Unipotent Flows on Homogeneous Spaces of Sl2

Home , Nilpotent group, Unipotent

UNIPOTENT FLOWS ON

HOMOGENEOUS SPACES OF SL2(C)

A THESIS

submitted to the Faculty of Science of the University of Bombay

for the degree of Master of Science

NIMISH A. SHAH

TATA INSTITUTE OF FUNDAMENTAL RESEARCH

BOMBAY 400 005

1992 Preface

In 1987, G.A. Margulis settled a 60 years old conjecture due to A. Oppen- heim about values of quadratic forms at integeral points, using ergodic theory on homogeneous spaces. This was achieved by proving a particular case of a conjecture of Raghunathan about orbit closures for unipotent actions on homogeneous spaces. In 1988, the National Board of Higher Mathematics or- ganized an instructional conference in ergodic theory, which provided me a good background to appreciate the significance of Raghunathan’s conjecture. After the conference S.G. Dani suggested me to read the work of Margulis and give lectures on it. He said that it would be a good idea to verify Raghu- nathan’s conjecture for the group SL2(C), using the new techniques developed by him and Margulis, and write my Master’s thesis on it. While the group structure of SL2(C) is easy to understand, the dynamics of unipotent flows on its homogeneous spaces illustrate many of the interesting ergodic theoretic phenomena. The study of this case provided me a good insight into Raghunathan’s conjecture in the general case. Some of the important results used in the proof are based on the work of Garland and Raghu- nathan about fundamental domains of noncocompact lattices in semisimple groups of real rank 1. In turn, their work depends on the theorem of Kazhdan and Margulis about existence of unipotent elements in noncompact lattices in semisimple groups. The main endeavor in writing this thesis is to present the whole theory for the group SL2(C), using only the language of linear algebra and topological groups. I wish to thank Professor S.G. Dani for introducing me the subject of ergodic theory on homogeneous spaces. Each discussion with him gave me new insights into the problems I was studying. I also thank him for suggesting several corrections in the thesis. Many of the papers about flows on homogeneous spaces use the machinery of algebraic groups and discrete subgroups. I was always helped by Professor Gopal Prasad in understanding the results used in these papers and learning the proofs. I express my thanks to him. I would like to thank Professor M.S. Raghunathan for always encouraging me, and telling me about many interesting results during several conversations in- cluding many at tea tables. His book on discrete subgroups of Lie groups was my most helpful source.

Nimish A. Shah

i Contents

Introduction 1

1 Topological groups and homogeneous spaces 8 1.1 Deﬁnitions and immediate consequences ...... 8 1.2 Invariant measures on topological groups ...... 13 1.3 Invariant measures on homogeneous spaces ...... 17 1.4 Closed subgroups of Rn and Rn/Zn ...... 22 1.5 Covolumes of discrete subgroups in euclidean spaces ...... 26

2 The group SL2(C) and its homogeneous spaces 30 2.1 Exponential map, Lie algebras, and Unipotent subgroups . . . 30 2.2 Subgroups of SL2(C)...... 37 2.3 The invariant measure on SL2(C)/SL2(Z[i]) ...... 44 2.4 Ergodic properties of transformations on homogeneous spaces 47

3 L-homogeneous spaces of SL2(C) 55 3.1 A result of Kazhdan and Margulis ...... 55 3.2 Non-divergence of unipotent trajectories ...... 62 3.3 Volumes of compact orbits of horospherical subgroups . . . . . 65 3.4 A lower bound for the relative measures of a large compact set on unipotent trajectories ...... 70 3.5 Description of noncompact L-homogeneous spaces of SL2(C). 76

4 Proof of Raghunathan’s conjecture for SL2(C) 81 4.1 Minimal invariant sets ...... 81 4.2 Topological limits and inclusion of orbits-I ...... 84 4.3 Specialization to the case of SL2(C)...... 87 4.4 Topological limits and inclusion of orbits-II ...... 91 4.5 Closures of orbits of N1 ...... 94 4.6 Closures of orbits of subgroups containing a nontrivial unipotent element ...... 101

Bibliography 110

ii Introduction

A subgroup action on a homogeneous space is a classical object of study in ergodic theory. In this thesis we shall study in detail the homogeneous spaces of G = SL2(C) admitting finite G-invariant measures, and verify Raghunathan’s conjecture about closures of orbits of unipotent subgroups of G acting on these spaces. A subgroup Γof a locally compact group G is called a lattice in G, if Γis discrete and the homogeneous space G/Γadmits a finite G-invariant measure. A linear transformation T of a finite dimensional vector space V is called n unipotent, if (T 1V ) = 0 for some n N. Let G be a Lie− group and let denote∈ its Lie algebra. An element u G is called an Ad-unipotent elementG of G, if the linear automorphism Ad u of∈ is unipotent. A subgroup U G is called a unipotent subgroup of G, if all ofG elements of U are unipotent⊂ elements of G.

Raghunathan’s conjecture : Let G be a Lie group, Γ be a lattice in G, and U be a unipotent subgroup of G. Then for any x G/Γ, there exists a closed subgroup L of G containing U such that ∈ Ux = Lx. Moreover, the orbit Lx admits a ﬁnite L-invariant measure.

For an abelian group G, the conjecture follows from elementary arguments. For a nilpotent group G, the conjecture follows from a result of Green [G] about unitary representations of nilpotent groups. For a solvable group G, a result due to Mostow [R, Sect.3.3] says that if N is the maximal connected normal unipotent subgroup of G, then every N-orbit in G/Γis compact. In this case every connected unipotent subgroup of G is contained in N, and hence the veriﬁcation of the conjecture reduces to the nilpotent case. In the case of semisimple groups the conjecture is of particular signiﬁcance. For G = SL2(R), it was proved by Hedlund [He] that if U is a nontrivial one- parameter unipotent subgroup of G then any orbit of U in G/Γis either dense or periodic; the periodic orbits exist if and only if G/Γis noncompact. In fact, for a cocompact lattice Γ, Furstenberg [F] proved that any U-invariant probability measure on G/Γis G-invariant, and hence it is unique.

1 Let G be a Lie group. For g G, deﬁne ∈ n n U(g)= u G : g ug− as n . { ∈ →∞ → ∞} Then U(g) is called the horospherical subgroup of G associated to g. Note that U(g) is a closed connected unipotent subgroup of G. Now let G be a semisimple group (with trivial center and no compact factors) and Γbe a cocompact lattice in G. It was proved by Veech [V] and Bowen [B] that every orbit of a nontrivial horospherical subgroup of G is dense in G/Γ. In the noncompact case it was proved by Margulis [M1] that if G = SLn(R), Γ=SLn(Z), and u(t) t R is a one-parameter unipotent subgroup of G, then ∈ for any x G/Γ,{ there} exists a compact set C G/Γsuch that the set (C) of return times,∈ deﬁned as ⊂ R

(C)= n N : u(n)x C , R { ∈ ∈ } is unbounded. In [D1, D3, D4] this result was strengthened by Dani, who proved, in particular, that for a suitable compact set C, the set (C) of return times has positive density in the set of positive integers. UsingR the strenghened verson, in [D6] he veriﬁed Raghunathan’s conjecture when Γis any lattice in a Lie group G and U is a horospherical subgroup of G. This generalizes Veech’s result for noncocompact lattices. A major source of interest in Raghunathan’s conjecture is the fact, observed by Raghunathan, that a very special case of the conjecture implies the validity of the following conjectue of A. Oppenheim.

Oppenheim’s conjecture : Let Q be a nondegenerate, indeﬁnite quadratic form on Rn, n 3. Suppose that Q is not multiple of a rational form. Then Q(Zn) is dense≥ in R.

The conjecture can be reduced to the case of n =3.LetQ be the quadratic form 2x x x2, and deﬁne 1 3 − 2 H = SO(Q)= g SL (R):Q(gx)=Q(x) for all x R3 . { ∈ 3 ∈ } In [M2], Margulis proved that every relatively compact orbit of H in the space SL3(R)/SL3(Z) is compact. From this result it is not diﬃcult to deduce Op- penheim’s conjecture using Mahler’s criterion and the standard techniques of algebraic groups. For an elementry proof of Oppenheim conjecture see [DM3, M5]. Note that H0 is not a unipotent subgroup, but it is generated by unipotent elements of G. The following conjecture was formulated by Margulis [M3].

2 Generalized Raghunathan conjecture: Let G be a Lie group and Γ be a lattice in G. Let U be a subgroup of G generated by unipotent elements of G contained in U. Then for any x G, there exists a closed subgroup L of G containing U such that Ux = Lx,∈ and the orbit Lx admits a ﬁnite L-invariant measure.

In his proof Margulis has developed new techniques to study orbit closures of unipotent subgroups. Generalizing these techinques and the results about returning of a unipotent trajectory to a compact set with positive density, in [DM1, DM2] Dani and Margulis proved that (i) any H-orbit in SL3(R)/Γis either dense or closed, and (ii) Raghunathan’s conjecture holds for the orbits of unipotent subgroups of H acting on SL3(R)/Γ, where Γis a lattice in SL3(R). In this thesis we shall verify the generalized conjecture for the Lie group G = SL2(C), using the methods of [M2], [DM1] and [DM2]. In this case it is possible to give a proof illustrating the ideas without having to go through the kind of technical work involved in the case of SL3(R). It would be appropriate to mention here that recently Ratner [Ra2] has settled Raghunathan’s conjecture by proving the following much stronger result.

Theorem. (Ratner) Let G be a Lie group, Γ be a lattice in G, and U = u(t):t R be a one-parameter unipotent subgroup of G. For a given x X, let{ L be the∈ smallest} closed subgroup of G containing U such that the orbit∈ Lx is closed, and it admits an L-invariant probability measure, say ν. Then the trajectory u(t)x : t>0 is uniformly distributed with respect to ν; that is, for every bounded{ continuous} function φ on G/Γ,

1 T lim φ(u(t)x) dt = φdν. T T →∞ 0 Lx It is hoped that this thesis will motivate the reader to study the recent work done in this area. For various related developements, the reader is referred to the survey articles by Dani [D5] and Margulis [M3, M6].

Chapterwise content of the thesis In Chapter 1, we give deﬁnitions and basic properties of topological groups and their homogeneous spaces. We give Fubini’s formula relating the invariant measures on a unimodular group G, a unimodular subgroup F , and the homogeneous space G/F . We begin Chapter 2 by introducing very basic concepts about linear Lie groups, their Lie algebras, and the exponential maps. We need to introduce some notations at this stage to decribe the furter contents.

3 Notation.LetG = SL2(C), Γbe any lattice in G, and µ be the G-invariant probability measure on the homogeneous space X = G/Γ. Let K = g G : gg¯t =1 , { ∈ } def α 0 D = d(α) = 1 : α C∗ , 0 α− ∈ M = d(eti):t R = D K, { ∈ } ∩ A = d(a):a>0 , { } A = d(a):a>0,a2 <η , where η>0, η { } def 1 z N = u(z) = : z C , 01 ∈ H = SL (R) and N = N H. 2 1 ∩ We study properties of these subgroups, and obtain Iwasawa and Bruhat decompositions of G. In the last section of this chapter, we study ergodic properties of the subgroup actions on X. Using Mautner phenomenon we prove the following result. Theorem 1 An element g G which is not contained in a compact subgroup ∈ of G, acts ergodically on (X, µ). In particular, the subgroups A, N1, N, and H act ergodically on (X, µ). As a consequence of the ergodicity of the N-action, we obtain the following important result.

Theorem 2 (Cf. [DM2, Sect. 1.7]) Let an C∗ be a sequence such that a as n . Then for every x X{, the}⊂ set | n|→∞ →∞ ∈

d(an)Nx n N ∈ is dense in X. In particular, every AN-orbit in X is dense. (Cf. [DR, Sect. 1.5]) In Chapter 3, we study the structural properties of a noncompact homogeneous space X of G admitting a ﬁnite G-invariant measure. We prove the following result about the returning of unipotent trajectories to compact sets with densities arbitrarily close to 1. Theorem 3 (cf.[D3]) Given a compact set C in X and >0, there exists a (larger) compact set C in X such that for any x C and any one-parameter ∈ unipotent subgroup v(t) t R of G, { } ∈ 1 ( t [0,T] v(t)x C ) > 1 T { ∈ | ∈ } − for all T>0, where denotes the Lebesgue measure on R.

4 We also obtain the following description of X.

Theorem 4 (cf. [GR]) Let X be a noncompact homogeneous space of G admitting a ﬁnite G-invariant measure. Then there exists a ﬁnite collection consisting of compact N-orbits in X such that the following holds. C

1. The set X(N) := MA O O∈C is the union of all compact N-orbits in X. (Note that NG(N)=MAN.)

2. There exists η0 > 0 such that

X = KA . η0 O O∈C (Note that G = KAN.)

3. There exists 0 <η <η such that for , , a, a A , and 1 0 O O ∈C ∈ η1 k, k K, if ∈ ka ka = O∩ O ∅ then = , a = a, kM = kM, and O O

ka = ka . O O For a subgroup F G and a closed F -invariant subset Z X, if every F -orbit in Z is dense in⊂Z, then the set Z is called F -minimal. ⊂ The following corollaries are consequences of Theorem 3.

Corollary 1 Let F be a subgroup of G containing a nontrivial unipotent element. Then any closed F -invariant subset of X contains a closed F -minimal subset.

Corollary 2 The set X(N) does not contain a closed AN1-invariant subset.

The Chapter 4 is devoted to proving the following result.

Theorem 5 (Main theorem) Let Y be the closure of an orbit of N1. Then one of the following holds.

(1) Y is a compact orbit of N1 or N.

1 (2) Y is a closed orbit of vHv− for some v N. ∈ (3) Y = X.

5 The Possibility (1) occurs only when X is noncompact.

The basic strategy of the proof is to ﬁnd, under certain conditions, orbits of larger subgroups in closed N1-invariant subset of X. The following result about the unipotent action on a linear space plays a crucial role in this regard.

Lemma 1 Let U be a one-parameter group of unipotent linear transformation on Rn. Put L = v Rn Uv = v . Suppose Y Rn L is such that Y L contains a point{ ∈p. Then| there exists} a nonconstant⊂ polynomial\ function ψ ∩: R L such that ψ(0) = p, and → ψ(R) UY. ⊂ The next result follows easily from our method.

Theorem 6 Any N-orbit in X is either compact or dense. The compact N- orbits exist if and only if X is noncompact.

The following result is the main step in our proof. In fact, its proof requires basically all the machinary developed before.

Lemma 2 Let Y be a closed N1-invariant subset of X. Suppose that Y contains a closed AN -minimal subset, say Z. Let z Z X(N). Deﬁne 1 0 ∈ \ M = g G NH : gz Y . { ∈ \ 0 ∈ } Now suppose if e M, then Y = X. ∈ We also prove the following result.

Theorem 7 (Cf. [M5, Thm. 4]) Any proper closed AN1-invariant subset of X is a ﬁnite union of closed H-orbits. In particular, any H-orbit in X is either closed or dense.

In the end we give a description of all closed N1-invariant subsets of X and prove the following result, which veriﬁes the generalized Raghunathan conjecture.

Theorem 8 Let X be a homogeneous space of G admitting a ﬁnite G-invariant measure. Let F be a closed subgroup of G containing a nontrivial unipotent element of G. Then for every x X, there exists a closed subgroup L of G such that the following holds: ∈

(a) The orbit Lx is closed and it admits a ﬁnite L-invariant measure;

6 (b) the subgroup L F has ﬁnite index in F ; and ∩ (c) (the main property) (F L)x = Lx. ∩ In fact, the group L can be chosen to contain F except in the following case: there exist g G, θ π/3,π/4,π/6 , and w C 0 such that ∈ ∈{ } ∈ \{ } kθi 2θi 1 F = g d(e ) k Z u(z):z Z[e ] w g− , { } ∈ ·{ ∈ · } 1 1 the orbit (gNg− )x is compact, (F gNg )x = Lx, and dim(L)=1. ∩ −

7 Chapter 1

Topological groups and homogeneous spaces

1.1 Deﬁnitions and immediate consequences

A group G which is also a topological space is called a topological group if the 1 map m : G G G, which is given by m(x, y)=x− y for all x, y G, is continuous. × → ∈ For any g G, let Lg : G G denote a left translation on G and Rg : G ∈ → →1 G denote a right translation on G, deﬁned by Lg(x)=gx and Rg(x)=xg− for all x G. The continuity of the map m is equivalent to the following conditions:∈

(a) L and R are homeomorphisms of G for every g G. g g ∈ (b) Given a neighbourhood Ωof the identity e in G, there exists a neigh- 1 bourhood Ω of e such that Ω− Ω Ω. 1 1 1 ⊂ To avoid pathological situations, we shall always work with the topological groups which are locally compact, hausdorﬀ, and second countable as topological spaces. Some examples of topological groups.

1. A countable group with discrete topology is a topological group, called a discrete group.

2. A ﬁnite dimensional real or complex vector space as an additive group is a topological group.

3. The multiplicative group of n n invertible real or complex matrices is × a topological group, denoted by GLn(k), where k = R or C. Note that the space of all n n matrices is naturally homeomorphic to kn2 and × GLn(k) is an open subset (see Sect. 2.1 for details).

8 4. Any closed subgroup of a topological group with the relative topology is a topological group. The following are some closed subgroups of GLn(k):

SLn(k)= A GLn(k) : det A =1 { ∈ } t U(n)= A GL (C):AA∗ =1 (where A∗ = A¯ ) { ∈ n } O(n)=GL(R) U(n) n ∩ SO(n) = SL (R) U(n) n ∩ 5. If G and H are topological groups then the direct product G H is a topological group with respect to the product topology. ×

The multiplicative group of elements of unit norm in C∗ is a topological group, called the circle group and denoted by T1.Forn>1, deﬁne Tn n 1 n 1 n inductively as T = T T − . The topological group T is called the n-torus. ×

6. Let G and V be topological groups. Suppose that there exists a continuous map φ : G V V with the following property: For every × → g G, the map φg : V V deﬁned by φg(v)=φ(g, v)( v V ) is an automorphism.∈ Then we→ can give a group structure to the∀ the∈ product space G V as follows: Identify G with G e , V with e V .For any g G× and v V , deﬁne ×{ } { }× ∈ ∈ 1 gvg− = φ(g, v).

Then for all g ,g G and v ,v V , we obtain 1 2 ∈ 1 2 ∈ 1 (g1v1)(g2v2)=(g1g2)((g2− v1g2)v2).

