Hausdorff Dimension of Singular Vectors

HAUSDORFF DIMENSION OF SINGULAR VECTORS

YITWAH CHEUNG AND NICOLAS CHEVALLIER

Abstract. We prove that the set of singular vectors in Rd, d ≥ 2, has Hausdorff dimension d2 d d+1 and that the Hausdorff dimension of the set of ε-Dirichlet improvable vectors in R d2 d is roughly d+1 plus a power of ε between 2 and d. As a corollary, the set of divergent t t −dt trajectories of the flow by diag(e , . . . , e , e ) acting on SLd+1 R/ SLd+1 Z has Hausdorff d codimension d+1 . These results extend the work of the first author in [6].

1. Introduction Singular vectors were introduced by Khintchine in the twenties (see [14]). Recall that d x = (x1, ..., xd) ∈ R is singular if for every ε > 0, there exists T0 such that for all T > T0 the system of inequalities ε (1.1) max |qxi − pi| < and 0 < q < T 1≤i≤d T 1/d admits an integer solution (p, q) ∈ Zd × Z. In dimension one, only the rational numbers are singular. The existence of singular vectors that do not lie in a rational subspace was proved by Khintchine for all dimensions ≥ 2. Singular vectors exhibit phenomena that cannot occur in dimension one. For instance, when x ∈ Rd is singular, the sequence 0, x, 2x, ..., nx, ... fills the torus Td = Rd/Zd in such a way that there exists a point y whose distance in the torus to the set {0, x, ..., nx}, times n1/d, goes to infinity when n goes to infinity (see [4], chapter V). While it is easy to see that the set Sing(d) of all singular vectors in Rd has zero Lebesgue measure, it is hard to compute its Hausdorff dimension. After a few partial results by Baker and Rynne (see [2], [3] and [20]), it has been proved in [6] by the first author that the 4 Hausdorff dimension of Sing(2) is 3 . In this paper we extend this result to higher dimensions.

d2 Theorem 1.1. The Hausdorﬀ dimension of Sing(d) is d+1 . Singular vectors x ∈ Rd are characterized by Mahler’s criterion d+1 lim inf{kwk : 0 6= w ∈ gthx } = 0 t→∞ Z

Date: February 28, 2014. 1991 Mathematics Subject Classiﬁcation. 11J13, 11K55, 37A17. Key words and phrases. singular vectors; self-similar coverings; multi-dimensional continued fractions; simultaneous Diophantine approximation; best approximations; divergent trajectories. The ﬁrst author is partially supported by NSF grant DMS 0956209.

1 1 −x where g = diag(et, . . . , et, e−dt) and h = d . Equivalently, {g h SL } is a t x 0 1 t x d+1 Z t∈R divergent trajectory of the flow on SLd+1 R/ SLd+1 Z induced by the left action of gt. d Corollary 1.2. Divergent trajectories of gt have Hausdorff codimension d+1 . Proof. The Hausdorff codimension of Sing(d) in Rd coincides with that for divergent trajectories in SLd+1 R/ SLd+1 Z (see [5]). In [12] Einsiedler-Kadyrov showed that the Hausdorff dimension of divergent trajectories of gt does not increase in the case of SL3 R/ SL3 Z even after enlarging it to include trajectories that diverge ”on average”. This result generalizes to higher dimensions; see [?]. Divergent trajectories and their Hausdorff dimension have been considered for other classes of ”hyperbolic” flows, starting with Masur’s Duke paper [18] which established codimension 1 ≥ 2 for Teichmüllerflows. Positive Hausdorff codimension for divergent trajectories in a product of real rank one spaces is also known; see [24] and also [5]. Let ε be a fixed positive real number. Recall that a vector x in Rd is ε-Dirichlet improvable if the system of inequalities (1.1) admits a solution for T large enough. H. Davenport and W. Schmidt have proved that the set DIε(d) of ε-Dirichlet improvable vectors has zero measure for all ε < 1 ([10] and [11]). This result has been generalized in several directions (see for instance [15]) but less is known about the Hausdorff dimension of DIε(d). The intersection of the sets DIε(d), ε > 0, is the set of singular vectors, so it is natural to expect that the Hausdorf dimension of DIε(d) decreases to the Hausdorff dimension of Sing(d) as ε goes to zero. When d = 2, the bounds on Hdim DIε(d) obtained in [5] implies this continuity at ε = 0 of the Hausdorff dimension. Our second result extends these bounds to higher dimensions. Theorem 1.3. Let d ≥ 2 be an integer and let t be any real number > d. There is a constant C such that for ε small enough, d2 d2 + εt ≤ Hdim DI (d) ≤ + Cεd/2. d + 1 ε d + 1 Actually, we give a slightly better lower bound; see corollary 6.10. In the two dimensional case, this result already improves the lower bound of [6] but there is a further improvement giving a precise rate of convergence when ε goes to zero. Theorem 1.4. For all real numbers t > 1, we have 4 + εt ≤ Hdim DI (2) 3 ε for ε small enough. Our proof of the above Theorem doesn’t work when the dimension is ≥ 3. In particular, our analysis stops short of establishing the existence of the exponent d2 log(Hdim DIε(d) − ) (1.2) lim d+1 ε→0+ log ε

2 for d ≥ 3. It seems that our approach could potentially be carried further to establish existence of the above limit, which we will guess to lie in the interval [d − 1, d[. At last we have to say that results about singular linear forms can be deduced using a transference Theorem ([4], chapter V, corollary section II). But a direct proof is not clear. The proofs rely on three main tools. The ﬁrst is lattice theory, especially Minkowski’s theorems and reduced bases. The second is the notion of self-similar covering introduced in [5] and [6]. The third is the notion of best Diophantine approximation. Section 2 is devoted to lattices theory and counting results. Section 3 is devoted to Hausdorﬀ dimension bounds using self-similar covers and Section 4 is devoted to best Diophantine approximations. The proofs of the above Theorems actually begin in Section 5. Proving the upper bounds is the most easy parts of the proofs. While the upper bounds are proved in Section 5, the lower bounds need the Sections 6, 7 and 8.

2. Counting lattice points 2.1. Notation. We shall use the standard notations , , in number theory: f(a, b, ...) g(a, b, ...) means that f(a, b, ...) ≤ Cg(a, b, ...) where C is a constant that does not depend on the parameters a, b, ... etc. Most of the time, C depends only the dimension or on the norm. The symbol is used for the reverse inequality and when both inequalities hold. Suppose that Rn is endowed with a norm k · k. For a lattice L in Rn and 1 ≤ k ≤ n, we denote by

λk(L) = inf{r ≥: B(0, r) contains at least k linearly independent vector of L} the kth minimum of L. The successive minima of a lattice L in Rn are the real numbers λ1(L), . . . , λn(L). This notation suppresses the dependence on the norm.

2.2. Successive minima and reduced basis. Assume that Rn is equipped with a norm. By the (second) Minkowski theorem, the successive minima of a lattice L in Rn satisfy (see [4]) 2n (M) vol(L) ≤ λ (L) ··· λ (L) vol(B(0, 1)) ≤ 2n vol(L). n! 1 n We shall use this fact many times and simply write

λ1(L) ··· λn(L) vol(L) where the constants involved in depend only on the dimension and on the volume of the unit ball.

3 We shall need reduced bases in a lattice. Minkowski reduced bases are well-suited to our needs. Other kinds of reduced bases could work as well. We shall use the following properties of a Minkowski reduced basis v1, . . . , vn. For all i, i (M’) λi(L) ≤ kvik ≤ 2 λi(L), A proof of this inequality can be found in [23]. Actually, Minkowski reduced bases are associated to Euclidean norms and the above inequality holds only in these cases. When dealing with a non Euclidean norm, we consider a reduced basis associated to the standard Euclidean norm and by the norm equivalence, one has kvik λi where the constants involved depend not only on the dimension but also on the norm. An important property of a reduced basis is that the angles between a vector of the basis and the subspace spanned by other vectors cannot be too small. More precisely, there is positive constant cn depending only on the dimension such that ⊥ (M”) vi ≥ cnλi(L) ⊥ where vi is the orthogonal projection of vi on the line orthogonal to the subspace spanned by the other basis vectors. Once again, the above inequality holds for Euclidean norms and for other norms, one has ⊥ vi λi(L) where the implicit constants depend on the norm. To prove (M”), note that by (M’) one has ⊥ ⊥ λ1(L) ··· λi−1(L) vi λi+1(L) ··· λn(L) | det(v1, . . . , vi−1, vi , vi+1, . . . , vn)| = vol(L) and then use the Minkowski theorem. For sake of simplicity, the reader can suppose that all the results are stated for the standard Euclidean norm. However, it is clear that the results of Section 2 hold with any norm with constants possibly depending on norms. The results in Section 4 hold for any norm.

2.3. Codimension one sublattice of minimal volume. Fix a norm k · k on Rn. For a subset X ⊂ Rn, we denote the radius of the largest ball disjoint from X by d e(X) = sup{r > 0 : ∃p s.t. B(p, r) ∩ X = ∅} = sup{d(p, X): p ∈ R } n = inf{r > 0 : ∀x ∈ R ∃p ∈ X s.t. d(x, y) < r}.

Lemma 2.1. The successive minima satisfy λn(L) ≤ 2e(L) ≤ λ1(L) + ··· + λn(L). Proof. There is a parallelepiped P with edge lengths given by the successive minima of the lattice L, the translates of which cover Rn, i.e. P + L = Rn. Therefore, 2e(L) ≤ λ1(L) + ··· + λn(L). Furthermore, the intersection of L with the (open) ball B(0, λn(L)) is contained in an (n − 1)-dimensional subspace. It follows that there exists a ball of radius λn(L) λn(L) 2 contained in B(0, λn(L)) that does not meet L; therefore, e(L) ≥ 2 . Lemma 2.2. Let L be a lattice in Rn, L0 ⊂ L a sublattice of codimension one with minimal volume, and H0 the real span of L0. Then 0 0 0 2e(L + H ) = inf{kvk : v ∈ (L + H ) \ H } λn(L).

