# Geodesic path on the unit sphere with the sup norm

Let $X$ be the unit sphere of $\mathbb{R}^n$ with the sup norm, i.e. $X=\{x\in\mathbb{R}^n: \|x\|_{\infty}=1\}$. Let the metric $d$ on $X$ be the geodesic metric induced by the sup norm, i.e for any two points $x,y\in X$, $$d(x,y)=\inf \{\ell(\alpha): \alpha \mbox{ a continuous curve on }X \mbox{ joining }x,y \},$$ where $\ell(\alpha)$ denotes the arc length of $\alpha$ measured with the sup norm. It is obvious that one only has to consider piecewise linear curves.

My question is: Has anyone obtained formulae for $d(x,y)$ and the geodesic paths (curves of minimum length) joining $x,y$?

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Also asked at MSE: math.stackexchange.com/questions/79008 . –  Joseph O'Rourke Apr 6 '12 at 14:29
@Joseph. No, the questions are different. –  Denis Serre Apr 6 '12 at 14:44
@Denis: Thanks for the correction. The MSE question was a specific version of this more general one, and the answer there does not answer it here. –  Joseph O'Rourke Apr 6 '12 at 15:03
Silly question: why can't one just develop the cube onto the $\ell_\infty$ plane ? Of course, this "development" wouldn't work for arbitrary polyhedra, or other norms, but offhand it seems it could work in this case (?). –  alvarezpaiva Apr 6 '12 at 19:57

Dear TCL,

If one does not confuse geodesics with minimal geodesics, then one can say that if you develop the cube onto the plane, then straight lines are geodesics. Of course, not all geodesics arise in this way and I've never considered the question of giving an explicit formula for the distance function.

If $P$ is an $(n-1)$-dimensional polyhedron in an $n$-dimensonal normed space, which we consider as a piecewise flat Finsler space, then a large class of geodesics that do not pass through the $(n-3)$-skeleton of $P$ can be easily described as follows:

(1) On each facet they are straight lines.

(2) They cross the $(n-2)$-dimensional faces transversely and at each crossing they satisfy the generalized Snell-Descartes law (of refraction).

This statement almost obvious because straight lines are geodesics in normed spaces (and hence on the facets, which I consider as pieces of normed spaces with the induce norm) and because the generalized Snell-Descartes law follows from Fermat's principle.

In the case of the cube with the piecewise flat $\ell_\infty$ metric, this gives that straight lines in the developed cube are geodesics on the cube.

## The generalized Snell-Descartes law

This is a good time to say that what I'm writing now is basically an annoucement of an essay I'm writing for a book with A.C. Thompson ("An invitation to Minkowski geometry"). The presentation here is a bit sketchy at some points (otherwise it would be too long), but if you reconstruct the pictures, I think everything will be clear.

A well-known secret in physics is that everything becomes simpler when you use momentum (a truly physical notion) rather than velocity (which is mere kinematics) and so it is with the laws of reflection and refraction. Moreover, once formulated in terms of momenta, the generalization to the Finsler (or normed-space) setting is obvious.

Correspondence between velocity and momentum or one minute with the Legendre transform

Let $v$ be a unit vector on a normed space $(X,\|\cdot\|)$, the momenta associated to $v$ is the set of unit covectors $\xi \in X^*$, the dual normed space, such that the hyperplane $\xi = 1$ supports the unit ball of $X$ at the point $v$. We extend the definition to all non-zero vectors by homogeneity. You can also write this in terms of the subdifferential of the square of the norm.

(Here I really should have inserted a nice picture!)

The same procedure allows us to associate to every non-zero momentum a set of velocities. When the unit spheres of $X$ and $X^*$ are strictly convex, the correspondence between velocities and momenta is a bijection.

The Snell-Descartes law

Consider an $(n-1)$-dimensional cooriented subspace $W \subset {\mathbb R}^n$ (the wall) and consider a norm on each half-space that describes the propagation properties of light in this anisotropic medium. The norms do not have to agree on the wall, although they do when we look at piecewise flat Finsler metrics on polyhedra and so for simplicity (I do not want to deal here with "critical angles" at which refaction becomes reflection) I will assume both norms agree on the subspace $W$.

Law of refraction: If the light ray hits the wall $W$ transversally at the origin with incoming unit momentum $\xi$ (i.e. $\|\xi\|^*_1 = 1$), then the outgoing momenta $\eta$ (there may be an infinity of light rays refracting from a single ray!) are characterized by the conditions:

(a) $\|\eta\|^*_2 = 1$,

(b) The restrictions of the covectors (= linear forms) $\xi$ and $\eta$ to the subspace $W$ agree,

(the third condition is easier to draw than to state and it distinguishes refraction from reflection)

(c) The points at which the hyperplane $\eta = 1$ supports the unit ball $B_2 = \{v : \|v\|_2 \leq 1\}$ lie on the same side of $W$ as the points at which the hyperplane $\xi = 1$ supports the unit ball $B_1 = \{v : \|v\|_2 \leq 1\}$.

