### Trip to Oxford; a little lattice problem

Yesterday I gave the relativity seminar at Oxford (the last one to be organized by Piotr Chrusciel, who will be moving to Vienna soon). And I committed one of the cardinal sins of talks: I “put away” information too quickly.

I fully intend to blame it on the technology.

Usually I don’t have this problem with board talks. In the case I have a set number of fixed blackboards, I will go from board 1 to board 2 to board 3 and then back to board 1. Sometimes using board 4 to keep a running tab of important details. In the case I have the sliding blackboards (the kind that has two/three layers of boards and you can slide a completed board up and write on the one underneath), I usually do top layer of board 1 to bottom layer to top layer of board 2 then bottom layer. After filling up all boards I will erase what I don’t need and recycle those boards. Oxford has a rather different system then what I am used to. Firstly, they use whiteboards. While it is more comfortable to hold a marker than a piece of chalk, my handwriting is actually slower and more legible with chalk on blackboard. But most importantly, they have an interesting design of whiteboards. The writing surface is essentially a belt. Imagine three whiteboards looped together so that the bottom of board 1 is connected to top of board 2, bottom of board 2 to top of board 3, and bottom of board 3 to top of board 1. Now mount this belt on a frame that is 1 and a half boards tall. So you can scroll up and down. Now put three of these monsters side by side.

In hindsight, the correct motion should’ve been: board 1 of the first set, board 1 of the second set, board 1 of the third set, then board 2 of the first set, and so on. This way what I just wrote will remain visible on the previous boards. But my habit with the sliding blackboard kicked in, so I did board 1, then board 2, then board 3 of the first set, completely ignoring the fact that after I scroll up, what I just wrote disappears to behind the whiteboard!.

That was a rather big failure on my part. I reflected upon this on the train ride home, and I shall not commit the same mistake again. (There was also the usual problem of my notes containing twice as much information as what I can physically go through in 1 hour, but that I have already accepted as a fact of life.) I had some nice discussions with Piotr and Luc Nguyen. The trip, overall, was very enjoyable.

On the train ride home, I also spent some time reading Misha Gromov’s “Spaces and Questions”. There are some really deep thoughts in there, and much of it I have no hope of understanding. But one of the few elementary things that I did understand was the proof of the following statement

Let $\Lambda$ be any fixed lattice on the Euclidean plane. Let $P$ be a regular polygon whose vertices are all points in the lattice. Then the number of sides $P$ has is either 3, 4, or 6.

This is closely related to the fact that only the triangle, and square, and the hexagon can be used to get a regular tiling of the plane. Before I state the proof, I should remind you what a lattice is in this context:

A discrete subset $\Lambda$ of the Euclidean plane is a lattice if for any three points $a, b, c$ in $\Lambda$, the point obtained by $c + (a - b)$ is also a point in $\Lambda$. Here $(a-b)$ is the vector in the Euclidean plane pointing from $b$ to $a$.

To prove the claim, let us first label the vertices of the polygon $P$. We will start with a fixed vertex, which we call $a_1$, and continue clockwise labeling them $a_2, a_3, \ldots, a_n$, where $n$ is the number of sides of the polygon. Write $v_1 = a_2 - a_1$ for the vector representing the edge between $a_1$ and $a_2$, and similarly for $v_i = a_{i+1} - a_i$, and lastly $v_n = a_1 - a_n$. So now we have a set of edges.

Pick an arbitrary point $O$ in the lattice $\Lambda$. By the definition of a lattice, we see that the points $b_i = O + v_i$ are all points in $\Lambda$. A bit of basic geometry will convince you that the $n$ points $b_1, \ldots, b_n$ form a regular $n$-gon with vertices in the lattice.

Now we use the following fact: for a regular $n$-gon with $n \geq 7$, the length of each of the edges is smaller than the radius of the circumscribed circle. (One notes that for $n = 6$, the length of an edge is the same as the radius.) So $b_i$ represent a smaller $n$-gon than $P$ if $n\geq 7$. Now we repeat the construction for $b_i$ and get a even smaller $n$-gon. Repeating this indefinitely we show that the distance between two points in the lattice $\Lambda$ can be made arbitrary small, and hence $\Lambda$ is actually not a discrete set, and so $\Lambda$ cannot be a lattice. This takes care of the case for large number of sides.

To finish, we need to show that the pentagon is also not allowed. The same construction won’t work, since using this method we get a bigger pentagon. But what we can do is the following: pick a point in the lattice and call it $b_1$. And let $b_2 = b_1 + v_1$, $b_3 = b_2 + v_3$, $b_4 = b_3 + v_5$, $b_5 = b_4 + v_2$. And we can check that indeed, $b_1 = b_5 + v_4$. This is because $v_1 + \cdots + v_n = 0$ by definition. And what is the shape that we have drawn? It is in fact a pentagram! The shape given by $b_1, b_4, b_2, b_5, b_3$ is the pentagon circumscribing it, and it is easy to see that this pentagon is smaller than the original one. Now arguing as before, we also obtain a contradiction.

Obviously, this second construction doesn’t apply to the triangle, square, or the hexagon (and this is related to the number of automorphisms of the cyclic groups of orders 3, 4, and 6). Q.E.D.

The next question to ask is whether this statement has analogous for the other two-dimensional space-forms (models of non-Euclidean geometry). The two cases we need to consider are the round sphere (model with positive curvature) and the hyperbolic plane (negative curvature).

For the sphere, we note that it is in fact possible to have 2-gons: for two arbitrary points, take the great circle between them, and it is composed of two segments. A regular 2-gon, on the other hand, is formed by two points at the antipodes. So any lattice on the sphere with antipodal points admits 2-gons.

But similarly, we can also take an arbitrary great circle and chop it in $n$ even parts. This set of points form a lattice, and they admit an $n$-gon. As one can imagine, the problem we are considering is intimately tied to the problem of what lattices are permitted on the sphere. As it turns out, this problem is the same as the question about what are the possible discrete subgroups of the three-dimensional rotation group $SO(3)$. A bit of thought shows that, aside from the trivial example obtained by chopping up a fixed great circle, the only admissible subgroups are those corresponding to regular polyhedrons, and those corresponding to the dihedral groups (obtained by taking the evenly chopped up great circle [equator] and adding to it points at the north and south poles). Which means that for a lattice on the sphere, the only admissible regular polygons which are not great circles are those with sides 2, 3, 4, and 5.

How about hyperbolic plane? For this, we will use the property that, in hyperbolic geometry, the sums of the interior angles of a polygon is strictly less than the values in flat space. And in fact, the sums of the interior angles can be made arbitrarily small by making the polygon bigger and bigger. So for any $n \geq 3$, we can make a regular $n$-gon such that its interior angle divides $2\pi$ evenly. Then we can obtain a tiling, and generate a lattice from there. This is somewhat related to the notion of Fuschian groups and fundamental polygons.