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?
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 $c < \pi / 2$.