On the correlations of directions in the Euclidean plane

We study the correlations of directions $P_0 P$ in the Euclidean plane, where $P_0$ is a point in a fixed disc, $P$ is an integer lattice point in the square $[-Q,Q]^2$, and $Q\to \infty$.

(ν) (x,y),Q denote the repartition of the ν-point correlation measure of the finite set of directions P (x,y) P , where P (x,y) is the fixed point (x, y) ∈ [0, 1) 2 and P is an integer lattice point in the square [−Q, Q] 2 . We show that the average of the pair correlation repartition R (2) (x,y),Q over (x, y) in a fixed disc D0 converges as Q → ∞. More precisely we prove, for every λ ∈ R+ and each 0 < δ < 1 10 , the estimate 1 Area(D0) We also prove that for each individual point (x, y) ∈ [0, 1) 2 , the 6-level correlation R (6) (x,y),Q (λ) diverges at any point λ ∈ R 5 + as Q → ∞, and give an explicit lower bound for the rate of divergence.

Introduction
In many problems one is led to consider in the Euclidean plane lines joining a fixed point P 0 (which is not necessarily an integer lattice point) with a set of integer lattice points.
A natural way of measuring the distribution of directions P 0 P , P ∈ Z 2 , is via correlations and consecutive spacings. When the fixed point is the origin, the problem is related to the distribution of Farey fractions with multiplicities, each fraction a q in F Q being counted Q q times. The consecutive h-level spacing measures of customary Farey fractions were computed for h = 1 in [6] and for h ≥ 2 in [1]. Limiting correlations of Farey fractions were shown to exist and computed recently in [5].
When the fixed point is not an integer lattice point, the problem of existence of limiting correlations/consecutive spacings is considerably more difficult. It is therefore natural to try to prove first some averaging results, letting the fixed point to vary in a given region. In the first part of this paper we derive such a result for the limiting pair correlation measure.
The limiting average pair correlation function is constant, as in the Poisson case. What is striking however is that this constant is not 1, as in the Poisson case, but π 3 . We now give a mathematical formulation of the problem. For each Q ≥ 1, let Q denote the set of integer lattice points in the square [−Q, Q] 2 , and set N = N Q = # Q = (2Q+1) 2 .
Date: April 5, 2004. 1 For each point P (x,y) = (x, y), we consider the finite sequences θ P (x, y) P ∈ Q , Q large integer, of angles between the line P (x,y) P and the horizontal direction. The pair correlation of this finite sequence is defined as R (2) (x,y),Q (λ) = # (P, P ′ ) ∈ 2 Q : P = P ′ , N 2π |θ P,P ′ (x, y)| ≤ λ N , where θ P,P ′ (x, y) denotes the measure of the angle ∠P P (x,y) P ′ .
Throughout the paper we shall consider a fixed disc D 0 of center (x 0 , y 0 ) ∈ [0, 1) 2 and radius r 0 . We are interested in the asymptotic behavior of the average (x,y),Q (λ) over D 0 , for fixed λ > 0 and Q → ∞.
The first three sections are concerned with the proof of the following result. If one replaces D 0 by a vertical or horizontal segment of length one, an identical asymptotic formula as in (1.1) turns out to be true. This can be proved by similar techniques as in this paper or using Erdös-Turán type discrepancy estimates, and suggests that (1.1) may be true regardless of the shape of the range of the fixed point.
The phenomenon is similar to the one encountered in the problem of the distribution of fractional parts of polynomials. There, one can handle the pair correlation problem generically (see [8], [3]). Moreover, in the case of the sequence n 2 α (mod 1) one is able to solve the problem for all m-level correlations for a large class of irrational numbers α (see [9], [10]). However, as shown in [9], there are irrational numbers α for which the 5-level correlation of fractional parts of n 2 α, 1 ≤ n ≤ N , diverges to infinity as N → ∞. This occurs as a result of the presence of large clusters of such fractional parts. In the case of Theorem 1.2 above, large clusters of elements of the given sequence are responsible, too, for the divergence of the 6-level correlations, and hence of any other higher level correlations.