This defines a topological group called the semidirect product of G and V , and it is denoted by G φ V . Note that V is a normal subgroup of G φ V . n Let G =GLn(k) and V = k ,wherek = R or k = C, and define φ(g, v)=gv for all (g, v) G V . Then Gφ V is the group of all affine ∈ × transformations of kn. Its subgroup O(n) Rn is the group of all rigid motions of Rn.

The following properties of topological groups are direct consequences of the deﬁnition. (See [MZ] as a general reference on topological groups.) Let G be a (locally compact) topological group.

1. Any open subgroup of G is closed.

2. Any locally compact subgroup of G is closed.

9 3. Let Ωbe a symmetric neighbourhood of the identity in G; that is, Ω= 1 n Ω− . Then the set n NΩ is an open subgroup of G. In particular, if ∈ G is connected then∪ n G = n NΩ . ∪ ∈ (For subsets A and B of G, we define AB = ab G : a A, b B 1 1 { ∈ ∈ ∈ } and A− = a− G : a A . Note that for any neighbourhood Ωof e, {1 ∈ ∈ } the set ΩΩ− is a symmetric neighbourhood of e.) 4. Let N be a discrete normal subgroup of G.IfG is connected then N is contained in the center of G. (To see this, fix n N and consider the 1 ∈ map G g gng− N.) → ∈ 5. Let H be a topological group and let φ : G H be a group homomorphism. Then φ is continuous if and only→ if φ is continuous at the identity. If a topological group G acts continuously on a locally compact, hausdorff, second countable topological space X, then X is called a G-space. More precisely, there exits a map ρ : G X X such that the following conditions are satisfied: × → 1. ρ is a group action; that is, for any g ,g G, and x X, we thave that 1 2 ∈ ∈ ρ(g1,ρ(g2,x)) = ρ(g1g2,x).

When the action ρ is understood, we denote ρ(g, x) by simply g x or gx. · 2. ρ is a continuous map. Some examples of G-spaces: n 1. Let G =GLn(C). Then C is a G-space with respect to the standard action of G via linear transformations. 2. Let X be a G-space and F be a closed subgroup of G. If a locally compact subset Y X is invariant under the action of F , then Y becomes an F -space with⊂ respect to the relative topology. Thus for every x X, the set Y = Fx is an F -space, where Fx = gx : g G , and the∈ bar denotes the closure. { ∈ } 3. Let F be a closed subgroup of G. Consider the left action of G on the quotient space G/F .Letq : G G/F be the natural quotient map. Equip the space G/F with the ﬁnest→ topology with respect to which q is continuous. For this topology a subset S G/F is open if and only 1 ⊂ if q− (S) is open in G. With this topology, the quotient space G/F becomes a G-space. Note that G/F is hausdorﬀbecause F is closed.

10 Definition.AG-space X is called a homogeneous space of G if the following holds: G acts transitively on X and for every x X, the map ρx : G X, defined by ρ (g)=gx for all g G, is open. ∈ → x ∈ As a consequence of Baire’s category theorem, the condition in this definition can be weakned. Recall that the category theorem says that any locally compact, hausdorff, second countable topological space is of second category; that is, the space cannot be expressed as a countable union of closed subsets with empty interiors.

Lemma 1.1.1 If G acts transitively on a (locally compact, hausdorﬀ, second countable) G-space X then X is a homogeneous space of G.

Proof. It is enough to show that given x X and a neighbourhood Ωof e in G, ∈ the set Ωx is a neighbourhood of x in X.LetΩ1 be a compact neighbourhood 1 of e such that Ω− Ω Ω. As G is second countable, there exists a sequence 1 1 ⊂ gi G such that G = i NgiΩ1. By transitivity of the G-action, { }⊂ ∪ ∈

X = giΩ1x. i N ∈

As Ω1 is compact, so is Ω1x. Since X is of second category, the interior of g Ω x is non-empty for some i N. Since the elements of G act on X i 1 ∈ as homeomorphisms, the interior of Ω1x is nonempty. Therefore there exists 1 ω Ω1 such that Ω1x contains a neighbourhood of ωx. But then ω− Ω1x Ωx ∈ 1 ⊂ and ω− Ω1x contains a neighbourhood of x. This completes the proof.

Notation.LetX be a homogeneous space of G, F be a closed subgroup of G, and x X.Let ∈ Fx = gx X : g F { ∈ ∈ } denote the orbit of F through the point x. Deﬁne

F = g F : gx = x . x { ∈ } Then F is closed subgroup of G.Letπ : F X be the map deﬁned by x → π(g)=gx, for all g F .Letq : F F/Fx be the natural quotient map. Then there exists a map∈ π ¯ : F/F →X such that π =¯π q. Clearly,π ¯ is x → ◦ an injective map. Now π is continuous and the topology of F/Fx is the ﬁnest topology such that q is continuous. Thereforeπ ¯ is continuous map.

Corollary 1.1.2 1. The map π¯ is a homeomorphism onto Fx if and only if the orbit Fx is locally compact with respect to the relative topology.

11 2. π¯ is a proper map if and only if Fx is a closed orbit. (A map is called proper if the inverse image of every compact set is compact.) In particular, if Fx is closed then F/Fx ∼= Fx and the F -actions on both the spaces are preserved under the isomorphism.

Proof. First suppose that Fx is locally compact. Since F acts transitively on Fx, by Lemma 1.1.1, Fx is a homogeneous space of F . Therefore π maps open subsets of F onto open subset of Fx. Thereforeπ ¯ maps open subsets of F/Fx onto open subsets of Fx. Thusπ ¯ is a homeomorphism onto Fx. We know that F/Fx is locally compact, therefore ifπ ¯ is a homeomorphism onto Fxthen Fxis locally compact. This completes the proof of the ﬁrst part. Next suppose that Fx is closed. Then the intersection of any compact subset of X with Fx is compact. In particular, Fx is locally compact. Now 1 1 for a compact set C X,¯π− (C)=¯π− (C Fx). Sinceπ ¯ is a homeomorphism 1⊂ ∩ onto Fx, the setπ ¯− (C Fx) is compact. This shows thatπ ¯ is a proper map. ∩ The converse follows from that fact that a proper map is a closed map. The following observations about homogeneous spaces can be veriﬁed by using the standard arguments of general topology. Let X be a homogeneous space of G. 1. Given x X and a compact set C X, there exists a compact set K G such∈ that C Kx. ⊂ ⊂ ⊂ 2. Let C be a compact subset of X. Then given any open subset Ψof X containing C, there exits a neighbourhood Ωof e in G such that ΩC Ψ. ⊂ 3. For compact sets K G and C X, the set KC is compact. ⊂ ⊂ 4. For a compact subset K of G and a closed subset Y of X, the set KY is closed in X.

5. Let F be a close subgroup of G and x X. Suppose if the orbit Fx is closed then the orbit F 0x is a component∈ of Fx. In particular, F 0x is closed. (Here F 0 denots the component of the identity in F .)

6. Every function φ Cc(X) is uniformly continuous in the following sense: Given >0 there∈ exists a neighbourhood Ωof e such that for all x X and all g Ω, we have that ∈ ∈ φ(gx) φ(x) <. | − |

(Here Cc(X) denotes the space of all complex valued continuous functions on X with compact support; that is, thoes vanishing outside a compact subset.)

12 The following applications of Corollary 1.1.2 are recorded here for future reference.

Lemma 1.1.3 Let X be a homogeneous space of G such that Gx is a discrete subgroup of G for every x X. Let F and H be closed subgroups of G. Suppose that for a point x ∈X, the orbits Fx and Hx are closed. Then the orbit (F H)x is an open∈ and a closed subset of Fx Hx. In particular, if Fx = Hx∩then F H is an open subgroup of F as well∩ as H. ∩

Proof. Put Z = Fx Hx.Letz Z. Then the orbits Fz = Fx and Hz = ∩ ∈ Hx are closed. Therefore by Corollary 1.1.2, F/Fz ∼= Fz and H/Hz ∼= Hz. Therefore there exists a neighbourhood Ωof the identity e in G such that 1 ΩΩ− Gz = e ,(Fz Ωz)=(F Ω)z, and (Hz Ωz)=(H Ω)z. This implies∩ that (Fz{ } Hz ∩ Ωz)=(F ∩ H Ω)z. Hence∩ (F H)z∩ is open in Fz Hz = Z for every∩ z∩ Z. Now (F∩ H∩)x is closed, because∩ its complement in Z∩ is the union of open∈ (F H)-orbits∩ in Z and Z is closed. ∩ The next observation is another appliction of Corollary 1.1.2. This result is not used later and the proof is left as an exercise.

Lemma 1.1.4 Let F and H be closed subgroups of G such that the set FH is closed. Then the map φ : G/(F H) G/F G/H, given by φ(g(F H)) = (gF, gH) for all g G, is proper.∩ → × ∩ ∈

1.2 Invariant measures on topological groups

Let X and Y be topological spaces and T : X Y be a continuous map. Given a borel measure µ on X, we define its projection→ T µ to be a measure ∗ on Y such that for every borel set E Y , we have that ⊂ 1 T µ(E)=µ(T − E). ∗ If X = Y and T µ = µ, we say that T is a measure preserving transforma- ∗ tion on the mesure space (X, µ). A borel measure on a topological space is called locally finite if it is finite on compact subsets. It is a fact that any locally finite measure on a locally compact, hausdorff, second countable topological space is regular (see [Ru2, Sect. 2.8]). To avoid pathological situations, we always work with the measures which are locally finite and regular. Note that if µ is locally finite and T is a proper map then T µ is also locally finite. ∗ Some examples of measure preserving transformations:

13 1. Let m be the lebesgue measure on Rn. Then the left and the right translations on Rn preserve the measure m. For any invertible linear transformation T on Rn,

1 T m = det T − m ∗ | | (see [Ru2, Sect. 8.28]). Thus if T SL (R) then T preserves m. ∈ n 2. Let µ be the standard ‘arc length measure’ on the circle group T1. Then the rotation on T1 by any angle preserves µ.Foreveryn N, the homomorphism z zn of T1 also preserves µ. ∈ → Definition.LetG be a topological group. A (nontrivial) locally finite measure µ on G is called a left haar measure if it is preserved by all the left translations on G. That is, for every g G, µ(gE)=µ(E) for all borel measurable subsets E G. Equivalently, for all∈ φ C (G) and all g G, ⊂ ∈ c ∈ φ(gx) dµ(x)= φ(x) dµ(x). G G Similarly one defines a right haar measure on G. Note that if T is the 1 transformation on G defined by T (g)=g− for all g G, and if µ is a left (or right) haar measure on G then T µ is a right (resp. left)∈ haar measure on G. ∗ Theorem 1.2.1 Every (locally compact, hausdorff, second countable) topological group G admits a left haar measure. Moreover, any two left haar measures on G are constant multiples of each other.

Proof. See [H, Chap.9] for a proof of the existence of a left haar measure on G. For the applications we shall explicitly construct left haar measures on the groups we shall work with. For ‘uniqueness’ of the left haar measure, we give a proof which is based on the proof in [H, Chap. 9]. To prove the ‘uniqueness’ assertion, let µ and ν be any two left haar measures on G. Then the product measure ν µ on the group G G is also a left haar measure. × × Let S and T be continuous transformations on G G deﬁned as follows: S(x, y)=(x, xy) and T (x, y)=(yx, y) for all x, y G.× First we verify that the transformations S and T preserve the measure µ ∈ν.Foreveryφ C (G G), × ∈ c × φ Sd(µ ν)= φ(x, xy) dν(y) dµ(x) G G ◦ × G G × = φ(x, y) dν(y) dµ(x) G G = φd(µ ν), G G × ×

14 where the ﬁrst and the third equalities hold due to Fubini’s theorem, and the second equality holds because ν is a left haar measure on G. This shows that the measure µ ν is preserved by S. Similarly by changing the order of integration and using× the left invariance of µ one proves that T preserves µ ν. ×Let Ωbe an open relatively compact subset of G. Now the measure µ ν 1 1 1 × is preserved by S− T . Note that S− T (x, y)=(yx, x− ) for all x, y G. For any nonnegative◦ measurable function◦ψ on (G,ν), using Fubini’s theorem,∈ we obtain the following chain of equalities:

µ(Ω) ψdν = χΩ(x)ψ(y) dµ(x)dν(y) G G G 1 = χΩ(yx)ψ(x− ) dµ ν(x, y) G G × × 1 1 = ν(Ωx− )ψ(x− ) dµ(x). (1.1) G 1 By Fubini’s theorem, x ν(Ωx− ) is a measurable function on (G, µ) (see [Ru2, Sect.7.8]). Since µ and→ν are arbitrary left haar measures on G, by putting µ = ν, we obtain that y ν(Ωy) is a measurable function on (G,ν). Since Ωy is an open and a relatively→ compact subset of G,0<ν(Ωy) < for every y G. For any positive borel measurable function φ on G, deﬁne∞ ∈ 1 ψ(y)=φ(y− )/ν(Ωy) for all y G. Then ψ is a measurable function on (G,ν). Now by Eq. 1.1, ∈ 1 ψdν = φdµ. (1.2) µ(Ω) G G By putting µ = ν in Eq. 1.2, we get 1 ψdν = φdν. (1.3) ν(Ω) G G Thus by Eq. 1.2 and Eq. 1.3, for every positive borel measurable function φ on G, 1 1 φdµ= φdν. µ(Ω) ν(Ω) G G Now for any borel measurable subset F of G, by putting φ = χF in this equality, we get µ(F ) ν(F ) = . µ(Ω) ν(Ω) This shows that µ and ν are multiples of each other.

15 Let µ be a left haar measure on G. Since a left translation and a right translation on G commute with each other, (Rg) µ is a left haar measure on ∗ G for every g G. Therefore by the ‘uniqueness’ of left haar measures there ∈ exists a positive constant, say ∆(g), such that (Rg) µ =∆(g)µ. Again due to ∗ the ‘uniqueness’, the function ∆(g) does not depend on the choice of µ. Thus for any measurable subset E G, ⊂

(Rg) µ(E)=µ(Eg)=∆(g)µ(E). ∗ Equivalently, for every φ C (G), ∈ c φ R dµ =∆(g) φdµ. ◦ g G G

Lemma 1.2.2 The map ∆:G R∗ is a continuous homomorphism. →

Proof. Since Rg1 Rg2 = Rg1g2 for all g1,g2 G, the map ∆is a group homomorphism. ◦ ∈ To prove the continuity, let >0 be given. Let C be a compact subset of G such that µ(C) > 0. By the regularity of µ, there exists an open subset U of G containing C such that µ(U C) µ(C). Let Ωbe a neighbourhood 1 \ ≤ · of e in G such that C(Ω Ω− ) U. Then for every g Ω, ∪ ⊂ ∈ ∆(g)=µ(Cg)/µ(C) µ(U)/µ(C) 1+ ≤ ≤ and 1 ∆(g)=µ(C)/µ(Cg− ) µ(C)/µ(U) 1/(1 + ). ≥ ≥ This shows that ∆is continuous at the identity. Now the continuity of ∆ follows from a general fact that, a homomorphism between topological groups is continuous if it is continuous at the identity. The function ∆: G R∗ is called the modular character of G. → By putting ν = µ in Eq. 1.2 one obtains the following formula.

Lemma 1.2.3 For every φ C (G), ∈ c

1 1 φ(g) dµ(g)= φ(g− )∆(g− ) dµ(g). G G

If∆is a trivial then G is called a unimodular group. Thus G is unimodular if and only if G admits a measure which is invariant under all the left and the right translations. Some examples:

16 1. Any discrete group Γis unimodular. Because the ‘counting measure’ on Γis invariant under all the left and the right translations.

2. Any abelian topological group is unimodular.

3. Any compact group is unimodular.

4. If G = [G, G] then ∆(G)=∆([G, G]) = 1, and hence G is unimodular, 1 1 where [G, G] is the subgroup generated by xyx− y− : x, y G . The { ∈ } group G = SLn(k) has this property, where k = R or k = C.

5. The following is a typical example of a nonunimodular group: Let

ab B = ρ(a, b)= 1 SL2(R):a R∗,b R . { 0 a− ∈ ∈ ∈ } Let m be the labesgue measure on R2, restricted to (R 0 ) R, and µ = ρ (m) be its projection on B. Using Fubini’s theorem\{ one} can× verify ∗ the following:

(a) µ is a right haar measure on B.

(b) For any measurable set E B, and elements a R∗ and b R, ⊂ ∈ ∈ µ(ρ(a, b)E)= a 2µ(E). | | (c) Let G be a compact group and let µ be a left haar measure on G. Let T : G G be a continuous and a surjective homomorphism. Then T µ is→ a left haar measure. Now since T µ(G)=µ(G) < , ∗ ∗ we have T µ = µ. ∞ ∗ Let µ to be the ‘arc length’ measure on T1 and for any n N, let µn be the product measure on Tn. Observe that any invertible∈ n n n n n × matrix A with integral entries acts on T ∼= R /Z (see Sect. 1.4) as follows: for every x Rn, A(x + Zn)=Ax + Zn. This action of A is a surjective homomorphism∈ of Tn. Therefore A µ = µ. ∗

1.3 Invariant measures on homogeneous spaces

Let G be a topological group and X be a homogeneous space of G.Let x0 X and π : G X be the map deﬁned by π(g)=gx0 for all g G. Put F =∈ G . Then F→is a closed subgroup of G. Deﬁne the mapπ ¯ : G/F∈ X x0 → asπ ¯(gF)=gx0 for all g G. Thenπ ¯ is a homeomorphism preserving the G-actions. ∈

17 1 Fix a left haar measure λ0 on F = π− (x0). Then for every x X there ex- 1 ∈ ists a unique measure λx supported on the ﬁber π− (x) such that the following conditions are satisﬁed:

λx0 = λ0 and λgx =(Lg) λx for every g G. (1.4) ∗ ∈

Deﬁne a linear map I : Cc(G) Cc(X) as follows: For every φ Cc(G) and every x X, → ∈ ∈ I(φ)(x)= φdλx. (1.5) 1 π− (x) Note that supp I(φ) π(supp φ) = (supp φ)x0. Also since φ is uniformly continuous on G, it follows⊂ that the function I(φ) is uniformly continuous on X. Thus I(φ) C (X). Due to Eqs. 1.5 and 1.4, for every g G, ∈ c ∈

I(φ)(gx0)= φ(gh) dλ0(h) F and I(φ L )=I(φ) L . ◦ g ◦ g Lemma 1.3.1 ([R, Sect. 1.1]) I is surjective. In fact,

+ + I(Cc (G)) = Cc (X),

+ where Cc (X) denotes the set of all nonnegative functions in Cc(X).

