Geometric Actions Svarc Milnor and Growth Examples

Translations preserve absolute-value distance, so the action is isometric. If bounded sets $B,C\subseteq \mathbb{R}$ are contained in intervals of lengths $M_B,M_C$, then only finitely many integers $n$ can make $(B+n)\cap C\mathrm{\ne }\mathrm{\varnothing }$. Also every real number lies within distance at most $1$ of some integer, so the action is cobounded. Thus the action is geometric by [L2].

The real line is geodesic under its usual metric, and step 1.1 gives a geometric action. Hence [L3] makes the orbit map $n\mapsto n\cdot 0=n$ a quasi-isometry from $\mathbb{Z}$ to $\mathbb{R}$.

Translations preserve Euclidean distance, so the action is isometric. If bounded sets $B,C\subseteq {\mathbb{R}}^n$ meet after translation by $m\in {\mathbb{Z}}^n$, then each coordinate of $m$ lies in a bounded interval, so only finitely many such integer vectors occur. Also every point of ${\mathbb{R}}^n$ lies within Euclidean distance at most $\sqrt{n}$ of some integer lattice point, so the action is cobounded. Hence the action is geometric by [L2].

Euclidean space is geodesic, and step 1.1 gives a geometric action. Therefore [L3] makes the orbit map $m\mapsto m\cdot 0=m$ a quasi-isometry from ${\mathbb{Z}}^n$ to ${\mathbb{R}}^n$.

By [L1], the geometric realization $T_X$ is a tree with unit-length edges, so the unique arc between any two points is a geodesic in its path metric. Left translation extends linearly across edges and preserves lengths. If bounded sets $B,C\subseteq T_X$ satisfy $gB\cap C\mathrm{\ne }\mathrm{\varnothing }$, choose $x\in B$ with $gx\in C$. Then $d(x_0,g{x}_0)\le d(x_0,x)+d(x_0,gx)$, which is bounded by constants depending only on $B,C$. Finite valence makes the vertex ball of that radius finite, and the free vertex action has a distinct orbit vertex for each $g$; hence only finitely many $g$ can carry $B$ into $C$. Thus the action is proper. The vertex orbit is all vertices, and every point of an edge lies within $1\mathrm{/}2$ of a vertex, so the action is cobounded. It is geometric by [L2].

Applying [L3] to the action on the geodesic metric tree in step 1.1 shows that the orbit map from $F_r$ into $T_X$ is a quasi-isometry.

Horizontal translation preserves Euclidean distance, so the action is isometric. If bounded sets $B,C\subseteq {\mathbb{R}}^2$ meet after translation by $n\in \mathbb{Z}$, then the $x$-coordinates of points in $B$ and $C$ differ by $n$, so only finitely many integers occur. Thus the action is proper in the sense of [L2].

Every orbit is a horizontal line $\mathbb{Z}+x$ at fixed $y$-coordinate. So the distance from $(0,m)$ to every orbit point of $(0,0)$ is at least $\mathrm{\mid }m\mathrm{\mid }$, and these distances are unbounded as $m\to \mathrm{\infty }$. Hence no bounded set of translates covers ${\mathbb{R}}^2$, so the action is not cobounded.

Therefore the action is proper but not cobounded.

In the standard word metric on ${\mathbb{Z}}^n$, the radius-$m$ ball is $\{a\in {\mathbb{Z}}^n:\mathrm{\mid }{a}_1\mathrm{\mid }+\cdots +\mathrm{\mid }{a}_n\mathrm{\mid }\le m\}$. It is contained in the cube $\{-m,\dots ,m{\}}^n$, so its cardinality is at most $(2m+1{)}^n$.

For each $k\in \mathbb{N}$, the cube $\{-k,\dots ,k{\}}^n$ is contained in the radius-$nk$ ball, because $\mathrm{\mid }{a}_1\mathrm{\mid }+\cdots +\mathrm{\mid }{a}_n\mathrm{\mid }\le nk$ there. So ${\beta }_{{\mathbb{Z}}^n}(nk)\ge (2k+1{)}^n$. Together with step 1.1, this shows the growth function is equivalent to $m^n$.

Step 2.1 gives polynomial growth of degree $n$ for the standard generators, and [L2] transports the same growth type to every finite generating set.

Write $H={\mathbb{Z}}^3$ with multiplication $(a,b,c)(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})=(a+a^{\mathrm{\prime }},b+b^{\mathrm{\prime }},c+c^{\mathrm{\prime }}+a{b}^{\mathrm{\prime }})$. A direct commutator calculation gives $[(a,b,c),(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})]=(0,0,a{b}^{\mathrm{\prime }}-a^{\mathrm{\prime }}b)$, so ${\gamma }_2(H)=[H,H]=\{(0,0,c):c\in \mathbb{Z}\}$ and ${\gamma }_3(H)=1$.