4 Proof. Since L + H0 is a regularly spaced net of hyperplanes, we have 2e(L + H0) = inf{kvk : v ∈ (L + H0) \ H0}. Let L00 be a codimension one sublattice of L spanned by linearly independent vectors v1, . . . , vn−1 ∈ L such that kvik = λi(L). Making use of the Minkowski theorem, we obtain

λ1(L) ··· λn−1(L) = kv1k · · · kvn−1k 00 00 ≥ λ1(L ) ··· λn−1(L ) vol(L00) ≥ vol(L0). 0 0 Since L is a subset of L, λi(L ) ≥ λi(L) for i = 1, ..., n − 1. Again with the Minkowski 0 theorem, we obtain vol(L ) λ1(L) ··· λn−1(L). Combining this with the above and using the Minkowski theorem once more, we get vol(L) λ (L). vol(L0) n Since L0 is a primitive sublattice of L, there exists a vector w in L such that L = L0 +Zw. We have vol(L) vol(L0)kw⊥k where w⊥ is the orthogonal projection onto the line orthogonal 0 ⊥ 0 0 0 to H . Hence kw k λn(L). Now a vector v = kw + v with k in R and v in H has a norm |k|kw⊥k. Therefore, 0 0 ⊥ inf{kvk : v ∈ (L + H ) \ H } kw k λn(L). ⊥ 0 0 For the reverse inequality, just observe that w ∈ (L + H ) \ H . 2.4. Coarse asymptotics. We shall need several counting estimates in lattices. All these estimates are coarse asymptotics that are well known to specialists. The proofs of these estimates, which are dispersed in many papers, are included for the convenience of the reader. Lemma 2.3. Let L be a lattice in Rn and let x ∈ Rn. Then for all r > 0, n X ri card L ∩ B(x, r) 1 + . λ (L) . . . λ (L) i=1 1 i rn In particular, card L ∩ B(x, r) vol(L) provided r λn(L). Proof. It is enough to prove that n X ri card L ∩ B(x, r) 1 + ke k ... ke k i=1 1 i

where e1, . . . , en is a reduced basis of L. We prove by induction on n that the above inequality holds for any ball B(x, r). If n = 1 the result is clear. Suppose n > 1 and denote Γ = d Ze1 + ··· + Zen−1. By induction hypothesis for any x, y ∈ R , n−1 X ri card(y + Γ) ∩ B(x, r) 1 + . ke k ... ke k i=1 1 i

5 Moreover since the basis is reduced, kPn a e k ≤ r implies |a | ≤ c r where c is a i=1 i i n n kenk n constant depending only on the norm. It follows that [ L ∩ B(x, r) ⊂ (aen + Γ) ∩ B(x, r) r a∈ , |a−xn|≤cn Z kenk Pn where x = i=1 xiei, thus n−1 X X ri card L ∩ B(x, r) (1 + ) r ke1k ... keik a∈ , |a−xn|≤cn i=1 Z kenk n−1 r X ri (1 + )(1 + ) ke k ke k ... ke k n i=1 1 i n X ri 1 + . ke k ... ke k i=1 1 i n n Lemma 2.4. Let L be a lattice in R and let x ∈ R . There exists a real constant Cn rn depending only on the dimension n such that for all r > Cnλn(L), card L ∩ B(x, r) vol(L) . rn Moreover, when x = 0, card L ∩ B(x, r) vol(L) for all nonnegative r. Pn Proof. Let e1, . . . , en be a basis for L, and P = { i=1 aiei : ai ∈ [0, 1[} the fundamental parallelepiped associated with the basis. Each parallelepiped y + P contains exactly one point of L. The number of parallelepipeds of the form y + P, y ∈ L that are contained in B(x, r) is at least the number parallelepipeds intersecting B(x, r) minus the number of parallelepipeds intersecting the boundary of B(x, r). It follows that vol(B(x, r)) vol(B(x, r + diam P)) − vol(B(x, r − diam P)) card L ∩ B(x, r) ≥ − . vol(L) vol(L)

If the basis is reduced, then diam P λn(L) and the ﬁrst inequality of the lemma follows. For i ≤ n, consider the sublattice Li = Zv1 + ··· + Zvi where the vi’s are linearly independent vectors with kvik = λi(L). The ﬁrst inequality gives ri card L ∩ B(0, r) ≥ card Li ∩ B(0, r) vol(Li) for all r ≥ Cnλi(L) (we can assume that the sequence (Cn)n≥1 is nondecreasing). Now, by the Minkowski theorem, if Cnλi(L) ≤ r ≤ Cnλi+1(L) then ri riλ (L) ··· λ (L) rn = i+1 n , vol(Li) vol(Li)λi+1(L) ··· λn(L) vol(L) rn and if r < Cnλ1(L) then L ∩ B(0, r) contains at least one point (x = 0) while vol(L) 1.

6 Lemma 2.5. Let L be a lattice in Rn, t a real number > n, and R a positive real number. Then n X 1 2nt X 1 t t−k . kxk t − n λ1(L) . . . λk(L)R x∈Λ,kxk≥R k=1

In particular, if R λn(L) then

X 1 2nt . kxkt (t − n) vol(L)Rt−n x∈Λ,kxk≥R

Proof. Let e1, . . . , en be a reduced basis of L. Since keik λi(L), the lemma will follow from the inequality

n X 1 2nt X 1 S := t t−k . kxk t − n ke1k ... kekkR x∈Λ, kxk≥R k=1

We will use the elementary inequality: if a > 0 and s > 1 then

X 1 2s 1 ≤ . ms s − 1 as−1 m≥a

We shall drop the factors of 2s in the proof. Pn Let x = i=1 xiei be in L. Consider for each nonempty subset I of {1, . . . , n} and each bijection σ : {1,..., |I| = card I} → I, the subsets of L

X R L = { x e ∈ L : ≥ |x |ke k for i ∈ IC }, I,σ i i n i i i∈IC X R L0 = { x e ∈ L : ≤ |x |ke k ≤ |x |ke k ≤ · · · ≤ |x |ke k}, I,σ σ(i) σ(i) n σ(1) σ(1) σ(2) σ(2) σ(|I|) σ(|I|) i≤|I| and the sum X 1 SI,σ = t . 0 kxk x∈LI,σ⊕LI,σ

P P P R Since k xieik ≤ |xi|keik, we have k xieik ≥ R implies max |xi|keik ≥ . Therefore, S n L \ B(0,R) ⊂ I,σ LI,σ, so it is enough to bound each sum SI,σ from above. Let Vi be the subspace spanned by the ej except ei. Since the basis is reduced, kxk |xi|keik for all i. It follows that for x ∈ LI,σ, kxk |xσ(|I|)|keσ(|I|)k, thus

7 X 1 SI,σ t t 0 |xσ(|I|)| keσ(|I|)k x∈LI,σ⊕L (I,σ) Y 2R X 1 ≤ 1 + nke k |x |tke kt C i 0 σ(|I|) σ(|I|) i∈I x∈LI,σ Y 2R X 1 1 1 + , t−1 nkeik t − 1 |xσ(|I|−1)|keσ(|I|−1)k t i∈IC x∈L00 keσ(|I|)k keσ(|I|)k Y 2R X 1 1 = 1 + t−1 , nkeik t − 1 i∈IC x∈L00 |xσ(|I|−1)|keσ(|I|−1)k keσ(|I|)k where X R L00 = { x e ∈ L : ≤ |x |ke k ≤ |x |ke k ≤ · · · ≤ |x |ke k}. σ(i) σ(i) n σ(1) σ(1) σ(2) σ(2) σ(|I|−1) σ(|I|−1) i<|I| Continuing inductively, we have 1 Y 2R 1 SI,σ 1 + t−|I| , t − |I| nkeik R i∈IC n keσ(1)k ... keσ(|I|)k 1 Y R 1 1 + t−|I| , t − |I| keik R keσ(1)k ... keσ(|I|)k i∈IC 1 X 1 = . t−|J| Q t − |I| R kejk J⊃I j∈J The estimates of this section will be enough for the lower bound calculations. For the lower bound calculations, we need to consider sums that are over restricted subsets (introduced in Section 6) of the lattice, a counting problem that we shall deal with in Section 8.

3. Self-similar coverings 3.1. Upper estimates. Deﬁnition 3.1. Let Y be a metric space. A self-similar structure on Y is a triple (J, σ, B) where J is countable, σ is a subset of J 2 and B is a map from J into the set of bounded subsets of Y .A σ-admissible sequence is a sequence (xn) in J such that

(i) for all integers n,(xn, xn+1) ∈ σ. Let X be a subset of Y .A self-similar covering of X is a self-similar structure (J, σ, B) such

that for all θ in X, there exists a σ-admissible sequence (xn)n∈N in J such that (ii) limn→∞ diam B(xn) = 0,

8 (iii) T B(x ) = {θ}. n∈N n The set covered by a self-similar structure (J, σ, B) is the set all θ in Y with the above property. Notation. σ(x) denotes the set of y in J such that (x, y) ∈ σ. There is an easy procedure for constructing self-similar coverings. For simplicity, let us assume that a countable dense subset J of Y is given together with a function r : J →]0, ∞[ going to zero at infinity. Suppose we have a map η : X → J N that associates to each θ ∈ X a sequence (xn) in J tending to infinity with d(xn, θ) ≤ r(xn) for all n. For each x in J, let Bη(x) be the set of all θ in X whose associated sequence contains x. Define ση by the criterion that y ∈ ση(x) if there is a sequence associated to some θ in X that contains both x and y, with y following immediately after x. Then (J, ση,Bη) is a self-similar covering of X. Theorem 3.2. ([6]) Let Y be a metric space, let X be a subset of Y that admits a self-similar covering (J, σ, B) and let s be a positive real number. If for all x in J, X (3.1) diam B(y)s ≤ diam B(x)s, y∈σ(x) then Hdim X ≤ s. Remark. This Theorem has been proved in the case X = Y in [5]. In [6], it has already been noted that in the more general case of a subset X of Y , the arguments of the proof given in [5] apply with essentially no changes. Definition 3.3. Let s(J, σ, B) be the best possible upper bound provided by Theorem 3.2

applied to a fixed self-similar structure; in other words, s(J, σ, B) = supx∈J s(x) where s(x) is the infimum over all positive real numbers satisfying (3.1). It is possible to have s(J, σ, B) = ∞, which is a situation we want to avoid. Note that if the self-similar coverings (J, σ1,B) and (J, σ2,B) are such that σ1(x) ⊂ σ2(x) for all x in J, then s(J, σ1,B) ≤ s(J, σ2,B). The problem of computing s(J, ση,Bη) reduces to the problem of enumerating the set ση(x) for which an exact description is most likely impossible. Also, the need to use estimates on diam Bη(x) (as opposed to an exact formula) can further complicate calculations. To overcome these difficulties, observe that if σ ⊂ J 2 and a map B from X to the set of bounded subsets of Y are such that for all x ∈ J, σ(x) ⊃ ση(x) and B(x) ⊃ Bη(x), then (J, σ, B) is also a self-similar covering of X. The choice of σ and B may considerably simplify the evaluation of s(J, σ, B) (for example, by choosing B(x) a metric ball of some specified radius) and one may reasonably expect s(J, σ, B) to be close to s(J, ση,Bη) if σ is a sufficiently good 1 approximation to ση. Our initial approach is to let η be the encoding by the sequence of best approximates. There are two natural choices for B(x), one given by Lemma 4.2 below and one given by

1 It is not always optimal to choose B approximating Bη as eﬃciently as possible. See Remark 5.3.

9 diam B(y) Dirichlet theorem; but regardless of the choice, the ratio diam B(x) may be close to one as y ranges over ση(x), which leads to ineffective upper bounds. This obstacle can easily be overcome by using an acceleration of η instead, i.e. an encoding η0 : X → J N such that for each x ∈ J, η0(x) is a subsequence of η(x). A potential drawback of using the acceleration is that ση0 may be more difficult to approximate than ση. In Section 5.1, we specify the particular approximation to ση0 that we will use. We do not specify η0 explicitly, although it can easily be inferred from the proof of Lemma 5.2. 3.2. Lower estimates. Theorem 3.4. Let Y be a metric space and let X be a subset of Y . Suppose that there is a self-similar structure (J, σ, B) that covers a subset of X (see Definition 3.1) and constants c, s ≥ 0 and ρ ∈]0, 1[ such that • for all x ∈ J, σ(x) is finite, B(x) is a nonempty compact subset of Y , and for all y ∈ σ(x) we have B(y) ⊂ B(x), • for each admissible sequence (xn), limn→∞ diam B(xn) = 0, • for each x ∈ J, and each y ∈ σ(x), there are at most c points z in σ(x) \{y} such that d(B(y),B(z)) ≤ ρ diam B(x), • for every x ∈ J, X diam B(y)s ≥ (c + 1) diam B(x)s, y∈σ(x) • for all x ∈ J, and each y ∈ σ(x), diam B(y) < diam B(x). Then X contains a subset of positive s-dimensional Hausdorff measure. Proof. By Theorem 3.2 of [5], it is enough to construct for each x in J, a subset σ0(x) of σ(x) such that - σ0(x) contains at least two points, - for any pair of distinct y, z in σ0(x), d(B(y),B(z)) > ρ diam B(x), P s s - y∈σ0(x) diam B(y) ≥ diam B(x) .

