AA FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES
ALAN HAYNES, JENS MARKLOF
Abstract.
The three distance theorem (also known as the three gap theorem orSteinhaus problem) states that, for any given real number α and integer N , there areat most three values for the distances between consecutive elements of the Kroneckersequence α, α, . . . , N α mod 1. In this paper we consider a natural generalisation of thethree distance theorem to the higher dimensional Kronecker sequence (cid:126)α, (cid:126)α, . . . , N (cid:126)α modulo an integer lattice. We prove that in two dimensions there are at most fivevalues that can arise as a distance between nearest neighbors, for all choices of (cid:126)α and N . Furthermore, for almost every (cid:126)α , five distinct distances indeed appear forinfinitely many N and hence five is the best possible general upper bound. In higherdimensions we have similar explicit, but less precise, upper bounds. For instance inthree dimensions our bound is 13, though we conjecture the truth to be seven. Wefurthermore study the number of possible distances from a point to its nearest neighborin a restricted cone of directions. This may be viewed as a generalisation of the gaplength in one dimension. For large cone angles we use geometric arguments to produceexplicit bounds directly analogous to the three distance theorem. For small cone angleswe use ergodic theory of homogeneous flows in the space of unimodular lattices to showthat the number of distinct lengths is (a) unbounded for almost all (cid:126)α and (b) boundedfor (cid:126)α that satisfy certain Diophantine conditions. Introduction
Consider a finite set S N comprising N distinct points ξ , . . . , ξ N on the unit torus T = R / Z . The points in S N partition T into N intervals, representing the gaps of S N . We denote by δ n,N the size of the n th gap, i.e., the distance between ξ n andits nearest neighbor to the right . We denote by g N = |{ δ n,N | ≤ n ≤ N }| thenumber of distinct gap sizes. For a generic choice of S N one has g N = N , since all gaplengths are generically distinct. A striking observation, known as the three distance(or three gap) theorem, is that for the Kronecker sequence ξ n = nα + Z , one has g N ≤
3, for any α ∈ R and N ∈ N ; see [36, 37, 38, 39] for the original proofs and[23, 25, 29, 31, 33, 34, 35] for alternative approaches. Natural extensions to returnmaps for billiards in rectangles and interval exchange transformations are discussed in[16, 19, 20, 21] and [40], respectively. There are various generalisations of the threegap theorem to higher dimensions, several of which require Diophantine conditions onthe choice of parameter. In the present paper we discuss natural extensions of thethree distance theorem to higher dimensional Kronecker sequences, which represent Date : 14 September 2020/13 November 2020.
Key words and phrases.
Steinhaus problem, three gap theorem, homogeneous dynamics.AH: Research supported by NSF grant DMS 2001248.JM: Research supported by EPSRC grant EP/S024948/1.MSC 2020: 11J71, 37A44. a r X i v : . [ m a t h . N T ] N ov ALAN HAYNES, JENS MARKLOF translations of a multidimensional torus by a vector (cid:126)α . We will here consider nearestneighbor distances as well as distances to neighbors in restricted directions. The formermay be viewed as a special case of the setting of Biringer and Schmidt [2], who consideredthe number of distinct nearest neighbor distances for an orbit generated by an isometryof a general Riemannian manifold. If distances are measured by the maximum normrather than a Riemannian metric, Chevallier [8, Corollaire 1.2] showed that there are atmost five distinct distances for Kronecker sequences on two-dimensional tori. Relatedstudies in this context include the papers by Chevallier [9, 10, 12] and by Vijay [41].For other higher dimensional variants of the three distance theorem, see [1, 3, 4, 5, 11,13, 18, 22, 24, 25, 26, 27, 30].Our setting is as follows. Let L be a unimodular lattice in R d (one example to keepin mind is Z d ) and consider a point set S N = { ξ , . . . , ξ N } on the d -dimensional torus T d = R d / L . The point set S N that we are interested in is the d -dimensional Kroneckersequence,(1.1) S N = S N ( (cid:126)α, L ) = { ξ n = n(cid:126)α + L | ≤ n ≤ N } ⊂ T d , for given (cid:126)α ∈ R d . Note that the ξ n are not necessarily distinct if (cid:126)α ∈ Q L ; in this case weremove all multiple occurrences, and as a result the number of elements in S N remainsbounded as N → ∞ .Let δ n,N be the distance of ξ n to its nearest neighbor with respect to the standardflat Riemannian metric on T d . As above, g N = g N ( (cid:126)α, L ) denotes the number of distinctnearest neighbor distances δ n,N . Biringer and Schmidt [2] proved that, for all choices of L , (cid:126)α, and N ,(1.2) g N ( (cid:126)α, L ) ≤ d + 1 . Our first theorem improves this bound, and also gives the best possible result in dimen-sion d = 2. Theorem 1.
For every unimodular lattice L , (cid:126)α ∈ R d and N ∈ N we have that (1.3) g N ( (cid:126)α, L ) ≤ (cid:40) if d = 2 ,σ d + 1 if d ≥ , where σ d is the kissing number for R d . Recall that the kissing number σ d for R d is the maximum number of non-overlappingspheres of radius one in R d which can be arranged so that they all touch the unitsphere in exactly one point. The study of kissing numbers has a long and interestinghistory, and is connected to many areas of mathematics (see [6] and [32] for surveys ofresults). It is interesting to note that the three distance theorem in dimension d = 1 iscompatible with the bound g N ≤ σ + 1 = 3, but that already in dimension d = 2 thisbecomes suboptimal, since here σ + 1 = 7 >
5. A table of known bounds for kissingnumbers in dimensions d ≤
24 is provided in Figure 1. Our corresponing bounds for g N in dimensions d = 3 , . . . ,
10 are therefore 13, 25, 46, 79, 135, 241, 365, 555. We donot claim that these bounds are optimal for any dimension d ≥
3. In particular, weconjecture that g N ( (cid:126)α, L ) ≤ d = 3. Figures 2 and 3 show examples where 5 (for d = 2) and 7 (for d = 3) distinct distances are obtained. FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 3 d Upper boundfor σ d d Upper boundfor σ d
13 206914 318315 486616 735517 1107218 1657219 2481220 3676421 5458422 8234023 12441624
Figure 1.
A table of known bounds for kissing numbers. All bounds aretaken from [6], and bounds listed in bold face are known to be best pos-sible.For general d , it follows from Theorem 1 together with an estimate for σ d due toKabatiansky and Levenshtein (see [28, Theorem 4, Corollary 1] or [14, Chapter 9]) that(1.4) g N ( (cid:126)α, L ) ≤ . d (1+ o ( d )) as d → ∞ . The rate of convergence of the o ( d ) term in this estimate can be made more precise andexplicit (non-asymptotic) upper bounds can also be obtained by applying [28, Equation(52)].Values of (cid:126)α for which g N ( (cid:126)α, L ) = 5 are surprisingly rare. Our first computer search,which took 1000 randomly and uniformly selected numbers (cid:126)α ∈ [0 , and checked g N ( (cid:126)α, Z ) for all N ≤ , found only five values of (cid:126)α for which there was an N with g N ( (cid:126)α, Z ) = 5.Nevertheless, as we will now see, every numerical example gives rise to a lower boundfor an infinite sequence of N and for almost every (cid:126)α . In what follows, we say a sequence N < N < N < . . . of integers is sub-exponential if(1.5) lim i →∞ N i +1 N i = 1 . Theorem 2.
Let L and L be unimodular lattices. There is a P ⊂ R d of full Lebesguemeasure, such that for every (cid:126)α ∈ P , (cid:126)α ∈ R d , and for every sub-exponential sequence ( N i ) i , we have (1.6) lim sup i →∞ g N i ( (cid:126)α, L ) ≥ sup N ∈ N g N ( (cid:126)α , L ) . In dimension d = 2, the choice (cid:126)α = (cid:0) , (cid:1) , N = 9 and L = Z (this is the examplein Figure 2, left) produces precisely five distinct distances given by √ ≈ . , √ ≈ . , √ ≈ . , √ ≈ . , √ ≈ . . ALAN HAYNES, JENS MARKLOF (cid:126)α = (0 . , . N = 9 (cid:126)α = (0 . , . N = 12 Figure 2.
The nearest neighbor graph for the Kronecker sequence n(cid:126)α + Z ( n = 1 , . . . , N ) in the torus R / Z (the unit square with opposite sidesidentified), for N = 9 (left) and N = 12 (right) and different choices of (cid:126)α . The vertex representing n(cid:126)α + Z is labeled by n and colored in pink.The blue directed edges point from a vertex to its nearest neighbour(s).The blue edge labels correspond to the indices of each of the five distinctdistances. Note that in each example there is a vertex with two nearestneighbours: vertex n = 5 on the left and n = 6 on the right.Moreover, for d = 3, (cid:126)α = ( , , ), N = 15 and L = Z (cf. Figure 3), we haveseven distinct distances, √ ≈ . , √ ≈ . , √ ≈ . , √ ≈ . , √ ≈ . , √ ≈ . , √ ≈ . . Applying these data to Theorem 2, combined with the upper bound of 5 in Theorem 1,we immediately obtain the following result.
Theorem 3.
Let d = 2 or . For any unimodular lattice L in R d , there is a set P ⊂ R d of full Lebesgue measure, such that for every (cid:126)α ∈ P , and for every sub-exponentialsequence ( N i ) i , we have that (1.7) lim sup i →∞ g N i ( (cid:126)α, L ) (cid:40) = 5 if d = 2 , ≥ if d = 3 . Thus the upper bound of 5 in Theorem 1 is indeed optimal for every L , almost every (cid:126)α and infinitely many N .In dimension d = 1 nearest neighbour distances do not necessarily coincide with theset of gap lengths, as gaps are the distance to the nearest neighbor in a fixed direction. FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 5
Figure 3.
