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](https://en.wikipedia.org/wiki/Amenable_group). 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](https://arxiv.org/abs/math/0601432). 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.



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


