Sparse Equidistribution of Unipotent Orbits in Finite-Volume Quotients of PSL(2, R)

Transcription

1 Sparse Equidistribution of Unipotent Orbits in Finite-Volume Quotients of PSL(2, R) Dissertation Presented in Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy in the Graduate School of the Ohio State University By Cheng Zheng, B.S. Graduate Program in Mathematics The Ohio State University 206 Dissertation Committee: Nimish A. Shah, Advisor Vitaly Bergelson James Cogdell

2 Copyright by Cheng Zheng 206

3 Abstract We consider the orbits {pu(n +γ ) n N} in Γ\ PSL(2, R), where Γ is a nonuniform lattice in PSL(2, R) and {u(t)} is the standard unipotent one-parameter subgroup in PSL(2, R). Under a Diophantine condition on the intial point p, we can prove that the trajectory {pu(n +γ ) n N} is equidistributed in Γ\ PSL(2, R) for small γ > 0, which generalizes a result of Venkatesh [V0]. In Chapter 2, we will compute Hausdorff dimensions of subsets of non-diophantine points in Γ\ PSL(2, R), using results of lattice counting problem. In Chapter 3 we will use the exponential mixing property of a semisimple flow to prove the effective equidistribution of horospherical orbits. In Chapter 4, we will give a definition of Diophantine points of type γ for γ 0 in a homogeneous space Γ\G and compute the Hausdorff dimension of the subset of points which are not Diophantine of type γ when G is a semisimple Lie group of real rank one. As an application, we will deduce a Jarnik-Besicovitch Theorem on Diophantine approximation in Heisenberg groups. ii

4 To my parents iii

5 Acknowledgements I would like to thank my advisor Professor Nimish Shah for his helps and encouragements during my PhD study. He is always patient to explain things that I need to know, which greatly broaden my view in mathematics, and many ideas in this thesis come from discussions with him. Also I would like to thank Davis Buenger for many discussions in our joint work [BZ6]. I would like to thank the Committee members Professor Vitaly Bergelson and Professor James Cogdell. And I would like to thank all the people in Mathematics Department for providing such a great academic environment. I am also grateful to Nicolas de Saxce who suggested Lemma.3.4 in this thesis. I would like to thank Professor Andreas Strömbergsson and Samuel Edwards for many discussions in Chapter 3. I was told that they had results about the effective equidistribution of horocycle orbits in homogeneous spaces using number theoretic tools. In Chapter 3 our purpose is just to show how to use mixing property only to get such results. I thank the MSRI for hospitality during spring 205. Finally I would like to thank my parents and all of my friends for their support and help. iv

6 Vita B.S., Mathematics, Peking University 200 to present Graduate Teaching Associate Department of Mathematics The Ohio State University Fields of Study Major Field: Mathematics v

7 Table of Contents Abstract Dedication Acknowledgements Vita ii iii iv v Chapter Sparse Equidistribution of Unipotent Orbits in Γ\ PSL(2, R).... Introduction Prerequisites The Diophantine Condition (C, α; ρ, ɛ 0 )-Good Functions in Presence of Diophantine Condition Calculations Proof of the Main Theorem Further Discussions Chapter 2 Hausdorff Dimension of Diophantine Points in Γ\ PSL(2, R) Introduction and Preliminaries Some Properties of Lattices Points in R Hausdorff Dimension Chapter 3 Effective Equidistribution of Abelian Horospherical Orbits Introduction Preliminaries Some Lemmas vi

8 3.4 Effective Equidistribution of Abelian Horospherical Orbits Chapter 4 Diophantine Points in Rank One Homogeneous Spaces Introduction Notations and Preliminaries Reductions Some Lie Group Facts Counting Rational Points Diophantine Points Proof of Theorem Proof of Theorem Further Discussions The Hyperbolic Case The Case of P SL(2, R) Diophantine Approximation in Heisenberg Groups Bibliography vii

9 Chapter Sparse Equidistribution of Unipotent Orbits in Γ\ PSL(2, R). Introduction The theory of equidistribution of unipotent flows on homogeneous spaces has been studied extensively over the past few decades. Furstenberg [F73] first proved that the unipotent flow on Γ\ PSL(2, R), where Γ is a uniform lattice, is uniquely ergodic. In [D78] Dani classified ergodic invariant measures for unipotent flows on finite volume homogeneous spaces of PSL(2, R), and using this result Dani and Smillie [DS84] proved that any non-periodic unipotent orbit is equidistributed on Γ\ PSL(2, R) for any lattice Γ. The proof of the Oppenheim Conjecture due to Margulis [M89] by proving a special case of Raghunathan s conjecture drew a lot of attention to this subject. Soon afterwords, Ratner published her seminal work [R90a, R90b, R9a] proving measure classification theorem for unipotent actions on homogeneous spaces as conjectured by Raghunathan and Dani [D8]. Using these results, Ratner [R9b] proved that any unipotent orbit in a finite volume homogeneous space is equidistributed in its orbit closure; see also Shah [Sh9] for the case of Rank- semisimple groups. Ratner s work has led to many new extensions and number theoretic applications of ergodic theory of unipotent flows. One of these results, which is related to this paper, was the work by Shah [Sh94]. In that paper, Shah asked whether {pu(n 2 ) n N} is equidistributed in a sub-homogeneous space of PSL(2, Z)\ PSL(2, R), where

10 u : R PSL(2, R) is the standard unipotent -parameter subgroup u(t) = t 0. In this direction, Venkatesh published a result about sparse equidistribution ([V0], Theorem 3.). There he introduced a soft technique of calculations by using a discrepancy trick, and proved that if Γ is a cocompact lattice in PSL(2, R) and γ > 0 is a small number depending on the spectral gap of the Laplacian on Γ\ PSL(2, R), then for any point p Γ\ PSL(2, R) we have N f(pu(n +γ )) N n=0 Γ\ PSL(2,R) fdµ. In other words, in the case of Γ\ PSL(2, R) being compact, the equidistribtion holds for the sparse subset {n +γ n N}. It is worth noting that recently Tanis and Vishe [TV5] improve some results of Venkatesh [V0] and they obtain an absolute constant γ > 0 which does not depend on the spectral gap. In this paper, we will consider the sparse subset {n +γ n N} and orbits of {u(n +γ ) n N} in Γ\ PSL(2, R), where Γ is a non-uniform lattice. We want to prove a sparse equidistribution theorem similar to Shah s conjecture [Sh94] and the work of Venkatesh [V0] and that of Tanis and Vishe [TV5]. To deal with the complexity caused by initial points of unipotent orbits, we introduce a Diophantine condition for points in Γ\ PSL(2, R) as follows. Let G = PSL(2, R) and we consider the Siegel sets N Ω A α K where N Ω = t 0 t is in a bounded subset Ω R 2

11 A α = s 0 0 s s α and K = SO(2). For the non-uniform lattice Γ, there exist σ j G and bounded intervals Ω j R ( j k) with the following property ([GR70], [DS84]). For some α > 0, G = k Γσ j N Ωj A α K. j= 2. σ j Γσ j N is a cocompact lattice in N. 3. N Ωj is a fundamental domain of σ j Γσ j N\N. We will fix σ j ( j k) in such a way that in the upper half plane H, each σ j corresponds to a cusp η j, i.e. lim t σ j it = η j, and η, η 2,..., η k are the inequivalent cusps of Γ\H. Let Γ j = Γ σ j Nσ j. Let π j be the covering map π j : Γ j \G Γ\G. Now consider the usual action of G on R 2 and let e = define a map m j : Γ j \G R 2 /± 0. For each j, we can by m j (q) = g σ j e for q = Γ j g Γ j \G, where R 2 /± means that we identify every v R 2 with its opposite v. In this way, we obtain k maps m j (j =, 2,..., k) whose images are all in R 2 /±. Using these notations, we can give the following definition of Diophantine condition of a point p Γ\G. 3

