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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?"

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    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 $\mathbb R^3$.2011-05-03
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    "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.2011-05-04
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    @Jacob: Perhaps that was bad terminology -- I meant a cylinder including top and bottom disks.2011-05-04
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    Ah, that makes sense.2011-05-04

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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.

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    Nice answer. Could you elaborate on the average distance formula for the flat case? Also, is there a simple argument that this is optimal?2011-05-03
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    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.2011-05-03
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    @kels it is a torus?2011-05-03
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    @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.2011-05-03
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    @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 $[0,a]^2$.2011-05-04
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    Thanks @joriki I maybe didn't do my due dilligence before I asked to see that calculation. However I've played with it a bit now, and I don't understand why $y$ doesn't appear in the limits of integration of the inner integral.2011-05-04
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    @yasmar: You beat me by 19 seconds. :-) I'd just noticed that error; I don't have the time to write out the correct calculation right now, but I'll do it later in the day; I think the result was nevertheless correct.2011-05-04
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    Since the hex lattice is optimal for disk packing as well as disk covering in the plane, I guess an argument could be made that that one is the optimal manifold for the flat metric. In terms of an optimal manifold sought be the OP, I would guess that there is no answer. This is just a guess though, based on the idea that if no curvature is better than positive curvature, negative curvature is better than none. Or, if one hole (genus) is better than no holes, two holes is better than one.2011-05-04
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    So I get a bunch more questions that I don't feel equipped to answer. E.g., if we fix the genus at two and the curvature at -1, can we guess at the group that would give us the optimal quotient? Is any metric with curvature -1 better than any flat metric? Does increasing the genus beyond two lead to better solutions? But these questions probably deserve a separate question post.2011-05-04
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    @joriki I was babbling away and didn't notice that you'd commented. I really appreciate what you're doing here.2011-05-04
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    @yasmar: I've updated the post with a (hopefully) correct derivation of the average distance in the case of the square lattice. I was also wondering about the case of negative curvature, but I'm not sure yet whether there's a lower bound in that case -- I'll try to post some ideas later about $n$-gons in hyperbolic geometry.2011-05-04