Choose y0 in σ0(x) = σ(x) such that diam B(y0) is maximal. Remove from σ(x) the subset τ(y0) consisting of y in σ(x) such that d(B(y0),B(y)) ≤ ρ diam B(x). This leaves a new set σ1(x) = σ(x) \ τ(y0). Note that y0 is in τ(y0). Next, choose y1 in σ1(x) such that diam B(y1) is maximal. Remove from σ1(x) the subset τ(y1) consisting of y in σ1(x) such that d(B(y1),B(y)) ≤ ρ diam B(x), leaving a new set σ2(x) = σ1(x) \ τ(y1). We continue this process until σn(x) is empty. Since σ(x) is a ﬁnite set, we get a ﬁnite sequence y0, . . . , yn−1 such that for i = 0, . . . , n−1,

- diam B(yi) ≥ diam B(y) for all y in σi(x), - d(B(yi),B(y)) > ρ diam B(x) for all y in σi+1(x). It follows that

- diam B(yi) ≥ diam B(y) for all y in τ(yi),

10 - d(B(yi),B(yj)) > ρ diam B(x) for all 0 ≤ i 6= j < n. 0 Set σ (x) = {y0, . . . , yn−1}. We have n−1 X X X diam B(y)s = diam B(y)s

y∈σ(x) i=0 y∈τ(yi) n−1 X s ≤ card τ(yi) × diam B(yi) i=0 n−1 X s ≤ (c + 1) diam B(yi) i=0 X = (c + 1) diam B(y)s, y∈σ0(x) therefore X diam B(y)s ≥ diam B(x)s. y∈σ0(x) Since diam B(y) < diam B(x), this last inequality ensures that σ0(x) contains at least two points. Definition 3.5. By a strictly nested self-similar structure we mean one that satisfies the first two and the last conditions of Theorem 3.4. To obtain the dimension lower bound for singular vectors, we shall require the following weighted analog of Theorem 3.4. We omit the proof which goes exactly the same way: just replace ρ diam B(x) by ρ(x) diam B(x) and diam B(y) by ρ(y) diam B(y) almost everywhere. Theorem 3.6. Let Y be a metric space and let X be a subset of Y . Suppose that there is a self-similar structure (J, σ, B) that covers a subset of X, a subset J0 ⊂ J that contains a tail of any σ-admissible sequence, a function ρ : J →]0, 1[ and two constants c, s ≥ 0 such that

• for all x ∈ J0, σ(x) is ﬁnite, B(x) is a nonempty compact subset of Y , and for all y ∈ σ(x) we have B(y) ⊂ B(x), • for each admissible sequence (xn), limn→∞ diam B(xn) = 0, • for each x ∈ J0, and each y ∈ σ(x), there are at most c points z in σ(x) \{y} such that d(B(y),B(z)) ≤ ρ(x) diam B(x),

• for every x ∈ J0, X (ρ(y) diam B(y))s ≥ (c + 1)(ρ(x) diam B(x))s, y∈σ(x)

• for all x ∈ J0, and each y ∈ σ(x), ρ(y) diam B(y) < ρ(x) diam B(x). Then X contains a subset of positive s-dimensional Hausdorﬀ measure.

11 4. Farey lattices Let the set of primitive vectors in Zd+1 corresponding to rationals in Qd in their ”lowest terms representation” be denoted by

d+1 Q = {(p1, . . . , pd, q) ∈ Z : gcd(p1, . . . , pd, q) = 1, q > 0}. Given x = (p, q) ∈ Q, where p ∈ Zd, we use the notation p |x| = q and x = . b q For x in Q, let d d+1 Λx := Z + Zxb = πx(Z ) d+1 d where πx : R → R is the ”projection along lines parallel to x” given by the formula d −1 πx(m, n) = m − nx for (m, n) ∈ R × R. Observe that vol Λx = |x| . b d Given a norm on R , we denote successive minima of Λx by λi(x) and the normalized successive minima by

1/d λbi(x) := |x| λi(x) for i = 1, . . . , d.

4.1. Inequalities of best approximation. We gather some ”well-known lemmas” about best simultaneous approximations and, for the convenience of the reader, reproduce their proofs in our notation. For example, the proof of Lemma 4.1 below is taken from [7]. Recall d d that a vector (p, q) in Z × Z>0 is a best simultaneous approximation vector of θ ∈ R with d respect to the norm k · k, if for all (u, v) in Z × Z>0,

|v| < |q| ⇒ kp − qθk < ku − vθk, |v| = |q| ⇒ kp − qθk ≤ ku − vθk.

d d The set of best approximation vectors of θ ∈ R \ Q is an inﬁnite sequence (pn, qn)n. This sequence can be arranged so that the denominators qn are increasing. See [7], [16], [17],[19] for more about best approximations. In what follows we shall often write best approximate instead of best simultaneous approximation vector. We drop the ﬁrst best approximation vectors in order that the inequality kqnθ − pnk < 1 d 2 inf{kxk : x ∈ Z \{0}} holds for every n ≥ 0. d Lemma 4.1. If xn = (pn, qn) and xn+1 = (pn+1, qn+1) ∈ Z × Z>0 are two consecutive best approximation vectors of θ ∈ Rd, then d (i) qn+1kqnθ − pnk 1, and 1 (ii) 2 λ1(xn+1) < kqnθ − pnk < 2λ1(xn+1).

Proof. By Minkowski’s theorem, λb1(xn+1) 1 so that (i) is a consequence of the second

inequality in (ii). For the ﬁrst inequality in (ii), note that πxn+1 (xn) is a nonzero element of

12 Λ whose length is kp − q x k = |x |d(x , x ). Therefore, xn+1 n nbn+1 n bn bn+1 λ1(xn+1) ≤ |xn|d(xbn, xbn+1) ≤ |xn|kxbn − θk + |xn|kxbn+1 − θk |xn| = kqnθ − pnk + kqn+1θ − pn+1k |xn+1|

< 2kqnθ − pnk. d For the second inequality in (ii), note that λ1(xn+1) = dist(qxbn+1, Z ) for some integer q ∈ [1, q − 1]. Since dist(qx , d) = dist((q − q)x , d), we may assume q ≤ qn+1 n+1 bn+1 Z n+1 bn+1 Z 2 by replacing q with qn+1 − q, if necessary. Therefore, d d λ1(xn+1) = dist(qxbn+1, Z ) = dist(qθ + q(xbn+1 − θ), Z ) d ≥ dist(qθ, Z ) − dist(qθ + q(xbn+1 − θ), qθ) q ≥ kqnθ − pnk − kqn+1θ − pn+1k qn+1 1 > kq θ − p k. 2 n n Lemma 4.2. For x ∈ Q, let ∆(x) = {θ : xb is a best approximate of θ}. Then λ (x) 2λ (x) B¯(x, 1 ) ⊂ ∆(x) ⊂ B(x, 1 ). b 2|x| b |x|

Proof. Let θ be in ∆(x) and let xn = (pn, qn) be the sequence of best approximation vectors of θ. Then x = xn for some n. By Lemma 4.1, kqnθ − pnk ≤ kqn−1θ − pn−1k < 2λ1(x), hence

pn 2λ1(x) θ − < qn qn giving the second inclusion. For the ﬁrst inclusion, suppose that θ ∈ B¯(x, λ1(x) ). For all b 2|x| positive integers q < |x| and all p ∈ Zd we have kqθ − pk + qkθ − xbk ≥ kqxb − pk ≥ λ1(x) so that λ (x) λ (x) kqθ − pk ≥ λ (x) − q 1 > 1 1 2|x| 2 whereas λ (x) k |x|θ − |x|xk = |x| kθ − xk ≤ 1 . b b 2 Therefore, by deﬁnition, x is a best approximation vector of θ. 1 (iii) of the next Lemma is taken from [5]. The notation x 2 y means 2 y ≤ x ≤ 2y.

13 Lemma 4.3. Let xn = (pn, qn), n ≥ 0 be the sequence of best approximation vectors of θ ∈ Rd. Then

4λ1(xn+1) (i) kxn − xn+1k < . b b |xn| 4λ1(xn) (ii) For all k ≥ 0, kxn − xn+kk < . b b |xn| d+1 (iii) For all y = (p, q) ∈ Z with 0 < q < |xn|, kp − qθk 2 kp − qxbnk.

Proof. (i) By Lemma 4.1, kqnθ − pnk < 2λ1(xn+1). Therefore,

kqnxbn+1 − pnk ≤ kqnθ − pnk + kqn(xbn+1 − θ)k qn = kqnθ − pnk + kpn+1 − qn+1θk qn+1

< 2kqnθ − pnk < 4λ1(xn+1).

(ii)

kxbn − xbn+kk ≤ kxbn − θk + kxbn+k − θk kq θ − p k kp − q θk = n n + n+k n+k qn qn+k 2kq θ − p k 2kq θ − p k 4λ (x ) ≤ n n < n−1 n−1 < 1 n . qn qn qn

0 0 (iii) Let (p , q ) = xn − y. Since

0 0 0 0 0 qnkxbn − θk = kpn − qnθk < kp − q θk ≤ kp − q xbnk + q kxbn − θk, we have

0 q 0 0 (1 − )kpn − qnθk < kp − q xbnk = kp − qxbnk, qn thus q kp − q θk < n kp − qx k. n n q bn

Using once again the triangle inequality, we get

kp − qθk ≤ kp − qxbnk + qkxbn − θk < 2kp − qxbnk.

14 The poof of the opposite inequality is straightforward:

kp − qxbnk ≤ kp − qθk + qkθ − xbnk q = kp − qθk + kqnθ − pnk qn q ≤ kp − qθk + kqθ − pk qn < 2kp − qθk.

5. Proofs of Theorems: upper bounds We want to use self-similar covers and Theorem 3.2 to give upper bounds on Hausdorff dimension. The main difficulty is to define the map σ of the self-similar structure (J, σ, B). We shall use the encoding via best approximates together with an acceleration as suggested in Section 3.1.

5.1. Acceleration. For each x in Q we ﬁx once and for all a codimension one sublattice 0 −1 0 0 0 of Λx of minimal volume and call it Λx. Let Hx = πx Hx where Hx is the real span of Λx. Thus, 0 0 Λx = Λx ∩ Hx. Deﬁnition 5.1. Let ε be a positive real number and let x be an element in Q. (1) D(x) is the set of y ∈ Q such that |x| ≤ |y|, y ∈ H , and kx − yk ≤ 4λ1(x) . x b b |x| (2) Let y be in D(x). E(x, y, ε) is the set of z ∈ Q such that |y| < |z|, z 6∈ Hx, and ky − zk < ε . b b |y||z|1/d S (3) σε(x) = σ(x, ε) = y∈D(x) E(x, y, ε).

Let Q = {x ∈ Q : λ (x) < ε}, B(x) = B(x, 2µd ) where µ is the supremum of λ (L) ε b1 b |x|1+1/d d b1 over all d-dimensional lattices L ⊂ Rd, and let d DIfε(d) = {θ ∈ R : the best approximates xn of θ are in Qε for n large enough} and ∗ DIfε(d) = {θ ∈ DIfε(d) : 1, θ1, . . . , θd are linearly independent over Q}. From Lemma 4.3(iii), it is clear that

DIfε/2(d) ⊂ DIε(d) ⊂ DIf3ε(d).

So Sing(d) = ∩ε>0DIfε(d) and we can deal with DIfε(d) instead of DIε(d). ∗ Lemma 5.2. (Qε, σε,B) is a self-similar covering of DIfε/4(d).

