The least distorted curves and surfaces

Every subset ${A\subset \mathbb R^n}$ inherits the metric from ${\mathbb R^n}$, namely ${d(a,b)=|a-b|}$. But we can also consider the intrinsic metric on ${A}$, defined as follows: ${\rho_A(a,b)}$ is the infimum of the lengths of curves that connect ${a}$ to ${b}$ within ${A}$. Let’s assume there is always such a curve of finite length, and therefore ${\rho_A}$ is always finite. All the properties of a metric hold, and we also have ${|a-b|\le \rho_A(a,b)}$ for all ${a,b\in A}$.

If ${A}$ happens to be convex, then ${\rho_A(a,b)=|a-b|}$ because any two points are joined by a line segment. There are also some nonconvex sets for which ${\rho_A}$ coincides with the Euclidean distance: for example, the punctured plane ${\mathbb R^2\setminus \{(0,0)\}}$. Although we can’t always get from ${a}$ to ${b}$ in a straight line, the required detour can be as short as we wish.

On the other hand, for the set ${A=\{(x,y)\in \mathbb R^2 : y\le |x|\}}$ the intrinsic distance is sometimes strictly greater than Euclidean distance.

For example, the shortest curve from ${(-1,1)}$ to ${(1,1)}$ has length ${2\sqrt{2}}$, while the Euclidean distance is ${2}$. This is the worst ratio for pairs of points in this set, although proving this claim would be a bit tedious. Following Gromov (Metric structures on Riemannian and non-Riemannian spaces), define the distortion of ${A}$ as the supremum of the ratios ${\rho_A(a,b)/|a-b|}$ over all pairs of distinct points ${a,b\in A}$. (Another term in use for this concept: optimal constant of quasiconvexity.) So, the distortion of the set ${\{(x,y) : y\le |x|\}}$ is ${\sqrt{2}}$.

Gromov observed (along with posing the Knot Distortion Problem) that every simple closed curve in a Euclidean space (of any dimension) has distortion at least ${\pi/2}$. That is, the least distorted closed curve is the circle, for which the half-length/diameter ratio is exactly ${\pi/2}$.

Here is the proof. Parametrize the curve by arclength: ${\gamma\colon [0,L]\rightarrow \mathbb R^n}$. For ${0\le t\le L/2}$ define ${\Gamma(t)=\gamma(t )-\gamma(t+L/2) }$ and let ${r=\min_t|\Gamma(t)|}$. The curve ${\Gamma}$ connects two antipodal points of magnitude at least ${r}$, and stays outside of the open ball of radius ${r}$ centered at the origin. Therefore, its length is at least ${\pi r}$ (projection onto a convex subset does not increase the length). On the other hand, ${\Gamma}$ is a 2-Lipschitz map, which implies ${\pi r\le 2(L/2)}$. Thus, ${r\le L/\pi}$. Take any ${t}$ that realizes the minimum of ${|\Gamma|}$. The points ${a=\gamma(t)}$ and ${b=\gamma(t+L/2)}$ satisfy ${|a-b|\le L/\pi}$ and ${\rho_A(a,b)=L/2}$. Done.

Follow-up question: what are the least distorted closed surfaces (say, in ${\mathbb R^3}$)? It’s natural to expect that a sphere, with distortion ${\pi/2}$, is the least distorted. But this is false. An exercise from Gromov’s book (which I won’t spoil): Find a closed convex surface in ${\mathbb R^3}$ with distortion less than ${ \pi/2}$. (Here, “convex” means the surface bounds a convex solid.)

Intrinsic vs. extrinsic; also Pogorelov vs. me

There are two common ways to measure distances between two points on a sphere, such as the (idealized) surface of the Earth:

• intrinsic: length of the shortest path along the surface (not coincidentally, such a path is called a geodesic)
• extrinsic: length of the chord between two points (never mind that traveling along this chord requires digging a tunnel).

The same choice between intrinsic and extrinsic metrics is present for any surface. It is not hard to recover the intrinsic distance from the extrinsic one:

$(*)\qquad\qquad\displaystyle d_i(p,q)=\lim_{\delta\to 0}\inf \left\{ \sum_{k=1}^n d_e(x_k,x_{k-1})\colon x_0=p, x_n=q, \; \forall k\; d_e(x_k,x_{k-1})<\delta \right\}$

The construction (*) mimics the Riemannian definition of length in which the integrals along curves from $p$ to $q$ are replaced with sums along chains of points from $p$ to $q$. In the process, the surface forgets about its particular embedding into Euclidean space. For example, the parabolic cylinder $z=x^2$ and the plane $z=0$ have different extrinsic metrics, but their intrinsic metrics are the same (in the sense that there is a bijection between the two surfaces which preserves the intrinsic metric).

However, I would be hard pressed to find such an example with the paraboloid $z=x^2+y^2$ or a sphere. In fact, Pogorelov’s uniqueness theorem states that the intrinsic metric determines extrinsic metric for any closed convex surface, as well as for unbounded closed surfaces of total curvature at least $2\pi$. In all likelihood, the problem was posed to Pogorelov by his advisor A.D. Alexandrov, and Pogorelov’s book Extrinsic Geometry of Convex Surfaces is obviously meant as a companion to Alexandrov’s Intrinsic Geometry of Convex Surfaces.

Pogorelov is not exactly a household name in the Western hemisphere, even though his work on geometry of surfaces and on the Monge–Ampère equation eventually came to be cited. (The 4th Hilbert problem story is complicated, and perhaps will make another blog post.) The situation is remarkably different in Russia where Pogorelov’s “Geometry 6-10” was the standard geometry textbook for a long time. The numbers 6-10 refer to the grades from 6th to 10th. With the switch to 11-grade system these became “7-11”.

Nostalgia alert. I got the geometry textbook from the school library after finishing 5th grade, and decided to read ahead in the summer. Ended up reading the entire 6th grade material over the summer (and did the exercises too, writing the proofs neatly in my notebook). Still remember some of the diagrams, and in particular the struggle to understand the proof of Theorem 3.4, seen above. My translation:

Theorem 3.4 (characterization of isosceles triangles). If a triangle has two equal angles, then it is isosceles.

Proof. Let $ABC$ be a triangle in which $\angle A=\angle B$ (Fig.50). We will prove that it is isosceles with base $AB$.
The triangle $ABC$ is equal to the triangle $BAC$ by the second criterion of equality of triangles. Indeed, $AB=BA$, $\angle B=\angle A$, $\angle A=\angle B$. The equality of triangles implies that $AC=BC$. Therefore, $ABC$ satisfies the definition of isosceles triangle. Theorem is proved.

This got me stumped. I thought that there was something wrong with Figure 50, or perhaps 50 was a typo and the proof referred to another figure. Why is the author talking about two equal triangles, where I see only one? I eventually kept going and understood the material that followed, but did not really get this particular proof until my teacher explained it in class.