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

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. $$|\partial A|\geq C|A|^{(d-1)/d}$$ 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 $A$.

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?

• In the general context of finitely generated groups, there is little reason to relate $d$ with the number of generators: for instance you can reduce or increase the minimal number of generators by embedding $\mathbf{Z}^k$ as a finite index subgroup of a suitable virtually abelian group.
– YCor
Dec 28, 2012 at 23:56

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.

• Could you please expand on what makes you think that non-integer values might be possible?
– R W
Dec 30, 2012 at 7:06
• it depends. if the conjecture is that nilpotent groups have isoperimetric dimension equal to the degree of growth, then this is an integer. But if this this is not true, then there is no particular reason to expect that this is an integer. The upper bound given by Breuillard and Le Donne for balls in (torsion free f.g.) nilpotent groups of nilpotency length $>2$ is not always an integer.
– YCor
Dec 30, 2012 at 12:15
• Thank you - so the evidence so far is rather circumstantial.
– R W
Dec 31, 2012 at 10:35
• Thanks for this. It seems that seeking a connection between the isoperimetric dimension and number of generators is indeed rather wrongheaded. What I was ultimately asking about, I suppose, is the existence of some kind of "dimension gap." So the issues you raise--of whether being finite-dimensional is equivalent to having polynomial growth, and of whether the isoperimetric dimension must be an integer--are quite interesting.
– Jan
Jan 5, 2013 at 20:44
• I'm not sure if I misread something, but the paper of Breuillard and LeDonne is about the correction to the dominant (integer exponent) part of the growth function. The growth as well as the isoperimetric exponents are always integers (these are much older results) if they are finite. The realm of intermediate growth is where things are mysterious.
– ARG
Oct 1, 2021 at 18:47

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.

• Why do you believe that the Heisenberg group is "simpler" than the group $\mathbf{Z}^k\rtimes C_k$ (the cyclic group permuting cyclically the factors in $\mathbf{Z}^k$)?
– YCor
Nov 29, 2019 at 23:20
• If I read the question correctly, it was asked that $d<\dim(G)<\infty$. So the wreath product does not actually answers the question (since $\dim(G) = \infty$). It's true that "simplest" has no meaning here, but for a group to answer the question it has to be [virtually-]nilpotent but not abelian. And I think the first example of such a group (which comes in many textbooks) is the Heisenberg group. It's also [one of] the group[s] with the smallest growth which falls in this category. No offence meant, your answer is very much on point and shows to the OP that this was not the good way to go.
– ARG
Nov 30, 2019 at 8:25
• No, of course this wreath product is virtually abelian has finite dimension.
– YCor
Nov 30, 2019 at 10:32
• oh sorry, I misread your answer. I automatically thought it would be the wreath product over the infinite cyclic group. Then indeed a smallest example (in terms of growth) is $\mathbb{Z}^3 \rtimes C_3$
– ARG
Nov 30, 2019 at 11:39