15 2λ1(x) Proof. For any best approximate x of θ, Lemma 4.2 implies ∆(x) ⊂ B(x, |x| ) ⊂ B(x) ∗ so that θ ∈ B(x). We need to verify that for any θ ∈ DIfε/4(d), the sequence of best approximates (xn) contains a σε-admissible subsequence. By deﬁnition, there exists an n0 such that xn ∈ Qε/4 for all n > n0. Given ni, let ni+1 be the smallest integer m > ni such that x 6∈ H . Since 1, θ , . . . , θ are linearly independent over , n is deﬁned for all i ∈ . m xni 1 d Q i Z≥0

Now suppose x = xni for some i. Let y = xni+1−1 and z = xni+1 . Then |x| ≤ |y| < |z|, y ∈ Hx, z 6∈ H , and by Lemma 4.3, kx − yk < 4λ1(x) and ky − zk < 4λ1(z) = 4λb1(z) < ε . x b b |x| b b |y| |y||z|1/d |y||z|1/d Remark 5.3. It could seem that B0(x) = B(x, 2λ1(x) ) is a better choice for the map B than b |x| B(x) = B(x, 2µd ) because the balls B0(x) are included in the balls B(x). However, the b |x|1+1/d 0 exponent s(Qε, σε,B ) is more diﬃcult to estimate and might be greater than s(Qε, σε,B). Indeed, λb1(x)/ε small does not imply λb1(z)/ε small for z in σε(x). The only constraint on z is λb1(z) < ε. Therefore, for some x in Qε, the optimal value of the exponent s may increase when using B0 instead of B.

For x ∈ Q we shall simply write ex for e(Λx) (see Section 2.3).

Lemma 5.4. For all y ∈ D(x), ey ≤ ex + 4λ1(x).

Proof. Assume r := ey − 4λ1(x) > 0 for otherwise the result is clear. Let p be a point such that B(p, ey) ∩ Λy = ∅. It is enough to prove that B(p, r) ∩ Λx = ∅. Suppose not. d Then there exists (a, b) ∈ Z × Z with 0 ≤ a < |x| such that d(p, axb + b) < r. But then d(axb + b, ayb + b) = akxb − ybk < 4λ1(x) so that ayb + b ∈ B(p, ey) ∩ Λy. 0 Lemma 5.5. Let x and y be in Q. Then y ∈ Hx if and only if yb ∈ xb + Hx . 0 −1 Proof. Note that y ∈ Hx if and only if πx(y) ∈ Λx and that yb − xb = |y| πx(y). 0 Lemma 5.6. Let y ∈ D(x) and α ∈ Λy \ Hx. Then λd(x) λd(y) kαk. 0 0 Proof. Since y − x ∈ Hx, we have Λy ⊂ Λx + R(y − x) ⊂ Λx + Hx. Hence, by Lemma 2.2, b b0 b b kαk e(Λx + Hx) λd(x). Now Lemma 5.4 implies ey ex λd(x), whereas e(Λy) ≥ 0 e(Λx +Hx) λd(x). And since ey λd(y), by Lemma 2.1, it follows that λd(x) λd(y). 5.2. Upper bound calculation and results. Lemma 5.7. For t > d we have t X |x| 1 . |y| t − d y∈D(x)

Proof. For each k > 0, let Dk := {y ∈ D(x): k|x| ≤ |y| < (k + 1)|x|}. Since t ∞ t ∞ X |x| X X |x| X card Dk = ≤ |y| |y| kt y∈D(x) k=1 y∈Dk k=1

16 it is enough to estimate card Dk. For all y = (p, q) ∈ Dk, 4λ (x) kπ (y)k = kp − qxk = qky − xk ≤ q 1 ≤ 8kλ (x). x b b b |x| 1 d Since Hx is a hyperplane which contains x, its projection is a subspace of R of dimension d+1 d − 1. Since x is a primitive element of Z , the projection πx induces a bijection from d+1 {(u, v) ∈ Z : k|x| ≤ v < (k + 1)|x|} onto Λx. It follows that πx induces a bijection from d+1 0 {(u, v) ∈ Hx ∩ Z : k|x| ≤ v < (k + 1)|x|} onto Λx = Λx ∩ πx(Hx). So we have to count 0 the number of elements of the lattice Λx inside the ball B(0, 8kλ1(x)). 0 Since rank Λx = d − 1, making use of Lemma 2.3, we ﬁnd that

d−1 i X (kλ1(x)) card D 1 + k λ (Λ0 ) . . . λ (Λ0 ) i=1 1 x i x d−1 i X (kλ1(x)) 1 + kd−1 λ (x) . . . λ (x) i=1 1 i 0 0 because Λx ⊂ Λx implies λi(Λx) ≥ λi(x) for i = 1, . . . , d − 1. Lemma 5.8. For t > d we have t X |y| εt . |z| t − d z∈E(x,y,ε)

0 Proof. Note that E(x, y, ε) is the set of z ∈ Q such that α := πy(z) ∈ Λy \ Hx, and kαk|y| < 1−1/d P P ε|z| . Hence, we have a double sum of the form α z where α ranges over the set 0 −1 Λy \ Hx and z ranges over the subset of πy α satisfying |z| 1 kαk|y|d/(d−1) kαk|y|1/d d/(d−1) > = |y| |y| ε ε −1 Since z ∈ πy α is uniquely determined by |z|/|y|, we have t t0 X |y| X ε 1/d |z| 0 kαk|y| z∈E(x,y,ε) α∈Λy\Hx 0 d(t−1) 0 where t = d−1 > t. Since α 6∈ Hx, Lemma 5.6 imply −1/d kαk λd(y) λd(x) |x| . Applying Lemma 2.5, we now get t 0 0 (t0−d)/d X |y| εt |x|(t −d)/d εt |x| εt 0 t0/d ≤ ≤ . |z| (t − d) vol(Λy)|y| (t − d) |y| (t − d) z∈E(x,y,ε)

17 d2 d/2 Corollary 5.9. Hdim DIε(d) ≤ d+1 + O(ε ). Proof. Applying the preceding two lemmas with t = (1 + 1/d)s we have for some C > 0 (1+1/d)s t t X |x| X |x| X |y| Cεt (5.1) ≤ ≤ |z| |y| |z| (t − d)2 z∈σ(x,ε) y∈D(x) z∈E(x,y,ε) √ √ t/2 d2 d+1 d/2 which is ≤ 1 provided t > d + Cε , which holds for any s > d+1 + d Cε . Hence, ∗ d2 d/2 by Lemma 5.2 and Theorem 3.2, Hdim DIfε/4(d) ≤ d+1 + O(ε ) and the same holds for Hdim DIε(d). d2 Corollary 5.10. Hdim Sing(d) ≤ d+1 .

6. Geometry of quotient lattices For the remainder of the paper, we assume the Euclidean norm on Rd. Given y ∈ Q and a primitive element α in Λy, we let ⊥ Λα⊥ = πα (Λy) ⊥ d ⊥ d where πα is the orthogonal projection of R onto α , the subspace of vectors of R orthogonal to α. For any z ∈ Q such that πy(z) = α, the (d − 1)-volume of Λα⊥ satisﬁes

vol(Λy) 1 1 vol(Λ ⊥ ) = = = . α kαk kαk|y| |y ∧ z| Here, the quantity |y ∧ z| is the 2-volume of the orthogonal projection of y ∧ z ∈ Λ2Rd+1 onto the subspace spanned by e1 ∧ ed+1, . . . , ed ∧ ed+1. Equivalently, (see §2 of [5])

|y ∧ z| = |y||z|d(y,b zb).

Denote the ﬁrst successive minimum of Λα⊥ by λ1(α) and its normalized form by 1/(d−1) λb1(α) := |y ∧ z| λ1(α). |y∧z| |y∧z| Note that for any z ∈ Q, we have πz(y) ∈ Λz and kπz(y)k = |z| . Thus, λ1(z) ≤ |z| provided y is in Q and y 6= z. Many of the estimates we need rely on a lower bound for the quantity λ1(z), which is the main point of the next lemma.

Lemma 6.1. Let y ∈ Q and let α be a primitive element of Λy. Suppose z is an element in |y∧z| |y∧z| Q such that πy(z) = α. Then |z| ≤ λ1(α) implies λ1(z) = |z| .

2 d+1 Proof. Note that since α is a primitive element of Λy, y ∧ z is a primitive element of Λ Z , ⊥ which in turn implies that β = πz(y) is a primitive element in Λz. Let Λβ⊥ = πβ (Λz) be the orthogonal projection onto β⊥. Note that α and β are proportional to z − y so that ⊥ ⊥ ⊥ d+1 b b πα (k − lyb) = πβ (k − lyb) = πβ (k − lzb) for all (k, l) ∈ Z . Hence, Λα⊥ = Λβ⊥ .

18 |y∧z| Now suppose |z| ≤ λ1(α) and consider a nonzero γ ∈ Λz. If γ is not a real multiple of |y∧z| β, then kγk ≥ λ1(Λβ⊥ ) = λ1(α) ≥ |z| = kβk. On the other hand, if γ = tβ for some t ∈ R then since β is primitive, 0 6= t ∈ Z so that kγk = |t|kβk ≥ kβk. In either case kγk ≥ kβk. |y∧z| Therefore, λ1(z) = kβk = |z| . Deﬁnition 6.2. Given a positive real number ε and y in Q, let

Λy(ε) := {α ∈ Λy : α is primitive and λb1(α) > ε}.

The sets Λy(ε) will be involved in the definition of the self-similar structure we construct for the lower bound argument. In Section 8 we study the distribution of Λy(ε) inside Λy. 0 6.1. Self-similar structure for Dirichlet set. Recall the subspace Hx defined as the real 0 span Λx, which is a distinguished codimension one sublattice of Λx of minimal volume. By a primitive coset of H0 we shall mean a coset H0 such that Λ ∩H0 6= and Λ \H0 ⊂ S nH0. x x ∅ x x n∈Z There are exactly two possibilities for H0 and we fix a choice once and for all. The unique 0 0 ⊥ element of H that is perpendicular to Hx will be denoted by αx . Note that, from the proof of Lemma 2.2, we have ⊥ vol(Λx) λd(x) kαx k = 0 . vol(Λx) ⊥ 0 0 0 0 The coset nαx + H will be denoted by Hx(n). Since, by Lemma 5.5, Λy + Hx = Λx + Hx for any y ∈ Hx ∩ Q, we have for all y ∈ D(x) 0 [ 0 Λy \ Hx ⊂ Hx(n). n∈Z, n6=0 d ⊥ −1 Let C(x) be the set of vectors in R whose angle with αx is at most tan Ad, where the constant Ad, which depends only on the dimension, will be determined later in the proof of Lemma 8.5. Let N 0 0 [ 0 Cn(x) = C(x) ∩ Hx(n) and CN (x) = Cn(x). n=1 Definition 6.3. Let N be a positive integer, ε be a positive real number, x in Q and y in |y∧z|d/(d−1) D(x). F (x, y, ε) is the set of z ∈ Q such that πy(z) ∈ C(x) ∩ Λy(ε) and |z| ∈ εd/(d−1) ]1, 2[. 0 The subset of F (x, y, ε) consisting of z ∈ Q such that πy(z) ∈ Cn(x) for some n ∈ {1,...,N} will be denoted by FN (x, y, ε). Equivalently, [ FN (x, y, ε) = ζ(y, α, ε)

α∈Λy(ε)∩CN (x)

|y∧z|d/(d−1) where ζ(y, α, ε) = {z ∈ Q : πy(z) = α, |z| ∈ εd/(d−1) ]1, 2[}.

