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I am trying to solve an exercise on page 13 of the book Metric structures on Riemannian and non-Riemannian spaces by Gromov.

Construct a closed, convex surface $X$ in $\mathbb R^3$ such that any two points $a,b\in X$ can be joined by a curve $\gamma\subset X$ of length $$\ell(\gamma)\le c|a-b| \tag1$$ where $c<\pi/2$. Here $c$ is independent of $a,b$.

Remarks

Here $|a-b|$ is the Euclidean norm of the vector $a-b$. The geometric meaning of inequality (1) is that the surface is not too twisted: a bug crawling from $a$ to $b$ along the surface does not have to travel much further than if it flew directly from $a$ to $b$.

A closed convex surface is precisely the boundary of a convex bounded set.

Gromov calls the smallest value of $c$ for the surface satisfies the above the distortion of $X$. Other authors call it the constant of quasiconvexity.

Some ideas

  • A sphere has distortion $\pi/2$. Indeed, any curve connecting antipodal points (distance $2r$) has length at least $\pi r$, where $r$ is the radius.
  • Ellipsoids are no good; they are distorted more than spheres. Look at the vertices of the shortest axis.
  • More generally, every centrally symmetric surface has distortion at least $\pi/2$. Indeed, let $a\in X$ be a nearest point to the center of symmetry, and $b$ its antipode. Any curve connecting $a$ to $b$ stays outside of a ball with diameter $ab$, and therefore has length at least $\frac{\pi}{2}|a-b|$.
  • One can consider closed curves instead of surfaces, hoping to get inspiration from there. But the distortion of a closed curve cannot be less than $\pi/2$; proof here. That is, a circle is the least distorted closed curve.
  • Among non-symmetric $X$, a natural candidate is the regular tetrahedron, but it does not work. The dihedral angles $\alpha=\cos^{-1}(1/3)$ are too small and difficult to get around: $c$ cannot be less than $1/\sin (\alpha/2) = \sqrt{3}>\frac{\pi}{2}$.
  • Minkowski sum of a tetrahedron and a sphere of sufficiently large radius might work, but the length estimates look scary.

Any better ideas?

Community
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  • I think a cube will do the job. Take two antipodal vertices; the distance between them is $\sqrt 3$ and the smallest travelling in the surface itself is $\sqrt 5$ (I think). But $\sqrt{5/3}\lt\pi/2$. It remains to prove that this $c$ is valid for every two points in the cube. – Ian Mateus Aug 10 '14 at 03:18
  • Unfortunately the cube does not work, two points in the middle of opposite sides require $c\geq 2$. But I strongly suspect cutting the cube will yield an example of such surface. – Ian Mateus Aug 10 '14 at 03:31
  • @IanMateus Cube is also centrally symmetric, so the symmetry would have to be destroyed. (See the third bullet point). –  Aug 10 '14 at 03:38
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    There is a more immediate reason why a tetrahedron does not work: it has a cross-section which is a totally geodesic "square" hence cannot have good distortion. The cross-section is the intersection with the plane through the origin parallel to a pair of disjoint edges of the tetrahedron. – Mikhail Katz Aug 11 '14 at 13:08
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    i suggest the egg – user66081 Aug 12 '14 at 19:20
  • @user66081, a pair of antipodal points on the shortest closed geodesic on the egg have distortion $\pi/2$. – Mikhail Katz Aug 13 '14 at 08:04
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    @user72694: I'm not sure that cross section argument all by itself is valid: even though a path within the cross section might be too long, there might be a shorter path somewhere outside that cross section. – MvG Aug 13 '14 at 08:23
  • @MvG, good point, but I interpreted the "egg" as being an ellipsoid of revolution with the axis of revolution longer than the others, but I see what you mean. – Mikhail Katz Aug 13 '14 at 08:25
  • @user72694: I was more referring to your point about the tetrahedron. But in general mostly worried that someone might find a valid solution then incorrectly discard it due to this “rule”. – MvG Aug 13 '14 at 08:31
  • @MvG, I see, but I am pretty sure the square is embedded strongly isometrically. Am I wrong? – Mikhail Katz Aug 13 '14 at 08:32
  • @user72694: the egg as it comes out of a hen, is not an ellipsoid of revolution. a simplified model is two "co"circular cones, one pointy and one blunt, patched at their base, such that they meet at an angle >90${}^\circ$. – user66081 Aug 13 '14 at 12:25
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    @MvG But user72694 said "totally geodesic square". I did not check this claim, but if true, it indeed implies $c\ge 2$ for the tetrahedron. –  Aug 13 '14 at 18:50

2 Answers2

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How about a sharp cone? Suppose the cone's lateral surface unrolls to a circular sector of angle $2\theta$ for some small positive $\theta$. Then:

$\bullet$ the base is flat, so any $a,b$ on the base are joined by a line of length $|a-b|$.

