Badly approximable points on curves and unipotent orbits in

BADLY APPROXIMABLE POINTS ON CURVES AND UNIPOTENT ORBITS IN HOMOGENEOUS SPACES

arXiv:1703.03461v1 [math.NT] 9 Mar 2017

LEI YANG Abstract. In this paper, we study the weighted n-dimensional badly approximable points on curves. Given an analytic non-degenerate curve ϕ : I = [a, b] → Rn , we will show that any countable intersection of the sets of the weighted badly approximable points on ϕ(I) has full Hausdorff dimension. This strengthens a result of Beresnevich [Ber15] by removing the condition on the weights. Compared with the work of Beresnevich, in this paper, we study the problem through homogeneous dynamics. It turns out that in order to solve this problem, it is crucial to study the distribution of long pieces of unipotent orbits in homogeneous spaces. The proof relies on the linearization technique and SL(2, R) representations.

1. Introduction 1.1. Badly approximable vectors. Given a positive integer n, the weighted version of Dirichlet’s approximation theorem says the following: Theorem 1.1 (Dirichlet’s Theorem, 1842). A vector r = (r1 , . . . , rn ) is called a n-dimensional weight if ri ≥ 0 for i = 1, . . . , n and r1 + · · · + rn = 1. For any n-dimensional weight r = (r1 , . . . , rn ), the following statement holds. For any vector x = (x1 , . . . , xn ) ∈ Rn and any N > 1, there exists an integer vector p = (p1 , . . . , pn , q) ∈ Zn+1 such that 0 < |q| ≤ N and |qxi + pi | ≤ N −ri , for i = 1, . . . , n.

This theorem is the starting point of study in simutaneous Diophantine approximation. Using this theorem, one can easily show the following: Corollary 1.2. For any vector x = (x1 , . . . , xn ) ∈ Rn , there are infinitely many integer vectors p = (p1 , . . . , pn , q) ∈ Zn+1 with q 6= 0 satisfying the following: (1.1)

|q|ri |qxi + pi | ≤ 1 for i = 1, . . . , n.

For almost every vector x ∈ Rn , the above corollary remains true if we replace 1 with any smaller constant c > 0 on the right hand side of (1.1). The exceptional vectors are called r-weighted badly approximable vectors. We give the formal definition as follows: 2010 Mathematics Subject Classification. 11J13; 11J83; 22E40. Key words and phrases. Diophantine approximation, badly approximable points, the linearization technique, unipotent orbits. The author is supported in part by ISF grant 0399180 and ERC grant AdG 267259. 1

Definition 1.3. Given a n-dimensional weight r = (r1 , . . . , rn ), a vector x ∈ Rn is called r-weighted badly approximable if there exists a constant c > 0 such that for any p = (p1 , . . . , pn , q) ∈ Zn+1 with q 6= 0, max |q|ri |qxi + pi | ≥ c.

1≤i≤n

For a n-dimensional weight r, let us denote the set of r-weighted badly approximable vectors in Rn by Bad(r). In particular, Bad(1) denotes the set of badly approximable numbers. The study of the size of Bad(r) has a long history and is active in both number theory and homogeneous dynamical systems. It is well known that the Lebesgue measure of Bad(r) is zero for any n-dimensional weight r. However, people have shown that Bad(r) has full Hausdorff dimension, cf. [Jar28], [Sch66], [PV02] and [KW10]. For the intersection of sets of different weighted badly approximable vectors, Wolfgang M. Schmidt makes the following famous conjecture in 1982: Conjecture 1.4 (Schmidt’s Conjecture, see [Sch83]). Bad(1/3, 2/3) ∩ Bad(2/3, 1/3) 6= ∅. In 2011, Badziahin, Pollington and Velani [BPV11] settle this conjecture by showing the following: for any countable collection of 2-dimensional weights {(it , jt ) : t ∈ N}, if lim inf t→∞ min{it , jt } > 0, then ! ∞ \ Bad(it , jt ) = 2. dimH t=1

Here dimH (·) denotes the Hausdorff dimension of a set. An [An16] later strengthens their result by removing the condition on the weights. In fact, An proves√the following much stronger result: for any 2-dimensional weight (r1 , r2 ), Bad(r1 , r2 ) is (24 2)−1 -winning. Here a set is called α-winning if it is a winning set for Schmidt’s (α, β)-game for any β ∈ (0, 1). This statement implies that any countable intersection of sets of weighted badly approximable vectors is α-winning. The reader is refered to [Sch66] for more details of Schmidt’s game. For n ≥ 3, Beresnevich [Ber15] proves the following theorem: Theorem 1.5 (see [Ber15, Corollary 1]). Let n ≥ 2 be an integer and U ⊂ Rn be an analytic and non-degenerate submanifold in Rn . Here a submanifold is called non-degenerate if it is not contained in any hyperplane of Rn . Let W be a finite or countable set of ndimensional weights such that inf r∈W {τ (r)} > 0 where τ (r1 , . . . , rn ) := min{ri : ri > 0} for an n-dimensional weight (r1 , . . . , rn ). Then ! \ dimH Bad(r) ∩ U = dim U. r∈W

1.2. Notation. In this paper, we will fix the following notation. For a set S, let |S| denote the cardinality of S. For a measurable subset E ⊂ R, let m(E) denote its Lebesgue measure. For a matrix M, let M T denote its transpose. For integer k > 0, let Ik denote the k-dimensional identity matrix. 2

Let k · k denote the supremum norm on Rn and Rn+1 . Let k · k2 denote the Euclidean norm on Rn and Rn+1 . For x ∈ Rn+1 (or ∈ Rn ) and r > 0, let B(x, r) denote the closed ball in Rn+1 (or Rn ) centered at x of radius r, with respect to k · k. For every i = 1, . . . , n + 1, V there is a natural supremum norm on i Rn+1 . Let us denote it by k · k. Throughout this paper, when we say that c is a constant, we always mean that c is a constant only depending on the dimension n. For quantities A and B, let us use A ≪ B to mean that there is a constant C > 0 such that A ≤ CB. Let A ≍ B mean that A ≪ B and B ≪ A. For a quantity A, let O(A) denote a quantity which is ≪ A or a vector whose norm is ≪ A. 1.3. Main results. In this paper, we will strengthen Theorem 1.5 by removing the condition on weights: Theorem 1.6. Let n ≥ 2 be an integer and U ⊂ Rn be an analytic and non-degenerate submanifold in Rn . Let W be a finite or countable set of n-dimensional weights. Then ! \ dimH Bad(r) ∩ U = dim U. r∈W

By the reduction argument in [Ber15], to prove the above theorem, it suffices to prove the theorem for analytic curves:

Theorem 1.7. Let ϕ : I = [a, b] → Rn be an analytic and non-degenerate curve in Rn . Let W be a finite or countable set of n-dimensional weights. Then ! \ dimH Bad(r) ∩ ϕ(I) = 1. r∈W

By [Ber15], Theorem 1.6 has the following corollary:

Corollary 1.8. Let m, n ∈ N, B be a ball in Rm , W be a finite or countable set of ndimensional weights and Fn (B) be a finite family of analytic non-degenerate maps f : B → Rn . Then   \ \ dimH  f −1 (Bad(r)) = m. f ∈Fn (B) r∈W

Compared with [Ber15], in this paper, we study this problem through homogeneous dynamics and prove Theorem 1.7 using the linearization technique.

1.4. Bounded orbits in homogeneous spaces. Let us briefly recall the correspondence between Diophantine approximation and homogeneous dynamics. The reader may see [Dan84], [KM98] and [KW08] for more details. Let G = SL(n + 1, R), and Γ = SL(n + 1, Z). The homogeneous space X = G/Γ can be identified with the space of unimodular lattices in Rn+1 . The point gΓ is identified with the lattice gZn+1 . For ǫ > 0, let us define Kǫ := {Λ ∈ X : Λ ∩ B(0, ǫ) = {0}} .

It is well known that every Kǫ is a compact subset of X and every compact subset of X is contained in some Kǫ . 3

For a weight r = (r1 , . . . , rn ), let us define the diagonal subgroup Ar ⊂ G as follows:    r1 t  e       ..   .   Ar := ar (t) :=  . : t ∈ R  rn t   e     e−t For x ∈ Rn , let us denote



 In x V (x) := . 1

Proposition 1.9. x ∈ Bad(r) if and only if {ar (t)V (x)Zn+1 : t > 0} is bounded. Proof. The proof is well known and standard. We give the proof here for completeness. On the one hand, if x ∈ Bad(r), then there exists a constant c > 0 such that for any integer vector p = (p1 , . . . , pn , q)T with q 6= 0, we have that max1≤i≤n |q|ri |qxi + p| ≥ c. Now let us consider the lattice Λ(t) = ar (t)V (x)Zn+1 . We claim that for any t > 0, every nonzero vector in Λ(t) has norm at least c. In fact, for any nonzero integer vector p = (p1 , . . . , pn , q)T , we have that ar (t)V (x)p = (er1 t (qx1 + p1 ), . . . , ern t (qxn + pn ), e−t q)T . If q = 0, then the claim is obvious since kar (t)V (x)pk = max eri t |pi | > 1. 1≤i≤n

Let us assume that q 6= 0. For et < |q|, we have that

kar (t)V (x)pk ≥ e−t |q| > 1.

