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 graph $T_X$ is a tree, hence geodesic in its path metric. Left translation sends edges to edges, so the action is isometric. It is free and transitive on vertices, hence proper and cobounded. Therefore the action is geometric by [L2].

Applying [L3] to the action of step 1.1 shows that the orbit map from $F_r$ to the vertex set of $T_X$ 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 $D(H)=1\cdot 2+2\cdot 1=4$. Applying [A1], the growth function of $H$ is equivalent to $m^4$.

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.

By [A1], some finitely generated group has intermediate growth, meaning subexponential growth but not polynomial growth.

Such a group satisfies the hypothesis of the statement and fails its conclusion, so the statement is false.

By [A1], the theorem is explicitly marked as not proved here.

Therefore the statement is false.