# Følner sequences with weird shapes

Let $$G$$ be a discrete and finitely generated group. Recall that $$\{F_n\}_{n \in \mathbb{N}}$$ is a Følner sequence if $$|g F_n \cup F_n|/|F_n| \rightarrow 1$$ for every $$g \in G$$. As is well known, existence of a Følner sequence is equivalent to amenability of $$G$$.

It is often said that Følner sequences have strange shapes. My soft question is: which examples do we have that support this claim? Of course, if $$G$$ is of subexponential growth then a subsequence of balls forms a Følner sequence, and this does not have a weird shape. Hence, more specifically: which examples of groups of exponential growth do we know that have explicit Følner sequences not made of balls?

As instances of the examples I am asking for, Star-shaped Folner sequence asks for Følner sets of a certain form, while an answer of Folner sets and balls gives explicit sequences made of rectangles (as opposed to balls). Likewise, the ax + b group has a Følner sequence made of rectangles where one side is exponentially larger than the other.

• Maybe Fölner sequences of group extensions fit the bill, see e.g. terrytao.wordpress.com/2009/04/14/some-notes-on-amenability – Ville Salo Jan 13 at 10:05
• It's quite subjective... you could have a "weird" subsequence of "weird"-shaped balls. Also on $\mathbf{Z}$ the union of $\{1,\dots,n\}$ with any "weird" subset of cardinal $o(n)$ is Følner. Also in polycyclic groups or the $ax+b$ group the rectangles you mention look all but weird for me. The exponential size is with respect to matrix coordinates, but it's of linear size in the word length. – YCor Jan 13 at 10:07
• I'm not sure I agree with the premise of the question though, I have more often (= at least once) heard "there is a unique choice for the Fölner sequence", which does not imply they are not strange, but does imply they are natural (I guess this statement refers to mutual tileability). – Ville Salo Jan 13 at 10:07
• I do disagree with the statement that balls are natural shapes. For example on the lamplighter group, balls look pretty strange, while Fölner sequences are beautiful. – Ville Salo Jan 13 at 10:13
• @YCor you are indeed right that "weird" is subjective. I just meant Folner sequences that are not balls. Hence your example of $\mathbb{Z}$ fits the question, but I'd like some other examples. – Diego Martínez Jan 13 at 10:22

The algebra is more useful here than pictures, but the pictures are fun, so here goes. To substantiate my comment about lamplighter, quick renderings of a typical ball and Følner set of lamplighter. Actually I don't know which of these is prettier, but the Følner set is actually the one that looks more like a ball.

The two pictures are taken from different angles and thus form a stereogram, so if you look at the leftmost picture with your right eye and vice versa your stereopsis should kick in. I find this helpful, if you don't you can ignore one of the pictures.

First, the ball or radius $$3$$ with the generators where the head moves. When the head moves to the right, you go up the diagram. I'm using some conventions, which are hopefully guessable. Here's a typical Følner set with the same generators. • These are great, thanks! These are precisely what I was asking for. However, what do you mean from "different angles"? Each pair is identical, right? Mind you, I think it's more or less clear without the need from another angle, but I spent several minutes looking for the 7 differences. – Diego Martínez Jan 13 at 11:44
• @DiegoMartínez: note that if you spent several minutes looking for the 7 differences, then you are using a suboptimal algorithm for that game. A much better algorithm is what is suggested in the post, looking at one picture with one eye, and at the other with the other eye. – Ville Salo Jan 14 at 9:09
• I was looking at these again yesterday and you are right. Again, thanks for these. – Diego Martínez Jan 14 at 10:57

This question was popular in the 50s and 60s after Folner theorem was proved. Many examples of weird Folner sets were constructed. The typical examples of groups where Folner sets are nor balls are lamplighter groups and the wreath products of infinite cyclic gtoups. For a more recent papers see Anna Erschler. On isoperimetric profiles of finitely generated groups. Geom. Dedicata, 100:157–171, 2003 and the references therein.

• What's the precise meaning of "groups where Følner sets are not balls"? – YCor Jan 14 at 9:42
• @YCor I guess dodd means amenable groups where no subsequence of balls forms a Folner sequence. I don't know whether that helps. – Diego Martínez Jan 14 at 10:59
• With such a definition, this might precisely be the class of amenable groups with exponential growth (and still I disagree that these are weird! rather I'd say that balls are complicated subsets). – YCor Jan 14 at 11:16
• @YCor you're right, but let me rephrase into "amenable groups where Følner sets are known explicitly, but balls are not known to be Følner". These are necessarily of exponential growth, but may be a strictly smaller class of groups. Regarding the use of "weird" I guess it's mostly personal taste, but I get your point. – Diego Martínez Jan 14 at 11:40

An answer to your non-soft question is that the following groups all have [at least one] generating set where balls are known not to be Folner, but some other ("rectangular") sequence is: solvable Baumslag-Solitar, some wreath products (including the lamplighter), some extensions of $$\mathbb{Z}^d$$ by $$\mathbb{Z}$$ (those given by a matrix with no eigenvalues of norm 1), some $$ax+b$$ groups and basically nearly any amenable groups of exponential growth whose growth series is rational and has been computed (see below for details).