The sets σ(x) = FN (x, x, ε) provide the main example of a self-similar structure for the lower bound argument, enough for the proofs of Theorems 1.1 and 1.3. Only the proof of Theorem 1.4 will require y 6= x. We need to verify that the set covered by this self-similar

19 structure is contained in the Dirichlet set (see Proposition 6.6 below). For that, we ﬁrst need a couple of lemmas.

|y∧z| ε Lemma 6.4. Let 0 < ε ≤ 1 and z ∈ F (x, y, ε). Then λ1(z) = |z| and 2 < λb1(z) < ε. Moreover |z| > |y| when ε is small enough.

|y∧z|d/(d−1) 1/(d−1) Proof. Let α = πy(z). Since |z| ≥ εd/(d−1) and λb1(α) = |y ∧ z| λ1(α) > ε, |y ∧ z| εd/(d−1) (6.1) ≤ < ε1/(d−1)λ (α). |z| |y ∧ z|1/(d−1) 1

|y∧z| Since α is primitive and ε ≤ 1, it follows from Lemma 6.1 that λ1(z) = |z| . Therefore, 1/d |y∧z| ε ε λb1(z) = |z| λ1(z) = |z|1−1/d ∈ 21−1/d , ε ⊂ 2 , ε . By Lemma 5.6, kαk λd(y). Hence

d d d |y| |y|ε d−1 ε d−1 ε d−1 d d−1 (6.2) ≤ d 1 d = d ε , |z| d−1 d−1 d−1 (|y|kαk) (|y| d λd(y)) λbd(y) and therefore |z| > |y| when ε is small enough. 6.2. Nestedness and containment.

Lemma 6.5. Let ε be a positive real number and let x ∈ Q and y ∈ D(x) satisfy λb1(y) ε. Then for all z ∈ F (x, y, ε), the ball B(z, 2λ1(z) ) is contained in the ball B(y, λ1(y) ) when ε is b |z| b 2|y| small enough.

Proof. Let α = πy(z). Step 1. kαk > λ1(y) when ε is small enough. d−1 By Minkowski’s theorem, 1 λb1(y)λbd(y) and since λb1(y) ε, we have

1 1 d − d−1 − d−1 λbd(y) 1 > ε ε λb1(y). λb1(y) d−1

d 0 − d−1 Since α 6∈ Hx, it follows from Lemma 5.6 that kαk λd(y) ε λ1(y) so that kαk > λ1(y) when ε is small enough. Step 2. λ1(α) ≤ λ1(y). Let β ∈ Λy be such that kβk = λ1(y). Then β is a primitive element whose length is strictly smaller than that of α, by Step 1. Hence, it cannot be an integer multiple of α. Thus, πα(β) is a nonzero vector in Λα⊥ , whose length is bounded above by λ1(y). Step 3. Estimate d(y, z) in terms of λ1(y) . b b |y| From (6.1) in the proof of Lemma 6.4 together with Step 2, we have |y ∧ z| λ (α) λ (y) d(y, z) = < ε1/(d−1) 1 ≤ ε1/(d−1) 1 . b b |y||z| |y| |y|

20 λ1(z) λ1(y) Step 4. Estimate |z| in terms of |y| . Applying Lemma 6.4 together with Step 3, we have λ (z) |y ∧ z| |y| λ (y) 1 = = d(y, z) < ε1/(d−1) 1 . |z| |z|2 |z| b b |z|

|y∧z|d/(d−1) Since |z| ≥ εd/(d−1) , we get d+1 λ (z) ε d−1 λ (y) 1 < 1 . |z| |y ∧ z|d/(d−1) 0 1−1/d (d−1)/d Since α 6∈ Hx, we have |y ∧ z| = kαk|y| λd(y)|y| = λbd(y)|y| |y| so that

λ1(z) d+1 λ1(y) ε d−1 . |z| |y| Step 5. B(z, 2λ1(z) ) ⊂ B(y, λ1(y) ) when ε is small enough. b |z| b 2|y| This is a consequence of the preceding two steps by the triangle inequality. Proposition 6.6. Let σ (x) = F (x, x, ε), B(x) = B(x, λ1(x) ) and Q = S σ (x). ε,N N b 2|x| ε,N x∈Q ε,N Then (Qε,N , σε,N ,B) is a strictly nested self-similar structure (see Deﬁnitions 3.1 and 3.5) covering a subset of DI2ε(d) provided ε is small enough.

Proof. We need to verify that for each σε,N -admissible sequence (xk),

(i) for all k, B(xk+1) ⊂ B(xk) and diam B(xk+1) < diam B(xk), T (ii) limk→∞ diam B(xk) = 0 and B(xk) is a point contained in DI2ε(d).

Note that Lemma 6.5 implies (i) when ε is small enough. Moreover, by Lemma 6.4, |xk| → ∞, T hence limk→∞ diam B(xk) = 0 and k B(xk) consists in a single point θ. Since θ ∈ B(xk) ⊂ ∆(xk), each element xk of the σε,N -admissible sequence is a best approximate to θ. It remains to show that θ ∈ DI2ε(d). Let |xk| ≤ q < |xk+1|. By Lemma 4.3 (iii), we have d min d(lθ, Z ) ≤ kqkθ − pkk 0

Observe that the latter inequality only uses the fact that xk+1 is a best approximation vector. On the other hand, by deﬁnition of FN (xk, xk, ε)

|xk ∧ xk+1| −1/d kqkxbk+1 − pkk = < ε|xk+1| . |xk+1| Therefore, d −1/d min d(lθ, Z ) < 2εq . 0

21 6.3. Spacing and local ﬁniteness. In order to estimate the distance between two balls 0 λ1(z) 0 λ1(z ) 0 B(z, 2|z| ) and B(z , 2|z0| ) where z and z are in Q, we compare the distance between the 0 points zb and zb with the diameters of the balls. So we introduce the ratio 0 0 d(z,b zb) R(z, z ) = 0 . λ1(z) λ1(z ) |z| + |z0| 0 0 Lemma 6.7. Let 0 < ε ≤ 1 and y ∈ D(x). For all z, z in FN (x, y, ε), R(z, z ) > 1 implies

d+1 2d λ (z) λ (z) 1 d−1 ε d−1 λ (y) d B(z, 1 ),B(z0, 1 ) 1 . b b 0 d 2|z| 2|z | N d−1 |y| λbd(y) 0 0 Proof. Let z, z be in F (x, y, ε) and α = πy(z). By deﬁnition of R(z, z ), the distance between 0 the balls B and B0 centered at z and z0 with respective radii r = λ1(z) and r0 = λ1(z ) is b b 2|z| 2|z0| 0 0 0 0 0 d(B,B ) = d(z,b zb) − r − r = (2R(z, z ) − 1) (r + r ). λ1(z) λ1(y) Hence, it is enough to bound the ratio ( |z| ) ( |y| ) from below. |y∧z|d/(d−1) Applying Lemma 6.4 together with |z| ≤ 2 εd/(d−1) we have 2d 2d λ1(z) |y ∧ z| ε d−1 ε d−1 = 2 ≥ d+1 = d+1 d+1 |z| |z| 4|y ∧ z| d−1 4kαk d−1 |y| d−1 so that 2d λ (z) |y| ε d−1 1 |y| 1 × d+1 d+1 − 1 |z| λ1(y) kαk d−1 |y| d−1 λb1(y)|y| d 2d ε d−1 1 1 = d+1 d+1 kαk d−1 |y| d(d−1) λb1(y) d+1 2d d−1 λd(y) ε d−1 1 = 1 d+1 . kαk d−1 λb1(y) (λd(y)|y| d ) d−1 By Minkowski’s theorem, λb1(y) λbd(y) 1, hence

d+1 2d d−1 λ1(z) |y| λd(y) ε d−1 d . |z| λ (y) kαk d−1 1 λbd(y) 0 Since the angle α makes with the normal to Hx is 1, we have kαk Nλd(y) so that d+1 2d d−1 λ1(z) |y| 1 ε d−1 d . |z| λ (y) N d−1 1 λbd(y)

22 Lemma 6.8. Let ε be a positive real number and y ∈ D(x). For all z ∈ F (x, y, ε), the number of z0 ∈ F (x, y, ε) with R(z, z0) ≤ 1 is 1 when ε is small enough.

0 0 0 Proof. Let α = πy(z) and α = πy(z ) be the projections of z and z in F (x, y, ε). We ﬁrst kα0k d−1 0 0 show that if kαk ≥ 4 then R(z, z ) > 1. Observe that it implies that R(z, z ) > 1 when kαk d−1 |y∧z|d/(d−1) kα0k ≥ 4 . Since |z| ∈ εd/(d−1) [1, 2],

d d kαk kαkε d−1 ε d−1 ky − zk = ≥ = , b b d d 1 |z| 2|y ∧ z| d−1 2|y| d−1 kαk d−1 and similarly,

d d kα0k kα0kε d−1 ε d−1 (6.3) y − z0 = ≤ = , b b 0 d d 1 |z | |y ∧ z0| d−1 |y| d−1 kα0k d−1 It follows that

0 0 zb − zb ≥ kyb − zbk − yb − zb d d d ε d−1 ε d−1 ε d−1 ≥ d 1 − d 1 ≥ d 1 . 2|y| d−1 kαk d−1 |y| d−1 kα0k d−1 4|y| d−1 kαk d−1

|y∧z| 0 Since λ1(z) ≤ |z| , the denominator of R(z, z ) is bounded above by kαk kα0k |y| + |z|2 |z0|2 which can be bounded above as before: dividing (6.3) by |z0|, we get the upper bound

2d ! 2d ε d−1 1 1 ε d−1 2 d+1 d+1 + d+1 ≤ d+1 × d+1 . |y| d−1 kαk d−1 kα0k d−1 |y| d−1 kαk d−1 It follows that 1 d d−1 d−1 0 |y| kαk R(z, z ) ≥ d . 8ε d−1

Since z ∈ F (x, y, ε), Lemma 5.6 implies kαk λd(y), so that, by Minkowski’s theorem, 1 d |y| d−1 kαk d−1 1. Therefore R(z, z0) > 1 for ε small enough. 1 0 d−1 |y∧z|d/(d−1) Suppose now that 4d−1 kαk ≤ kα k ≤ 4 kαk. Again, since |z| ∈ εd/(d−1) [1, 2], and 0 |y∧z0|d/(d−1) |z | ∈ εd/(d−1) [1, 2] we have

d d |z0| |y ∧ z0| d−1 kα0k d−1 = 1. |z| |y ∧ z| kαk

23 |y∧z| |y|kαk 0 |y|kα0k Furthermore, by lemma 6.4, λ1(z) = |z| = |z| and λ1(z ) = |z0| , hence

λ (z0) λ (z) 1 1 . |z0| |z|

Assume now that z is ﬁxed. We wanted to count the number of z0 = (u0, v0) ∈ F (x, y, ε) such that R(z, z0) ≤ 1. With the above inequality R(z, z0) ≤ 1 implies kz0 − zk λ1(z) . It b b |z| follows that ku0 − v0zk |z0| λ1(z) λ (z). On the one hand, by Lemma 2.3 the number b |z| 1 d of points of the lattice Λz in a ball B(0, rλ1(z)) is 1 + r . On the other hand, since z 0 00 d+1 0 00 0 00 is primitive, if z and z are in the lattice Z and if πz(z ) = πz(z ) then z − z ∈ Zz; together with the inequality |z0| |z| this implies that there are at most O(1) z0s for a given 0 0 0 value of z ∧ z. It follows that the number of z in F (x, y, ε) with R(z, z ) ≤ 1 is 1.

6.4. Lower bound calculations. The next proposition shows that to get lower bounds on Hausdorﬀ dimension, it is enough to control the distribution of Farey lattices with bounded distortion.

