# Convex surface on which any two points $a,b$ can be joined by a curve of length $(\pi/2-\epsilon)|a-b|$

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?

• 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 3:18
• 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
• i suggest the egg – user66081 Aug 12 '14 at 19:20
• @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 8:23
• @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. – user147263 Aug 13 '14 at 18:50

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 $c < \pi / 2$.

• 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. – user147263 Aug 13 '14 at 18:53
• @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

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. 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. 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.

• 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. – user147263 Aug 15 '14 at 2:19