Proof. Given ψ C+(X), put ψ = ψ π C+(G). Then (supp ψ )x = ∈ c 1 ◦ ∈ 1 0 supp ψ.LetK be a compact subset of G such that supp ψ Kx0.Let + ⊂ θ1 Cc (G) be such that θ1 1onK. Now for every x supp ψ, we have 1∈ ≡ ∈ π− (x) K = , and hence I(θ )(x) > 0. Deﬁne a function θ on G as follows: ∩ ∅ 1 ψ (g)θ (g)/I(θ )(gx ) if g supp ψ θ(g)= 1 1 1 0 1 0 otherwise.∈ Then θ is continuous and supp θ supp θ . Thus θ C+(G). It is straight ⊂ 1 ∈ c forward to verify that I(θ)=ψ.

Lemma 1.3.2 Suppose that the groups G and F are unimodular. Let µ be a haar measure on G and let Λ:C (G) R be the positive linear functional c → Λ(φ)= φdµ. G Then ker I ker Λ. ⊂ 18 + Proof.Letφ ker I; i.e. I(φ) = 0. By Lemma 1.3.1, there exists θ Cc (G) such that I(θ)∈ 1onπ(supp φ). Now ∈ ≡

0= I(φ)(gx0)θ(g) dµ(g) G

= φ(gh) dλ0(h) θ(g) dµ(g) G F

= φ(gh)θ(g) dµ(g) dλ0(h) F G 1 = φ(g)θ(gh− ) dµ(g) dλ0(h) F G 1 = φ(g) θ(gh− ) dλ0(h) dµ(g) G F

= φ(g)I(θ)(gx0) dµ(g) G = φ(g) dµ(g)=Λ(φ), G where the fourth equality holds because µ is right invariant, the sixth equality holds because F is unimodular (see Lemma 1.2.3), and the last equality holds because I(θ) 1 on (supp φ)x . Thus we have shown that ker I ker Λ. ≡ 0 ⊂ Theorem 1.3.3 (Cf. [R, Sect. 1.4]) Let the notations be as above. Further suppose that the groups G and F are unimodular. Then there exists a unique locally ﬁnite G-invariant measure ν on X such that for all φ C (G), ∈ c

φdµ= I(φ) dν. (1.6) G X Moreover, any locally ﬁnite G-invariant measure on X is a constant multiple of ν.

Proof.LetΛ:Cc(G) R be the positive linear functional deﬁned by Λ(φ)= → + + G φdµ for all φ Cc(G). Since I : Cc (G) Cc (X) is surjective and ker I ker Λ, there∈ exists a unique positive linear→ functional Ψ: C (X) R c such that⊂ → Λ=Ψ I. (1.7) ◦ By Riesz representation theorem ([Ru2, Sect. 2.14]), there exists a unique locally ﬁnite measure ν on X such that for all φ C (X), ∈ c

Ψ(φ)= φdν. X 19 By Eqs. 1.4 and 1.7, for every φ C (G) and every g G, ∈ c ∈ Ψ(I(φ) L )=Ψ(I(φ L )) = Λ(φ L )=Λ(φ)=Ψ(I(φ)). ◦ g ◦ g ◦ g

Therefore ν =(Lg) ν for all g G. This proves the ﬁrst part of the theorem. ∗ ∈ Now let ν be another G-invariant measure on X. Deﬁne a positive linear functional Λ : C (G) R such that for all φ C (G), c → ∈ c

Λ(φ)= I(φ) dν. (1.8) X

By Riesz representation theorem, there exists a unique measure µ on G such that for all φ C (G), ∈ c φdµ =Λ(φ). (1.9) G Now for any g G, ∈

Λ(φ L )= I(φ L ) dν = I(φ) L dν = I(φ) dν =Λ(φ). ◦ g ◦ g ◦ g X X X

Thus µ is a left haar measure on G. By the ‘uniqueness’ of left haar measures on G there exists c>0 such that µ = cµ. Now since I is surjective, Eqs. 1.6, 1.8, and 1.9 imply that ν = cν. This completes the proof of the ‘uniqueness’ assertion of the theorem. We shall be mainly interested in the case when G is a unimodular group and the stabilizer of every point of the homogeneous space X is a discrete subgroup of G. Fix a point x X, then Γ= G is a discrete subgroup of G 0 ∈ x0 and X ∼= G/Γ. The counting measure on Γis a natural choice for a left and a right haar measure on Γ. Now for every φ C (G) and x = gx X, ∈ c 0 ∈ I(φ)(x)= φ(δ)

δ G:δx0=x { ∈ } = φ(gγ). γ Γ ∈ Fix a haar measure µ on G. Then by Theorem 1.3.3, there exists a unique G-invariant measure ν on G/Γsuch that for all φ C (G), ∈ c

φdµ = I(φ)(y) dν(y) G G/Γ = φ(gγ) dν(gΓ). G/Γ γ Γ ∈

20 Lemma 1.3.4 For every x X and a measurable subset E G, ∈ ⊂ ν(Ex) µ(E). ≤

Further if the map πx : G X, given by πx(g)=gx for all g G, is injective on the set E then → ∈

ν(Ex)=µ(E).

Proof. Suppose x = gx for some g G. Since G is unimodular, replacing E 0 ∈ by Eg there is no loss of generality in proving the theorem for x0 in place of x. Take >0. By regularity of µ and ν, there exists a compact set C E such that µ(E C) <and ν(Ex Cx ) <. Also there exists an open⊂ set \ 0 \ 0 D C such that µ(D C) <and ν(Dx0 Cx0) <. ⊃Let φ C (G) be such\ that φ 1onC\and supp φ D. Then ∈ c ≡ ⊂ I(φ)(y) 1 if y Cx and I(φ)(y) = 0 if y X Dx . ≥ ∈ 0 ∈ \ 0 Therefore ν(Cx ) I(φ) dν = φdµ µ(D). 0 ≤ ≤ X G Thus, ν(Ex ) ν(Cx )+ µ(D)+ µ(E)+3. 0 ≤ 0 ≤ ≤ Since is arbitrary, ν(Ex ) µ(E). 0 ≤ This completes the proof of the ﬁrst part. Now suppose that π is injective on E. In this case we can choose D as above satisfying one more condition, that π is injective on D. Then I(φ)(y) 1for y Dx . Therefore ≤ ∈ 0 µ(C) φdµ= I(φ) dν ν(Dx ). ≤ ≤ 0 G X Hence

µ(E) µ(C)+ ν(Dx )+ ν(Cx )+2 ν(Ex )+2. ≤ ≤ 0 ≤ 0 ≤ 0 Since is arbitrary, µ(E) ν(Ex ). ≤ 0 Therefore by the ﬁrst part, ν(Ex0)=µ(E). Notation. For a neighbourhood Ωof e in G, deﬁne

X = x X : G Ω = e . Ω { ∈ x ∩ { }}

21 Lemma 1.3.5 For any compact set Y X, there exists a neighbourhood Ω ⊂ of e in G such that Y XΩ = . Thus if Ωn is a sequence of neighbourhoods of e in G such that∩Ω e∅ as n {, then} X X X as n . { } n ↓{ } →∞ \ Ωn ↑ →∞ Proof. Fix x X. There exists a compact set C G such that Y Cx . 0 ∈ ⊂ ⊂ 0 Let Ω0 be a neighbourhood of e in G such that Gx0 Ω0 = e . Note that for 1 ∩ { } any x Cx0, Gx = cGx0 c− for some c C. Since C is compact and Ω0 is ∈ ∈ 1 open, there exists a neighbourhood Ωof e in G such that c− Ωc Ω0 for all c C. Therefore, G Ω= e for all x Cx . Hence Y X =⊂ . ∈ x ∩ { } ∈ 0 ∩ Ω ∅ Deﬁnition. A discrete subgroup Γof G is called a lattice in G if the homogeneous space X = G/Γadmits a ﬁnite G-invariant measure.

Lemma 1.3.6 Suppose that the homogeneous space X = G/Γ admits a ﬁnite G-invariant measure. Then for any neighbourhood Ω of the identity in G, the set X X is relatively compact. \ Ω Proof. Without loss of generality we may assume that Ωis relatively compact. Now suppose that X X is not relatively compact. Then there exists a \ Ω sequence xi X XΩ such that Ωxi Ωxj = for all i = j.LetΩ1 be a { }⊂ \ ∩ ∅ 1 relatively compact neighbourhood of e in G such that Ω− Ω Ω. Then for 1 1 ⊂ every x X XΩ, the map Ω1 g gx X is injective. Due to Lemma 1.3.4, ν(Ω x )=∈ µ(\Ω ) for all i N. Therefore→ ∈ 1 i 1 ∈

∞ ∞ ν(X) ν( ∞ Ω x )= ν(Ω x )= µ(Ω )= . ≥ ∪i=1 1 i 1 i 1 ∞ i=1 i=1 This is a contradiction to our assumption.

Deﬁnition. LetΓbe a discrete subgroup of G and X = G/Γ. We say that Γ is an L-subgroup of G if for every neighbourhood Ωof e in G, the set X XΩ is relatively compact in X. In this case we call X an L-homogeneous space\ of G.

Thus by Lemma 1.3.6, every lattice is an L-subgroup. In the next chapter we shall prove that every L-subgroup of G = SL2(C) is a lattice in G.

1.4 Closed subgroups of Rn and Rn/Zn

The results of this section will not be used later in the notes, but they are of interest on their on right.

22 Lemma 1.4.1 Let V be a ﬁnite dimensional real vector space and ∆ be a nontrivial discrete subgroup of V . Then there exist linearly independent vectors x1,...,xk in V such that ∆=Zx Zx Zx . 1 ⊕ 2 ⊕···⊕ k The dimension of the subspace spanned by ∆ is k, and it is called the rank of ∆. Note that the quotient group V/∆ is compact if and only if dim V = rank ∆.

Proof. Without loss of generality we may assume that span ∆= V .Take linearly independent vectors y1,...,yk ∆such that V = span y1, ,yk . Let ∈ { ··· } ∆ = Zy Zy ∆. 1 ⊕···⊕ k ⊂ And let π : V V/∆ be the quotient map. If → k I = a y V :0 a < 1for1 i k i i ∈ ≤ i ≤ ≤ i=1 then V = I +∆ and π(I)=V/∆. Since I is compact, V/∆ is compact. Now π(∆) is a closed, and hence a discrete subgroup of V/∆. Therefore ∆/∆ is finite. Now since ∆ is finitely generated, so is ∆. Since ∆is abelian and torsion free, there exist x1,...,xn ∆such that ∈ ∆=Zx Zx 1 ⊕···⊕ n (see [L, Chap. 1, Sect. 10, Theorem 7]). By a rearrangement, we can assume that x1,...,xk are linearly independent. Let ∆ be the free abelian group generated by x1,...,xk. Then span(∆)=V . Repeating the above argument for ∆ in place of ∆,we obtain that [∆:∆] < . Therefore k = n. This completes the proof. ∞ Theorem 1.4.2 Let V be a finite dimensional vector space and F be a closed subgroup of V . Then F = F 0 ∆, where F 0 is a subspace of V and ∆ is a discrete subgroup of V . ⊕

Proof.IfF is discrete, there is nothing to be proved. Otherwise there exists a sequence v F 0 such that v 0asi .Foreachi N, { i}⊂ \{ } i → →∞ ∈ put xi = vi/ vi S(1), where S(1) is the unit sphere in V . Since S(1) is compact, by passing∈ to a subsequence we may assume that as i , x x →∞ i → for some x S(1). Let W1 be the one dimensional subspace passing through x. ∈ We claim that W1 F . To prove the claim, let λ R and let >0be given. For each i N,⊂ there exists n Z such that ∈ ∈ i ∈ n v λx v . i i − i≤ i 23 Choose i N such that for all i i , 0 ∈ ≥ 0 λ x x </2 and v /2. | |· i − i≤ Therefore n v λx n v λx + λ x x . i i − ≤ i i − i | |· i − ≤ Since F is a closed subgroup of V , λx F . This proves the claim. ∈ Let V1 be a subspace of V such that V = W1 V1 and put F1 = F V1. Since W F , we have F = W F . Now since⊕F is a closed subgroup∩ of 1 ⊂ 1 ⊕ 1 1 V1, the conclusion of the theorem follows from the induction on dim V . Lemma 1.4.3 Let F be a closed subgroup of Rn. Then the following conditions are equivalent: (a) span(F Qn) = span(F ). ∩ (b) The group F/(F Zn) is compact. ∩ (c) F = F 0 ∆, where F 0 Qn is dense in F 0 and ∆ is a discrete subgroup contained⊕ in Qn. ∩

(d) The subgroup F Qn is dense in F . ∩ Proof. Put F = F Zn and F = F Qn. Then Z ∩ Q ∩

V := span(FZ) = span(FQ).

By Lemma 1.4.1, V/FZ is a compact group. Now if condition (a) holds then F/FZ is a closed subgroup of V/FZ, and hence the condition (b) holds. Let π : F F/FZ denote the natural quotient homomorphism. Then 0 → π(F ) is an open, and hence a closed subgroup of F/FZ. 0 n Now suppose that the condition (b) holds. Then the group F /(F0 Z ) ∼= π(F 0) is compact. By Theorem 1.4.2, the group F 0 is a subspace of R∩n. Now by Lemma 1.4.1, F 0 = span(F 0 Zn) and hence F 0 Qn is dense in F 0. Therefore there is a decomposition∩ of Qn as ∩

Qn =(F 0 Qn) (W Qn), ∩ ⊕ ∩ where W is a subspace of Rn. Also F = F 0 (F W ) and by Lemma 1.4.2, the subgroup ∆:= F W is discrete. Now ⊕ ∩ ∩ F := F = F 0 Qn (∆ Qn). 1 Q ∩ ⊕ ∩ Since F F , the quotient group F/F is compact. Therefore the group Z ⊂ 1 1 ∆/(∆ Qn) = F/F is ﬁnite. Now for any δ ∆, there exists n N such ∩ ∼ 1 ∈ ∈ 24 that nδ ∆ Qn. But then δ Qn. Hence ∆ Qn. Thus the condition (c) holds. ∈ ∩ ∈ ⊂ It is obvious that the condition (c) implies the condition (d), and the condition (d) implies the condition (a). This completes the proof.

Definition. A closed subgroup F of Rn is called rationally defined, if it satisfies any one of the equivalent conditions of Lemma 1.4.3.

Let T1 denote the circle group and Tn denote the n-torus as before. There is a homomprphism φ : R T1 given by φ(t)=e2πit for all t R. Then ker φ = Z and we have a→ topological group isomorphism φ¯ : R∈/Z T1. n n n n n n → Clearly, R /Z ∼= (R/Z) . Therefore R /Z ∼= T as topological groups. Thus we have realized Tn as a homogeneous space of Rn.

Theorem 1.4.4 Let F be a closed subgroup of Rn. Let L be the smallest rationally deﬁned closed subgroup of Rn containing F . Then the closure of n n n any orbit of F on R /Z ∼= T is a compact orbit of L. Proof.Letπ : Rn Rn/Zn be the quotient homomorphism. Since Rn is abelian, it is enough→ to show that π(F )=π(L). Since Rn/Zn is a group, π(F ) is a closed subgroup of Rn/Zn. By Lemma 1.4.3, π(L) is compact, therefore π(F ) π(L). ⊂ 1 n Let F = π− (π(F )). Then F is a closed subgroup of R containing F and π(F )=π(F ) is compact. Therefore by Lemma 1.4.3, F is a rationally n deﬁned closed sugroup of R . By minimality, L F , and hence π(L) π(F ). ⊂ ⊂ Therefore π(L)=π(F ).

n Lemma 1.4.5 Let v =(a1,...,an) R and L be the smallest closed rationally deﬁned subgroup of Rn containing∈ v. Think of R as a vector space over Q. Then dim Q- span 1,a ,...,a = 1 + dim L0. Q { 1 n} 0 Proof.Letm = dim L and k = dimQ Q- span 1,a1,...,an . There exists a n 0 { } n basis e1,...,em Q of L and a discrete subgroup ∆ Q such that L = L{0 ∆. Since}⊂v L, there exist x ,...,x R and δ ⊂∆such that ⊕ ∈ 1 m ∈ ∈ m

v = xiei + δ. i=1 Therefore 1,a ,...,a Q- span 1,x ,...,x . { 1 n}⊂ { 1 m} Hence k m +1. ≤ 25 Without loss of generality we may assume that the reals 1,a1,...,ak 1 are − linearly independent over Q, and for every 1 j n and 0 i k 1, there exists r Qn, such that ≤ ≤ ≤ ≤ − ij ∈ k 1 − aj = r0j + rijai. (1.10) i=1 For 0 i k 1, put f =(r ,...,r ). Let ≤ ≤ − i i1 in E = R- span f :1 i k 1 Z f . { i ≤ ≤ − }⊕ · 0 Then E is a rationally deﬁned closed subgroup Rn, and by Eq. 1.10, we have that v E. Therefore by minimality L E and hence m = dim L0 dim E0 =∈ k 1. Thus we have showed that ⊂k = m +1. ≤ − From Theorem 1.4.4 and Lemma 1.4.5, we deduce the following theorem due to Kronecker.

n Corollary 1.4.6 Let v =(a1,...,an) R . If the elements 1,a1,...,an are linearly independent over Q then every∈ orbit of the cyclic subgroup generated n n n by v is dense in R /Z ∼= T .

1.5 Covolumes of discrete subgroups in euclidean spaces

Let W be a finite dimensional vector space with a euclidean metric. Let ∆ be a discrete subgroup of W such that W = span ∆. Then W/∆is compact. By Theorem 1.3.3, the standard lebesgue measure m on W induces a unique finite measurem ¯ on W/∆. By Lemma 1.3.4, if the map π : W W/∆is injective on a measurable set E W thenm ¯ (π(E)) = m(E). We define→ ⊂ Vol(W/∆) =m ¯ (W/∆),

which can be computed as follows. Let x1,...,xk ∆beaZ-basis of ∆. Then ∆= Zx1 Zxk and W = Rx R∈x . If we deﬁne ⊕···⊕ 1 ⊕···⊕ k k I = a x W :0 a < 1for1 i k , i i ∈ ≤ i ≤ ≤ i=1 then W = I +∆and I ∆= 0 ; that is π is injective on I. Therefore ∩ { } m¯ (W/∆) = m(I).

