Ellipse/Gaußabbildung/Bijektiv/Umkehrabbildung/Aufgabe/Lösung

1. Als Einheitsnormalenfeld können wir vom Gradienten der beschreibenden Funktion, also von

   ${}G(x,y)={\begin{pmatrix}2\alpha x\\2\beta y\end{pmatrix}}\,$

   ausgehen, durch Normalisierung erhalten wir

   ${}N(x,y)={\frac {1}{\sqrt {\alpha ^{2}x^{2}+\beta ^{2}y^{2}}}}{\begin{pmatrix}\alpha x\\\beta y\end{pmatrix}}\,.$

   Die Gauß-Abbildung ist also

   $\varphi \colon Y={\left\{{\begin{pmatrix}x\\y\end{pmatrix}}\in \mathbb {R} ^{2}\mid \alpha x^{2}+\beta y^{2}=1\right\}}\longrightarrow S^{1},\,{\begin{pmatrix}x\\y\end{pmatrix}}\longmapsto {\frac {1}{\sqrt {\alpha ^{2}x^{2}+\beta ^{2}y^{2}}}}{\begin{pmatrix}\alpha x\\\beta y\end{pmatrix}}.$

2. Wir behaupten, dass

   $\psi \colon S^{1}\longrightarrow Y,\,{\begin{pmatrix}u\\v\end{pmatrix}}\longmapsto {\frac {1}{\sqrt {{\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}}}}{\begin{pmatrix}{\frac {u}{\alpha }}\\{\frac {v}{\beta }}\end{pmatrix}},$

   eine Umkehrfunktion ist. Wegen

   ${}{\frac {1}{{\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}}}{\left(\alpha {\frac {u^{2}}{\alpha ^{2}}}+\beta {\frac {v^{2}}{\beta ^{2}}}\right)}=1\,$

   liegt das Bild von ${}\psi$ in ${}Y$. Wegen

   ${}{\begin{aligned}\varphi {\left(\psi {\begin{pmatrix}u\\v\end{pmatrix}}\right)}&=\varphi {\left({\frac {1}{\sqrt {{\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}}}}{\begin{pmatrix}{\frac {u}{\alpha }}\\{\frac {v}{\beta }}\end{pmatrix}}\right)}\\&=\varphi {\begin{pmatrix}{\frac {1}{\alpha {\sqrt {{\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}}}}}u\\{\frac {1}{\beta {\sqrt {{\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}}}}}v\end{pmatrix}}\\&={\frac {1}{\sqrt {\alpha ^{2}{\frac {u^{2}}{\alpha ^{2}{\left({\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}\right)}}}+\beta ^{2}{\frac {v^{2}}{\beta ^{2}{\left({\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}\right)}}}}}}{\begin{pmatrix}{\frac {1}{\sqrt {{\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}}}}u\\{\frac {1}{\sqrt {{\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}}}}v\end{pmatrix}}\\&={\frac {1}{\sqrt {{\frac {u^{2}}{\left({\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}\right)}}+{\frac {v^{2}}{\left({\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}\right)}}}}}{\begin{pmatrix}{\frac {1}{\sqrt {{\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}}}}u\\{\frac {1}{\sqrt {{\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}}}}v\end{pmatrix}}\\&={\frac {\sqrt {{\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}}}{\sqrt {u^{2}+v^{2}}}}{\begin{pmatrix}{\frac {1}{\sqrt {{\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}}}}u\\{\frac {1}{\sqrt {{\frac {u^{2}}{\alpha }}+{\frac {v^{2}}{\beta }}}}}v\end{pmatrix}}\\&={\begin{pmatrix}u\\v\end{pmatrix}}\end{aligned}}$

   und

   ${}{\begin{aligned}\psi {\left(\varphi {\begin{pmatrix}x\\y\end{pmatrix}}\right)}&=\psi {\begin{pmatrix}{\frac {\alpha x}{\sqrt {\alpha ^{2}x^{2}+\beta ^{2}y^{2}}}}\\{\frac {\beta y}{\sqrt {\alpha ^{2}x^{2}+\beta ^{2}y^{2}}}}\end{pmatrix}}\\&={\frac {1}{\sqrt {{\frac {\alpha x^{2}}{\alpha ^{2}x^{2}+\beta ^{2}y^{2}}}+{\frac {\beta y^{2}}{\alpha ^{2}x^{2}+\beta ^{2}y^{2}}}}}}{\begin{pmatrix}{\frac {x}{\sqrt {\alpha ^{2}x^{2}+\beta ^{2}y^{2}}}}\\{\frac {y}{\sqrt {\alpha ^{2}x^{2}+\beta ^{2}y^{2}}}}\end{pmatrix}}\\&={\frac {\sqrt {\alpha ^{2}x^{2}+\beta ^{2}y^{2}}}{\sqrt {\alpha x^{2}+\beta y^{2}}}}{\begin{pmatrix}{\frac {x}{\sqrt {\alpha ^{2}x^{2}+\beta ^{2}y^{2}}}}\\{\frac {y}{\sqrt {\alpha ^{2}x^{2}+\beta ^{2}y^{2}}}}\end{pmatrix}}\\&={\begin{pmatrix}x\\y\end{pmatrix}}\end{aligned}}$

   sind die Abbildungen invers zueinander.

Zur gelösten Aufgabe