12 Definition... Let p Γ\G. We say that p is Diophantine of type (κ, κ 2,..., κ k ) for some κ j > 0 (j =, 2,..., k) if for each j, there exist µ j, ν j > 0 such that for every point a b m j (π j (p)), we have either b µ j or a κ j b ν j. Remark... This notion of Diophantine type on p Γ\G has been studied well in an equivalent form; it can be connected to the excursion rate of the geodesic orbit {g t (p)} t>0. We will prove this in section.3. It is straightforward to verify that if g AN then the Diophantine types of p and pg are the same; although the choices of µ j, ν j > 0 in the above definition may differ. The hausdorff dimension of the complement of the set of points of the Diophantine type (κ, κ 2,..., κ k ) will be discussed in section.7. We will see that almost every point satisfies the Diophantine condition of type (κ, κ 2,..., κ k ) when κ, κ 2,..., κ k >. When min{κ, κ 2,..., κ k } =, the set of points of the Diophantine type (κ, κ 2,..., κ k ) has zero Haar measure but has full Hausdorff dimension. Now we state the main theorem in this paper. Theorem.. (Main theorem). Let Γ be a non-uniform lattice in PSL(2, R) and k the number of inequivalent cusps of Γ\ PSL(2, R). Suppose that p Γ\ PSL(2, R) is Diophantine of type (κ, κ 2,..., κ k ). Then there exists a constant γ 0 > 0 such that for any 0 < γ < γ 0, we have N f(pu(n +γ )) N n=0 Γ\ PSL(2,R) fdµ. Here the constant γ 0 depends on κ, κ 2,..., κ k and Γ, and f is any bounded continuous function on Γ\ PSL(2, R). 4

13 Remark..2. From the proof of the main theorem, we will see that the constant { s 2 } γ 0 = min j =, 2,..., k. (s + 4)(κ j + 4) Here s is defined as follows: if we let λ > 0 denote the smallest eigenvalue in the discrete spectrum of the Laplacian on Γ\H then 4λ, if 0 < λ < 2 s = ; 4, otherwise. 2 Now let Γ be a subgroup of finite index of PSL(2, Z). Then we have the following corollary of the main theorem, which will be explained in section.3. Corollary... Let Γ be a subgroup of finite index of PSL(2, Z). Let p = Γg Γ\ PSL(2, R) with g = a c b d. If a/c R is a Diophantine number of type ζ; that is, there exists C > 0 such that for all m/n Q, we have ( n ζ n a ) m C, c then the orbit {pu(n +γ ) n N} is equidistributed in Γ\ PSL(2, R) for 0 < γ < γ 0 := s 2 (4+s)(ζ+4). To prove the main theorem, we shall use the technique of Venkatesh in [V0] and Strömbergsson s result in [S3] about effective version of Dani and Smillie s result [DS84] on Γ\ PSL(2, R). In fact, an immediate consequence of the technique of [V0] and result of [S3] is obtained in the following theorem. Before stating the theorem, we need some notations. For f C k (Γ\G) we let f p,k be the Sobolev L p -norm 5

14 involving all the Lie derivatives of order k of f. Note that f,0 is the supremum norm of f. We know that G acts on the upper half plane H by the action a c b d z = az + b cz + d and we have the standard projection of Γ\G to the fundamental domain of Γ in H π : Γ\G Γ\H by sending Γg to Γg(i). We define the geodesic flow on Γ\G by g t (Γg) = Γg et/2 0 0 e t/2. Fix, once for all, a point p 0 Γ\H. For p Γ\G let dist(p) = d H (p 0, π(p)) where d H (, ) is the hyperbolic distance on Γ\H. Theorem..2 (Cf.[V0] Theorem 3.). Let T > K > 2 and f C (Γ\G) satisfying Γ\G fdµ = 0 and f,4 <. Suppose that q Γ\G satisfies r = r(q, T ) = T e dist(g log T (q)). Then we have for β = T/K j Z 0 Kj<T f(qu(kj)) K 2 ln 3 2 (r + 2) f r β,4 2 sκ. Here κ is the constant in the mixing property of the unipotent flow 2(8+κ) 6

15 (see Theorem.2. and Remark.2.) and s is defined as in Remark..2. This theorem gives an estimate for the average of the unipotent action along an arithmetic progression with gap K, which is crucial in our proof of the main theorem. This was proved first in [V0] and later in [TV5], both in the case of Γ\G being compact. The strategy of the paper is the following: note that the bound in Theorem..2 depends on the initial point, and hence when we combine the results with different arithmetic progressions and different initial points, the outcomes would get out of control. To overcome this difficulty, we need the Diophantine condition. With the help of this Diophantine condition along with the notion of (C, α; ρ, ɛ 0 )-good functions, we will be able to control the rates of these effective results. In section.2, we list the concepts and theorems that we need in this paper. In section.3, we study the Diophantine condition and deduce Corollary.. from the main theorem. In section.4, we will study dynamics of a special class of orbits in Γ\G. The dynamical properties of these orbits will help us control the rates of the effective results in this paper. Since we are dealing with the noncompact case of Γ\G, and also for the sake of completeness, we include the technique of [V0] and prove Theorem..2 in section.5. We will finish the proof of the main theorem in section.6. Further discussions will be included in section.7. It may be interesting to explore the relation between the techniques used in this work and those developed in the work of Sarnak and Ubis [SU5], where they have described the limiting distribution of horocycles at primes. 7

16 .2 Prerequisites Throughout this note, if there exists an absolute constant C > 0 such that f Cg, then we write f g. If f g and g f, then we use the notation f g. We denote G = PSL(2, R) and Γ a non-uniform lattice in G. Let N = {u(t) t R}, A = s 0 0 s s R +. For any element a A, we denote α(a) = s. One of the ingredients in our calculations is the effective version of the mixing property of unipotent flows in Γ\G. The following effective version is proved by Kleinbock and Margulis [KM99]. Theorem.2. (Kleinbock and Margulis [KM99]). There exists κ > 0 such that for any f, g C (Γ\G), we have Γ\G f(xu(t))g(x)dµ(x) f Γ\G Γ\G g ( + t ) κ f, g,. Here µ is the Haar measure on Γ\G. Remark.2.. Note that when G = PSL(2, R), we can calculate κ explicitly. Indeed, let λ > 0 denote the smallest eigenvalue in the discrete spectrum of the Laplacian on Γ\H, then it follows from [V0] formula (9.7) and the technique of Lemma 2.3 in [R87] that κ = 2s ɛ for any ɛ > 0. Here s is defined as in Remark..2. Another ingredient in the calculations is the effective version of Dani and Smillie s result [DS84] proved by Strömbergsson [S3]. 8

17 Theorem.2.2 (Strömbergsson [S3]). For all p Γ\G, T 0, and all f C 4 (Γ\G) such that f,4 < we have T T f(pu(t))dt 0 Γ\G fdµ O( f,4)r s ln 3 (r + 2) provided that r. Here s > 0 is a number depending on the spectrum of the Laplacian on Γ\H and r = r(p, T ) = T e dist(g log T (p)). The implied constants depend only on Γ and p 0. Remark.2.2. Here we can take s as in Remark..2, i.e., let λ > 0 be the smallest eigenvalue in the discrete spectrum of the Laplacian on Γ\H, and 4λ, if 0 < λ < 2 s = ; 4, otherwise. 2 Readers may refer to [S3] for more details. We will prove a weaker version of this theorem in Chapter 3 using only mixing property, and Theorem.. could be proved by using techniques only from dynamical systems for many Diophantine points (at least for a subset of full Haar measure)..3 The Diophantine Condition First we deduce Corollary.. from the main theorem. Proof of Corollary... If Γ is a subgroup of finite index of PSL(2, Z), then we can pick σ j PSL(2, Z) ( j k). Now let p = Γg Γ\G with g = a c b d. 9