26 To compute m(I), let e ,...,e be the standard orthonormal basis of { 1 k} W , and let T : W W be the linear transformation such that T (ei)=xi for all 1 i k. → ≤ ≤ If k J = a e :0 a < 1for1 i k i i ≤ i ≤ ≤ i=1 then I = T (J). Note that for any linear automorphism A of W and a measurable set E W , ⊂ m(A(E)) = det A m(E) | |· (see [Ru2, Sect. 8.28]). Thefefore

m(I)= det T m(J)= det T . | |· | | Thus Vol(W/∆) = det T . (1.11) | | Now let V be a n-dimensional vector space equipped with a euclidean metric. Let , denote the associated positive definite inner product. · · Definition. Given a discrete subgroup ∆of rank k in V , we associate a quantity called the covolume to ∆in the following way: Let W = span ∆. Then W inherites the euclidean structure from V . Now define

covol(∆) = Vol(W/∆). (1.12)

We will express covol(∆) in terms of a basis of ∆.

Notation.Let e1,...,en be an orthonormal basis of V . Consider the k-th exterior power { kV of V .} Then the set ∧ B = e e :1 i

covol(∆) = x x . 1 ∧···∧ k

27 We recall some standard facts used for the proof. Asetofvectors x1,...,xk V is linearly independent if and only if { }⊂ k x1 xk is a nonzero vector in V . Also if y1,...,yk is another set of linearly∧···∧ independent vectors in V then∧ { }

span y ,...,y = span x ,...,x y ... y = λ(x ... x ) { 1 k} { 1 k}⇔ 1 ∧ ∧ k 1 ∧ ∧ k

for some λ R∗. In this case, let W = span x1,...,xk and T be the linear transformation∈ on W such that Tx = y for 1{ i k then} i i ≤ ≤ y ... y = det T (x ... x ). 1 ∧ ∧ k 1 ∧ ∧ k If A is a linear transformation on V then there exists a unique linear transformation kA on kV such that for any k vectors x ,...,x V , ∧ ∧ 1 k ∈ kA(x ... x )=(Ax ) (Ax ). ∧ 1 ∧ ∧ k 1 ∧···∧ k If B is another linear transformation on V then

k(A B)= k(A) k(B). ∧ ◦ ∧ ◦∧

Deﬁnition.LetE be an inner product space with the inner product , .For any linear transformation A on E, there exists a unique linear transformation· · A∗ on E, called the adjoint of A, such that for all x, y E, ∈

A∗x, y = x, Ay . A transformation A on E is called orthogonal if it preserves the innerproduct on E. Thus A is orthogonal AA∗ =1 A transforms one orthonormal basis into another orthonormal basis.⇔ ⇔

Lemma 1.5.2 For any linear transformation A on V ,

k k ( A)∗ = A∗ ∧ ∧ on kV . In particular, if A is an orthogonal transformation of V then kA is an∧ orhogonal transformation of kV . ∧ ∧ Proof.Let1 i

28 where X is the k k matrix whose (p, q)-th entry is ×

A∗e ,e = e , Ae . ip jq ip jq Therefore

det X = e e , Ae Ae i1 ∧···∧ ik j1 ∧···∧ jk = e e , kA(e e ) . i1 ∧···∧ ik ∧ j1 ∧···∧ jk Since B is a basis of kV ,weget ∧ k k A∗ =( A)∗. ∧ ∧

Proof of Theorem 1.5.1. Let W = span ∆and let w1,...,wk be an orthonormal basis of W . Extend this basis to an orthonormal{ basis} of V given by w ,...,w . { 1 n} Let T : W W be the linear transformation given by Twi = xi for 1 i k. Then→ by Eq. 1.11 and 1.12, ≤ ≤ covol(∆) = det T . | | Now x x = det T (w w ) kW kV. 1 ∧···∧ k 1 ∧···∧ k ∈∧ ⊂∧ Therefore x x = det T w w . 1 ∧···∧ k | |· 1 ∧···∧ k Note that there exits a unique orthogonal transformation A of V which takes e1,...,en to the orthonormal basis w1,...,wn . Therefore by Lemma 1.5.2, { } k { } w1 wk = A(e1 ek) is an element of an orthonormal basis of kV∧.··· Hence∧ w ∧ w∧···=∧ 1. Thus ∧ 1 ∧···∧ k covol(∆) = det T = x x . | | 1 ∧···∧ k

29 Chapter 2

The group SL2(C) and its homogeneous spaces

2.1 Exponential map, Lie algebras, and Unipo- tent subgroups

In this section we shall introduce some very basic concepts regarding Linear Lie groups. Part of the material in this section is taken from an excellent article on ‘Very Basic Lie theory’ by R. Howe (see [Ho]).

The exponential map on matrices Let V be a finite dimensional real or complex vector space. Let End(V ) denote the algebra of linear maps from V to itself, and let GL(V ) denote the group of invertible linear maps from V to itself. The usual name for GL(V ) is the general linear group for V .IfV = kn, where k = R or k = C, then End(V ) = M (k), the space of n n matrices with entries from k, and ∼ n × GL(V ) ∼= GLn(k), the matrices with nonvanishing determinants. Let be a norm on V . In the usual way there is induced an operator norm, also denoted· by , on End(V ) defined as follows: For every A End(V ), · ∈ A = sup Av / v : v V 0 . { ∈ \{ }} The norm on End(V ) makes End(V ) into a matric space. Thus GL(V )also becomes a metric space with respect to the restricted metric. For A End(V ) and r>0, define ∈

(A)= A End(V ): A A

(1V + A)(1V + B)=1V + A + B + AB. we obtain that for r<1/2,

1 (1 )− (1 ) (1 ). Br V Br V ⊂B4r V In view of the remarks made just after the definition of a topological group, GL(V ) is a topological group. For a topological group G, any continuous homomorphism ρ : R G is called a one-parameter subgroup of G. Note that ρ(0) is the identity element→ of G and ρ(r + s)=ρ(r)ρ(s) for all r, s R. All one-parameter subgroups of GL(∈V ) can be described using the exponential map on matrices. For A End(V ), define ∈ ∞ exp(A)= An/n!. n=0 Since An A n, we see by the standard estimates in the exponential series, that the series≤ defining exp A converges absolutely for all A and uniformly on r(0) for every r>0. Hence exp defines a smoodh, in fact an analytic, map fromB End(V ) to itself. For s, t R, ∈ ∞ ∞ exp(sA) exp(tA)= (sm/m!)Am (tn/n!)An · m=0 n=0 ∞ = (smtn/m!n!)Am+n m,n=0 ∞ = ((m + n)!/m!n!)smtn Al/l! l=0 m+n=l ∞ = ((s + t)A)l/l! = exp((s + t)A). l=0 31 In particular exp(A) exp( A)=1V . Hence exp(A) GL(V )foreveryA End(V ). Thus we have defined− a map ⊂ ∈ exp : End(V ) GL(V ). → Also for any A End(V ), the map ρ : R GL(V ) defined by ∈ → ρ(t)=exptA ( t R) ∀ ∈ is a one-parameter subgroup of GL(V ). It can be proved that any one- parameter subroup of GL(V ) is of the form t exp tA ( t R) for some A End(V ) (see [Ho, Thm. 10]). We will not→ use this fact∀ ∈ later in these notes.∈ We shall need the following result. Lemma 2.1.1 There exists r > 0 such that the function exp maps (0) 0 Br0 ⊂ End(V ) homeomorphically onto an open neighbourhood of 1V in GL(V ).

Proof.LetD expA be the differential of exp at A. It is a linear map from End(V ) to End(V ) defined by d D exp (B)= exp(A + tB) A dt t=0 = lim(exp(A + tB) exp A)/t. t 0 → − From the definition of exp, one verifies that

D exp0(B)=B.

That is, D exp0 is the identity map on End(V ). In particular D exp0 is invertible. Therefore the lemma follows from the Inverse Function Theorem (see [Ru1, Chap. 9]). Remark. If one deﬁnes,

∞ log(1 A)= An/n, V − − n=1 then just as in the case of real numbers, this series converges absolutely for A < 1. Now for all B (1 ), ∈B1 V exp(log B)=B. (2.1) The formula is known in the scalar case, and this implies that in fact Eq. 2.1 is an identity in the absolutely convergent power series, whence it follows in the matrix case. Thus we have an explicit proof of Lemma 2.1.1.

The following result is also useful.

32 Lemma 2.1.2 Let and be linear subspaces of End(V ) such that H1 H2 End(V )= . H1 ⊕H2 Let pi be the projection of End(V ) on i with respect to the above decomposition. Deﬁne a map φ : End(V )= H GL(V ) by H1 ⊕H2 → φ(A) = exp(p1(A)) exp(p2(A)). Then φ maps a neighbourhood of 0 in End(V ) homeomorphically onto a neighbourhood of 1V in GL(V ). Proof. From the formula of the exp map one computes that d exp(p (tA)) exp(p (tA)) = p (A)+p (A)=A dt 1 2 1 2 t=0 for every A End( V ). Therefore the diﬀerential of φ at 0 is the identity map on End(V ).∈ Now the conclusion of the lemma follows by the Inverse Function Theorem.

The Lie algebra associated to a matrix group Definition. A real Lie algebra is a real vector space equipped with a skew symmetric bilinear product [ , ]:G satisfying the Jacobi Identity: · · G×G→G [x, [y, z]] + [z,[x, y]] + [y, [z,x]] = 0 for all x, y, z . ∈G The first main example of a Lie algebra is End(V ) equipped with the product operation of commutator defined as follows: For all A, B End(V ), ∈ [A, B]=AB BA. − Any subspace of End(V ) which is closed under the [ , ] operation is a Lie algebra in its own right. · · Remark.LetA End(V ). If the one-parameter group exp tA : t R is regarded as a curve∈ inside the vector space End(V ), then{ this curve∈ passes} through the identity 1V at time t = 0. By differentiating the formula for exp tA, we see that the the tangent vector at the point 1V to this curve is just A. Thus A is called an infinitesimal generator of the one-parameter group exp tA . { } To any closed subgroup G GL(V ) one associates a set ⊂ = A End(V ):exptA G for all t R , G { ∈ ∈ ∈ } which is a collection of the infinitesimal generators of all one-parameter subgroups of G.

33 Theorem 2.1.3 The set is a Lie subalgebra of End(V ). In particular, is a vector subspace of End(VG). G The map exp : G maps a neighbourhood of 0 in bijectively onto a neighbourhood of 1 G→G. G V ∈ Due to this theorem the set is called the Lie algebra associated to G. G The theorem asserts that the set of tangent vectors at 1V to the curves defined by one-parameter subgroups of G actually fills out some linear subspace, namely , of End(V ), and further, if we make a smooth change of co-ordinates given byGA exp A, then this linear subspace is bent in such a way that it → G lies entirely in G, and fills up G around 1V . In other words, G is shown to be a multidimensional surface inside End(V ), and is simply its tangent space G at the point 1V . See [Ho, Thm. 17] for a proof of this theorem. We will directly verify this theorem for G = SLn(k) and the ‘horospherical subgroups’. These are essentially the only cases for which we will need to use this theorm. Note that if G =GLn(R) is considered as a closed subgroup of GLn(C) then the Lie algebra associated to G is Mn(R) Mn(C). The following formulaG is useful in describing Lie⊂ algebras of certain closed subgroups of GLn(C). Lemma 2.1.4 For A End(V ), ∈ exp(tr A) = det(exp A).

Proof. It is enough to prove the formula for A M (C). Note that if B is an ∈ n upper triangular matrix with the diagonal (b11,b22,...,bnn) then the diagonal of exp B is (exp b11,...,exp bnn). Therefore exp(tr B) = exp(b + + b ) 11 ··· nn =(expb ) (exp b ) 11 ··· nn = det(exp B).

1 Now given A Mn(C) there exists g GLn(C) such that gAg− is upper triangular. Therefore,∈ ∈

1 exp(tr A) = exp(tr(gAg− )) 1 = det(exp(gAg− )) 1 = det(g(exp A)g− )=det(expA).

The group G = SL(V )= g GL(V ) : det g =1 is a closed subgroup of GL(V ). If denotes the Lie algebra{ ∈ associated to G}then G = A End(V ) : det(exp tA)=1 t R . G { ∈ ∀ ∈ } 34 Now det(exp tA)=1 exp(tr tA)=1 ⇔ (tr A)t 2πiZ. ⇔ ∈ Now (tr A)t 2πiZ for all small enough t R if and only if tr A =0. Therefore ∈ ∈ = A End(V ):trA =0 . G { ∈ } Clearly is a vector subspace of End(V ). Since tr(AB)=tr(BA), we have tr([A, B])G = 0 for all A, B End(V ). Thus is a Lie subalgebra of End(V ). ∈ G Now let r0 > 0 be such that tr A < 1 for all A r0 (0). Then due to Lemma 2.1.1, there exists r (0| ,r )| such that the function∈B exp maps the set ∈ 0 r(0) homeomorphically onto an open neighbourhood of 1V in G. Thus weB have∩G veriﬁed Theorem 2.1.3 in this case.

The Adjoint representation For g GL(V ) and A End(V ), we can form the conjugate ∈ ∈ 1 Ad g(A)=gAg− .

Let A, B End(V ), a, b R, and g, g1,g2 GL(V ). Then the following identities can∈ be easily veriﬁed:∈ ∈ (i) Ad g(aA + bB)=a Ad g(A)+b Ad g(B), (ii) Ad g(AB) = Ad g(A) Ad g(B), (iii) Ad g([A, B]) = [Ad g(A), Ad g(B)],

(iv) Ad g1g2(A)=Adg1(Ad g2(A)),

1 (v) exp(Ad g(A)) = g(exp A)g− . The identities (i) and (iii) say that Ad g is a Lie algebra automorphism of End(V ), and formula (iv) says that the map Ad : g Ad g is a group homomorphism from GL(V ) to the automorphism group of→ End(V ). It is straight forward to verify that the map Ad : GL(V ) GL(End(V )) is continuous. → Let G be a closed subgroup of GL(V ) and be the corresponding Lie algebra. Then for any g GL(V ), the Lie algebraG associated to the group 1 ∈ gGg− is Ad g( )= Ad g(A):A . G { ∈G} In particular, for all g G, we have that Ad g( )= . Note that Ad : G Aut( ) is a continuous∈ homomorphism. ThusG Ad providesG a natural linear→ actionG of the group G on its Lie algebra. This action is called the Adjoint representation of G on . G 35 Horospherical subgroups and unipotent elements

Deﬁnition.LetG be a closed subgroup of GL(V ). For g G, deﬁne ∈ n n U(g)= u G : g ug− e as n . { ∈ → → ∞} Then U(g) is a closed subgroup of G and it is called the horospherical subgroup of G associated to g.

Let denote the Lie algebra of G. Then the Lie algebra associated to U(g) can be givenG by

(g)= A : Ad gn(A) 0asn . U { ∈G → → ∞} Since Ad g is a linear transformation on , the set (g) is a linear subspace of . The other part of Theorem 2.1.3 isG also straightU forward to verify for (gG). Moreover, in this case the map exp : (g) U(g) is a homeomorphism ofU (g)ontoU(g). U → U Definition. An element u GL(V ) is called unipotent if every eigenvalue of u is 1. ∈ An element N End(V ) is called nilpotent if every eigenvalue of N is 0. ∈ Observe the following: 1. N is a nilpotent element of End(V ) if and only if N n = 0 for some n dim(V ). ≤ 2. If N End(V ) is nilpotent then exp N GL(V ) is a unipotent. ∈ ∈ 3. Let u GL(V ) be a unipotent element. Then 1 is the only root of the ∈ characteristic polynomial P (λ) = det(u λ1V ). Therefore if N = u 1V then N n = 0 for some n dim V . Define− − ≤ ∞ k n 1 log(u)= N /k = (N + N/2+ + N − /(n 1)). (2.2) − − ··· − k=1 Then log(u) is nilpotent and u = exp(log u). This shows that every unipotent element is the exponential of a nilpotent element. If N is a nilpotent element then the one-parameter subgroup u(t) := exp tN : t R is called a one-parameter unipotent subgroup. { ∈ } Lemma 2.1.5 Let u(t):t R be a one-parameter unipotent subgroup of GL(V ). Then for every{ v V∈, the} co-ordinates of the curve c(t)=u(t)v, with respect to any basis of V ,∈ are polynomials in t. Moreover, if u(t )v = v for some t =0then u(t)v = v for all t R. 0 0 ∈ 36 Proof.LetN be a nilpotent element such that u(t)=exptN for all t R. Since N n = 0 for some n dim V , we have ∈ ≤ 2 2 n 1 n 1 c(t)=(exptN)v = v + tNv +(t /2)N v + +(t − /(n 1)!)N − v. ··· − This readily implies the first part.

Now suppose u(t0)v = v for some t0 = 0. Then u(kt0)v = v for all k Z. But a polynomial which takes a ﬁxed value inﬁnitely often must be a constant∈ polynomial. Therefore c(t)=v for all t R. ∈ The polymomial behaviour of the orbits of a one-parameter group of unipotent transformations on a vector space turns out to be a key property for studying orbits of a unipotent subgroup of G on a homogeneous space of G, where G is a closed subgroup of GL(V ). Note that the eigenvalues of an element and the eigenvalues of its conjugate are same. Also given an >0, there exists an r>0 such that the eigenvalues of every A r(0) are of absolute value less than . Therefore the Lie algebra of any horospherical∈B subgroup consists of nilpotent elements. And hence every horospherical subgroup consists of unipotent elements. Let g = diag(a ,...,a ) be a diagonal matrix, where 0

Any unipotent element of GLn(C) has a conjugate which is an uppertrian- gular matrix with all diagonal entries 1. Therefore every unipotent element of GLn(C) is contained in a horospherical subgroup of GLn(C).

Lemma 2.1.6 Let G be a closed subgroup of GLn(C) and U(g) be the horospherical subgroup of G associated to an element g G. Let ρ be a continuous representation of G on a ﬁnite dimensional vector∈ space V ; that is, ρ : G GL(V ) is a continuous homomorphism. Then ρ(U(g)) is contained in the→ horospherical subgroup of GL(V ) associated to ρ(g). In particular, all elements of ρ(U(g)) are unipotent linear transformations of V .

