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Martin Thoma 2014-02-12 07:51:55 +01:00
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@ -389,18 +389,18 @@ an $S$ in $s$.
Dann ist $\Set{D_P F(e_1), D_P F(e_2)}$ eine Basis von $T_s S$.
\item Bzgl. der Basis $\Set{D_P F(e_1), D_P F(e_2)}$ hat das
Standardskalarprodukt aus \cref{bem:19.1a} die Darstellungsmatrix
\[ I_S = \begin{pmatrix}
\begin{align*}
I_S &= \begin{pmatrix}
g_{1,1}(s) & g_{1,2}(s)\\
g_{1,2}(s) & g_{2,2}(s)
\end{pmatrix} =
\begin{pmatrix}
E(s) & F(s) \\
F(s) & G(s)
\end{pmatrix}\]
mit $\begin{aligned}
g_{i,j} &= g_s(D_P F(e_i), D_P F(e_j))\\
&= \langle \frac{\partial F}{\partial u_i} (p), \frac{\partial F}{\partial u_j} (p) \rangle \;\;\; i,j \in \Set{1,2}
\end{aligned}$.\\
\end{pmatrix}\\
\text{mit } g_{i,j} &= g_s(D_P F(e_i), D_P F(e_j))\\
&= \langle \frac{\partial F}{\partial u_i} (p), \frac{\partial F}{\partial u_j} (p) \rangle \;\;\; i,j \in \Set{1,2}
\end{align*}
Die Matrix $I_S$ heißt \textbf{erste Fundamentalform}\xindex{Fundamentalform!erste}
von $S$ bzgl. der Parametrisierung $F$.
\item $g_{i,j}(s)$ ist eine differenzierbare Funktion von $s$.
@ -408,7 +408,7 @@ an $S$ in $s$.
\end{bemerkung}
\begin{bemerkung}
\[\det(I_S) = \| \frac{\partial F}{\partial u_1}(p) \times \frac{\partial F}{\partial u_2}(p)\|^2\]
\[\det(I_S) = \left \| \frac{\partial F}{\partial u_1}(p) \times \frac{\partial F}{\partial u_2}(p) \right \|^2\]
\end{bemerkung}
\begin{beweis}
@ -424,12 +424,12 @@ an $S$ in $s$.
\begin{align*}
z_1 &= x_2 y_3 - x_3 - y_2\\
z_2 &= x_3 y_1 - x_1 y_3\\
z_3 &= x_1 y_2 - x_2 y_1
z_3 &= x_1 y_2 - x_2 y_1\\
\Rightarrow \|\frac{\partial F}{\partial u_1} (p) \times \frac{\partial F}{\partial u_2} (p)\| &= z_1^2 + z_2^2 + z_3^2\\
\end{align*}
$\Rightarrow \|\frac{\partial F}{\partial u_1} (p) \times \frac{\partial F}{\partial u_2} (p)\| = z_1^2 + z_2^2 + z_3^2$\\
\begin{align*}
\det(I_S) &= g_{1,1} g_{2,2} - g_{1,2}^2\\
&= \langle \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}\rangle \langle \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}, \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}\rangle - \langle \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}\rangle^2\\
&= \left \langle \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \right \rangle \left \langle \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}, \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} \right \rangle - \left \langle \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} \right \rangle^2\\
&= (x_1^2 + x_2^2 + x_3^2) (y_1^2 + y_2^2 + y_3^2) - (x_1 y_1 + x_2 y_2 + x_3 y_3)^2
\end{align*}
\end{beweis}