Geometric Actions Svarc Milnor and Growth

If the action is proper in the metric sense, then [L2] turns every finite set into a bounded set. So for every finite $F\subseteq X$, the set $\{g\in G:(g\cdot F)\cap F\mathrm{\ne }\mathrm{\varnothing }\}$ is finite by [L1].

Conversely, suppose the finite-set condition holds. Let $B,C\subseteq X$ be bounded. By [L2], the union $F:=B\cup C$ is finite. If $(g\cdot B)\cap C\mathrm{\ne }\mathrm{\varnothing }$, then certainly $(g\cdot F)\cap F\mathrm{\ne }\mathrm{\varnothing }$, so the transporter of $B$ into $C$ is contained in the finite set supplied for $F$. Hence the action is proper in the metric sense.

The finite-set condition implies the pointwise bound by taking $F:=\bar{B}(x,R)$; and the pointwise bound implies the finite-set condition because a finite set lies in some ball $\bar{B}(x,R)$. Thus all three formulations are equivalent.

Fix $x,y\in X$. If $d_X(x,y)\le 1$, the displayed hypothesis already gives $d_Y(f(x),f(y))\le C\le C{d}_X(x,y)+C$.

Suppose $d_X(x,y)>1$. By [L1], choose a geodesic $\gamma :[0,\mathrm{\ell }]\to X$ from $x$ to $y$ with $\mathrm{\ell }=d_X(x,y)$. By [L3] and [L4], let $m$ be the least natural number with $\mathrm{\ell }\le m$. Then $m-1<\mathrm{\ell }\le m$, so $m\le \mathrm{\ell }+1$.

Put $x_i:=\gamma (i\mathrm{\ell }\mathrm{/}m)$ for $0\le i\le m$. Consecutive points satisfy $d_X(x_{i-1},x_i)=\mathrm{\ell }\mathrm{/}m\le 1$, so the hypothesis gives $d_Y(f(x_{i-1}),f(x_i))\le C$ for every $i$. Summing along the chain yields $d_Y(f(x),f(y))\le mC\le C\mathrm{\ell }+C=C{d}_X(x,y)+C$.

Steps 1.1 and 2.1 give the displayed global bound for all $x,y$, so [L2] makes $f$ coarse Lipschitz.

Because $S\cup S^{-1}$ is finite by [L1], adjoining $0$ gives a nonempty finite set of real numbers, so the maximum $M$ exists.

Let $u:=g^{-1}h$, and write $u=s_1\cdots s_n$ with $n=\mathrm{\mid }u{\mathrm{\mid }}_S=d_S(g,h)$ and each $s_i\in S\cup S^{-1}$ by [L3]. Repeated use of the triangle inequality gives $d_X(x_0,u\cdot x_0)\le {\sum }_{i=1}^n{d}_X(x_0,s_i\cdot x_0)\le nM$. Applying the isometry $g$ and [L2] yields $d_X(g\cdot x_0,h\cdot x_0)=d_X(x_0,u\cdot x_0)\le M{d}_S(g,h)$.

The displayed estimate is a coarse-Lipschitz bound with multiplicative constant $M$ and additive constant $0$, so the orbit map is coarse Lipschitz by [L4].

The set $S$ is finite because the action is proper by [L1], the singleton $\{x_0\}$ and the ball $\{y:d_X(x_0,y)\le 2D+1\}$ are bounded, and $S$ is exactly the transporter set from the first to the second.

Let $g\in G$. By [L2], choose a geodesic $\gamma :[0,\mathrm{\ell }]\to X$ from $x_0$ to $g\cdot x_0$, where $\mathrm{\ell }=d_X(x_0,g\cdot x_0)$. By [L3] and [L4], let $m$ be the least positive natural number with $\mathrm{\ell }\le m$. Thus $m\ge 1$ even when $g\cdot x_0=x_0$. Put $p_i:=\gamma (i\mathrm{\ell }\mathrm{/}m)$ for $0\le i\le m$. Then $d_X(p_{i-1},p_i)=\mathrm{\ell }\mathrm{/}m\le 1$ for each $i$.

For each $i$, choose $g_i\in G$ with $d_X(p_i,g_i\cdot x_0)\le D$, and arrange $g_0=e$ and $g_m=g$. This is possible because $p_0=x_0$ and $p_m=g\cdot x_0$.

Put $s_i:=g_i^{-1}{g}_{i+1}$. Then $d_X(x_0,s_i\cdot x_0)=d_X(g_i\cdot x_0,g_{i+1}\cdot x_0)\le D+1+D=2D+1$, so every $s_i$ lies in $S$. Since $g=s_0{s}_1\cdots s_{m-1}$, the set $S$ generates $G$.

