Math-1713 · mathematics · Graph theory

Math-1713

Roughly transitive graphs versus homogeneous spaces

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Statement

Definitions and mathematical statement

For metric spaces $X,Y$, a map $f:X\to Y$ is an $(A,B)$-quasi-isometry, with $A\ge1$ and $B\ge0$, if \[ A^{-1}d_X(x,x')-B\le d_Y(f(x),f(x'))\le A d_X(x,x')+B \] for all $x,x'\in X$, and every point of $Y$ is within distance $B$ of $f(X)$. The term rough isometry in this problem permits these multiplicative and additive constants.

Call a connected graph $G$, with its graph metric, uniformly roughly transitive if some fixed finite $A,B$ have the following property: for every pair of vertices $x,y$, there is an $(A,B)$-quasi-isometry $f_{x,y}:G\to G$ with $f_{x,y}(x)=y$. A metric space $M$ is homogeneous if its group of distance-preserving bijections acts transitively on its points.

Does there exist an infinite uniformly roughly transitive graph which is not quasi-isometric to any homogeneous metric space? Equivalently, is every such graph quasi-isometric to some homogeneous space, with no requirement that this space itself be a graph or a Cayley graph?

Uniformity of $A,B$ over the choice of $x,y$ is part of the hypothesis. Allowing constants to depend arbitrarily on that pair would give a different and much weaker notion. A proposed homogeneous comparison space must admit actual isometries between all pairs of points, while the original graph only supplies approximate self-symmetries. The question asks whether the approximate symmetry can always be replaced, at bounded metric distortion, by exact homogeneity.

Short English statement

Is every graph with uniform approximate transitivity coarsely equivalent to a space with exact transitive isometries?

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