The nearest neighbor graph for the Kronecker sequence n(cid:126)α + Z ( n = 1 , . . . ,
15) in the torus R / Z (the unit cube with opposite facesidentified), for (cid:126)α = ( , , ). The vertex representing n(cid:126)α + Z islabeled by n and colored in pink. The blue directed edges point from avertex to its nearest neighbour(s). The blue edge labels correspond to theindices of each of the seven distinct distances.To generalise this interpretation of a gap to higher dimensions, fix a subset D of theunit sphere S d − , and denote by δ n,N ( D ) the distance from ξ n to its nearest neighbor inthe direction of D . More precisely, denote by (cid:126)ξ n ∈ R d a fixed representative of the coset ξ n mod L so that ξ n = (cid:126)ξ n + L and define(1.8) δ n,N ( D ) = min {| (cid:126)ξ m − (cid:126)ξ n + (cid:126)(cid:96) | | (cid:126)ξ m − (cid:126)ξ n + (cid:126)(cid:96) ∈ R > D , ≤ m ≤ N, (cid:126)(cid:96) ∈ L} , where | · | is the standard Euclidean norm in R d . For D = S d − we recover the nearestneighbor distance(1.9) δ n,N ( S d − ) = δ n,N = min {| (cid:126)ξ m − (cid:126)ξ n + (cid:126)(cid:96) | > | ≤ m ≤ N, (cid:126)(cid:96) ∈ L} . In particular, note that the nearest neighbor of ξ n might be ξ n itself; in this case δ n,N = | (cid:126)(cid:96) | for suitable non-zero (cid:96) ∈ L .As an illustration consider dimension d = 1. The circle of radius one is S = {− , } and the choice D = { } produces the distance to the nearest neighbor to the right (whichleads to the classical three distance theorem). On the other hand D = {− , } yieldsthe distance to the nearest neighbor (in both directions). Since we are in dimension one,the set of distances to nearest neighbors is contained in the set of distances to nearestneighbors to the right, but it is not necessarily equal to it. However, for the case of theKronecker sequence, the three distance theorem also holds also for nearest neighbors.In other words, there are examples of α and N for which the number of distinct nearestneighbor distances is equal to three (e.g. take α = e and N = 5). ALAN HAYNES, JENS MARKLOF
The central object of our study is the number g N ( D ) of distinct nearest neighbordistances in direction D ⊆ S d − ,(1.10) g N ( D ) = |{ δ n,N ( D ) | ≤ n ≤ N }| , for d ≥
2. We will also write g N ( D ) = g N ( D , (cid:126)α, L ) to highlight the dependence onvector and lattice.We first present detailed results for dimension d = 2. In this case S d − is the unitcircle in R . The following theorem deals with the case when D ⊆ S is an interval ofarclength τ > π . (The case τ = 2 π has already been covered in Theorem 1 and weinclude it here for completeness.) Theorem 4.
Let d = 2 , and assume D ⊆ S is a half-open interval of arclength τ > π .Then for any unimodular lattice L , (cid:126)α ∈ R and N ∈ N we have that (1.11) g N ( D , (cid:126)α, L ) ≤ if τ = 2 π, if π/ < τ < π, if τ = 5 π/ , if π/ < τ < π/ , if τ = 4 π/ ,
12 + (cid:106) sin( τ/ π/ τ − π ) (cid:107) if π < τ < π/ . Related results in various settings have been obtained independently by Chevallier[9, 10, 12] and Vijay [41], but the precise description of the nearest neighbor problemwhich we give here does not appear to have been considered. However, the problemstudied by Vijay in his paper “Eleven Euclidean distances are enough” [41] is roughlycomparable to the special case of τ = 3 π/ D is essential. Notethat the upper bound for g N in the final case of Theorem 4 tends to infinity as τ → π + .In fact, in dimension d = 2 intervals D ⊂ S of lengths τ < π produce unboundednumbers of distinct distances, for almost every (cid:126)α . This is part of the content of thefollowing theorem, which also deals with analogous regions D in higher dimensions. Theorem 5.
Let d ≥ and L a unimodular lattice. There exists a set P ⊂ R d of full Lebesgue measure, such that for every D ⊂ S d − with non-empty interior andclosure contained in an open hemisphere, for every (cid:126)α ∈ P , and for every sub-exponentialsequence ( N i ) i , we have (1.12) sup i g N i ( D , (cid:126)α, L ) = ∞ , lim inf i g N i ( D , (cid:126)α, L ) < ∞ . Our final observation is that the finite distance phenomenon is recovered in anydimension, for general test sets D , if we impose Diophantine conditions on (cid:126)α . Wesay that (cid:126)α ∈ R d is badly approximable by Q L if there is a constant c > | n(cid:126)α − (cid:126)(cid:96) | ∞ > cn − /d for all (cid:126)(cid:96) ∈ L , n ∈ N . Here | · | ∞ denotes the maximum norm. FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 7
Theorem 6.
Let d ≥ , L a unimodular lattice and (cid:126)α ∈ R d badly approximable by Q L .For D ⊆ S d − with non-empty interior, we have that (1.13) sup N ∈ N g N ( D , (cid:126)α, L ) < ∞ . For comparison, the more precise bounds in Theorems 1 and 4 hold for all (cid:126)α , butonly for a restricted class of D .To relate the above Diophantine condition on (cid:126)α to the standard notion of badlyapproximable by Q d (which corresponds to the special case L = Z d ), take M ∈ SL( d, R )so that L = Z d M . We then see that (cid:126)α ∈ R d is badly approximable by Q L if and only if (cid:126)α = (cid:126)αM − is badly approximable by Q d . (The positive constants c appearing in bothdefinitions are not necessarily the same.) Furthermore, by Khintchine’s transferenceprinciple (see the Corollary to Theorem II in [7, Chapter V]), the vector (cid:126)α is badlyapproximable by Q d if and only if there is c > (cid:107) (cid:126)m · (cid:126)α (cid:107) R / Z > c | (cid:126)m | − d forall non-zero (cid:126)m ∈ Z d . Here (cid:107) x (cid:107) R / Z = min k ∈ Z | x + k | denotes the distance to the nearestinteger.The key strategy of the proofs of the above theorems is to express the quantities S N ( (cid:126)α ), δ n,N ( D ) and g N ( D ) in terms of functions on the space SL( d + 1 , Z ) \ SL( d + 1 , R )of unimodular lattices in R d +1 . This is explained in detail in Section 2. Once thisconnection is established, the proofs of Theorems 1 and 4 reduce to geometric argumentsinvolving lattices and sphere coverings, which are laid out in Sections 3-5 in dimension d = 2 and in Section 6 for dimensions d ≥
3. The proofs of Theorems 2, 5, and 6 requireupper and lower bounds for the relevant functions on the space of lattices, combinedwith the same ergodic-theoretic arguments used in [26]. This material is presented inSection 7.Acknowledgments: We would like to thank Nicolas Chevallier for helpful comments,and Felipe Ramirez and Carl Dettmann for discussions that led to an improvement ofour bounds in Theorem 1 in dimension d ≥
3. The images in Figures 2-7 were gener-ated using the computer software packages SageMath, Jmol, GeoGebra, and Inkscape.Finally, we would like to thank Timothy Haynes for his help in optimizing our Pythoncode, which aided in the discovery of the examples illustrated in Figures 2 and 3.2.
Reformulation in terms of lattices
This section follows the approach developed for the three gap theorem [31] and higherdimensional variants concerning gaps in values taken by linear forms, and hitting timesfor toral rotations [26].By substituting k = m − n in equation (1.8), we find that the distance from ξ n to itsnearest neighbor in the direction of D is given by δ n,N ( D ) = min {| k(cid:126)α + (cid:126)(cid:96) | | k(cid:126)α + (cid:126)(cid:96) ∈ R > D , − n < k ≤ N − n, (cid:126)(cid:96) ∈ L} = min {| k(cid:126)α + (cid:126)(cid:96) | | k(cid:126)α + (cid:126)(cid:96) ∈ R > D , − n < k < N + − n, (cid:126)(cid:96) ∈ L} , (2.1)where N + := N + . Select M ∈ SL( d, R ) so that L = Z d M , and let(2.2) A N ( (cid:126)α ) = A N ( (cid:126)α, L ) = (cid:18) M (cid:19) (cid:18) (cid:126)α d (cid:19) (cid:18) N − N /d d (cid:19) . ALAN HAYNES, JENS MARKLOF
Then we have that(2.3) δ n,N ( D ) = N − /d + min (cid:26) | (cid:126)v | (cid:12)(cid:12)(cid:12)(cid:12) ( u, (cid:126)v ) ∈ Z d +1 A N + ( (cid:126)α ) , − nN + < u < − nN + , (cid:126)v ∈ R > D (cid:27) , for all 1 ≤ n ≤ N . To cast this in a more general setting, let G = SL( d + 1 , R ) andΓ = SL( d + 1 , Z ). Then for general M ∈ G and t ∈ (0 , Q D ( M, t ) = (cid:8) ( u, (cid:126)v ) ∈ Z d +1 M (cid:12)(cid:12) − t < u < − t, (cid:126)v ∈ R > D (cid:9) and(2.5) F D ( M, t ) = min (cid:8) | (cid:126)v | (cid:12)(cid:12) ( u, (cid:126)v ) ∈ Q D ( M, t ) (cid:9) . In this notation it is clear that(2.6) δ n,N ( D ) = N − /d + F D (cid:18) A N + ( (cid:126)α ) , nN + (cid:19) . Before proceeding further, we first establish the following basic result.
Proposition 1. If D ⊆ S d − has non-empty interior, then F D is well-defined as afunction Γ \ G × (0 , → R > .Proof. We first show that the set Q D ( M, t ) is non-empty for all M ∈ G and t ∈ (0 , (cid:126)w ∈ D ◦ , and denote by Σ ⊥ ⊂ R d the ( d − (cid:126)w . Denote by (cid:126)v ⊥ the orthogonal projection of (cid:126)v to Σ ⊥ . Given ( M, t ), let (cid:15) > (cid:15) < min { t, − t } , (ii) there is no non-zero lattice pointin Z d +1 M within (cid:15) -distance to the origin. Furthermore, fix δ > { (cid:126)v ∈ R d | (cid:126)w · (cid:126)v > (cid:15), | (cid:126)v ⊥ | < δ } ⊂ R > D . Such a δ exists since (cid:126)w ∈ D ◦ . By construction(2.8) { ( u, (cid:126)v ) ∈ Z d +1 M | | u | < (cid:15), (cid:126)w · (cid:126)v > (cid:15), | (cid:126)v ⊥ | < δ } ⊂ Q D ( M, t )In view of Minkowski’s theorem, the symmetric, convex set { ( u, (cid:126)v ) ∈ R d +1 | | u | <(cid:15), | (cid:126)v ⊥ | < δ } contains a non-zero element of Z d +1 M . By construction there is no non-zero lattice point in Z d +1 M within (cid:15) -distance to the origin, and furthermore Z d +1 M is symmetric under reflection at the origin. We can therefore conclude that the set in(2.8), and therefore also Q D ( M, t ), is non-empty. Hence by the uniform discreteness of Z d +1 M the minimum value in the definition of F D exists.Finally, note that F D ( γM, t ) = F D ( M, t ) for γ ∈ Γ, and hence F is well-defined onΓ \ G × (0 , (cid:3) Note that for R ∈ SO( d ) we have(2.9) F D R (cid:18) M (cid:18) R (cid:19) , t (cid:19) = F D ( M, t ) . FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 9
Let us define(2.10) Q D ( M ) = (cid:91) t ∈ (0 , Q D ( M, t ) = (cid:8) ( u, (cid:126)v ) ∈ Z d +1 M (cid:12)(cid:12) | u | < , (cid:126)v ∈ R > D (cid:9) . The set(2.11) M D ( M ) = (cid:8) | (cid:126)v | (cid:12)(cid:12) ( u, (cid:126)v ) ∈ Q D ( M ) (cid:9) contains the set of values taken by the function t (cid:55)→ F D ( M, t ). It is a locally finitesubset of R > , i.e., there are at most finitely many points in any bounded interval. Itfollows that for fixed M , the function t (cid:55)→ F D ( M, t ) is piecewise constant.We denote by(2.12) G D ( M ) = |{ F D ( M, t ) | < t < }| the number of distinct values attained by the function t (cid:55)→ F D ( M, t ). For
N >
0, let(2.13) G D ,N ( M ) = |{ F D ( M, nN + ) | ≤ n ≤ N }| . We have G D ,N ( M ) ≤ G D ( M ), and so in particular(2.14) g N ( D ) = G D ,N ( A N + ( (cid:126)α )) ≤ G D ( A N + ( (cid:126)α )) . Geometric lemmas in dimension d = 2To fix notation for our subsequent discussion, we define a representative set of vectors( u i , (cid:126)v i ) ∈ Q D ( M ), for which the lengths | (cid:126)v i | are distinct, and each of which correspondsto an element in the set(3.1) F D ( M ) = { F D ( M, t ) | < t < } . To be specific, for each M ∈ Γ \ G we fix vectors ( u , (cid:126)v ) , . . . , ( u K , (cid:126)v K ) ∈ Q D ( M ) with K = G D ( M ), so that the following conditions hold:(V1) 0 < | (cid:126)v | < | (cid:126)v | < · · · < | (cid:126)v K | .(V2) For each δ ∈ F D ( M ) there exists an 1 ≤ i ≤ K such that δ = | (cid:126)v i | .(V3) For each 1 ≤ i ≤ K , there exists a t ∈ (0 ,
1) such that ( u i , (cid:126)v i ) ∈ Q D ( M, t ) and | (cid:126)v i | = F D ( M, t ).Let us now focus on the case d = 2. In the following we identify the unit circle S with the interval [ − π, π ) mod 2 π , so that 0 corresponds to direction (1 , ∈ S . In viewof the rotation invariance (2.9) we may assume without loss of generality that D ⊆ S is centered at θ = 0, i.e., D = [ − τ / , τ / S (not R ).Our proof of Theorem 4, or rather the proofs of the more general Theorems 7 and8 below, will be divided into three main cases, which together cover all possible angles τ ∈ ( π, π ] described in (1.11). In each of these cases we will partition D into subsets,consisting of a symmetric set S ⊂ D (symmetric with respect to the rotation θ (cid:55)→ θ + π mod 2 π ) and up to three asymmetric subsets. For notational convenience, let usset(3.2) ψ = 2 π − τ ∈ [0 , π ) and φ = τ − π ∈ (0 , π ] . First we specify our definitions of the asymmetric subsets in each of the three maincases.
