The equilateral dimension of Riemannian manifolds
My recent preprint on integer distances in various metrics appears to have solved a 40-year-old open problem posed by Richard Guy in an interestingly-titled work, “An olla-podrida of open problems, often oddly posed” (Amer. Math. Monthly 1983, doi:10.1080/00029890.1983.11971188, jstor:2975549; both links are paywalled but maybe you can access one or the other). I’d read his paper before but forgotten it; after noticing it again I updated the preprint to make the connection explicit.
Guy’s problem concerns the equilateral dimension of metric spaces. This is a long phrase for a simple concept: how many points can be at unit distance from each other? For Euclidean spaces of dimension \(d\), for instance, it is \(d+1\): you can embed an equilateral triangle in the plane, but you need three-dimensional space for a tetrahedron, etc. For the \(L_\infty\) metric on the same space, it is \(2^d\), as shown (at least as a lower bound) by the vertices of a unit hypercube. In problem 2 of his paper, Guy asks how this number behaves in Riemannian manifolds. He gives an example showing that arbitrarily many points on a Riemannian 2-sphere can be very nearly equilateral, and suggests that maybe they can be exactly equilateral. But if not, he asks for an upper bound on the equilateral dimension of arbitrary Riemannian manifolds.
My paper gives a general bound on complete Riemannian manifolds of bounded genus, which in the special case of Riemannian 2-spheres implies that they cannot have equilateral dimension more than five. I think applying the same proof ideas more directly to this case would reduce this upper bound to four. (Essentially, this is because you would get a non-crossing drawing of \(K_5\) from the shortest geodesics of the equilateral set, but \(K_5\) is non-planar; my paper uses the non-planarity of \(K_{3,3}\) instead, so its bound is off by one.) It is easy to place four points at unit distance: just inscribe a regular tetrahedron onto a Euclidean sphere. So Guy’s almost-equilateral sets can never be made exact. More generally, as my paper shows, complete Riemannian 2-manifolds of genus \(g\) have equilateral dimension \(O(g)\). On the other hand, it also includes examples showing that, without these restrictions, Riemannian manifolds can indeed have arbitrarily many points at distance exactly one from each other.
The first example is the reason for the “complete” when I say “complete Riemannian manifolds”. This means that they are complete as metric spaces: every Cauchy sequence of points converges. The Euclidean plane is complete in this sense: sequences of points that eventually get arbitrarily close to each other have a limit point. On the other hand, other sequences of points are allowed to diverge to infinity without having a limit; it’s ok, because those sequences are not Cauchy sequences. My example of an incomplete Riemannian 2-manifold with unbounded equilateral dimension is very simple: topologically it is just a disk, with a locally Euclidean metric. You’re probably familiar with making a cone from paper, by cutting out a wedge and gluing the cut ends together; at the apex of the cone, the total angle is less than \(2\pi\) by the angle of the wedge you cut out. It doesn’t matter how you bend or fold it in space; its internal geometry is still that of a cone, locally Euclidean except at the apex. But you can also make cones with total angle more than \(2\pi\), by cutting slits in more than one sheet of paper and gluing them together at the slits. You may have trouble getting the glued parts to stay smoothly joined, if you’re making them out of physical paper, but they’re still well-behaved metric spaces. Now do this with infinitely many sheets of paper, giving a cone with apex angle \(\infty\), and remove the apex point. This cone has been seen before, for instance in Weber, Hoffman, and Wolf, “An embedded genus-one helicoid”, PNAS 2005. I like to think of it as looking sort of like an infinite pencil shaving. If you had a pencil of infinite length and radius, and you shaved it into an infinite-length conical shaving without ever breaking it off, you would get something like this space.
In the paper, I describe it differently: parameterize it by a pair of numbers \((r,\theta)\), just like polar coordinates in the plane, but for all real values \(\theta\) rather than taking \(\theta\) modulo \(2\pi\), and for \(r>0\) rather than \(r\ge 0\). Then, just lift the familiar Euclidean metric on polar coordinates to the space of these parameter pairs. In this space, the points \(\{(\tfrac12,k\pi)\mid k\in\mathbb{Z}\}\) form an infinite set in which all distances are one. If the cone were completed by adding its apex, the shortest curve between any of these points would follow a radius of the cone from one point up to the apex and then a second radius back down to the other point. We cannot include the apex point and continue to have a Riemannian manifold, so we omit it; its removal does not change the distances, but it does leave the space incomplete.
The second example is easiest to describe as a graph. Just chain together cliques of all different sizes into a path, to make an infinite graph with no finite bound on the size of its largest clique. The reason for chaining them together in this way rather than just making one big infinite clique is to make sure that every vertex has finite degree, something that will be important when constructing a Riemann surface. Viewed as a metric space (either a discrete one using unweighted graph distance, or as a complex of points and unit-length line segments) its equilateral dimension is the same as its clique number, unbounded.
To make this into a Riemannian 2-manifold, replace each vertex by a small sphere, replace each edge by a thin tube connecting two spheres, and smooth the joints. Choose a representative point on each sphere, and adjust the length of the tubes so that, within each clique, each two of these representative points are at unit distance. The resulting surface has infinite genus; that’s why there is a requirement for bounded genus in the theorem of my preprint.
To make it into a Riemannian metric on \(\mathbb{R}^3\), intuitively, drop the surface onto a stretchy sheet of rubber 3-manifold material so that it stretches the sheet downwards all around it. Less intuitively, but maybe more rigorously, embed the 2-manifold smoothly into \(\mathbb{R}^3\) (using small enough spheres and thin but wiggly tubes so that each clique can be packed into its own separate unit ball), thicken it slightly, and use a metric for which traveling perpendicular to the surface through the thickened part is much more expensive than traveling along the surface. Then smooth out this metric to match the normal Euclidean metric away from the thickened surface. Since this space has the same topology as the usual metric on \(\mathbb{R}^3\), no purely topological invariant like genus can be used to bound the equilateral dimension of Riemannian 3-manifolds.
There is another work in this area that I should also mention: a 2013 paper by then-undergraduate Jeremy Mann, “Equilateral dimension of Riemannian manifolds with bounded curvature”. Mann shows that, regardless of dimension, the equilateral dimension of a complete Riemannian manifold can be bounded in terms of the curvature of the manifold. But this is a geometric invariant of the manifold, not a topological one.