# Cayley graphs of finitely generated infinite groups quasi-isometrically embeddable in R^3

Dear friends,

I am only a theoretical physicist. However, the answer to this question is relevant for emergence of space-time from a quantum cellular automaton (in the future I will pose a much more interesting and difficult problem). But, for now:

Consider the Cayley graph of a finitely generated infinite group. Is there a theorem stating that the only graphs of this kind quasi-isometrically embeddable in R^3 are the Bravais lattices?

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Mauro: Incidentally, what are the references for the relevance of quasi-isometries to quantum cellular automation? – Misha May 17 '13 at 19:41
The reference is still under writing. As soon as it will be posted on the web (within a month), I will pose the more interesting (and I suppose more difficult) problem mentioned in the question, with a link to the preprint. – Mauro May 18 '13 at 11:11

Yes, the theorem is:

Suppose that $G$ is a finitely-generated group whose Cayley graph quasi-isometrically (qi) embeds in $R^3$. Then $G$ is commensurable to a free abelian group of rank $\le 3$. (The converse is, of course, also true.)

In other words, $G$ contains a free abelian subgroup $A$ of finite index, so that $A$ has rank $\le 3$.

Here is a proof. First, you note that $R^3$ has polynomial growth, equivalent to $x^3$. Thus, $G$ also has growth at most $x^3$ since growth of a qi embedded subspace can only be lower than the one for the ambient space. By Gromov's polynomial growth theorem, it follows that $G$ is virtually nilpotent. For nilpotent groups there is a precise formula for growth in terms of their derived series, due to Bass and Guivarch. This formula implies that the group has to be virtually abelian of rank $\le 3$.

You can find definitions and proofs of most of the results in this book.

There is a bit more general result, due to Scott Pauls that if a nilpotent group $G$ qi embeds in, say, a Hilbert space, then $G$ is virtually abelian. However, in the 3d case you are interested in, you do not need this.

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