Proposition 6.9. Let ε be a positive real number and let y be in D(x). If ε is small enough, then for all N ∈ Z>0 and all real numbers s and t,

s s+(t−1) d X λb1(z) ε d−1 S1(N, t) t (t−d) d |z| λ (y) d−1 |y|t z∈FN (x,y,ε) bd

where

N 0 X 1 card Cn(x) ∩ Λy(ε) S1(N, t) = d 0 . 1+(t−d) d−1 card C (x) ∩ Λ n=1 n n y

Proof. Since ||α|| λd(y),

d d ! d−1 d−1 1 ||α|||y| λbd(y) ≥ |y| ε ε which is > 1 when ε is small enough. This implies both card ζ(y, α, ε) ≥ 1 and

d ||α|||y|1/d d−1 card ζ(y, α, ε) . ε

24 0 By Lemma 6.4, λb1(z) ε. Also, for any α ∈ Λy ∩ Cn(x), kαk nλd(x) nλd(y). Thus, 0 splitting the sum according to α = πy(z) ∈ Cn(x), we get

d s s t d−1 X λb1(z) X X ε ε |z|t |y|t kαk|y|1/d z∈FN (x,y,ε) α∈Λy(ε)∩CN (x) z∈ζ(y,α,ε) N (t−1) d X X εs ε d−1 t 1/d 0 |y| kαk|y| n=1 α∈Λy(ε)∩Cn(x) N d ! s+(t−1) d−1 0 X ε card Λy(ε) ∩ Cn(x) d d . (t−1) t (t−1) n=1 n d−1 |y| λbd(y) d−1 By Lemma 2.4,

d−1 d−1 d 0 (nλd(y)) n λd(y) card Λy ∩ Cn(x) 0 = 0 vol(Λy ∩ Hx) vol(Λy ∩ Hx)λd(y) d−1 d d−1 d n λd(y) |y| = n λbd(y) , thus

s N s+(t−1) d ! 0 X λb1(z) X ε d−1 1 card Λy(ε) ∩ C (x) × n t 1+(t−1) d −d (t−1) d −d 0 |z| n d−1 |y|t λ (y) d−1 card Λy ∩ Cn(x) z∈FN (x,y,ε) n=1 bd s+(t−1) d S1(N, t)ε d−1 = d . (t−d) t λbd(y) d−1 |y| Section 8 is devoted to the proof of Proposition 8.1 and by the Corollary 8.7 of this Proposition, we have

N X 1 (6.4) S1(N, t) 1+(t−d) d n=1 n d−1

1 t−d which is t−d when t > d provided N is large enough, e.g. N ≥ 2. For the rest of this section, we carry out the lower bound calculations assuming (6.4). Corollary 6.10. Let ε be a positive real number and assume d ≥ 2. For any function d2 d f :]0, 1] →]0, ∞[ with limu→0 f(u) = 0, Hdim DIε(d) ≥ d+1 + f(ε)ε for ε is small enough.

Proof. Let (Qε,N , σε,N ,B) be the self-similar structure as in Proposition 6.6. Let x ∈ Qε,N 0 0 and z, z ∈ FN (x, x, ε) with z 6= z . By Lemma 6.4, λb1(z) ε, so that, by deﬁnition of Qε,N , d−1 −1 d−1 d−1 λb1(x) ε. By Minkowski’s theorem, λb1(x) λbd(x) 1 so that λbd(x) λb1(x) ε .

25 By Lemma 6.7, R(z, z0) > 1 implies d(B(z),B(z0)) > ρ diam B(x) for a constant

d+1 d−1 1 2d +d ρ ε d−1 . N By Lemma 6.8, there is a constant c > 0 which depends neither on N nor on ε, such 0 that for each z ∈ FN (x, x, ε) there are at most c points z ∈ FN (x, x, ε) \{z} such that d(B(z),B(z0)) ≤ ρ diam B(x). Now Proposition 6.9 together with (6.4) implies that, with an N depending only on t − d,

d2(t−d) s s+(t−1) d d(t−d) s+(t−1) d +d(t−d) s+d+ X λb1(z) ε d−1 λb1(x) ε d−1 ε d−1 = |z|t (t − d)|x|t (t − d)|x|t (t − d)|x|t z∈FN (x,x,ε) so that, setting t = (1 + 1/d)s, we have

2 d+ d (t−d) X diam B(z)s ε d−1 . diam B(x)s t − d z∈FN (x,x,ε) If t−d = f(ε)εd then the right side in the above tends to inﬁnity when ε goes to zero, so that the sum on the left side is ≥ (c+1) when ε is small enough. By Theorem 3.4, Hdim DI2ε(d) ≥ d2 d d+1 + f(ε)ε , from which we deduce the desired statement for Hdim DIε(d).

d2 Proposition 6.11. For d ≥ 2, Hdim Sing(d) ≥ d+1 . d2 Moreover Sing(d) has positive d+1 -dimensional Hausdorﬀ measure.

Proof. Let (εi)i≥0 be a decreasing sequence of positive real numbers and (Ni)i≥0 a increasing sequence of positive integers such that εi+1 εi, Ni+1 Ni, d lim εi = 0 and lim εi log Ni = ∞. i→∞ i→∞

−1 For example, Ni = i + 10 and εi = (log log Ni) d .

Let x0 ∈ FNi (x−1, x−1, 2ε0) for some arbitrary x−1 in Q. Deﬁne Qi, i ≥ 0 recursively by [ Q0 = {x0},Qi+1 = FNi (x, x, εi).

x∈Qi 0 S 2 0 0 3 Let Q = i≥0 Qi be a formal disjoint union and for each x ∈ Q , let σ (x) = FNi (x, x, εi). Let B(x) = B(x, λ1(x) ) as before. The same argument as in the proof of Proposition 6.6 b 2|x| shows that (Q0, σ0,B) is a strictly nested self-similar structure covering a subset of Sing(d).

2 0 S 0 For example, Q = i≥0 Qi × {i} and σ (x) = FNi (˜x, x,˜ εi) × {i} where x = (˜x, i). 3We represent the ﬁrst coordinate of an element in Q0 by the same name (e.g. in the case of x = (˜x, i), we drop the tilde) and let the context decide whether an element of Q0 or one of Q is intended.

26 By Lemma 6.7, we can ﬁx a constant cρ > 0 depending only the dimension such that the function ρ : Q0 →]0, 1[ given by (x ∈ Q0)

d+1 d−1 2d 1 d−1 +d ρ(x) = cρ εi Ni 0 0 has the property that for any z, z ∈ FNi (x, x, εi) such that R(z, z ) > 1, we have d(B(z),B(z0)) > ρ(x) diam B(x).

By Lemma 6.8, there is a constant c > 0 such that for each z ∈ FNi (x, x, εi) there are at 0 0 most c points z ∈ FNi (x, x, εi) \{z} such that d(B(z),B(z )) ≤ ρ(x) diam B(x). Moreover, since ρ(x) decreases with i, Lemma 6.5 implies both B(z) ⊂ B(x) and ρ(z) diam B(z) < ρ(x) diam B(x)

0 for all z ∈ σ (x) when ε0 is small enough. Corollary 8.7 implies S1(N, d) log N (see (6.4)) so that Proposition 6.9 with t = (1 + 1/d)s = d now implies, as in the proof of Corollary 6.10

s s+d X λb1(z) ε i log N |z|t |x|t i z∈FNi (x,x,εi) so that X (ρ(z) diam B(z))s εd log N . (ρ(x) diam B(x))s i i z∈FNi (x,x,εi)

d2 For i large enough, the sum exceeds c + 1, so that by Theorem 3.6, Hdim Sing(d) ≥ d+1 .

7. Disjoint spheres We wish to enhance the lower bound of Theorem 1.3 in the two dimensional case. The two next Lemmas are important ingredients of this improvement. Observe that these Lemmas are optimal when d = 2. This is the reason we shall assume that d = 2 after their proofs.

Lemma 7.1. Let x and y be in Q and ε > 0. Suppose that λ (x) ≤ ε, kx − yk ≤ λ1(x) , bd−1 b b 10|x| |x| ≤ |y|, and y ∈ Hx. Then for all q ∈ {|x|, |x| + 1,..., |y|},

d+1 −1/d d({y,b . . . , qyb}, Z ) εq when ε is small enough.

Proof. Since kx − yk ≤ λ1(x) , by Lemma 4.2, x is a best approximation vector of the rational b b 10|x| vector θ = y. Let xj = x, . . . , xk = y be all the intermediate best approximation vectors of 0 d y. Consider the net H of parallel hyperplanes Λx + Hx in R . Since y ∈ Hx,Λy ⊂ H.

27 Let us show that the lattices Λxi , i = j+1, ..., k−1 are also included in H. Let 1 ≤ q < |xi|. Writing xl = (pl, ql), we have q kqxbi − qybk = kqiyb − pik ≤ kqjyb − pjk ≤ 2λ1(x) qi by Lemma 4.1. Making use of the Minkowski theorem, we see that the assumption λbd−1(x) ≤ λbd−1(x) ε implies λbd(x) εd and therefore the same inequality holds without the hat. Now by Lemma 2.2, e(H) λd(x), hence e(H) > 10λ1(x) when ε is small enough. With the above

inequality, it follows that the entire lattice Λxi is within e(H)/5-neighborhood of H. Looking d 0 at the picture in the quotient R /Hx, we see that Λxi ⊂ H. 1 Let i be in {j, ..., k}. We have vol(Λx ∩ Hx) × e(H) = vol(Λx ) = , and the ﬁrst d − 1 i i |xi| minima of the lattice Λxi ∩ Hx are larger than those of Λi. By Minkowski theorem, it follows that 1 λ1(xi) ··· λd−1(xi) , e(H)|xi| 1 λ1(xj) ··· λd−1(xj) . e(H)|xj|

Since |xi| ≥ |xj|, we obtain

1/d λb1(xi) λbd−1(xi) |xj| × · · · × 1/d ≤ 1. λb1(xj) λbd−1(xj) |xi| Therefore, one of the d − 1 ratios is 1. This implies that for some l,

λb1(xi) ≤ λbl(xi) λbl(xj) ≤ λbd−1(xj) ≤ ε.

Let q be an integer between |xi−1| and |xi|. By Lemma 4.1, d d({y,b ..., qyb}, Z ) ≤ kqi−1yb − pi−1k ≤ 2λ1(xi) −1/d = 2λb1(xi)|xi| εq−1/d. Lemma 7.2. For all x in Q and y in D(x) (see Deﬁnition 5.1 for D(x))

1 2 λb1(y) λb1(x) (d−1) . Proof. By Lemma 5.6, 1 |y| d λbd(y) 1 λbd(x) |x| d and by Minkowski’s theorem we have both d−1 d−1 λb1(y) λbd(y) 1 and λbd(x) λb1(x) 1;

28 therefore, 1 1 d(d−1) 2 |x| λb1(y) λb1(x) (d−1) . |y| We assume until the end of this section that d = 2. Deﬁnition 7.3. Let ε be a positive real number and x in Q. i D0(x) is the set of y ∈ Q such that |x| ≤ |y|, y ∈ H , kx − yk < λ1(x) . x b b 10|x| ii The subset of D0(x) consisting of y ∈ Q such that |x| ≤ |y| < N|x| will be denoted 0 by DN (x). Set 0 [ σε,N (x) = FN (x, y, ε) 0 y∈DN (x) (F (x, y, ε) is deﬁned Section 6.1), Q0 = S σ0 (x) and B(x) = B(x, λ1(x) ). N ε,N x∈Q ε,N b 2|x| 0 0 Lemma 7.4. For ε small enough, (Qε,N , σε,N ,B) is a strictly nested self-similar structure covering of a subset of DIcε(d) for some absolute constant c > 0 (recall that d = 2). Proof. We need to verify the conditions (i) and (ii) in the proof of Proposition 6.6 for 0 0 each σε,N -admissible sequence (xk). Let yk be the element in DN (xk) such that xk+1 ∈ 0 FN (xk, yk, ε). First observe that by Lemma 6.4, for all x ∈ Qε,N , λb1(x) ≤ ε, hence by Lemma 7.2, λb1(yk) λb1(xk) ≤ ε. Next, by Lemma 6.5, B(xk+1) ⊂ B(yk) so that to prove (i) it is enough to verify that B(yk) ⊂ B(xk). We can assume yk 6= xk. Since

λ1(xk) kπxk (yk)k λ1(xk) ≤ = d(xbk, ybk) < |yk| |yk| 10|xk|

we have |yk| > 10|xk|. Similarly,

λ1(yk) kπyk (xk)k λ1(xk) ≤ = d(xbk, ybk) < |xk| |xk| 10|xk|

λ1(xk) so that λ1(yk) < 10 . Therefore, λ (y ) λ (x ) 1 k < 1 k . |yk| 100|xk|

Now, given z ∈ B(yk) we have

kxbk − zbk ≤ kxbk − ybkk + kybk − zbk λ (x ) λ (y ) λ (x ) < 1 k + 1 k < 1 k 10|xk| 2|yk| 2|xk|

so that z ∈ B(xk).

