Large gaps in Singer planar difference sets?

By a classical result of Singer (1938), for a prime number $p$ the cyclic group $C_n$ of order $n=1+p+p^2$ contains a subset $D$ of cardinality $|D|=1+p$ such that $DD^{-1}=C_n$. Such set $D$ is called a difference set. The cardinality restrictions imply that each non-unit element $x\in C_n$ can be uniquely written as the difference $x=a-b$ for $a,b\in D$.

Identify $C_n$ with a subgroup of the unit circle $\mathbb T=\{z\in C_n:|z|=1\}$ on the complex plane. Let $D$ be a subset of $C_n\subset \mathbb T$. A 2-element set $\{a,b\}\subset D$ is called a gap in $D$ if there exists an arc $A\subset \mathbb T$ with end-points $a,b$ such that $A\cap D=\{a,b\}$. The real number $|a-b|$ is called the diameter of the gap $\{a,b\}$.

Problem 1. What is the largest possible diameter of a gap in a difference set in a cyclic group $C_n$?

More precisely:

Problem 2. Is there $\epsilon>0$ such that for every prime number $p$ the cyclic group $C_n\subset\mathbb T$ of order $n=1+p+p^2$ contains a difference set $D$ with a gap of diameter $>\varepsilon$?

1 Answer

No you cannot have very big gaps in a perfect difference sets -- the elements in a perfect difference set will be pretty uniform. Suppose $A$ is a perfect difference set $\mod n$ with $n=1+p+p^2$ as above. Put $${\hat A}(k) = \sum_{a \in A} e(ak/n).$$ Since $A$ is a perfect difference set, for any $k\neq 0$ one has $$|{\hat A}(k)|^2 = |A| + \sum_{a \neq b \in A} e(k(a-b)/n) = |A|-1.$$ In other words, all the non-trivial Fourier coefficients of $A$ are small.

Now suppose $I$ is an arc $\mod n$, and let us count the number of elements of $A$ on $I$. By orthogonality we can write this as $$\frac{1}{n} \sum_{k \mod n} {\hat A}(k) {\hat I}(-k),$$ and using our knowledge of the Fourier coefficients above, this is $$\frac{|I||A|}{n} +O\Big( \frac{\sqrt{|A|}}{n} \sum_{k\neq 0} |{\hat I}(-k)|\Big).$$ It is a simple calculation to bound the $L^1$ norm of the Fourier coefficients of an interval -- it is $\ll n \log n$. So from the above analysis we get $$\frac{|I||A|}{n} + O(\sqrt{|A|} \log n).$$

Since $|A| = p+1$ is about $\sqrt{n}$, this means that any arc with size a bit more than $n^{\frac 34}$ will contain many points of $A$. This settles Problem 2; I don't really have a guess as to the right size of the maximum gap.

• Thank you very much for the answer. In fact, the above question was motivated by another question: mathoverflow.net/questions/261920/… – Taras Banakh Feb 13 '17 at 0:18
• I think the argument I gave here basically shows that the answer to your Problem 2 in that question is negative. If I have some time, I'll post an answer there (or maybe this is enough for you). – Lucia Feb 13 '17 at 2:11