# A Polish Milnor Schwarz Lemma

November 06, 2022

Perhaps the fundamental observation in geometric group theory, the Milnor–Schwarz lemma (an independent discovery of both Milnor and Schwarz, sometimes romanized Švarc, hence sometimes titled with the names transposed) says, very roughly, that if you have a group of symmetries of a geometric object that satisfies certain properties, then the group of symmetries may be regarded as a geometric object in its own right and if you “squint,” the geometry of the group and the object are the same.

More formally, suppose that $X$ is a geodesic metric space, so the distance between any two points is realized by the length of some shortest path between them. (In fact this can be weakened somewhat) Let’s assume further that the metric on $X$ is proper, so that closed balls are compact. (It follows that $X$ is locally compact and that the metric is complete.)

An action of a (discrete) group $G$ on the proper metric space $X$ is properly discontinuous if for every compact subset $K \subset X$ , the set

$\{g \in G : g.K \cap K \ne \varnothing\}$

is finite.

Lemma (Milnor–Schwarz). Suppose $G$ acts properly discontinuously and cocompactly by isometries on the proper, geodesic metric space $X$ . Then $G$ is finitely generated, and the word metric on $G$ with respect to any finite, symmetric generating set is quasi-isometric to $X$ via any orbit map.

Here cocompactly means the quotient is compact. We could replace cocompactness with coboundedness with no loss. I defined quasi-isometry quite a while ago now; it makes precise what I said about “squinting” above.

## For Polish groups

I wanted to see whether such a Milnor–Schwarz lemma exists for Polish groups, since it provides a nice method for computing the quasi-isometry type of groups. Indeed, such a lemma does exist in Rosendal’s book! I’d like to state and prove it here.

Lemma (Rosendal). Suppose the Polish group $G$ acts continuously, coarsely properly and couboundedly by isometries on the proper geodesic metric space $X$ . Then $G$ is generated by a coarsely bounded set and the word metric on $G$ with respect to any analytic, symmetric, coarsely bounded generating set is quasi-isometric to $X$ via any orbit map.

Here we say that an action of a Polish group on a proper metric space $X$ is coarsely proper if for every compact subset $K \subset X$ , the set

$\{g \in G : g.K \cap K \ne \varnothing\}$

is coarsely bounded in $G$ .

Proof. First of all, and this bugged me for a bit, because $X$ is proper, cobounded implies cocompact. Indeed, there exists by coboundedness a ball $B$ of finite radius whose translates by $G$ cover $X$ . The closure of this ball $\bar B$ is compact by properness of $X$ and its translates manifestly still cover. It follows that the image of $\bar B$ under the quotient map is the whole space $G\backslash X$ , which is therefore compact.

Anyway, we have our compact set, which I’ll switch to calling $K$ , whose translates cover $X$ . Crucially we shall assume as above that actually the translates of the interior of $K$ cover $X$ . I claim that $G$ is generated by the coarsely bounded set $S$ of elements in $G$ satisfying $g.K \cap K \ne \varnothing$ . To prove the claim, take a basepoint $x_0 \in K$ , take $g \in G$ arbitrary, and let $\gamma$ be a geodesic from $x_0$ to $g.x_0$ in $X$ . Since translates of $\operatorname{int}(K)$ cover $X$ , the geodesic $\gamma$ can be covered by translates of $\operatorname{int}(K)$ , and by compactness of $\gamma$ , finitely many suffice. Label the translating group elements as $g_0,\ldots,g_n$ such that $g_0 = 1$ and for $1 \le i \le n$ , we have $g_i.K\cap g_{i-1}.K \ne \varnothing$ . It follows that each $s_i = g_{i-1}^{-1}g_i$ belongs to $S$ , and that $g = s_1\ldots s_n$ , proving that $S$ generates $G$ .

Now suppose that $S$ is a coarsely bounded generating set for $G$ . By Rosendal’s criterion, saying that $S$ is coarsely bounded is equivalent to saying that there exists a finite set $F$ , an open neighborhood of the identity $U$ and a natural number $k$ such that $S$ is contained in $(FU)^k \subset G$ . If $S$ is not already analytic, replacing $S$ with $F \cup U$ provides a coarsely bounded, analytic generating set. If the resulting $S$ is not symmetric, replacing it with $S \cup S^{-1}$ produces an analytic, symmetric coarsely bounded generating set.

We aim to show that $G$ equipped with the word metric from $S$ is quasi-isometric to $X$ via any orbit map. So take $x_0 \in X$ . Since there is a bounded set containing $x_0$ whose translates cover $X$ , the whole space $X$ is quasi-isometric to the orbit $G.x_0$ with the metric inherited from $X$ . What we aim to show then is that there exists constants $K \ge 1$ (different $K$ , sorry) and $C \ge 0$ such that

$\frac 1K d_G(g,h) - C \le d_X(g.x_0, h.x_0) \le Kd_G(g,h) + C.$

Suppose $d_G(g,h)$ , the word length of $h^{-1}g$ , is $n$ , and take a word $s_1\cdots s_n$ representing $h^{-1}g$ . Now, since $S$ is coarsely bounded, the orbit $S.x_0$ is bounded; in particular there is an upper bound $D$ to $d_X(s.x_0,x_0)$ as $s \in S$ varies. Repeated application of the triangle inequality therefore shows that $d_X(g.x_0,h.x_0) \le Dd_G(g,h)$ .

For the other inequality, take a ball $B$ of radius $r$ centered at $x_0$ whose translates cover $X$ . The set $S'$ of group elements such that $g.\bar B \cap \bar B \ne \varnothing$ is coarsely bounded in $G$ and therefore has word lengths bounded by some constant $M$ . Write $k$ for the floor of $\frac{d_X(g.x_0,h.x_0)}{r}$ . We can find a sequence $h.x_0 = y_0, y_1,\ldots,y_k, y_{k+1} = g.x_0$ of points on a geodesic from $h.x_0$ to $g.x_0$ such that the distance between successive points from the sequence is bounded by $r$ . Associated to these points we can choose group elements $h = h_0, h_1,\ldots h_k, h_{k+1} = g$ such that $y_i \in h_i.\bar B$ . Since $d_X(y_{i-1},y_i) \le r$ , it follows that $h_{i-1}^{-1}h_i \in S$ , and thus that $h^{-1}g$ has word length at most $Mk + M \le M\frac{d_X(g.x_0,h.x_0)}{r} + M$ , from which we conclude that $\frac{r}{M}d_G(g,h) - r \le d_X(g.x_0,h.x_0).$

Taking $K = \max\{1,\frac{M}{r}, D\}$ and $C = r$ , we see that $G$ is quasi-isometric to $X$ . $\blacksquare$

Something that I find satisfying about the proof is that it really is “the same” proof that one sees applied to properly discontinuous, cocompact actions of discrete groups!