Re: “How many sides does a circle have?”

The post is inspired by this story told by JDH at Math.SE.


My third-grade son came home a few weeks ago with similar homework questions:

How many faces, edges and vertices do the following
have?

  • cube
  • cylinder
  • cone
  • sphere

Like most mathematicians, my first reaction was that for the latter objects the question would need a precise definition of face, edge and vertex, and isn’t really sensible without such definitions.

But after talking about the problem with numerous people, conducting a kind of social/mathematical experiment, I observed something intriguing. What I observed was that none of my non-mathematical friends and acquaintances had any problem with using an intuitive geometric concept here, and they all agreed completely that the answers should be

  • cube: 6 faces, 12 edges, 8 vertices
  • cylinder: 3 faces, 2 edges, 0 vertices
  • cone: 2 faces, 1 edge, 1 vertex
  • sphere: 1 face, 0 edges, 0 vertices

Indeed, these were also the answers desired by my son’s teacher (who is a truly outstanding teacher). Meanwhile, all of my mathematical colleagues hemmed and hawed about how we can’t really answer, and what does “face” mean in this context anyway, and so on; most of them wanted ultimately to say that a sphere has infinitely many faces and infinitely many vertices and so on. For the homework, my son wrote an explanation giving the answers above, but also explaining that there was a sense in which some of the answers were infinite, depending on what was meant.

At a party this past weekend full of mathematicians and philosophers, it was a fun game to first ask a mathematician the question, who invariably made various objections and refusals and and said it made no sense and so on, and then the non-mathematical spouse would forthrightly give a completely clear account. There were many friendly disputes about it that evening.


Let’s track down this intuitive geometric concept that non-mathematicians possess. We are given a set {E\subset \mathbb R^n} and a point {v\in E}, and try to figure out whether {v} is a vertex, a part of an edge, or a part of a face. The answer should depend only on the shape of the set near {p}.

It is natural to say that a vector {v} is tangent to {E} at {p} if going along {v} we stay close to the set. Formally, the condition is {\lim_{t\to 0+} t^{-1}\,\mathrm{dist}\,(p+tv,E)=0}. Notice that the limit is one-sided: if {v} is tangent, {-v} may or may not be tangent.

The set of all tangent vectors to {E} at {p} is denoted by {T_pE} and is called the tangent cone. It is indeed a cone in the sense of being invariant under scaling. This set contains the zero vector, but need not be a linear space. Let’s say that the rank of point {p} is {k} if {T_pE} contains a linear space of dimension {k} but no linear space of dimension {k+1}.

Finally, define a rank {k} stratum of {E} as a connected component of the set of all points of rank {k}.

If {E} is the surface of a polyhedron, we get the familiar concepts of vertices (rank 0 strata), edges (rank 1) and faces (rank 2). For each of the homework solids the answer agrees with the opinion of non-mathematical crowd. Take the cone as an example:

Cone
Cone

At the vertex the tangent cone to the cone is… a cone. It contains no nontrivial linear space, hence the rank is 0. This is indeed a vertex.

Along the edge of the base the tangent cone is the union of two halfplanes:

Tangent cone at an edge point
Tangent cone at an edge point

Here the rank is 1: the tangent cone contains a line, but no planes.

Finally, at every point of smoothness the tangent cone is the tangent plane, so the rank is 2. The set of such points has two connected components, separated by the circular edge.

So much for the cone. As for the circle mentioned in the title, I regrettably find myself in agreement with Travis.


More seriously: the surface of a convex body is a classical example of an Alexandrov space (metric space of curvature bounded below in the triangle comparison sense). Perelman proved that any Alexandrov space can be stratified into topological manifolds. Lacking an ambient vector space, one obtains tangent cones by taking the Gromov-Hausdorff limit of blown-up neighborhoods of {p}. The tangent cone has no linear structure either — it is also a metric space — but it may be isometric to the product of {\mathbb R^k} with another metric space. The maximal {k} for which the tangent cone splits off {\mathbb R^k} becomes the rank of {p}.

Recently, Colding and Naber showed that the above approach breaks down for spaces which have only Ricci curvature bounds instead of triangle-comparison curvature. More precisely, their examples are metric spaces that arise as a noncollapsed limit of manifolds with a uniform lower Ricci bound. In this setting tangent cones are no longer uniquely determined by {p}, and they show that different cones at the same point may have different ranks.

