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How can I prove that if $S,S'$ are two different finite generating sets of a group $G$ , then the metric spaces induced by the "word metric" are quasi-isometric?

The definition of quasi-isometry is: Let $X,Y$ be compact metric spaces, and $f:X \to Y$ a map. We say that $f$ is a quasi-isometry if there exist $L,A>0$ such that for every $x_1 ,x_2 \in X$ , $y \in Y$ : $ \frac{1}{L} d(x_1, x_2) -A \leq d(f(x_1), f(x_2)) \leq Ld(x_1,x_2) +A$ and $d(y, f(X))\leq A$ .

[The first condition is the problematic one... How can we find proper constants L, A ? ] Thanks in advance !

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  • $\begingroup$ I like t.b.'s answer below, but another approach is to prove that if $X$ and $Y$ are two metric spaces on which $G$ acts freely, properly, and cocompactly by isometries then $X$ and $Y$ are quasi-isometric. $G$ acts on the Cayley graphs of $(G,S)$ and $(G,S')$ with one point quotients, so the result follows. $\endgroup$
    – Paul Siegel
    Commented May 20, 2012 at 17:10
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    $\begingroup$ @Paul: I think you need to assume that $X$ and $Y$ are geodesic. Consider the natural action of $\mathbb{Z}$ given by $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$ on the "horosphere" $H_t = \{1+it\,:\,t \in \mathbb{R}\}$ in the upper half plane. But in any case, your argument works if you let $G$ act on the Cayley graph for a counterexample to your statement. .@joshua See here for a proof of the fact Paul is alluding to. $\endgroup$
    – t.b.
    Commented May 21, 2012 at 9:06
  • $\begingroup$ Yeah, you're right. I've become accustomed to the metric geometers' convention that all spaces are length spaces unless otherwise specified. $\endgroup$
    – Paul Siegel
    Commented May 26, 2012 at 21:06

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For compact spaces the notion of quasi-isometry is rather meaningless: simply take $A$ larger than the diameter of both $X$ and $Y$, so I'll assume that you don't want this answer...

I'm used to the convention that the word metric $d_S$ is obtained from the word length $$ \ell_S (x) = \min\left\{n \in \mathbb{N}\,:\, x= s_{1}\cdots s_n \text{ for some } s_{1},\ldots,s_n \in S^{\pm1}\right\} $$ (allowing the empty product, so that $\ell_S(e) = 0$) by setting $d_S(x,y) = \ell_S(x^{-1}y)$, but there are some variations which need some modifications of what I'm saying here, but nothing essential.

The subadditivity property $\ell_{S}(xy) \leq \ell_{S}(x) + \ell_{S}(y)$ of the length function $\ell_S$ is responsible for the triangle inequality of $d_S$.

Let $C' = \max\left\{\ell_{S}(s')\,:\,s' \in (S')^{\pm 1}\right\} \geq 1$. Then it follows from the definitions and subadditivity that $\ell_{S}(x) \leq C'\ell_{S'}(x)$, so $d_S(x,y) \leq C'd_{S'}(x,y)$.

Symmetrically $\ell_{S'}(x) \leq C \ell_{S}(x)$, so $d_{S'}(x,y) \leq Cd_{S}(x,y)$ which means that $$ \frac{1}{C'} d_{S}(x,y) \leq d_{S'}(x,y) \leq Cd_{S}(x,y), $$ so that the identity map $(G,d_{S}) \to (G,d_{S'})$ is a quasi-isometry (in fact a bi-Lipschitz map) with constants $L = \max\{C,C'\}$ and $A = 0$.

If instead you are interested in the Cayley graphs of $G$ with respect to $S$ and $S'$, you can take $A = 1$ and the same $L$ as above.

Also, notice that I didn't use that $G$ is finite (so that $(G,d_S)$ is compact), only that the generating sets $S$ and $S'$ are finite. For the reason I mentioned at the beginning of my answer, compactness isn't usually included in the definition of quasi-isometries, and it is common to require $L \geq 1$ and allow $A = 0$.

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  • $\begingroup$ Thanks a lot both of you ! $\endgroup$
    – joshua
    Commented May 20, 2012 at 18:19

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