On the arithmetic Kakeya conjecture of Katz and Tao

The arithmetic Kakeya conjecture, formulated by Katz and Tao (Math Res Lett 6(5–6):625–630, 1999), is a statement about addition of finite sets. It is known to imply a form of the Kakeya conjecture, namely that the upper Minkowski dimension of a Besicovitch set in \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbf {R}}^n$$\end{document}Rn is n. In this note we discuss this conjecture, giving a number of equivalent forms of it. We show that a natural finite field variant of it does hold. We also give some lower bounds.


Introduction and statement of results
The arithmetic Kakeya conjecture, sometimes known as the sums-differences conjecture, was formulated by Katz and Tao around fifteen years ago. It is a purely additive-combinatorial Ben Green is supported by a Simons Investigator Grant, and is very grateful to the Simons Foundation for their support. Imre Ruzsa is supported by ERC-AdG. 321104 [13][14][15]., Budapest 1053, Hungary statement which, if true, would have a deep geometric consequence-that the Minkowski dimension of Besicovitch sets in R n is n. This is the celebrated Kakeya conjecture, discussed at length in many places: for an introduction see [21].
The arithmetic Kakeya conjecture is mentioned explicitly 1 in [20]. One of the main aims of this paper is to give a number of equivalent forms of the conjecture. Here is probably the simplest formulation. It is not the original one of Katz and Tao, which is Conjecture 1.3 below. This conjecture was raised by the second author as [17,Conjecture 4.2], but no links to the Kakeya problem were mentioned there.
We turn now to arguably the most natural of our formulations, concerning the entropy of random variables. As usual, the entropy H of a random variable X with finite range is defined by where x ranges over all values taken by X.

Conjecture 1.2
For any ε > 0 there are 2 r 1 , . . . , r k ∈ Q, none equal to −1, such that for any two real-valued random variables X and Y taking only finitely many values we have Next we give the original form of the conjecture discussed by Katz and Tao. Let A ⊂ Z×Z be a finite set. For rational r we write π r (A) := {x + r y : (x, y) ∈ A}. We also write π ∞ (A) := {y : (x, y) ∈ A}. Conjecture 1. 3 Let ε > 0 be arbitrary. Then there are r 1 , . . . , r k ∈ Q ∪ {∞}, none equal to −1, such that #π −1 (A) sup i #π r i (A) 1+ε for all finite sets A ⊂ Z × Z.
Our fourth conjecture has not, so far as we are aware, appeared explicitly in the literature before. It is in fact a whole family of conjectures, one for each natural number n; however, we will later show that all of these are equivalent. 1 In the earlier paper [13, p. 234] of Katz and Tao, the authors only go so far as to suggest that it is "not too outrageous tentatively to conjecture" this statement. In fact, the conjecture made in [20] is over fields of "sufficiently large characteristic" (or characteristic zero) whereas this paper provides evidence that it is natural, and simpler, to work only in characteristic zero. We believe that in any case the statements are equivalent but have not bothered to check this carefully. 2 It is often convenient to "work projectively" and allow the r i to take values in Q ∪ {∞}, where we define X + ∞Y = Y. The two versions of Conjecture 1.2 are equivalent to one another, as may easily be seen by applying a projective transformation such as X = (a + 1)X, Y = aX + Y which preserves X − Y but moves other rational combinations around. Conjecture 1.4 (n) Let k be a positive integer. If p is a prime, let f k,n ( p) denote the size of the smallest set containing, for every d ∈ F n p \{0}, a k-term progression with common difference d. Then Remarks Note that f p,n ( p) is the size of the smallest Besicovitch set in F n p , that is to say a set containing a full line in every direction. Since f p,n ( p) f k,n ( p) whenever p k, Conjecture 1.4 (n) trivially implies that i.e. any Besicovitch set in F n p has size p n−o p→∞ (1) . This is known to be true, a celebrated result of Dvir [5]. However, the only known arguments use the "polynomial method" (see, for example, [11,22] for modern introductions). This very strongly hints that any proof of Conjecture 1.4 (and hence, by our main theorem, of the other conjectures) would have to use some form of the polynomial method.
Our fifth and final conjecture is included mainly for historical interest, as it relates very closely to a question asked by Erdős and Selfridge in the 1970s, well before the current wave of interest in the Kakeya problem and related matters. Conjecture 1.5 Fix a positive integer k. Then, uniformly for all positive integers N , all finite sets p 1 < · · · < p N of primes and all intervals I ⊂ N of length kp N , we have where γ k → 0 as k → ∞.

