Which groups are doubling?

Question

A metric space $(M,d)$ is doubling if there exists $n$ such that every ball of radius $r$ can be covered by $n$ balls of radius $r/2$, for all $r$. For which f.g. groups $G$ and finite symmetric generating sets $S$, is $\mathrm{Cay}(G, S)$ doubling under the path metric? Groups like this have polynomial growth, so they are virtually nilpotent by Gromov's theorem.

So which virtually nilpotent groups are doubling, and for which generating sets? All, I suppose, but I got cold feet trying to do it, it seemed quite difficult straight from the definitions and I don't really know the Lie group stuff well enough.

If $S$ is a finite symmetric generating set for a group $G$, is $\mathrm{Cay}(G, S)$ doubling precisely when $G$ is virtually nilpotent?

I'll note that in general (undirected) graphs, doubling implies polynomial growth, but not the other way around, consider for example the comb graph with vertices $\mathbb{Z} \times \mathbb{N}$ and edges $\{\{(m,n), (m,n+1)\}, \{(m,0), (m+1,0)\} \;|\; m \in \mathbb{Z}, n \in \mathbb{N}\}$. But could be true for vertex-transitive graphs.

Answer 1

I think this follows from a standard ball-packing argument.

Suppose that $G$ with the metric $\rho$ induced from the Cayley graph has growth $V(R)=|B_R(1)| \sim R^d$, i.e. $\exists\ 0<c< C$ such that $cR^d\leq V(R)\leq CR^d$, where $B_R(1)$ is the open ball of radius $R$ about the identity (indeed, this argument works for any metric-measure space with polynomial growth of balls about every point in this sense). This holds for finitely generated nilpotent groups by a result of Bass.

Then take a maximal $R/2$-packing $N_R$ of $B_R(1)$, i.e. $N_R\subset B_R(1)$ where $\rho(g_1,g_2)\geq R/2$ for $g_1,g_2\in N_R, g_1\neq g_2$. Then $B_{R/4}(g_1)\cap B_{R/4}(g_2) =\emptyset$.

By maximality, $B_R(1)\subseteq \cup_{g\in N_R} B_{R/2}(g)$: if not, then we could find another point $h\in B_R(1)$ whose distance $\rho(h,N_R)$ is at least $R/2$, so $N_R\cup \{h\}$ is an $R/2$-packing, a contradiction to maximality of $N_R$. Thus $N_R$ is an $R/2$-net of $B_R(1)$.

Moreover, the union of the balls of radius $R/4$ about points of $N_R$ lies in $B_{5R/4}(1)$. Hence $|N_R|V(R/4) \leq V(5 R/4)$. So we have $|N_R| \leq \frac{V(5R/4)}{V(R/4)} \leq \frac{C (5R/4)^d}{c(R/4)^d} =C/c5^d $. Hence the space is doubling.

Answer 2

Yes: a f.g. discrete, and more generally compactly generated locally compact group is doubling iff it has polynomial growth.

For f.g. groups, you mentioned $\Rightarrow$, and asked $\Leftarrow$, which I justify below.

Define $X$ to be large-scale doubling if for some $R_0,M_0$, every ball of radius $R\ge R_0$ is finite union of $M_0$ balls of radius $R/2$.

To be large-scale doubling is a QI-invariant. For a metric space in which balls of given radius have bounded cardinal, it's obviously equivalent to doubling.

So for a f.g. group, being doubling doesn't depend on a choice of finite generating subset. Since every f.g. nilpotent group is QI to some simply connected nilpotent Lie group (Malcev), it is enough to check that every simply connected nilpotent Lie group $G$ is large-scale doubling. (More generally every compactly generated locally compact group of at most polynomial growth is QI to such a $G$.)

Indeed Pansu proved in 1983 that every asymptotic cone of such a Lie group $G$ is homeomorphic to $G$ and is a proper metric space. This implies that $G$ is large-scale doubling, by the following fact:

If a space $X$ is not large-scale doubling, then there exists a sequence of points $(x_n)$, and radii $r_n\to\infty$ and $M_n\to\infty$ such that the $2r_n$-ball around $x_n$ contains $M_n$ points at distance $\ge r_n$. It easily follows that the ultralimit of rescaled metric spaces $(X,x_n,\frac{1}{r_n}d)$, which has a natural basepoint $o$ has infinitely many points in the $2$-ball around $o$ at pairwise distance $\ge 1$, so is not a proper metric space.