18 Note that for each m j, we have m j (π j (p)) g Z 2 \ {0} = dm bn cm + an m n Z 2 \ {0}. If there exist constants ζ > 0, µ, ν > 0 such that for any (m, n) Z 2 \ {0} an cm µ or dm bn ζ an cm ν, (.) then p is Diophantine of type (ζ,..., ζ) by the definition above. In particular, if a/c R is a Diophantine number of type ζ, i.e. there exists C > 0 such that for m/n Q, n ζ n a c m C, then condition (.) holds because when an cm is sufficiently small, dm bn = cdm bcn c = cdm (ad )n c = d(cm an) + n c n. Hence, Corollary.. follows from the main theorem. In order to prove the main theorem, we have to analyze the map m j : Γ j \G R 2 /± for each j. The following lemma is well known. The reader may refer to [DS84]. We will denote B d the ball of radius d around the origin in R 2. Lemma.3. ([DS84] Lemma 2.2). For each j with the maps π j : Γ j \G Γ\G and m j : Γ j \G R 2 /±, there exists a constant d j > 0 such that for any p Γ\G there exists at most one point of m j (π j (p)) which lies in B dj. Remark.3.. We will fix these d j s for j =, 2,..., k throughout this note. 0

19 Lemma.3.2. If p Γ\G is Diophantine of type (κ, κ 2,..., κ k ), then the orbit {g t (p) t 0} is non-divergent. Proof. Suppose that {g t (p) t 0} is divergent. Let η j be the cusp where {g t (p) t 0} diverges. By Lemma.29 in [EW0], we know that {g t (p) t 0} is divergent if and only if {pu(t)} is periodic in Γ\G. Combined with Lemma 2. in [DS84], this would imply that there is a point x y m j (π (p)) lying on the x-axis in R 2, i.e. j y = 0, which contradicts the Diophantine condition. Therefore, {g t (p) t 0} is non-divergent. Definition.3.. For p Γ\G, we define p j := min a b a b m j (π j (p)) where denotes the standard Euclidean norm in R 2. Moreover, we define d(p) = min{ p j j =, 2,..., k}. Lemma.3.3. For any p Γ\SL(2, R), we have e dist(p) d(p) 2. Proof. Recall that η j ( j k) are the inequivalent cusps of Γ\H. For each j k, we fix a small neighborhood C j of η j in Γ\G such that C, C 2,..., C k are pairwise disjoint. Also we fix a point q j C j for each j k. We observe that it suffices to prove the lemma for p C j (j =, 2,..., k) since the complement of Cj is compact. Let p C j for some j {, 2,..., k}. Let α j > 0 be such that

20 π j maps σ j N Ωj A αj K isomorphically to C j. Then we can pick a representative for p in σ j N Ω A αj K, say σ j n p a p k p, i.e. p = Γσ j n p a p k p = π j (Γ j σ j n p a p k p ). By definition we know that d(p) = p j = kp a p e = α(a p ). On the other hand, in the fundamental domain of Γ\H, the point corresponding to p = π j (Γ j σ j n p a p k p ) C j is equal to Γσ j n p a p k p i = Γσ j (n p a p i) = Γσ j (n p (α(a p ) 2 i)). Since σ j is fixed and n p is in the compact set N Ωj of N, we obtain d H (π(q j ), π(p)) ln α(a p ) 2 C j for some constant C j > 0. Since q j is fixed, we have dist(p) d H (π(q j ), π(p)) C j for some C j > 0. Therefore we get dist(p) ln α(a p ) 2 C for C = max{c + C, C 2 + C 2,..., C k + C k } and hence e dist(p) = e (dist(p) ln α(ap)2 )+ln α(a p) 2 e ln α(ap)2 = α(a p ) 2 = d(p) 2. Now we prove that the Diophantine condition on p Γ\G can be defined by 2

21 the excursion rate of the geodesic orbit {g t (p)} t>0. We need some notations. As in the proof of Lemma.3.3, let η, η 2,..., η k be the inequivalent cusps of Γ\H and we choose the neighborhood C j of η j ( j k) in Γ\G such that C, C 2,..., C k are pairwise disjoint. For each j k, we define a function on Γ\G by dist(p), if p C j ; dist (j) (p) = 0, otherwise. Lemma.3.4. A point p Γ\G has Diophantine type (κ, κ 2,..., κ k ) if and only if κ j and ( lim sup dist (j) (g t (p)) κ ) j t κ j + t < for each j {, 2,..., k}. Proof. As in the proof of Lemma.3.3, we know that there exists α j > 0 such that π j maps σ j N Ωj A αj K isomorphically to C j. Also using the same argument in the proof of Lemma.3.3, we can get that for any q C j, e dist(j) (q). (.2) q 2 j If p Γ\G is Diophantine of type (κ, κ 2,..., κ k ), then for each j, any a b m j (π j (p)) satisfies b µ j or a κ j b ν j. 3

22 Since m j (π j (g t (p))) = e t/2 0 0 e t/2 m j (π j (p)), (.3) this implies that any x y m j (π (g t (p))) satisfies j y e t/2 µ j or x κ j y ν j e ( κ j)t/2. (.4) Note that this holds for all t > 0. By Lemma.3.2, we know that {g t (p) t 0} is nondivergent. So there exists a compact subset S Γ\G such that g ti (p) remains in S for infinitely many t i. By the compactness of S, we can find a constant C 0 > 0 such that for each t i, there exists x i y i m j (π (g ti (p))) satisfying j x i y i C 0. This implies via equation (.4) that κ j. Now fix t and for any x y m j (π j (g t (p))) we have (x, y) y e t/2 µ j or (x, y) max{ x, y } x κ j κ j + κ y j + ν κ j + j κ j t +κ e j 2. Therefore, g t (p) j e t/2 µ j or ν κ j + j κ j t/2 +κ e j. By equation (.2) and κj, we get 4

23 that ( lim sup dist (j) (g t (p)) κ ) j t κ j + t <. Conversely, if the above inequality holds for each j with κ j, then by equation (.2) there exists a constant C > 0 such that for all t > 0 we have κ j t +κ g t (p) j Ce j 2. This implies via equation (.3) that for any a b m j (π (p)) we have j κ j e t a 2 + e t b 2 t +κ Ce j. (.5) By discreteness of m j (π j (p)) in R 2, there exists a constant µ j > 0 such that if a b m j (π (p)) satisfies b < µ j, then b < a. Now for such j a b, we take t > 0 such that e t a 2 = e t b 2. By equation (.5), this implies that b and hence a κ j b = e t b κ j b ( ) κj + C. 2 C 2 e κ j t +κ j This implies that p is Diophantine of type (κ, κ 2,..., κ k )..4 (C, α; ρ, ɛ 0 )-Good Functions in Presence of Diophantine Condition This section will be important in the proof of the main theorem. First, we need a modified version of the concept of (C, α)-good functions (see [KM98] for the definition of (C, α)-good functions). 5