Case (C1):
If 5 π/ ≤ τ ≤ π then we define one asymmetric subset A ⊂ D by(3.3) A = [ − ψ/ , ψ/ . (Note that this is the empty set if τ = 2 π .) Case (C2):
If 4 π/ ≤ τ < π/ A − and A by(3.4) A − = [ − ψ/ , , A = [0 , ψ/ . Case (C3): If π < τ < π/ A − , A , and A by(3.5) A − = [ − ψ/ , − π/ , A = [ − π/ , π/ , A = [ π/ , ψ/ . In all three cases, we define the symmetric subset S by(3.6) S = [ − τ / , − ψ/ ∪ [ ψ/ , τ / . It is clear that S is the largest symmetric subset of D , that D is the disjoint union of S and its asymmetric subsets, and that each asymmetric subset is a half-open interval oflength at most π/ d ≥ Proposition 2. If d ≥ and if the angle between two non-zero vectors (cid:126)w , (cid:126)w ∈ R d isless than π/ , then (3.7) | (cid:126)w − (cid:126)w | < max {| (cid:126)w | , | (cid:126)w |} . Furthermore, this inequality also holds if the angle between (cid:126)w , (cid:126)w ∈ R d is equal to π/ ,as long as | (cid:126)w | (cid:54) = | (cid:126)w | .Proof. For the first part of the proposition, suppose without loss of generality that | (cid:126)w | ≤ | (cid:126)w | . Then the vectors (cid:126)w / | (cid:126)w | and (cid:126)w / | (cid:126)w | lie on the unit sphere and areseparated by an angle less than π/
3. Therefore(3.8) (cid:12)(cid:12)(cid:12)(cid:12) (cid:126)w | (cid:126)w | − (cid:126)w | (cid:126)w | (cid:12)(cid:12)(cid:12)(cid:12) < , and the result follows. Furthermore, under the assumptions of the second part of theproposition, we draw the same conclusion. (cid:3) Next we will prove several propositions which place various restrictions on the integer K = G D ( M ) defined at the start of this section; recall (V1)–(V3). Proposition 3.
If for some integer ≤ i ≤ K , we have that u i ∈ ( − / , / and (cid:126)v i ∈ R > S , then we must have that i = K . FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 11
Proof.
Suppose first that u i ∈ [0 , / < t < − u i , we havethat ( u i , (cid:126)v i ) ∈ Q D ( M, t ) and thus | (cid:126)v i | ≥ F D ( M, t ). By the symmetry of S we have( − u i , − (cid:126)v i ) ∈ Q D ( M ). Thus for any u i < t <
1, we have that ( − u i , − (cid:126)v i ) ∈ Q D ( M, t ) andthus | (cid:126)v i | ≥ F D ( M, t ). Since u i < /
2, we conclude that | (cid:126)v i | ≥ F D ( M, t ) for all 0 < t < i = K . The case u i ∈ ( − / ,
0] follows from the same argument. (cid:3)
Proposition 4.
If for some i and j with ≤ i, j ≤ K, we have that u i ∈ ( − / , and u j ∈ [0 , / , then i = K or j = K .Proof. Under the hypotheses of the proposition, we have | (cid:126)v i | ≥ F D ( M, t ) for − u i < t < | (cid:126)v j | ≥ F D ( M, t ) for 0 < t < − u j . This covers all possible values of t ∈ (0 , i or j must equal K . (cid:3) Proposition 5. If ≤ i, j ≤ K and u i = u j , then i = j .Proof. Suppose by way of contradiction that i (cid:54) = j , and without loss of generality that i < j . Then for any t ∈ (0 ,
1) satisfying − t < u i = u j ≤ − t , we would have by (V1)that(3.9) F D ( M, t ) ≤ | (cid:126)v i | < | (cid:126)v j | . However this contradicts condition (V3), so we must have that i = j . (cid:3) Proposition 6.
Let ≤ i ≤ K and ( u, (cid:126)v ) ∈ Q D ( M ) . If (3.10) − < u i ≤ u ≤ or ≤ u ≤ u i < then | (cid:126)v i | ≤ | (cid:126)v | .Proof. Suppose that 0 ≤ u ≤ u i <
1; the other case follows by symmetry. By (V3)there exist t i ∈ (0 , − u i ) such that F D ( M, t i ) = | (cid:126)v i | . Furthermore F D ( M, t ) ≤ | (cid:126)v | forall t ∈ (0 , − u ). Thus, taking t = t i ∈ (0 , − u ), we have | (cid:126)v i | ≤ | (cid:126)v | . (cid:3) Proposition 7.
Let ≤ i, j ≤ K . If (3.11) − < u i < u j ≤ or ≤ u j < u i < then | (cid:126)v i | < | (cid:126)v j | and i < j .Proof. In view of (V1), this is a direct consequence of Proposition 6. (cid:3)
Proposition 8.
Let τ ≥ π , ≤ j ≤ K and ( u, (cid:126)v ) ∈ Q D ( M ) such that (cid:126)v j (cid:54) = (cid:126)v and < | (cid:126)v | ≤ | (cid:126)v j | . Suppose the angle between the vectors (cid:126)v j and (cid:126)v is less than π/ . If (3.12) − < u ≤ u j ≤ − / or / ≤ u j ≤ u < , then j = K .Proof. We consider the case 1 / ≤ u j ≤ u <
1; the proof for the alternative followsfrom the same argument by symmetry. By the assumption on the angle between thevectors (cid:126)v j and (cid:126)v , we have by Proposition 2 that | (cid:126)v j − (cid:126)v | < | (cid:126)v j | . Since (cid:126)v j (cid:54) = (cid:126)v and τ ≥ π ,we have that at least one of (cid:126)v j − (cid:126)v , (cid:126)v − (cid:126)v j is in R > D .First suppose that (cid:126)v − (cid:126)v j ∈ R > D . Then, since0 ≤ u − u j < / ≤ u j < , we have ( u − u j , (cid:126)v − (cid:126)v j ) ∈ Q D ( M ) and by Proposition 6 that | (cid:126)v − (cid:126)v j | ≥ | (cid:126)v j | . This is acontradiction, so we conclude that (cid:126)v − (cid:126)v j / ∈ R > D .The only other possibility is that (cid:126)v j − (cid:126)v ∈ R > D . In this case, ( u j − u, (cid:126)v j − (cid:126)v ) ∈ Q D ( M ) , (3.13) u j − u ≤ u − u j < − u j . It follows from this that, for u − u j < t <
1, we have F D ( M, t ) ≤ | (cid:126)v − (cid:126)v j | < | (cid:126)v j | andfor 0 < t < − u j (which in particular holds for all t with 0 < t ≤ u − u j ), we have F D ( M, t ) ≤ | (cid:126)v j | . Therefore j = K . (cid:3) Proposition 9.
Let τ ≥ π and ≤ i, j ≤ K . Suppose the angle between the vectors (cid:126)v i and (cid:126)v j is less than π/ . If (3.14) − < u i < u j ≤ − / or / ≤ u j < u i < , then j = K .Proof. This is a direct corollary of Proposition 8 (take u = u i ). (cid:3) Proposition 10.
Let τ ≥ π and ≤ i, j ≤ K . Suppose the angle between the vectors (cid:126)v i and (cid:126)v j is less than π/ . If (3.15) − / < u i < u j ≤ or ≤ u j < u i < / , then (cid:126)v j − (cid:126)v i / ∈ R > D and (3.16) | (cid:126)v i | ≤ | (cid:126)v i − (cid:126)v j | < | (cid:126)v j | . Proof.
We assume 0 ≤ u j < u i < /
2; the other case follows by symmetry. It followsfrom Proposition 7 that i < j and | (cid:126)v i | < | (cid:126)v j | , and it follows from Proposition 2 that | (cid:126)v i − (cid:126)v j | < | (cid:126)v j | .Suppose, contrary to what we are trying to prove, that (cid:126)v j − (cid:126)v i ∈ R > D . Then, since0 < u i − u j < /
2, we have ( u j − u i , (cid:126)v j − (cid:126)v i ) ∈ Q D ( M ) and hence for u i − u j < t < F D ( M, t ) ≤ | (cid:126)v i − (cid:126)v j | < | (cid:126)v j | . Furthermore, for 0 < t < − u i we have F D ( M, t ) ≤| (cid:126)v i | < | (cid:126)v j | . Now u i − u j < − u i , so F D ( M, t ) < | (cid:126)v j | for all t ∈ (0 , t ∈ (0 ,
1) such that F D ( M, t ) = | (cid:126)v j | . This is a contradiction, so we concludethat (cid:126)v j − (cid:126)v i / ∈ R > D .It remains to show that | (cid:126)v i | ≤ | (cid:126)v i − (cid:126)v j | . Since (cid:126)v j − (cid:126)v i / ∈ R > D and τ ≥ π we have (cid:126)v i − (cid:126)v j ∈ R > D . Then ( u i − u j , (cid:126)v i − (cid:126)v j ) ∈ Q D ( M ) with 0 < u i − u j < /
2. Thusfor 0 < t < − ( u i − u j ) we have F D ( M, t ) ≤ | (cid:126)v i − (cid:126)v j | . Note that by (V3) thereexists a t i ∈ (0 , − u i ) such that F D ( M, t i ) = | v i | . Now 1 − u i ≤ − ( u i − u j ) and so F D ( M, t i ) = | v i | ≤ | (cid:126)v i − (cid:126)v j | as needed. (cid:3) The previous proposition will be used in the proof of Theorem 7 in conjunction withthe following two elementary geometric propositions.