Indeed if we change (c) by

(c') The points at which the hyperplane $\eta = 1$ supports the unit ball $B_2 = \{ v : \|v\|_2 \leq 1 \}$ and the points at which the hyperplane $\xi = 1$ supports the unit ball $B_1 = \{ v : \|v\|_2 \leq 1 \}$ lie on different sides of $W$.

we obtain a slight generalization of the Law of reflection for Minkowski spaces given in Corollary 3.2 of the paper by Gutkin and Tabachnikov).

Remarks.

1. The laws of refraction and reflection admit a very pretty synthetic interpretation in terms of pencils of hyperplanes. Have fun!

2. When condition (b) cannot be met is exactly the case where we have a "critical angle" and light is reflected instead of refracted. A good exercise is to check these constructions in the standard case and rediscover the usual laws of reflection and refraction (and the condition for critical angles) that you find in any physics textbook.

3. When the normed spaces are such that their unit balls and their duals are strictly convex, then to each incoming light ray corresponds one and only and outgoing light ray.

4. On the upper half-space take the usual Euclidean norm, on the lower half-space take the norm whose unit ball is a double cone with base equal to the unit disc on the wall and apexes $(0,0,\dots,0,1)$ and $(0,0,\dots,0,-1)$. You can verify that all light rays coming from the upper-half plane get refracted into the sheaf of vertical rays.

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The unit ball in the sup norm is, as everybody else says, a cube. In dimension 3, the boundary is geometrically a flat manifold with singularities at the corners, so the easiest way I can see to compute the distance between any two points is to unfold the faces of the boundary and just draw a straight line segment between the two points. It should be straightforward to work out formulas for all the possibilities. The two points must be on the same face, on adjacent faces, or on opposite faces.

I suspect that something similar can be done in higher dimensions, but I am less confident of the details.

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However this would work for very special reasons (in this case I think it would). For example, if you want to compute the length metric induced on the suface of the cube by the $3$-dimensional $\ell_1$ space (where the unit ball is the octahedron) I don't think you could use this idea. –  alvarezpaiva Apr 6 '12 at 20:12
Juan-Carlos, that sounds right! Any chance you can articulate the special reasons? –  Deane Yang Apr 6 '12 at 20:19
Think of the cube as a Finsler surface. The unit tangent discs are all nice $\ell_infty$ discs (unit squares with sides parallels to the sides of the faces) and developing does not seem to screw that up. –  alvarezpaiva Apr 6 '12 at 20:43

Using TLC's example (in comment to Igor), $x=(1,−0.3,0.2)$, $y=(0.5,1,−0.7)$, I compute the path should turn at $z=(1,1,-0.61)$ ($\frac{11}{18}=0.61$), whence the path length is 1.8. Unless I am mistaken, this does not unfold to a straight line. [True but irrelevant; see below.]

Update. Now showing the range of $z=(1,1,t)$, $t\in(-1,-0.2)$ (green), which yields the same distance $d(x,y)=1.8$.

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Could you post a picture of the geodesic with the two faces unfolded? –  Deane Yang Apr 6 '12 at 20:17
@Deane: I remain a bit uncertain of my computations, and I'm too rushed to verify more carefully at the moment. So consider this tentative... –  Joseph O'Rourke Apr 6 '12 at 20:34
Let $0.8\le a\le b\le 0.9$. Then the path: x to (1,b,-1) to (a,1,-1) to y has length 1.7, which is the shortest. –  TCL Apr 6 '12 at 20:44
@alvarezpaiva. The path that I show above has lenght 1.7. I don't know how to make it a straightline. But for some values of a,b, say a=4/5, b=57/70, the path is planar, i.e. it lies on a plane. One of my conjectures is that a planar shortest path always exists. –  TCL Apr 6 '12 at 21:29
I don't see how TCL gets a distance of 1.7. It appears that for any pair of points on a vertical face, where the horizontal distance (along the faces) is greater than the difference in heights, then any path that is monotone horizontally and where the vertical distance traveled is less than the horizontal distance is a geodesic. And the distance is of course just the horizontal distance traveled, which for this example is 1.8. So although the straight line is not the only geodesic, it is always one of them. And therefore, you really can compute the distance by unfolding the faces. –  Deane Yang Apr 7 '12 at 9:16

Isn't the $l\infty$ unit ball the cube? And aren't geodesics just curves keeping all but one coordinates constant piecewise? (just modify the coordinate of greatest difference between $x, y?$ From here it seems that measuring the distance is just an arithmetic exercise.

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I'm not sure. What you say about geodesics in the $\ell _ \infty$ norm does not seem quite correct: I'd say there are more geodesics than the ones you mention. For instance, in a normed space, the affine segment between two points is always a geodesics... –  Pietro Majer Apr 6 '12 at 17:17
@Pietro: I agree that these are not ALL the geodesics, but I am saying that you can compute distance by looking at these. –  Igor Rivin Apr 6 '12 at 17:21
@Igor: Say $x=(1,-0.3,0.2), y=(0.5,1,-0.7)$ in R^3. The greatest difference in coordinates is 1.3. How do you proceed from here? –  TCL Apr 6 '12 at 19:11