24 Definition.4.. A function f(x) is said to be (C, α; ρ, ɛ 0 )-good if for any 0 < ɛ < ɛ 0 and any I = (x, x 2 ) [, ) with f(x ) = ρ, we have m({x I f(x) ɛ}) C ( ) α ɛ m(i) ρ where m denotes the Lebesgue measure on R. Now we shall begin to study a special class of functions and prove that they are (C, α; ρ, ɛ 0 )-good for some C, α, ρ and ɛ 0 > 0. Note that we restrict these functions to the domain [, ). Lemma.4.. Let κ, µ, ν > 0 and 0 < γ < κ+4. Let a b R 2 \ {0} be such that b µ or a κ b ν. Then there exist C, ɛ 0 > 0 such that f(x) = (bx 3 4 +γ ax 4 ) 2 (x 4 κ+4 ) 2 + (bx 4 ) 2 (x 4 κ+4 ) 2 is (C, 2 ; ρ, ɛ 0)-good on [, ), where ρ is any fixed constant f(). Here the constants C, ɛ 0 depend only on ρ, κ, µ, ν and γ. Proof. We observe that if b µ, then f(x) (bx 4 ) 2 (x 4 κ+4 ) 2 b 2 µ 2 and f(x) is automatically (C, α; ρ, ɛ 0 )-good for any C, α, ρ and ɛ 0 = µ 2 /2. Therefore, in the following we assume that b < µ and hence a κ b ν. We have two cases: ab < 0 and ab > 0. 6

25 Case : ab < 0. Our function f(x) then becomes f(x) = ( b x 3 4 +γ + a x 4 ) 2 (x 4 κ+4 ) 2 + (bx 4 ) 2 (x 4 κ+4 ) 2. We have f(x) ) (( b x 3 4 +γ + a x 2 4 )x 4 κ+4 ) 2 (max{ b x +γ κ+4, a x κ+4 } ) (( b x +γ κ+4 ) κ+ ( a x κ 2 κ+4 ) κ+ ) (( b a κ κ+ 2 +γ x κ+4 ) κ+ ν 2 κ+. This implies that f(x) is (C 2, α 2 ; ρ, ɛ 0 )-good for any C 2, α 2 > 0 and ɛ 0 = 2 ν 2 κ+. Case 2: ab > 0. Without loss of generality, we assume that a > 0, b > 0. Now let I = (x, x 2 ) [, ) be an interval (x, x 2 ) where f(x ) = ρ. Since f(x ) = ρ, we know that either (bx 3 4 +γ ax 4 ) 2 (x 4 κ+4 ) 2 ρ 2 or (bx 4 ) 2 (x 4 κ+4 ) 2 ρ 2. If (bx 4 ) 2 (x 4 κ+4 ) 2 ρ/2, then there is nothing to prove because f(x) ρ/2 for all x x. Otherwise, we have (bx 3 4 +γ ax 4 ) 2 (x 4 κ+4 ) 2 ρ 2. 7

26 Note that {x I f(x) ɛ} {x I (bx 3 4 +γ ax 4 ) 2 (x 4 κ+4 ) 2 ɛ}. Therefore, to finish the proof of the lemma, it suffices to show that there exist C, ɛ 0 > 0 depending only on ρ, κ, µ, ν, γ such that for any 0 < ɛ < ɛ 0 we have x 2 x m({x (x, x 2 ) g(x) ɛ}) C ( ) ɛ 2 ρ (.6) where g(x) = (bx 3 4 +γ ax 4 )x 4 κ+4 = bx +γ κ+4 ax κ+4. Note that g(x) is increasing and g(x ) ρ/2. Without loss of generality, we may assume that g(x ) = ρ/2. If g(x ) = ρ/2, since g(x) is increasing, the (C, α; ρ, ɛ 0 )-good property automatically holds in this case with ɛ 0 = 2 ρ/2. Therefore we assume that g(x ) = ρ/2. In this case, we will prove that the inequality (.6) holds with ɛ 0 = 2 ρ/2. Let 0 < ɛ < ɛ 0. Since g(x) is increasing with g(x ) = ρ/2, if we fix x and let x 2 vary as a variable, then the ratio x 2 x m({x (x, x 2 ) g(x) ɛ}) would attain its maximum when g(x 2 ) = ɛ. So we will assume that g(x 2 ) = ɛ. To compute this maximal ratio, let z (x, x 2 ) such that g(z) = ɛ, and then by the mean value theorem we obtain x 2 x m({x (x, x 2 ) g(x) ɛ}) = x 2 z x 2 x = g (ξ 2 ) g (ξ ) g(x 2) g(z) g(x 2 ) g(x ) 8

27 = g (ξ 2 ) g (ξ ) 2 ɛ (.7) ɛ + ρ/2 where ξ is between x 2 and z, ξ 2 is between x and x 2. Let x 3 [, ) such that g(x 3 ) = ρ/2. Then (x, x 2 ) (x, x 3 ). According to equation (.7), to prove formula (.6), it suffices to prove that for any x, y (x, x 3 ) the ratio g (x) g (y) is bounded above by constants depending only on ρ, κ, µ, ν and γ. Observe that g (x) = ( + γ ) bx γ a κ+4 + κ + 4 κ + 4 x κ+5 κ+4 is decreasing since γ < κ+4. Therefore, to get an upper bound for g (x)/g (y) (x, y (x, x 3 )), we only need to estimate g (x )/g (x 3 ). By the condition that g(x ) = ρ/2 and g(x 3 ) = ρ/2, we have g (x ) g (x 3 ) = = ( + γ κ+4 ( + γ κ+4 ( + γ κ+4 ( + γ κ+4 ( + γ κ+4 ( + γ κ+4 ) bx γ κ+4 + a κ+4 ) bx γ κ a κ+4 x κ+5 κ+4 x κ+5 κ+4 3 ) (ax κ+4 ρ/2)/x + a κ+4 ) (ax κ ρ/2)/x 3 + a κ+4 ) ax κ+4 /x + a κ+4 ) ax κ+4 3 /x 3 + a κ+4 x κ+5 κ+4 x κ+5 κ+4 3 = x κ+5 κ+4 x κ+5 κ+4 3 ( x3 x ) κ+5 κ+4. Now let x 0 (x, x 3 ) such that g(x 0 ) = 0. (x 0 = (a/b) +γ by solving the equation g(x) = 0). We set x = δ x 0 and x 3 = δ 2 x 0 for some δ, δ 2. Then δ, δ 2 satisfy the following equation b(x 0 δ) +γ κ+4 a(x0 δ) κ+4 = ρ 2 9

28 since g(x ) = g(x 3 ) = ρ/2. By the fact that bx +γ κ+4 0 = ax κ+4 0 and ba κ ν, this equation becomes ax κ+4 0 δ +γ κ+4 ax κ+4 0 δ κ+4 = ρ 2 δ +γ κ+4 δ κ+4 ρ x κ+4 = 0 = 2 a ρ 2 ν (+γ)(κ+4) ρ 2 a (a κ+ /ba κ ) a κ+ (+γ)(κ+4) a. (+γ)(κ+4) Here κ+ (+γ)(κ+4) <. Since b µ and hence a κ ν/µ, the above inequality becomes δ +γ κ+4 δ κ+4 ρ 2 ν (+γ)(κ+4) ( κ ν µ ) κ+ (+γ)(κ+4) which holds for δ = δ and δ 2. This shows that δ, δ 2 are bounded above and below by constants depending only on ρ, κ, µ, ν and γ. Therefore g (x ) g (x 3 ) ( x3 x ) (κ+5)/(κ+4) = ( ) (κ+5)/(κ+4) δ2 δ is bounded above by a constant depending only on ρ, κ, µ, ν and γ. This completes the proof of the lemma. For the rest of this section, we turn to the dynamics on Γ\G. For later use, we give the following definition. Definition.4.2. For any δ > 0 and any j {, 2,..., k}, we define the subset of 20

