Given a closed simple curve in $\mathbb{R}^3$, find a disk $S$ bounding $C$ so that its geodesic curvature remains the same in $S$

Question

Assume we have a simple closed curve $C$ in $\mathbb{R}^3$ equipped with the Euclidean metric. Is it possible to find a Disk $D$ bounded by $C$ so that the geodesic curvature of $C$ in $D$ is equal to its geodesic curvature in $\mathbb{R}^3$. One instance would take $C$ to be a circle and then the usual flat disk satisfies this. Although it does not satisfy the property mentioned in this question, if you take $C$ to be one of the boundaries of the cylinder then its geodesic curvature is $0$. My main motivation for asking this question comes from something along these lines: Given a curve $C$ with "big total curvature",to find a disk so that its total mean curvature is also "big" (if total curvature of the disk is big enough then the term coming from Euler characteristic will not make too much of a difference). If it is not possible to do in general what would be some approaches to try to do it in specific cases?

Answer 1

Motivated by the answer above it seems it is easy to put the answer into a form (following Cartan) which relates the Frenet frame and the Darboux frame of the curve. I assume the curve $C(t)$i s arc length parameterized with no where $0$ derivative and it lies on a surface $S \subset \mathbb{R}^3$. By $K_e$, I denote its curvature in $\mathbb{R}^3$ (i.e Euclidean curvature) and by $K_g$, I denote its geodesic curvature, i.e curvature in $S$.

The Frenet frame of $C$ is denoted by $e_1(t), e_2(t), e_3(t)$ and the Darboux by $d_1(t), d_2(t) , d_3(t)$. Here $d_1(t)=e_1(t)$ are the directions tangent to the curve, so I will denote both by $e_1(t)$.$e_2$ is a continuous choice of unit vectors parallel to derivative of $e_1$ (sometimes known as principal normal to the curve) and $d_2$ is a continuous choice of unit vectors in $TS$ orthogonal to $e_1$. Then $f_3$ and $d_3$ are fixed so that the the frames have positive orientation.

We know that Frenet and Darboux frames satisfiy the equations $$ \frac{de_1}{dt}= K_e e_2 $$ $$ \frac{de_1}{dt}= K_gd_2 +K_{n}d_3 $$ for some function $K_{n}$ (which is also known as normal curvature). Finally $e_2$ lies on the plane spanned by $d_2$ and $d_3$, i.e $$ e_2 = sin(\theta)d_2 + cos(\theta)d_3. $$ Combining these equations we get $$ K_gd_2 +K_{n}d_3= K_e ( sin(\theta)d_2 + cos(\theta)d_3) $$ from which we get $sin(\theta)K_e = K_g$. In particular if one has a surface where $\theta= \frac{\pi}{2}$ or in other words the principal normal of the curve and the normal of the surface are in right angles so that $e_2 = d_2$, then the geodesic and Euclidean curvature coincides. One such example is the example given above by Ted where you form the surface by flowing the curve by a vector field that is in the principal normal direction along the curve.

From a geometrical point of view, once you put a curve on a surface the Euclidean curvature can be written as a "sum" of geodesic curvature (how much the curve bends within the surface) and normal curvature (how much the curve bends due to the bending of the surface) using the following relation: $K_e^2 = K_g^2 + K_n^2$ (this is directly deduced from the results above). If you put a curve in a surface where its principal normal is a tangent direction in the surface, then all the bending of the curve is "within" the tangent plane of the surface and it has no normal curvature. Therefore none of Euclidean curvature of the curve comes from the bending of the surface itself. So geodesic and Euclidean curvature coincides.