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How to show that if a curve C in a surface is both a line of curvature and a geodesic, then C is a plane curve.


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If you've seen the Frenet-Serret apparatus, you've probably seen that a curve is planar if and only if its torsion is zero, which is a condition on the 2nd derivatives of the curve. Similarly, the "line of curvature" and "geodesic" conditions are conditions on the 2nd derivative of the curve. So those are probably good things to start comparing. – Ryan Budney Apr 21 '11 at 21:00
Where is this problem from? What have you tried so far? – Jesse Madnick Apr 21 '11 at 21:02

As Ryan Budney mentioned in the comments, a curve (with non-vanishing curvature) is planar if and only if its torsion is zero. Therefore, we will show that the torsion is zero.

Notation: Let $t, n, b$ denote the Frenet frame, and let $N$ denote the normal vector to the surface. Let $\kappa_n$ denote the normal curvature, and $\kappa$ the (extrinsic) curvature.

Edit: Hm, I'm thinking that it might be better to write the solution as a series of hints.

  1. Show that being a geodesic implies that $\kappa = \kappa_n$ and $n = N$.

  2. Show that being a line of curvature implies that $$\frac{dN}{ds} = -\kappa_nt.$$

  3. Conclude that the torsion is zero by using $\tau = \frac{dn}{ds} \cdot b$.

Note: It may be helpful to assume that $\kappa \neq 0$ and $\kappa_n \neq 0$. This is an okay assumption because if either $\kappa = 0$ or $\kappa_n = 0$, then the curve reduces to a line (why?) and the result is trivial.

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Here is another way showing explicitely the plane containing $C$. Let me call $S$ the surface whose line of curvature $C$ is geodesic. Let $p \in C$ and let $v_p$ a non-zero tangent vector to $S$ perpendicular to $C$ at $p \in S$. By using the intrisic parallel transport of $S$ extend $v_p$ to a parallel vector field $v$ along $C$. Since $C$ is geodesic $v$ is always perpendicular to $C$ hence $v$ is always an eigenvector of the shape operator of $S$. We claim that $v$ regarded as vector field in $\mathbb{R}^3$ (along $C$) is constant. That is to say, $D_{C'} v$ the derivative of $v$ along $C$ in $\mathbb{R}^3$ is identically zero. Indeed, the tangent component (to the surface $S$) of $D_{C'}v$ is zero due to the construction of $v$ i.e. such tangent component is the intrinsic covariant derivative of $S$. The normal component of $D_{C'} v$ is the second fundamental form of $S$ w.r.t. $v$ and $C'$ which is also zero since both $C'$ and $v$ are eigenvectors of the shape operator of $S$. Thus $C$ is contained in the plane through $p$ normal to the vector $v_p$.

An even simpler proof : consider the cross $C' \times N$. The derivative of the cross along $C$ is zero hence the cross is constant along $C$. Then $C$ is contained in the plane through any of its points perpendicular to the cross.

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