Gromov-Hausdorff convergence

The Hausdorff distance {d_{ H}(A,B)} between two subsets {A,B} of a metric space {X} is defined by {d_{ H}(A,B)=\inf\{r>0: A\subset B_r \text{ and } B\subset A_r \}}, where {A_r,B_r} are (open/closed does not matter) {r}-neighborhoods of the sets. Informally: {d_{ H}(A,B)<r} if no matter where you are in one set, you can jump into the other by traveling less than {r}. For example, the distance between letters S and U is about the length of the longer green arrow.

Hausdorff distance
Hausdorff distance

Generally, one assumes the sets to be closed (to avoid zero distance) and bounded (to avoid infinite distance). But I will not; in this post I’m not really interested in verifying all the axioms of a metric.

The Gromov-Hausdorff distance is defined between metric spaces {X,Y} as follows: it is the infimum of {d_{ H}(f(X),g(Y))} taken over all isometric embeddings {f\colon X\rightarrow Z} and {g\colon Y\rightarrow Z} into some metric space {Z}.

The infimum over all pairs of embeddings into all conceivable metric spaces does not sound like something you would want to compute in practice. Of course, the matter boils down to equipping the abstract union {X\sqcup Y} with pseudometrics that are compatible with the original metrics on {X} and {Y}.

A more directly computable notion of distance (not necessarily a metric) can be given as follows: {\rho_{GH}(X,Y)} is the infimum of all {\epsilon>0} for which there exist two maps {f\colon X\rightarrow Y} and {g\colon Y\rightarrow X} such that:

  1. {d_X(g\circ f(x),x)\le \epsilon} for all {x\in X}
  2. {d_Y(f\circ g(y),y) \le \epsilon} for all {y\in Y}
  3. {|d_Y(f(x_1), f(x_2)) - d_X(x_1,x_2)| \le \epsilon } for all {x_1,x_2\in X}
  4. {|d_X(g(y_1), g(y_2)) - d_Y(y_1,y_2)| \le \epsilon } for all {y_1,y_2\in Y}

This is not as elegant as “infimize over all metric space”, but is more practical. For example, it is easy to check that the sequence of one-sheeted hyperboloids {H_n = \{x^2+y^2=z^2+1/n\}}

One hyperboloid,...
One hyperboloid,…
... another hyperboloid, ...
… another hyperboloid, …

converges to the cone {C = \{x^2+y^2=z^2\}}.

... and we get a cone.
… and we get a cone.

Using cylindrical coordinates, define {f_n\colon H_n\rightarrow C} by {f_n(r,\theta,z) = (\sqrt{r^2-1/n}, \theta,z) } and {g\colon C\rightarrow H_n} by {g_n(r,\theta,z) = (\sqrt{r^2+1/n}, \theta,z)}, with an arbitrary choice of {\theta} at the point {g(0,0,0)}. Now check the items one by one:

  1. {f_n\circ g_n} is the identity map on {C}
  2. {g_n\circ f_n} fixes all points of {H_n} except for those with {z=0}. The latter are displaced by at most {2/\sqrt{n}}.
  3. follows from {|\sqrt{r^2-1/n} - r|\le 1/\sqrt{n}} on {[1/n,\infty)}
  4. follows from {|\sqrt{r^2+1/n} - r|\le 1/\sqrt{n}} on {[0,\infty)}

Note that the Gromov-Hausdorff convergence of manifolds is understood with respect to their intrinsic metrics. Although both {H_n} and {C} are naturally identified with subsets of {\mathbb R^3}, it would be a mistake to use the Hausdorff distance based on the Euclidean metric of {\mathbb R^3}. Even though this extrinsic metric is bi-Lipschitz equivalent to the intrinsic metric on both {H_n} and {C}, bi-Lipschitz equivalence is too coarse to preserve the GH convergence in general. In general, the intrinsic metric on a manifold cannot be realized as the extrinsic metric in any Euclidean space.

In the example of hyperboloids we are lucky to have a Gromov-Hausdorff convergent sequence of unbounded spaces. Normally, bounded sets are required, and boundedness is imposed either by an exhaustion argument, or by changing the metric (moving to the projective space). For instance, the parabolas {y=x^2/n} do not converge to the line {y=0} as unbounded subsets of {\mathbb R^2}, but they do converge as subsets of {\mathbb R\mathbb P^2}. In the process, the degree jumps down from {2} to {1}.

It seems that the Gromov-Hausdorff limit of algebraic varieties of degree at most {d} is also an algebraic variety of degree at most {d} (provided that we use the intrinsic metric; otherwise flat ellipses {x^2+n^2y^2=1} would converge to a line segment). If this is true, I’m sure there’s a proof somewhere but I never saw one.