For et ≥ |q|, we have that

kar (t)V (x)pk ≥ max1≤i≤n eri t |qxi + pi | ≥ max1≤i≤n q ri |qxi + pi | ≥ c.

This proves one direction of the statement. On the other hand, if {ar (t)V (x)Zn+1 : t > 0} is bounded, we want to show that x ∈ Bad(r). In fact, there exists a constant c > 0 such that ar (t)V (x)Zn+1 ∈ Kc for all t > 0. Then for any integer vector p = (p1 , . . . , pn , q)T with q 6= 0, we have that kar (t)V (x)pk ≥ c, for any t > 0.

Let t = t0 such that et0 = 2|q|/c. Then the above inequality tells that max ξ ri |q|ri |qxi + pi | ≥ c, where ξ = 2/c.

1≤i≤n

Let ǫ = max1≤i≤n ξ ri , then the above inequality implies that max |q|ri |qxi + pi | ≥ cǫ−1 .

1≤i≤n

This proves the other direction.

 4

Therefore our main theorem is equivalent to saying that for any analytic submanifold U ⊂ Rn and any countable collection of one-parameter diagonal subgroups {Ars : s ∈ N}, the set of x ∈ U such that {ars (t)V (x)Zn+1 : t > 0}

is bounded for all s ∈ N has full Hausdorff dimension. The study of bounded trajectories under the action of diagonal subgroups in homogeneous spaces is a fundamental topic in homogeneous dynamics and has been active for decades. The basic set up of this type of problems is the following. Let G be a Lie group and Γ ⊂ G be a nonuniform lattice in G. Then X = G/Γ is a noncompact homogeneous space. Let A = {a(t) : t ∈ R} be a one-dimensional diagonalizable subgroup and let Bd(A) be the set of x ∈ X such that A+ x is bounded in X, where A+ := {a(t) : t > 0}. Then one can ask whether Bd(A) has full Hausdorff dimension. For a submanifold U ⊂ X, one can also ask whether Bd(A) ∩ U has Hausdorff dimension dim U. In 1986, Dani [Dan86] studies the case where G is a semisimple Lie group with R-rank one. In this case, he proves that for any non-quasi-unipotent one parameter subgroup A ⊂ G, Bd(A) has full Hausdorff dimension. His proof relies on Schmidt’s game. In 1996, Kleinbock and Margulis [KM96] study the case where G is a semisimple Lie group and Γ is a irreducible lattice in G. In this case, they prove that Bd(A) has full Hausdorff dimension for any nonquasi-unipotent subgroup A. Their proof is based on the mixing property of the action of A on X. Recently, An, Guan and Kleinbock study the case where G = SL(3, R) and Γ = SL(3, Z). They prove that for any countable T∞ one-parameter diagonalizable one-parameter subgroups {Fs : s ∈ N}, the intersection s=1 Bd(Fs ) has full Hausdorff dimension. Their proof closely follows the argument in the work of An [An16] and uses a variantion of Schmidt’s game. 1.5. The linearization technique. In [Ber15], the proof relies on a very careful study of the distribution of integer points in Rn+1 and the argument is elementary. In this paper, we study this problem through homogeneous dynamics and tackle the technical difficulties using the linearization technique. We study the Diophantine properties using homogeneous dynamics. It turns out that in order to get the Hausdorff dimension, it is crucial to study distributions of long pieces of unipotent orbits in the homogeneous space G/Γ. To be specific, for a particular long piece C of a unipotent orbit, we need to estimate the length of the part in C staying outside a large compact subset K of G/Γ. In homogeneous dynamics, the standard tool to study this type of problem is the linearization technique. The linearization technique is a standard and powerful technique in homogeneous dynamics. Using the linearization technique, we can transform a problem in dynamical systems to one in linear representations. Then we can study this problem using tools and results in representation theory. Let us briefly describe the technical difficulty when we apply the linearization technique. Let V be a finite dimensional linear representation of SL(n + 1, R) with a norm k · k and Γ(V) ⊂ V be a fixed discrete subset of V. Let U = {u(r) : r ∈ R} be a one dimensional unipotent subgroup of G. Given a large number T > 1, our main task is to estimate the measure of r ∈ [−T, T ] such that there exists v ∈ Γ(V) such that ku(r)vk ≤ ǫ where ǫ > 0 is a small number. By Dani non-divergence theorem (see [Dan84]), the measure is very small 5

compared with T given that for any such v ∈ Γ(V) max{ku(r)vk : r ∈ [−T, T ]} ≥ ρ where ρ > 0 is some fixed number. The difficulty is that there exists some v ∈ Γ(V), such that max{ku(r)vk : r ∈ [−T, T ]} ≤ ρ.

Let us call such intervals T -bad intervals. In this paper, we will use the representation theory to get some nice properties of such v’s. We then use these nice properties to show that in a longer interval, say [−T 2 , T 2 ], the number of T -bad intervals is ≪ T 1−µ for some constant µ > 0. This result is sufficient to prove Theorem 1.7. V In this paper, V is the canonical representation of SL(n + 1, R) on i Rn+1 and Γ(V) = Vi n+1 Z \ {0} where i = 1, . . . , n. The main technical results in this paper are proved in §4, §5.3 and §5.4. We refer the reader to [Rat91], [MT94], [MS95], [Sha09b], [Sha09a] and [LM14] for more applications of the linearization technique. 1.6. The organization of the paper. The paper is organized as follows: • In §2, we will recall some basic facts on Diophantine approximation, linear representations and lattices in Rn+1 . • In §3, we will recall a theorem on computing the Hausdorff dimension of Cantor like sets. We will also construct a Cantor-like covering of the set of weighted badly approximable points. • In §4, we will prove two technical results on counting lattice points. This is one of main technical contributions in this paper. Our proof relies on the linearization technique and SL(2, R) representations. • In §5, we will finish the proof of Theorem 3.5. Our proof relies on the KleinbockMargulis non-divergence theorem (Theorem 5.1) and the linearization technique.

Acknowledgements. The author would like to thank Professor Elon Lindenstrauss and Professor Barak Weiss for sharing many insightful ideas on this problem. He also thanks Professor Shahar Mozes for helpful conversations on this problem. He appreciates their encouragements during the process of this work. He also wants to thank Professor Jinpeng An for some early discussion on related problems. 2. Preliminaries 2.1. Dual form of approximation. We first recall the following equivalent definition of Bad(r): Lemma 2.1 (see [Ber15, Lemma 1]). Let r = (r1 , . . . , rn ) ∈ Rn be a weight and x ∈ Rn . The following statements are equivalent: (1) x ∈ Bad(r). (2) There exists c > 0 such that for any integer vector (p1 , . . . , pn , q) such that q 6= 0, we have that max |q|ri |qxi + pi | ≥ c. 1≤i≤n

6

(3) There exists c > 0 such that for any N ≥ 1, the only integer solution (a0 , a1 , . . . , an ) to the system |a0 + a1 x1 + · · · + an xn | < cN −1 , |ai | < N ri for all 1 ≤ i ≤ n is a0 = a1 = · · · = an = 0. Proof. The reader is referred to [Mah39], [BPV11, Appendix] and [Ber15, Appendix A] for the proof.  Later in this paper we will use the third statement as the definition of Bad(r). Given a weight r = (r1 , . . . , rn ), let us define    t  e         e−r1 t   Dr := dr (t) :=  ..  : t ∈ R .  .     e−rn t

For x ∈ Rn , let us define

  1 xT U(x) := . In

If we use the third statement in Lemma 2.1 as the definition of Bad(r), then it is easy to show that x ∈ Bad(r) if and only if U(x)Zn+1 ∈ Bd(Dr ). In fact, the statement can be proved using the same argument as in the proof of Proposition 1.9. 2.2. The canonical representation. Let V = Rn+1 . There is a canonical V representation of G = SL(n + 1, R) on V . It induces a canonical representation of G on i V for every i = 1, 2, . . . , n. For g ∈ G and ^i V, v = v1 ∧ · · · ∧ vi ∈

gv = (gv1 ) ∧ · · · ∧ (gvi ). For i = 1, . . . , n, let ei ∈ Rn denote the vector with 1 in the ith component and 0 in other components. Let us fix a basis for V as follows. Let w+ := (1, 0, . . . , 0)T . For i = 1, . . . , n, let wi := (0, . . . , 1, . . . , 0)T with 1 in the i + 1st component and 0 in other components. Then {w+ , w1 , . . . , wn } is a basis for V . Let W denote the subspace of V spanned by {w1 , . . . , wn }. For j = 2, . . . , n, let Wj the subspace of W spanned by {wj , . . . , wn }. Let us define     1 Z := z = : k(z) ∈ SO(n) . k(z) There is a canonical action of SO(n) on Rn . For k ∈ SO(n) and x ∈ Rn , letus denote  by 1 k · x the canonical action of k on x. It is straightforward to check that for z = ∈Z k(z) and x ∈ Rn , zU(x)z −1 = U(k(z) · x). 7

