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The following questions might be trivial, however, I couldn't solve them:

Let $G$ be generated by a finite symmetric set $S$. Suppose that $\Gamma(G,S)$ is the corresponding right Cayley graph of $G$. Let $X$ be a metric space (or, maybe, a topological space with some nice structure).

(1) Is there a way to check the following property: $X$ is not quasi-isometric to a space $Z$ which is quasi-isometric to a (hence, every) Carley graph $\Gamma(G,S)$ of some f.g. group $G$.

I.e., if we partition the space of spaces up to quasi-isometric equivalence, then does every equivalence class contain a space which is quasi-isometric to a Cayley graph of some f.g. group $G$?

(2) By Stalling's theorem, the number of ends is a geometric property of the group. Does this mean that the number of ends is a quasi-isometric invariant of the spaces which are quasi-isometric to Cayley graphs?

If the answer of question (2) is affirmative and if the question (1) about equivalence classes has a negative answer, i.e., there is an equivalence class whose elements are not quasi-isomorphic to any Cayley graph, then what is an example of spaces $W_1, W_2$ which are not quasi-isometric to any Cayley graph, but $W_1$ is quasi-isometric to $W_2$ ,however, the number of ends of $W_1$ is different from the number of ends of $W_2$?

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    $\begingroup$ There are a lot of conditions missing in order for these questions to make a little sense. Make the spaces geodesic, "almost homogeneous" and of bounded geometry for a start. $\endgroup$
    – Theo Buehler
    Feb 25, 2011 at 7:43
  • $\begingroup$ Could you precise what definition of the number of ends of a metric space you use? $\endgroup$
    – Benoît Kloeckner
    Feb 25, 2011 at 7:56
  • $\begingroup$ @Buehler: Yes, I couldn't figure out the minimal conditions on $X.$ I am interested in the invariance of number of ends. And the end is the number of components of $X-B(n)$ as n->\infity. I know that it doesn't make sense for general metric spaces. But, according to answer given below, this number should make sense for a large collection of spaces also. $\endgroup$
    – Niyazi
    Feb 25, 2011 at 9:32
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    $\begingroup$ @Niyazi: the problem is that the number of ends defined in your way is infinite for an unbounded discrete space (in particular it sis far from being a quasi-isometric invariant, even for spaces quasi-isometric to a Cayley graph), so you should really make explicit the conditions on $X$, as asked by Theo Buehler. $\endgroup$
    – Benoît Kloeckner
    Feb 25, 2011 at 15:43
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    $\begingroup$ Niyazi, Stallings's Theorem does not say that the number of Ends is a quasi-isometric invariant of the group. It gives a condition under which a fg group has infinitely many ends. $\endgroup$
    – HJRW
    Feb 25, 2011 at 16:07

2 Answers 2

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I guess that a star (a tree with $n$ infinite branches issued from a single vertex) should answer at least your first question. It should have $n$ ends, whatever meaningful definition you use, an we know that a group has $1$, $2$ or an infinity of ends.

Since quasi-isometry is an equivalence relation, you do not need to invoke a space $Z$ in your first question and the answer of your second question is obviously positive.

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    $\begingroup$ My comment was intended to exclude such examples as you give in the first paragraph (among other things). With any of the definitions I know, $n$ parallel lines in $\mathbb{R}^2$ with the Euclidean metric will have $2n$ ends. $\endgroup$
    – Theo Buehler
    Feb 25, 2011 at 8:24
  • $\begingroup$ @Kloeckner: So, if $Z$ is quasi-isometric to 5-star then can we conclude that Z has 5 ends? I know that the number of ends is a quasi-iso. invariant only for cayley graphs, is it again an invariant of non-Cayley but quasi-isometric spaces? Also, according to your answer examples of $W_1$ and $W_2$ exist. Is it correct? Thank you. $\endgroup$
    – Niyazi
    Feb 25, 2011 at 9:24
  • $\begingroup$ Benoit answers both questions. E.g. the union of the $x$ and $y$ axes in the plane is a metric space with 4 ends. It is therefore not q-i to any group (which must have 0, 1, 2, or infinity ends). It is also not q-i to the union of the $x$, $y$, and $z$ axes in $R^3$, which has 6 ends. In other words, yes, the number of ends is a q-i invariant of metric spaces, not just of groups. (It is easy to prove: a q-i induces a bijection on the ends.) $\endgroup$
    – aaron
    Feb 25, 2011 at 12:45
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    $\begingroup$ @aaron: Either you're working with a coarse notion of ends which I don't know or you're a bit simplistic here. I agree with your argument for proper geodesic spaces. Could you please elucidate your argument with a precise definition of your notion of an end? With the usual topological definition of an end, a quasi-geodesic doesn't need to lie in a single end, see my first comment to this answer. $\endgroup$
    – Theo Buehler
    Feb 25, 2011 at 14:10
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    $\begingroup$ Oh right, certainly one needs a coarse notion of ends. The integers should have 2 ends regardless of whether or not you draw in the edges connecting $n$ to $n+1$. $\endgroup$
    – aaron
    Feb 25, 2011 at 21:39
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There was a conjecture by Woess that every infinite vertex-transitive graph is quasi-isometric to a Cayley graph. A slightly more sophisticated counter-example for your first question is the counter-example to this conjecture that was proposed by R.Diestel and I. Leader in "A conjecture concerning a limit of non-Cayley graphs". It was later proved by A. Eskin, D. Fisher, and K. Whyte in "Quasi-isometries and rigidity of solvable groups".

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  • $\begingroup$ @Zaimi: Does it make sense to define number of edges for Diestel-Leader graphs,or, for their quasi-isometric copies? Is this number again invariant? $\endgroup$
    – Niyazi
    Feb 25, 2011 at 12:40
  • $\begingroup$ Do you mean "ends" for D-L graphs? If so then they have infinitely many ends, which is clear from their construction as horocycle products of homogeneous trees. $\endgroup$
    – Gjergji Zaimi
    Feb 25, 2011 at 13:24

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