathbb {R} ^{n}\rightarrow \mathbb {R}$ gilt

${\begin{aligned}d(dw)=&d\left(\sum _{j=1}^{n}{\frac {\partial w}{\partial x_{j}}}dx_{j}\right)\\=&\sum _{j=1}^{n}d\left({\frac {\partial w}{\partial x_{j}}}\right)\wedge dx_{j}\\=&\sum _{j=1}^{n}\left(\sum _{i=1}^{n}{\frac {\partial ^{2}w}{\partial x_{i}\partial x_{j}}}dx_{i}\wedge dx_{j}\right)\end{aligned}}$

Mit

${\begin{aligned}{\frac {\partial ^{2}w}{\partial x_{i}\partial x_{j}}}=&{\frac {\partial ^{2}w}{\partial x_{j}\partial x_{i}}}\\dx_{i}\wedge dx_{j}=&-dx_{j}\wedge dx_{i}\end{aligned}}$

folgt:

${\begin{aligned}d(dw)=&\sum _{i<j}\left(\underbrace {{\frac {\partial ^{2}w}{\partial x_{i}\partial x_{j}}}-{\frac {\partial ^{2}w}{\partial x_{j}\partial x_{i}}}} _{=0}\right)dx_{i}\wedge dx_{j}\\=&0\end{aligned}}$

2. Fall: Sei $w=\sum a_{I}dx_{I}$. Wegen b) reicht es, $w=a_{I}dx_{I}$ zu betrachten.

Nach c) gilt

${\begin{aligned}&d(dx_{j_{1}}\wedge \ldots \wedge dx_{j_{k}})\\{\stackrel {c)}{=}}&(\underbrace {ddx_{j_{1}}} _{=0})\wedge dx_{j_{2}}\wedge \ldots \wedge dx_{j_{k}}\\&+(-1)^{k-1}dx_{j_{1}}\wedge d(dx_{j_{2}}\wedge \ldots \wedge dx_{j_{k}})\\{\stackrel {c)}{=}}&(-1)^{k-1}(\underbrace {ddx_{j_{2}}} _{=0})\wedge dx_{j_{3}}\wedge \ldots \wedge dx_{j_{k}}\\&+(-1)^{k-1}(-1)^{k-2}dx_{j_{1}}\wedge dx_{j_{2}}\wedge d(dx_{j_{3}}\wedge \ldots \wedge dx_{j_{k}})\\{\stackrel {c)}{=}}&\ldots =\sum _{i=1}^{k}(-1)^{i-1}dx_{j_{1}}\wedge \ldots \wedge dx_{j_{k-1}}\wedge \underbrace {d(dx_{j_{k}})} _{=0}\\=&0\end{aligned}}$

Somit

${\begin{aligned}d(dw)=&d(da_{I}\wedge dx_{I}+a_{I}d(dx_{I}))\\=&d(da_{I}\wedge dx_{I})\\=&\underbrace {d(da_{I})} _{=0}\wedge dx_{I}+da_{I}\wedge \underbrace {d(dx_{I})} _{=0}\\=&0\end{aligned}}$

e) 1. Fall: Sei $g:\mathbb {R} ^{n}\rightarrow \mathbb {R}$ differenzierbar, also eine 0-Form.

${\begin{aligned}f^{\ast }(dg)=&f^{\ast }\left(\sum _{i=1}^{n}{\frac {\partial g}{\partial y_{i}}}dy_{i}\right)\\=&\sum _{i=1}^{n}{\frac {\partial g}{\partial y_{i}}}f^{\ast }(dy_{i})\\=&\sum _{i,j=1}^{n}{\frac {\partial g}{\partial y_{i}}}{\frac {\partial f_{i}}{\partial x_{j}}}dx_{j}\\=&\sum _{j}{\frac {\partial (g\circ f)}{\partial x_{j}}}dx_{j}\\=&d(g\circ f)\\=&d(f^{\ast }g)\end{aligned}}$

2. Fall: Sei $w=\sum _{I}a_{I}dx_{I}$ eine k-Form.

${\begin{aligned}d(f^{\ast }w)=&d\left(\sum _{I}f^{\ast }(a_{I})f^{\ast }(dx_{I})\right)\\=&\sum _{I}d(f^{\ast }(a_{I}))\wedge f^{\ast }(dx_{I})\\=&\sum _{I}f^{\ast }(da_{I})\wedge f^{\ast }(dx_{I})\\=&f^{\ast }\left(\sum _{I}da_{I}\wedge dx_{I}\right)\\=&f^{\ast }(dw)\end{aligned}}$