For any x ∈ Rn , U(x) can be embedded into a subgroup SL(2,   x) of G isomorphic to 1 kxk2 SL(2, R). In this SL(2, R) copy, U(x) corresponds to . For r > 0, let ξx (r) ∈ 1   r ∈ SL(2, R). SL(2, x) denote the element corresponding to r −1 Let us consider the representation of SL(2, x) on V . Let us first consider the case x = e1 . Let us denote U1 := {u1 (r) := U(re1 ) : r ∈ R},

and

Ξ1 := {ξ1 (r) := diag{r, r −1 , 1, . . . , 1} : r > 0}. It is easy to see that ξ1 (r)w+ = rw+ , u1 (r)w+ = w+ , ξ1 (r)w1 = r −1 w1 , u1 (r)w1 = w1 +rw+ , and for any w ∈ W2 , w is fixed by SL(2, e1 ). For general x ∈ Rn , we have that x = kxk2 k · e1 for some k ∈ SO(n). Then SL(2, x) = z(k)SL(2, e1 )z −1 (k)

where z(k) =



1 k



∈ Z. In particular, we have that U(x) = z(k)u1 (kxk2 )z −1 (k)

and ξx (r) = z(k)ξ1 (r)z −1 (k). Since z(k)w+ = w+ and z(k)W = W , we have that ξx (r)w+ = rw+ , U(x)w+ = w+ , ξx (r)z(k)w1 = r −1 k · w1 , U(x)z(k)w1 = z(k)w1 + kxk2 w+ and for any w ∈ z(k)W2 , w is fixed by SL(2, x). V Let us consider the action of SL(2, x) on i V for i = 2, . . . , n. Let us denote x = kxk2 k·e1 Vi−1 as above. For any w ∈ z(k)W2 , we have that ξx (r)((z(k)w1 ) ∧ w) = r −1 ((z(k)w1 ) ∧ w)

,

U(x)((z(k)w 1 ) ∧ w) = (z(k)w1 ) ∧ w + kxk2 (w+ ∧ w), ξx (r)(w+ ∧ w) = r(w+ ∧ w)

and

U(x)(w+ ∧ w) = w+ ∧ w. V For any w ∈ z(k)W2 and any w′ ∈ i−2 z(k)W2 , we have that w and w+ ∧ (z(k)w1 ) ∧ w′ are fixed by SL(2, x). Vi

2.3. Lattices in Rn+1 . In this subsection let us recall some basic facts on lattices and sublattices in Rn+1 . For a discrete subgroup ∆ of Rn+1 , let SpanR (∆) denote the R-span of ∆. Let Λ ∈ X = G/Γ be a unimodular lattice in Rn+1 . For i = 1, . . . , n + 1, let Λi ⊂ Λ be a i-dimensional sublattice of Λ. We say that Λi is primitive if SpanR (Λi ) ∩ Λ = Λi . Given a i-dimensional primitive sublattice Λi of Λ, let us choose a basis {v1 , . . . , vi } of Λi . Let us denote d(Λi ) = kv1 ∧ · · · ∧ vi k. For j = 1, . . . , i, let λj (Λi ) := inf{r ≥ 0 : B(0, r) contains at least j linearly independent vectors of Λi }. 8

By the Minkowski Theorem (see [Cas57]), we have the following: λ1 (Λi ) · · · λi (Λi ) ≍ d(Λi ).

(2.1)

Moreover, there exists a basis (called Minkowski reduced basis) of Λi , {vj : j = 1, . . . , i}, such that kvj k ≍ λj (Λi ) for every j = 1, . . . , i. We will need the following result on counting sublattices: Proposition 2.2. There exists a constant N > 1 such that the following statement holds. For any 0 < ǫ < ρ and any i = 1, . . . , n, let Λ ∈ Kǫ be a unimodular lattice in Rn+1 such that λ1 (Λ) ≥ ǫ. For ρ > 0, let Ci (Λ, ρ) denote the collection of i-dimensional primitive sublattices Λi of Λ with d(Λi ) ≤ ρ. Then we have that, |Ci (Λ, 1)| ≤ ǫ−N .

Proof. First note that there exists a constant N1 > 1 such that for any i = 1, . . . , n and ρ > 0, |Ci (Zn+1 , ρ)| ≤ ρN1 . It is a standard fact that there exists a constant N2 > 1 such that for any Λ ∈ Kǫ , there eixsts g ∈ SL(n + 1, R) with kg −1 k ≤ ǫ−N2 such that Λ = gZn+1 . Let us fix ρ > ǫ and i = 1, . . . , n. Then for any Λi ∈ Ci (Λ, 1), then we have that g −1Λi ⊂ Zn+1 and Therefore, we have that

d(g −1Λi ) ≤ kg −1 ki d(Λi ) ≤ ǫ−(n+1)N2 .

|Ci (Λ, 1)| ≤ |Ci (Zn+1 , ǫ−(n+1)N2 )| ≤ ǫ−N

where N = N1 N2 (n + 1). This completes the proof.



3. A Cantor like construction In this section, we will introduce a Cantor like construction which will help us to compute Hausdorff dimension. Definition 3.1 (See [Ber15, §5]). For an integer R > 0 and a closed interval J ⊂ R, let us denote by ParR (J) the collection of intervals obtained by dividing J into R equal closed subintervals. For a collection I of closed intervals, let us denote [ ParR (I) := ParR (I). I∈I

A sequence {Iq }q∈N of collections of closed intervals is called a R-sequence if for every q ≥ 1, Iq ⊂ ParR (Iq−1 ). For a R-sequence {Iq }q∈N and q ≥ 1, let us define Iˆq := ParR (Iq−1 ) \ Iq and \ [ Iq . K({Iq : q ∈ N}) := q∈N Iq ∈Iq

Then every R-sequence {Iq }q∈N gives a Cantor like subset K({Iq }q∈N ) of R. For q ≥ 1 and a partition {Iˆq,p }0≤p≤q−1 of Iˆq , let us define q−1  q−p X 4 dq ({Iˆq,p }0≤p≤q−1 ) := max F (Iˆq,p , Ip ), Ip ∈Ip R p=0 9

where F (Iˆq,p , Ip ) := |{Iq ∈ Iˆq,p , Iq ∈ Ip }|. Let us define dq (Iq ) := min dq ({Iˆq,p }0≤p≤q−1 ), {Iˆq,p }0≤p≤q−1

where {Iˆq,p }0≤p≤q−1 runs over all possible partitions of Iˆq . Let us define d({Iq }q∈N ) := max dq (Iq ). q∈N

Definition 3.2 (See [Ber15, §5]). For M > 1 and a compact subset X ⊂ R, we say that X is M-Cantor rich if for any ǫ > 0 and any integer R ≥ M, there exists a R-sequence {Iq }q∈N such that K({Iq }q∈N ) ⊂ X and d({Iq }q∈N ) ≤ ǫ. Our proof relies on the following two theorems:

Theorem 3.3 (See [Ber15, Theorem 6]). Any M-Cantor rich set X has full Hausdorff dimension. Theorem 3.4 (See [Ber15, Theorem 7]). Let I0 be a compact interval. Then any countable intersection of M-Cantor rich sets in I0 is M-Cantor rich. To show Theorem 1.6, it suffices to find a constant M > 1 and show that for any weight r, ϕ−1 (Bad(r) ∩ ϕ(I)) is M-Cantor rich. We will determine M > 1 later.

Theorem 3.5. There exists a constant M > 1 such that for any weight r, ϕ−1 (Bad(r)∩ϕ(I)) is M-Cantor rich. Our main task is to prove Theorem 3.5. Let us fix R ≥ M. We will show that for any ǫ > 0, we can construct a R-sequence {Iq }q∈N such that K({Iq }q∈N ) ⊂ Bad(r) and d({Iq }q∈N ) < ǫ. We will follow the construction in [Ber15] with some modifications. Standing Assumption 3.6. Let us make some assumptions to simplify the proof. A.1 Without loss of generality, we may assume that r1 ≥ r2 ≥ · · · ≥ rn . We may also assume that rn > 0. By [Ber15], if rn = 0, we can reduce the problem to the n − 1 dimensional case. A.2 Since ϕ = (ϕ1 , . . . , ϕn )T : I → Rn is analytic and nondegenerate, we may assume that for any s ∈ I and any i = (1) 1, . . . , n, ϕi (s) 6= 0. If this is not the case, we can choose a closed subinterval I ′ ⊂ I satisfying this condition. Then since I is closed, there exist constants C1 > c1 > 0 (1) such that for any s ∈ I and any i = 1, . . . , n, c1 ≤ |ϕi (s)| ≤ C1 .

Let us fix some notation. Let m > 0 be a large integer which we will determine later. Let κ = R−m . Let b > 0 be such that b1+r1 = R. For t > 0, let us denote  t  b   b−r1 t  . gr (t) :=  ..  . b−rn t

10

1+ri . Then we have that 1 = λ1 ≥ λ2 ≥ · · · ≥ λn . Without loss of For i = 1, . . . , n, let λi = 1+r 1 generality, we may assume that m(I) = 1. Let us define the R-sequence as follows. Let I0 = {I}. Suppose that we have defined Iq−1 for q ≥ 1 and every Iq−1 ∈ Iq−1 is a closed interval of length R−q+1 . Let us define Iq ⊂ ParR (Iq ) as follows. For any Iq ∈ ParR (Iq ), Iq ∈ Iˆq if and only if there exists s ∈ Iq such that gr (q)U(ϕ(s))Zn+1 ∈ / Kκ . That is to say, there exists a ∈ Zn+1 \ {0} such that kgr (q)U(ϕ(s))ak ≤ κ. Let us define Iq = ParR (Iq−1 ) \ Iˆq . This finishes the construction of {Iq }q∈N . It is easy to see that K({Iq }q∈N ) ⊂ Bad(r).

