Let $\bf{p}$ be a point on a regular surface $S$. Let $\boldsymbol{\alpha}(s)$ and $\boldsymbol{\beta}(s)$ be two curves parametrized by arc length on the surface $S$ such that $\boldsymbol{\alpha}(0) = \boldsymbol{\beta}(0) = \bf{p}$.

Denote the Frenet trihedron of $\boldsymbol{\alpha}(s)$ at $\bf{p}$ by $\bf{t}_{\alpha}$, $\bf{n}_{\alpha}$, $\bf{b}_{\alpha}$; the Frenet trihedron of $\boldsymbol{\beta}(s)$ at $\bf{p}$ by $\bf{t}_{\beta}$, $\bf{n}_{\beta}$, $\bf{b}_{\beta}$.

Suppose both curves have the same osculating plane at $\bf{p}$ and the osculating plane is not equal to the tangent plane at $\bf{p}$.

i) Show that $\bf{t}_{\alpha}$ and $\bf{t}_{\beta}$ are parallel, i.e. $\bf{t}_{\alpha} = \pm\bf{t}_{\beta}$.

ii) Show that $\bf{n}_{\alpha}$ and $\bf{n}_{\beta}$ are parallel, i.e. $\bf{n}_{\alpha} = \pm\bf{n}_{\beta}$.

iii) Show that both curves have the same curvature at $\bf{p}$.

I have only succeeded in showing i), where the strategy was to first assume they are not parallel. Taking the cross product of the two tangent vectors would then produce the normal of the osculating plane, a non-zero vector parallel to the Gauss map at $\bf{p}$. Since the tangent plane at $\bf{p}$ is defined by the normal vector at $\bf{p}$, i.e. the Gauss map, this implies that the osculating plane is the same as the tangent plane, which contradicts the setup.

This question was from an exam I took recently. Thank you.


You've done the hard part. Since the osculating planes of $\alpha$ and $\beta$ at $\mathbf p$ are the same and the tangent vectors are ($\pm$) the same, ii) follows. Here's a major hint for iii): Apply Meusnier's formula, which relates the normal curvature of $S$ in the direction of $\mathbf t_\alpha = \pm\mathbf t_\beta$ and the curvature of $\alpha$ (or $\beta$). What's the missing ingredient?

EDIT: Meusnier tells us that $\kappa_n = \kappa \mathbf n\cdot \nu = \kappa\cos\phi$, where $\nu$ is the surface normal and $\phi$ is the angle between the curve's principal normal and the surface normal. So if $\kappa_n$ is the same for both curves (as it depends only on the tangent direction of the curve), now what do you have?

  • $\begingroup$ Oh I see what's going on for ii). Since the osculating planes are the same, there's only one perpendicular direction(up to a sign) in which both normals can be in. As for iii), I googled Meusnier's formula which didn't give me anything(there was a Meusnier's theorem though, which doesn't seem right). The only relation I know of is one relating curvature, normal curvature and geodesic curvature. Clearly the normal curvature is the same since it comes from the second fundamental form of the same tangent vector. I have to think about why the geodesic curvature should be the same though. $\endgroup$ – Repainted May 7 '14 at 17:56
  • $\begingroup$ I've added a bit more :) The geodesic curvatures only turn out to be the same up to sign, but that's not intrinsically obvious. $\endgroup$ – Ted Shifrin May 7 '14 at 18:05
  • $\begingroup$ Nice I see it now. Thanks. That was actually directly from the definition of normal curvature. $\endgroup$ – Repainted May 7 '14 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.