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The amenable graph $G=(V, E)$ is a graph that satisfies the following

$$ \inf\limits_{K \subset V,\, |K|< \infty} \frac{\partial K}{|K|}=0$$

I know for example that $\mathbb{Z}^2$ is amenable and the Bethe lattice (Cayley tree) is non-amenable.

I am looking for more examples of amenable and non-amenable graphs.

Honeycomb graph, isoradial graph are amenable?

Jir
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1 Answers1

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The honeycomb, the triangular lattice and any other graph with finite isoperimetric dimension is amenable. The most standard examples of the class of non-amenable graphs are rooted regular trees of degree at least $d\ge 2$: take $K$ to be the neighborhood of the root and radius $n$. You have that $|\partial K|$ and $|K|$ have order $d^n$. You could also get the $\mathbb{Z}^d$ and adding edges between any two vertexes to get a non-amenable graph.

Just from those example you can generate a variety of graphs in either class, by noticing that if you have a non-amenable subgraph, your graph is also non-amenable. And that if you "clue" different amenable graphs together, without adding "too many" edges, you will still have a amenable graph.

Another classical way of generating either class of graphs is looking at Caley graphs.

Hope it was useful.

Kernel
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  • As a follow-up question, I would ask how to make an amenable graph non amenable? For exemple how many connection I have to add to make $\mathbb{Z}^d$ non-amenable? adding the edges with points at distance 2d would suffice? – percojazz May 11 '19 at 08:50
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    Suppose add edges for some "finite-range" $r$ of this graph. That is, for $x$ and $y$ in $\mathbb{Z}^d$ with $|x-y|\infty \le r$, ${x,y}$ becomes and edge. The box $[-n,n]^d$ will still have $(2n+1)^d$ vertices, however its new "external boundary" will have $(2n+1+r)^d - (2n+1)^d =O(n^{d-1})$ vertexes. Which would still lead to amenable graph. I took $|\cdot|\infty$ norm, but it could be any other norm. Therefore, you need to add a truly long-range set of edges. This is why amenable and non-amenable behave in very distinct ways. Hope it helped. – Kernel May 16 '19 at 16:22
  • Thanks, indeed any non-amenable graph has to have exponential growth. – percojazz May 19 '19 at 17:49