Rotated Corners in the Integer Lattice Hans Parshall Abstract In this paper we prove the existence of rotated corners within positive density subsets of the lattice [, N] for sufficiently large N. We further show that the number of these corners is some positive proportion of the total possible number of corners. Preliminaries Roth in [] proved that for N sufficiently large, a subset of [, N] with positive density will necessarily contain a non-trivial 3-term arithmetic progression. In the integer lattice, we are interested in finding three points that are vertices of an isosceles triangle. To do this, we look for rotated corners. Definition. A rotated corner is a triple x,y, z [, N] such that y x = R(z x), where R = ( 0 0 ). We will eventually show that either N is bounded above in terms of the density of our subset, or our subset has increased density in some lattice. Definition. A lattice is a set P [, N] along with some x [, N], q, k [, N] such that P = {(x + iq, x + q) : i, k}. The approach will be similar to Roth s Fourier analytic proof for 3-term arithmetic progressions. For this we will require some notation and definitions. For a in some -dimensional set A, we will denote its coordinates as (a, a ). Definition. For x, y Z N, we define the dot product x y := x y + x y. Definition. Let f : Z N C and define e(α) := exp(πiα). We define the Fourier transform of f to be bf(ξ) := f(x)e( x ξ/n). For convenience, we will rely on context and hereafter denote the dot product by uxtaposition. Definition. Let f, g : Z N C. We define the convolution of f with g to be f g(x) = f(y)g(x y). y Z N Definition. Given a set A Z N, we denote its characteristic function if x A A(x) = 0 otherwise Finally, we state without proof some fairly standard but useful identities. Proposition. Let f, g : Z N C and let A Z N. Then the following are true. (i) f(x) = f(ξ) N b (ii) f g(ξ) = b f(ξ)bg(ξ) (iii) max f(ξ) b f(x) ξ Z N (iv) A(0) b = A N Identity (i) is frequently referred to as Plancherel s identity. For our proof of existence of rotated corners, we need to consider two cases. The first is when our set is sufficiently random and the second is when it can be shown to be somewhat structured.
The Random Case A sufficient notion of randomness for our purposes is that of ε-uniformity. Definition. We say that a set A Z N is ε-uniform if b A(ξ) ε for all ξ (0,0). Let A [, N] with A = δ. If we consider A as a subset of Z N and let B := A (N/3, N/3], it s easily verified that every corner counted by (A) := {(x, y,z) B A : y x R(z x) (modn)}. is a genuine corner in [, N] as well. That is, (A) provides a lower bound on the number of corners contained in A. We can count (A) by (A) = e((y x R(z x))ξ/n) = x B y B z A = = N 4 e(yξ/n) y B z Ae( R(z)ξ/N) x B y B e(yξ/n) z Ae( zr (ξ)/n) x B e( (x R(x))ξ/N) e( x(ξ R (ξ))/n) bb( ξ) bb(ξ R (ξ)) ba(r (ξ)) () With this count for (A), we can make the following observation. Lemma. If A is ε-uniform, then (A) δ B ε B. Proof. By pulling out the terms when ξ = 0, the summation in () provides (A) δ B = N 4 B( ξ) b B(ξ b R (ξ)) b A(R (ξ)) \{0} Utilizing ε-uniformity of A and the Cauchy-Schwarz inequality, we have (A) δ B εn 4 bb( ξ) B(ξ b R (ξ)) 0 εn 4 @ B( ξ) b A / 0 @ B(ξ b R (ξ)) A / Finally, noting that both ξ and ξ R (ξ) vary completely over Z N, Plancherel s identity yields ««B (A) δ B εn 4 / B / = ε B N N In particular, Lemma implies that if A is ε-uniform with ε δ B /( ), then A contains at least δ B / corners. This is useful to prove the main result of this section. Lemma. Let A [, N] with A = δ be ε-uniform for some ε δ B /( ). Suppose that A contains no nontrivial rotated corners. Then either N < 5/δ or there exists some lattice P [, N] such that δ + δ 80