29 Γ\G S j,δ := {q Γ\G q j δ}. Moreover, we define S δ := j S j,δ = {q Γ\G d(q) δ}. Lemma.4.2. Let p Γ\G be Diophantine of type (κ, κ 2,..., κ k ). We fix j {, 2,..., k} and let 0 < γ < /(κ j + 4). Then for sufficiently small ɛ > 0 and T, we have T m p x [, T ] x 0 x 4 4 x 3 4 +γ Sj,ɛx 4 + κ j +4 Cɛ where C is a constant only depending on p and γ. Proof. We will use the notations, the maps m j and π j in section 3. Then the image of π j p x 4 x 3 4 +γ under m j is equal to 0 x 4 = x 4 x 3 4 +γ 0 x 4 x 4 x 3 4 +γ 0 x 4 m j (π j (p)) a b = ax 4 bx 3 4 +γ bx 4 where a b runs over all points in m j (π (p)). By definition, what we need to j 2

30 prove is equivalent to the following m x [, T ] p x 0 x 4 4 x 3 4 +γ ɛx j 4 + κ j +4 CɛT. which is equivalent to the following m x [, T ] a point in mj π j }) with length ɛx 4 + κ j +4 p x 0 x 4 4 x 3 4 +γ CɛT. Let ρ = min d p, d, d 2,..., d k 0, where d j s are as in Remark.3.. We denote by P the subset m j (π j (p)). For (a, b) P, let I(a,b) l (l =, 2,... ) be all the maximal connected subintervals in [, T ] such that for any x I(a,b) l the point of π j p x 4 x 3 4 +γ corresponding to (a, b); that is 0 x 4 x 4 x 3 4 +γ 0 x 4 a b = ax 4 bx 3 4 +γ bx 4 has norm ρx 4 + κ j +4. Since x, Lemma.3. implies that all the intervals {I l (a,b) (a, b) P, l =, 2,... } 22

31 are pairwise disjoint. Therefore, we have a m x [, T ] point in mj = (a,b) P π j }) with length ɛx 4 + κ j +4 m x I(a,b) l ax l 4 bx 3 4 +γ bx 4 p x 0 x 4 4 x 3 4 +γ ɛx 4 + κ j +4 Because of this, to prove the lemma, it suffices to prove the following m x Il (a,b) ax 4 bx 3 4 +γ bx 4 ɛx 4 + κ j +4 Cɛm(I(a,b)), l for some C, ɛ 0 with 0 < ɛ < ɛ 0, or to prove that the function f(x) = (bx 3 4 +γ ax 4 ) 2 (x 4 κ j +4 ) 2 + (bx 4 ) 2 (x 4 κ j +4 ) 2 has (C, /2; ρ 2, ɛ 2 0)-good property for some C = Cρ. This follows immediately from Lemma.4.. To conclude this section, we give the following proposition, which is crucial in our proof of the main theorem. It is the discrete version of Lemma.4.2. Proposition.4.. Let p Γ\G be Diophantine of type (κ, κ 2,..., κ k ). Let 0 < γ < min{/(κ j + 4) : j =, 2,..., k}. Then there exists a constant C 0 > 0 depending 23

32 only on p and γ such that for sufficiently small ɛ > 0 and any N N, N p n [, N] N n 0 n 4 4 n 3 4 +γ S θ(n,ɛ) C 0 ɛ where θ(x, ɛ) = ɛ min{x 4 + κ j +4 j =, 2,..., k}. (.8) Proof. By the definition of S δ, it suffices to prove that there exists a constant C 0 > 0 depending only on p and γ such that for each j and any ɛ > 0, N p n [, N] N n 0 n 4 4 n 3 4 +γ Sj,ɛn 4 + κ j +4 C 0 ɛ. We compute that for any δ (, ) and n = = = n 4 n 3 4 +γ 0 n 4 n 4 n 3 4 +γ 0 n 4 (n + δ) 4 (n + δ) 3 4 +γ 0 (n + δ) 4 (n + δ) 4 (n + δ) 3 4 +γ 0 (n + δ) 4 ( + δ/n) 4 n 4 (n + δ) 3 4 +γ n 3 4 +γ (n + δ) 4 0 ( + δ/n) 4 ( + δ/n) 4 ((n + δ) +γ n +γ )(n(n + δ)) 4 0 ( + δ/n) 4 which lies in a compact neighborhood U of identity in PSL(2, R). Let L = max{ g g U} 24

33 where g denotes the operator norm of g on R 2. Then by the computations above, we know that p n [, N] N n 4 n 3 4 +γ 0 n 4 N m p x [, N] x 4 x 3 4 +γ 0 x 4 N Sj,ɛn 4 + κ j +4 Sj,Lɛx 4 + κ j +4. Now the proposition follows immediately from Lemma Calculations In this section, we shall apply the technique of Venkatesh to obtain some effective results about averaging over arithmetic progressions. It is very similar to [V0], where Venkatesh proved the sparse equidistribution theorem for Γ being cocompact. Since in our setting Γ is non-uniform, and for the sake of self-containedness, we include the details of the calculations in this section. We will follow the notations in [V0]. Throughout this section, we fix an arbitrary point q Γ\G. For a character ψ : R S, we define for f on Γ\G. µ T,ψ (f) = T T 0 ψ(t)f(qu(t))dt Lemma.5. (Cf. [V0, Lemma 3.]). There exists a constant β > 0 which only depends on Γ such that for any f C (Γ\G) satisfying f,4 < and Γ\G fdµ = 0, any character ψ : R S, any T and any q Γ\G satisfying r = r(q, T ) = T e dist(g log T (q)), (.9) 25

34 we have µ T,ψ (f) r β ln 3 (r + 2) f,4 and the implicit constant is independent of ψ. Proof. The proof is almost the same as that of [V0, Lemma 3.] combined with [S3]. We define σ H (f)(x) = H H 0 ψ(s)f(xu(s))ds. First it is easy to get that µ T,ψ (f) µ T,ψ (σ H (f)) H T f,0 H r f,0. Now we estimate µ T,ψ (σ H (f)). By Cauchy-Shwartz inequality, we have ( T µ T,ψ (σ H (f)) T ( T ( H 2 0 T ) ( ψ(t) 2 2 T dt ) σ H (f)(qu(t)) 2 2 dt 0 H H 0 0 T T 0 0 ) σ H (f)(qu(t)) 2 2 dt ) f y f z 2 (qu(t))dt dydz. Here f y and f z denote the right translation of f by u(y) and the right translation of f by u(z), respectively. Therefore, by Strombergsson s effective equidistribution Theorem.2.2, we have µ T,ψ (σ H (f)) ( H H O( f H y f z 2,4 )r s ln 3 (r + 2)dydz H 2 H 0 H 0 (f y z, f) dydz ) 2. for some s > 0 depending only on Γ (the spectral gap) and r is as in (.9) (see Theorem.2.2). 26

35 By mixing property of unipotent flows (Theorem.2.), we know that (f h, f) ( + h ) κ f 2,. Also by product rule and chain rule in Calculus (see [V0] Lemma 2.2 for details), we know that O( f y f z,4 ) O( f y,4 f z,4 ) y 4 z 4 O( f 2,4). Therefore, combining all the computations above, we obtain µ T,ψ (f) µ T,ψ (f) µ T,ψ (σ H (f)) + µ T,ψ (σ H (f)) H r f + (H 8 r s ln 3 (r + 2) + H κ ) 2 f,4. Let H = r s 8+κ and we get β = sκ/2(8 + κ). This completes the proof of the lemma. We deduce Theorem..2 from Lemma.5.. It will be crucial in the proof of the main theorem in section.6. Proof of Theorem..2. The proof is almost the same as that of [V0, Theorem 3.]. Let δ > 0 and g δ (x) = max{δ 2 (δ x ), 0}. Let g(x) = j Z g δ (x + Kj). 27