2.2 Subgroups of SL2(C)

Now onwards we shall deal with the group G = SL2(C) and the actions of its subgroups on the homogeneous spaces of G. In this section we shall obtain some useful properties of this group. Consider the standard action of G on E = C2. Fix a point p =(1, 0) E. ab ∈ For any g = G, gp =(a, c). Thus Gp = E 0 and the stabilizer cd ∈ \{ } 37 of p in G is the closed subgroup

def 1 z N = u(z) = : z C . 01 ∈ By Lemma 1.1.2, G/N ∼= E 0 and the isomorphism preserves the standard actions of G on both the spaces.\{ } a 0 Note that N is the horospherical subgroup of G associated to 1 , 0 a− for every a C∗ with a < 1. The Lie subalgebra of associated to N is ∈ | | G 0 z = : z C N 00 ∈ and the exponential map exp : N is an isomorphism of topological groups. Thus N is a complex one-parameterN→ subgroup of G via the map

0 z C z exp N. → 00 ∈ Deﬁne L = v E : Nv = v . Then L = Cp = (x, 0) E : x C .For any g G, { ∈ } { ∈ ∈ } ∈ gp L N(gp)=gp ∈ ⇔ 1 (g− Ng)p = p ⇔ 1 g− Ng N ⇔ ⊂ g N (N). ⇔ ∈ G Thus

N (N)= g G : gp L G { ∈ ∈ } ab = 1 : a C∗,b C . 0 a− ∈ ∈ Define a 0 D = d(a) := 1 : a C∗ . 0 a− ∈ Then N (N)=DN. Also N (N)p = L 0 = Dp. G G \{ } Let denote the euclidean norm on C2. Define · K = g G : gv = v for all v C2 . { ∈ ∈ } Then K = U(2) := g G : gg∗ = g∗g =1 . { ∈ } 38 By definition, K acts on the unit sphere S(1) in E.Let(z ,z ) S(1). Then 1 2 ∈ 2 2 z1 z¯2 z1 + z2 =1.Ifg = − then g K and gp =(z1,z2). Thus S(1) | | | | z2 z¯1 ∈ is a homogeneous space of K. NowK N = e . Therfore the map K S(1) given by k kp is a homeomorphism∩ and it{ preserves} the standard actions→ of K on both→ spaces. Thus topologically K is a 3-sphere. Define M = D K = d(eθi):θ R . ∩ ∈ Then M is isomorphic to the circle group and Mpis a circle, which is the intersection of L with S(1). Put

A = d(a):a>0 . { } Then D = MA.

Iwasawa decomposition of G Lemma 2.2.1 The map φ : K A N G given by × × → φ(k, a, n)=kan ( (k, a, n) K A N) ∀ ∈ × × is a homeomorphism. The decomposition of G as G = KAN is called an Iwasawa decomposition of G.

Proof. Given v C2 0 , put a = d( v ) A and let k K be such that kp = v/ v . Then∈ (ka\{)p =} v. Thus the map ∈ψ : K A ∈C2 0 given by ψ(k, a)=( ka)p ( (k, a) K A) is a homeomorphism.× → \{ } ∀ ∈ × 1 Now given g G, let (k, a)=ψ− (gp). Then gp =(ka)p. Hence n = 1 ∈ g− ka N. Thus the map G g (k, a, n) K A N is continuous, and ∈ → ∈ × × it is the inverse of φ. Therefore φ is a homeomorphism.

1 Lemma 2.2.2 (1) Given g G, there exits k K such that kgk− DN. ∈ ∈ ∈ 1 (2) Given h DN C(G)N, there exits v N such that vhv− D. (Here C(G)= ∈d(1),d\( 1) is the center of G∈.) ∈ { − }

0 1 1 1 (3) Let w = − . Then for every α C∗, wd(α)w− = d(α− ). 10 ∈ 1 2 (4) For any z C and α C∗, d(α)u(z)d(α− )=u(α z). ∈ ∈ 1 In particular, given g G there exists x G such that either xgx− = d(α) 1 ∈ ∈ with α 1 or xgx− = u(1). | |≤ 39 Proof.(1)Letv C2 be an eigen vector of g with unit norm. Let k K be ∈ 1 ∈ 1 such that kv = p. Therefore p is an eigen vector of kgk− . Hence kgk− ∈ NG(N)=DN. (2) Any h DN is of the form d(a)u(z) for some a C∗ and z C.For any y C, ∈ ∈ ∈ ∈ 2 u(y)hu( y)=d(a)u(a− y + z y). − − 2 1 If h C(G)N then a = 1. If we choose y = (a− 1)− z and v = u(y) N, ∈ 1 ± − − ∈ then vhv− = d(a) D. ∈ The statements (3) and (4) are evident.

Corollary 2.2.3 Any nontrivial horospherical subgroup of G is of the form 1 kNk− for some k K. ∈ Proof.Letg G be such that U(g) is the given horospherical subgroup. By ∈ 1 Lemma 2.2.2, there exists x G such that if we put h = xgx− then either ∈ 1 h = u(1) or h = d(α) with α 1. Now U(g)=xU(h)x− . | |≤ If h = u(1) or h = d(α) with α = 1 then U(h)= e and hence U(g)= e ; a contradiction. Therefore h|=| d(α) with α < 1{ and} hence U(h)=N. { } 1 | | Due to Lemma 2.2.1, we can express x− = kb, where k K and b AN 1 ∈ ∈ ⊂ NG(N). Then U(g)=kNk− .

Closed subgroups of DN Lemma 2.2.4 Let L be a nontrivial closed subgroup of the additive group C and let B be a subgroup of the multiplicative group C∗. Suppose that B preserves L under the action by multiplications and B 1, 1 . Then one of the following holds: ⊂ { − }

(1) L = C.

(2) L = Rw for some w C 0 and B R∗. ∈ \{ } ⊂ θi kθi (3) L = Z[e ] w and B = e k Z, where θ π/3,π/2, 2π/3 and ∈ w C 0 ·. { } ∈{ } ∈ \{ }

0 0 Proof.IfL = C then (1) holds. If L is one-dimensional then B R∗. Since B 1, 1 ,foranyz C, the additive group generated by Bz⊂is dense in Rz⊂. Hence { − } in particular∈L is connected and (2) holds. Now suppose that L0 = 0 . Then L is a discrete subgroup of C. Put { } r = inf z > 0 z L 0 | | ∈ \{ } 40 and let S(r) denote the circle of radius r in C centered at 0. Let w L S(r). ∈ ∩ If α C∗ and α < 1 then 0 < αw

Proposition 2.2.5 Let F be a closed subgroup of DN. Then there exists v 1 v N such that for F := vFv− , one of the following holds. ∈ (1) F v d(1),d( 1) N. ⊂{ − }· (2) F v D. ⊂ v k v (3) F = ( d(i ) k Z A) F (F N). { } ∈ · ∩ ∩ (4) F v =(D F v)N. ∩ v kθi 2iθ (5) F = d(e ) k Z u(z):z Z[e ] w , where θ π/6,π/4,π/3 ∈ and w{ C }0 . ·{ ∈ · } ∈{ } ∈ \{ } In particular, if L is a closed subgroup of DN containing N and Γ is a lattice in L, then L MN and N Γ is a lattice in N. ⊂ ∩

Proof.Letπ : DN D be the natural quotient homomorphism. For a C∗, x C, and y C,→ if h = d(a)u(x) F and u(y) F , then ∈ ∈ ∈ ∈ ∈ 1 2 1 hu(y)h− = u(a y)=π(h)u(y)π(h)− F. ∈ Therefore if we deﬁne L = z C : u(z) F { ∈ ∈ } and 2 B = a C∗ : a C∗ and u(a) π(F ) { ∈ ∈ ∈ } then L is invariant under the action of B via complex multiplications. There exists h F such that if π(h)=d(a), a C∗, then the following statements hold: ∈ ∈

kθi k (i) For θ π/2,π/3 , if a e k Z, then B = a k Z. ∈{ } ∈{ } ∈ { } ∈ 41 (ii) If a R∗, then B R∗. ∈ ⊂ If B = 1 then (1) holds. Now suppose that B = 1 . Then by Lemma 2.2.2,{ there} exists v N such that d(a) F v. { } If F v D then (2) holds.∈ Now suppose that∈F v D. Then there exists ⊂ v⊂ z C 0 and b C∗ such that h = d(b)u(z) F . Now ∈ \{ } ∈ 1 ∈ 1 1 v h− d(a)h d(a)− = u(z ) F , 1 1 1 ∈ 2 where z1 =(a 1)z = 0. Hence L = 0 . If B = 1, −1 then a = i and (3) { holds.} Now suppose that B 1, 1 . Then by applying{ − } Lemma 2.2.4,± we can conclude the remaining possibilities.⊂ { − }

Horospherical subgroups of G Given a complex line L in C2, the group

U(L) def= u G : uv = v, v L { ∈ ∀ ∈ } is a horospherical subgroup of G. To see this let k G be such that L = 1 ∈ k Cp. Then U(L)=kNk− , and it is the horospherical subgroup associated · 1 to kd(a)k− for any a C∗ with a < 1. Note that ∈ | | N (U(L)) = g G : gL = L . G { ∈ } 2 Let L1 and L2 be two distinct complex lines in C ; that is, L L = C2. 1 ⊕ 2 For i =1, 2, if we put U = U(L ) then for any v C2 L , we have i i ∈ \ i Uiv = v + Li. (2.3) Lemma 2.2.6 The group G is generated by unipotent elements. In fact

G = U2U1U2U1.

2 Proof.Letq L1 0 . Since the map φ : G/U1 C 0 deﬁned by φ(gU )=gq (∈g G\{), is} a homeomorphism, it is enough→ to prove\{ } that 1 ∀ ∈ U U U q = C2 0 . 2 1 2 \{ } Now due to Eq. 2.3,

U2q = q + L2, U (U q)=q +(L 0 )+L , and 1 2 2 \{ } 1 U (U U q)=C2 0 . 2 1 2 \{ } This completes the proof. 

42 Lemma 2.2.7 (Cf. [R, Sect. 12.6]) Let U1 and U2 be two distinct nontrivial horospherical subgroups of G then U U = e . 1 ∩ 2 { }

Proof.Letu U1 U2. Then u ﬁxes the lines L1 and L2 pointwise. Since 2∈ ∩ L L = C , the linear transformation u =1 2 . 1 ⊕ 2 C 

Lemma 2.2.8 Let U1, U2, and U3 be three distinct nontrivial horospherical subgroups of G. Then

N (U ) N (U ) N (U )=C(G). G 1 ∩ G 2 ∩ G 3 Proof.Fori 1, 2, 3 , suppose U ﬁxes the complex line L pointwise. If ∈{ } i i g NG(U1) NG(U2) NG(U3) then gLi = Li for all i. Since the three lines are pairwise∈ independent∩ ∩ and any two of them span C2, the linear transformation g acts as a scaler on C2. Hence g C(G). ∈

0 1 2 Notation.Letw = − K. Then w = d( 1). Put L = wL and 10 ∈ − 1 N − = U(L)=wNw− . Then N − is the horospherical subgroup associated to d(a)foranya C∗ with a > 1. Note that ∈ | | 10 N − = : z C . z 1 ∈ Bruhat decomposition Lemma 2.2.9 The group G admits a decomposition as

G = DN NwDN, ∪ called a Bruhat decomposition of G.

Proof. It is enough to prove that

Dp NwDp = C2 0 . (2.4) ∪ \{ }

Now Dp = C∗p = L 0 . Therefore wDp = L 0 . Hence by Eq. 2.3, \{ } \{ } 2 NwDp = N(L 0 )=L 0 + L = C L. \{ } \{ } \ Thus Eq. 2.4 holds and the proof is complete. Corollary 2.2.10 Let F be any closed subgroup of G containing a nontrivial 1 unipotent element. Then there exists g G such that either gFg− NG(N) 1 1 ∈ ⊂ or gFg− N = e and gFg− N − = e . Moreover,∩ in { the} second case,∩ F U{=} e for inﬁnitely many distinct horospherical subgroups U of G. ∩ { }

43 1 Proof. By Lemma 2.2.2, there exists k k such that kFk− N = e . 1 ∈ ∩ { } Put F = kFk− .IfF N (N), then by the Bruhat decomposition, there 1 1 ⊂ G exits g1 F1 such that g1 = nwh for some n N and h DN. Therefore 1 ∈ 1 1 ∈ 1 ∈ 1 g1Ng1− F1 = e . But g1Ng1− = nN −n− . Put g = n− k then gFg− N = ∩ 1 { } ∩ e and gFg− N − = e . { } ∩ { } 1 v 1 v 1 Now every any v gFg− N −, if we put N = vNv− then N gFg− = ∈ ∩ v1 v2 ∩1 e .Foranyv = v in N −, we have N = N . Now since gFg− N − is { } 1 2 ∩ an inﬁnite group, the second conclusion follows.

2.3 The invariant measure on SL2(C)/SL2(Z[i])

The haar measure on SL2(C) First we construct a haar meausure on G. The space X = C2 0 is a homogeneous space of G and for the point p =(1, 0) X, the\{ stabilizer} 2 ∈ Gp = N. The lebesgue meausure on C restricted to X,saym, is G-invariant. Since N is isomorphic to C as a topological group, the lebesgue measure, say λ,onC corresponds to a haar measure on N under the isomorphism. Now as in Section 1.3, there exists a linear map I : Cc(G) Cc(X) with the following properties: →

+ + (i) I(Cc (G)) = Cc (X). (ii) For any g G and ξ C (G), ∈ ∈ c I(ξ L )=I(ξ) L . (2.5) ◦ g ◦ g Therefore there exists a unique locally ﬁnite measure µ on G such that for every ξ C (G), ∈ c ξdµ= I(ξ) dm. G X Due to Eq. 2.5, the measure µ is left G-invariant. Thus we have explicitly constructed a haar measure on G. Note that due to Lemma 2.2.6 and Lemma 2.1.6, there does not exist any nontrivial continuous homomorphism of G into GL1(C) ∼= C∗. Therefore the modular character of G must be trivial. Hence G is a unimodular group and µ is right invariant. Due to Lemma 2.2.1, there is a continuous function ψ : X KA G → ⊂ such that for any x X, k K, and a A if ka p = x then ψ(x)=ka. Then for any ξ C (∈G) and ∈x X, ∈ · ∈ c ∈

I(ξ)(x)= ξ(ψ(x)u(z)) dλ(z). C 44 Hence for all ξ C (G), ∈ c

ξdµ= ξ(ψ(x)u(z)) dm(x)dλ(z). G C2 0 C \{ } As a consequence we obtain the following.

Lemma 2.3.1 For measurable subsets F K, F A and F N, if 1 ⊂ 2 ⊂ 3 ⊂ F = F1F2F3 then, µ(F )=m(F F p)λ(F ). 1 2 · 3 

For η>0, deﬁne A = d(a):a>0,a2 <η η { } and for t>0, deﬁne

N(t)= u(z):z C, (z) t, (z) t. . { ∈ | |≤ | |≤ } Definition.Forη>0 and a compact set C N, define ⊂ (η,C)=KA C. S η A subset of G of the form (η,C) is called a Siegel set. S Lemma 2.3.2 The haar measure of a Siegel set is finite.

Proof.

µ( (η,C)) = m(KA p)λ(C) S η · = m(B(√η))λ(C) < , ∞ where B(η) denotes the ball of radius η in C2 centered at the origin. 

A Siegel domain for SL2(Z[i]) Now we shall give an explicit example of a naturally arising lattice in G.Let Z[i] denote the ring of Gaussian integers in C. Then ∆= Z[i] Z[i] is a lattice in C2. If we deﬁne × ab Γ=SL(Z)= G : a, b, c, d Z[i] 2 cd ∈ ∈ then Γis a discrete subgroup of G and

Γ= g G : g∆=∆ . { ∈ } 45 Recall that Z[i] is a euclidean domain ([J, Sect.2.15]), and by euclidean al- ax gorithm, given any a, b Z[i], there exists an element Γif and ∈ by ∈ only if gcd(a, b) = 1 ([J, Sect.2.15,Exercise 11]). Also for every g G, the discrete subgroup g∆is a Z[i]-module with respect to the action of Z∈[i]onC2 by complex multiplication. Moreover, by Eq. 1.11,

covol(g∆) = det g covol(∆) = 1. (2.6) | |· Therefore G-acts transitively on the set of all discrete Z[i]-submodules of C2 with covolume one. Since elements Lare lattices, we can topologize by declaring that two lattices are ‘nearby’ ifL they admit basises which are ‘nearby’.L With this topology, becomes a homogeneous space of G and it is isomorphic to G/Γ, via the mapLgΓ g∆( g G). We claim that admits→ a ﬁnite∀ ∈G-invariant meausre; in other words, Γis a lattice in G. DueL to Lemma 1.3.4 and Lemma 2.3.2, it is enough to show the following:

Proposition 2.3.3 ([R, Sect. 10.4]) G = (η,N(1/2)) Γ for some η>0. S · Proof. Since N Γ= u(z):z Z[i] , we have N = N(1/2) (N Γ), and hence NΓ=N∩(1/2)Γ.{ Therefore∈ to} prove the proposition, it· is enough∩ to show that G =(KAηN)Γ for some η>0. By Iwasawa decomposition, G = KAN. Therefore given h G, h KAηN if and only if hp B(η). Therefore it is enough to prove that∈ there∈ exists η>0 such that for every∈ g G, ∈ g Γp B(η) = . · ∩ ∅ By the euclidean algorithm, it is straight forward to verify that ∆= Z Γp. Therefore it is enough to show that ·

g∆ B(η) = 0 . (2.7) ∩ { } Suppose for some η>0, g∆ B(2η)= 0 . Then for any δ ,δ ∆with ∩ { } 1 2 ∈ δ1 = δ2, (B(η)+gδ ) (B(η)+gδ )= . 1 ∩ 2 ∅ Therefore by Lemma 1.3.4 and Eq. 2.6,

λ(B(η)) covol(g∆) = covol(∆). ≤ Hence if η>0 is such that λ(B(η/2)) > covol(∆), then Eq. 2.7 holds for all g G. This completes the proof. ∈ 46 2.4 Ergodic properties of transformations on homogeneous spaces

Let G be a (locally compact, hausdorﬀ, second countable) topological group and let X be a G-space admitting a locally ﬁnite G-invariant borel measure, say ν.

Deﬁnition. We say that a subgroup F G acts ergodically on the measure space (X,ν) if the following condition is⊂ satisﬁed: For any measurable subset E X, if ν(gE∆E)=0foreveryg F , then either ν(E)=0orν(X E)=0, where⊂ A∆B =(A B) (B A). ∈ \ We say that an\ element∪ g\ G acts ergodically on (X,ν) if the subgroup generated by g acts ergodically∈ on (X,ν).