Proof. First suppose B δ /0. Then, by our lower bound from Lemma, it must be that > δ B δ3 N 4 00 N < 5/δ. Suppose instead B < δ /0. Then the other 8 N/3 by N/3 lattices comparable to (N/3,N/3] must contain at least 9δ /0 elements of A. By the pigeonhole principle, we can conclude that one of them, call it P, has 9δ /80. But then we have = 9 8δ 80 δ + δ 80. The Structured Case For x R, let x denote the distance to the nearest integer from x. Lemma 3. Fix ξ Z N\{0} and let Q [, N / ]. Then there exists some q [, Q ] such that qξ /N /Q, qξ /N /Q Proof. Consider the lattice [, N] modulo. Split this into Q lattices of the form [i/q, (i + )/Q) [/Q, ( + )/Q) for 0 i, Q. If there existed a q such that qξ/n was in one of the four corner lattices, we would be done. Suppose instead that there was no such q. Then there are Q 4 lattices with Q choices of q, so there must exist q < q Q such that q ξ/n and q ξ/n lie within the same lattice. Then we know (q q )ξ /N /Q (mod), (q q )ξ /N /Q (mod). () But q q [, Q], and () implies that (q q )ξ/n lies within the bottom left corner lattice. This contradicts our assumption that there were no such q, and therefore one must exist. Let L N / /8π. By Lemma 3, we can now fix q 64π L such that qξ /N /8πL and qξ /N /8πL for all nonzero ξ. Define P q := {( iq, q) : i,, L} and set η := L /. We can now show that P q has relatively large Fourier coefficients. Fixing ξ Z N \{0}, c P q(ξ) = = = η = η/ e((qξ + kqξ )/N) k= e(qξ /N) e(kqξ /N) k= Re(e(qξ /N)) Re(e(kqξ /N)) k= ( πqξ /N ) ( πkqξ /N ) L π qξ /N «L πl 8πL 4πL3 8πL k=! L π qξ /N! k k= 3
Define the balanced function of a set A Z N as f A := A(x) δ. A simple computation shows f A P q(x) = A (x Pq) δη (3) Suppose that A is not ε-uniform. Then we have f A P q(x) max cf A(ξ) cp q(ξ) = max b A(ξ) cp q(ξ) εη ξ Z ξ Z N N \{0} If we let (f A P q) + := max{f A P q,0}, then we see 0 (f A P q) +(x) = @ f A P q(x) + f A P q(x) A But noting P x Z f A P N q(x) = 0, we now have (f A P q) +(x) εη 4 (4) In particular, together with (3) and the pigeonhole principle, we have that there exists some x Z N such that A (x P q) δη εη 4 This allows us to prove the main result of this section. Lemma 4. Suppose A [, N] with A δ is not ε-uniform. Then there exists a genuine lattice P [, N] with p εn/5π such that δ + ε 8 Proof. To begin, we count x Z N where x P q would be an adequate lattice in Z N. Let k := x Z A (x Pq) N : δ + ε P q 8 ff Equivalently, we know k = x Z N : A (x Pq) δη εη 8 ff and hence by (5), we know k. Note that f A P q(x) η always. By (4), we have εη 4 f A P q(x) ( εη k) + kη εn k ε k 8 8 8 In order to show that one of these k starting points is for a genuine solution, it would suffice to show that the total size of the lattice is less than k. That is, we need q L ε /8. But we know q L 64π NL 4, so it is sufficient to have 64π NL 4 εn 8 L Thus we can choose our L such that p εn/5π. Existence of a Rotated Corner r εn 5π We now have sufficient machinery to prove the existence of a rotated corner in a positive density subset of the integer lattice. Theorem (Roth for Rotated Corners). Let δ > 0. Then there exists some absolute constant C such that, provided N exp exp(c/δ), any subset A [, N] with A δ will necessarily contain a non-trivial rotated corner. (5) 4
