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Let ${\bf r}(u,v)$ be a parameterization for the surface $M$ in $\mathbb{R}^3$. If we denote by ${\bf N}$ the unit normal vector field of $M$, we can define a parallel surface $M_{d}$ in the following way: \begin{equation} {\bf r}_{d}(u,v) = {\bf r}(u,v) + d{\bf N}(u,v) \end{equation} where $d \neq 0$ is a fixed real number.

Now, I'm trying to show that the singular points of the parameterization ${\bf r}_{d}$ correspond to those points of $M$ at which $1/d$ is a principal curvature. My current approach is rather ugly, and involves computing eigenvalues of the shape operator of ${\bf r}_{d}(u,v)$. Is there a more computationally simple way to do this?

Ben Perez
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  • You want to compute the derivative of $\mathbf r_d$ itself. – Ted Shifrin Nov 04 '13 at 02:16
  • I did that, if by derivative you mean partial derivatives. I'm trying to show that their cross product is equal to zero if and only if $1/d$ is a principal curvature of ${\bf r}(u,v)$. – Ben Perez Nov 04 '13 at 08:46

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Let me write $\mathbf x(u,v) = \mathbf r_d(u,v)$. Let's assume that $u$- and $v$-curves are (near the point in question) lines of curvature with principal curvatures $k_1$ and $k_2$, respectively. Then we have \begin{align*} \mathbf x_u &= \mathbf r_u + d\mathbf N_u = \mathbf r_u + d(-k_1\mathbf r_u) \\ \mathbf x_v &= \mathbf r_v + d\mathbf N_v = \mathbf r_v + d(-k_2\mathbf r_v)\,. \end{align*} So, we see that if $d=1/k_j$ for either $j=1$ or $j=2$, then one of the partial derivatives of $\mathbf x$ vanishes and the parametrization is not regular at such a point.

Ted Shifrin
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  • Sorry if I misunderstand, but why can assume that the $u$ and $v$ curves are lines of curvature with principal directions? Do you basically just reparametrize the surface? – Ashwin Iyengar Nov 04 '15 at 01:47
  • @AshwinIyengar: You can always reparametrize this way away from umbilic points. The basic differential equations theorem is that if you have linearly independent vector fields $X$ and $Y$, locally there are coordinates $(u,v)$ so that the $u$-curves are tangent to $X$ and the $v$-curves are tangent to $Y$. – Ted Shifrin Nov 04 '15 at 02:12