Let $G$ be an amenable (countable, discrete) group and let $F_1,F_2,...,F_n,...$ and $G_1,G_2,...,G_n,...$ be two Følner sequences. Is the product sequence (i.e. the sequence $(H_n)$ where $H_n$ is all elements of the form $f_ng_n$ for $f\in F_n, g\in G_n$) necessarily also a Følner sequence? If not, is this at least true for some nice class of groups (say, locally virtually nilpotent groups, or solvable groups or when the $F_i=G_i$ for all $i$?

  • $\begingroup$ A left Folner sequence. Though I would be very interested even if it only held for two sided Folner sequences. $\endgroup$
    – Josh F
    Jan 4, 2020 at 0:47
  • $\begingroup$ There are at least 3 possible defintions of left Folner seqiences. What definition do you use here? $\endgroup$
    – user6976
    Jan 4, 2020 at 1:47
  • $\begingroup$ A sequence of finite subsets of subsets of $G$, $F_n$, so that $F_{i-1}$ is a subset of $F_i$, the union of all $F_n$ is $G$ and, for each $g\in G$, the limit as $n\to \infty$ of $|gF_n-F_n|/|F_n|$ goes to 0. (Where - represents the symmetric difference) $\endgroup$
    – Josh F
    Jan 4, 2020 at 7:06
  • 1
    $\begingroup$ Ok. Do you know any non-virually abelian amenable group where the product of two Folner sequences is always Folner? $\endgroup$
    – user6976
    Jan 4, 2020 at 11:59
  • $\begingroup$ @Mark Sapir - What are the other two definitions you mean? - just out of curiosity $\endgroup$
    – R W
    Jan 5, 2020 at 2:14

1 Answer 1


some details added January 7th

It seems that the answer is "no" even for virtually abelian groups, and even if $G_i$ is chosen by your arch enemy. The sequence $(F_i)_i$ can also be chosen rather freely in the proof. This answers some subquestions of the question, but not all of them. In particular do not know what happens if $F_i = G_i$ is required or if there are strong constraints on how fast the sets become invariant.

Suppose $G$ is a countable amenable group where some conjugacy class is infinite. Then for every Følner sequence $(G_i)_i$ there exists a Følner sequence $(F_i)_i$ such that $(F_iG_i)_i$ is not a Følner sequence.

Proof. Let $h$ have infinite conjugacy class, and $(G_i)_i$ be a Følner sequence. Since $h$ has infinite conjugacy class, for each $i \in \mathbb{N}$ there exists infinitely many $f_i \in G$ such that $f_i^{-1}hf_i \notin G_iG_i^{-1}$. Then $hf_iG_i \cap f_iG_i = \emptyset$, since otherwise $hf_ig_1 = f_ig_2$ for some $g_1, g_2 \in G_i$ and thus $f_i^{-1}hf_i = g_2g_1^{-1} \in G_iG_i^{-1}$. Since there are infinitely many ways to choose $f_i$, we can also ensure $f_iG_i \cap G_i = \emptyset$.

Let now $(F_i')_i$ be any Følner sequence for $G$. For each $i \in \mathbb{N}$ pick maximal $\alpha(i) \leq i$ so that, writing $\chi_A$ for the characteristic function of a set $A \subset G$, and $| \cdot |$ for the $\ell^1$-norm, we have $$ |\chi_{F_{\alpha(i)}' G_i} - \chi_{G_i}| < 1/(\alpha(i) + 1) |G_i|. $$ Since $(G_i)_i$ is Følner, $\alpha(i)$ tends to infinity with $i$, so $(F_{\alpha(i)}')_i$ is a Følner sequence (using the definition of the asker).

Define $F_i = F_{\alpha(i)}' \cup \{f_i\}$. Obviously $(F_i)_i$ is a Følner sequence since the new element can contribute at most $2$ to the symmetric differences $|\chi_{gF_i} - \chi_{F_i}|$. With the shorthand $o(|G_i|)$ for $\ell^1$ functions $x$ with $|x/|G_i|| \rightarrow 0$ with $i$, and also for its usual meaning for reals, we have (by the properties of $f_i$ chosen in the first paragraph) $$ \chi_{F_iG_i} = \chi_{f_iG_i} + \chi_{F_{\alpha(i)}' G_i} - \chi_{f_iG_i \cap F_{\alpha(i)}' G_i} = \chi_{f_i G_i} + \chi_{G_i} + o(|G_i|) $$ By $f_iG_i \cap G_i = \emptyset$, we then have $|\chi_{F_iG_i}| = 2|G_i| + o(|G_i|)$.