Note that ergodicity is the condition of irreducibility of an action in measure theoretic sense. It has the following topological consequence. Lemma 2.4.1 (Hedlund) If a subgroup F acts ergodically on (X,ν) then for almost all x X, the orbit Fx is dense in supp ν. Similarly,∈ if an element g G acts ergodically on (X,ν) then for alomst all x X, the trajectory gnx ∈: n 0 is dense in supp ν. ∈ { ≥ } Proof.Let be a countable basis of open subsets of X. Replacing X by supp ν, we can assumeB that supp ν = X. Then for every E , ν(E) > 0. Now ν(FE) > 0 and the set FE is F -invariant. Therefore∈B\{ by the∅} ergodicity of the F -action, ν(X FE) = 0. Now if \ Y = FE E ∈B\{∅} then ν(Y )=ν(X). Now take any y Y . Then Fy E = for every E . Thus Fy is dense in X. ∈ ∩ ∅ ∈InB\{ the∅ second} part, since we consider only the positive trajectory of gnx : n Z , we need to argue as follows: For any E , deﬁne { ∈ } ∈B\{∅} ∞ n X(E)= g− E. n=0 Then g(X(E)) X(E). Since g preserves the measure ν, ⊃ ν(gX(E)∆X(E)) = 0. Since ν(E) > 0 and g acts ergodically on (X,ν), we have ν(X X(E)) = 0. If \ Y = X(E) E ∈B\{∅} 47 then ν(X Y ). Now take any y Y . Then for every nonempty open subset E of X, there\ exists n 0 such∈ that gny E. Therefore the trajectory gny : n 0 is dense in ≥X. ∈ { ≥ } Ergodicity and unitary representations Now we reformulate the condition of ergodicity in terms of unitary representations of G.LetL2(X,ν) denote the space of complex valued square integrable functions on the measure space (X,ν). It is well known that L2(X,ν) is a 2 hilbert space with respect to the L -norm, which is denoted by L2 . The group G acts on L2(X,ν) by the following formula: For every g· G and ξ L2(X,ν), ∈ ∈ 1 (g ξ)(x)=ξ(g− x) · for almost all x X. Since the measure ν is G-invariant, g ξ L2 = ξ L2 for every g G and∈ every ξ L2(X,ν). Thus G acts on L2(X,· ν) by unitary automorphisms.∈ ∈

Lemma 2.4.2 The above unitary representation of G on L2(X,ν) is continuous; that is, the map G L2(X,ν) (g, ξ) g ξ L2(X,ν) is continuous. × → · ∈ 2 Proof. Since Cc(X) is a dense subset of L (X,ν), it is enough to show that for every ψ C (X), the map (g, ξ) g ξ,ψ is continuous. ∈ c → · Since g acts by a unitary automorphism on L2(X,ν),

1 g ξ,ψ = ξ,g− ψ = ξ(x)ψ¯(gx) dν(x). · · X

Since ψ Cc(X), given (0, 1) there exists a neighbourhood Ωof the identity in G such∈ that for all h∈ Ωand x X, ψ(hx)ψ(x) <. In particular, 1 ∈ ∈ 2 | | h− ψ ψ 2 . Now for any ξ L (x, ν) such that ξ ξ 2 <, · − L ≤ ∈ − L 1 1 1 g ξ,ψ hg ξ,ψ = ξ,g− ψ ξ,g− h− ψ · − · · −1 1· 1 1 = ξ ξ,g− ψ + ξ,g− ψ g− h− ψ − · · − 1 · ξ ξ 2 ψ 2 + ξ 2 ψ h− ψ 2 ≤− L L L − · L ( ψ 2 + ξ 2 +1). ≤ L L This shows that the map (g, ξ) g ξ,ψ is continuous. → · This representation is called the regular representation of G on L2(X,ν).

Lemma 2.4.3 Suppose further that ν is a probability measure. Then a subgroup F of G acts ergodically on (X,ν) if and only if every F -invariant function in L2(X,ν) is constant ν-almost everywhere on X.

48 Proof. First suppose that F acts ergodically on (X,ν). Let ξ L2(X,ν)be an F -invariant function. Then for any measurable subset D C∈, ⊂ 1 1 ν(g(ξ− (D))∆ξ− (D)) = 0

1 for every g F , where ξ− (D)= x X : ξ(x) D . By ergodicity of the ∈ 1 { ∈1 ∈ } F -action, either ν(ξ− (D)) = 0 or ν(ξ− (C D)) = 0. Since D is an arbitrary \ 1 measurable subset of C, there exists z C such that ν(X ξ− (z )) = 0. 0 ∈ \ 0 Thus ξ(x)=z0 for almost all x (X,ν). Next suppose that every F -invariant∈ function in L2(X,ν) is constant almost everywhere. Let E be a measurable subset of G such that ν(gE∆E)=0for every g F .LetχE denote the charasteristic function of E on X. Then 2∈ χE L (X,ν) and χE is F -invariant. Therefore it is constant ν-a.e. and hence∈ either ν(E)=0orν(X E)=0. \ Mautner phenomenon - a criterion for ergodicity Deﬁnition.LetG be a locally compact group. For subsets F G and L G, we say that the triple (G, F, L) has the Mautner property ⊂if the following⊂ condition is satisﬁed: For any continuous unitary representation of G on a hilbert space and an element ξ , if Fξ = ξ then Lξ = ξ. H ∈H Remark. Thus if the triple (G, F, G) has the Mautner property and if G acts ergodically on the probability space (X,ν), then due to Lemma 2.4.3, F acts ergodically on (X,ν).

For a subgroup F G, deﬁne ⊂ L(F )= g G : there exist sequences f F , f F , { ∈ { i}⊂ { i }⊂ and m e in G such that f m f g as i . i → i i i → → ∞} Let denote the closed subgroup generated by L(F ).

Example.LetG be a closed subgroup of GLn(C). For g G, let U(g) denote the horospherical subgroup of G associated to g.IfF =∈ gk : k Z then 1 { ∈ } U(g) L(F ) and U(g− ) L(F ). ⊂ ⊂ Lemma 2.4.4 The triple (G, F, < L(F ) >) has the Mautner property.

Proof. Given a continuous unitary representation of G on a hilbert space H and an element ξ , suppose that Fξ = ξ. Then for any f,f F and m G, ∈H ∈ ∈ 1 fmf ξ ξ = mf ξ f − ξ = m ξ ξ . · − · − · · − Now since the representation of G on is continuous, the lemma follows H from the deﬁnition of L(F ).

49 Lemma 2.4.5 Let G = SL2(C). Let g G be such that g is not contained in a compact subgroup of G. Then the∈ triple (G, g ,G) has the Mautner property. { }

Proof. (Cf. [M3, Sect. 2]) First observe that the triple (G, g ,G) has the 1 { } Mautner property if and only if the triple (G, hgh− ,G) has the Mautner {1 } property for any h G.Forg ξ = ξ (hgh− )(h ξ)=h ξ, and Gh ξ = h ξ G ξ = ξ. ∈ · ⇔ · · · · Now⇔ due· to Lemma 2.2.2, it is enough to prove this proposition for g = d(a), 0 < a < 1, and for g = u(1) N.LetF = gk : k Z . | | ∈ { ∈ } 1 First suppose that g = d(a), 0 < a < 1. Then U(g)=N and U(g− )= | | N −. Therefore N L(F ) and N − L(F ). Since G is generated by N and ⊂ ⊂ N − (see Lemma 2.2.6), G =. Therefore by Lemma 2.4.4, (G, F, G) has the Mautner property. This completes the proof for the case at hand. Next suppose that g = u(1). If we show that d(a) L(F ) for some a C∗ with a > 1 then we are through due to the previous∈ case. Now for k ∈ N, | | 10 ∈ put m = . Then k 1/(2k)1

2k k 12k 101 k g m g− = k 01 1/(2k)1 01− 20 = d(2) as k . 1/(2k)1/2 → →∞ Hence d(2). This completes the proof. Notation. Till the end of this chapter, let X denote a homogeneous space of G = SL2(C) admitting a (unique) G-invariant borel probability measure, say ν. Clearly G acts ergodically on (X,ν).

Due to Lemma 2.4.3 and Lemma 2.4.5, we have the following result.

Theorem 2.4.6 Let g G be such that g is not contained in a compact subgroup of G. Then g acts∈ ergodically on (X,ν). In particular, the subgroups A, N1, N, and H act ergodically on (X,ν). 

Some consequences of ergodicity of subgroup actions

The next result is crucial for our proof of Raghunathan’s conjecture for SL2(C).

Theorem 2.4.7 (Cf. [DM2, Sect. 1.7]) Let an C∗ be a sequence such that a as n . Then for every x X{, the}⊂ set | n|→∞ →∞ ∈

d(an)Nx n N ∈ 50 is dense in X. In particular, every AN-orbit in X is dense. (Cf. [DR, Sect. 1.5])

Proof.Let

Y = d(an)Nx. n N ∈ Then Y is N-invariant. For r ,r 0, deﬁne 1 2 ≥ ab U(r ,r )= G : c r , a 1 r . 1 2 cd ∈ | |≤ 1 | − |≤ 2 

Take r1 > 0 and r2 (0, 1/2). Then the set U(r1,r2)x contains a neighbourhood of x in X, and∈ hence

ν(U(r1,r2)x)=α>0. (2.8)

Now for any a C∗, n ∈ 1 2 d(a )U(r ,r )d(a− )=U( a − r ,r ). n 1 2 n | n| 1 2 Therefore,

ν(U(r1,r2)x)=ν(d(an)U(r1,r2)x) 2 = ν(U( an − r1,r2)d(an)x) | | 2 ν(U( a − r ,r )Y ). (2.9) ≤ | n| 1 2 Since a ,foreveryn N there exists m n such that | n|→∞ ∈ ≥ n 2 2 U( a − r ,r )Y U( a − r ,r )Y. (2.10) | k| 1 2 ⊃ | m| 1 2 k=0 Note the following statements. (i) Y is closed and N-invariant.

(ii) U(r1,r2)/N is a compact subset of G/N.

2 (iii) U( a − r ,r )/N U(0,r )/N in G/N as k . | k| 1 2 ↓ 2 →∞ Therefore ∞ 2 U( a − r ,r )Y = U(0,r )Y. (2.11) | k| 1 2 2 k=0 Now since ν is ﬁnite, we conclude that

0 <α ν(U(0,r )Y ) ν(DY ). (2.12) ≤ 2 ≤ 51 Since N acts ergodically on (X,ν), by Lemma 2.4.1 there exists ω D and y Y such that X = N(ωy)=ωNy. Hence Y Ny = X. This completes∈ the∈ proof. ⊃ The last part can also be argued alternatively as follows: Since U(0,r ) 2 ⊂ NG(N), the set U(0,r2)Y is N-invariant. From Eq. 2.12 and the ergodicity of the N-action, ν(U(0,r2)Y )=1.

Since this holds for all r2 > 0, we have that ν(Y ) = 1. Now X Y is open and ν(X Y ) = 0. Therefore X = Y . \ \ Now we prove a version of the Borel density theorem for SL2(C). Although it is not essential for the proof of our main theorem, it is of interest in its own right.

Lemma 2.4.8 ([S1, Sect. 2.11]) Let V be a ﬁnite dimensional vector space and ρ : G GL(V ) be a continuous homomorphism. Fix an element x X → ∈ and let Gx denote its stabilizer in G. Then any Gx-invariant subspace W of V is G-invariant.

Proof.Letk = dim W and define a homomorphism kρ : G GL( kV )as follows: For every g G and every subset v ,...,v ∧ V , → ∧ ∈ { 1 k}⊂ kρ(g)(v v )=(ρ(g)v ) (ρ(g)v ). ∧ 1 ∧···∧ k 1 ∧···∧ k One can verify that kρ is a continuous homomorphism. ∧ Now kW is a one-dimensional subspace of kV . Note that any g G, ∧ ∧ ∈ ρ(g)W = W kρ(g)( kW )= kW. ⇔∧ ∧ ∧ Thus replacing V by its suitable exterior power, without loss of generality, we can assume that W is one dimensional. Let P1(V ) denote the set of all one-dimensional subspaces of V .Letπ : V 0 P1(V ) be the map defined such that for every nonzero vector v V , π(\{v) is}→ the one-dimensional subspace of V spanned by v. Equip P1(V )∈ with the finest topology for which the map π is continuous. The linear G action ρ on V induces a ‘projective linear’ actionρ ¯ of G on P1(V ) as follows: For every g G and every v V 0 , we define ∈ ∈ \{ } ρ¯(g)π(v)=π(ρ(g)v).

It is straight forward to verify that the actionρ ¯ : G P1(V ) P1(V ) is continuous. × → By Theorem 2.4.6, the element u(1) N acts ergodically on X.By Lemma 2.4.1, there exists y X such that∈ the set u(n)y : n N is dense ∈ { ∈ } 52 in X.Leth G be such that y = hx. Put W = ρ(h)W . Then W is Gy- ∈ 1 invariant. Now if we show that W is G-invariant then W = ρ(h− )W = W . Therefore replacing x by y, without loss of generality we may assume that

X = u(n)x : n N . { ∈ } Fix w W 0 . Then for every δ G ,wehave ∈ \{ } ∈ x ρ¯(δ)π(w)=π(w),

where π andρ ¯ are deﬁned as above. Fix an orthonormal basis e1,...,en of V with respect to some inner product on V . By Lemma 2.1.6,{ ρ(u(t))} : t R is a one-parameter group of unipotent transformations of V .{ Therefore∈ by Lemma} 2.1.5 there exist real polynomials φ ,...,φ such that for every t R, 1 n ∈ n

φ(t)=ρ(u(t))w = φi(t)ei. i=1 2 n 2 Now for every 1 i n, the rational function φi (t)/ j=1 φj (t)convergesas t . Therefore≤ there≤ exists p V 0 such that →∞ ∈ \{ } lim φ(t)/ φ(t) = p. t →∞

Since G = u(n):n N Gx, given any g G there exist sequences nk N and δ {G such that∈ }n and u(n∈)δ g as k . Now{ }⊂ { k}⊂ x k →∞ k k → →∞

ρ¯(g)π(w) = lim ρ¯(u(nk)δk)π(w) k →∞ = lim ρ¯(u(nk))π(w) k →∞ = lim π(φ(nk)) k →∞ = π(p).

Putting g = e,wegetπ(p)=π(w). Henceρ ¯(g)π(w)=π(w) for all g G. ∈ Thus W is G-stable. 

Corollary 2.4.9 Let X be a nontrivial homogeneous space of G = SL2(C) admitting a ﬁnite G-invariant measure. Then the stabilizer of every point of X is a discrete subgroup of G. In other words, there exists a lattice Γ in G such that X ∼= G/Γ.

Proof. Fix x X.LetF be the connected component of e in Gx. Thus F is a closed connected∈ subgroup of G.

53 Let denote the Lie algebra of F . Then is also the Lie algebra of G .By F F x Theorem 2.1.3, is a Vector subspace of and there exists r0 > 0 such that the map exp : F (0) G is a homeomorphismG onto a neighbourhood F∩Br0 → x of e in Gx. Thus F is an open subgroup of Gx. To prove that Gx is discrete, we should show that F = e . Now consider the Adjoint representation Ad : { } G GL( ). Then Ad(Gx) = . →By LemmaG 2.4.8, Ad(g)F = F for every g G. Therefore by Lemma 2.2.2 and Corollary 2.2.10, eitherF =0orF = . Since∈ F = G and F is connected, we have F = e . Hence G Fis discrete.F G { } x Corollary 2.4.10 Let Γ be a lattice in G and F be a proper closed subgroup of G containg Γ. Then F is a lattice in G, and hence Γ is a subgroup of ﬁnite index in F .

Proof.Letπ : G/Γ G/F be the natural quotient map which preserves the G-actions on both the→ spaces. Let µ be the unique G-invariant probability measure on G/Γ. Then π (µ) is a G-invariant probability measure on G/F . ∗ Now since F is a proper closed subgroup of G, by Corollary 2.4.9, F is a discrete subgroup of G. Hence F is a lattice in G. Applying the arguments of this section to the group SL2(R) one can prove the following:

Theorem 2.4.11 Let Y be a homogeneous space of H = SL2(R) admitting a ﬁnite H-invariant measure. Then

(1) Any subgroup of H which is not relatively compact acts ergodically on Y .

(2) Y ∼= H/Γ for a lattice Γ in H. (3) For any x Y and a sequence a in R, the set ∈ i →∞

d(ai)N1x i N ∈ is dense in Y .

In particular, every AN1-orbit in Y is dense.

54 Chapter 3

L-homogeneous spaces of SL2(C)

Let G = SL2(C), let Γbe a discrete subgroup of G, and denote by X the homogeneous space G/Γ. Our main objective is to study closures of orbits of one-parameter unipotent subgroups of G acting on X, when X admits a finite G-invariant measure; i.e. when Γis a lattice in G. In our method we often need to construct convergent sequences in X with some ‘control’ over their limit points. It turns out that if we assume X to be compact or more generally, the orbit closure to be compact, then such constructions can be easily justified by general topological arguments. But when X is noncompact, one needs to use the fact that X admits a finite G-invariant measure in order to make the constructions. We shall first study in detail the geometric structure of the complement of a large compact subset in X. The construction of convergent sequences will then be made using the structural properties of the complement of the compact set (see Corollaries 3.4.6 and 3.4.7). Note that by Lemma 1.3.6, X is an L-homogeneous space of G; in other words, Γis an L-subgroup of G. Being an L-homogeneous space is a topological condition on a homogeneous space which is, a priori, a weaker condition than the measure-theoretic condition of admitting a finite invariant measure. For the results of this chapter it is enough to assume that X is an L-homogeneous space of G. In fact we show that as a consequence of these results, any L- homogeneous space of G admits a finite G-invariant measure (Corollary 3.4.3).

3.1 A result of Kazhdan and Margulis

The following obeservation is important in understanding the contrast between the noncompact and the compact L-homogeneous spaces of G.

Lemma 3.1.1 Suppose X is compact. Then X does not contain any compact orbit of a nontrivial unipotent subgroup of G; in other words, the stabilizer of a point in X does not contain a nontrivial unipotent element.

55 Proof.LetU be a nontrivial unipotent subgroup of G. Suppose that for some x X, the orbit Ux is compact. Since Ux = Ux = Ux, replacing U by U,we ∈ may assume that U is closed. Then by Corollary 1.1.2, the map φ : U/Ux X given by φ(gU )=gx ( g U) is proper. Hence U/U is compact. Since→U is x ∀ ∈ x noncompact, Ux contains a nontrivial element, say u. Therefore there exists an n n n n element g G such that g ug− e as n . But g ug− Ggnx and X is an L-homogeneous∈ space of G.→ Therefore→∞ the sequence gnx∈ is divergent, { } a contradiction to the hypothesis that X is compact. In the remaining chapter we shall prove the following particular case of a more general theorem due to Kazhdan and Margulis, by specializing their method for G = SL2(C). The result, in particular, shows that every point of a noncompact L-homogeneous space X is ﬁxed by a nontrivial unipotent element of G.