Proposition 11.
Suppose that • we are in case (C1) or (C3) and (cid:126)w , (cid:126)w ∈ R > A , or • we are in case (C2) and (cid:126)w , (cid:126)w ∈ R > A − , or • we are in case (C2) and (cid:126)w , (cid:126)w ∈ R > A . FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 13 If (3.17) | (cid:126)w | ≤ | (cid:126)w − (cid:126)w | < | (cid:126)w | , then (cid:126)w − (cid:126)w ∈ R > D .Proof. First suppose that we are in case (C2), that (cid:126)w and (cid:126)w point in direction A , and that (3.17) holds. We will argue using Figure 4. Figure 4.
Diagram of A in case (C2)In the figure, angle AOB measures ψ/ A , and the line through Hand C is parallel to the x -axis. The vector (cid:126)w is shown, and the angles HCF and HCGalso measure ψ/
2. By condition (3.17), vector (cid:126)w has to lie outside of both the circleof radius | (cid:126)w | centered at O, and the circle of radius | (cid:126)w | centered at C. The circle ofradius | (cid:126)w | centered at C intersects the boundary of R > A at the three points D, O,and E, and the points F and G have also been chosen so that they lie on this circle.We will show (as indicated in the figure) that F and G lie outside of R > A . This willcomplete the proof in this sub-case since, if (cid:126)w − (cid:126)w were not in R > D then (cid:126)w wouldhave to lie in the cone swept out by angle FCG, above the ray originating from C andpassing through G, and on or below the ray originating from C and passing through F.This, together with the condition that it lies outside of the circle of radius | (cid:126)w | centeredat C, would force it to lie outside of R > A , which is contradictory to our hypotheses.It is clear from the fact the ψ/ ≤ π/ x -axis, so it cannot bein R > A (in fact we only need ψ/ < π/ R > A , first note that angle OCH has measure θ , from which it follows that angle CIOhas measure π − ψ (to avoid circular reasoning, the point I is defined as the intersection of the line through C and F with the line through O and A). On the other hand, angleCDO has measure ψ/ − θ and, since 3 ψ/ ≤ π , we have that(3.18) ψ/ − θ ≤ π − ψ. This implies that the point F lies on or to the left of the line through O and D, thereforeit is not in R > A . The proof for case (C2) when (cid:126)w and (cid:126)w point in direction A − follows by symmetry.Next suppose that we are in case (C1), that (cid:126)w and (cid:126)w point in direction A , andthat (3.17) holds. Here the proof is similar, and we will argue using Figure 5. Figure 5.
Diagram of A in case (C1)First assume that 0 ≤ θ < ψ/
2. Since ψ/ ≤ π/
6, the argument given above impliesagain that F lies outside of R > A . Angle CIO has measure π − ψ , while angle CEOhas measure ψ/ θ , and since(3.19) ψ/ θ < π − ψ, this implies that G lies outside of R > A . This argument actually works for all θ and ψ satisfying 0 ≤ θ < ψ/ ≤ π/ , and a symmetrical argument applies when − π/ ≤ − ψ/ < θ <
0. The proof for case (C3), when (cid:126)w and (cid:126)w point in direction A , follows from the same argument. (cid:3) Proposition 12.
Suppose that we are in case (C3) and that (cid:126)w , (cid:126)w , . . . , (cid:126)w n are anyvectors which all point in direction A − , or which all point in direction A . If, for each ≤ i ≤ n − , we have that (cid:126)w i +1 − (cid:126)w i (cid:54)∈ R > D and that (3.20) | (cid:126)w i | ≤ | (cid:126)w i − (cid:126)w i +1 | < | (cid:126)w i +1 | , FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 15 then we must have that (3.21) n ≤ (cid:22) sin( τ / π/ τ − π ) (cid:23) . Proof.
Suppose (cid:126)w , (cid:126)w , . . . , (cid:126)w n point in direction A and consider Figure 6. Figure 6.
Diagram of A in case (C3)In the figure, angle AOB measures ψ/ − π/ < π/ A , and theline through G and C is parallel to the x -axis. For each 1 ≤ i ≤ n −
1, we have that (cid:126)w i +1 − (cid:126)w i (cid:54)∈ R > D and(3.22) | (cid:126)w i +1 − (cid:126)w i | ≥ | (cid:126)w | , so it follows that(3.23) n − ≤ (cid:22) | CF || (cid:126)w | (cid:23) . Line segment CF is longest when θ = 0, which gives the bound(3.24) | CF | ≤ | (cid:126)w | sin( ψ/ − π/ π − ψ ) . Substituting ψ = 2 π − τ gives(3.25) | CF || (cid:126)w | ≤ sin( τ / π/ τ − π ) . Combining this with (3.23) completes the proof of the proposition. (cid:3)
Finally, to obtain the bounds reported in some of the cases of Theorem 4, we willneed to gather together a few more facts.
Proposition 13.
Let ≤ i, j ≤ K , i (cid:54) = j , and let (cid:126)v j ∈ R > S . If (3.26) − < u i ≤ − u j ≤ or ≤ − u j ≤ u i < , then | (cid:126)v i | < | (cid:126)v j | and i < j .Proof. The vector − (cid:126)v j is in S , and hence ( − u j , − (cid:126)v j ) ∈ Q D ( M ). Proposition 6 thenyields the statement. (cid:3) Proposition 14.
Let ≤ i, j ≤ K , and let (cid:126)v j ∈ R > S . Assume the angle between (cid:126)v i and − (cid:126)v j is less than π/ . If (3.27) − < u i ≤ − u j ≤ − / or / ≤ − u j ≤ u i < , then j = K .Proof. The proof is similar to that of Proposition 8, which would directly apply if wehad assumed (cid:126)v i ∈ R > S rather than (cid:126)v j ∈ R > S .The assumption on the angle implies i (cid:54) = j . We consider the case 1 / ≤ − u j ≤ u i < (cid:126)v i and − (cid:126)v j , we have by Proposition 2 that | (cid:126)v i + (cid:126)v j | < | (cid:126)v j | . Since (cid:126)v i (cid:54) = (cid:126)v j and τ ≥ π , we have that at least one of ± ( (cid:126)v i + (cid:126)v j ) is in R > D . Suppose (cid:126)v i + (cid:126)v j ∈ R > D .Then, since − ( u j , (cid:126)v j ) ∈ Q D ( M ) and0 ≤ u i + u j < / ≤ − u j < , we have ( u i + u j , (cid:126)v i + (cid:126)v j ) ∈ Q D ( M ) and by Proposition 6 that | (cid:126)v i + (cid:126)v j | ≥ | (cid:126)v j | , acontradiction. Therefore (cid:126)v i + (cid:126)v j / ∈ R > D and we must have − ( (cid:126)v i + (cid:126)v j ) ∈ R > D . Thismeans − ( u i + u j , (cid:126)v i + (cid:126)v j ) ∈ Q D ( M ) and(3.28) 0 ≤ u i + u j < u j . It follows from this that, for all t with u i + u j < t <
1, we have F D ( M, t ) ≤ | (cid:126)v i + (cid:126)v j | < | (cid:126)v j | and for 0 < t < u j (which in particular holds for all t with 0 < t ≤ u i + u j ), we have F D ( M, t ) ≤ | (cid:126)v j | . Therefore j = K . (cid:3) The previous proposition allows us to deduce the following simple and useful result.Recall that φ = τ − π , so that τ = ψ + 2 φ . Proposition 15.
Let S denote the number of integers i with ≤ i ≤ K, (cid:126)v i ∈ R > S ,and u i ∈ ( − , − / ∪ [1 / , . Then (3.29) S ≤ (cid:108) φπ/ (cid:109) if (cid:126)v K ∈ R > S and u K ∈ ( − , − / ∪ [1 / , , (cid:108) φπ/ (cid:109) otherwise.Proof. The quantity φ is the angle swept out by the part of S which lies above the x -axis. The maximum number of vectors which can be placed in this region, so thatthe angles between any two vectors is at least π/
3, is (cid:100) φ/ ( π/ (cid:101) . The upper bound in(3.29) therefore follows from combining the results of Propositions 9 and 14. (cid:3) FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 17 Explicit upper bounds in dimension d = 2 , part 1 Throughout this section we take d = 2. In view of (2.14), the following statementdirectly implies all cases of Theorem 4, except the case when D = S (which is handledin the next section). Theorem 7.
