# Surface curves equidistant from a simple closed geodesic

Let $S \subset \mathbb{R}^3$ be a surface embedded in $\mathbb{R}^3$, let's say (to keep it simple) of genus zero. Let $\gamma$ be a simple, closed, oriented geodesic on $S$. Because $\gamma$ is oriented, it partitions $S$ into two "halves," $S^+$ and $S^-$. I am interested in learning properties of the curves that are equidistant from $\gamma$.

Define $$\gamma^+(\delta) = \{ x \in S^+ \;|\; d(x, \gamma) = \delta \} \;,$$ where $d(x,A)$ is the length of the shortest path on $S$ from $x$ to any point in set $A$.

Q1. Does $\gamma^+(\delta)$ have a name in the literature? Has it been studied?

Q2. Under what conditions is $\gamma^+(\delta)$ a geodesic? Presumably rather stringent conditions on $S$.

Q3. Under what conditions is $\gamma^+(\delta)$ a simple, closed curve? It might partition into several disconnected components. But perhaps for convex $S$, it is always a simple, closed curve?

Q4. Under what conditions could $\gamma^+(\delta)$ be a single point? [Revised to reflect Will Jagy's comment.] (Analogous to the north pole with respect to the equator on a (geometric) sphere.)

Thanks for pointers and help!

(Image below suggestive only!)

• similar to Morse functions, really, and some similar to the cut locus of a point. From an ellipsoid, note that the farthest point (Q4) need not be the same distance from all points of $\gamma,$ although perhaps from two points, otherwise a small movement could take it a hair farther. In general, though, thinking of hydra heads, en.wikipedia.org/wiki/… – Will Jagy Mar 22 '14 at 23:42
• Alright, one approach with a reference, en.wikipedia.org/wiki/… – Will Jagy Mar 22 '14 at 23:58
• Right. for Q2, cannot imagine all geodesics unless it is a cylinder over a plane curve. – Will Jagy Mar 23 '14 at 18:53
• You get geodesics for small $\delta$ if the Gaussian curvature is $0$ around $\gamma$. For example, take a tetrahedron with a closed geodesic separating two vertices from the other two. If you want a smooth example, you can smooth the tetrahedron keeping the Gaussian curvature $0$ around the geodesic. – Douglas Zare Mar 23 '14 at 19:25
• I think these are sometimes called wave fronts, but sometimes families of equidistants. Imagine that light is emitted simultaneously everywhere along $\gamma$. $\gamma^+(\delta)$ shows the points being illuminated on one side at time $\delta$. This should be a smooth Lagrangian curve until you hit a center of curvature. Arnold said a few times that Cayley investigated these for a triaxial ellipsoid but I haven't found the original yet. – Douglas Zare Mar 27 '14 at 22:27

To address question 2, consider perpendicular geodesics connecting $\gamma$ with $\gamma^+(\delta)$. These must be perpendicular with $\gamma^+(\delta)$, too, or else moving slightly in one direction or the other will decrease the distance below $\delta$. Consider the quadrilateral formed by a small piece of $\gamma$, perpendicular geodesics, and a piece of $\gamma^+(\delta)$. The Gauss Bonnet theorem implies that the Gaussian curvature of the interior equals the curvature of the piece of $\gamma^+(\delta)$. If the piece of $\gamma^+(\delta)$ is a geodesic its curvature is $0$, so the total Gaussian curvature in the quadrilateral must be $0$.

The Gaussian curvature doesn't have to be identically $0$ between $\gamma$ and $\gamma^+(\delta)$. For example, you can revolve $y=2+\sin x$ about the $x$-axis and there are parallel geodesics where $x$ is a multiple of $\pi$.

Along any geodesic perpendicular to $\gamma$ at $v$ we can construct a function of the Gaussian curvature at distance $d$. Here is a construction that shows the function can change with $v$, unlike the surfaces of revolution: Choose a cylinder $\mathbb{R} \times C$ where $C$ is some plane curve. Choose a smooth function $f$, let $S$ be the surface of points $(x,y,z)$ which are of distance $f(x)$ away from the cylinder. If $f$ has a critical point at $x_0$ then the points of $S$ where $x=x_0$ form a geodesic. We can choose $C$ so that it has flat parts and curved parts, and then the Gaussian curvature does not depend only on $x$ because it is $0$ parallel to the flat parts of $C$.
• I like your $2+\sin x$ example! I took the liberty of adding an image of it to your answer. – Joseph O'Rourke Mar 25 '14 at 1:09
• Oops, I stated the locations of the critical points of $2+\cos x$, of course. – Douglas Zare Mar 25 '14 at 1:59