Can the isoperimetric dimension of a d-generated group attain any value?

Question

Background

The isoperimetric dimension of a finitely generated group $G$, which we denote by $\dim(G)$, is the largest number $d$ such that any Cayley graph $\Gamma$ of $G$ (constructed with respect to a finite generating set) satisfies a $d$-dimensional isoperimetric inequality, i.e. \begin{equation} |\partial A|\geq C|A|^{(d-1)/d} \end{equation} for all finite subsets $A\subseteq\Gamma$, where $C$ is some constant (which depends on $\Gamma$ and $d$ but not on $A$). Here $\partial A$, the bounday of $A$, is the set of vertices in $\Gamma\backslash A$ which have a neighbor in $A$.

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Suppose now that $G$ is a $d$-generated group, i.e. a quotient of $\mathbb{F}_d$, the free group of rank $d$. Then provided $d>1$, $\dim(G)$ may attain any value in the set $\{0,\ldots,d\}\cup\{\infty\}$, as is evidenced, for instance, by the free Abelian groups $\mathbb{Z}^k$, where $0\leq k\leq d$ (since $\dim(\mathbb{Z}^k)=k$), and the free group $\mathbb{F}_d$ itself (since $\dim(\mathbb{F}_d)=\infty$). My question is:

What are examples of $d$-generated groups $G$ that satisfy $d<\dim(G)<\infty$?

Going a bit further:

Can the isoperimetric dimension of a $d$-generated group attain any value?

Answer 1

Denoting by $C_k$ a cyclic group of order $k$, the wreath product $\mathbf{Z}\wr C_k=\mathbf{Z}^k\rtimes C_k$ is 2-generated (hence $d$-generated for any $d\ge 2$) and has isoperimetric dimension (in the above sense) $k$.

It's likely that the "isoperimetric dimension" is finite only for f.g. groups with polynomial growth. In this case the computation is not easy and might (?) give rise to non-integral values. I do not know whether the terminology "$d$-dimensional isoperimetric inequality" is motivated by any example beyond the Euclidean setting. A natural question is whether it can be greater than the polynomial degree of growth. The results of Breuillard and Le Donne about volumes of spheres might suggest it can be greater if the nilpotency length is greater than 2.

Answer 2

The simplest Another group answering the question is the Heisenberg group (over $\mathbb{Z}$): $$ H_3(\mathbb{Z}) = \left\lbrace \left(\begin{smallmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{smallmatrix}\right) \bigg| x, y, z \in \mathbb{Z} \right\rbrace $$ This group is generated by two elements (the matrices $\left(\begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{smallmatrix}\right)$ and $\left(\begin{smallmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix}\right)$ (since the matrix $\left(\begin{smallmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix}\right)$ is the commutator of the two other.

On the other hand it has volume growth of the type $V(n) \simeq n^4$ and satisfies a 4-dimensional isoperimetric inequality. So this answers the first question.

As for the second question, note first that your inequality implies a lower bound on the volume growth of the type $n^d$. Indeed, by looking at $B_n$ the ball of radius $n$, one gets that $|\partial B_n| \geq C|B_n|^{1-1/d}$. Since the vertices are of bounded degree, $|B_n| -|B_{n-1}| \geq C'|B_n|^{1-1/d}$. The easiest way to conlcude is to introduce a piecewise affine extension $b(x)$ of the function $n \mapsto |B_n|$. Then your inequality reads $b'(x) \geq C'b(x)^{1-1/d}$ which integrates to a bound of $b(x)^{1/d} \geq C'' x$

On the other hand there is a reverse inequality (see Coulhon, Thierry, and Saloff-Coste, Laurent. "Isopérimétrie pour les groupes et les variétés.." Revista Matemática Iberoamericana 9.2 (1993): 293-314. There is a survey written in English and availiable online by Pittet and Saloff-Coste, see here. Look at section 1 (the function you are interested in is $\tfrac{1}{J}$, but not $J$ or $I$) and section 7 (more precisely Theorem 7.0.10): one has that $V(n) \succeq n^d$ implies $J(t) \preceq t^{1/d}$ (which implies $\tfrac{1}{J(t)} \succeq t^{-1/d}$, i.e. a $d$-dimesional inequality).

So this means the answer to your second question is: the isoperimetric dimension of a group is an integer or $+\infty$ (see Yves' answer for an example of a group with infinite dimension which is amenable [in fact solvable] a group which is amenable [in fact solvable] and has infinite isoperimetric dimension is $\mathbb{Z}^\infty \rtimes \mathbb{Z}$, basically the limit as $k \to \infty$of the groups in Yves' answer).

Indeed, if a group had an isoperimetry of $k +\epsilon$ for some $k \in \mathbb{N}$ and $\epsilon \in ]0,1[$ then, its growth would be at least $k+\epsilon$. Since the growth exponent is an integer, then its growth is actually at least $k+1$. In turns this implies the group has a $(k+1)$-dimensional isoperimetry.