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The question is: prove that an infinite finitely generated group $G$ contains an isometric copy of $\mathbb{R}$, i.e., contains a bi-infinite geodesic ($G$ is equipped with the word metric).

I do not even know what I have to prove. It does not make sense to me. The word metric of $G$ assumes values in the natural numbers. How could there be an isometry between a subgraph of the Cayley graph of $G$ and the real line $\mathbb{R}$.

I am really confused.

I found this question here (sheet 6, ex. 1).

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  • $\begingroup$ Also, a finitely generated group is countable, so can't contain a copy of $\mathbb{R}$ in any sense. Where have you encountered this question? $\endgroup$ Apr 26, 2012 at 14:17
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    $\begingroup$ Well, if $f: \mathbb{R} \to \operatorname{Ca}{(\Gamma)}$ you want $d(f(x),f(y)) = |x-y|$. Think about an infinite cyclic group generated by an element of the generating set for example (this won't work in general, of course). @Chris: the question is whether there is an isometric copy of $\mathbb{R}$ in the Cayley graph of $G$. The choice of title is somewhat unfortunate (but a common abuse in geometric group theory). $\endgroup$
    – t.b.
    Apr 26, 2012 at 14:19
  • $\begingroup$ @t.b.: Right. So the real question is to find a bi-infinite path in the Cayley graph, and the metric is just a distraction. $\endgroup$ Apr 26, 2012 at 14:22
  • $\begingroup$ @Chris: exactly. $\endgroup$
    – t.b.
    Apr 26, 2012 at 14:23
  • $\begingroup$ By the way, the question rla linked to doesn't have the same 'unfortunate' phrasing. It just asks for a bi-infinite geodesic. $\endgroup$
    – Tara B
    Apr 26, 2012 at 14:36

2 Answers 2

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Here's the general answer. First, show that any connected infinite locally finite graph contains a geodesic ray. Indeed, fix a vertex $x$ and find a vertex $x_n$ at distance $n$ to $x$ and a geodesic segment $S_n$ joining $x$ to $x_n$. By a compactness argument, the sequence $(S_n)$ accumulates on a geodesic ray emanating from $x$.

Now assume in addition that your graph is isometry-transitive (e.g. the Cayley graph of a f.g. group). Then by the previous case for each $n$ there a geodesic segment $T_n$ of length $2n$, which, using homogeneity, can be supposed to be centered at the fixed point $x$. The same compactness argument shows that $(T_n)$ accumulates to a bi-infinite geodesic. This is an isometrically embedding of the Cayley graph of $\mathbf{Z}$.

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I'm just going to focus on what you've said you are confused about, namely:

"How could there be an isometry between a subgraph of the Cayley graph of G and the real line $\mathbb{R}$?".

We can extend the word metric on $G$ to a metric on the Cayley graph in a natural way, with each edge being an isometric copy of a unit interval. Under this metric, the Cayley graph of $\mathbb{Z}$ with respect to the generator $1$ is isometric to $\mathbb{R}$.

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  • $\begingroup$ You seem to only answer the question in case $G$ is $\mathbf{Z}$ with the standard generating set. $\endgroup$
    – YCor
    Feb 23, 2013 at 18:39
  • $\begingroup$ @Yves: I didn't even answer the question in the title; I was only explaining how it is possible for a (subgraph of a) Cayley graph to be isometric to $\mathbb{R}$ (as I said in my answer). $\endgroup$
    – Tara B
    Feb 23, 2013 at 19:33
  • $\begingroup$ I see, you're right. $\endgroup$
    – YCor
    Feb 24, 2013 at 13:47

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