We need to prove the following:

Proposition 3.7. For any ǫ > 0, there exists an integer m > 0 such that the R-sequence {Iq }q∈N constructed as above with κ = R−m satisfies that (3.1)

d({Iq }q∈N ) ≤ ǫ.

Let N > 1 be the constant from Proposition 2.2. Let us give the partition {Iˆq,p }0≤p≤q−1 of Iˆq for each q ∈ N. Definition 3.8. Let us fix a small constant 0 < ρ < 1. We will modify the choice of ρ later in this paper according to the constants coming from our technical results. For q ≤ 106 n4 Nm, let us define Iˆq,0 := Iˆq and Iˆq,p = ∅ for other p’s. For q > 106 n4 Nm and l = 2000n2 Nm, let p = q −2l. Let us define Iˆq,p to be the collection of Iq ∈ Iˆq with the following property: there exists x ∈ Iq such that for any j = 1, . . . , n and V any w = w1 ∧ · · · ∧ wj ∈ j Zn+1 \ {0}, max{gr (q)U(ϕ(x′ ))w : x′ ∈ [x − R−q+l , x + R−q+l ]} ≥ ρj .

η 1 ′ 6 4 2 ′ Let η = 100n 2 and η = 1+r . For q > 10 n Nm and 2000n Nm ≤ l ≤ 2η q, let p = q − 2l. 1 For i = 1, . . . , n, let us define Iˆq,p (i) to be the collection of Iq ∈ Iˆq satisfying that there exists V s ∈ Iq and v = v1 ∧ · · · ∧ vi ∈ i Zn+1 \ {0} such that

kgr (q)U(ϕ(s′ ))vk ≤ ρi ,

for any s′ ∈ [s − R−q+l , s + R−q+l ] and for any j = 1, . . . , n and any w = w1 ∧ · · · ∧ wj ∈ Vj n+1 Z \ {0},  max kgr (q)U(ϕ(s′ ))wk : s′ ∈ [s − R−q+l , s + R−q+l ] ≥ ρj . S Let us define Iˆq,p = ni=1 Iˆq,p (i). For i = 1, . . . , n, let us define Iˆq,0 (i) to be the collection of Iq ∈ Iˆq satisfying that there V exists s ∈ Iq and v = v1 ∧ · · · ∧ vi ∈ i Zn+1 \ {0} such that n o ′ −q(1−2η′ ) −q(1−2η′ ) max kgr (q)U(ϕ(s))vk : s ∈ [s − R ,s+ R ] ≤ ρi . S Let us define Iˆq,0 = ni=1 Iˆq,0 (i). Let us define Iˆq,p := ∅ for other p’s. It is easy to see that {Iˆq,p }0≤p≤q−1 is a partition of Iˆq , cf. [Ber15, Proposition 3]. 11

Besides the definition of {Iˆq,p }0≤p≤q−1 , let us also introduce the notion of dangerous intervals and extremely dangerous intervals: Definition 3.9. For q ≥ 106 n4 Nm, 1000n2 Nm ≤ l ≤ η ′ q, and a ∈ Zn+1 \ {0}, the (q, l)dangerous interval associated with a, which is denoted by ∆q,l (a), is a closed interval of the form ∆q,l (a) = [x − R−q+l , x + R−q+l ] ⊂ I such that Iq ⊂ ∆q,l (a) for some Iq ∈ Iˆq and max{kgr (q)U(ϕ(s′ ))ak : s′ ∈ ∆q,l (a)} = cρ,

for some c ∈ [1/2, 1]. For q ≥ 106 n4 Nm and a ∈ Zn+1 \ {0}, the q-extremely dangerous interval associated with ′ ′ a, which is denoted by ∆q (a), is a closed interval of the form ∆q (a) = [x − R−q+l , x + R−q+l ] with l′ > η ′ q such that Iq ⊂ ∆q (a) for some Iq ∈ Iˆq and ′

′

max{kgr (q)U(ϕ(s′ ))ak : s′ ∈ [x − 2R−q+l , x + 2R−q+l ]} = cρ

for some c ∈ [1/2, 1].

Remark 3.10. Note that for any q ≥ 106 n4 Nm, there are only finitely many a’s such that ∆q,l (a) or ∆q (a) exist. 4. Counting dangerous intervals In this section we will count dangerous intervals and extremely dangerous intervals. Proposition 4.1. Let q ≥ 106 n4 Nm, 1000n2Nm ≤ l ≤ η ′ q and p = q − 2l. For Ip ∈ Ip , let Dq,l (Ip ) denote the collection of (q, l)-dangerous intervals which intersect Ip . Then for any Ip ∈ Ip , 1 |Dq,l (Ip )| ≪ R(1− 10n )l .

Proposition 4.2. . Let q ≥ 106 n4 Nm. Let Eq ⊂ I denote the union of q-extremely dangerous intervals contained in I. Then Eq can be covered by a collection of Nq closed intervals of length δq and K0 (ρn+1 b−ηq )α Nq ≤ δq 1 −q(1−η′ ) . where δq = R , K0 > 0 is a constant, and α = (n+1)(2n−1) Proposition 4.2 is a rephrase of the following theorem due to Bernik, Kleinbock and Margulis: Theorem 4.3 (See [Ber15, Proposition 2] and [BKM01, Theorem 1.4]). Let q > 106 n4 Nm. Let us define Eq ⊂ I to be the collection of s ∈ I satisfying that there exists a = (a0 , a1 , . . . , an )T ∈ Zn+1 \ {0} such that |ai | < ρbri q for i = 1, . . . , n, |f (s)| < ρb−q and |f ′ (s)| < b(r1 −η)q where f (x) = a0 + a1 ϕ1 (x) + · · · + an ϕn (x).

Then Eq can be covered by a collection Eq of intervals such that and

m(∆) ≤ δq for all ∆ ∈ Eq ,

K0 (ρn+1 b−ηq )α , δq ′ 1 where K0 > 0 is a constant, δq = R−q(1−η ) and α = (n+1)(2n−1) . |Eq | ≤

12

The theorem we quote here is the version used in [Ber15] with some minor modifications. The original version proved in [BKM01] is more general. Proof of Proposition 4.2. In fact, for every q-extremely dangerous interval ∆q (a) where l′ ≥ η ′ q and a = (a0 , a1 , . . . , an )T , we have that kgr (q)U(ϕ(x))ak ≤ ρ

(4.1)

Let us fix x ∈ ∆q (a). By direct computation, we have that

gr (q)U(ϕ(x))a = (v0 (x), v1 (x), . . . , vn (x))T

where v0 (x) = bq (a0 + a1 ϕ1 (x) + · · · + an ϕn (x)), and vi (x) = b−ri q ai for i = 1, . . . , n. Following the notation in Theorem 4.3, let us denote f (x) = a0 + a1 ϕ1 (x) + · · · + an ϕn (x).

Then (4.1) implies that |ai | ≤ ρbri q for i = 1, . . . , n, and |f (x)| ≤ ρb−q . Moreover, for any ′ ′ x′ ∈ [x − R−q(1−η ) , x + R−q(1−η ) ], we have that kgr (q)U(ϕ(x′ ))ak ≤ ρ.

By direct calculation, this implies that ′

′

|f (x′ )| ≤ ρb−q

′

for any x′ ∈ [x − R−q(1−η ) , x + R−q(1−η ) ]. Let x′ = x + rR−q(1−η ) for some r ∈ [−1, 1]. Then ′

′

f (x′ ) = f (x) + f ′ (x)rR−q(1−η ) + O(R−2q(1−η ) ).

Therefore, we have that for any r ∈ [−1, 1], ′

′

|f ′ (x)rR−q(1−η ) | = |f (x′ ) − f (x) − O(R−2q(1−η ) )| ′ ≤ |f (x′ )| + |f (x)| + O(R−2q(1−η ) ) ≤ ρb−q + ρb−q + ρb−q ≤ b−q .

By letting r = 1, we have that

′

|f ′ (x)| ≤ Rq(1−η ) b−q = bq(r1 −η) .

η . This shows that x ∈ Eq for The last equality above holds because b1+r1 = R and η ′ = 1+r 1 any x ∈ ∆q (a), i.e., ∆q (a) ⊂ Eq . Therefore, we have that Dq ⊂ Eq . Then the conclusion follows from Theorem 4.3. 

The rest of the section is devoted to the proof of Proposition 4.1. This is one of the main technical results of this paper. Proof of Proposition 4.1. Let us fix Ip ∈ Ip . Let Ip = [x − R−q+2l , x + R−q+2l ]. We claim that we may assume that ϕ(Ip ) is a straight line. In fact, for any x′ ∈ Ip , let us write x′ = x + rR−q+2l for some r ∈ [−1, 1]. By Taylor’s expansion, we have that gr (q)U(ϕ(x′ )) = gr (q)U(ϕ(x) + rR−q+2l ϕ(1) (x) + O(R−2q+4l )) = gr (q)U(O(R−2q+4l ))gr (−q)gr (q)U(ϕ(x) + rR−q+2l ϕ(1) (x)) = U(O(R−q+4l ))gr (q)U(ϕ(x) + rR−q+2l ϕ(1) (x)).

