Caps with free pairs of points

Abstract

We say that two points x, y of a cap C form a free pair of points if any plane containing x and y intersects C in at most three points. For given N and q, we denote by m + 2 (N, q) the maximum number of points in a cap of PG(N, q) that contains at least one free pair of points. It is straightforward to prove that m + 2 (N, q) ≤ (q N−1 + 2q − 3)/(q − 1), and it is known that this bound is sharp for q = 2 and all N . We use geometric constructions to prove that this bound is sharp for all q when N ≤ 4. We briefly survey the motivation for constructions of caps with free pairs of points which comes from the area of statistical experimental design. (2000): 51E22, 62K15 Mathematics Subject Classification

Large caps with free pairs in dimensions five and six

“…, 4, then we can make {P 3 , P 5 } a free pair in T 0 and in the entire 42-cap in PG (5,3). As mentioned in the introduction, this construction gives the same cap as was found by computer search in [8].…”

“…In [8] the bound in Theorem 2.3 is shown to be sharp for N ≤ 4. In this section we show that the bound is attained asymptotically in projective dimensions 5 and 6.…”

“…Throughout we assume that q is a prime power, q > 2. The problem which we study is solved for q = 2 [8,Theorem 2.2]. An n-cap C ⊂ PG(N, q) is a set of n points no three of which are collinear.…”

“…One way to find large caps with free pairs is to use geometric or other arguments to construct a cap while ensuring that one pair remains free; this is the method employed in [8]. Another approach is to take known large caps (not necessarily possessing a free pair) and to delete points from the cap until a pair becomes free.…”

“…This is the best known lower bound for odd q except in the case of q = 3. In [8] we mentioned that a 42cap in PG(5, 3) with a free pair was found via a computer search. Hence, by Theorem 2.3: The construction of this 42-cap was not presented in [8], and we present all details of the construction in this section.…”