# Is a product of Følner sets Følner?

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 $$i$$?

• A left Folner sequence. Though I would be very interested even if it only held for two sided Folner sequences. – Josh F Jan 4 '20 at 0:47
• There are at least 3 possible defintions of left Folner seqiences. What definition do you use here? – user6976 Jan 4 '20 at 1:47
• 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) – Josh F Jan 4 '20 at 7:06
• Ok. Do you know any non-virually abelian amenable group where the product of two Folner sequences is always Folner? – user6976 Jan 4 '20 at 11:59
• @Mark Sapir - What are the other two definitions you mean? - just out of curiosity – R W Jan 5 '20 at 2:14

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.

original

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.

• 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. – Josh F Jan 7 '20 at 18:59
• Surely true for abelian. Well, I was sure about virtually abelian before doing the algebra. – Ville Salo Jan 7 '20 at 19:16