Rough isometries

An isometry is a map {f\colon X\rightarrow Y} between two metric spaces {X,Y} which preserves all distances: {d_Y(f(x),f(x')) = d_X(x,x')} for all {x,x'\in X}. (After typing a bunch of such formulas, one tends to prefer shorter notation: {|f(x)f(x')| = |xx'|}, with the metric inferred from contexts.) It’s a popular exercise to prove that every isometry from a compact metric space into itself is surjective.

A rough isometry allows additive distortion of distances: {|\,|f(x)f(x')| - |xx'|\,|\le \epsilon}. (As contrasted with bi-Lipschitz maps, which allow multiplicative distortion). Rough isometries ignore small-scale features (in particular, they need not be continuous), but accurately capture the large scale structure of a space. This makes them convenient for studying hyperbolic spaces (trees and their relatives), where the large-scale structure is of interest.

If {f} is an {\epsilon}-rough isometry, then the Gromov-Hausdorff distance {d_{GH}} between {X} and {f(X)} is at most {\epsilon/2}. This follows from a convenient characterization of {d_{GH}}: it is equal to {\frac12 \inf_R \textrm{dis}\,(R)} where the infimum is taken over all subsets {R\subset X\times Y} that project surjectively onto each factor, and {\textrm{dis}\,(R) = \sup \{|\,|xx'|-|yy'|\,| \colon xRy, \ x'Ry' \} }.


Since small-scale features are ignored, it is not quite natural to talk about rough isometries being surjective. A natural concept for them is {\epsilon}-surjectivity, meaning that every point of {Y} is within distance {\le \epsilon} of {f(X)}. When this happens, we can conclude that {d_{GH}(X,Y)\le 3\epsilon/2}, because the Hausdorff distance between {Y} and {f(X)} is at most {\epsilon}.

Recalling that an isometry of a compact metric space {X} into {X} is automatically onto, it is natural to ask whether {\epsilon}-rough isometries of {X} into {X} are {\epsilon}-surjective. This, however, turns out to be false in general.


First example, a finite space: {X=\{0,1,2,\dots,n\}} with the metric {d(i,j) = i+j} (when {i\ne j}). Consider the backward shift {f(i)=i-1} (and {f(0)=0}). By construction, this is a rough isometry with {\epsilon=2}. Yet, the point {n} is within distance {n} from {f(X) = \{0,1,2,\dots, n-1\}}.

The above metric can be thought of a the distance one has to travel from {i} to {j} with a mandatory visit to {0}. This makes it similar to the second example.


Second example, a geodesic space: {X} is the graph shown below, a subset of {\mathbb R^2} with the intrinsic (path) metric, i.e., the length of shortest path within the set.

Comb space
Comb space

Define {f(x,y) = ((x-1)^+, (y-1)^+)} where {{\,}^+} means the positive part. Again, this is a {2}-rough isometry. The omitted part, shown in red below, contains a ball of radius {5} centered at {(5,5)}. Of course, {5} can be replaced by any positive integer.

Omitted part in red
Omitted part in red

Both of these examples are intrinsically high-dimensional: they cannot be accurately embedded into {\mathbb R^d} for low {d} (using the restriction metric rather than the path metric). This raises a question: does there exist {C=C(d)} such that for every compact subset {K\subset \mathbb R^d}, equipped with the restriction metric, every {\epsilon}-rough isometry {f\colon K\rightarrow K} is {C(d)\epsilon}-surjective?

1 thought on “Rough isometries”

  1. The last question is very nice. As you probably already considered, one might guess that it holds for all doubling metric spaces K, with the constant C depending only on the doubling constant. Both your examples get worse and worse doubling constants as they grow. But I don’t see a proof (of either the Euclidean case or the general doubling case) at the moment.

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