Math-1712 · mathematics · Graph theory

Math-1712

Grid-or-tree embeddings in superlinear Cayley graphs

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Statement

Definitions and mathematical statement

Let $\Gamma$ be a finitely generated infinite group and $S=S^{-1}$ a finite generating set not containing its identity. Let $G=\operatorname{Cay}(\Gamma,S)$ be the undirected graph with edges $\{g,gs\}$ for $g\in\Gamma$, $s\in S$. Give $G$ its graph metric. Suppose it has superlinear volume growth: \[ \frac{|B_G(1,n)|}{n}\longrightarrow\infty. \] Let $Q$ be the nearest-neighbor square lattice on $\mathbb Z^2$, and let $T$ be the infinite rooted binary tree, with each vertex having two children.

Must at least one of $Q,T$ admit a rough isometric embedding in $G$? Here, as in the source's coarse-geometric terminology, this means a map $f:V(H)\to V(G)$ for $H=Q$ or $H=T$, and constants $A\ge1$, $B\ge0$, such that for all $x,y\in V(H)$, \[ A^{-1}d_H(x,y)-B\le d_G(f(x),f(y))\le A d_H(x,y)+B. \] This is a quasi-isometric embedding; coarse surjectivity onto $G$ is not required. The constants may depend on the group and its generating set. The map need not carry each edge to one edge, so the formulation is a metric embedding problem rather than a demand for a literal lattice or binary-tree subgraph.

The two possible targets have different growth: a binary-tree embedding forces exponential growth in the ambient graph, while the lattice alternative also allows intermediate or polynomial growth. The question concerns every finite generating set of every superlinear Cayley graph.

Short English statement

Does every Cayley graph growing faster than a line contain a coarse metric copy of a square grid or a binary tree?

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