Since l ≤ η ′ q, we have that O(R−q+4l ) is exponentially small. Therefore, we may assume that ϕ(x′ ) = ϕ(x) + (x′ − x)R−q+2l ϕ(1) (s) for any x′ ∈ Ip . 13

Let us take a typical (q, l)-dangerous interval ∆q,l (a) that intersects Ip . Let us take x ∈ ∆q,l (a) ∩ Ip such that x ∈ Iq−1 for some Iq−1 ∈ Iq−1 . It is easy to see that either [x, x + R−q+l ] ⊂ ∆q,l (a), or [x − R−q+l , x] ⊂ ∆q,l (a). Without loss of generality, we may assume that [x, x + R−q+l ] ⊂ ∆q,l (a). Let us write a = (a0 , a1 , . . . , an )T ∈ Zn+1 \ {0}. For x′ ∈ [x, x + R−q+l ], let us denote gr (q)U(ϕ(x′ ))a = v(x′ ) = (v0 (x′ ), v1 (x′ ), . . . , vn (x′ ))T . Then we have that max{kv(x′ )k : x′ ∈ [x, x + R−q+l ]} = cρ for some c ∈ [1/2, 1]. 1+r Recall that for j = 1, . . . , n, λj = 1+r1j . Let 1 ≤ n′ ≤ n be the largest index j such that (1 − λj )q ≤ l. For x′ ∈ [x, x + R−q+l ], let us write x′ = x + rR−q+l for r ∈ [0, 1]. By our assumption we have that ϕ(x′ ) = ϕ(x) + rR−q+l ϕ(1) (x). Let us write (1)

T ϕ(1) (x) = (ϕ1 (x), . . . , ϕ(1) n (x)) . (1)

By our standing assumption on ϕ (Standing Assumption A.2), we have that c1 ≤ |ϕj (x)| ≤ C1 for j = 1, . . . , n. By direct calculation, we have that gr (q)U(ϕ(x′ ))a = gr (q)U(ϕ(x′ ) − ϕ(x))gr (−q)gr (q)U(ϕ(x))a = gr (q)U(rR−q+l ϕ(1) (x))gr (−q)v(x) = gr (q)U(rR−q+l ϕ(1) (x))gr (−q)v(x). Let us write gr (q)U(rR−q+l ϕ(1) (x))gr (−q)v(x) = v(x′ ) = (v0 (x′ ), v1 (x′ ), . . . , vn (x′ ))T , where x′ = x + rR−q+l . By direct calculation, we have that gr (q)U(rR−q+l ϕ(1) (x))gr (−q) = U

rRl

n X

(1) R−(1−λi )q ϕi (x)ei

i=1

!

.

By our assumption, for i ≥ n′ + 1, we have that |rRl R−(1−λi )q | ≤ 1. Therefore, if we write ! n X (1) R−(1−λi )q ϕi (x)ei v(x′ ) = v′ (x′ ) = (v0′ (x′ ), v1′ (x′ ), . . . , vn′ (x′ ))T , U −rRl i=n′ +1

P (1) where v0′ (x′ ) = v0 (x′ ) −r ni=n′ +1 Rl R−(1−λi )q ϕi (x)vi (x′ ) and vi′ (x′ ) = vi (x′ ) for i = 1, . . . , n. Then |v0′ (x′ )| < C = (n + 1)C1 ρ, and |vi′ (x′ )| < ρ for i = 1, . . . , n. Let ′

h=

n X

(1)

R−(1−λi )q ϕi (x)ei

i=1

and

′

hW =

n X i=1

(1)

R−(1−λi )q ϕi (x)wi ∈ W.

Then khk2 = khW k2 ≍ 1. For r ∈ [−1, 1], and x′ = x + rR−q+l , our discussion above shows that U(rRl h)v(x) = (v0′ (x′ ), v1′ (x′ ), . . . , vn′ (x′ ))T , where |v0′ (x′ )| < C, and |vi′ (x′ )| < ρ for i = 1, . . . , n. Let En′ be the subspace of Rn spanned by {e1 , . . . , en′ } and Wn′ ′ be the subspace of W spanned by {w1 , . . . , wn′ }. Then h ∈ En′ . Let 14

′ k ∈ SO(n) be  theelement such that k·e1 = h, k·En′ = En′ , and k·ei = ei for i = n +1, . . . , n. 1 Let z(k) = ∈ Z. It is easy to see that z(k)w+ = w+ , z(k)w1 = hW , z(k)Wn′ ′ = Wn′ ′ , k and z(k)wi = wi for i = n′ + 1, . . . , n. By the definition of z(k) and our discussion in §2.2, we have that U(h) = z(k)U(khk2 e1 )z −1 (k). Therefore, we have that U(h)hW = hW + khk2 w+ . Moreover, we have that U(h)w+ = w+ ; for i = 2, . . . , n′ , U(h)z(k)wi = z(k)wi ; and for i = n′ + 1, . . . , n, U(h)wi = wi . Let us write ′

v(x) = a+ (x)w+ +

n X

ai (x)z(k)wi +

n X

ai (x)wi .

i=n′ +1

i=1

Then the above discussion shows that ′

l

l

U(rR h)v(x) = (a+ (x) + rR a1 (x))w+ +

n X

ai (x)z(k)w i +

n X

ai (x)wi .

i=n′ +1

i=1

By our previous argument, we have that there exists a constant C > 0 such that |ai (x)| < C for i = 1, . . . , n and |a+ (x) + rRl a1 (x)| < C for any r ∈ [0, 1]. This implies that |a+ (x)| < C, and |a1 (x)| < CR−l . Therefore, we have that v(x) ∈ z(k)([−C, C] × [−CR−l , CR−l ] × [−C, C]n−1 ). Now let us estimate |Dq,l (Ip )|. Suppose that Dq,l (Ip ) = {∆q,l (au ) : 1 ≤ u ≤ L}. For each u = 1, . . . , L, let us take xu ∈ ∆q,l (au ) ∩ Ip such that xu ∈ Iq−1,u for some Iq−1,u ∈ Iq−1 . Let us denote vu = gr (q)U(ϕ(xu ))au . Then by our previous argument, we have that ′

vu = au,+ w+ +

n X

au,i z(k)wi +

n X

au,i wi ,

i=n′ +1

i=1

where |au,+ | < C, |au,1| < CR−l , and |au,i | < C for i = 2, . . . , n. Now let us consider gr (q)U(ϕ(x1 ))au . Let us write xu = x1 − rR−q+2l for some r ∈ [−1, 1]. Using our assumption that ϕ(Ip ) is a straight line, we have that = = = = Let us denote h = have that

Pn′

gr (q)U(ϕ(x1 ))au gr (q)U(ϕ(x1 ) − ϕ(xu ))gr (−q)gr (q)U(ϕ(xu ))au gr (q)U(ϕ(x1 ) − ϕ(xu ))gr (−q)vu −q+2l (1) gr (q)U(rR ϕ (x))gr (−q)vu  P (1) n U rR2l i=1 R−(1−λi )q ϕi (x)ei vu .

i=1

(1)

R−(1−λi )q ϕi (x)ei as before. Then by our previous argument, we

P R−(1−λi )q ϕ(1) (x)ei )vu . gr (q)U(ϕ(x1 ))au = U(rR2l h + rR2l ni=n′ +1P
n 2l 2l −(1−λi )q (1) = au,+ + rR au,1 + rR ϕ (x)au,i w+ ′ +1 R i=n P P ′ + ni=1 au,i z(k)wi + ni=n′ +1 au,i wi . 15

(1)

Since |au,1| ≤ CR−l , and since for i = n′ + 1, . . . , n, (1 − λi )q > l, |au,i | < C, and |ϕi (x)| ≤ C1 , we have that P au,+ + rR2l au,1 + rR2l n ′ R−(1−λi )q ϕ(1) (x)au,i i=n +1 P ≤ |au,+ | + |r|R2l |au,l | +P|r|R2l ni=n′ +1 R−(1−λi )q |ϕ(1) (x)||au,i | ≤ C + R2l CR−l + R2l ni=n′ +1 R−l C1 C ≤ C + R2l CR−l + R2l nR−l C1 C ≤ C2 R l where C2 = 2C + nC1 C > 0. This implies that for any u = 1, . . . , L, we have that gr (q)U(ϕ(x1 ))au ∈ z(k)([−C2 Rl , C2Rl ] × [−CR−l , CR−l ] × [−C, C]n−1 ). Let us consider the range of gr (q − l)U(ϕ(x1 ))au = gr (−l)gr (q)U(ϕ(x1 ))au . Let us write gr (−l) = d2 (l)d1 (l) where  −l  b   br1 l In1   rn′ +1 l , b d1 (l) =    .   .. rn l b

and



Then we have that

   d2 (l) =    

1



1 b−(r1 −r2 )l ..

. b−(r1 −rn′ )l In−n′

   .   

gr (q − l)U(ϕ(x1 ))au ∈ d2 (l)d1 (l)z(k)([−C2 Rl , C2 Rl ] × [−CR−l , CR−l ] × [−C, C]n−1 ). By the definition of z(k), we have that d1 (l)z(k) = z(k)d1 (l). Therefore, we have that d1 (l)z(k)([−C2 Rl , C2 Rl ] × [−CR−l , CR−l ] × [−C, C]n−1 ) = z(k)d1 (l)([−C2 Rl , C2 Rl ] × [−CR−l , CR−l ] × [−C, C]n−1 ) Q n1 −1 = z(k)([−C2 br1 l , C2 br1 l ] × [−Cb−l , Cb−l ] × [−Cbr1 l , Cbr1 l ]Q × ni=n′ +1 [−Cbri l , Cbri l ]) ′ ⊂ z(k)([−C2 br1 l , C2 br1 l ] × [−1, 1] × [−Cbr1 l , Cbr1 l ]n −1 × ni=n′ +1 [−Cbri l , Cbri l ]).