A first approximation for R
(2) For obvious practical reasons, we try to replace from the beginning θ P,P ′ (x, y) by one of its trigonometric functions in the definition of R (x,y),Q (λ). Suppose that two distinct points P = (q, a), P ′ = (q ′ , a ′ ) ∈ Q , are such that q, q ′ ≥ 0 and max{a, a ′ } > 0 > min{a, a ′ }.
Then for sufficiently large Q (depending only on λ) we have As a result, we may only consider in the definition of R (x,y),Q points from the same quadrant. Thus if we set 2 Q = (P, P ′ ) ∈ 2 Q : P = P ′ and P, P ′ belong to the same quadrant and (2.1) For P = (q, a), P ′ = (q ′ , a ′ ), (x, y) ∈ R 2 , we define For each P, P ′ ∈ Q , consider the weight and define for every µ > 0 In the remainder of this section we show that the asymptotic of R For fixed P, P ′ , denote by θ the angle between the line ℓ determined by P and P ′ and the horizontal direction. Consider also the lines ℓ ± , parallel to ℓ and such that dist(ℓ, ℓ ± ) = µγ cos θ |q ′ −q| . The equation of ℓ is given by while the equation of ℓ ± is given by We see that The set whose area defines A P,P ′ (Q, µ) is the intersection of the strip bounded by ℓ + and ℓ − and the disc D 0 , thus We also have Lemma 2.1. Let α ∈ (0, 1]. Let C be a compact set in R + . Then for all ε > 0 and all Proof. The estimate (2.5) reads as A P,P ′ (Q, µ) = O C,D 0 1 P P ′ ). Combining it with (2.6) we see that it suffices to show that Taking P ′′ = (q ′′ , a ′′ ) = (q ′ − q, a ′ − a) ∈ 2Q , we gather The two conditions on (q, a) above yield that (q, a) should belong to the intersection of a strip of width ≪ C OP ′′ OP ′′ = 1 bounded by the lines y = a ′′ q ′′ x ± α C with the square [−Q α , Q α ] 2 . The number of integer lattice points inside this region is of order O C (Q α ), Since r 2 (k) = {(m, n) ∈ Z 2 : m 2 + n 2 = k} = O ε (k ε ), this gives as desired.

A formula for G Q (µ)
An immediate consequence of (2.5) and (2.6) is that the contribution to G Q (µ) of pairs of points (P, P ′ ) ∈˜ 2 Q with a ′ = a or with q ′ = q is negligible. Indeed, we see from (2.6) that when a ′ = a = 0, the term A P,P ′ (Q, µ) is zero unless |q ′ − q| ≤ 2µ + 1 |a| ≤ 2µ + 1, thus the total contribution of such points to G Q (µ) is The contribution of pairs of points (P, Similar estimates in the case q ′ = q show that As a result, we shall subsequently assume that a ′ = a and q ′ = q. We now set The remainder of this section is elementary and is concerned with putting G Q (µ) in a tidy form, suitable for a precise estimation which will be completed in the next section.
Let C 0 denote the center of D 0 , let ℓ ′ be the line passing through C 0 and perpendicular to ℓ, and denote by A + and A − the intersections of ℓ ′ with the circle ∂D 0 , by E 0 the intersection of ℓ ′ and ℓ, and by E ± the intersection of ℓ ′ with ℓ ± . Direct computation gives While ordering the points x E + < x E − and x A + < x A − the following situations may occur: This gives µγ 0 > r 0 √ α 2 + 1, hence Suppose first that |a ′ − a| ≤ |q ′ − q|. By (2.5) we know that for fixed (q, q ′ ), the expression D = aq ′ − a ′ q will only take values between −2µ − 4µ r 0 and 2µ + 4µ r 0 . Hence the number of solutions (a, a ′ ) of aq ′ − a ′ q = D is of order O C (d), where d is the greatest common divisor of q and q ′ . But d ≤ 2µ r 0 , hence this is actually O C (1) and the contribution to G Q is The case |q ′ − q| ≤ |a ′ − a| is settled similarly by first summing over (a, a ′ ).
or equivalently So, keeping a ′′ and q ′′ fixed, the range of a ′′ q − aq ′′ has cardinality O C (1). Now the equation

The contribution of terms
We infer as in Case 2 that the contribution of A P,P ′ is in this case too O δ (Q − 1 2 +δ ).
Hence we find that the contribution of terms A P,P ′ for One shows similarly that the contribution of points P, P ′ for which −r 0 P P ′ + µγ < L P,P ′ (x 0 , y 0 ) < r 0 P P ′ is of the same order. Therefore by (3.1) and the previous considerations we infer that