Theorem. Let X be a noncompact L-homomogeneous space of G. Then there exists a neighbourhood Ω of e in G such that for any x X for which { } ∈ Gx Ω = e , then there exists a unique horospherical subgroup U of G such that∩G Ω{ } U. x ∩ ⊂ First we introduce the concept of Zassenhaus neighbourhood, which plays a crucial role in the proof.

Notation.Fors1,s2 GLn(C), we denote by [s1,s2], the commutator of s1 ∈1 1 and s , namely s s s− s− . For subsets S ,S GL (C), deﬁne 2 1 2 1 2 1 2 ⊂ n [S ,S ]= [s ,s ]:s S ,i=1, 2 . 1 2 { 1 2 i ∈ i } (k) (0) (k) For S GLn(C) and k 0, deﬁne S inductively as: S = S and S = (k ⊂1) ≥ [S, S − ]fork 1. ≥

Lemma 3.1.2 Let A, B Mn(C) be such that A < 1 and B < 1. Put g =1 A and h =1 B∈. Then − − 1 1 [g, h] 1 2 A B (1 A ) − (1 B )− . (3.1) − ≤ · · − · − Proof.

1 1 ghg− h = g(1 B)g− (1 B) − − 1 − − = B gBg− − 1 = B gB(1 + Ag− ) − 1 = B (1 A)B gBAg− − − 1− = AB gBAg− . − 56 Therefore

1 1 [g, h] 1 A B (1 + g g− ) h− − ≤· · · · 1 1 A B (1 + (1 + A )(1 A )− ) (1 B )− ≤· · 1 − 1 · − =2A B (1 A )− (1 B )− . · · − · − This completes the proof.

Lemma 3.1.3 There exists a neighbourhood Ω of e in GLn(C) such that the set Ω(k) e as k . ↓{ } ↑∞ Proof. Deﬁne Ω= g GL (C): g 1 < 1/8 . { ∈ n − } It is enough to prove that for any k 0 and g Ω(k), ≥ ∈ k g 1 (1/8)2− . (3.2) − ≤ For k = 0 the Eq. 3.2 obviously holds. Now suppose it holds for some k 0. By Lemma 3.1.2, for any g Ωand h Ω(k), ≥ ∈ ∈ k 1 [g, h] 1 (1/8)2− − . − ≤ Thus by induction, Eq. 3.2 holds for all k 0. ≥ Deﬁnition. A group F is called nilpotent if F (k) = e for some k N. { } ∈

Deﬁnition.LetG be a closed subgroup of GLn(C). A neighbourhood Ωof the identity e in G is called a Zassenhaus neighbourhood if for every discrete subgroup Γof G, the set Ω Γis contained in a closed connected nilpotent subgroup of G. ∩

It is a fact that every closed subgroup of GLn(C) admits a Zassenhaus neighbourhood; (see [R, Sect. 8.16] for a proof). Since we shall deal only with G = SL2(C), here we give a proof just for this case.

Proposition 3.1.4 The group G = SL2(C) admits a Zassenhaus neighbourhood. Moreover, if Ω is a Zassenhaus neighbourhood then for any discrete subgroup Γ G, the set Ω Γ is contained in a conjugate of either D or N. ⊂ ∩ Proof. LetΩbe as in the proof of Lemma 3.1.3. Given a discrete subgroup Γ G, put ∆=Γ Ω. Then ∆(k) =Ω(k) Γ. If∆ = e then there is nothing to⊂ be proved. Otherwise,∩ since Ω(k) e∩ as k {and} since Γis discrete, (k 1) ↓{} ↑∞(k) there exists a k 1 such that ∆ − = e and ∆ = e . ≥ { } { } 57 (k 1) Let h ∆ − e . Then h commutes with every element of ∆. By ∈ \{ } 1 1 Lemma 2.2.2 there exists g G such that ghg− D or ghg− C(G)N. ∈ ∈ ∈ 1 Since h Ω, h is not in the center of G. Therefore the centralizer of ghg− is contained∈ in either D or C(G)N. Note that Ω C(G)U =Ω U for any ∩1 1∩ horospherical subgroup U of G. Hence either g∆g− D or g∆g− N. This ⊂ ⊂ completes the proof. We need to setup some notations and deﬁnitions before we proceed to the proof of the main theorem of this section.

Notation.Let denote the Lie algebra associated to G. Then G = A M (C):tr(A)=0 . G { ∈ 2 } The space inherits the operator norm, denoted by ,fromM2(C)= 2 G · End(C )(see 2.1). For r>0, let r denote the open ball of radius r in centered at§ the origin. Put B =expB G. By Lemma 2.1.1, there G r Br ⊂ exists r0 > 0 such that the map exp : r0 Br0 is a homeomorphism. Let log : B denote its inverse. For gB B→ , deﬁne r0 →Br0 ∈ r0 g = log(g) . | | Now let X be an L-homogeneous space of G. Recall that for any neighbourhood Ωof e in G, X = x X : G Ω = e . Ω { ∈ x ∩ { }} Deﬁne a function R : XB (0,r0) in the following way: For every x XB , r0 → ∈ r0 R(x) = inf δ : δ G B e . {| | ∈ x ∩ r0 \{ }} Note the following properties of the map R: (i) R is continuous.

(ii) For every k K and x XB ,wehaveR(kx)=R(x). ∈ ∈ r0 (iii) R vanishes at the inﬁnity.

For convenience we choose r0 > 0 small enough so that Br0 is a Zassenhaus neighbourhood in G. Now by Proposition 3.1.4, for any x XB , the set ∈ r0 Gx Br0 is contained in a conjugate of either D or N. Therefore Gx Br0 exp(∩ ), where = Ad g( )or = Ad g( ) for some g G; (in fact∩ every⊂ maximalL abelianL Lie subalgebraD L is ofN this form). ∈ L⊂G Lemma 3.1.5 ([R, Sect. 11.6]) Given c>1, there exists a compact set E = E(c) G such that the following condition is satisﬁed: For any abelian Lie subalgebra⊂ , there exists g E such that for every x , L⊂G ∈ ∈L Ad(g)x c x . ≥

58 Proof. Given any abelian Lie subalgebra , there exists k K such that either Ad k( ) or Ad k( ) Ad v( ) forL some v N. ∈ SinceL ⊂ AdNK preservesL the⊂ normD on and since∈ K is compact, without loss of generality we may assume that eitherG = or = Ad v( ). If = then for any x and a>L0, N L D L N ∈L Ad(d(a))x = a2x. (3.3)

1 z Now suppose = Ad v( ), where v = and z C.Letξ C L D 01 ∈ ∈ be such that ξ = 1 and z + ξ = z + 1; that is ξ= z/ z if z =0.Forany a>0, put g =| |d(a)u(ξ) | N (N| ).| Now| any x is of the| | form ∈ G ∈L y 0 1 2z x = Ad v = y 0 y 0 − 1 − − for some y C. Therefore ∈ 10 Ad g(x)=y Ad(d(a)) Ad(u(z + ξ)) 0 1 − 1 2a2(z + ξ) = y . 0 − 1 − Note that for any t C, ∈ 1 t ( t 1) t +1. | | |− |≤ 0 1 ≤|| − Therefore if a>1 then Ad g(x) 2a2( z +1) 1 | | − a2. (3.4) x ≥ 2 z +1 ≥ | | Therefore due to Eq. 3.3 and Eq. 3.4, the compact set

E = E(c)=d(√c)N(1)K G ⊂ has the required properties, where

N(1) = u(z) N : z C, (z) 1, (z) 1 . { ∈ ∈ | |≤ | |≤ } Fix c>1. Let E = E(c) be a compact subset of G as in Lemma 3.1.5. Since Ad E is a compact subset of End( ), the operator norm of Ad g is bounded for G all g E. Therefore there exist 0

cR(x)

(3) For any t (0,r ], ∈ 2 1 G B g(G B 1 )g− . gx ∩ t ⊂ x ∩ c− t Proof. (cf. [R, Sect. 11.7]) There exists an abelian subalgebra such L⊂G that Gx Br0 exp . Now by Lemma 3.1.5 there exists a g E such that ∩ ⊂ L 1 ∈ for every γ exp B , gγg− B and ∈ L∩ r1 ⊂ r0 1 gγg− c γ . | |≥ | | This veriﬁes (1). 1 Note that g(Gx Br3 )g− Ggx Br2 . Therefore R(gx) 0. ·

Proposition 3.1.7 For any x X with R(x) r . { ∈ ∪{ } 3} We prove the proposition by induction on n(x). If n(x) = 0 then R(x) >r3 and the proposition holds for n =0andg = e.

60 Now suppose n(x) 1. Then R(x) r3. By Proposition 3.1.6, there exists h E such that if we≥ put x = hx then≤ ∈ 1

cR(x)

and for every t (0,r ), ∈ 2 1 G B h(G B 1 )h− . (3.5) x1 ∩ t ⊂ x ∩ c− t Hence n(x )

and 1 G B g (G B n1 )g− . (3.6) g1x1 ∩ r2 ⊂ 1 x1 ∩ c− r2 1 Now from Eq. 3.5 and Eq. 3.6, the conclusion of the proposition holds for n = n1 + 1 and g = g1h. The following result is an immediate interesting consequence of Proposi- tion 3.1.7.

Theorem 3.1.8 ([R, Sect. 11.8]) There exists a neighbourhood Ω of e in G such that for every discrete subgroup Γ G, there exists g G such that ⊂ ∈ 1 gΓg− Ω= e . ∩ { } Proof. Put X = G/Γand Ω= B .Letx = eΓ X.IfΓ Ω = e r3 ∈ ∩ { } then by Proposition 3.1.7, there exists g G such that R(gx) >r3. Thus 1 ∈ gΓg− Ω=G = e . ∩ gx ∩Br3 { } Theorem 3.1.9 Let X be an L-homomogeneous space of G. Then there exists δ0 > 0 such that for any x X with R(x) <δ0, there exists a unique horospherical subgroup U of G such∈ that

G B U. x ∩ r0 ⊂ Proof. (cf. [R, Sect. 11.10]) Fix a point x X. Since X is an L-homogeneous 0 ∈ space, there exists a compact set C G such that X XB Cx0.Forany ⊂ \ r3 ⊂ y X XB take h C such that y = hx0. Then ∈ \ r2 ∈ 1 1 h− (G B )h G C− B C =: S. (3.7) y ∩ r2 ⊂ x0 ∩ r2 1 Now S is a ﬁnite set, because Gx0 is discrete and C− Br2 C is relatively compact in G.LetS be the set of all unipotent elements in S and put S = S S . 1 2 \ 1 61 Since every non-unipotent element of G has an eigenvalue diﬀerent from 1 and since eigenvalues are preserved under conjugation, there exists δ (0,r2) such that no element of S has a conjugate in B .Letp N be such∈ that 2 δ ∈ p c δ>r2.

Since E is compact, there exists δ (0,r ) such that for all g Ep, 0 ∈ 3 ∈ 1 gB g− B . (3.8) δ0 ⊂ r3

Take x X such that R(x) <δ0. From Proposition 3.1.7, it follows that there exists∈n N and g En such that ∈ ∈

Because R(x) <δ0, due to Eq. 3.8 and Eq. 3.9, we have n>p. Therefore n c− r2 <δ. By Eq. 3.7, the element γ has a conjugate in S,sayγ. Then by 1 Eq. 3.10, γ has a conjugate g− γg Bδ. Therefore γ S2. Therefore γ S1. 1 ∈ ∈ ∈ Hence γ G and g− γg G are nontrivial unipotent elements of G. ∈ gx ∈ x Since a conjugate of D cannot contain a nontrivial unipotent element, by Proposition 3.1.4, there exists a horospherical subgroup U of G such that G B U. The uniqueness of U follows from Corollary 2.2.7. x ∩ r0 ⊂ 3.2 Non-divergence of unipotent trajectories

In this section we shall show that unipotent trajectories on L-homogeneous spaces are nondivergent. As a consequence we show that every horospherical subgroup of G admits a compact orbit on X.

Lemma 3.2.1 Let X be a noncompact L-homogeneous space of G. Suppose that for g G and x X, the sequence gnx : n N has no convergent subsequence.∈ Then there∈ exists a unique horospherical{ ∈ subgroup} U of G such that for all large n N, ∈ n n G n B g (G U)g− . g x ∩ r0 ⊂ x ∩ 1 Proof. ([GR, Lemma 3.5]) Let δ (0,δ0) be such that gBδg− Br0 , where δ0 is deﬁned as in Theorem 3.1.9. Since∈ X is an L-homogeneous space,⊂ R(gnx) 0asn . Without loss of generality we may assume that R(gnx) <δfor all→ →∞ 62 n 0. By Theorem 3.1.9, for each n 0 there exists a unique horospherical ≥ ≥ subgroup Un such that G n B U . g x ∩ r0 ⊂ n n Since R(g x) <δ, we have that G n B = e . Also g x ∩ δ { } 1 g(G n B )g− G n+1 B . g x ∩ δ ⊂ g x ∩ r0 1 1 Therefore gUng− Un+1 = e . By Corollay 2.2.7, gUng− = Un+1. Thus for all n 0, ∩ { } ≥ n n Un = g U0g− and hence n n G n B g (G U )g− . g x ∩ r0 ⊂ x ∩ 0 This completes the proof. Lemma 3.2.2 Let g G and U be any nontrivial horospherical subgroup of ∈ G. Suppose that there exits a sequence un : n N U Ω such that n n { ∈ }⊂ \ g ung− e as n , where Ω is a neighbourhood of the identity in U. Then g is→ a semisimple→∞ element and U = U(g), the horospherical subgroup of G associated to g.

Proof. ([GR, Lemma 3.6]) First observe that there is no loss of generality, if we replace g with its conjugate by an element of G. Due to Lemma 2.2.2, the element g has a conjugate in either D or C(G)N. Therefore we may assume that only one of the following is possible: Either g = d(a) for some a C∗ with a 1org = u(z) for some z C. Since g is not contained in a compact∈ subgroup| |≤ of G, in first case we have∈ a < 1. Define | | xz = : x, z C B 0 x ∈ − and 00 − = : z C . N z 0 ∈ Then = − . LetG Nbe the⊕B Lie algebra associated to U. Since U is a horospherical U subgroup of G, the map exp : U is a homeomorphism. Now put xn = 1 U→ exp− (un) for every n 0. Since un Ω, there exits c>0 such that ∈U ≥ n n ∈ x c for all n 0. Since g u g− e as n , we have that n≥ ≥ n → →∞ n lim Ad g (xn) =0. (3.11) n →∞ Suppose that = 0 . Since every element of is nilpotent, . Therefore by CorollaryU∩B 2.2.7,{ } = . Now due to Eq.U 3.11, g N.U Therefore∩B⊂N U N ∈ 63 g = d(a)forsomea C∗, a < 1 and hence U = N = U(g). This completes the proof for the case∈ at hand.| | Now suppose that = 0 .Letp : = − − be the projection on the first factor.U∩B According{ } to the hypothesis,G N ⊕pBis→ injectiveN when restricted to . Therefore there exists µ>0 such that for all x , U ∈U p(x) µ x . ≥ 2 For every y −, Ad(d(a))y = a− y and p(Ad(u(z))y)=y. Therefore ∈N p(Ad g(x)) = p(Ad g(p(x))) for all x , and ∈G p(Ad g(y)) y ≥ for all y −. Thus for every x and n 1, ∈N ∈U ≥ p(Ad gn(x)) = p(Ad gn(p(x))) p(x) µ x µc, ≥ ≥ ≥ a contradiction to Eq. 3.11. This completes the proof. We will combine the above two lemmas to obtain the main result of this section.

Theorem 3.2.3 Let X be an L-homogeneous space of G. Suppose that for an element g G and a point x X, the sequence gnx : n N diverges ∈ ∈ { ∈ } as n . Then g is semisimple and there exists γ Gx e such that n →∞n ∈ \{ } g γg− e as n . In particular,→ a→∞ unipotent element u G does not have a divergent trajectory on X; that is, for any x X, there∈ exists a compact set C = C(x) X such that unx C for inﬁnitely∈ many n N. ⊂ ∈ ∈ Proof. By Lemma 3.2.1, there exists a unique nontrivial horospherical subgroup U such that for all large n N, ∈ n n G n B g (G U)g− . g x ∩ δ0 ⊂ x ∩ n Since R(g x) 0asn , there exist un Gx U e such that n n → →∞ ∈ ∩ \{} g ung− e as n . Therefore by Lemma 3.2.2, g is a semisimple element → →∞ n n and U = U(g). Hence for any γ G U, g γg− e as n . ∈ x ∩ → →∞ Compact orbits of horospherical subgroups Corollary 3.2.4 (Cf. [GR, Lemma 3.16]) Let X be a noncompact L-homogeneous space of G. Let x X and a horospherical subgroup U of G be such ∈ that U Gx = e . Then U/Ux is compact; equivalently, the orbit Ux is compact.∩ { }

64 Proof.Letu Gx U e . Because U is a horospherical subgroup, there exists ∈ ∩ \{ } 1 1 g NG(U) such that gug− <δ0. Also gug− Ggx U. Since g(Ux)=Ugx, the∈ orbit Ux is compact| if| and only if the orbit∈ Ugx∩ is compact. Therefore replacing x by gx, we may assume that there exists

γ G B e U. ∈ x ∩ δ0 \{ }⊂ Let Ωbe an open neighbourhood of e in G such that for every ω Ω, ∈ 1 ωB ω− B δ0 ⊂ r0 and 1 ΩγΩ− G = γ . ∩ x { } Note that for any y Ux, G U = G U. Since the set of ﬁxed points ∈ y ∩ x ∩ of γ in X is a closed subset, γ Gz for every z Ux.Takez Ux. There exists an ω Ωsuch that y = ω∈z Ux. Thus ∈ ∈ ∈ ∈ 1 ωγω− G B G U G . ∈ y ∩ r0 ⊂ y ∩ ⊂ x 1 1 Therefore ωγω− = γ, and hence ω U and in turn z = w− y Ux. This proves that Ux is closed. (Notice that∈ to prove that Ux is closed,∈ we only use the fact that Gx is discrete; the L-homogeneous condition is not used). Now by Lemma 1.1.2, the map π : U/U X deﬁned by π(uU )= x → x ux ( u U) is proper. Recall that U/Ux is a topological group and it is ∀ ∈ 2 isomorphic to R /∆, where ∆is a discrete subgroup. Therefore if U/Ux is not n compact, there exists u U such that the sequence u Ux : n N U/Ux ∈ { n∈ }⊂n has no convergent subsequence. Since the map π is proper, π(u Ux)=u x in X as n . This contradicts Theorem 3.2.3. Hence U/U is compact.→ ∞ →∞ x We combine Theorem 3.1.9 and Corollary 3.2.4 in the following result.