It is easy to see that

′

z(k)([−C2 br1 l , C2 br1 l ] × [−1, 1] × [−Cbr1 l , Cbr1 l ]n −1 ×

n Y

[−Cbri l , Cbri l ])

i=n′ +1

can be covered by a collection B of O(bλl ) balls of radius 1 where λ = n′ r1 + Then we have that S gr (q − l)U(ϕ(x1 ))au ∈ S d2 (l) B∈B B = B∈B d2 (l)B. 16

Pn

i=n′ +1 ri .

Since d2 (l) is a contracting map, for every B ∈ B, there exists a ball B ′ of radius C such that d2 (l)B ⊂ B ′ . Let B′ denote the collection of all such B ′ ’s. Then we have that [ gr (q − l)U(ϕ(x1 ))au ∈ B′. B ′ ∈B′

Since gr (q − l)U(ϕ(x1 ))au ∈ gr (q − l)U(ϕ(x1 ))Zn+1 , we have that [ B ′ ∩ Λ, gr (q − l)U(ϕ(x1 ))au ∈ B ′ ∈B′

where Λ = gr (q − l)U(ϕ(x1 ))Zn+1 . By our assumption, x1 ∈ Iq−1,1 for some Iq−1,1 ∈ Iq−1 . This implies that x1 ∈ Iq−l for some Iq−l ∈ Iq−l . Therefore, Λ = gr (q −l)U(ϕ(x1 ))Zn+1 ∈ Kκ , i.e., Λ does not contain any nonzero vectors with norm ≤ κ. Therefore, there exists a constant C4 such that every ball of radius 1 contains at most C4 κ−n−1 = C4 R(n+1)m points in Λ. Thus, we have that P |Dq,l (Ip )| = |{gr (q − l)U(ϕ(x1 ))au : 1 ≤ u ≤ L}| ≤ PB′ ∈B′ |B ′ ∩ Λ| (n+1)m ≤ B ′ ∈B′ C4 R 1 ≤ C5 bλl+4nm ≤ C5 b(λ+ 200n )l , P where C5 = C3 C4 and λ = n′ r1 + ni=n′ +1 ri . Now let us estimate λ. In fact, P ′ P λ = ni=1 ri + ni=1 (r1 − ri ) P ′ = 1 + ni=1 (r1 − ri ). 1 By our assumption, for i = 1, . . . , n′ , we have that r1 − ri ≤ ql ≤ 100n 2 . Therefore, we have that 1 1 λ≤1+n =1+ . 2 100n 100n Thus, we have that 1 1 1 |Dq,l (Ip )| ≤ C5 b(1+ 100n + 200n )l ≤ C5 R(1− 10n )l . 1

n

The last inequality above holds because b = R 1+r1 ≤ R n+1 . This completes the proof.  5. Proof of the main result In this section we will finish the proof of Proposition 3.7. By our discussion in §1 and §3, Proposition 3.7 implies Theorem 3.5 and Theorem 1.7. The structure of the section is as follows. In the first subsection, we will prove Proposition 3.7 for the case q ≤ 106 n4 Nm. The second, third and fourth subsections are devoted to the proof for the case q > 106 n4 Nm. The key point is to estimate F (Iˆq,p , Ip ) for Ip ∈ Ip . The second subsection deals with the case p = q − 4000n2 Nm. The third subsection deals with the case p = q − 2l where 2000n2 Nm < l < 2η ′ q. The fourth subsection deals with the case p = 0. The third and fourth subsections contain some technical results on the canonical represenVi tation of SL(n + 1, R) on V for i = 2, . . . , n. They are also main technical contributions of this paper. Our basic tool is the following non-divergence theorem due to Kleinbock and Margulis: 17

Theorem 5.1 (see [KM98, Theorem 5.2]). There exist constants C, α > 0 such that the following holds: ForVany subinterval J ⊂ I and any t ≥ 0, if for any i = 1, 2, . . . , n and any v = v1 ∧ · · · ∧ vi ∈ i Zn+1 \ {0}, then for any ǫ > 0,

max{kgr (t)U(ϕ(x))vk : x ∈ J} ≥ ρi ,

 α ǫ ∈ / Kǫ } ≤ C m(J). ρ

n+1




m {x : gr (t)U(ϕ(x))Z

Remark 5.2. The original version of Theorem 5.1 is more general. The version above is tailored for our setting. For the case where ϕ(x) is a polynomial, the statement is proved by Dani [Dan84]. We will also need the following result due to Nimish Shah [Sha10] on SL(n + 1, R) representations. Theorem 5.3 (see [Sha10, Proposition 4.9]). Let V be any finite dimensional representation of SL(n + 1, R) with a norm k · k and let r be any n-dimensional weight. There exists a constant c > 0 such that for any nonzero vector v ∈ V and any t ≥ 0, max{kgr (t)U(ϕ(x))vk : x ∈ I} ≥ ckvk.

Remark 5.4. The exact statement in [Sha10, Proposition 4.9] is different from the above version, but it easily implies the above statement. The proof makes use of the fact that ϕ is not contained in any proper affine subspace in n R . We may choose 0 < ρ < 1 small enough such that ρn+1 < c where c denotes the constant from Theorem 5.3. 5.1. The case where q is small. In this subsection, let us assume that q ≤ 106 n4 Nm. Then only Iˆq,0 = Iˆq is nonempty. Proposition 5.5. F (Iˆq,0 , I) ≪ Rq−αm . Proof. By Theorem V 5.3 and our assumption on ρ, we have that for any i = 1, 2, . . . , n and v = v1 ∧ · · · ∧ vi ∈ i Zn+1 \ {0}, max{kgr (q)U(ϕ(x))vk : x ∈ I} ≥ ckvk ≥ c ≥ ρi .

Then by Theorem 5.1, we have that

n+1

m({x ∈ I : gr (q)U(ϕ(x))Z

∈ / K2κ }) ≤ C



2κ ρ

α

.

On the other hand, it is easy to see that gr (q)U(ϕ(Iq ))Zn+1 ⊂ X \ K2κ for any Iq ∈ Iˆq . Therefore, we have that S  F (Iˆq,0, I)R−q = m I Iq ∈Iˆq q = m({x ∈ I : gr (q)U(ϕ(x))Zn+1 ∈ / K2κ }) ≤ C6 κα = C6 R−αm  α  where C6 = C ρ2 . This finishes the proof. 18

6 n4 N

Let us choose M > 1 such that M α > 100010

.

Proof of Proposition 3.7 for q ≤ 106 n4 Nm. It suffices to show that  q 4 F (Iˆq,0 , I) R can be arbitrarily small. In fact, by Proposition 5.5, we have that

q 4 q F (Iˆq,0 , I) = 4 O(Rq−αm) R

R

106 n4 Nm

q

= O( R4αm ) = O( 4 Rαm ) = O(
q Then it is easy to see that R4 F (Iˆq,0 , I) → 0 as m → ∞. This completes the proof for q ≤ 106 n4 Nm.


106 n4 N m 4 ). 1000 

5.2. The generic case. The rest of the section is devoted to the proof of Proposition 3.7 for q > 106 n4 Nm. In the following subsections, we will estimate F (Iˆq,p , Ip ) for different p’s. In this subsection we will estimate F (Iˆq,p , Ip ) for p = q − 4000n2 Nm. We call it the generic case. Proposition 5.6. Let q > 106 n4 Nm and p = q − 4000n2 Nm. Then for any Ip ∈ Ip , we have that F (Iˆq,p , Ip ) ≪ Rq−p−αm . 2

2

Proof. Let us fix Ip ∈ Ip . Let Ip = [a, b] and Ip′ = [a − R−q+2000n N m , b + R−q+2000n N m ]. It is easy to see that Ip ⊂ Ip′ and m(Ip′ ) < 2m(Ip ). If F (Iˆq,p , Ip ) = 0, then the statement apparently holds. Suppose F (Iˆq,p , Ip ) > 0, let us take Iq ∈ Iˆq,p and x ∈ Iq ∩ Ip . Then for any i = 1, . . . , n V and v = v1 ∧ · · · ∧ vi ∈ i Zn+1 \ {0}, we have that max{kgr (q)U(ϕ(x′ ))vk : x′ ∈ [x − R−q+2000n

It is easy to see that [x − R−q+2000n

2N m

2Nm

, x + R−q+2000n

, x + R−q+2000n

2N m

2N m

]} ≥ ρi .

] ⊂ Ip′ . Therefore, we have that

max{kgr (q)U(ϕ(x′ ))vk : x′ ∈ Ip′ } ≥ ρi .

By Theorem 5.1, we have that m({x ∈

Ip′

n+1

: gr (q)U(ϕ(x))Z

∈ / K2κ }) ≤ C

This implies that n+1

m({x ∈ Ip : gr (q)U(ϕ(x))Z

∈ / K2κ }) ≤ C



2κ ρ

α



2κ ρ

m(Ip′ )

α

m(Ip′ ).

≤ 2C



2κ ρ

α

m(Ip ).