Step 3.1 shows that $S$ is a generating set, and step 1.1 shows that it is finite. Hence [L5] makes $G$ finitely generated.

By [L1], the set $S$ is finite and generates $G$, so $d_S$ is a word metric on $G$. Step [L2] gives the coarse-Lipschitz upper bound for ${\varphi }_{x_0}$.

The orbit ${\varphi }_{x_0}(G)=G\cdot x_0$ is $D$-dense in $X$ by the choice of $D$, so it is coarsely dense in the sense of [L3].

For $g\in G$, the proof of [L1] writes $g$ as a product of at most $m$ elements of $S$, where $m$ is the least positive integer with $d_X(x_0,g\cdot x_0)\le m$. Therefore $\mathrm{\mid }g{\mathrm{\mid }}_S\le d_X(x_0,g\cdot x_0)+1$. Applying this to $g^{-1}h$ and using isometricity gives $d_S(g,h)=\mathrm{\mid }{g}^{-1}h{\mathrm{\mid }}_S\le d_X(g\cdot x_0,h\cdot x_0)+1$.

The finite generating set $S$ gives a fixed enumeration of $G$: order words in the finite alphabet $S$ first by length and then lexicographically, and retain the first word representing each group element. The empty word represents the identity. For each $x\in X$, step 1.2 makes the set of words whose represented element $g$ satisfies $d_X(x,g\cdot x_0)\le D$ nonempty. Let $r(x)$ be the element represented by its first word. This defines one function $r:X\to G$ without choosing independently over $X$, and $d_X(x,r(x)\cdot x_0)\le D$ for every $x$.

For $x,y\in X$, step 1.3 with $g=r(x)$ and $h=r(y)$ gives $d_S(r(x),r(y))\le d_X(r(x)\cdot x_0,r(y)\cdot x_0)+1\le d_X(x,y)+2D+1.$ So $r:X\to G$ is coarse Lipschitz.

For every $x\in X$, step 2.1 gives $d_X({\varphi }_{x_0}(r(x)),x)\le D$. For every $g\in G$, step 1.3 and step 2.1 with $x=g\cdot x_0$ give $d_S(r(g\cdot x_0),g)\le d_X(r(g\cdot x_0)\cdot x_0,g\cdot x_0)+1\le D+1.$ Thus $r\circ {\varphi }_{x_0}$ and ${\mathrm{id}}_G$, and also ${\varphi }_{x_0}\circ r$ and ${\mathrm{id}}_X$, are at bounded distance.

Step 1.1 shows that ${\varphi }_{x_0}$ is coarse Lipschitz, and step 3.2 gives a coarse Lipschitz quasi-inverse. Hence the orbit map is a quasi-isometry by [L3].

Choose points $x_G,x_H\in X$. By [L1], the orbit maps based at $x_G$ and $x_H$ make the groups $G$ and $H$ each quasi-isometric to the common space $X$.

By transitivity of quasi-isometry from [L2], the spaces $G$ and $H$ are quasi-isometric to each other.

Reflexivity holds with $C=1$, since $f(n)\le f(n+1)+1$ for every $n$ and $f$ is nondecreasing. So $f\preccurlyeq f$.

Suppose $f\preccurlyeq g$ via $C_1$ and $g\preccurlyeq h$ via $C_2$. Put $C:=C_1{C}_2+C_1+C_2$, which is again a natural number with $C\ge 1$. Then $f(n)\le C_{1}g(C_{1}n+C_1)+C_1\le C_1{C}_{2}h(C_2(C_{1}n+C_1)+C_2)+C_1{C}_2+C_1+C_2$, and the argument of $h$ is at most $Cn+C$. Since $h$ is nondecreasing and $C\ge C_1{C}_2$, this gives $f(n)\le Ch(Cn+C)+C$ for every $n$. Hence $f\preccurlyeq h$.

Steps 1.1 and 1.2 make $\preccurlyeq$ a preorder. The relation $\simeq$ is therefore an equivalence relation by definition.

By [L3], choose a natural number $C\ge 1$ such that $d_T(g,h)\le C{d}_S(g,h)$ and $d_S(g,h)\le C{d}_T(g,h)$ for all $g,h\in G$.

