aa r X i v : . [ m a t h . C A ] A p r ON THE SIZE OF NIKODYM SETS IN FINITE FIELDS
LIANGPAN LI
Abstract.
Let F q denote a finite field of q elements. Define a set B ⊂ F nq to be Nikodymif for each x ∈ B c , there exists a line L such that L ∩ B c = { x } . The main purpose of thisnote is to show that the size of every Nikodym set is at least C n · q n , where C n dependsonly on n . Introduction
The finite field Kakeya problem, posed by Wolff in his influential survey [13], asks forthe smallest subset of F nq that contains a line in each direction, where F q denotes a finitefield of q elements. A subset containing a line in each direction is called a Kakeya set. Inanalogy with the Euclidean Kakeya problem, Wolff conjectured that ♯K ≥ C n q n holds forany Kakeya set K ⊂ F nq , where C n depends only on the dimension n . For n = 2 Wolffimmediately proved the bound ♯K ≥ q ( q + 1) /
2, and it is best possible when q is even.To the author’s knowledge, Blokhuis and Mazzocca [1] studied the finite field Kakeyaproblem in two dimensions and proved the sharp bound ♯K ≥ q ( q + 1) / q − / q is odd, as conjectured by Faber in [6]. The higher dimensional finite field Kakeyaproblem has been extensively investigated in [2, 8, 10, 12, 13] such as proving the bound ♯K ≥ C n q ( n +2) / or ♯K ≥ C n q (4 n +3) / . Recently, using the polynomial method in algebraicextremal combinatorics, Dvir [5] completely confirmed this conjecture by proving ♯K ≥ (cid:18) n + q − n (cid:19) . On the other hand, Nikodym [9] proved that there exists a null set in the unit square suchthat every point of the complement is “linearly accessible through the set”, which means itlies on a line that is otherwise included in the set. Falconer [7] extended Nikodym’s resultto higher dimensions proving there exists a set N ⊂ R n of zero Lebesgue measure suchthat for each x ∈ N c , there is a hyperplane P satisfying P ∩ N c = { x } . In the Euclideanspaces Nikodym sets are closely related to Kakeya sets through Carbery’s transformation[4, 11].Motivated by the above works, we shall define a set B in F nq to be Nikodym if for each x ∈ B c there exists a line L such that L ∩ B c = { x } . The main purpose of this note is toprove the lower bound ♯B ≥ (cid:18) n + q − n (cid:19) . Slightly different with the two dimensional finite field Kakeya problem, this bound is notbest possible in two dimensions. General dimensions
Theorem 2.1.
Any Nikodym set B ⊂ F nq satisfies | B | ≥ (cid:18) n + q − n (cid:19) , where F q denotes a finite field of q elements.Proof. We argue by contradiction and suppose | B | < (cid:18) n + q − n (cid:19) . A basic result in combinatorics [3] says that the number of monomials in F [ x , . . . , x n ] ofdegree at most d is (cid:18) n + dn (cid:19) , hence there exists a nonzero polynomial g ∈ F [ x , . . . , x n ] of degree at most q − g ( y ) = 0 ( ∀ y ∈ B ) . For each x ∈ B c , there exists a line L such that L ∩ B c = { x } . The restriction of g to this line is a univariate polynomial of degree at most q −
2, andsince it has at least q − L . Considering x belongsto this line, it follows that g ( x ) = 0 . This would mean g is the zero polynomial, a contradiction. (cid:3) Two dimensions
Theorem 3.1.
Any Nikodym set B ⊂ F q satisfies ♯B ≥ q O ( q ) ( q → ∞ ) , where F q denotes a finite field of q elements.Proof. Write s = ⌊ q ⌋ . First, assume that ♯B c ≤ s ( q −
1) + 2 q, then(3.1) ♯B ≥ q − s ( q − − q ≥ q − q q − − q = 2 q − q . Else suppose that ♯B c ≥ s ( q −
1) + 2 q. Since B is a Nikodym set, for each x ∈ B c there exists a line L x such that L x ∩ B c = { x } . IKODYM SETS 3
Obviously, all of these lines are distinct from each other. Noting that there are in total q +1directions in F q , we partition { L x } x ∈ B c into classes { G i } qi =0 according to their directions.Without loss of generality we may assume that ♯G ≥ ♯G ≥ ♯G ≥ · · · ≥ ♯G q . Thus q + q + ♯G · ( q − ≥ q X i =0 ♯G i = ♯B c ≥ s ( q −
1) + 2 q, from which yields ♯G ≥ s. Choose s parallel lines { X l } sl =1 from G , s parallel lines { Y m } sm =1 from G and s parallellines { Z n } sn =1 from G , then it follows that ♯B ≥ s X l =1 ( ♯X l −
1) + s X m =1 ( ♯Y m − − s ) + s X n =1 ( ♯Z n − − s )= s ( q −
1) + s ( q − − s ) + s ( q − − s ) = 3 s ( q − − s ) ≥ q −
23 ( q − − q q − q . (3.2)Combining (3.1) and (3.2) yields the desired result. (cid:3) Question : How small can the Nikodym sets really be in two dimensions?4.
Acknowledgements
The author thanks Yaokun Wu for clarifying the proof. He also thanks Aart Blokhuisand Qing Xiang for kindly pointing out the recent progresses on the finite field Kakeyaproblem to the author.
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LIANGPAN LI [12] T. Tao, A new bound for finite field Besicovitch sets in four dimensions, Pacific J. Math. 222 (2005)337–363.[13] T. Wolff, Recent work connectecd with the Kakeya problem, Prospects in Mathematics (Princeton,NJ, 1996), Amer. Math. Soc. (1999) 129–162.
Department of Mathematics, Shanghai Jiaotong University, Shanghai 200240, People’sRepublic of China
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