Let d = 2 , and assume D ⊆ S is a half-open interval of arclength τ > π .Then for any M ∈ SL(3 , R ) we have that (4.1) G D ( M ) ≤ if π/ < τ < π, if τ = 5 π/ , if π/ < τ < π/ , if τ = 4 π/ ,
12 + (cid:106) sin( τ/ π/ τ − π ) (cid:107) if π < τ < π/ . The remainder of this section is dedicated to the proof of this theorem. For theproof, we will apply the propositions from the previous section to each of the five casesdescribed in (4.1). To summarize the main points of our arguments:(i) Let S denote the number of 1 ≤ i ≤ K with u i ∈ ( − / , /
2) and (cid:126)v i ∈ R > S .Proposition 3 guarantees that S ≤
1, and that if S = 1 then the correspondingvalue of i equals K .(ii) Let S denote the number of 1 ≤ i ≤ K with u i ∈ ( − , − / ∪ [1 / ,
1) and (cid:126)v i ∈ R > S . Proposition 15 gives an upper bound for S .(iii) Let A denote the number of 1 ≤ i ≤ K with u i ∈ ( − / , /
2) and with (cid:126)v i indirection of any asymmetric subset. Propositions 4 and 10-12 give upper boundsfor A + S .(iv) Let A denote the number of 1 ≤ i ≤ K with u i ∈ ( − , − / ∪ [1 / ,
1) and with (cid:126)v i in direction of any asymmetric subset. Propositions 5-9 give upper bounds for A + S .In all cases, we have that G D ( M ) = K = S + A + S + A . In what follows, recall that ψ = 2 π − τ and φ = τ − π . Case (C1) , τ = 2 π : This is actually not one of the cases considered in Theorem 7,but we include it as a demonstration of the proof technique, and to provide an easyargument that G D ( M ) ≤
6. This bound will be improved in the next section, by aslightly more complicated argument, to show that G D ( M ) ≤ ψ = 0 and φ = π , so it is clear that A = A = 0. We claim that S ≤ S ≥
6. Since τ = 2 π , we may assume withoutloss of generality that the vectors ( u i , (cid:126)v i ) have been chosen so that u i ∈ [0 ,
1) for each i . Then we can find 1 ≤ i < j ≤ K with 1 / ≤ u i , u j <
1, and for which the anglebetween (cid:126)v i and (cid:126)v j is less than or equal to π/
3. By the second part of Proposition 2, wethen have that(4.2) | (cid:126)v i − (cid:126)v j | < | (cid:126)v j | . However, since | u i − u j | < / (cid:126)v i − (cid:126)v j ∈ R > S , this implies that(4.3) F D ( M, t ) ≤ | (cid:126)v i − (cid:126)v j | < | (cid:126)v j | , for all t ∈ (0 , S ≤
5. Since S ≤ , this gives the bound G D ( M ) ≤ Case (C1) , 5 π/ < τ < π : In this case 0 < ψ < π/ π/ < φ < π . There isonly one asymmetric subset, namely A .A. Assume S = 1. Then by Proposition 3 we must have that(4.4) u K ∈ ( − / , / . Proposition 4 implies that we cannot have i and j with 1 ≤ i, j < K, u i ∈ ( − / , , and u j ∈ [0 , / . Therefore, by Propositions 10 and 11, we have that A ≤ (cid:100) φπ/ (cid:101) = 3 and (4.4) holds, Proposition 15 implies that S ≤ . We claimthat, in this sub-case, S + A ≤ . It is clear from Proposition 9 that, since (4.4)holds, we must have that A ≤
2, and that if A = 2 then one element has its firstcomponent in ( − , − /
2] and the other in [1 / , S ≥
5. If S ≥ ≤ i < K with u i ∈ ( − , − /
2] and (cid:126)v i ∈ R > S , or atleast three values with u i ∈ [1 / ,
1) and (cid:126)v i ∈ R > S (or possibly both, if S = 6).The argument in both cases is the same, so suppose without loss of generality thatthe former condition holds. Then, there are at least 2 values 1 ≤ i < j < K with u i , u j ∈ ( − , − / , (cid:126)v i , (cid:126)v j ∈ R > S , and with (cid:126)v i and (cid:126)v j either both above or bothbelow the x -axis. If there were also a value of 1 ≤ k ≤ K with u k ∈ ( − , − / (cid:126)v k ∈ A then, since φ + ψ = π , at least one pair of the vectors (cid:126)v i , (cid:126)v j , and (cid:126)v k would be separated by an angle of less than π/
3. This, together with Proposition9, would contradict (4.4), therefore such a k cannot exist. This means A ≤ , and S + A ≤ G D ( M ) ≤ S = 0. Then by Propositions 10 and 11, we have that A ≤
2. If A = 2then by Proposition 4 we again have that (4.4) holds. The same argument as in theprevious Case A yields S + A ≤ G D ( M ) ≤ A ≤
1. Since (cid:100) φπ/ (cid:101) = 3 we have by Proposition 15 that S ≤
7, andthe case S = 7 can only arise if (cid:126)v K ∈ R > S and u K ∈ ( − , − / ∪ [1 / , S = 7. Then there exist vectors (cid:126)v i , . . . , (cid:126)v i ∈ R > S with indices(4.5) 1 ≤ i < i < · · · < i < K chosen so that the vectors have the smallest possible lengths. There must be atleast 2 of these vectors, say (cid:126)v i and (cid:126)v j , which both lie above or below the x -axis,and with corresponding u i values either both in ( − , − /
2] or both in [1 / , ≤ k ≤ K with u k ∈ ( − , − /
2] and (cid:126)v k ∈ A then, since φ + ψ = π , at least one pair of the vectors (cid:126)v i , (cid:126)v j , and (cid:126)v k would beseparated by an angle of less than π/
3. Proposition 9 implies that one of the
FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 19 indices has to be equal to K . Since i, j < K , we have k = K , contradicting thefact that (cid:126)v K ∈ R > S . We have therefore A ≤
1, so G D ( M ) ≤ S = 5 or 6. In this case we claim that A ≤
2. To see why this istrue, suppose that A ≥
3. Then, by Proposition 9, we have 1 ≤ i < j < K suchthat (cid:126)v i , (cid:126)v j , (cid:126)v K ∈ A and u i , u j , u K / ∈ ( − / , /
2) with u i and u j having oppositesign. Since S ≥
5, there are at least five values of (cid:96) with 1 ≤ (cid:96) < K, (cid:126)v (cid:96) ∈ R > S ,and u (cid:96) / ∈ ( − / , / x -axis, and have u (cid:96) values bothin ( − , − /
2] or both in [1 / , i or j, or one of these (cid:96) values, equals K , which is a contradiction.Therefore A ≤ S ≤ G D ( M ) ≤ S ≤
4. By Proposition 9 we have that A ≤
3, so G D ( M ) ≤ Case (C1) , τ = 5 π/
3: In this case ψ = π/ φ = 2 π/
3, and there is only oneasymmetric subset, namely A . In this case the argument is similar to the previouscase. The key improvement is in the application of Proposition 15, since now (cid:100) φπ/ (cid:101) = 2.A. If S = 1 then (4.4) holds and we conclude as in the previous Case A that A ≤ ,S ≤
4. Proposition 9 and (4.4) imply that A ≤
2, and so G D ( M ) ≤ S = 0 then A ≤ A = 2 then (4.4) holds, so S ≤ G D ( M ) ≤ A ≤
1. Proposition 15 gives S ≤ . If S = 5 then, by the argument inCase B (b) above, A ≤
2. We conclude G D ( M ) ≤ S ≤ A ≤ G D ( M ) ≤ Case (C2) , 4 π/ < τ < π/
3: In this case π/ < ψ < π/ π/ < φ < π/ A − and A . The argument here is similar to thatgiven above, except that now when we use Propositions 10 and 11, they must be appliedto both asymmetric subsets.A. Assume S = 1. Then (4.4) holds, and Propositions 10 and 11 imply that A ≤ (cid:100) φπ/ (cid:101) = 2 and thus Proposition 15 yields S ≤ S ≥ , then A ≤
2. Suppose by way of contradiction that S ≥
3, that 1 ≤ i, j, k < K, u i , u j , u k / ∈ ( − / , / , and that (cid:126)v i , (cid:126)v j , (cid:126)v k aredistinct vectors in direction A − ∪ A . It could not be the case that either u i , u j , u k ∈ ( − , − /
2] or that u i , u j , u k ∈ [1 / , i, j, or k equals K . Therefore two of the numbers u i , u j , and u k lie in one of the intervals ( − , − /
2] or [1 / , u i , u j ∈ ( − , − /
2] and that u k ∈ [1 / , ≤ (cid:96) < K with u (cid:96) ∈ ( − , − /
2] and (cid:126)v (cid:96) ∈ R > S . If there were then, again since ψ + φ = π , at least one pair of thethree vectors (cid:126)v i , (cid:126)v j , and (cid:126)v (cid:96) would be separated by an angle of less than π/ giving the contradiction that i, j, or (cid:96) equals K . This means that there are atleast three values of 1 ≤ (cid:96) < K for which u (cid:96) ∈ [1 / ,
1) and (cid:126)v (cid:96) ∈ R > S . At leasttwo of these vectors lie either above or below the x -axis. Since u k is also in[1 / , u k , or one ofthese u (cid:96) values, is u K . This is also a contradiction, so we conclude that A ≤ . Putting this all together, we have that G D ( M ) ≤ S = 1 or 2, then we must have that A ≤
3. To see why, suppose A ≥ ≤ i < i < i < i < K so that u i , . . . , u i / ∈ ( − / , / (cid:126)v i , . . . , (cid:126)v i / ∈ S . By the same argument as before, we cannot have threeof these u i values in either ( − , − /
2] or in [1 / , ≤ (cid:96) < K with u (cid:96) / ∈ ( − / , /
2) and (cid:126)v (cid:96) ∈ R > S , and as before, this implies that one of the vectors we have justlisted is (cid:126)v K ; a contradiction with (4.4). Therefore A ≤
3. This shows that G D ( M ) ≤ S = 0. By Proposition 9, we have that A ≤
5, and so G D ( M ) ≤ S = 0. Then A ≤ A = 4 contradictsProposition 4, so in fact A ≤
3. If A = 3 then by Proposition 4 we have that(4.4) holds and, for the problem of bounding A + S , we are in the same positionas we just were in Case A. By exactly the same arguments, we therefore have that A + S ≤
6, and that G D ( M ) ≤ A ≤ S ≤ S = 5 then (cid:126)v K ∈ R > S , and we claim that A ≤
2. To see why, suppose that1 ≤ u < · · · < u < K are chosen so that (cid:126)v i , . . . , (cid:126)v i are in direction S . Supposethat at least three of these vectors all have their u i values either in ( − , − /
2] orin [1 / ,
1) and without loss of generality, suppose these values lie in ( − , − / x -axis, there cannot be a value of 1 ≤ j < K with u j ∈ ( − , − /
2] and with (cid:126)v j / ∈ R > S . Furthermore, by Proposition 9, there can be at most two valuesof 1 ≤ j < K with u j ∈ [1 / ,
1) and (cid:126)v j / ∈ S , which gives A ≤
2. The otherpossibility is that two of values of u i , . . . , u i lie in ( − , − / / , ≤ j < K with (cid:126)v j / ∈ R > S and with u j ∈ ( − , − / u j ∈ [1 / , A ≤
2. This showsthat if A ≤ S = 5 then G D ( M ) ≤ S = 3 or 4 then we claim that A ≤
3. To see why this is true, suppose byway of contradiction that S = 3 or 4 and that (cid:126)v i , . . . , (cid:126)v i / ∈ R > S are distinctvectors with corresponding u i values all in ( − , − / ∪ [1 / , − , − /
