If $(F_n)_n$ is a Følner sequence satisfying Tempelman's condition, is $(F_n^{-1}F_n)_n$ also Følner?

Question

Let $G$ be a countable group. A Følner sequence is a sequence of finite subsets $(F_n)_n$ such that

$$\lim_{n\to\infty} \frac{|KF_n \mathbin\triangle F_n|}{|F_n|} = 0$$

for each fixed finite subset $K \subset G$. Such a sequence exists if and only if the group is amenable. The sequence is said to satisfy Tempelman's condition if there is a constant $C \geq 1$ such that $|F_n^{-1}F_n| \leq C|F_n|$ for every $n$. Let us say $(F_n)_n$ is a Tempelman sequence if it is a Følner sequence satisfying Tempelman's condition. Not every countable amenable group admits a Tempelman sequence; see Hochman - Averaging sequences and abelian rank in amenable groups. Every countable nilpotent group has a Tempelman sequence.

My question is:

If $(F_n)_n$ is a Tempelman sequence, is the sequence $(F_n^{-1}F_n)_n$ necessarily a Følner sequence? Or: can a Tempelman sequence $(F_n)_n$ always be found such that $(F_n^{-1}F_n)_n$ is Følner?

This is obviously true in $\mathbb{Z}^d$, where if $F_n = [0,n)^d \subset \mathbb{Z}^d$ then $F_n^{-1}F_n = (-n,n)^d$. It seems like it should be true in general; if $K$ is fixed and $n$ is large then intuitively $|KF^{-1}_nF_n \mathbin\triangle F_n^{-1}F_n| \lesssim C|KF_n \mathbin\triangle F_n|$ should follow from $|F_n^{-1}F_n| \leq C|F_n|$. Yet, I cannot find a proof.

I can see how it is done under a modified condition: suppose for every $n$ there is a subset $C_n \subset G$ such that $|C_n| \leq C$ and $F_n^{-1}F_n \subset F^{-1}_nC_n$. If it makes things easier, we can assume that $F_n^{-1} = F_n$ (in which case $(F_n)_n$ is both left Følner and right Følner).

Proof: assume that $e\in K$ and $F = F_n = F_n^{-1}$ is large. Any element $g \in KFF \setminus FF$ decomposed as $g = kf_1f_2$ in any way necessarily has $kf_1 \notin F$. Then, $FF \subset FC$ implies each $g \in KFF\setminus FF$ may be decomposed as $g = kfc$. There is therefore a map $g \mapsto (kf,c)$ which embeds $KFF \setminus FF$ in $(KF \setminus F) \times C$; the result follows.

Answer 1

The answer to the first question is negative, even for the integers $G = \mathbb{Z}$. The point is that Følner sequences $F_n$ are stable under modification by small sets, but the difference set $F_n^{-1} F_n$ can be significantly affected by such a modification. For a concrete example, take

$$ F_n = \{0,\dots,n^2\} \cup \{ 2n^2 + 2nj: 0 \leq j < n \} \cup \{ -2n^2 - 2i: 0 \leq i < n\}.$$ This is a modification of $\{0,\dots,n^2\}$ by $O(n)$ elements and is thus a Følner sequence; it is also a dense subset of $\{-4n^2,\dots,4n^2\}$ and thus obeys the Tempelman condition. On the other hand, $F_n^{-1} F_n$ has cardinality $\asymp n^2$ and contains the even elements of $\{4n^2,\dots,6n^2-2\}$ but not the odd elements, so will not be Følner. (This counterexample is not symmetric, but it is not difficult to modify the construction to give a symmetric counterexample; we leave this as an exercise to the interested reader.)

On the other hand, given a Følner sequence $F_n$ obeying the Tempelman condition, one can use a standard pigeonholing argument to find a sequence of subsets $F'_n$ of $F_n$ with $|F_n \backslash F'_n| = o(|F_n|)$ (so that $F'_n$ is also a Følner sequence and will also obey the Tempelman condition) such that ${F'}_n^{-1} {F'}_n$ is Følner; in the above example, this would amount to "trimming" the unwanted portions $\{ 2n^2 + 2nj: 0 \leq j < n \} \cup \{ -2n^2 - 2i: 0 \leq i < n\}$ of $F_n$ and only keeping the "core" portion $\{0,\dots,n^2\}$. A sketch of the construction is as follows. Let $K \subset G$ be a finite symmetric neighborhood of the identity and let $k \geq 2$ be a natural number. For $n$ large enough one can use the Følner condition to find $F''_n \subset F_n$ with $|F_n \backslash F''_n| \leq \frac{1}{k} |F_n|$ and $F''_n \cdot K^k \subset F_n$. The sets $K^i \cdot (F''_n)^{-1} \cdot F''_n \cdot K^{i}$ for $i=0,\dots,k$ are then increasing with cardinality at most $|F_n^{-1} \cdot F_n| \leq C |F_n|$, so by the pigeonhole principle one can find $0 \leq i < k$ such that $$ |K^{i+1} \cdot (F''_n)^{-1} \cdot F''_n \cdot K^{i+1}| \leq |K^{i} \cdot (F''_n)^{-1} \cdot F''_n \cdot K^{i}| + \frac{C}{k} |F_n|.$$ If we then set $F'_n := F''_n \cdot K^i$ then $F'_n$ is a subset of $F_n$ with $|F_n \backslash F'_n| \leq \frac{1}{k} |F_n|$ and $$ |(K \cdot (F'_n)^{-1} \cdot F'_n) \backslash ((F'_n)^{-1} \cdot F'_n)| \leq \frac{C}{k} |F_n|.$$ From this it is an easy matter to verify that $(F'_n)^{-1} \cdot F'_n$ will obey the Følner condition (letting $K, k$ grow sufficiently slowly with $n$).