29 T For (ii), let θ be in k B(xk) and consider |xk| ≤ q < |xk+1|. In the case |yk| ≤ q < |xk+1| we have d 0 0 0 0 min d(lθ, Z ) ≤ kqkθ − pkk < 2kqkxk+1 − pkk 1≤l≤q b 0 0 by Lemma 4.3(iii) where yk = (pk, qk). On the other hand, by Lemma 6.4,

0 0 |yk ∧ xk+1| −1/d kqkxbk+1 − pkk = < ε|xk+1| |xk+1| so that d −1/d min d(lθ, Z ) < 2εq . 1≤l≤q

In the case |xk| ≤ q < |yk| we have d d min d(lθ, Z ) ≤ 2 min d(lyk, Z ) 1≤l≤q 1≤l≤q b

by Lemma 4.3(iii) and because θ ∈ B(xk+1) ⊂ B(yk). Since λb1(xk) ≤ ε, we can use Lemma 7.1 to conclude that d −1/d min d(lθ, Z ) εq . 1≤l≤q

λ1(y) Deﬁnition 7.5. Let x, y ∈ Q with y ∈ Hx. The sphere of diameter |y| that is tangent to 0 ⊥ Hx at yb, with inward normal αx will be denoted S(x, y). 0 0 Lemma 7.6. Let x ∈ Q, y 6= y ∈ Hx ∩ Q. Then S(x, y) and S(x, y ) have disjoint interiors. Proof. The condition for S(x, y) and S(x, y0) to have disjoint interiors is

2 0 2 0 2 λ1(y) λ1(y ) λ1(y) λ1(y ) y − yb0 + − ≥ + b 2|y| 2|y0| 2|y| 2|y0| or equivalently, 0 λ1(y)λ1(y ) ky − yb0k2 ≥ . b |y||y0| Since 0 0 0 0 2 |y ∧ y | |y ∧ y | 0 0 |y||y |ky − ybk = = kπy(y )kkπy0 (y)k ≥ λ1(y)λ1(y ) b |y| |y0| the lemma follows. Lemma 7.7. Let χ be a chord of a circle S with diameter one that makes an angle φ with each of the radial segments joining its endpoints to the center. Let ρ(δ) be the radius of the largest disk contained in S centered at a point on the chord a distance δ from the endpoints. Then δ ρ(δ) ≥ cos φ. 2

30 Proof. Let ` be the distance of the point from the center of the circle. By Pythagoras’ theorem, `2 = (cos φ − δ)2 + sin2 φ = 1 − δ(2 cos φ − δ). Thus, 1 ρ(δ) = 1 − ` ≥ 1 − (1 − δ(2 cos φ − δ)) 2 and since δ ≤ cos φ, δ ρ(δ) ≥ cos φ. 2 Lemma 7.8. Let y ∈ D(x) for some x, y ∈ Q and let z ∈ F (x, y, ε). Then the ball B(z, 2λ1(z) ) b |z| is contained in the interior of the sphere S(x, y) when ε is small enough. Proof. Let ρ(z) be the radius of the largest ball centered at z that is contained in S(x, y). b ⊥ −1 Since the point zb is in the cone yb + C(x), the angle between zb − yb and αx is tan Ad. By Lemma 7.7, considering the chord going through yb and zb, we see that 1 λ (y) d(y, z) d(y, z) ≤ 1 =⇒ ρ(z) ≥ b b . b b p 2 2|y| p 2 1 + Ad 2 1 + Ad Now, by Lemma 6.4, λ (z) |y ∧ z| |y| 1 = = d(y, z) |z| |z|2 |z| b b |y| d 2|y|d(y,z) and by inequality (6.2), ε d−1 , hence the ball B(z, b b ) is contained in the interior |z| b |z| of the sphere S(x, y) when ε is small enough. Corollary 7.9. Let y, y0 ∈ D(x) for some x, y, y0 ∈ Q and let z ∈ F (x, y, ε) and z0 ∈ 0 0 0 0 λ1(z) F (x, y , ε). If y 6= y then R(z, z ) > 1 and d(B(z),B(z )) > |z| when ε small enough.

0 Proof. The balls B(z, 2λ1(z) ) and B(z0, 2λ1(z ) ) have same centers as B(z) and B(z0) with b |z| b |z0| quadruple radii, and by Lemmas 7.6 and 7.8, they do not meet.

Lemma 7.10. Let ε be small enough, N a positive integer and x ∈ Q. Suppose that λb1(x) 0 ε. There are constants c > 0 and ρ > 0 such that for all z ∈ σε,N (x), there are at most c 0 0 0 elements z ∈ σε,N (x) \{z } such that ε6 λ (x) (7.1) d(B(z),B(z0)) < ρ 1 . N 6 |x| 0 Proof. Step 1. For any y ∈ DN (x), λ (y) 1 λ (x) 1 1 . |y| N 2 |x|

31 Since by Lemma 5.6, λ2(y) λ2(x), Minkowski’s theorem implies

|y|λ1(y) |x|λ1(x). Therefore, we have λ (y) |x|λ (x) 1 λ (x) 1 1 ≥ 1 . |y| |y|2 N 2 |x| Step 2. R(z, z0) > 1 implies (7.1) does not hold if ρ is small enough. In the case z, z0 belongs to the same set F (x, y, ε) and R(z, z0) > 1, making use of Lemma 6.7 we obtain 1 3 ε4 λ (y) d (B(z),B(z0)) 1 2 N λb2(y) |y| |y| 1 1 2 2 −1 −1 Noticing that λb2(y) λb2(x)( |x| ) ≤ λb2(x)N and that λb2(x) λb1(x) ε , together with Step 1, we see that ε6 λ (x) d (B(z),B(z0)) 1 . N 6 |x| In the case z, z0 belongs to two diﬀerent sets F (x, y, ε) and F (x, y0, ε), Corollary 7.9 and the λ1(z) ε4 λ1(y) inequality 3 2 from the proof of Lemma 6.7, imply |z| N λb2(y) |y| λ (z) ε6 λ (x) d(B(z),B(z0)) > 1 1 . |z| N 6 |x| 0 0 0 Step 3. For all z ∈ σε,N (x), the number of z ∈ σε,N (x) such that (7.2) holds is 1. This is a consequence of Lemma 6.8, Corollary 7.9 and Step 2. Lemma 7.11. Let L = Zu ⊕ Zv be a two dimensional lattice, let n be a positive integer and let I1, ..., In be n intervals in Z of same length l ≥ n. Then the number of primitive points of the lattice L in the region

R = {au + bv : a ∈ {1, ..., n}, b ∈ Ia} is ln. Proof. Denote φ(n) the Euler function, i.e. φ(n) = the number of positive integers ≤ n P 2 prime with n, and Φ(n) = k≤n φ(k). It is well known that Φ(n) n (see [13]). Since each interval of length k contains at least φ(k) integers prime with k, the number of primitive elements in R is at least n n X l X φ(k) φ(k) l k k k=1 k=1 n X 1 1 ≥ l Φ(k) − ln k k + 1 k=1 where bxc stands for the integer part of the real number x.

32 Proposition 7.12. For x ∈ Q such that λb1(x) ε, s s+2(t−1)+δ X λ1(z) ε t S2(N, t) t 0 |z| |x| z∈σε,N (x) where δ = 2(t − 2) and

log2 N X (2−t−δ/2)k S2(N, t) = S1(N, t) × 2 . k=1

s+(t−1) 2 Proof. Set ν(ε, N, s, t) = S1(N, t)ε 2−1 and D0(l, k) = {y ∈ D0(x): l|x| ≤ |y| ≤ k|x|}. d+1 From the proof of Lemma 5.7 we know that πx induces a bijection from {(u, v) ∈ Hx ∩ Z : 0 k|x| ≤ v < (k + 1)|x|} onto Λx = Λx ∩ πx(Hx). Making use of Lemma 7.11, we obtain k card{y ∈ Q :(k + 1)|x| ≤ |y| ≤ 2k|x|, kπ (y)k ≤ λ (x)} k2, x 10 1 and therefore card D0(k + 1, 2k) k2. 0 Observe that for each z ∈ σε,N (x), Lemma 6.4 implies |λ1(z) = ||πz(y)|| so that πz(y) is a shortest vector of Λx. And since |y| < |z| (by Lemma 6.4 again) for each of the O(1) 0 possibilities for πz(y) there are at most one y ∈ D (x) such that z ∈ FN (x, y, ε). Therefore, by Proposition 6.9, s s X λ1(z) X X λ1(z) X ν(ε, N, s, t) t t δ t+δ/2 0 |z| 0 |z| 0 λ2(y) |y| z∈σε,N (x) y∈DN (x) z∈FN (x,y,ε) y∈DN (x)

δ δ 0 k k+1 and by Lemma 5.6, λ2(y) λ2(x) , thus, grouping the y in the sets D (2 + 1, 2 ), 0 ≤ k ≤ log2 N, we obtain

s log2 N 2k X λ1(z) X ν(ε, N, s, t) 2 t δ k t+δ/2 0 |z| λ2(x) (2 |x|) z∈σε,N (x) k=1 log N X2 ν(ε, N, s, t) 2(2−t−δ/2)k = . δ |x|t k=1 λb2(x) −1 The Proposition follows making use of the lower bound λb2(x) λb1(x) . Corollary 7.13. Let ε be a positive real number. For any function f :]0, 1] →]0, ∞[ with 4 limu→0 f(u) = 0, Hdim DIε(2) ≥ 3 + f(ε)ε for ε is small enough.

33 0 0 Proof. By Lemma 7.4, (Qε,N , σε,N ,B) is a strictly self-similar structure that cover a susbset 0 of DIc0ε(d). Let x ∈ Qε,N . By Lemma 7.10, there are constants c > 0 and ρ > 0 such that 0 0 0 0 for all z ∈ σε,N (x), there are at most c elements z ∈ σε,N (x) \{z } such that ε6 λ (x) (7.2) d(B(z),B(z0)) < ρ 1 . N 6 |x| Now Proposition 7.12 together with (6.4) implies

s s+2(t−1)+δ X λb1(z) ε t S2(N, t) t 0 |z| |x| z∈σε,N (x) log N εs+2(t−1)+δ 1 X2 × × 2(2−t−δ/2)k |x|t t − 2 k=1 εs+2(t−1)+δ 1 × |x|t (t − 2)(t + δ − 2) provide that N (t−2) ≥ 2. So that, setting t = (1 + 1/d)s, we have X diam B(z)s ε2(t−1)+δ s 2 . 0 diam B(x) (t − 2) z∈σε,N (x) If t − 2 = f(ε)ε then the right side in the above tends to inﬁnity when ε goes to zero, so that the sum on the left side is ≥ (c + 1) when ε is small enough. By Theorem 3.4, 4 Hdim DI2ε(2) ≥ 3 + f(ε)ε, from which we deduce the desired statement for Hdim DIε(2).

