6

This is an exercise in do Carmo's Differential Geometry:

Let $\alpha : I \longrightarrow S$ be a curve parametrized by arc length $s$, with nonzero curvature. Consider the parametrized surface \begin{align}\textbf{x}(s,v)=\alpha(s)+vb(s), & s \in I, -\epsilon < v < \epsilon, \epsilon > 0\end{align} where $b$ is the binormal vector of $\alpha$. Prove that if $\epsilon$ is small, $\textbf{x}(I \times (-\epsilon, \epsilon)) = S$ is a regular surface over which $\alpha(I)$ is a geodesic (thus, every curve is a geodesic on the surface generated by its binormals).

An errata online says that the first conclusion of this exercise is wrong:

p. 262. Exercise 17: The first conclusion is false: It can happen that for all $\epsilon > 0$, the set $\textbf{x}(I \times (-\epsilon,\epsilon))$ fails to be a regular surface. (Consider a curve $\alpha : (0,1) \rightarrow \mathbb{R}^3$ such that $\alpha(s)$ approaches $(0,0,0)$ from the same direction as $s \rightarrow 0^+$ or $s \rightarrow 1^-$, and such that the part of $\alpha$ near $s=0$ is contained in a plane, and the part of $\alpha$ near $s=1$ is contained in a different plane.)

I don't quite understand the counterexample in the errata. Can somebody help to explicitly construct the curve $\alpha$?

  • Maybe something like this: the two parabolas parameterized as $(t,t^2,0)$ and $(t,0,t^2)$ are both tangent to the $x$-axis at the origin, but they approach the origin in different planes. Not that this is an example of the curve he mentions, but this is how it could happen. – MPW Apr 23 '14 at 01:44

1 Answers1

2

In the original exercise it is assumed that the curve $\alpha$ is a Frenet curve, that is, its second derivative by the arc length does never vanish. Otherwise the binormal vector is not defined. If there is a jump in the binormal vector then there will be a bend in the surface. So it is not really a differentiable surface.

Chappers
  • 67,606
user145673
  • 21