On the other hand, it is easy to see that gr (q)U(ϕ(Iq ))Zn+1 ⊂ X \ K2κ for any Iq ∈ Iˆq . Therefore we have that F (Iˆq,p , Ip )R−q n+1 ≤ m({x ∈ / K2κ })  ∈Ip : gr (q)U(ϕ(x))Z ≤ 2C = 2C

2κ ρ

α

 α 2 ρ

m(Ip )

κα R−p = C7 R−p−κm 19

where C7 = 2C

 α 2 ρ

. This proves the statement.



By Proposition 5.6, we have that for p = q − 4000n2 Nm and any Ip ∈ Ip , the following holds:

q−p q−p−αm 4 q−p F (Iˆq,p , Ip ) ≪ R4 R R 2 (5.1)
4000n2 N m 44000n Nm 4 = = . αm R 1000
q−p F (Iˆq,p , Ip ) → 0 as m → ∞. Then it is easy to see that R4

5.3. Dangerous case. In this subsection, we will consider the case where 2000n2 Nm < l < 2η ′ q and p = q − 2l. We call this case the (q, l)-dangerous case. Proposition 5.7. For any Ip ∈ Ip , we have that

l F (Iˆq,p , Ip ) ≪ Rq−p− 20n .

Let us recall that for 1000n2 Nm < l′ < η ′ q, a (q, l′ )-dangerous interval ∆q,l′ (a) associated with a nonzero integer vector a ∈ Zn+1 is a closed interval of the form ′

′

∆q,l′ (a) = [x − R−q+l , x + R−q+l ] such that Iq ⊂ ∆q,l′ (a) for some Iq ∈ Iˆq and

max{kgr (q)U(ϕ(x′ ))ak : x′ ∈ ∆q,l′ (a)} = cρ

for some c ∈ [1/2, 1]. The following lemma is crucial to prove Proposition 5.7 and is one of the main technical contributions of this paper: Lemma 5.8. For any i = 1, . . . , n and Iq ∈ Iˆq,p (i) intersecting Ip , either there exists a (q, l′ )-dangerous interval ∆q,l′ (a) containing Iq for some l/2 ≤ l′ ≤ l, or there exists x ∈ Iq and ^i v = v1 ∧ · · · ∧ vi ∈ Zn+1 \ {0}

such that if we write

gr (q)U(ϕ(x))v = w+ ∧ w(i−1) + w(i) V V where w(i−1) ∈ i−1 W and w(i) ∈ i W , then we have that kw+ ∧ w(i−1) k = kw(i−1) k ≤ ρi and kw(i) k ≤ ρi R−l/2 .

Proof. If i = 1, then the first statement apparently holds. We may that i ≥ 2. Vi assume n+1 ˆ By the definition of Iq,p (i), there exists v = v1 ∧ · · · ∧ vi ∈ Z \ {0} such that for any x ∈ Iq , max{kgr (q)U(ϕ(x′ ))vk : x′ ∈ [x − R−q+l , x + R−q+l ]} = cρi for some c ∈ [1/2, 1]. Without loss of generality, we may assume that the sublattice Li generated by {v1 , . . . , vi } is a primitive i-dimensional sublattice of Zn+1 . Then Λi = gr (q)U(ϕ(x))Li is a primitive i-dimensional sublattice of Λ = gr (q)U(ϕ(x))Zn+1 . For simplicity, let us denote g = gr (q)U(ϕ(x)). Let us choose the Minkowski reduced basis {gv′1 , . . . , gv′i } of Λi . Since d(Λi ) = kgvk ≤ ρi ,

we have that kgv′1 k ≤ ρ by the Minkowski Theorem. 20

Let us follow the argument in the proof of Proposition 4.1. Recall that for j = 1, . . . , n, 1+r λj = 1+r1j . Let 1 ≤ n′ ≤ n be the largest index j such that (1 − λj )q ≤ l. Let us write ϕ(s) = (ϕ1 (s), . . . , ϕn (s))T . (1)

By Standing Assumption A.2, we have that c1 ≤ |ϕi (s)| ≤ C1 for any i = 1, . . . , n and P ′ s ∈ I. Let h = ni=1 R−(1−λi )q ϕ(1) (s)ei . For any x′ ∈ [x − R−q+l , x + R−q+l ], let us write x′ = x + rR−q+l where r ∈ [−1, 1]. By the same argument as in the proof of Proposition 4.1, we have that gr (q)U(ϕ(x′ )) = U(O(1))U(rRl h)gr (q)U(ϕ(x)) = U(O(1))U(rRl h)g. Therefore, we have that kU(rRl h)gvk ≤ ρi

for any r ∈ [−1, 1]. Following the notation in the proof of Proposition 4.1, let us denote h = k·e1 for k ∈ SO(n) and   1 z(k) = ∈ Z. k For j = 1, . . . , i, let us write gv′j = a+ (j)w+ + a1 (j)z(k)w1 + w′ (j) where w′ (j) ∈ z(k)W2 . Then

gv = (gv′1 ) ∧ · · · ∧ (gv′i ) Vi a1 (j)z(k)w1 + w′ (j)) = j=1 (a+ (j)w + +   P V ′ ′ ′ = w+ ∧ (z(k)w1 ) ∧ ǫ (j, j )a (j)a (j ) w (k) + 1 j R−l/2 . Then we have that   V V ǫ1 (1)a1 (1) ij=1 w′ (j) = w′ (1) ∧ ǫ1 (1)a1 (1) k6=1 w′ (k) P  V i ′ = w′ (1) ∧ ǫ (j)a (j) w (k) . 1 1 j=1 k6=j

Therefore, we have that

V

P  V



i ′ |a1 (1)| ij=1 w′ (j) = w′ (1) ∧ ǫ (j)a (j) w (k)

1 1 k6=j

P j=1

V

i ≤ kw′ (1)k j=1 ǫ1 (j)a1 (j) k6=j w′ (k) ≤ ρ · ρi R−l = ρi+1 R−l .

Since |a1 (1)| > R−l/2 and ρ < 1, we have that

i

^

′ w (j)

≤ ρi R−l/2 .

j=1

If we write

where w(i−1) ∈

Vi−1 w

W and w(i)

(i)

gv = w ∧ w(i−1) + w(i) V ∈ i W , then

= (z(k)w1 ) ∧

i X

ǫ1 (j)a1 (j)

j=1

^

′

w (k)

k6=j

By our previous argument, we have that

!

+

i ^

w′ (j).

j=1

kw(i) k ≤ ρi R−l/2 .

This proves the second part of the statement.



The following lemma takes care of the second case of Lemma 5.8. Lemma 5.9. Let i ∈ {2, . . . , n}. Let Dq,p (Ip , i) denote the collection of Iq ∈ Iˆq,p intersecting Ip and not contained in any (q, l′ )-dangerous interval for any l/2 ≤ l′ ≤ l. Let [ Dq,p (Ip , i) := Iq . Iq ∈Dq,p (Ip ,i)

1

Then for any closed subinterval J ⊂ Ip of length R−q+(1+ 2n )l , we have that l

m(Dq,p (Ip , i) ∩ J) ≪ R− 20n m(J). 22

1

Proof. Let us fix a closed subinterval J ⊂ Ip of length R−q+(1+ 2n )l . V For any x ∈ Iq ∈ Dq,p (Ip , i), there exists v = v1 ∧ · · · ∧ vi ∈ i Zn+1 \ {0} such that max{kgr (q)U(ϕ(x′ ))vk : x′ ∈ [x − R−q+l , x + R−q+l ]} ≤ ρi .

Let us denote the interval [x − R−q+l , x + R−q+l ] by ∆q,l (v, i). Then every Iq ∈ Dq,l (Ip , i) is contained in some ∆q,l (v, i) and every ∆q,l (v, i) contains at most O(Rl ) different Iq ∈ Dq,l (Ip , i). We will follow the notation used in the proof of Lemma 5.8. Let g = gr (q)U(ϕ(x)), h = k · e1 and   1 z(k) = ∈Z k be as in the proof of Lemma 5.8. . For j = 1, . . . , i, let us write

gvj = a+ (j)w+ + a1 (j)z(k)w1 + w′ (j) = a+ (j)w+ + w(j) where w′ (j) ∈ z(k)W2 and w(j) = a1 (j)z(k)w1 + w′ (j) ∈ W . Then P  V ′ ′ ′ gv = w+ ∧ (z(k)w1 ) ∧ ǫ (j, j )a (j)a (j ) w (k) ′ ′ +,1 + 1 j 0 such that for any q > 106 n4 Nm, we have that F (Iˆq,0 , I) ≪ R(1−ν)q . Similar to Lemma 5.8, we have the following: Lemma 5.12. For any i = 1, . . . , n and Iq ∈ Iˆq,0 (i), either there exists a q-extremely V dangerous interval ∆q (a) such that Iq ∈ ∆q (a), or there exists v = v1 ∧· · ·∧vi ∈ i Zn+1 \{0} such that the following holds: for any x ∈ Iq , if we write where w(i−1)

gr (q)U(ϕ(x))v = w+ ∧ w(i−1) + w(i) V V ′ ∈ i−1 W and w(i) ∈ i W , then kw+ ∧ w(i−1) k ≤ ρi and kw(i) k ≤ ρi R−η q .

Proof. The proof is the same as the proof of Lemma 5.8.