Remark
Erdős and Selfridge [8, §6] in fact asked whether or not one can take γ k = 0. The second-named author [16] showed that the answer is negative, and in fact we must have γ k 1 k . We note that Proposition 4.1 and Theorem 1.7 combine to give the much better bound γ k 1 log log k .
As previously stated, our main result is the equivalence of the five conjectures stated above. Let us make some further remarks.
(1) Once Theorem 1.6 is proven, it seems natural to use the term "arithmetic Kakeya conjecture" to refer to any one of the five conjectures. (2) It is known that Conjecture 1.3 (and hence all the other conjectures) implies that the upper Minkowski dimension of any Besicovitch set 3 in R n is n, a statement often referred to as the Kakeya conjecture. This follows by a straightforward generalisation of the "slicing" argument of Bourgain [4]: a sketch of this may be found in [21]. However, Bourgain [2,3] observed that, in the notation of Conjecture 1.1, if the statement is true for all η > 0 then the Kakeya conjecture follows. Since F k (N ) is a nondecreasing function of k, (1.1) is immediately implied by Conjecture 1.1, whilst an implication in the reverse direction seems very unlikely without resolving both conjectures. In this sense, the arithmetic Kakeya conjecture should be considered a strictly harder problem than the Kakeya conjecture. (3) The equivalence of Conjectures 1.2 and 1.3 was proven by the second author in [18] (see also [14]). We are not aware of any references for the other implications.
Now we discuss the other results in the paper. First, we establish a lower bound showing that the convergence in Theorem 1, if it occurs, is very slow. Theorem 1.7 In the notation of Conjecture 1.1, we have where the constant c > 0 is absolute.
Here, the constant in the O() notation is absolute.
The O( 1 log p ) term is best possible, as we remark in Sect. 6. We neither discuss nor make progress on partial results towards any of Conjectures 1.1, 1.2, 1.3, 1.4 or 1.5. We believe that the best value of ε for which Conjecture 1.2 is known is ε ≈ 0.67513 . . . , which is equivalent to a result obtained in [13]. (The precise value here is α − 1, where α solves α 3 − 4α + 2 = 0.) This bound is now 15 years old.
Notation. Most of our notation is quite standard. We use # X for the cardinality of a set X . Occasionally, if A is a set in some abelian group and k is an integer we will write k · A to mean {ka : a ∈ A}.

Progressions, projections and entropy
In this section we establish around half of Theorem 1.6 by proving that the first three conjectures mentioned in the introduction are equivalent. Whilst at a local level the arguments are a mix of fairly unexciting linear algebra and standard tools such as Freiman isomorphisms, random projections and taking tensor powers, the large number of them makes the proof of Theorem 1.6 somewhat lengthy.
It is convenient to proceed by first showing that Conjectures 1.1, 1.3 and 1.2 are equivalent. In the course of doing so, and for later use, it is convenient to introduce a further conjecture, apparently stronger than Conjecture 1.1 but, as it turns out, equivalent to it.

