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Consider a group $G$; let call $A\subset G$ a weak Besicovitch subset whenever every element of $G$ can be written under the form $gh^{-1}$, where $g,h\in A$.

General question: how large must a weak Besicovitch set be?

The motivation for this question and this definition was initially in the Kakeya conjecture; a Besicovitch set of $\mathbb{R}^n$ is a subset $A$ containing unit segments in all directions, and the conjecture is that such a set should have Hausdorff dimension $n$. This question have seen recent important developments when $\mathbb{R}^n$ is replaced by a vector space (or module) over certain finite fields (or rings); then one replaces unit segments with lines. It is easy to see that the conjecture on $\mathbb{R}^n$ is equivalent to the statement where $A$ is a subset containing arbitrarily long segments on all direction. Such a subset in particular is weak Besicovitch, hence the choice of words.

Edited question 1: what is the least Hausdorff dimension of a weak Besicovitch set of $\mathbb{R}^n$?

There is an affine feel to this question: we seek for small sets of point which pairwise define all possible vectors. Comments by Anton Petrunin on the original wording show that for $n=1$, the Hausdorff dimension of a weak Besicovitch set can be as small as $\log(2)/\log(3)$, but its box-counting dimension cannot be smaller than $1/2$.

We can also make sense of the question for any group, not just modules. When $G$ is finite, denote by $K(G)$ the smallest cardinal of a weak Besicovitch set A. A simple counting argument shows that $|K(G)|(|K(G)|-1) \ge |G| - 1$.

Question 2. Does there exist groups $(G_n)$ with small weak Besicovitch sets, i.e. $|G_n|\to\infty$ and $K(G_n) \lesssim \sqrt{|G_n|}$? does there exist groups with large Besicovitch sets, i.e. $|G_n|\to\infty$ and $K(G_n) \gtrsim |G_n|$?

(Here $a_n \lesssim b_n$ means that there exist a constant $C$ such that for all $n$, $a_n\le C b_n$.)

Even the case of the cyclic groups seems not completely obvious.

Question 3. Find the value of $K(\mathbb{Z}/n\mathbb{Z})$, or at least its asymptotic behavior.

A quick search seems to indicate that $K(\mathbb{Z}/21\mathbb{Z})=6$, above the trivial bound given above (which is attained for $n\le 13$ if I am not mistaken).

One can also consider one's favorite non-commutative Lie group, and ask the question of the Hausdorff dimension of weak Besicovitch set. Maybe restricting to closed sets makes sense, but maybe measurability plays an interesting role if we do not make this restriction.

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    $\begingroup$ Let $B$ be the set of real numbers that can be written using $0$ and $2$ in the base-3 system. Note that $B$ is a Besicovitch set in $\mathbb{R}$ and its Hausdorff dimension $<1$. $\endgroup$
    – Anton Petrunin
    Sep 3, 2022 at 19:58
  • $\begingroup$ @AntonPetrunin right, the connection to Kakeya conjecture was an illusion. Do you know whether the dimension can be brought down further? $\endgroup$
    – Benoît Kloeckner
    Sep 3, 2022 at 20:34
  • $\begingroup$ No, but the dimension cannot be less than ½ --- the real line is a projection of $B\times B$. $\endgroup$
    – Anton Petrunin
    Sep 3, 2022 at 20:42
  • $\begingroup$ @AntonPetrunin: if by $B$ you mean a weak Besicovitch set whose dimension we want to bound below, it seems to me that the Hausdorff dimension of $B\times B$ could be higher than twice the Hausdorff dimension of $B$. The box-counting dimension can indeed not be lower than $1/2$ since $1\le \dim_H(B\times B)\le \dim_H(B)+\dim_B(B) \le 2\dim_B(B)$. $\endgroup$
    – Benoît Kloeckner
    Sep 3, 2022 at 21:41
  • 2
    $\begingroup$ The case of cyclic groups is well-studied as 'planar difference sets' - these correspond to those $n$ such that the trivial lower bound you mention is achieved. It is known that these exist whenever $n=q^2+q+1$ with $q$ a prime power, and conjectured that these are the only such $n$. $\endgroup$
    – Thomas Bloom
    Sep 3, 2022 at 22:32

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