Definition 5.13. For i = 2, . . . , n, let Dq (i) denote the collection of Iq ∈ Iˆq,0 (i) such that the second case in Lemma 5.12 holds and let [ Iq . Dq (i) := Iq ∈Dq (i)

V Moreover, for Iq ∈ Dq (i), let v = v1 ∧ · · · ∧ vi ∈ i Zn+1 \ {0} be the vector given in the second case of Lemma 5.12. Then for x ∈ Iq , we can write gr (q)U(ϕ(x))v = w+ ∧ w(i−1) + w(i)

′ as in the second case of Lemma 5.12. For l ≥ η ′ q, let Dq,l (i) denote the collection of Iq ∈ Dq (i) such that ρi R−l+1 ≤ kw(i) k ≤ ρi R−l , and let [ ′ Dq,l (i) := Iq . ′ (i) Iq ∈Dq,l

Lemma 5.14. There exists a constant ν > 0 such that for any q > 106 n4 Nm and any i = 2, . . . , n, we have that m(Dq (i)) ≪ R−νq . 26

Proof. For any η ′ q ≤ l ≤ 2η ′q, using the same argument as in the proof of Lemma 5.9, we can prove that l ′ m(Dq,l (i)) ≪ R− 20n . Therefore, we have that S ′  P2η′ q 2η q ′ ′ m ≤ l=η′ q Dq,l (i) l=η′ q m(Dq,l (i)) ′ P2η′ q − l −η q 20n ≪ R 20n . R ≪ ′ l=η q Let us denote

[

Dq′ (i) :=

l>2η′ q

and Dq′ (i) :=

[

′ Dq,l

Iq .

Iq ∈Dq′ (i)

Then it is enough to show that

m(Dq′ (i)) ≪ R−νq .

V For any Iq ∈ Dq′ (i) and x ∈ Iq , there exists v = v1 ∧ · · · ∧ vi ∈ i Zn+1 \ {0} such that if we write gr (q)U(ϕ(x))v = w+ ∧ w(i−1) + w(i) V V where w(i−1) ∈ i−1 W and w(i) ∈ i W , then we have that kw+ ∧ w(i−1) k ≤ ρi and ′ kw(i) k ≤ ρi R−2η q . Recall that η = (1 + r1 )η ′ . Let us deal with the following two cases separately: (1) rn ≥ nη . (2) There exists 1 < n1 ≤ n such that for ri ≥ nη for 1 ≤ i < n1 and ri < nη for n1 ≤ i ≤ n. Let us first deal with the first case. For this case, let us define   −ηt b η ∈ SL(n + 1, R) g (t) := bηt/n In and gr,η (t) := g η (t)gr (t). It is easy to see that

′

g η (t)w+ = b−ηt w+ = R−η t w+ , and

′

g η (t)w = bηt/n w = Rη t/n w for any w ∈ W . Then we have that kgr,η (q)U(ϕ(x))vk = kg η (q)(w+ ∧ w(i−1) + w(i) )k ≤ kg η (q)(w+ ∧ w(i−1) )k + kg η (q)w(i) k ηqi i−1 = b−ηq(1− n ) kw+ ∧ w(i−1) k + b n kw(i) k ηq

′

≤ b− n ρi + bηq R−2η q ρi ≤ R−

η′ q n

ρi .

By the Minkowski Theorem, the above inequality implies that the lattice gr,η (q)U(ϕ(x))Zn+1 η′ q

contains a nonzero vector with norm ≤ R− n2 ρ. Therefore, for any Iq ∈ Dq′ (i) we have that gr,η (q)U(ϕ(Iq ))Zn+1 6∈ Kσ 27

η′ q

where σ = R− n2 ρ. By Theorem 5.3, for any j = 1, . . . , n and any v = v1 ∧ · · · ∧ vj ∈ Vj n+1 Z \ {0}, we have that max{gr,η (q)U(ϕ(x))v : x ∈ I} ≥ ρi .

Then by Theorem 5.1, we have that

This proves that


αη ′ q m {x ∈ I : gr,η (q)U(ϕ(x))Zn+1 6∈ Kσ } ≪ σ α = R− n2 . αη ′ q

m(Dq′ (i)) ≪ R− n2 . This finishes the proof for the first case. Now let us take care of the second case. Let us denote  −βt  b   1   ..   .   ξ(t) :=   ∈ SL(n + 1, R) brn1 t     . ..   rn t b Pn where β = j=n1 rj < η and  χt  b   b−r1 t   .   ..     ′ −rn1 −1 t b g (t) := ξ(t)gr (t) =     1     . ..   1 Pn1 −1 rj . Then it is easy to see that where χ = j=1 ξ(t)w+ = b−βt w+ , ξ(t)wj = wj for j = 1, . . . , n1 − 1, and

ξ(t)wj = brj t wj

for j = n1 , . . . , n. Then we have that kg ′ (q)U(ϕ(x))vk = ≤ ≤ ≤ ≤

kξ(q)(w+ ∧ w(i−1) + w(i) )k kξ(q)(w+ ∧ w(i−1) )k + kξ(q)w(i) k kw+ ∧ w(i−1) k + bβq kw(i) k ′ ρi + bβq R−2η q ρi ′ ′ ρi + bηq R−2η q ρi ≤ ρi + R−η q ρi < (2ρ)i . ′

′

Moreover, for any x′ ∈ ∆(x) := [x − R−q(1−2η ) , x + R−q(1−2η ) ], we also have that kg ′ (q)U(ϕ(x′ ))vk < (2ρ)i . 28

Let C > 0 and α > 0 be the constants given in Theorem 5.1. We can choose 0 < ρ < 1 small at beginning such that there exists another small constant 0 < ρ1 < 1 with  enough α 2ρ 1 C ρ1 . Then by the Minkowski Theorem, the inequality above implies that for < 1000

any x′ ∈ ∆(x), the lattice g ′(q)U(ϕ(x′ ))Zn+1 contains a nonzero vector of length < 2ρ. Let vx′ ∈ Zn+1 \ {0} be the vector such that kg ′ (q)U(ϕ(x′ ))vx′ k < 2ρ. Let us write vx′ = (vx′ (0), vx′ (1), . . . , vx′ (n)).

Then for j = n1 , . . . , n, we have that |vx′ (j)| < 2ρ. Therefore, vx′ (j) = 0 for any j = n1 , . . . , n. In other words, vx′ is contained in the subspace spanned {w+ , w1 , . . . , wn1 −1 }. For notational simplicity, let us denote this subspace by Rn1 and denote the set of integer points contained in the subspace by Zn1 . Accordingly, let us denote by SL(n1 , R) the subgroup    X : X ∈ SL(n1 , R) ⊂ SL(n + 1, R) In+1−n1

and denote by SL(n1 , Z) the subgroup of integer points in SL(n1 , R). Note that g ′(g) ∈ SL(n1 , R) and U(ϕ(x′ )) can be considered as an element in SL(n1 , R) since it preserves Rn1 . Then kg ′ (q)U(ϕ(x′ ))vx′ k < 2ρ implies that for any x′ ∈ ∆(x), the lattice g ′(q)U(ϕ(x′ ))Zn1 contains a nonzero vector of length < 2ρ. Let K2ρ (n1 ) ⊂ X(n1 ) = SL(n1 , R)/SL(n1 , Z) denote the set of unimodular lattices in Rn1 which do not contain any nonzero vector of length < 2ρ. Then the claim above implies that m({x′ ∈ ∆(x) : g ′(q)U(ϕ(x′ ))Zn1 6∈ K2ρ (n1 )}) = m(∆(x)). V By Theorem 5.1, there exist j ∈ 1, . . . , n1 − 1 and v′ = v′1 ∧ · · · ∧ v′j ∈ j Zn1 \ {0} such that ′

′

max{kg ′(q)U(ϕ(x′ ))v′ k : x′ ∈ [x − R−q(1−2η ) , x + R−q(1−2η ) ]} < ρj1

since otherwise we will have that

α 2ρ 1 m({x ∈ ∆(x) : g (q)U(ϕ(x ))Z 6∈ K2ρ (n1 )}) ≤ C m(∆(x)). m(∆(x)) < ρ1 1000 This reduces the second case in dimension n + 1 to the first case in dimension n1 < n + 1. Then we can finish the proof by induction.  ′

′

′



n1

Now we are ready to prove Proposition 5.11. Proof of Proposition 5.11. Recall that in Theorem 4.3, we denote by Eq the union of all q-extremely dangerous intervals. By Lemma 5.12, we have that n [[ Iq ⊂ Eq Dq (i). i=2

By Theorem 4.3, we have that

m(Eq ) ≪ R−νq for some constant ν > 0. On the other hand, by Lemma 5.14, we have that m(Dq (i)) ≪ R−νq

for any i = 2, . . . , n. Therefore, we have that S  F (Iˆq,0 , I)R−q = m I ˆ Iq ∈Iq,0 q S P ≤ m (Eq ni=2 Dq (i)) ≤ m(Eq ) + ni=2 m(Dq (i)) ≪ R−νq . 29

This completes the proof.



Now we are ready to prove Proposition 3.7 for q > 106 n4 Nm. Proof of Proposition 3.7 for q > 106 n4 Nm. We can choose M such that M ν > 1000. Recall that R ≥ M. By Proposition 5.11, we have that

q
q
q 4 q (5.4) F (Iˆq,0 , I) ≪ 4 R(1−ν)q = 4ν < 4 . R

R

R

Combining (5.1), (5.2) and (5.4), we have that q−1  q−p X 4 max F (Iˆq,p , Ip ) → 0 Ip ∈Ip R p=0

as m → ∞. This proves the statement.

1000



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