The quotient ${\gamma }_1(H)\mathrm{/}{\gamma }_2(H)$ is generated by the images of $(1,0,0)$ and $(0,1,0)$ and is isomorphic to ${\mathbb{Z}}^2$, while ${\gamma }_2(H)\mathrm{/}{\gamma }_3(H)\cong \mathbb{Z}$. Therefore [L1] gives $D(H)=1\cdot 2+2\cdot 1=4$.

Put $x=(1,0,0)$, $y=(0,1,0)$ and $z=[x,y]=(0,0,1)$, using $[x,y]=xy{x}^{-1}{y}^{-1}$. Every element has the unique form $x^a{y}^b{z}^d=(a,b,ab+d)$, so $x,y$ generate $H$. A word of length at most $m$ in $x^{\pm 1},y^{\pm 1}$ has $\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid }\le m$ and $\mathrm{\mid }c\mathrm{\mid }\le m^2$: multiplication by $x^{\pm 1}$ changes only $a$, and multiplication by $y^{\pm 1}$ changes $b$ by $\pm 1$ and $c$ by $\pm a$. Thus its ball has at most $(2m+1{)}^2(2m^2+1)$ elements.

For an integer $d>0$, put $q=\lfloor \sqrt{d}\rfloor \ge 1$ and write $d=q\mathrm{\ell }+r$ with $0\le r<q$. Then $\mathrm{\ell }\le q+2$ and $z^d=[x^q,y^{\mathrm{\ell }}][x^r,y]$, a word of length at most $2q+2\mathrm{\ell }+2r+2\le 6q+6$. For $d<0$ invert a word for $z^{-d}$, and for $d=0$ use the empty word. Hence, for every integer $k\ge 1$, all the distinct elements $x^a{y}^b{z}^d$ with $\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid }\le k$ and $\mathrm{\mid }d\mathrm{\mid }\le k^2$ lie in the ball of radius $8k+6$. That ball has at least $(2k+1{)}^2(2k^2+1)$ elements.

The upper bound in step 2.2 is $O(m^4)$ for $m\ge 1$. For $m\ge 28$, take $k=\lfloor (m-6)\mathrm{/}8\rfloor \ge m\mathrm{/}16$ in step 3.1 to obtain a lower bound by a positive constant times $m^4$. These bounds prove equivalence with $m^4$ for $\{x,y\}$, and [L2] transfers that growth type to every finite generating set. Together with step 2.1 this proves the example.

Let $p:X\to Y$ collapse every attached leaf at $n$ to the spine vertex $n$, and let $i:Y\to X$ be the inclusion of the spine. Both maps are $1$-Lipschitz, $p\circ i={\mathrm{id}}_Y$, and every vertex of $X$ lies at distance at most $1$ from $i(Y)$. So $p$ is a quasi-isometry by [L1].

The radius-one ball about the spine vertex $n\ge 1$ in $X$ contains the two neighboring spine vertices, the center $n$, and the $n$ attached leaves, so it has cardinality $n+3$. The radius-one ball about $n$ in $Y$ always has cardinality $3$. These ball sizes are not uniformly comparable as $n\to \mathrm{\infty }$.

Thus $X$ and $Y$ are quasi-isometric by step 1.1, while step 1.2 refutes the stated ball-count conclusion.

By [L1], the action is proper.

The orbit of $0$ is $\mathbb{Z}$, which is unbounded in $\mathbb{R}$. So the conclusion of the statement fails.

Steps 1.1 and 1.2 refute the statement.

The interval $(0,1)$ is bounded, so for the trivial action its single orbit already covers the space. Thus the action is cobounded by [L1].

The orbit space is again $(0,1)$, which is not compact: the open cover $U_n:=(0,1-1\mathrm{/}n)$ for $n\ge 2$ covers it, but every finite subfamily misses points sufficiently close to $1$. So the action is not cocompact.

Step 1.1 gives coboundedness while step 1.2 denies cocompactness, refuting the statement.

With respect to $S$, the radius-one ball is $\{-1,0,1\}$, so ${\beta }_{\mathbb{Z},S}(1)=3$. With respect to $T$, the radius-one ball is $\{-2,-1,0,1,2\}$, so ${\beta }_{\mathbb{Z},T}(1)=5$.

Thus the two growth functions are not equal pointwise, even though [L2] says they have the same growth type. This refutes the statement.