Let $G$ be an infinite, countable, finitely generated group. Let $H$ be a finite index subgroup of $G$. Let $S$ be a finite, symmetric set of generators of $G$, and let $d(\cdot,\cdot)$ be the word length metric on $G$ induced by $S$. Then $d$ is also a metric on $H$. The question is the following: is $d$ in the same bilipschitz equivalence class as all the word length metrics on $H$? That is, if $d'(\cdot,\cdot)$ is a word length metric on $H$, does there exist a constant $K$ such that $\frac{1}{k}d'(g,h) \leq d(g,h) \leq K d'(g,h)$ for all $g,h \in H$?

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    $\begingroup$ Yes, and the proof is straightforward: First check the bilipschitz inequality for generators of H. The same constants will work for all other elements. $\endgroup$ – Misha Sep 12 '12 at 22:57
  • $\begingroup$ Thanks, Misha, but I'm not sure I see it. Let's assume that $S$ includes a set $S'$ of generators of $H$. It is not in general true that the $S'$ distances on $H$ are the same as the $S$ distances. Intuitively, you could make big jumps outside $H$ using $S$, and then return to $S$ at a further distance than would have been possible using just the $S'$ generators. $\endgroup$ – Vladimir Sep 13 '12 at 0:09
  • $\begingroup$ Of course, distances are different. Consider the identity map from H to itself with different norms and check that they are Lipschitz. As I said, it suffices to compute Lip constants only on generators. $\endgroup$ – Misha Sep 13 '12 at 1:04

Even more is true. A metric space is called D-separated if the distance between any distinct points is at least D.

Claim. Let $X, X'$ be $1$-separated metric spaces. Let $f: X\to X'$ be a bijective quasi-isometry. Then $f$ is bilipschitz.

Proof. Inequality $d(f(x), f(y))\le L d(x,y)+A$ implies that $d(f(x), f(y))\le M d(x,y) + M\le (M+1)d(x, y)$. Here $M$ is the maximum of $L, A$. The same argument applies to the inverse of $f$. Thus, $f$ is bilipschitz. Qed

In your case the identity map is a quasi-isometry (by Milnor-Schwarz Lemma), so the above claim applies. Of course, one can give a more direct argument as I explained in my comments: let $S$ and $S'$ be finite generating sets for $G$ and $H$ respectively. Without loss of generality we can assume that $S'\subset S$. We can also assume that representatives of all the cosets in $G/H$ are in $S$. Then the embedding of $H$ in $G$ is 1-lipschitz. Conversely, let $p: G\to H$ be nearest point projection with respect to the metric $d$. Then, since $H$ is 1-dense in $G$, the map $p$ is 3-lipschitz. This gives you the inequality $$ d\le d'\le 3d $$ on the group $H$.

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  • $\begingroup$ Thanks a lot - very interesting. Sounds like this means that all G invariant metrics on a finitely generated group G are in the same quasi-isomorphism class as the word length metrics. $\endgroup$ – Vladimir Sep 13 '12 at 17:36
  • $\begingroup$ That's not true: restrict the standard distance function on $\mathbb{R}$ to the embedding of $\mathbb{Z}^2$ generated by $1$ and an irrational number. In order for that to be true you need an additional ``$\epsilon$-separated'' hypothesis for some $\epsilon>0$. $\endgroup$ – Lee Mosher Sep 13 '12 at 18:00
  • $\begingroup$ @Lee and @Vladimir : Lee is right, but one needs even more assumptions than separation, otherwise you can take metric taking exactly two values on the group. One way to state the assumption is to say that for some finite $r$, the $r$-Rips complex of $G$ is connected and locally finite. Then the assertion follows from Milnor-Schwartz Lemma. $\endgroup$ – Misha Sep 13 '12 at 19:06
  • $\begingroup$ @user56097: Did you actually read my answer? Where did I claim that the identity map $H\to G$ is a bijection? I apply the initial observation to the identity map $H\to H$, where $H$ is equipped with two different metrics, which is what the question was about. $\endgroup$ – Misha Feb 1 '17 at 6:57
  • $\begingroup$ @Misha Indeed, you did not claim it at all, my bad. (I read your answer well, but was actually not so careful with the original question, and that's why I did not guess correctly which identity map you were talking about. Sorry about that.) $\endgroup$ – user56097 Feb 1 '17 at 22:05

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