2] or all in [1 / , − , − / i , i , and i andthat the largest of these indices is i . We must then have that i = K . Wehave that u i lies in [1 / , FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 21 values of 1 ≤ (cid:96) < K with u (cid:96) ∈ [1 / ,
1) and (cid:126)v (cid:96) ∈ R > S . Then, there must beat least one value of (cid:96) with u (cid:96) ∈ ( − , − /
2] and (cid:126)v (cid:96) ∈ R > S . However, we thenconclude by previous arguments that either this u (cid:96) value, or one of u i or u i , must equal u K . This is a contradiction. We are left with the possibility thattwo of the numbers u i , . . . , u i lie in ( − , − / / , ≤ (cid:96) ≤ K with (cid:126)v (cid:96) ∈ R > S and u (cid:96) ∈ ( − , − / (cid:126)v (cid:96) ∈ R > S and u (cid:96) ∈ [1 / , S = 3 or 4, so we concludethat in this case A ≤
3. This gives that G D ( M ) ≤ S ≤ A ≤ G D ( M ) ≤ . Case (C2) , τ = 4 π/
3: In this case ψ = 2 π/ φ = π/
3, and so (cid:100) φπ/ (cid:101) = 1. Thereare two asymmetric cones, A − and A .A. Assume S = 1. Then (4.4) holds, and as in the previous Case A we have A ≤ S ≤
2. If S = 2 then choose 1 ≤ i, j ≤ K − i (cid:54) = j and (cid:126)v i , (cid:126)v j ∈ R > S . If u i and u j both lie in ( − , − /
2] then there can be at most one value of 1 ≤ k ≤ K − (cid:126)v k (cid:54)∈ R > S and with u k ∈ ( − , − / u k ∈ [1 / , A ≤
3. Similarly if u i and u j both lie in [1 / , u i ∈ ( − , − /
2] and u j ∈ [1 / , ≤ k ≤ K − (cid:126)v k (cid:54)∈ R > S and with u k ∈ ( − , − / u k ∈ [1 / , A ≤
2. This gives the bound G D ( M ) ≤ . If S = 0 or 1 we use the bound A ≤
4, which follows from Proposition 9 and (4.4).Thus G D ( M ) ≤ S = 0. Then A ≤
3. If A = 3 then (4.4) holds and, for the problem ofbounding A + S , we may follow exactly the same arguments just used in Case Ato obtain the bound A + S ≤
5. Therefore G D ( M ) ≤ A ≤ S ≤ S = 3 then A ≤ G D ( M ) ≤ S = 1 or 2 then A ≤ G D ( M ) ≤ S = 0 then A ≤ G D ( M ) ≤ Case (C3) , π < τ < π/
3: In this case 2 π/ < ψ < π and 0 < φ < π/
3. Now thereare three asymmetric subsets, A − , A , and A . The argument is similar to those givenpreviously, except that to bound the number of vectors (cid:126)v i ∈ A with u i ∈ ( − / , / S = 1 then (4.4) holds, and Propositions 10-12 imply that(4.6) A ≤ (cid:22) sin( τ / π/ τ − π ) (cid:23) . Since (cid:100) φπ/ (cid:101) = 1, Proposition 15 implies that S ≤
2. Also, Proposition 9 (appliedto each of the three asymmetric sets) implies that A ≤
6, so we have that(4.7) G D ( M ) ≤
12 + (cid:22) sin( τ / π/ τ − π ) (cid:23) . B. Finally, suppose that S = 0. In this case, again by Propositions 10-12, there canbe at most(4.8) 3 + (cid:22) sin( τ / π/ τ − π ) (cid:23) indices 1 ≤ i ≤ K with (cid:126)v i in an asymmetric set and with u i ∈ [0 , / (cid:126)v j in an asymmetric set, with u j ∈ ( − / , j = K . Therefore therecan be at most one such vector (cid:126)v j . The same argument applies with the intervals[0 , /
2) and ( − / ,
0) replaced by ( − / ,
0] and (0 , / A ≤ (cid:22) sin( τ / π/ τ − π ) (cid:23) . If equality holds in this inequality then we know that (4.4) holds, we are in thesame situation just encountered in Case A, and we again use the bounds S ≤ A ≤
6, arriving at the same bound (4.7) for G D ( M ).If there is strict inequality in (4.9) then Proposition 15 implies that S ≤ , and italso tells us that if S = 3 then (cid:126)v K lies in a symmetric set. Similarly, we have fromProposition 15 that A ≤
7, and that if A = 7 then (cid:126)v K lies in an asymmetric set.Therefore we cannot have both S = 3 and A = 7. This gives that S + A ≤ Explicit upper bounds in dimension d = 2 , part 2 Throughout this section we set d = 2 and D = S . In this case the upper bound of 6obtained in the previous section falls just short of establishing the claimed five distancetheorem. We will first deduce a little more information about the possible distanceswhich can occur in this case. The goal of this section is to prove the following theorem,which will thereby complete the proof of Theorem 4 and the d = 2 case of Theorem 1. Theorem 8.
For any M ∈ SL ( R ) we have that (5.1) G S ( M ) ≤ . Since we are dealing with the case D = S we can assume, by replacing each vector( u i , (cid:126)v i ) by its negative if necessary, that u i ≥ ≤ i ≤ K . For simplicity we makethis assumption for the duration of this section. It is convenient at this point to gathertogether some additional properties which must be satisfied by the vectors (cid:126)v i . Proposition 16. If G S ( M ) = K then, for all ≤ i < j ≤ K − , | (cid:126)v i − (cid:126)v j | > | (cid:126)v j | , (5.2) FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 23 and for all ≤ i < j < k ≤ K − , | (cid:126)v i − (cid:126)v j − (cid:126)v k | ≥ | (cid:126)v k | . (5.3) Proof.
Let 1 ≤ i < j ≤ K −
1. To prove (5.2), first note that(5.4) 0 < u i − u j < / . Since G S ( M ) = K we must have that u j ≥ /
2. Therefore, applying Proposition 6 tothe vector ( u i − u j , (cid:126)v i − (cid:126)v j ) ∈ Q D ( S ) gives that(5.5) | (cid:126)v i − (cid:126)v j | ≥ | (cid:126)v j | . If there were equality in this inequality then, by the argument used in the proof ofProposition 3, we would have that j = K , which is a contradiction. Therefore, thestrict inequality (5.2) holds.Next, to prove (5.3), let 1 ≤ i < j < k ≤ K −
1. We have that 1 / ≤ u k < u j < u i < < u k + u j − u i < u k . Applying Proposition 6 to the vector ( u k + u j − u i , (cid:126)v k + (cid:126)v j − (cid:126)v i ) ∈ Q D ( S ) gives that(5.7) | (cid:126)v k + (cid:126)v j − (cid:126)v i | ≥ | (cid:126)v k | , which proves the result. (cid:3) As a corollary of Proposition 16, we also deduce the following result.
Proposition 17. If G S ( M ) = K then, for all ≤ i < j ≤ K − , the angle between (cid:126)v i and (cid:126)v j must be greater than π/ . Also, for all ≤ i < j < k ≤ K − , the vector (cid:126)v i does not lie in the positive cone determined by (cid:126)v j and (cid:126)v k .Proof. For the first part of the proposition, if 1 ≤ i < j ≤ K −
1, then the fact that theangle between (cid:126)v i and (cid:126)v j must be greater than π/ ≤ i < j < k ≤ K − (cid:126)v i does lie in the positive cone determined by (cid:126)v j and (cid:126)v k . Note that | (cid:126)v i | < | (cid:126)v j | < | (cid:126)v k | and that, by the first part of the proposition, the angle between (cid:126)v j and (cid:126)v k is greater than 2 π/
3. With these observations in mind, consider Figure 7. The figureis rotated so that the vector (cid:126)v k is aligned with the positive x -axis. Depending on theorientation of the vectors involved, it may also be reflected about the x -axis. The vector − (cid:126)v j − (cid:126)v k must lie in the sector indicated in red. Once (cid:126)v j is chosen, the vector (cid:126)v i − (cid:126)v j − (cid:126)v k must lie in a sector of the circle of radius | (cid:126)v i | centered at − (cid:126)v j − (cid:126)v k , as indicated by theblue region in the figure. However, no matter what choice is made for (cid:126)v j , this sectorwill lie completely within the open disc of radius | (cid:126)v k | centered at the origin.Since this contradicts (5.3), we conclude that (cid:126)v i can not lie in the positive conedetermined by (cid:126)v j and (cid:126)v k . (cid:3) Proof of Theorem 8.
Suppose, contrary to the statement of the theorem, that G S ( M ) ≥
6. Consider the collection of vectors (cid:126)v i , with 1 ≤ i ≤
5. We will say that two vectorsfrom this collection are consecutive if there is no other vector from the collection whichlies in their positive cone. Every vector in the collection is consecutive to two others. (cid:126)v i (cid:126)v j (cid:126)v k (cid:126)v i − (cid:126)v j − (cid:126)v k π/ π/ Figure 7.
Illustration corresponding to the contradictory hypothesisthat (cid:126)v i lies in the positive cone determined by (cid:126)v j and (cid:126)v k .By Proposition 17, the angle between any pair of consecutive vectors is greater than π/
3. Since there are five vectors in the collection, it follows that if i, j, and k are distinctindices with 1 ≤ i, j, k ≤ (cid:126)v i is consecutive to both (cid:126)v j and (cid:126)v k , then the anglebetween (cid:126)v j and (cid:126)v k is less than π . In other words, in the situation just described, thevector (cid:126)v i lies in the positive cone determined by (cid:126)v j and (cid:126)v k . Therefore, by Proposition17, if i, j, and k are distinct and if (cid:126)v i is consecutive to both (cid:126)v j and (cid:126)v k , then it mustbe the case that i > min { j, k } . However, the vector (cid:126)v is consecutive to two vectors (cid:126)v j and (cid:126)v k , with 1 < j < k , and this gives a contradiction. Therefore we conclude that G S ( M ) ≤ (cid:3) Explicit upper bounds in dimension d > G = SL( d + 1 , R ) and Γ = SL( d + 1 , Z ). As in the proof of Theorem 7, for each M ∈ Γ \ G we suppose that K ∈ N and { ( u i , (cid:126)v i ) } Ki =1 ⊆ M are chosen so that conditions(V1)-(V3) hold. In this section we will prove the following statement, which by (2.14)implies Theorem 1 in dimension d ≥ Theorem 9.
Let d ≥ and D = S d − . Then for any M ∈ G we have that (6.1) G D ( M ) ≤ σ d + 1 . We will use the following analogues of Propositions 3 and 9.
FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 25
Proposition 18.
Let d ≥ and D = S d − . If ( u, (cid:126)v ) ∈ Q D ( M ) with u ∈ ( − / , / ,then | (cid:126)v K | ≤ | (cid:126)v | . It follows that if, for some integer ≤ i ≤ K we have that u i ∈ ( − / , / , then we must have that i = K .Proof. Suppose first that u ∈ ( − / , − u ∈ [0 , /
2) and ( − u, − (cid:126)v ) ∈ Q D ( M ).Therefore, for any t ∈ (0 , Q D ( M, t ) can have length at most | (cid:126)v | . If u ∈ [0 , /
2) the argument is symmetric, so this verifies the first claim of theproposition.For the second claim, apply the first with ( u, (cid:126)v ) = ( u i , (cid:126)v i ). Then it follows fromproperties (V1)-(V3) that i = K . (cid:3) Proposition 19.