Figure 2.
Next S P,P ′ is approximated by elementary calculus.
Lemma 3.1. The area of the region inside the circle of radius r 0 centered at the origin and inside the strip bounded by the vertical lines y = t 1 and y = t 2 , −r 0 < t 1 ≤ t 2 < r 0 , is given, for small t 2 − t 1 , by Proof. The error is seen to be given (see Figure 2) by It is ≪ (t 2 − t 1 ) 3/2 as a result of We take Notice now that and find that the contribution of the error provided by Lemma 3.1 in (3.2) is Therefore by (3.2), Lemma 3.1, (3.3) and (3.5) we find that where B P,P ′ (Q, µ) denotes the contribution of the main term in Lemma 3.1 to (3.2), that is (3.7) Finally we show that, if a ′ − a ≤ q ′ − q, one can replace γ by 1 This gives and similarly Hence one can replace B P,P ′ (Q, µ) in (3.6) by at the cost of an error which is In summary, we have shown for any µ in a fixed compact set C ⊂ R + , that

Estimating the sum S Q
By reflecting D 0 about the axes and about the line y = x, we see that it suffices to only Then we gather from (3.8) and the above formula for S Q that Changing q to q ′ − q and a to a ′ − a, we may write Putting D = aq ′ − a ′ q and taking J q,a as in (3.4), that is we get We will prove the following result.
From this and (4.1) we infer the following We now start the proof of Proposition 4.1. We first lay out some notation and prove an elementary calculus lemma. Fix α 0 , β 0 ∈ R and consider the function and the domain   Define the function ψ = ψ α 0 ,β 0 : pr 2 D → [0, ∞) by Proof. (i) By the definition of Φ it is seen that I x is the union of one or two intervals [a(x), b(x)], where a(x) and b(x) are equal to 0, 1, or a root of Φ(t, x) = 0. In all these cases and as a result we get Using the triangle inequality and the change of variables (u, v) = T (t, x) = (t, tα 0 − β 0 − x) we obtain (ii) The same change of variable as in (i) gives We start to evaluate S Q . If D ∈ J q,a , then D ∈ Ω q = − (2 + r 0 √ 2)q, (2 + r 0 √ 2)q . The equality aq ′ − a ′ q = D is equivalent to a ′ = aq ′ −D q . Hence in the inner sum we should sum over q ′ ∈ [q, Q] such that aq ′ = D (mod q) and or equivalently Next we show that the bulk of the contribution to S Q of (4.5) only comes from To see this, notice first that for q ′ fixed, the relations D = aq ′ (mod q) and |D| ≤ (r 0 + √ 2)q imply that D takes at most 3 + [r 0 ] values. So the total contribution of terms with 0 ≤ When q − a ≥ 3 + [r 0 ], we get −D ≤ (3 + [r 0 ])Q ≤ (q − a)Q, thus Q ≤ qQ+D a . Suppose now that q ′ is between q and q + D a . Owing again to aq ′ = D (mod q), it follows that aq ′ = D + kq for some integer k. But the range of q ′ is an interval of length |D| a , hence the range of k = aq ′ q − D q is an interval of length ≪ 1 + a q · |D| a ≤ 4 + [r 0 ]. Thus k, and consequently q ′ , take at most O(1) values. Besides, in this case we have 0 ≤ q ′ − q ≤ |D| a . Hence the contribution to S Q of terms with q ′ between q and q + D a is Therefore we have shown that we can replace the summation conditions in the inner sum from (4.2) by q ′ ∈ [q, Q] and aq ′ = D (mod q).

Then (4.1) and the previous considerations lead to
where I x and J t are defined as in (4.3) and (4.4).
Take d = gcd(q, q ′ ) and write q = dq 0 , q ′ = dq ′ 0 . Then d divides D, so D = dD 0 . The congruence adq ′ 0 = D (mod dq 0 ) is equivalent to aq ′ 0 = D 0 (mod q 0 ), and we may write To estimate the inner sum above, we need some information about the distribution of solutions of the congruence xy = h(mod q). We shall employ the following result, which is a consequence of Proposition 6.4 from the Appendix.
Proposition 4.4. Assume that q ≥ 1 and h are two given integers, I and J are intervals, and f : I × J → R is a C 1 function. Then for every integer T > 1 and every δ > 0 a∈I, b∈J ab=h(mod q) where · ∞ = · ∞,I×J .
We now return to the formula for S Q given in (4.6) and first give an upper bound for the contribution to S Q of quadruples (d, q 0 , D 0 , a) for which with L = L q 0 > 1 to be chosen later. Proof. Using we see that it suffices to consider the case u > 0. In this case the statement follows from the fact that the double inequality K ≤ F (t) ≤ K + L 2 is equivalent to and from the inequality Suppose that (d, q 0 , D 0 ) is fixed and consider the following two cases: Case 1) r 0 = x 0 . By (4.7) and Lemma 4.5 the range of a is the union of at most two intervals of length Lq 0 r 0 . Hence a can only assume O(L q 0 ) values. But, for each a, q ′ 0 belongs to q 0 , Q d and is subject to the condition q ′ 0 a = D 0 (mod q 0 ). Hence q ′ 0 takes O 1 + Q dq 0 = O Q dq 0 values. Thus the contribution to S Q of quadruples (d, q 0 , D 0 , a) which satisfy (4.7) is Case 2) r 0 = x 0 thus α 0 = 1. In this case we collect directly from (4.7) Hence a can only assume O dD 0 values and we find, arguing as in Case 1, that the contribution to S Q of quadruples (d, q 0 , D 0 , a) which satisfy (4.7) is Next we investigate the case We consider the range of q ′ 0 : the range of a (which is the union of at most two intervals): , and the functions With this notation the following estimates hold on I q 0 ,d × J q 0 ,D 0 ,d,L : Applying Proposition 4.4 we find that LT .
we see that