$\bullet$ if $a$ is on the base and $b$ on the side, then we can choose $\gamma$ to go straight down from $b$ to the edge and thence straight to $a$; if these two segments have lengths $x,y$ then $|a-b| = \sqrt{x^2 - \epsilon(\theta) x y + y^2}$ for some $\epsilon(\theta)$ that approaches zero as $\theta \rightarrow 0$, so $\ell(\gamma) = x+y \leq (\sqrt 2 + \delta(\theta)) \, |a-b|$ for some small $\delta(\theta)$ that again tends to zero as $\theta \rightarrow 0$.

$\bullet$ Finally, if $a,b$ are both on the side then the shortest $\gamma$ is a path that unrolls to a straight line on a sector of angle at most $\theta$. At worst $a$ and $b$ are at the same height, separated by $\psi \leq \theta$ on the unrolled cone, and thus by $(\psi/\theta) \pi$ on a circular cross-section of the solid cone. Then $$ \ell(\gamma) = \frac {\mathop{\rm sinc} \frac\psi2} {\mathop{\rm sinc} \frac\pi2 \! \frac\psi\theta} |a-b| $$ where $\mathop{\rm sinc}(x) = \sin(x)/x$. Since $\mathop{\rm sinc}$ is logarithmically convex upwards, the ratio $\mathop{\rm sinc} \frac\psi2 \big/ \mathop{\rm sinc} \frac\pi2 \! \frac\psi\theta$ is an increasing function of $\psi$, so is maximized at $\psi = \theta$, where it equals $\frac\pi 2\mathop{\rm sinc} \frac\theta2$ $-$ which is still less than $\pi/2$ for any positive $\theta$, so we've attained $c < \pi / 2$.

Noam D. Elkies
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  • Would a sharp pyramid work as well? If so, then I'd expect that smoothly deforming the cone into the pyramid would produce a family of examples. – Semiclassical Aug 13 '14 at 13:05
  • I don't think a sharp pyramid works: choose a horizontal square cross-section near the vertex and let $a,b$ be midpoints of opposite sides; then $c$ is almost $2$. – Noam D. Elkies Aug 13 '14 at 13:12
  • Hmm, I think you're right; the only escape clause is whether the horizontal path in that case is the shortest path, but it seems like the sharpness of the pyramid ensures that. I would still imagine $c$ to change smoothly under deformations, though, so there should be a class of 'perturbed' sharp cones that satisfy $c<\pi/2$. – Semiclassical Aug 13 '14 at 13:29
  • @Semiclassical If a pyramid has a polygon with many edges as its base, the same $c<\pi/2$ argument should apply. I can't quite tell what kind of convergence of shapes is needed to justify the convergence of distortion constants; Hausdorff convergence seems too weak. –  Aug 13 '14 at 18:53
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    @900situpsaday: what I had in mind in particular was 'slanting' the cone slightly (i.e. tip slightly off-center) or slightly stretching the cone along an axis. In either case, a small enough distortion should be alright. (Though quantifying 'small enough' seems tricky.) – Semiclassical Aug 13 '14 at 19:33
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For completeness, I add some numerical details to the excellent answer by Noam D. Elkies. Let $C$ be the cone such that the lateral surface unrolls to a circular sector of angle $2\theta$. There are two sources of distortion:

  • Between two points on lateral surface (worst case: same level, diametrally opposite). This gives $c\ge (\pi/\theta) \sin(\theta/2)$, blue curve below.
  • Between points on lateral surface and on the base (worst case: they are equally distant from the edge). This gives $c\ge \sqrt{2\pi/(\pi-\theta)}$, red curve below.

distortion

The distortion is minimized when the curves intersect, $\theta\approx 0.53378$. This corresponds to the cone of radius $\approx 0.1675$ at height $1$; pretty sharp indeed.

cone

This least-distorted cone has $c\approx 1.5522$, beating the sphere for which $c=\pi/2 \approx 1.5708$.

Lower bounds, for comparison

  • [Gromov] Every subset of $\mathbb R^n$ with $c<\pi/(2\sqrt{2}) \approx 1.11$ is contractible.
  • [Pansu] Every subset of $\mathbb R^n$ with $c<\frac{\alpha}{2\sin(\alpha/2)}$, $\alpha=\cos^{-1}(-1/n)$, is contractible. For $n=3$ this evaluates to $\approx 1.17$.

The estimate of Pansu is in Appendix A of the same book. I don't know of any better lower bounds.

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    The cone can be improved by gluing a spherical cap to its base. Then one can look at general surfaces of revolution, but minimization looks hard... and I'm still unsure whether the least-distored closed surface overall should have rotational symmetry or tetrahedral symmetry. –  Aug 15 '14 at 02:19