We similarly have $$ \chi_{hF_iG_i} = \chi_{h f_i G_i} + \chi_{h G_i} + o(|G_i|). $$ Using these equalities, the fact $G_i$ is a Følner sequence, and the inverse triangle inequality, we get $$ |\chi_{hF_iG_i} - \chi_{F_iG_i}| = |\chi_{h f_i G_i} + \chi_{hG_i} + o(|G_i|) - \chi_{f_i G_i} - \chi_{G_i} - o(|G_i|)| \geq |\chi_{h f_i G_i} - \chi_{f_i G_i}| - |\chi_{h G_i} - \chi_{G_i}| - o(|G_i|) = 2|G_i| - o(|G_i|). $$

This gives $\frac{|\chi_{hF_iG_i} - \chi_{F_iG_i}|}{|\chi_{F_iG_i}|} = \frac{2|G_i| - o(|G_i|)}{2|G_i| + o(|G_i|)} \rightarrow 1$, and thus $(F_iG_i)_i$ is not a Følner sequence.

The infinite dihedral group admits Følner sequences $(F_i)_i$ and $(G_i)_i$ such that $(F_iG_i)_i$ is not a Følner sequence.

Proof. Every element of finite order has an infinite conjugacy class.


Not a solution but too long for a comment.

I think it's not necessarily Følner, as long as you can find a Følner sequence $G_i$ such that for some fixed $\epsilon > 0$ and $h \in G$, for all $i$ there exist arbitrarily large $g$ such that $|hgG_i \setminus gG_i|/|gG_i| > \epsilon$. I imagine can happen in the non-abelian case, and it happens on the Heisenberg group.

Here's the idea of the construction. Pick $G_i$ such a Følner sequence. For each i construct $F_i$ like so: Pick a tiny set $F_i'$ around the identity, containing the identity, in such a way that $F'_iG_i = G_i \cup A_i$ where $|A_i|$ is much smaller than $G_i$ (using that $G_i$ is Følner) and so that $F_i'$ itself is a Følner sequence.

Now, recall our assumption was that $gG_i$ is not a very left Følner-ish set for infinitely many g and a fixed translation $h$. So pick a couple of such $g$ and add them to $F_i'$ to get $F_i$. As long as we pick much fewer than the cardinality of $F'_i$, $F_i$ will also be a Følner sequence.

But if we pick at least $\ell$ such $g$, then if we pick them very separated from each other, $F_iG_i$ will actually consist of $F'_iG_i \approx G_i$ plus some disjoint $gG_i$'s. If you pick a random element of this $F_iG_i$, it's going to be one of the $gG_i$ with probability roughly $(\ell-1)/\ell$ once $i$ is large (and $\ell$ can grow to infinity with $i$), and then with probability at least $\epsilon > 0$ the $h$-translate gets outside $gG_i$ (and by picking $g$ separated enough, we can make sure they don't hit any of the other $g'G_i$ either). So we don't even get a left Følner sequence for $h$-translations.

On the Heisenberg group, $G_i$ can be any Følner sequence: Consider Heisenberg with generators $x$, $y$ and $z = [x, y]$, and any Følner sequence $G_i$. Consider $g = x^j$ and $h = y$. Since $h g G_i = g h z^j G_i$ we have $h g G_i \cap g G_i \neq \emptyset \implies h z^j G_i \cap G_i \neq \emptyset$ which means $hz^j \in G_iG_i^{-1}$, which happens only for finitely many $j$. So for the Heisenberg group, the answer is that the product is not necessarily Følner.

  • $\begingroup$ Thank you! I'd still be interested in special cases (is it even true when the group is abelian, for example) but you've definitely answered much of what I'm interested in. $\endgroup$
    – Josh F
    Jan 7, 2020 at 18:59
  • $\begingroup$ Surely true for abelian. Well, I was sure about virtually abelian before doing the algebra. $\endgroup$
    – Ville Salo
    Jan 7, 2020 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.