The book "The World is Flat" uses flatness as a metaphor for a global economy. In fact, a spherical world would seem to be better than a flat world in terms of reducing the distances between two random points on the surface of the world. The shorter the distance between any two points, the easier it is for information and objects to travel between different places. While it may be obvious that a spherical world is better than a flat world, it's far from obvious that a spherical world is optimal in this regard, which brings me to the question: what should the book have been titled? More precisely:

Question: Define a world to be a 2-manifold with some fixed surface area S and a metric d that calculates distance on the surface of the manifold. What shaped world minimizes average distance between any two randomly selected points in the world?

Does the answer depend on whether the world can be embedded in $\mathbb{R}^3$?

Does it depend on the specific metric used?

Does it depend on how we define "average distance" or "randomly selected?"

  • $\begingroup$ I think there's only one natural definition of "randomly selected" -- if you can define the total surface area of the manifold, you must have a notion of area, and then "randomly selected" should mean that equal areas have the same chance of being selected. I'm sure it does depend on the metric used -- a cylinder and a sphere are homeomorphic, but I'd be rather surprised if they had the same average distance with the metrics induced by their respective embeddings in $\mathbb R^3$. $\endgroup$
    – joriki
    May 3 '11 at 18:01
  • $\begingroup$ "a cylinder and a sphere are homeomorphic" This isn't true, is it? A cylinder is a 2-manifold with boundary and further more has infinite cyclic fundamental group? A sphere is a closed simply-connected manifold. $\endgroup$
    – JSchlather
    May 4 '11 at 2:31
  • $\begingroup$ @Jacob: Perhaps that was bad terminology -- I meant a cylinder including top and bottom disks. $\endgroup$
    – joriki
    May 4 '11 at 18:13
  • $\begingroup$ Ah, that makes sense. $\endgroup$
    – JSchlather
    May 4 '11 at 18:45

The average distance on a sphere of radius $r$ is $\frac{\pi}{2}r$ (which we can get without any integrations because there's as much area at a distance $\frac{\pi}{2}+\theta$ from a given point as there is at $\frac{\pi}{2}-\theta$); so this is

$$\bar{d}=\frac{\pi}{2}\sqrt{\frac{A}{4\pi}}=\frac{\sqrt{\pi}}{4}\sqrt{A}\approx 0.443 \sqrt{A}\;.$$

The average distance on the flat manifold $(\mathbb R / a\mathbb Z)^2$ is $\frac{1}{6}(\sqrt{2}+\sinh^{-1} 1)a$ (the average Euclidean distance between the point $(a/2,a/2)$ and a random point in $[0,a]^2$), and the area of that manifold is $a^2$, so that makes

$$\bar{d}=\frac{1}{6}(\sqrt{2}+\sinh^{-1} 1)\sqrt{A}\approx 0.383 \sqrt{A}\;.$$

So it's not true that a flat manifold has higher average distance than a sphere; it's the other way around. I suspect the comparison you mean is between a sphere and a flat disk; but a disk is a manifold with boundary, not a manifold, and the higher average distance is due to the boundary, not to the flatness.

[Edit:] Here's the calculation of the average distance in the flat case, as requested. [This is a corrected version.]

We'll need an integral of the form $\sqrt{x^2+c^2}$ several times, so I'll do that in general form first, using the substitution $x=c\sinh u$:

$$ \begin{eqnarray} \int\sqrt{x^2+c^2}\mathrm dx &=& \int c\sqrt{1+\sinh^2 u}\;c\cosh u\mathrm du \\ &=& c^2\int\cosh^2u\mathrm du \\ &=& \frac{c^2}{2}(u+\sinh u\cosh u) \\ &=& \frac{c^2}{2}\left(\sinh^{-1}\frac{x}{c}+\frac{x}{c}\sqrt{1+\left(\frac{x}{c}\right)^2}\right) \\ &=& \frac{1}{2}\left(c^2\sinh^{-1}\frac{x}{c}+x\sqrt{x^2+c^2}\right)\;. \end{eqnarray} $$

