# Homometric $\Rightarrow$ isometric?

Suppose you know that there is a mapping between two Riemmanian manifolds $M_1$ and $M_2$ such that, for each $x_1 \in M_1$, the (codimension-1) measure of the set of points at distance $d$ from $x_1$ is the same as the measure of the set of points at distance $d$ from the corresponding $x_2 \in M_2$, for all $d>0$. (Theo J-F provides in the comments a precise definition of the appropriate "measure.") Say then the two manifolds are homometric. Does this imply that they are isometric?

On a 2-manifold, the homometric condition means that the circumference of radius-$d$ circles matches. When only a discrete set of points are at distance $d$ from $x_1$, I would like the cardinality of that set to match that from $x_2$, although perhaps this is too much to ask.

The notion of a homometric set comes from discrete geometry (e.g., in crystallography), where indeed there are homometric but incongruent sets. See, e.g., (1), or the earlier MO question (2).

It may be that for manifolds, the two concepts coincide. However, I cannot see a direct implication. I'd prefer to make as few assumptions as possible (e.g., my manifolds are not smooth), but any insight under any assumptions would be welcomed. Thanks!

Addendum. James Cranch suggested that perhaps discrete examples of pairs of homometric sets can be mimicked by pairs of manifolds, which would answer my question negatively. But he subsequently deleted his sketch, as it was not clear it could be turned into a proof. As I mentioned in the comments, there is a follow-up question by Anton Petrunin, "Voronoi cell of lattices with the same profile," also currently (18 Aug 2012) unanswered.

(1) Joseph Rosenblatt and Paul D. Seymour. "The Structure of Homometric Sets." SIAM. J. on Algebraic and Discrete Methods, 3, 343-350, 1982.

• How are you defining distances and measures? Ordinarily one would do this with a Riemannian structure, and that ordinarily would require smoothness. – James Cranch May 17 '11 at 23:14
• @James: Good point. I meant to use shortest paths, which work on, say, polyhedral manifolds, which have a finite number of "cone" points (vertices). But I would be quite happy to understand smooth Riemannian manifolds. – Joseph O'Rourke May 17 '11 at 23:32
• I'm sure you do not mean "measure", as this usually refers to the top-dimensional integral. Rather, you mean the following codimension-1 "measure". Let $M$ be a Riemannian manifold, and $C\subset M$ a closed subset. Let $C_\epsilon$ denote the subset of $M$ of all points that are within distance $\epsilon$ of $C$. Then your "measure" of $C$ is $\frac{d}{d\epsilon}|_{\epsilon=0}{\rm Vol}(C_\epsilon)$. In general, the $n$th Taylor coefficients in $\epsilon$ of ${\rm Vol}(C_\epsilon)$ is (up to some global constant) the "codimension-$n$ measure of $C$". – Theo Johnson-Freyd May 18 '11 at 0:31
• See Anton's follow-up question, whose positive resolution would answer my question: "Voronoi cell of lattices with the same profile" mathoverflow.net/questions/65491 – Joseph O'Rourke May 21 '11 at 0:13
• A few thoughts in the (smooth) Riemannian case. First, you can equivalently define homometricity to be the matching of ball's volume, by integration. Second, the volume of small balls is asymptotically euclidean, and the second term is governed by scalar curvature. So a homometric map preserves scalar curvature. – Benoît Kloeckner Aug 18 '12 at 10:04