8. Bounded distortion Notation. For a positive integer n, denote φ(n) the number of positive integers j ≤ n P prime with n, i.e. the Euler function, and D1(n) the sum of the divisors of n: D1(n) = k|n k. Proposition 8.1. Let x be in Q, let y be in D(x) and let n be a positive integer. Then

0 card Cn(x) ∩ Λy(ε) φ(n) d−1 D1(n) 0 1 − Cε card Cn(x) ∩ Λy n n where C depends only on the dimension.

When d = 2, the proof of the above Proposition is rather simple because the lattice Λα⊥ is one dimensional. It implies λb1(α) = 1 and therefore α is in Λy(ε) provided that α is primitive and ε < 1. In that case the proof reduces to a lower bound on the number of primitive points 0 in Cn(x), which is easy. We give the proof for all d ≥ 2; it needs a few lemmas. Lemma 8.2. Let Λ be a lattice in Rd and let α be a nonzero vector in Λ. Let ψ : Rd → Rd be a linear bilipschitz map and consider Γ = ψ(Λ) and β = ψ(α). Then λb1(Λα⊥ ) λb1(Γβ⊥ ).

34 Proof. Observe that Γβ⊥ can be obtained in two steps with projections parallel to β. First project Γ onto the space ψ(α⊥) and then project the image L of Γ onto β⊥. Since L = ⊥ ⊥ ψ(Λα⊥ ), λ1(Λα⊥ ) λ1(L). Next, the orthogonal projection on β restricted to ψ(α ) is also bilipschitz, hence

λ1(Γβ⊥ ) λ1(L) λ1(Λα⊥ ).

Since vol(Γ) vol(Λ) and kβk kαk, it follows that vol(Γβ⊥ ) vol(Λα⊥ ).

Lemma 8.3. Let n and k be positive integers. Let l1, l2, . . . , lk be positive real numbers. k Qk Consider the box B in (Z/nZ) deﬁned by the product of real intervals i=1[−li, li]. Then the number of points z in (Z/nZ)k of order n such that the subgroup hzi generated by z intersects the box B only at 0 is at least ! D1(n) Y φ(n) nk−1 − 3k l . n i 1≤i≤k

Proof. We can suppose that the li’s are arranged in increasing order: l1 ≤ l2 ≤ · · · ≤ lk. Set k−1 T = {(x, y) ∈ (Z/nZ) × (Z/nZ) : order(x) = n, h(x, y)i ∩ B = {0}} k−1 Bad = {(x, y) ∈ (Z/nZ) × (Z/nZ) : order(x) = n, h(x, y)i ∩ B 6= {0}} Since card T + card Bad = φ(n)nk−1, it is enough to bound the number of elements of Bad from above. Let x be an element of order n in Z/nZ. Set k−1 Bad(x) = {y ∈ (Z/nZ) : h(x, y)i ∩ B 6= {0}}. Note that Bad(x) = xBad(1) so that card Bad = φ(n)Bad(1). Hence, we have to bound card Bad(1) from above. Since

y ∈ Bad(1) ⇔ ∃(a, b) ∈ B \{0}, ∃ k ∈ Z, a = k, b = ky, ⇔ ∃(a, b) ∈ B \{0}, b = ay, we have X card Bad(1) ≤ card E(a, b) (a,b)∈B\{0} where E(a, b) = {y ∈ (Z/nZ)k−1 : b = ay}. Let (a, b) be in (Z/nZ) × (Z/nZ)k−1 such that a 6= 0 and order(a) = p (we assume a 6= 0 because a = 0 and b 6= 0 implies E(a, b) = ∅). The number of solutions to the equation b = az is (n/p)k−1 if b ∈ (hai)k−1 and 0 otherwise. Hence card E(a, b) is (n/p)k−1 or 0. Since hai = hn/pi, the number of b in (hai)k−1 that are in the box Y [−li, li] 2≤i≤k

35 j k Q li is at most 2≤i≤k(2 n/p + 1). Therefore, the number of pairs (a, b) in B such that a 6= 0 and order(a) = p is at most l1 Y li 2 (2 + 1). n/p n/p 2≤i≤k Thus, X l1 Y li card Bad(1) ≤ (n/p)k−12 (2 + 1). n/p n/p p|n 2≤i≤k j k j k j k j k l1 li l1 Q li Since n/p 6= 0 implies n/p ≥ 1 for i ≥ 1, the product 2 n/p 2≤i≤k(2 n/p + 1) is j k li li either 0 or, for each i = 2, . . . , k, 2 n/p + 1 ≤ 3 n/p . Therefore, l1 Y li Y li 2 (2 + 1) ≤ 3k . n/p n/p n/p 2≤i≤k 1≤i≤k Hence,

X Y li card Bad(1) ≤ (n/p)k−13k n/p p|n 1≤i≤k

Y X p D1(n) Y = 3k l = 3k l i n n i 1≤i≤k p|n 1≤i≤k and the lemma follows. 0 Lemma 8.4. Let y be in D(x), let α1, . . . , αd−1 be a Minkowski-reduced basis for Λxy = 0 0 Λy ∩ Hx and let αd be the shortest vector in Λy \ Hx. Then kαik λi(y) for i = 1, . . . , d. 0 0 0 Proof. Since λ1(Λxy) ··· λd−1(Λxy) vol(Λxy) and λ1(y) ··· λd(y) vol(Λy), we have for i < d 0 Q ! 0 λi(Λ ) ˆ λj(y) λd(y) vol(Λ ) xy j∈{1,...,i,...,d−1} xy . Q 0 λi(y) j∈{1,...,ˆi,...,d−1} λj(Λxy) vol(Λy) ⊥ Recall that αx , which is deﬁned in Section 6.1, is a vector orthogonal to Hx such that ⊥ 0 ⊥ 0 Hx + Zαx = Hx + Λx. Since Λx + Hx = Λy + Hx, vol(Λxy)kαx k = vol Λy. Clearly λj(Λxy) ≥ λj(y), j = 1, . . . , d − 1, hence 0 λi(Λxy) λd(y) ⊥ . λi(y) kαx k ⊥ 0 By Lemmas 5.6 and 2.2, λd(y) λd(x) kαx k, thus λi(Λxy) λi(y), so that kαik 0 ⊥ 0 λi(Λxy) λi(y) for i = 1, . . . , d − 1. Since αd lies in B(αx , e(Λxy)) we have kαdk ≤ ⊥ 0 0 kαx k + e(Λxy) λd(x) + λd−1(Λxy) λd(y).

36 Lemma 8.5. The constant Ad can chosen large enough so that the following property holds: P λd(y) Let n be a positive integer and let α = miαi + nαd be in Λy. Then |mi| ≤ n for i

⊥ 0 Proof. Observe ﬁrst that kαx k kαdk. Also, the condition α ∈ Cn(x) is equivalent P ⊥ ⊥ P ⊥ to k i

1 X mi ∗ Uα(ed) = ed − αi . σd n i

λd(y) Since |mi| n, Uα is bilipschitz with constant λi(y)

1 X |mi| 1 X |mi| 1 + σi 1 + λi(y) 1. σd n λd(y) n i

∗ It follows that the map ψ : x → Uα(x ) is bilipschitz as well. By Lemma 8.2, λb1(Λα⊥ ) λb1(ψ(Λy)ψ(α)⊥ ) and since

ψ(Λy) = Zψ(α1) + ··· + Zψ(αd)

∗ ∗ 1 X mi ∗ = Zα1 + ··· + Zαd−1 + Zσd(ed − αi ) σd n i

37 Proof of Proposition 8.1. By Lemma 8.6, we are reduced to ﬁnding a lower bound, for a P 0 given n, on the number of primitive α = i

λb1(Lα) > Cε where C is a constant depending only on the dimension. σ1···σd−1 Observe that, when α is primitive, vol(Lα) = n . Hence, the condition λb1(Lα) > Cε is equivalent to σ ··· σ 1/(d−1) λ (L ) > Cε 1 d−1 . 1 α n Changing C, we may assume that we deal with the sup norm instead of the Euclidean norm. Set m = (m1, . . . , md−1) and X L(m, n) = Zne1 + ··· + Zned−1 + Z miei. i

L(m, n) is the image of Lα by the linear map that sends σiei to nei, i = 1, . . . , d − 1. So λb1(Lα) > Cε holds if the lattice L(m, n) has no nonzero points in the box

1/(d−1) σ1 ··· σd−1 Y n n B(n, y, ε) = Cε − , n σi σi i

d and if α is primitive, i.e., (m1, ..., md−1, n) is a primitive element of Z . Since the lattice L(m, n) depends only on m1, . . . , md−1 mod n, we have a lower bound on the number of α with λb1(Lα) > Cε by counting the number of α with 0 ≤ m1, . . . , md−1 < n such that L(m, n) has no nonzero point in B(n, y, ε) and then multiplying by the number of disjoint translates of the parallelepipeds ( ) X t = tiαi + nαd : ti ∈ [0, n[, i = 1, . . . , d − 1 i

0 λd(y) that are contained in C (x). Since the system of inequalities |ti| ≤ n for i = 1, . . . , d − 1, n λi(y) d−1 implies t ∈ C0 (x), by Lemma 8.5, the number of translates is λd(y) . n λ1(y)···λd−1(y) 0 Finally, by Lemma 8.3, the number of α in Cn(x) ∩ Λy(ε) is

d−1 ! λd d−2 d−1 D1(n) Y × φ(n) n − 3 li λ1 ··· λd−1 n i

38 σ1···σd−1 1/(d−1) n where li = Cε and λi is an abbreviation for λi(y). Hence, with a new large n σi enough constant C, it is

λd−1 D (n) λ ··· λ nd−1 d × φ(n) nd−2 − C 1 εd−1 1 d−1 λ1 ··· λd−1 n n λ1 ··· λd−1 d−1 λd d−2 d−1 d−3 = × φ(n) n − Cε D1(n)n . λ1 ··· λd−1

0 The last thing to do is bound card Cn(x) ∩ Λy from above. By Lemma 2.3,

0 d−1 d−1 0 diam Cn(x) (nλd) card Cn(x) ∩ Λy 0 vol(Λy) λ1 ··· λd−1

hence,

d−1 λd d−2 d−1 d−3 0 × φ(n)(n − Cε D1(n)n ) card Cn(x) ∩ Λy(ε) λ1···λd−1 d−1 0 (nλd) card Cn(x) ∩ Λy λ1···λd−1 φ(n) D (n) = 1 − Cεd−1 1 . n n

Corollary 8.7. If ε is small enough, then

N X 1 S1(N, t) . 1+(t−d) d n=1 n d−1 Proof. It is well known (see [13]) that

2 Φ(n) Ds(n) n P P where Φ(n) = k≤n φ(k) and Ds(n) = k≤n D1(k). Therefore,

N N X 1 φ(n) X 1 = (Φ(n) − Φ(n − 1)) n1+δ n n2+δ n=1 n=1 N−1 X 1 1 Φ(N) = − Φ(n) + n2+δ (n + 1)2+δ N 2+δ n=1 N−1 X 1 n2 n3+δ n=1

39 and N N X 1 D1(n) X 1 = (D (n) − D (n − 1)) n1+δ n n2+δ s s n=1 n=1 N−1 X 1 1 Ds(N) = − D (n) + n2+δ (n + 1)2+δ s N 2+δ n=1 N−1 X 1 n2. n3+δ n=1 Remark 8.8. Precise asymptotics are known for counting problems such as card Λ(ε) ∩ B(0,T ) lim T →∞ card Λ ∩ B(0,T ) which can be computed using the results of Wolfgang Schmidt in [21]. Such estimates give good control when T is larger than some T0 (which may depend on Λ), but provide little information for smaller values of T .

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