LT .
Taking T = [q 1 5 0 ] and L = q 9 10 0 we find that Thus the total contribution of E q 0 ,D 0 ,d,L to S Q is Moreover, the quantities in (4.8) and (4.9) become and respectively Thus we gather (4.10) and as a result the error that results from replacing J q 0 ,D 0 ,d,L by dq 0 ID 0 On the other hand we find that In particular, the integral of G q 0 on q 0 , Q d is ≪ Q 3 d 3 , and we find that the total cost of replacing J q 0 ,D 0 ,d,q 9/10 0 by dq 0 ID 0 Inserting this back into (4.11) we obtain (4.12) Proposition 4.1 now follows from (4.10) and (4.12).

On the 6-level correlations
In this section we prove Theorem 1.2. We first prove a counting result.
Proof. Using Möbius inversion we express the left-hand side of (5.1) as which proves the lemma.
For any positive integers a, b, q, consider the set Lemma 5.2. For q large and 1 ≤ a, b ≤ q such that gcd(a, b, q) = 1 #N a,b,q ≫ ϕ(q) ln q ≫ q ln q ln ln q .
Remark. If in the definition of N a,b,q we took the range of A and B to be [1, q] instead of [1, 2q], the cardinality of N a,b,q would be much smaller. For example, if 1 ≤ a ≤ q, gcd(a, q) = 1, and b = a, then q| A − B, and in the range 1 ≤ A, B ≤ q this forces A = B.
Then the only pair (A, B) with gcd(A, B) = 1 is (1, 1), so N a,b,q will only contain one element. Lemma 5.2 shows a sudden increase in the cardinality of N a,b,q when the range of A and B increases by a factor 2.
Proof of Lemma 5.2. To make a choice, assume in what follows next that d = gcd(a, q) ≤ gcd(b, q). Then gcd(d, b) = 1 since gcd(q, a, b) = 1. It follows that for any solution (A, B) to the congruence Ab = Ba (mod q), A has to be divisible by d. Write q = dq 1 and a = da 1 , so gcd(a 1 , q 1 ) = 1. Denote byā 1 the multiplicative inverse of a 1 (mod q 1 ) in the interval Note that since q is divisible by the product gcd(a, q) gcd(b, q), we have d < √ q. Therefore q 1 > √ q. So q 1 is large for large q, and by the Prime Number Theorem we know that For any fixed prime p with q 1 < p ≤ 2q 1 , we count the solutions 1 ≤ B ≤ 2q of the congruence (5.3) dpb = Ba (mod q).
Since gcd(q 1 , a 1 ) = 1, this congruence has a unique solution modulo q 1 , namely B = a 1 pb (mod q 1 ). Denote by B 0 the solution to (5.4) which belongs to the interval [1, ≤ q 1 ]. Then (5.3) will have 2d solutions, given by It remains to be seen how many of the numbers B from (5.5) are relatively prime with A = dp. Note first that at most two such numbers B are divisible by p. Assume that (5.6) p| B 0 + q 1 m 1 , p| B 0 + q 1 m 2 , and p| B 0 + q 1 m 3 , with 0 ≤ m 1 < m 2 < m 3 ≤ 2d − 1. Since p > q 1 , p does not divide q 1 . Then it follows from Here at least one of the differences m 2 − m 1 , m 3 − m 2 is less than d, and d < √ q < q 1 < p, so it is impossible that both divisibilities in (5.7) hold true. Therefore at most two numbers from (5.5) are divisible by p. Note also, by the same reasoning, that for smaller values of d -more precisely for d < q 2 , that is d < q 1 2 -one has 2d < p. Then one concludes that at most one number B from (5.5) can be a multiple of p.
We now count the numbers B from (5.5) which are relatively prime with d. We claim that gcd(B 0 , q 1 , d) = 1. Indeed, let us assume this fails and choose a prime divisor p 1 of gcd(B 0 , q 1 , d). Since p 1 | d, we have p 1 | a and p 1 | q. Recall that B 0 satisfies (5.4), hence Here p 1 | B 0 , p 1 | q 1 , so p 1 must also divide the left side of (5.8). The It follows in particular that there are always at least two numbers B as in (5.5) for which gcd(B, d) = 1, and as soon as d 2 ≥ 2 or d 1 ≥ 3, there are at least four such numbers. Since at most two numbers B as in (5.5) are divisible by p, and for small values of d we know that at most one number B as in (5.5) is divisible by p, we conclude that in all cases we have (5.9) #{1 ≤ B ≤ 2d : gcd(B, dp) = 1, dpb = Ba (mod q)} ≥ ϕ(d 1 )d 2 ≥ ϕ(d).