Then the integral over the distance in the primitive cell of a square lattice with $a=2$ is

$$ \begin{eqnarray} \int_{-1}^1\int_{-1}^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy &=& 4\int_0^1\int_0^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy \\ &=& 4\int_0^1\left[ \frac{1}{2}\left(y^2\sinh^{-1}\frac{x}{y}+x\sqrt{x^2+y^2}\right) \right]_0^1 \mathrm dy \\ &=& 2\int_0^1 \left(y^2\sinh^{-1}\frac{1}{y}+\sqrt{y^2+1}\right) \mathrm dy\;. \end{eqnarray} $$

We can deal with the first term by integrating by parts twice:

$$ \begin{eqnarray} \int y^2\sinh^{-1}\frac{1}{y}\mathrm dy &=& \frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+\int y\frac{1}{\sqrt{1+(1/y)^2}}\mathrm dy\right) \\ &=& \frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+\int y\frac{y}{\sqrt{y^2+1}}\mathrm dy\right) \\ &=& \frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1}-\int \sqrt{y^2+1}\mathrm dy\right)\;. \end{eqnarray} $$

Putting everything together, we get

$$ \begin{eqnarray} \int_{-1}^1\int_{-1}^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy &=& 2\left\{\frac{1}{3}\left[ y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1} \right]_0^1 +\left(1-\frac{1}{3}\right) \int_0^1\sqrt{y^2+1}\mathrm dy\right\} \\ &=& \frac{2}{3}\left[ y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1} +\sinh^{-1}y+y\sqrt{y^2+1}\right]_0^1 \\ &=& \frac{4}{3}\left(\sinh^{-1}1+\sqrt{2}\right)\;. \end{eqnarray} $$

This has to be divided by the area $2^2$ to get the average distance for $a=2$, and then by $2$ to get the average distance for $a=1$, since the average distance scales linearly with $a$; that leads to the above result.

For the average distance on the quotient of the plane with respect to a hexagonal lattice, we can use symmetry to restrict the calculation to half of one of the $6$ equilateral triangles. The integrals are rather complicated to work out by hand, but can be done analytically; WolframAlpha gives

$$ \int_0^{\sqrt{3}/2} \int_0^{1-y/\sqrt{3}} \sqrt{x^2+y^2}\mathrm dx \mathrm dy =\frac{4+\log 27}{32\sqrt{3}} $$

for side length $a=1$. The area of one of these half-triangles is $\frac{\sqrt{3}}{8} a^2$, so the total area of the manifold is $\frac{3\sqrt{3}}{2}a^2$, and the average distance comes out as

$$\bar{d}=\left(\frac{4+\log 27}{32\sqrt{3}}/\frac{\sqrt{3}}{8}\right)a=\frac{4+\log 27}{12}\sqrt{\frac{A}{3\sqrt{3}/2}}\approx 0.370 \sqrt{A}\;,$$

so yasmar's manifold indeed slightly improves on the one using a square lattice.

  • $\begingroup$ Nice answer. Could you elaborate on the average distance formula for the flat case? Also, is there a simple argument that this is optimal? $\endgroup$
    – yasmar
    May 3 '11 at 18:52
  • $\begingroup$ I doubt that the flat case is optimal, since I can turn it into a torus, say, and allow to jump from one edge to the other. $\endgroup$
    – kels
    May 3 '11 at 19:10
  • $\begingroup$ @kels it is a torus? $\endgroup$
    – yasmar
    May 3 '11 at 20:01
  • $\begingroup$ @joriki If your flat example corresponds to a Cartesian lattice, which has a square Voronoi cell on which you did your average distance calculation, then I expect we could do better on the hexagonal lattice, i.e., if you're fundamental polygon has an angle $60^\circ$ between the basis vectors. I'm curious about the average distance calculation though ... I don't understand it. $\endgroup$
    – yasmar
    May 3 '11 at 20:03
  • $\begingroup$ @kels: $\mathbb R / a\mathbb Z$ means the (additive) quotient of $\mathbb R$ with respect to the integer multiples of $a$; that does exactly what you were proposing, identifying opposite edges in $[0,a]^2$. $\endgroup$
    – joriki
    May 4 '11 at 0:32

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