Let d ≥ and D = S d − . If ≤ i, j ≤ K and (6.2) − < u i < u j ≤ − / or / ≤ u j < u i < , then the angle between (cid:126)v i and (cid:126)v j is greater than π/ .Proof. Suppose the hypotheses of the proposition are satisfied and that u i , u j ∈ ( − , − / u i < u j . Then conditions (V1) and (V3) imply that i < j and that | (cid:126)v i | < | (cid:126)v j | . Ifthe angle between (cid:126)v i and (cid:126)v j were less than or equal to π/ | (cid:126)v i − (cid:126)v j | < | (cid:126)v j | . Since ( u i − u j , (cid:126)v i − (cid:126)v j ) ∈ Q D ( M ) satisfies u i − u j ∈ ( − / , / | (cid:126)v K | ≤ | (cid:126)v i − (cid:126)v j | < | (cid:126)v j | , which contradicts (V1). Therefore the angle between (cid:126)v i and (cid:126)v j is greater than π/
3. Theargument is symmetric if u i , u j ∈ [1 / ,
1) with u i > u j . (cid:3) Proof of Theorem 9.
By Proposition 18, there is at most one value of 1 ≤ i ≤ K with u i ∈ ( − / , / K .For each 1 ≤ i ≤ K − (cid:126)x i = (cid:126)v i | (cid:126)v i | ∈ S d − . By Proposition 19 any points (cid:126)x i and (cid:126)x j with i (cid:54) = j are separated by an angle greaterthan π/
3. Therefore the collection of spheres of radius 1 / (cid:126)x i ,for 1 ≤ i ≤ K −
1, do not overlap, and they are all tangent to the sphere of radius 1 / K − ≤ σ d , and this completes the proof of Theorem 9. (cid:3) Continuity and local upper/lower bounds
We now turn to the case of general
D ⊆ S d − . We will show in this section thatthere are choices of D for which F D ( M, t ) is unbounded. First we establish local upperbounds (i.e. upper bounds for M on compacta) that hold for general D . Proposition 20.
Suppose that
D ⊆ S d − has non-empty interior and that C ⊂ Γ \ G × (0 , is compact. Then the following must hold: (i) There exists a number κ ( C ) > such that F D ( M, t ) < κ ( C ) if (Γ M, t ) ∈ C , and (ii) F is continuous at every point (Γ M, t ) ∈ C with (7.1) ( Z d +1 M \ { } ) ∩ ∂ (( − t, − t ) × (0 , κ ( C )] D ) = ∅ . We emphasise that in (ii), relation (7.1) needs to be verified only for one specific butarbitrary κ ( C ) > F D ( M, t ). Proof.
This is analogous to the proof of [26, Prop. 2]: The existence of a constant κ ( C ) in (i) is guaranteed by the lattice point constructed via Minkowski’s theorem inthe proof of Proposition 1, where uniformity over C follows from Mahler’s compactnesscriterion. Claim (ii) follows from the uniform discreteness of Z d +1 M , and in particularfrom the fact that the packing radii of the sets Z d +1 M are uniformly bounded away from0 whenever M is restricted to lie in a bounded subregion of Γ \ G (again, a consequenceof Mahler’s criterion). (cid:3) Let us now extend the above uniform upper bound to all t ∈ (0 , M in acompact set. Given a bounded subset A ⊂ R d +1 with non-empty interior, and M ∈ G ,we define the covering radius (also called inhomogeneous minimum )(7.2) ρ ( M, A ) = inf { θ > | θ A + Z d +1 M = R d +1 } . Because A has non-empty interior, ρ ( M, A ) < ∞ . We will in the following take A to be of the form A = (0 , × (0 , r ] D . Then A ⊂ λ A for any λ >
1, and thus θ A + Z d +1 M = R d +1 for every θ > ρ ( M, A ). Therefore, for such θ, the set θ A + (cid:126)x intersects Z d +1 M in at least one point, for every (cid:126)x ∈ R d +1 .For a given set C ⊂ Γ \ G , we define(7.3) ρ ( C , A ) = sup Γ M ∈C ρ ( M, A ) . It is well known that ρ ( C , A ) < ∞ for every compact C ⊂ Γ \ G . For θ >
0, set(7.4) D ( θ ) = (cid:18) θ d θ − d (cid:19) ∈ G. Proposition 21.
Let
D ⊆ S d − with non-empty interior. Assume C ⊂ Γ \ G is compact,and θ > ρ ( C , (0 , × (0 , D ) . Then (7.5) F D ( M, t ) ≤ θ d +1 for Γ M ∈ C D ( θ ) − and t ∈ (0 , .Proof. This is almost identical to the proof of [26, Prop. 3]. The set A t,θ = ( − t, − t ) × (0 , θ d +1 ] D ⊂ R d +1 has non-empty interior. The task is to show that A t,θ intersects Z d +1 M in at least one point, for every Γ M ∈ C D ( θ ) − and every t ∈ (0 , A t,θ ∩ Z d +1 M (cid:54) = ∅ is equivalent to θ d A t, ∩ Z d +1 M D ( θ ) (cid:54) = ∅ . The latter holds becausethe assumption that θ > ρ ( C , (0 , × (0 , D ) implies that θ A t, ∩ Z d +1 M (cid:48) (cid:54) = ∅ for everyΓ M (cid:48) ∈ C , and Γ M (cid:48) = Γ M D ( θ ) ∈ C by assumption. (cid:3) We have the following corollary.
FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 27
Proposition 22.
Let
D ⊆ S d − have non-empty interior. Assume C ⊂ Γ \ G is compact,and θ > ρ ( C , (0 , × (0 , D ) . Then there is a constant C θ, D < ∞ such that (7.6) G D ( M ) ≤ C θ, D for Γ M ∈ C D ( θ ) − .Proof. This follows from an upper bound, uniform in Γ M ∈ C D ( θ ) − , on the numberof lattice points in a given bounded set (via Mahler’s criterion); see the proof of [26,Prop. 4]. (cid:3) The key point is now that, in addition to the above upper bounds, we can find opensets
U ⊂ Γ \ G , on which G D ,N ( M ) can exceed any given value. This requires howeverthat D is contained in a hemisphere. Let H ⊂ S d − be an (arbitrary) open hemisphere,and D ⊂ S d − with non-empty interior so that D cl ⊂ H . Choose d + 1 row vectors (cid:126)e , (cid:126)e , . . . , (cid:126)e d ∈ S d − with the properties(i) (cid:126)e ∈ D ◦ and (cid:126)e ∈ H \ D such that (cid:126)e · (cid:126)e > (cid:126)e , . . . , (cid:126)e d / ∈ D such that ( (cid:126)e , . . . , (cid:126)e d ) forms an orthonormal basis of R d withdet (cid:126)e ... (cid:126)e d = 1 . Given (cid:15) >
0, define the matrix(7.7) M (cid:15) = (cid:15) − (cid:15)(cid:126)e (cid:126)e (cid:15) − / ( d − (cid:126)e ... ...0 (cid:15) − / ( d − (cid:126)e d . Note that det M (cid:15) = 1, and thus the vectors (cid:126)b = ( (cid:15), − (cid:15)(cid:126)e ) , (cid:126)b = (0 , (cid:126)e ) , (cid:126)b = (0 , (cid:15) − / ( d − (cid:126)e ) , . . . , (cid:126)b d = (0 , (cid:15) − / ( d − (cid:126)e d )form a basis of the unimodular lattice L (cid:15) = Z d +1 M (cid:15) .Let L ( x ) denote the largest integer strictly less than x . That is, in terms of the floorfunction L ( x ) = (cid:98) x (cid:99) if x / ∈ Z and L ( x ) = x − x ∈ Z . Proposition 23.
Let
D ⊂ H and M (cid:15) be as above. Then there exist λ ∈ (0 , and (cid:15) > such that, for any (cid:15) ∈ (0 , (cid:15) ] and t ∈ ( λ, , (i) F D ( M (cid:15) , t ) = | (cid:126)e − (cid:15)L ( (cid:15) − (1 − t )) (cid:126)e | and (ii) F D is continuous at (Γ M (cid:15) , t ) ∈ Γ \ G × (0 , if t / ∈ (cid:15) Z .Proof. (i) Fix s ∈ (0 , (cid:126)e · (cid:126)e ), and denote by s − ≤ ≤ s + the infimum and supremumof all s such that (cid:126)e − s(cid:126)e ∈ R > D ; since e ∈ D ◦ we have s − < < s + . Note also that(7.8) dds | (cid:126)e − s(cid:126)e | = s − (cid:126)e · (cid:126)e | (cid:126)e − s(cid:126)e | < s ≤ s . We are interested in the lattice points from Z d +1 M (cid:15) contributing to (2.4), i.e.,(7.9) Q D ( M (cid:15) , t ) = (cid:8) ( u, (cid:126)v ) ∈ Z d +1 M (cid:15) (cid:12)(cid:12) − t < u < − t, (cid:126)v ∈ R > D (cid:9) , and in particular those with minimal | (cid:126)v | . We begin with those elements of the form(7.10) ( u, (cid:126)v ) = m (cid:126)b + m (cid:126)b = ( (cid:15)m , m (cid:126)e − (cid:15)m (cid:126)e ) , m ∈ Z , m ∈ Z ≥ . By construction, we have m (cid:126)e − (cid:15)m (cid:126)e ∈ R > D ◦ if and only if s − < (cid:15)m /m < s + ; and m (cid:126)e − (cid:15)m (cid:126)e ∈ R > D cl if and only if s − ≤ (cid:15)m /m ≤ s + . In view of (7.8), the length | (cid:126)e − (cid:15)m /m (cid:126)e | is strictly decreasing (as a function of m /m ) if (cid:15)m /m ≤ s . Letus restrict our attention to those t for which 1 − t < min { s , s + } . Then the condition (cid:15)m < − t implies (cid:15)m /m < s + and (cid:15)m /m ≤ s for all m ≥
1. Hence the smallestvalue of | m (cid:126)e − (cid:15)m (cid:126)e | is obtained for m = L ( (cid:15) − (1 − t )) and m = 1. In summary,we have shown thus far that for t > − min { s , s + } ,(7.11) F D ( M (cid:15) , t ) ≤ | (cid:126)e − (cid:15)L ( (cid:15) − (1 − t )) (cid:126)e | . Note that(7.12) | (cid:126)e − s(cid:126)e | = (cid:112) − s (cid:126)e · (cid:126)e + s < < s ≤ s and (cid:126)e · (cid:126)e > s , and hence F D ( M (cid:15) , t ) < Q D ( M (cid:15) , t ) that are notof the form (7.10) have larger | (cid:126)v | . Consider first the set of vectors(7.13) ( u, (cid:126)v ) = m (cid:126)b + m (cid:126)b = ( (cid:15)m , m (cid:126)e − (cid:15)m (cid:126)e ) , m ∈ Z , m ∈ Z ≤ , Since (cid:126)e ∈ H \ D and (cid:126)e ∈ D ◦ we have m (cid:126)e − (cid:15)m (cid:126)e / ∈ R > D for all m ≤