36 On the one hand, since g(x) has most mass on the points {Kj j Z}, we know that T 0 g(t)f(qu(t))dt j Z 0 Kj<T f(qu(kj)) 2 f + T K δ f,. On the other hand, since g(x) is periodic, we have the Fourier expansion g(x) = k Z a k e 2πikx/K. A simple calculation shows that a k = g(0) = δ. k Z By Lemma.5. with characters ψ k = e 2πikx/K, we have T 0 g(t)f(qu(t))dt a k k Z T 0 e 2πikt/K f(qu(t))dt T ln3 (r + 2) f δr β,4. Combining the calculations above, we have T/K j Z 0 Kj<T f(qu(kj)) Note that K < T, r T and β <. Let δ = of the theorem. ( K T + δ + K ) ln3 (r + 2) f δr β,4. K ln 3 (r+2) r β and we complete the proof 28

37 .6 Proof of the Main Theorem Proof of the main theorem. By a standard approximation argument, we may assume that f C (Γ\G) with f,4 < and fdµ = 0. Γ\G We want to find γ 0 > 0 depending on κ,..., κ k such that for any 0 < γ < γ 0, the main theorem holds. Note that by Taylor expansion, for any M N and k N, (M + k) +γ = M +γ + ( + γ)m γ k + O(M γ k 2 ). Therefore, if M is sufficiently large and γ < /2, then the sequence { } (M + k) +γ 0 k + γ M 2 γ (k N) is approximately equal to the arithmetic progression { M +γ + ( + γ)m γ k 0 k } + γ M 2 γ (k N) since O(M γ k 2 ) O(M γ (M 2 γ ) 2 ) = O(M γ ) 0 as M. 29

38 By Proposition.4., we know that for any ɛ > 0 and any N > 0, N p n [, N] n 0 n 4 4 n 3 4 +γ S θ(n,ɛ) C 0 ɛ, where θ(n) = ɛ min{n 4 + κ j +4 j =, 2,..., k}. Set p B = n N n 0 n 4 4 n 3 4 +γ Sθ(n,ɛ) c. We proceed as follows. Fix ɛ > 0. We pick the first element M N which lies in B. Then we take { } P = M + k 0 k + γ M 2 γ (k N). Next we pick the first element M 2 N which appears after P and lies in B, and we take { P 2 = M 2 + k 0 k } + γ M 2 γ 2 (k N). Then we pick the first element M 3 N which appears after P 2 and lies in B, and so on. In this manner, we get pieces P, P 2,... in N and by our choices of M, M 2,..., we know that B P P 2... and hence for any N > 0 N [, N] \ (P P 2... ) C 0 ɛ. (.0) Now we consider each of the pieces P i. From the discussion above, we know that 30

39 {n +γ n P i } is approximated by the arithmetic progression { } P i = M +γ i + ( + γ)m γ i k 0 k + γ M 2 γ i (k N). We would like to apply Theorem..2 with T = M /2 i, K = ( + γ)m γ i and q = q i := pu(m +γ i ) for sufficiently large i. So first we have to check that r i := r(q i, M 2 i ) for sufficiently large i. We compute that g (log Mi )/2(q i ) = p M +γ i M 4 i M 4 i = p M 4 i 0 M 4 i i M 3 4 +γ S c θ(m i,ɛ), by our choice of M i B. By definition.4.2 and equation (.8), we have d(g (log Mi )/2(q i )) θ(m i, ɛ) = ɛ min{m 4 + κ j +4 i j =, 2,..., k}. By Lemma.3.3, e dist(q) d(q) 2. Hence r i = M /2 i e dist(g (log M i )/2(q i )) ɛ 2 min{m 2/(κ j+4) i j =, 2,..., k}. (.) This implies that r i as i, since M i by our choices of M i s. By Theorem..2 with T = M /2 i, K = ( + γ)m γ i, q i = pu(m +γ i ) and r i = r(m 2 i, q i ), we have P f(pu(n)) i n P i = M /2 i /( + γ)m γ i (( + γ)m γ i ) 2 ln 3 2 (ri + 2) 0 (+γ)m γ i k<m 2 i γ f(q i u(( + γ)mi k)) r β 2 i f,4. (.2) 3

40 Since M i, according to inequalities (.) and (.2), as long as γ < min{2β/(κ j + 4) j =, 2,..., k}, we have P f(pu(n)) i n P i 0 and hence by the fact that {n +γ n P i } is approximated by P i, i.e., for 0 k M 2 γ +γ i, f((m i + k) +γ ) f(m +γ i + ( + γ)m γ i k) M γ i f, and f, < we obtain f(pu(n +γ )) P i 0 (.3) n P i as i. By formula (.0), the proportion in [, N] which is not covered by P i s is small relative to N. Also observe that for the P i s which intersect [, N], their lengths are small relative to N. Therefore, by (.3) we have lim sup N lim sup N lim sup N N N N N n=0 f(pu(n +γ )) f(pu(n +γ )) f(pu(n +γ )) n [,N]\( P i ) n [,N]\( P i ) C 0 ɛ f,0 + 0 = C 0 ɛ f,0. + lim sup N + lim sup N N N n [,N] ( P i ) [,N] P i f(pu(n +γ )) f(pu(n +γ )) n P i 32

41 Let ɛ 0 and we complete the proof of the main theorem with γ 0 = min{2β/(κ j + 4) j =, 2,..., k}..7 Further Discussions In the introduction, we have defined Diophantine points in Γ\ PSL(2, R). Let S κ,κ 2,...,κ k = {p Γ\ PSL(2, R) p is Diophantine of type (κ, κ 2,..., κ k )}. Then we can calculate the Hausdorff dimension of the complement of S κ,κ 2,...,κ k. In fact, we have the following Theorem.7.. We have dim H S c κ,κ 2,...,κ k = min{κ j + j k}. If min{κ, κ 2,..., κ k } =, then S κ,κ 2,...,κ k has zero Lebesgue measure but has full Hausdorff dimension. Remark.7.. Note that the Diophantine type remains constant on any weak unstable leaf of {g t } t>0. Therefore the set of non Diophantine points on any strong stable leaf has zero Hausdorff dimension. We will give a different proof of this theorem in Chapter 2. Proof. For each cusp η j ( j k), we define S j,κ to be the subset of points p Γ\ PSL(2, R) satisfying the condition that there exist µ, ν > 0 such that for every point a b m j (π j (p)), either b µ or a κ b ν. Here µ and ν depend on p. 33

42 Then by definition, we have S κ,κ 2,...,κ k = S,κ S 2,κ2 S k,κk and hence S c κ,κ 2,...,κ k = S c,κ S c 2,κ 2 S c k,κ k. Let κ 0 = min{κ, κ 2,..., κ k }. Note that by Lemma.3.4, S κ0,κ 0,...,κ 0 S κ,κ 2,...,κ k. Therefore we get k k k Sj,κ c S2,κ c 2 Sk,κ c k Sκ c,κ 2,...,κ k Sκ c 0,κ 0,...,κ 0. j= j= j= By Theorem 2 and Theorem 3 in [MP93], for any κ we have This implies that dim H S c κ,κ,...,κ = κ + and dim H dim H S c κ,κ 2,...,κ k = dim H S c κ 0,κ 0,...,κ 0 = max k j= { S c j,κ = κ +. dim H k j= = κ 0 + = min{κ j + j k}. S c j,κ i i k } For the second statement, if min{κ, κ 2,..., κ k } =, then by Lemma.3.4 and the ergodicity of the geodesic flow on Γ\ PSL(2, R), we know that S κ,κ 2,...,κ k has zero 34