Conjecture 1'.
Let k be a positive integer. Write F k (N ) for the cardinality of the smallest set A ⊂ Z which contains an arithmetic progression of length k and common difference d, It is obvious that Conjecture 1' implies Conjecture 1.1, because F k (N ) F k (N ). It turns out that the reverse implication holds as well. In fact, we claim that the following is true.
Proof Suppose we have a set where the d i are distinct. We claim that there is a set A 1 , #A 1 k 3 log N · # A 0 , containing an arithmetic progression of length k and common difference d for all d ∈ {1, . . . , N }. This obviously implies the result.
Pick θ ∈ (0, 1) uniformly at random, and define the function It follows that the expected number of pairs By linearity of expectation, there is some choice of θ for which, and therefore there are at least N /3 values of n for which f (n) = 0, or in other words there are at least N /3 distinct values amongst the d i . Now consider the set A 2 := φ θ (A 0 ). Obviously # A 2 # A 0 . Whilst A 2 itself does not obviously contain any long progressions, we observe that By taking random translates (see Lemma 6.2 for details) and the fact that there are N /3 distinct d i , there is some set T of integers, #T log N , such that every element of {1, . . . , N } can be written as d i + t with t ∈ T . Set It is easy to see that A 1 contains an arithmetic progression of length k and common difference d i + t, for all i and for all t ∈ T , and hence contains an arithmetic progression of length k and common difference d for all d ∈ {1, . . . , N }. This concludes the proof of Proposition 2.1. Now we turn to the proof that Conjectures 1', 1.2 and 1.3 are equivalent.
where H k denotes the set of rationals with height at most k, that is to say Our first step is to use a "tensor power" argument to show that there are arbitrarily large sets with the same property; in fact, we shall argue that for every j there is a set This is simple if the A k, j are allowed to be subsets of Z n . Indeed we may define A (n) k to be the set (a 1 , a 2 , . . . , a n ), (a 1 , a 2 , . . . , a n ) ∈ Z n × Z n : (a i , a i ) ∈ A k for all i .
Then, writing π (n) r : Z n × Z n → Z n , for the map sending (x, y) to x + r y (or, when r = ∞, to y) we have n for all r , n. In particular, by choosing n large enough (depending on j) we have, since the inequality (2.1) is strict, To create a subset of Z×Z from A (n) k , we take an integer t and apply a map ψ t : Z n ×Z n → Z × Z of the form , t 2 , . . . , t n ) · x, (t, t 2 , . . . , t n ) · y)), where the dot denotes the usual inner product. Set A := ψ t (A (n) k ). Choose t such that for r ∈ H k and (x, y), y ). There is such a t, because for each of the finite number of choices of x, y, x , y , r the left-hand side of (2.4) is a nontrivial polynomial equation in t. For such a choice of t it follows that π r (ψ t (x, y)) = π r (ψ t (x , y )) if and only if π (n) r (x, y) = π (n) r (x , y ), and so k ) for all r . This establishes the existence of the sets A k, j satisfying (2.2).
For each j, k, consider the set S k, j ⊂ Q defined by On the other hand, suppose that d ∈ −π −1 (A k, j ). This means that d = y − x for some (x, y) ∈ A k, j . If 0 i k − 1, we have , it follows that x + id k ∈ S k, j for i = 0, 1, . . . , k − 1, that is to say S k, j contains a progression of length k and common difference d k . Thus, writing N j := #π −1 (A k, j ), we see that S k, j is a set of size ( j −1 N j ) 1/(1+ε) containing progressions of length k with at least N j distinct common differences. Since, evidently, #S k, j k, the presence of the factor j −1 forces N j → ∞ as j → ∞. By multiplying through by an appropriate integer, we may find setsS k, j ⊂ Z with the same property, contrary to Conjecture 1'. Conjecture 1.3 implies Conjecture 1.2. This implication is essentially given in [18]. The notation there takes a little unpicking and the proof is short, so we repeat the argument.