Proof. Let A [, N] with A δn such that A contains no non-trivial rotated corners. Set ε := δ /0. By Lemmas and 4, since we are assuming N is sufficiently large, we have that there exists some lattice P [, N] such that cδ N and δ + δ 80 Setting N 0 := N, A 0 := A, δ 0 := δ, we can iterate this argument as follows. Suppose for A we had a lattice P = {(x + iq, x + kq) : i, k p P } that satisfied the density increment above. Then we could set N + p P and «ff a x a x A + =, : a, a A P [, N +] q q First notice that if A + contains some rotated corner, then we have a,b, c A P such that a x b x c x = R b x «a b = R(c b) q q q q Since we have assumed A 0 contains no rotated corners, this implies no A contains a rotated corner. Now, note A + δ + + where δ + δ + δ /80. This iteration cannot continue indefinitely. Indeed, a simple induction argument demonstrates that any δ will double after 80/δ iterations. A finite amount of doubling will cause some δ k+ >, where certainly k i=0 80 = 60. i δ 0 δ 0 But δ k+ > is absurd, so we must have N k < 5/δ k. Letting c vary to denote the appropriate absolute constant, we have 5 5 > N cδ k N / k δ 0 δ cδ k k δ k N / k / cδ 0 0 N 0 < cδ 3 0. An easy calculation provides that log log N 0 < log log(cδ 3 0 ) C/δ for some absolute constant C, contradicting our choice of N 0. Therefore, A must contain a non-trivial rotated corner. Existence of Many Rotated Corners Any two points chosen in [, N] give rise to at least one rotated corner; that is, there are approximately N 4 rotated corners in [, N]. We can apply an argument of Varnavides from [], originally applied to Roth s theorem on 3-term arithmetic progressions, to guarantee that a positive proportion of the total number of rotated corners are in any set with positive density. Theorem (Varnavides for Rotated Corners). Let δ > 0 and let N be sufficiently large. Then any subset A [, N] with A δ will necessarily contain c(δ)n 4 rotated corners, where c(δ) is a constant depending only on δ. Proof. We first choose an integer k := k(δ) sufficiently large to apply Roth for Rotated Corners on a subset of [, k] with density δ/. For u [, N] and d N, we consider lattices P u,d = {(u + id, u + d) : i, k} [, N]. We call P u,d good if A P u,d δk /. Fix d < δn/k and let g d denote the number of good lattices with step size d. For every a A with kd < a, a < N kd, we can choose u {(a id, a d) : i, k} so that a is contained in exactly k lattices. The number of these a is bounded below by A 4Nkd δ 4δ /k = δ( 4/k). Provided k > 6, we have 4 «3 k 4 δk (6) u,d:p u,d [,N] A P u,d k δ 5
The most A could intersect with any good lattice is precisely the size of the lattice, k. The most A could intersect with any other lattice is δk /, and certainly the total number of such lattices is less than. Thus, we have u,d:p u,d [,N] A P u,d g d k + δk (7) Then by (6) and (7), we have g d (/4)δ. But then we have at least δ /4 δ /k = c (δ)n 4 good lattices, each of which gives rise to a rotated corner by applying Roth for Rotated Corners. The only concern now is that each corner is being counted multiple times. To account for this, fix some solution in A. Note in the three points in the solution, there must be at least two distinct horizontal coordinates and two distinct vertical coordinates. For the solution to lie in a lattice P u,d, we have at most (k ) choices for u and u, hence at most (k ) choices for u. Further observe that the difference between the two distinct horizontal coordinates, call it d, must be divisible by d. That is, d = d /t for some t. But we must have kd > d or the lattice is not large enough to span our corner. Thus we have at most k choices for t, and so at most k choices for d. Our P u,d is completely determined by our choices of u and d, hence our solution can be contained in at most k(k ) lattices. Therefore we have at least (/k(k ) )c (δ)n 4 = c(δ)n 4 rotated corners in A. References [] K. Roth, On certain sets of integers, J. Lond. Math. Soc., 8 (953), pp. 04 09. [] P. Varnavides, On certain sets of positive density, J. Lond. Math. Soc., 34 (959), pp. 358 360. 6