43 Haar measure. Since S,,..., S κ,κ 2,...,κ k and by Theorem. in [KM96] S,,..., has full Hausdorff dimension, this implies that S κ,κ 2,...,κ k has full Hausdorff dimension. Finally, using the same argument as in section 4, we can actually prove that if p is Diophantine of type (κ, κ 2,..., κ k ) with all κ j < 3 and 0 γ < /4, then for any ɛ > 0, there exists a compact subset K ɛ Γ\ PSL(2, R) such that for all T 0, T p x [, T ] x 4 x 3 4 +γ 0 x 4 K ɛ ɛ. Then using the arguments of [DS84] and [Sh94, Proposition 4.], we get Theorem.7.2. If p is Diophantine of type (κ, κ 2,..., κ k ) with all κ j < 3 and 0 γ < /4, then the trajectory p x 0 x 4 4 x 3 4 +γ x is equidistributed in Γ\ PSL(2, R). 35

44 Chapter 2 Hausdorff Dimension of Diophantine Points in Γ\ PSL(2, R) 2. Introduction and Preliminaries Here we will give a a different proof of Theorem.7. using results of lattice counting problem, that is, Theorem 2... dim H S c κ,κ 2,...,κ k = min{κ j + j k}. To prove Theorem 2.. we need some preliminaries. Readers may refer to [KM96]. Let X be a Riemannian manifold, m a volume form and E a compact subset of X. We will denote the diameter of a set E by diam(e). A countable collection A of compact subsets of E is said to be tree-like if A is the union of finite subcollections A j such that. A 0 = {E}. 2. For any j and A, B A j, either A = B or A B =. 3. For any j and B A j+, there exists A A j such that B A. 4. d j (A) := sup A Aj diam(a) 0 as j. 36

45 We write A j = A A j A and define A = j N A j. Moreover, we define m(a j+ B) j (A) = inf. B A j m(b) The following theorem gives a way to estimate the Hausdorff dimension of A. Theorem 2..2 ([M87], [U9] or [KM96]). Let (X, m) be a Riemannian manifold. Assume that there exist constants D > 0 and k > 0 such that m(b(x, r)) Dr k for any x X. Then for any tree-like collection A of subsets of E dim H (A ) k lim sup j j i=0 log( i (A) ) log( d j+ (A) ) 2.2 Some Properties of Lattices Points in R 2 In this section, we will show some lemmas which will be used in the proof of Theorem 2... For each cusp η j ( j k), we define S j,κ to be the subset of points p Γ\ PSL(2, R) satisfying the condition that there exist µ, ν > 0 such that for every point a b m j (π j (p)), either b µ or a κ b ν. Here µ and ν depend on p. Then by definition, we have S c κ,κ 2,...,κ k = S c,κ S c 2,κ 2 S c k,κ k and hence dim H S c κ,κ 2,...,κ k = max{dim H S c j,κ j j k}. 37

46 Therefore, to prove Theorem 2.., it suffices to prove dim H S c j,κ j = κ j +. In the rest of this part, we will consider S c j,κ for a fixed cusp η j. Without loss of generality, we may assume that σ j = e, η j = i and that 0 Γ. Since Γ N {e}, this implies that Γe is a discrete subset in R 2. The following lemmas concern some properties of lattice points in Γe R 2. Lemma There exists a constant C > 0 such that for any (α, β) Γe we have β C or β = 0. Proof. We know that Γe is discrete in R 2. So there is a constant C > 0 such that for any point (α, β) Γe we have (α, β) 2C where is the standard Euclidean norm. Suppose that there exists (α 0, β 0 ) Γe with 0 < β 0 < C. Then there exists an integer n Z such that α 0 + nβ 0 < β 0. 38

47 Since 0 Γ, we have n 0 α 0 β 0 = α 0 + nβ 0 β 0 Γe and (α 0 + nβ 0, β 0 ) 2C < 2C which contradicts the definition of C. This completes the proof of the lemma. Lemma There exists a constant C > 0 such that for any two distinct points (α, β ) and (α 2, β 2 ) in Γe we have α β 2 α 2 β C. Proof. Now let γ, γ 2 Γ be such that γ = α β γ 2 = α 2 β 2. Then we have γ γ 2 e = α β 2 α 2 β. Note that (α, β ) and (α 2, β 2 ) are distinct and hence α β 2 α 2 β 0. By Lemma 2.2., we conclude that α β 2 α 2 β C for some C > 0. 39

48 Remark We will fix this constant C for later use. Note that by the definition of C, for any point (α, β) Γe we have (α, β) 2C. Definition For l > 0 and 0 θ θ 2 < 2π, we define the subset of R 2 S(l, θ, θ 2 ) := {(x, y) R 2 l r 2l, θ < θ < θ 2 } where (r, θ) are the polar coordinates of (x, y). Theorem 2.2. ([EM93], [GOS0]). We have Γe S(l, θ, θ 2 ) l 2 (θ 2 θ ) as l. Lemma Fix C > 0 as in Lemma and let κ. There exists a constant C 0 > 0 with the following property: for any (α, β) Γe with 0 < α β <, there exists a large constant L (α,β) > 0 such that for any l > L (α,β) the interval [ α β κ+ ] contains at least C 0 l 2 /β κ+ many disjoint subintervals C, α + C β 8 β κ+ β 8 [ ] α β C, 8 β α β + C where ( α, β) Γe κ+ 8 β κ+ S(l, π, π). 4 2 Proof. Suppose that [ ] α C, α + C contains two subintervals β 8 β κ+ β 8 β κ+ [ α β C 8 β κ+, α β + C 8 ] β κ+ and [ γ δ C 8 δ κ+, γ δ + C 8 ] δ κ+ where ( α, β) and ( γ, δ) are two distinct points in Γe S(l, π, π ). By Lemma 2.2.2, 4 2 we have α β γ δ = α δ β γ β δ C β δ C 4l 2 = C ( ) 6 (l/ 2) + 2 (l/ C ( 2) 2 6 β + δ ) 2 2 C ( 6 β + ). κ+ δ κ+ 40

49 This implies that any two such subintervals are disjoint, and hence to prove the lemma [ ] α it suffices to prove that in the interval C, α + C there are at least β 8 β κ+ β 8 β κ+ C 0 l 2 /β κ+ many points of the form α/ β where ( α, β) Γe S(l, π, π ). We have 4 2 [ α α β β C 8 β, α κ+ β + C 8 [ ( arg( α, β) α arccot β + C 8 β κ+ β κ+ ] ) ( α, arccot β C 8 )]. β κ+ Since ( α arccot + C β 8 β κ+ ) arccot know that the number of points in S ( α ) C β 8 β, by Theorem 2.2. we κ+ β ( κ+ ) ( )) α + C α, arccot C β 8 β κ+ β 8 β κ+ ( l, arccot is asymptotically equal to l 2 /β κ+ up to a constant. Note that the implicit constant is absolute since 0 < α/β <. This completes the proof of the lemma. 2.3 Hausdorff Dimension In this section, we will give a proof of Theorem 2... We need some preparations. Definition We say that x R is Diophantine of type κ with respect to Γe if there exists a constant C > 0 such that for any (α, β) Γe with β 0 we have β κ xβ α C. We denote by S κ the subset of R of all Diophantine numbers of type κ with respect to Γe. Lemma Let p = Γ a c and only if a/c Sκ. c b d Γ\ PSL(2, R) with c 0. Then p S c j,κ if 4

