0
$\begingroup$

Let $S$ be a regular surface with no parabolic or umbilical points. Let $\mathbf{x}: U \longrightarrow V$ be a parametrization of $S$ such that all the coordinates curves are also curvature lines. The parametrized surfaces:

$\begin{align*} \mathbf{y}(u,v) & = \mathbf{x}(u,v) + \rho_1 N(u,v), \\ \mathbf{z}(u,v) & = \mathbf{x}(u,v) + \rho_2 N(u,v), \end{align*}$

where $\rho_1=\dfrac{1}{k_1}$ and $\rho_2=\dfrac{1}{k_2}$ are called Focal Surfaces.

If $(k_1)_u$ and $(k_2)_v$ never vanish, how do I show that $\mathbf{y}$ and $\mathbf{z}$ are regular parametrized surfaces?

  • 0
    What is a simple example among focal surfaces?2015-01-26

1 Answers 1

1

Note that $\mathbf{y}_u=\mathbf{x}_u-\frac{(k_1)_u}{k_1^2}N+\frac{1}{k_1}N_u\mbox{ and }\mathbf{y}_v=\mathbf{x}_v-\frac{(k_1)_v}{k_1^2}N+\frac{1}{k_1}N_v.$ Since the coordinates curves are also curvature lines, we have $N_u=-k_1\mathbf{x}_u,\mbox{ and }N_v=-k_2\mathbf{x}_v.$ Substituting these into the first equation, we obtain $\mathbf{y}_u=-\frac{(k_1)_u}{k_1^2}N\mbox{ and }\mathbf{y}_v=(1-\frac{k_2}{k_1})\mathbf{x}_v-\frac{(k_1)_v}{k_1^2}N,$ which implies that $\mathbf{y}_u\times \mathbf{y}_v=-\frac{(k_1)_u}{k_1^2}(1-\frac{k_2}{k_1})N\times \mathbf{x}_v.$ Note that $N\times\mathbf{x}_v\neq\mathbf{0}$ becasue they are linearly independent. $k_1\neq 0$ since $S$ has no parabolic point, and $k_1\neq k_2$ since $S$ has no umbilical points. Hence, $\mathbf{y}_u\times \mathbf{y}_v\neq\mathbf{0}$.

Similarly, we have $\mathbf{z}_u=\mathbf{x}_u-\frac{(k_2)_u}{k_2^2}N+\frac{1}{k_2}N_u=(1-\frac{k_1}{k_2})\mathbf{x}_u-\frac{(k_2)_u}{k_2^2}N,$ $\mathbf{z}_v=\mathbf{x}_v-\frac{(k_2)_v}{k_2^2}N+\frac{1}{k_2}N_v=-\frac{(k_2)_v}{k_2^2}N,$ which implies that $\mathbf{z}_u\times \mathbf{z}_v=-\frac{(k_2)_v}{k_2^2}(1-\frac{k_1}{k_2})\mathbf{x}_u\times N.$ Again, $\mathbf{x}_u\times N\neq\mathbf{0}$ becasue they are linearly independent. $k_2\neq 0$ since $S$ has no parabolic point, and $k_1\neq k_2$ since $S$ has no umbilical points. Hence, $\mathbf{z}_u\times \mathbf{z}_v\neq\mathbf{0}$.