Y).)
We begin with a couple of observations. The first is that (2.5) is automatically true for sets A ⊂ Z n × Z n , for any n. This follows from the case n = 1 by applying a suitable map ψ t : Z n × Z n → Z × Z, exactly as in the argument following (2.3) above.
The second observation is that, by a simple limiting argument, we may assume that there is some q such that qP((X, Y) = (x, y)) ∈ Z for all (x, y): if we can prove the result for such (X, Y) the same inequality for arbitrary (X, Y) with finite range follows by letting q → ∞. Now let m be very large, and construct a set A ⊂ Z mq × Z mq as follows. Let it consist of all pairs ((x 1 , . . . , x mq ), (y 1 , . . . , y mq )) ∈ Z mq × Z mq for which #{i : (x i , y i ) = (x, y)} = mqP((X, Y) = (x, y)).
(Here, we interpret P(X + ∞Y = z) as P(Y = z).) Writing n = mq and p z = P(X + r Y = z) for short, it follows that Note that the product over z is finite, and that each np z is an integer. Taking logs and using the fact that log We may assume that the o(n) term is uniform in r ∈ {r 1 , . . . , r k } (since this is a finite set); of course, it also depends on X, Y, but we are thinking of these as fixed for the duration of the argument. Taking logs of (2.5) (which is valid for A ⊂ Z n × Z n , as remarked), we conclude that Now we may simply divide through by n and let n → ∞ to conclude the claim (2.6). Conjecture 1.2 implies Conjecture 1.1. This is relatively easy. Assume Conjecture 1.2. Let ε > 0 be arbitrary, and select r 1 , . . . , r m ∈ Q ∪ {∞}\{−1} so that we have On the other hand, By choosing Q and then M suitably, we may ensure that all the Qr j /(1 + r j ) are integers of magnitude < Q M, which means that That is, X + r j Y takes values in (1 + r j ) · A. Since H(W) log m for any random variable W taking values in a set of size m, this implies that is the natural map. Since {0, 1, . . . , k( p − 1)} is covered by k discrete intervals of length p, on each of which the projection map π : Z → F p is injective, we see that # A 2 k n # A 1 .
By construction, A 2 contains a progression of length k and common difference d for p n distinct values of d. Whilst A 2 is a subset of Z n , we can create a subset of Z with the same properties by looking at the image of A 2 under the map f : Z n → Z defined by f (x 1 , . . . , x n ) = n i=1 (10kp) i x i . It follows that # A 2 F k ( p n ), and hence # A 1 k −n F k ( p n ). In the notation of Conjecture 1.4, this means that f n,k ( p) k −n F k ( p n ). It follows that Assuming Conjecture 1' (taking N = p n ), the right hand side here is precisely n. This implies Conjecture 1.4. Conjecture 1.4 implies Conjecture 1.1. Suppose we have a set A 1 ⊂ Z containing a progression of length k and common difference d for each d ∈ {1, . . . , N }. Partition Z into intervals I j := 10k j N + {1, . . . , 10k N }, j ∈ Z. Any progression of length k and common difference d ∈ {1, . . . , N } is either wholly contained in some I j , or else is split into two progressions, one in I j and the other in I j+1 , with one of these having length at least k/2. It follows that the set A 2 ⊂ Z defined by 4 contains a progression of length at least k/2 and common difference d, for all d ∈ {1, . . . , N }. Manifestly # A 2 # A 1 , and by construction A 2 has the additional property that (3.1) Using A 2 , we construct a set A 3 ⊂ Z n . We will later use this to construct a further set A 4 ⊂ F n p , for a suitable prime p, by projection. To define A 3 , let M := N 1/n . Select t ∈ {−10k N , . . . , 20k N − 1} uniformly at random, and define If it so happens that t ∈ −x(d) + S then A 3 (t) contains a progression of length k and common difference (d 1 , . . . , d n ), namely {(s 1 , . . . , s n and moreover A 4 contains a progression of length k and common difference d for all d ∈ π (n) (D), that is to say for n,k N n,k p n values of d. By a standard argument (taking random translations of π (n) (D), see Corollary 6.4 for details) there is a further set A 5 ⊂ F n p , containing a progression of length k and common difference d, for all d ∈ F n p \{0}. Tracing back through (3.4), (3.3), (3.2) we see that where p = p(N ) ∼ N 1/n is some prime. It follows that Assuming Conjecture 1.4 (n), the limit on the right is 1. This concludes the proof that Conjecture 1.4 (n) implies Conjecture 1.1. Before leaving this topic, we remark that it is quite possible that very strong bounds such as f p η ,1 ( p) p/2 (3.5) are true, provided p p 0 (η) is large enough. This issue is strongly hinted at, if not explicitly conjectured, in [1]. It is pointed out there that such bounds imply vastly more than is currently known about the purely arithmetic problem of bounding the least quadratic nonresidue modulo p. Whilst a bound of the form (3.5) is not known to imply the arithmetic Kakeya conjecture (the progressions are of length p η , rather than of bounded size), the arguments of Bourgain may be adapted to show that it does imply the Euclidean Kakeya conjecture. Further details may be found in lecture notes of the first author [9,Section 10].
It is quite interesting that the innocent-looking statement (3.5) implies two famous unsolved problems in completely different mathematical areas.

A problem of Erdős and Selfridge
Finally, we turn to Conjecture 1.5. In fact, we prove the following rather tight connection between Conjecture 1' and Conjecture 1.5. Proof Suppose first we have a set of primes p 1 < · · · < p N and an interval I of length kp N so that # A = G k (N ), where A = N i=1 {x ∈ I : p i |x}. Note that A obviously contains a progression of length k and common difference p i , for each i, and therefore F k (N ) G k (N ).
In the other direction, suppose we have a set A attaining the bound F k (N ), that is to say # A = F k (N ) and A contains, for i = 1, . . . , N , a progression {a i + jd i : j = 0, 1, . . . , k−1}. By translating if necessary, we may assume that A consists of positive integers. Let δ ∈ (0, 1 2 ) be a quantity to be specified shortly. By the theorem of the first author and T. Tao provided that δ is chosen sufficiently small. Note also that if δ is small enough. In particular if δ is small enough then we have It remains to prove the claim. To prove the if implication, it suffices in view of (4.4) to show that w + A ⊂ I . However it is obvious that min(w + A ) min I (since all elements of A are positive) and moreover the last step being a consequence of the fact that p N (1 − δ)X 50. Thus (4.6) also holds, and this completes the proof of the claim.

Remark
The use of the theorem of the first author and Tao is a little excessive. One could do without it using simpler arguments if one was prepared to settle for logarithmic losses.

Small unions of progressions
In this section we prove Theorem 1.  Combining Proposition 6.1 with the construction of A satisfying (6.3) immediately gives the desired upper bound ε = O( 1 log p ), thereby concluding the proof of Theorem 1.8. It remains to prove Proposition 6.1.

Corollary 6.4
Suppose that A ⊂ F n p is a set containing a k-term arithmetic progression with common difference d, for all d lying in some set D of size δ p n . Then there is a set A , # A k,n log p · # A, containing a k-term arithmetic progression with every common difference.