50 Proof. We have m j (π j (p)) = d b c a Γe = dα bβ cα + aβ α β Γe. By the definition of S j,κ, if p S c j,κ, then there exist infinitely many (α, β) Γe such that aβ cα 0 and dα bβ κ aβ cα 0. By the discreteness of Γe, this implies that β. Note that dα bβ = cdα cbβ c = cdα (ad )β c = d(cα aβ) + β. c Therefore we have dα bβ β and a/c S c κ. Here the implicit constant in depends on p. Conversely, if a/c S c κ, then there exist infinitely many (α, β) Γe with β 0 such that β κ a c β α 0. By Lemma 2.2., this implies that aβ cα 0 and consequently β and dα bβ = d(cα aβ) + β c β. Hence we have aβ cα 0, dα bβ κ aβ cα 0 42

51 and p S c j,κ. This completes the proof of the lemma. Proof of Theorem 2... From the discussions above, we know that in order to prove Theorem 2.. it is enough to show that dim H S c j,κ = κ +. By Lemma.4. and the fact that the subset Γ a b 0 a a, b R Γ\ PSL(2, R) has dimension 2, it suffices to prove that dim H S c κ = 2 κ +. In the rest of this section we will prove this formula. Since 0 Γ, for any n Z we have S c κ (n, n + ) = n + S c κ (0, ). Therefore, we only need to compute the Hausdorff dimension of S c κ (0, ). For the upper bound, by the definition of S κ, we can construct an open cover { I (α,β) = ( α β β, α κ+ β + ) } β κ+ (α, β) Γe, α/β (0, ) Sκ c (0, ). 43

52 For δ > 0 by Theorem 2.2. we have diam(i (α,β) ) δ (α,β) Γe α/β (0,) β δ(κ+) n= (α,β) Γe S(2 n C, π 4, π 2 ) 2 2n 2 = nδ(κ+) n= n= 2 n(δ(κ+) 2). If δ > 2/(κ + ), then (α,β) Γe diam(i (α,β) ) δ converges and hence by properties of α/β (0,) Hausdorff dimension we have dim H S c κ (0, ) 2 κ +. For the lower bound, let ɛ > 0 be fixed and we construct a tree-like set in S c κ (0, ) as the intersection of closed subsets in [0, ] by induction. Let A 0 = {[0, ]} and A 0 = [0, ]. Let l be a sufficiently large number and define A = { [α β C 8 β, α κ+ɛ+ β + C ] } 8 β κ+ɛ+ (α, β) Γe S(l, π 4, π 2 ) and A = A. Suppose that we find l < l 2 < < l j and construct families A j, A j,..., A 0 and closed subsets A j A j A A 0. Now by Lemma 2.2.3, we can find a sufficiently large l j+ > 0 such that. log l j+ j 2 log(l j l j... l ). 2. For every [ α β C 8 β κ+ɛ+, α β + C 8 β κ+ɛ+ ] subintervals (since β l j ) of the form ( α, β) Γe S(l j+, π 4, π 2 ). lj+ A j, it contains at least C 2 0 l [ ] κ+ɛ+ j α β C, 8 β α β + C with κ+ɛ+ 8 β κ+ɛ+ 44

53 We denote the family of all these new subintervals by A j+ as [ α β C 8 β κ+ɛ+, α β + C 8 ] β κ+ɛ+ runs through all the intervals in A j and let A j+ = A j+. Here C and C 0 are as in Lemma Now we take A = j=0 A j and A = j=0 A j. From the construction of A j s and the definition of S κ, we know that A S c κ (0, ). Also we have j (A) l2 j+ l κ+ɛ+ j l κ+ɛ+ j+ Therefore by Theorem 2..2, we have and d j (A) l κ+ɛ+ j. dim H S c κ (0, ) dim H A lim sup j = lim sup j j i= log(l2 i+/(l i l i+ ) κ+ɛ+ ) log l κ+ɛ+ j+ (κ + ɛ + ) log l + j i=2 2(κ + ɛ) log l i + (κ + ɛ ) log l j+ (κ + ɛ + ) log l j+ = κ + ɛ κ + ɛ + = 2 κ + ɛ +. Since this is true for any ɛ > 0, we obtain that dim H S c κ (0, ) 2 κ +. This completes the proof of Theorem

54 Chapter 3 Effective Equidistribution of Abelian Horospherical Orbits 3. Introduction In this part, we will consider the effective equidistribution of horospherical orbits in homogeneous spaces. This topic has been studied well, and the present work is motivated by [S3] and [V0]. To be precise, let {a t } = {exp(tx)} t R be a one parameter R-diagonalizable subgroup in a semisimple Lie group G, Γ a lattice in G and µ the Haar measure on Γ\G. Let Ad(g) be the adjoint action of G on Lie(G) induced by the action of conjugation x gxg (x G). Let U be the horospherical subgroup of {a t }, i.e. U = {g G a t ga t e}. The decomposition of Lie(U) with respect to {a t } under the adjoint action is Lie(U) = g α g α2 g αn where α i are the roots of {a t } in U, that is, Ad(a t )X i = α i (a t )X i for any X i g αi. Without loss of generality, we can assume that each g αi is onedimensional and some of these α i s may be identical. We denote the exponential map 46

55 from Lie(G) to G by exp and we fix a norm on the Lie algebra g. For each i, fix v i g αi with norm and let B(T, T 2,..., T n ) be the parametrized box in U, i.e. B(T, T 2,..., T n ) = {exp(t v + t 2 v t n v n ) 0 t i T i ( i n)}. For any t > 0 we can find a number α > 0 such that for any t > 0 e αt = α (a t )α 2 (a t ) α n (a t ) and then t α = α (a ln t )α 2 (a ln t ) α n (a ln t ). Also we define B(t) : = B(α (a ln t ), α 2 (a ln t ),, α n (a ln t )) = a ln t B(,,..., )a ln t. We will denote by B r the open ball of radius r > 0 around e in G. Here the distance on G is induced by the norm. Also we will write BC l (Γ\G) for the set of bounded smooth functions on Γ\G with bounded Lie derivatives up to order l. Definition 3... For any x Γ\G, we define the injectivity radius at x by the largest number η > 0 with the property that the map B η xb η Γ\G by sending g B η to xg Γ\G is injective. We will denote the injectivity radius at x by η(x). 47

56 We will follow the proof of Lemma 9.5 in [V0] and prove the following theorem. Theorem 3... Suppose that U is abelian. There exist constants a, b > 0 such that for any f BC l (Γ\G), we have T α f(xu)du B(T ) Γ\G fdµ T a η f b,l for some large constant l > 0. Here η = η(a ln T x) is the injectivity radius at a ln T x and,l is the L -Sobolev norm involving Lie derivatives of orders up to l. The implicit constant depends only on Γ\G. Remark 3... We will always assume that f,l is defined and finite, and l is large enough so that all the theorems and arguments in this note would hold. Readers may refer to [KM99] for more details about the Sobolev norm and the number l. Remark Theorem 3.. is weaker than the theorem proved by Strömbergsson [S3] in the case of Γ\ PSL(2, R). But the proof would involve only mixing property of a semisimple flow and give a result for a general homogeneous space. Readers may compare T η b a and the r-factor in the main theorem of [S3]. Using the same arguments as in the proof of Theorem 3.., we can prove the following Theorem Suppose that U is abelian. Let h(u) be a compactly supported smooth function on U. Then there exist constants a, b > 0 such that for any f BC l (Γ\G) we have T α U f(xu)h(a ln T ua ln T )du Γ\G fdµ U h(u)du T a η f,l. b Here η, l and,l are the same as in Theorem 3... The implicit constant depends only on h(u) and Γ\G. 48