Combining (5.9) with (5.2) we infer that and the lemma is completed using the inequalities Let now q be a large positive integer, let a, b ∈ {1, . . . , q} such that gcd(a, b, q) = 1, and let Q be a positive integer larger than q. In our applications Q will be at least of the order of magnitude of q Denote by u the unique integer satisfying where B is the multiplicative inverse of B modulo A, and put v = Bu+C A .
Then v is an integer. Also, from the inequalities since v is an integer. Let now s = Q A , and define M A,B to be the set of lattice points given by Note that the case A = B can only occur when a = b, and in this situation we also get Some properties of these sets are collected in the following lemma.
(iii) The sets M A,B are disjoint.
Proof. (i) Owing to (5.11) and to the inequality we have for any (A, B) ∈ N a,b,q and any point (u + mA, u + mB) ∈ M A,B Recall that Q is much larger than q. It follows that the distance between any two points P ∈ [0, 1] 2 and P ′ ∈ M a,b,q satisfies (ii) For any (A, B) ∈ N a,b,q with, say, A ≥ B, and any point P = (u + mA, v + mB) ∈ for some m ′ ∈ {s ′ − M, . . . , s ′ − 1}, with u, v, s, u ′ , v ′ , s ′ given by appropriate definitions.
We compute the ratio qn 2 −b qn 1 −a in two ways. First, by (5.13), (5.10) and the equality v = Bu+C A we have By a similar computation we also have Since (5.14) Next, using the equality v = Bu+C A , we rewrite (5.14) as Since |C| ≤ max{A, B} ≤ 2q and 0 ≤ x, y ≤ 1, we see that |C + Bx − Ay| ≪ q. If M satisfies (5.11), then M |C + Bx − Ay| ≪ Q. As a consequence of (5.15) we also have From (5.15) we also infer that By (5.10) and (5.16) we derive that As a consequence of (5.17) and of the inequalities 1 ≤ A, B ≤ 2q, we have , then all these points lie on the same line, that passes through P (x,y) . Then all the 6-tuples of distinct elements from A will contribute to R (x,y),Q (λ). Since #A = Q m 0 , it follows that Note that in this case the 3-level correlations already diverge as Q → ∞.
choose a large positive integer Q. Let 1 < T < Q be a parameter, whose precise value will be chosen later and will be the integer part of a fractional power of Q. By Minkowski's convex body theorem (see [7,Thm. 6.25] for the formulation used here), there exists an where · denotes here the distance to the closest integer. Let a and b denote the closest integers to qx and qy respectively. Then 0 ≤ a, b ≤ q, max{a, b} > 0, and (5.19) gives Dividing if necessary a, b and q by gcd(a, b, q), we may assume in what follows that gcd(a, b, q) = 1.
We will have T → ∞ as Q → ∞, and since at least one of x, y is irrational, this forces q → ∞ as Q → ∞. Then all our previous results valid for large q are applicable.
Let M be a positive integer satisfying (5.11), whose precise order of magnitude will be chosen later. Consider the disjoint subsets M A,B of Q , with (A, B) ∈ N a,b,q . By (5.18) we know that for any (A, B) ∈ N a,b,q and any P, P ′ ∈ M A,B , the measure of the angle ∠P P (x,y) P ′ satisfies (5.21) |θ P,P ′ | ≪ M (|b − qy| + |qx − a|) Q 2 .
If we take the order of magnitude of M to be slightly smaller than that of both Q q and √ T , for instance then M will satisfy (5.11) on the one hand, and on the other hand we will have |θ P,P ′ | ≪ 1 Q 2 ln Q .
We now choose T = [Q 3 4 ]. Then, no matter how small q might be, we have Also, since q ≤ T , it follows that