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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?

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This is Exercise 9 a) from section 3-5 Ruled Surfaces and Minimal Surfaces of do Carmo's Differential Geometry of Curves and Surfaces. It would be a nice treat if you mentioned that. In addition, if you showed some of your own thoughts or attempts, this might motivate some people more to write out an answer. –  t.b. Nov 8 '11 at 5:22
    
My idea was show that $ \mathbf{y_{u} \times y_{v}} $ never vanish,but It didn't work. –  Jr. Nov 8 '11 at 20:40

1 Answer 1

up vote 1 down vote accepted

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, $\mathbf{z}_u\times \mathbf{z}_v\neq\mathbf{0}$.

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