If $\mathrm{\mid }g{\mathrm{\mid }}_S\le n$, then step 1.1 gives $\mathrm{\mid }g{\mathrm{\mid }}_T\le Cn$. So every element counted by ${\beta }_{G,S}(n)$ is also counted by ${\beta }_{G,T}(Cn)$, and therefore ${\beta }_{G,S}(n)\le {\beta }_{G,T}(Cn)$. Exchanging $S$ and $T$ yields the reverse inequality.

Because $C\ge 1$ and the growth functions are nondecreasing, step 2.1 implies ${\beta }_{G,S}(n)\le {\beta }_{G,T}(Cn)\le C{\beta }_{G,T}(Cn+C)+C,$ and likewise with $S$ and $T$ interchanged. Thus [L2] gives ${\beta }_{G,S}\preccurlyeq {\beta }_{G,T}$ and ${\beta }_{G,T}\preccurlyeq {\beta }_{G,S}$. Hence ${\beta }_{G,S}\simeq {\beta }_{G,T}$, and the growth type is independent of the finite generating set.

By composing $f$ with left translation by $f(e{)}^{-1}$ in $H$, which is an isometry by [L4], we may assume $f(e)=e$ without changing any fiber cardinalities or quasi-isometry constants up to harmless enlargement.

By [L1], choose a coarse Lipschitz quasi-inverse $q:H\to G$ and a real $c\ge 0$ with $d_S(q(f(g)),g)\le c$ for all $g\in G$. By [L2], enlarge constants so that both $f$ and $q$ are Lipschitz, say $d_T(f(g),f(h))\le L{d}_S(g,h)$ and $d_S(q(u),q(v))\le L^{\mathrm{\prime }}{d}_T(u,v)$.

Let $A:=\lceil L\rceil$, $B:=\lceil L^{\mathrm{\prime }}\rceil$, $m_G:={\beta }_{G,S}(\lceil 2c\rceil )$, and $m_H:={\beta }_{H,T}(\lceil 2c\rceil )$. Then $A,B,m_G,m_H$ are natural numbers, and the Lipschitz bound on $f$ gives $f(B_S(e,n))\subseteq B_T(e,An)$ for every $n$. If $f(g_1)=f(g_2)$, then step 1.2 gives $d_S(g_1,g_2)\le 2c\le \lceil 2c\rceil$, so every fiber of $f$ has size at most $m_G$, finite by [L3]. Therefore ${\beta }_{G,S}(n)\le m_G{\beta }_{H,T}(An)$.

Applying the same argument to the quasi-inverse $q$ gives ${\beta }_{H,T}(n)\le m_H{\beta }_{G,S}(Bn)$ for all $n$.

Let $K$ be a natural number with $K\ge A,B,m_G,m_H$. Growth functions are nondecreasing and every radius-$n$ word-metric ball contains the identity, so step 2.1 gives ${\beta }_{G,S}(n)\le m_G{\beta }_{H,T}(An)\le K{\beta }_{H,T}(Kn+K)+K,$ and step 2.2 similarly gives ${\beta }_{H,T}(n)\le K{\beta }_{G,S}(Kn+K)+K\mathrm{.}$ These are the two comparison directions of [L5]. Hence ${\beta }_{G,S}\simeq {\beta }_{H,T}$, so $G$ and $H$ have the same growth type.

For each $n\ge 1$, the reduced words of length exactly $n$ on $X\cup X^{-1}$ number $2r(2r-1{)}^{n-1}$: there are $2r$ choices for the first letter and, after that, $2r-1$ choices at each step to avoid immediate cancellation.

By [L1] and [L5], those reduced words represent distinct elements of word length exactly $n$. Therefore the ball of radius $n$ contains at least $2r(2r-1{)}^{n-1}$ elements, so ${\beta }_{F_r,X}(n)\ge (2r-1{)}^n$ for every $n\ge 1$.

Because $r\ge 2$, the real number $a:=2r-1$ satisfies $a>1$. Step 2.1 gives $a^n\le {\beta }_{F_r,X}(n)$ for all $n$, so $a^n\preccurlyeq {\beta }_{F_r,X}$ and [L3] makes the growth exponential.

Fix a finite generating set $S$ and put $d=D(G)$. Choose an integer $C\ge \max (1,C_S)$. For $n\ge 1$, [L1] gives ${\beta }_{G,S}(n)\le C_S{n}^d\le C(Cn+C{)}^d+C$. At $n=0$, the word ball consists of the identity, so the same inequality holds. Thus ${\beta }_{G,S}\preccurlyeq n^d$ by [F1]; for $d=0$ use the constant polynomial $1$.

Therefore [L2] makes $G$ a group of polynomial growth.