0, and hencethe corresponding vectors are not in Q D ( M (cid:15) , t ).Next consider the remaining cases(7.14) ( u, (cid:126)v ) = m (cid:126)b + · · · + m d (cid:126)b d , m , m ∈ Z , ( m , . . . , m d ) ∈ Z d − \ { } . We need to understand whether any of these vectors can lie in Q D ( M (cid:15) , t ) and satisfy | (cid:126)v | <
1. We make the following observations:(a) The domain ( − , × (0 , D is bounded.(b) The vector m (cid:126)b has bounded length | m (cid:126)b | = √ | (cid:15)m | < √ − <(cid:15)m < m (cid:126)b + · · · + m d (cid:126)b d has length at least (cid:15) − / ( d − since ( m , . . . , m d ) ∈ Z d − \ { } .Therefore, as long as (cid:15) is sufficiently small, ( − , × (0 , D will not contain any vectorsof the form (7.14). Since we have now considered all vectors for our restricted values of t , this establishes claim (i) with λ = 1 − min { s , s + } .(ii) We need to establish that, for λ < t < t / ∈ (cid:15) Z , the function F D is continuousat (Γ M (cid:15) , t ). By Proposition 20 (with C = { (Γ M (cid:15) , t ) } ), it is sufficient to check that(7.15) ( Z d +1 M (cid:15) \ { } ) ∩ ∂ (( − t, − t ) × (0 , κ ] D ) = ∅ , for any fixed choice of κ > F D (Γ M (cid:15) , t ). We fix κ so that F D (Γ M (cid:15) , t ) < κ < κ / ∈ {| m (cid:126)e − (cid:15)m (cid:126)e | | ( m , m ) ∈ Z ≥ } . FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 29
Since vectors with ( m , . . . , m d ) ∈ Z d − \{ } are outside the bounded domain (( − t, − t ) × (0 , κ ] D ) cl for (cid:15) sufficiently small (by the argument in the proof of fact (i) above),what we are aiming to show is equivalent to(7.17) { ( (cid:15)m , m (cid:126)e − (cid:15)m (cid:126)e ) | ( m , m ) ∈ ( Z \ { } ) ∩ ∂ (( − t, − t ) × (0 , κ ] D ) } = ∅ . By the same argument as above, for m ≤ m (cid:126)e − (cid:15)m (cid:126)e / ∈ R ≥ D cl (unless( m , m ) = 0, which is excluded). So we can assume m ≥ m ≤
0, we have by the monotonicity (7.8) that | m (cid:126)e − (cid:15)m (cid:126)e | ≥ m | (cid:126)e | ≥ | (cid:126)e | = 1.Therefore m (cid:126)e − (cid:15)m (cid:126)e / ∈ [0 , κ ] D cl . What remains is to check that(7.18) { ( (cid:15)m , m (cid:126)e − (cid:15)m (cid:126)e ) | ( m , m ) ∈ Z ≥ ∩ ∂ (( − t, − t ) × (0 , κ ] D ) } = ∅ . The truth of relation (7.18) is equivalent to the truth of both(7.19) { ( (cid:15)m , m (cid:126)e − (cid:15)m (cid:126)e ) | ( m , m ) ∈ Z ≥ ∩ ( {− t, − t } × [0 , κ ] D cl ) } = ∅ and(7.20) { ( (cid:15)m , m (cid:126)e − (cid:15)m (cid:126)e ) | ( m , m ) ∈ Z ≥ ∩ ([ − t, − t ] × ∂ ((0 , κ ] D ))) } = ∅ . The first relation (7.19) is automatically satisfied since (a) by assumption (cid:15)m (cid:54) = 1 − t for any integer m , and (b) (cid:15)m (cid:54) = − t because t > m > m (cid:126)e − (cid:15)m (cid:126)e ∈ ∂ ((0 , κ ] D ) implies s − = (cid:15)m /m or s + = (cid:15)m /m or | m (cid:126)e − (cid:15)m (cid:126)e | = κ . The first option is not possiblesince s − <
0, and third is excluded by assumption. If the second option holds, then,since (cid:15)m ∈ [ − t, − t ], we have m s + ≤ − t and thus s + ≤ − t . But this contradictsour assumption 1 − t < s + . Hence (7.20) holds and the proof is complete. (cid:3) The following lower bound on the number of distinct values of t (cid:55)→ F D ( M, t ) is acorollary of the previous proposition.
Proposition 24.
Let
H ⊂ S d − be an arbitrary open hemisphere, and suppose that D ⊂ S d − has non-empty interior and satisfies D cl ⊂ H . Then there is a constant c D > such that, for any (cid:15) > , there exists an open subset U (cid:15) ⊂ Γ \ G and integer N (cid:15) with the property that, for all Γ M ∈ U (cid:15) and N ≥ N (cid:15) , (7.21) G D ,N ( M ) ≥ c D (cid:15) − . Proof.
Proposition 23 (i) shows that G D ( M (cid:15) ) ≥ c D (cid:15) − for a sufficiently small c D > { F D ( M (cid:15) , t ) | t ∈ ( λ, } by 0 < ϕ < · · · <ϕ G D ( M (cid:15) ) <
1. Let δ = max i ( ϕ i +1 − ϕ i ). Then by the continuity established in Proposition23 (ii), there exists a neighbourhood U (cid:15) ⊂ Γ \ G of Γ M (cid:15) and an η (cid:15) > t / ∈ (cid:15) Z , we have that(7.22) | F D ( M, t (cid:48) ) − F D ( M (cid:15) , t ) | < δ/ , whenever Γ M ∈ U (cid:15) and t (cid:48) ∈ ( t − η (cid:15) , t + η (cid:15) ). For N ≥ η − (cid:15) we can find an integer n sothat nN + ∈ ( t − η (cid:15) , t + η (cid:15) ). This implies G D ,N ( M ) ≥ G D ( M (cid:15) ) ≥ c D (cid:15) − for all Γ M ∈ U (cid:15) and N ≥ η − (cid:15) . (cid:3) Proof of Theorem 5.
The proof of Theorem 5 now follows from the same argument asthe proof of [26, Theorem 1]. For (cid:126)α ∈ P , we have that the set { Γ A N i + ( (cid:126)α ) | i ∈ N } is dense in Γ \ G ; see Section 2 of [26] for details. The claim on the limit inferior then follows from Proposition 21, since by density we have infinite returns to a given compactsubset. The claim on the limit superior follows from Proposition 24, since by densitythe above set intersects any given open neighbourhood U (cid:15) . (cid:3) Proof of Theorem 6.
This is analogous to the proof of to the proof of Theorem 2 in [26].Note that for Z d M = Z d M and (cid:126)α = (cid:126)α M (7.23) A N ( (cid:126)α, L ) = (cid:18) (cid:126)α d (cid:19) (cid:18) N − N /d d (cid:19) (cid:18) M (cid:19) . By assumption (cid:126)α is badly approximable by Q L , i.e., (cid:126)α is badly approximable by Q d .Hence, in view of Dani’s correspondence, the set { Γ A N + ( (cid:126)α M , Z d M ) | N ∈ N } iscontained in a compact subset of Γ \ G . The claim then follows from Proposition 21. (cid:3) We now return to the special case D = S d − , and provide the remaining ingredientsfor the proof of Theorem 2. Proposition 25.
Let D = S d − , and fix a matrix A N + ( (cid:126)α ) as in (2.2) with N + := N + , N ∈ N . Then the following hold. (i) For given n = 1 , . . . , N , the function t (cid:55)→ F D ( A N + ( (cid:126)α ) , t ) is constant on the interval I n = ( N − ( n − ) , N − n ] . (ii) F D is continuous at (Γ A N + ( (cid:126)α ) , t ) ∈ Γ \ G × (0 , if t, − t / ∈ N − Z .Proof. Throughout this proof set D = S d − . Then(7.24) F D ( A N + ( (cid:126)α ) , t ) = N /d + min (cid:8) | k(cid:126)α + (cid:126)(cid:96) | > (cid:12)(cid:12) − N + t < k < N + (1 − t ) , k ∈ Z , (cid:126)(cid:96) ∈ Z d M (cid:9) . The set ( − N + t, N + (1 − t )) ∩ Z is independent of the choice of t ∈ I n ; this proves (i). Inview of Proposition 20, claim (ii) holds if(7.25) ( Z d +1 A N + ( (cid:126)α ) \ { } ) ∩ ∂ (( − t, − t ) × (0 , κ ] D ) = ∅ , for a fixed choice of κ > sup t ∈ (0 , F D ( A N + ( (cid:126)α ) , t ). The set(7.26) ˜ M ( M ) = (cid:8) | (cid:126)v | > (cid:12)(cid:12) ( u, (cid:126)v ) ∈ Z d +1 M, | u | ≤ (cid:9) is locally finite for every fixed M ∈ G (cf. (2.11)), and so we can choose κ / ∈ ˜ M ( A N + ( (cid:126)α )).This means that the lattice Z d +1 A N + ( (cid:126)α ) does not intersect the set [ − , × κ D . Fur-thermore, by the assumption t, − t / ∈ N − Z , we have(7.27) ( Z d +1 A N + ( (cid:126)α ) \ { } ) ∩ ( {− t, − t } × (0 , κ ] D ) = ∅ , which establishes (7.25), and hence completes the proof of claim (ii). (cid:3) Proof of Theorem 2.
In the following D = S d − . We fix N ∈ N until the last step of theproof. By (2.6) and Proposition 25 (i) we have(7.28) δ n,N ( D ) = N − /d + F D (cid:0) A N + ( (cid:126)α , L ) , t n (cid:1) , for any t n ∈ I n , and hence(7.29) g N ( (cid:126)α , L ) = |{ F D ( A N + ( (cid:126)α , L ) , t n ) | n = 1 , . . . , N }| . FIVE DISTANCE THEOREM FOR KRONECKER SEQUENCES 31
Choose δ > { F D ( A N + ( (cid:126)α , L ) , t n ) | n = 1 , . . . , N } are separated by at least δ . Fix any t n ∈ I n such that t n , − t n / ∈ N − Z .Proposition 25 (ii) implies that F D is continuous at (Γ A N + ( (cid:126)α , L ) , t n ), for n = 1 , . . . , N .That is, there exists a neighbourhood U ⊂ Γ \ G of the point Γ A N + ( (cid:126)α , L ) and an η > | F D ( M, t (cid:48) ) − F D ( A N + ( (cid:126)α , L ) , t n ) | < δ/ , whenever Γ M ∈ U and t (cid:48) ∈ ( t n − η, t n + η ). For every integer ˜ N ≥ η − and n = 1 , . . . , N we can find a positive integer m n ≤ ˜ N so that m n ˜ N + ∈ ( t n − η, t n + η ). This implies that G D , ˜ N ( M ) ≥ G D ( A N + ( (cid:126)α , L )) = g N ( (cid:126)α , L ) for all Γ M ∈ U and ˜ N ≥ η − .We can now conclude the proof as for Theorem 5. Let (cid:126)α ∈ P . The density of theorbit { Γ A N i + ( (cid:126)α, L ) | i ∈ N } implies(7.31) lim sup i →∞ g N i ( (cid:126)α, L ) ≥ g N ( (cid:126)α , L ) . Theorem 2 now follows by taking the supremum over N ∈ N . (cid:3) References [1] V. Berth´e, D. H. Kim:
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