"strangeness" of Folner sets: As mentioned in the question, [a subsequence of the sequence of] balls form a natural Folner sequence in any group of subexponential growth. Now, as pointed out by others, balls (w.r.t. to some finite generating set) are fairly "ugly". This can be made precise if one considers the concept of an optimal Folner set:

Let $$I(n)= \displaystyle \inf_{|A| \leq n} \dfrac{|\partial A|}{|A|}$$ (the $$\inf$$ runs over all sets $$A$$ of size $$\leq n$$) be the isoperimetric profile. Then a set $$F$$ is optimal if $$I(|F|)=\dfrac{|\partial F|}{|F|}$$. In words: if a set $$E$$ is not larger [cardinality-wise] than $$F$$, then it's isoperimetric ratio $$\dfrac{|\partial E|}{|E|}$$, does not beat the isoperimetric ratio of $$F$$.

One can check (using the Loomis-Whitney inequality) that optimal Folner sets in $$\mathbb{Z}^d$$ (w.r.t. the usual generating set) are [hyper]cubes (or that they tend to have a rectangular form). This is an unambiguous way of saying that balls are "clumsy" Folner sets. By comparison optimal sets are not "weird" at all (since they must be extremely well-chosen).

For more on strangeness, see the side notes below.

Explicit examples: Next, given a group of exponential growth, it's an open question whether any subsequence of the sequence of balls is Folner. I gave a partial answer which shows this is not case when the group [together with the choice of generating set] has pinched exponential growth. This includes many wreath products, solvable Baumslag-Solitar groups and some extensions of $$\mathbb{Z}^d$$ by $$\mathbb{Z}$$ (see link for details).

These groups can all be written as semi-direct products. If $$G$$ and $$H$$ are amenable, then one can show that $$G \rtimes H$$ is amenable and that Folner sets are of the Form $$E_n \times F_n$$ (where $$E_n$$ [resp. $$F_n$$] is a Folner sequence of $$G$$ [resp. $$H$$]). In that sense, the Folner sets that we come across (lazily, in the sense that they are produced by a general proof) in such groups are "rectangular".

Hence the groups mentioned above [solvable Baumslag-Solitar, some metabelian groups, groups whose growth series is rational and do not have a two poles at the radius of convergence (which includes many wreath products and $$ax+b$$-groups)] are a direct answer to your second question (for some generating set). One knows that balls (w.r.t. generating sets) are not Folner but some "rectangular" set is (just to be precise: there could be groups with a single pole which are not semi-direct products or extensions of amenable groups; for these groups [if any are known] there are no "rectangular" sets).

For non-split extensions a description of the Folner sets was given over there by Ycor. Note one could adapt the meaning of "rectangular" for non-split extensions: by taking a preimage of the Folner set of the quotient times some Folner set of the subgroup.

So now one might think that "rectangular" (and no longer balls) sets are favourites. But then there are also simple groups of intermediate growth see this question. And (if not for such groups, then for other simple groups of subexponential growth) I guess that balls are the only candidates one has.

Basically, I think the problem has more to do with how we construct amenable groups. We always use the four properties of amenability (extension, subgroup, quotient and direct limit). So ones start with growth as basic criterium, and uses those four properties (there are possibly many ways to do it). This will give you the known Folner sets for a given group. As a silly example you could say that natural Folner sets in $$\mathbb{Z}^3$$ are cylinders (balls in $$\mathbb{Z}^2$$ times balls in $$\mathbb{Z}$$).

Side Note 1: it's an long-standing open question to prove what are such sets in the (continuous) Heisenberg group (although the conjectured shape is well-described). That was my motivation for this question.

Side Note 2: As pointed out by Ycor, given a Folner sequence $$F_n$$ you can make it "as weird as you want" by considering an arbitrary sequence of finite sets $$E_n$$ with $$\dfrac{|E_n|}{|F_n|} \to 0$$. One the advantage of considering optimal Folner sequences would be to avoid such set-ups (the obvious disadvantage, is that there are almost no groups where optimal sets are known). A further note is that adding such a set $$E_n$$ has no influence on the invariant measure one obtains (for a fixed ultrafilter). Note that translating the sets can have an effect on the limit measure.

Side Note 3: Here is another aspect of the "strangeness" of Folner sets. Consider the sequence $$P_n = [2^n,2^{n+1}]$$, $$M_n = [-2^{n+1},-2^n]$$, as well as $$A_n = (-1)^n \cdot P_n$$ of sets in $$\mathbb{Z}$$. Then consider the function $$f(n) = \mathrm{sign}(n)$$. The invariant mean one gets from $$P_n$$ on $$f$$ is 1 (whatever the ultrafilter you choose), the one you get with $$M_n$$ is $$-1$$ (again, whatever the ultrafilter) and finally the one you get with $$A_n$$ depends on the ultafilter you choose. And you could construct for any real number in $$[-1,1]$$ a sequence $$R_n$$ which converges to that number (indenpendently of the ultrafilter). It's not too hard to construct a sequence which can, depending on the ultrafilter, converge to any rational number in $$[-1,1]$$.