Theorem 3.2.5 Let X be a noncompact L-homogeneous space of G. Then there exists a neighbourhood Ω of the identity e in G such that for any { } x X for which Gx Ω = e , there exists a unique horospherical subgroup U ∈of G such that G ∩ Ω U{ ;} in this case the orbit Ux is compact. x ∩ ⊂

3.3 Volumes of compact orbits of horospherical subgroups

Deﬁne a positive deﬁnite innerproduct , on the Lie algebra as follows: · · G For x, y , x, y =tr(xy∗). Let denote the associated norm on . ∈G ·1 G 65 Then for any x =(x ) , ij ∈G

2 x = x 2. 1 | ij| i,j=1 Note that for every k K, Ad k(x) 1 = x 1. Recall that for any∈ nontrivial horospherical subgroup U of G and its associated Lie subalgebra , the map exp : U is an isomorphism of the topological groups. NowU⊂Ginherits a euclideanU→ metric from , and hence it admits the standard lebesgueU measure associated to the metric.G This measure on induces a unique haar measure on U,sayσ . U U Note that for any other nontrivial horospherical subgroup U of G, there 1 exists k K such that U = kUk− . It is straight forward to verify that for any measurable∈ subset E U, ⊂ 1 σU (kEk− )=σU (E).

Now suppose that for some x X the orbit Ux is compact. Then Ux = ∈ ∼ U/Ux admits a ﬁnite volume with respect to σU denoted by Vol(Ux). Now 1 if =exp− (U ) then is a lattice in , and Vol(Ux) = covol( )(see Ux x Ux U Ux 1.5 and Lemma 1.3.4). Let x1, x2 form a basis of x as a free abelian group.§ Then by Theorem 1.5.1,{ }⊂U U

Vol(Ux)= x x , 1 ∧ 21 2 where 1 denotes the euclidean norm on induced by the norm 1 on . · ∧ G · G Lemma 3.3.1 The volumes of compact orbits of horospherical subgroups have the following properties.

1. For every g G, ∈ def 1 2 Vol(gUx) = Vol((gUg− )gx)= Ad g(x x ) . ∧ 1 ∧ 2 1 2. For every k K, ∈ Vol(kUx) = Vol(Ux).

3. For every h N (U), ∈ G Vol(Uhx) = Vol(hUx)= det(Ad h ) Vol(Ux). | |U |·

(Recall that by Iwasawa decomposition G = KNG(U).)

66 Proof. Statement 1 follows from the above discussion. Statement 2 follows from Statement 1 and the equality

2 Ad k(x x ) = x x . ∧ 1 ∧ 2 1 1 ∧ 21 Statement 3 follows from Statement 1 and the equality

2 Ad h(x1 x2) = det(Ad h ) (x1 x2). ∧ ∧ |U · ∧ 

A function V on an L-homogeneous space

Let δ0 > 0 be as in Theorem 3.1.9. Then for every x X with R(x) <δ0, there exists a unique horospherical subgroup of G, denoted∈ by U(x), such that

Gx Br0 U(x). Let (x) denote the Lie subalgebra of associated to U(x∩). Recall⊂ that due toU Corollary 3.2.4, the orbit U(x)x is compact.G Deﬁne a function V : XB (0, ) such for every x XB , δ0 → ∞ ∈ δ0 V(x) = Vol(U(x)x).

Lemma 3.3.2 (1) is uniformly continuous. (2) is bounded on X X V V Bδ0 Bδ for any δ (0,δ ). \ ∈ 0 Proof. Given α>1, let Ωbe a neighbourhood of e G such that Ω= 1 1 ∈ Ω− and for every g Ω, (i) gBδ0 g− Br0 and (ii) the operator norm 2 1 ∈ ⊂ Ad g (α− ,α). Then for any x X and g Ω, if gx XB then 1 Bδ0 δ0 ∧ ∈ 1 ∈ ∈ ∈1 Ggx Br0 g(Gx Bδ0 )g− = e . Therefore U(gx)=gU(x)g− and hence by Lemma∩ ⊃ 3.3.1, ∩ { }

1 1 V(gx) = Vol((gU(x)g− )gx) = Vol(gU(x)x) (α− V(x),αV(x)). ∈

This equation implies (1). Since the set XB XB is relatively compact, the δ0 \ δ uniform continuity implies (2). 

Lemma 3.3.3 V vanishes at the point at inﬁnity (for X).

Proof. Take a sequence x : i N X which has no convergent subse- { i ∈ }⊂ quence. Since X is an L-homogeneous space, R(xi) 0asi .Wemay assume that R(x ) <δ for every i N. → →∞ i 0 ∈ Let λi 1 be such that λiR(xi) (δ0/2,δ0) and let gi NG(U(xi)) be such that Ad≥ g (x)=λ x for every x ∈ (x ). Then ∈ i i ∈U i R(g x )=λ R(x ) (δ /2,δ ), i i i i ∈ 0 0 67 1 U(gixi)=giU(xi)gi− , and 2 V(gixi)= det(Ad gi (xi)) V(xi)=λi V (xi). | |U |· Since R(xi) 0asi , we have that λi as i . Now V is → →∞ →∞ →∞ bounded on XBδ XBδ /2 . Therefore V(xi) 0asi . 0 \ 0 → →∞

Lemma 3.3.4 (Cf. [D3, Lemma 2.4]) Let u(t) t R be a unipotent one- ∈ parameter subgroup of G. For an x X and an{ interval} I R, suppose that R(u(t)x) <δ for all t I. Then ∈ ⊂ 0 ∈ U(u(t)x)=u(t)U(x)u( t) − for all t I and there exists a polynomial φ : R R of degree at most 4 such that for∈ all t I, → ∈ V 2(u(t)x)=φ(t). Moreover, if I is unbounded then t V(u(t)x) is a constant function on R, U(u(t)x)=U(x) for all t R, and →u(t):t R U(x). ∈ { ∈ }⊂

Proof.Let>0 be such that u(t)Bδ0 u( t) Br0 for all t <. Express I as a union of countably many open intervals− ⊂ of length at| | most . It is enough to prove the result for each interval separately. Also there is no loss of generality if we replace x by u(t)x and I by I t for any t R. Therefore, without loss of generality, we may assume that−I ( ,) and∈ 0 I. Thus G B u(t)(G B )u( t), and hence ⊂ − ∈ u(t)x ∩ r0 ⊃ x ∩ δ0 − U(u(t)x)=u(t)U(x)u( t) (3.12) − for all t I. ∈ 1 Let U = U(x), = (x) and x =exp− (Ux) .Let x1, x2 be a basis of the latticeU . LetU the functionU ψ : R ⊂2 Ube deﬁned{ by}⊂U Ux →∧G ψ(t)= 2 Ad(u(t))(x x )(t R). (3.13) ∧ 1 ∧ 2 ∀ ∈ Since 2 Ad(u(t)) : t R is a one-parameter group of unipotent linear transformations{∧ of 2 ,∈ the} co-ordinates of ψ with respect to a basis of 2 are polynomials of∧ degreeG at most 2 (see Lemma 2.1.5 and note that taking∧ G the exterior product simply adds the degrees). By Lemma 3.3.1, Eq. 3.12, and Eq. 3.13, we have V(u(t)x)= ψ(t) for all t I. Now the function φ(t) def= ψ(t) 2 is a polynomial of degree at most ∈ 2 dim( 2 ). Thus V2(u(t)x)=φ(t) for all t I. This proves the ﬁrst part. · Now∧ supposeG that I is unbounded. Since ∈V is a bounded function, φ is a constant polynomial. Hence ψ must also be a constant function; that is,

2 Ad(u(t))(x x )=(x x ) for all t R. ∧ 1 ∧ 2 1 ∧ 2 ∈ 68 Hence Ad(u(t)) = for all t R. Therefore U(u(t)x)=U(x) for all t R and by CorollaryU 2.2.7,U u(t):t∈ R U(x). This completes the proof. ∈ { ∈ }⊂ 1 Notation.Letδ1 (0,δ0) be such that B− Bδ Bδ .LetU be a nontrivial ∈ δ1 1 ⊂ 0 horospherical subgroup of G and σU denote the haar measure on U as deﬁned earlier. Put ξ = σ (B U). 0 U δ1 ∩

Lemma 3.3.5 1. ξ0 is independent of the choice of U.

2. Suppose for an x X and a nontrivial horospherical subgroup U of G, ∈ the orbit Ux is compact and Vol(Ux) <ξ0. Then U = U(x), and hence Vol(Ux)=V(x).

Proof. To verify (1), let U be any nontrivial horospherical subgroup of G. 1 1 Then there exists k K such that U = kUk− . Since kBδ1 k− = Bδ1 ,we have ∈ 1 σ (B U )=σ (k(B U)k− )=σ (B U)=ξ . U δ1 ∩ U δ1 ∩ U δ1 ∩ 0 For (2), by Lemma 1.3.4 and the deﬁnition of δ , we have G B U = e . 1 x∩ δ0 ∩ { } Therefore (2) holds by the deﬁnitions of U(x) and V(x).

Notation.Forξ (0,ξ ], deﬁne ∈ 0

Xξ = x X : V(x) ξ { ∈ ≤ } and C = X X . ξ \ ξ Then Xξ is closed and Cξ is open. By Lemma 3.3.3, Cξ is relatively compact and C X as ξ 0. ξ ↑ ↓ For a nontrivial horospherical subgroup U of G, deﬁne

X(U)= x X : the orbit Ux is compact { ∈ } and for η>0, deﬁne

X (U)= x X(U) : Vol(Ux) <η . η { ∈ } Lemma 3.3.6 Let U be a nontrivial horospherical subgroup of G. Then the following statements hold.

1. For any k K and η>0, ∈ 1 kXη(U)=Xη(kUk− ).

69 2. For any g N (U) and η>0, ∈ G

gXη(U)=Xη (U),

where η = det (Ad g ) η. | |U · 3. For any nontrivial horospherical subgroup U = U, X (U) X (U )= . ξ0 ∩ ξ0 ∅

4. For any ξ (0,ξ0], ∈ X = K X (U). ξ · ξ Proof. Statements 1 and 2 follow from the deﬁnitions and Lemma 3.3.1. State- ment 3 follows from Lemma 3.3.5. For Statement 4, take x X . Then there ∈ ξ exists a nontrivial horospherical subgroup U such that the orbit U x is compact and Vol(U x)=V(x) <ξ. Thus x Xξ(U ). Now take k K such that 1 ∈ ∈ U = kUk− . Then by Statement 1, x Xξ(U )=kXξ(U). From the above notations and Lemma∈ 3.3.4, the following result is immediate.

Proposition 3.3.7 For any x X, a nontrivial horospherical subgroup U of G, and any ξ (0,ξ ], either∈ (a) x X (U) or (b) for every nontrivial ∈ 0 ∈ ξ one-parameter subgroup u(t):t R U and every T0 R, there exists T>T, such that u(T )x{ C . ∈ }⊂ ∈ 0 ∈ ξ 3.4 A lower bound for the relative measures of a large compact set on unipotent trajectories

To study asymptotic behaviour of trajectories of one-parameter unipotent ub- group of G, the following observation about the growths of polynomials of bounded degree is very useful.

Notation. For a polynomial φ, a subset I R, and α>0, deﬁne ⊂ I (φ)= t I : φ(t) α . α { ∈ | |≤ } Lemma 3.4.1 ([DM4, Lemma 4.1]) Let m N, (0, 1), and α>0 be given. Put β = m/((m +1)mm)α. Then for a polynomial∈ ∈ φ of degree at most m and a closed interval I R, if ⊂ sup φ(s) α, s I | |≥ ∈ 70 then (I (φ)) (I (φ)), β ≤ α where denotes the lebesgue measure on R.

Proof. Put Iα = Iα(φ) and Iβ = Iβ(φ). Note that for any component [a, b]of I either φ(a) = α or φ(b) = α. Since it is enough to prove the lemma for α | | | | each component of Iα, without loss of generality we can assume that Iα = I. We may also assume that (Iβ) > 0.

Claim. There exist t0

ti ti 1 (1/m)(Iβ) (3.14) − − ≥ and ((0,t) I ) (i/m)(I ) (3.15) i ∩ β ≤ β for all i 1,...,m . ∈{ }

We shall prove the claim by induction. Put t0 = inf Iβ. Suppose that for some i 0,...,m 1 , we have chosen t0 <...

((0,t +(1/m)(I )) I ) ((0,t) I )+(1/m)(I ) i β ∩ β ≤ i ∩ β β ((i +1)/m)(I ) ≤ β (I ). ≤ β Now put t = inf (I (0,t + (I)/m)) . i+1 β \ i Then t I , t t (1/m)(I ), and i+1 ∈ β | i+1 − i|≥ β ((0,t ) I ) ((0,t +(1/m)(I )) I ) ((i +1)/m)(I ). i+1 ∩ β ≤ i β ∩ β ≤ β

Thus ti+1 satisﬁes Eq. 3.14 and Eq. 3.15 and the induction is complete. This proves the claim. Now since deg φ m, by Lagrange’s interpolation formula, for any t R, ≤ ∈ m t tj φ(t)= φ(ti) − .  ti tj  i=0 j 0,...,m i ∈{ }\{ } −   Take t I such that φ(t) α. Since φ(ti) β and tj ti (1/m)(Iβ), we have∈ | |≥ | |≤ | − |≥ (I) m α φ(t) (m +1)β . ≤| |≤ (1/m)(I ) β Since β =(m/(m +1)mm)α, we have (I ) (I ). β ≤ α 71 Theorem 3.4.2 ([D3, Thm. 0.2]) Given ξ (0,ξ ) and >0, there exists ∈ 0 ξ (0,ξ) such that for every one-parameter unipotent subgroup u(t):t R of ∈G, every point x C , and every T>0, { ∈ } ∈ ξ 1 ( t [0,T]:u(t)x C ) > 1 . T { ∈ ∈ ξ } −

Proof.Let E = t (0,T):V(u(t)x) <ξ . { ∈ } Then E is a disjoint union of open intervals. Let I =(a, b) be a component of E. Since x C , we have a = 0. Hence by continuity of V , we have ∈ ξ V(u(a)x)=ξ. By Lemma 3.3.4, there exists a polynomial φ of degree at most m = 4 such that V2(u(t)x)=φ(t) for all t I. m m 1/2 ∈ Put ξ =( /(m +1)m ) ξ. Then by Lemma 3.4.1,

(I 2 (φ)) (Iξ2 (φ)). ξ ≤ Hence ( t I : u(t)x C ) (I). { ∈ ∈ ξ } ≤ This completes the proof. 

Corollary 3.4.3 Let X be an L-homogeneous space of G. Then X admits a ﬁnite G-invariant measure. In other words, every L-subgroup of G is a lattice in G.

Proof. (cf. [D1, Thm. 3.2]) Let U be a nontrivial horospherical subgroup of G and u(t):t R be a nontrivial one-parameter subgroup of U.Forany { ∈ } ξ (0,ξ ), deﬁne ∈ 0 T f(x) = lim inf(1/T ) χC (u(t)x) dt T ξ →∞ 0 for every x X. Then using Proposition 3.3.7 and Theorem 3.4.2, we can ∈ choose ξ (0,ξ ) such that ∈ 0 f(x) 1/2 ≥ for all x X X (U). ∈ \ ξ0 By Theorem 1.3.3, there exists a locally ﬁnite G-invariant measure µ on X. Now fdµ (1/2)µ(X X (U)). ≥ \ ξ0 X 72 By Fatou’s Lemma ([Ru2, Sect. 1.28]),

T fdµ lim inf(1/T ) χC (u(t)x) dt dµ(x) ≤ T ξ X →∞ X 0 T = lim inf(1/T ) χC (u(t)x) dµ(x) dt T ξ →∞ 0 X T = lim inf(1/T ) µ(Cξ ) dt T →∞ 0 = µ(C ) < . ξ ∞ Therefore µ(X X (U)) < . \ ξ0 ∞ Let U be a nontrivial horospherical subgroup of G such that U = U. Then similarly we have µ(X Xξ0 (U )) < . By Lemma 3.3.6, Xξ0 (U) Xξ0 (U )= . Therefore \ ∞ ∩ ∅ µ(X) µ(X X (U)) + µ(X X (U )) < . ≤ \ ξ0 \ ξ0 ∞

Corollary 3.4.4 (Cf. [M3, Thm. 3.10]) Let X be an L-homogeneous space of G and let H = SL2(R). Then every closed H-orbit in X admits a ﬁnite H-invariant measure.

Proof. Suppose that for some x X the orbit Hx is closed. Then by Lemma 1.1.1, Hx is a homogeneous∈ space of H. Therefore by Theorem 1.3.3, Hx admits a locally ﬁnite H-invariant measure, say µH . Note that H contains atleast two nontrivial one-parameter unipotent subgroups which are not contained in the same horospherical subgroup of G. Now arguing as in the proof of Corollary 3.4.3, with µH in place of µ, we obtain that

µ (X)=µ (Hx) < . H H ∞ Remark. Given an L-homogeneous space Y of H, all the results of this chapter go through if everywhere one replaces X by Y and G by H.

Lemma 3.4.5 Let U be a nontrivial horospherical subgroup of G associated to an element g G. Deﬁne ∈ D(g)= x X : gnx as n . { ∈ →∞ → ∞} Then D(g)=X(U).

73 Proof. Observe that g NG(U) and det(Ad g ) < 1. Let x D(g). Then ∈ | |U | ∈ by Theorem 3.2.3, Gx U = e . Therefore by Corollary 3.2.4, x X(U). If x X(U), an argument∩ { } in the proof of Lemma 3.1.1 shows∈ that x ∈ ∈ D(g). We deduce the next result from Theorem 3.4.2; for this purpose we follow the argument as in the proof of Lemma 2.1 in [Ra2].

Corollary 3.4.6 (Cf. [DM2, Thm. A.2]) Let U be a horospherical subgroup of G and let u(t):t R be a nontrivial one-parameter subgroup of U. Let { ∈ } a sequence xi X be such that for every η>0, xi Xη(U) for all large i N. Then{ there}⊂ exists a sequence t R such∈ that a subsequence of ∈ { i}⊂ >0 u(ti)xi converges to a point in X X(U). { In particular,} any closed u(t):\t R -invariant subset of X(U) is con- { ∈ } tained in Xη(U) for some η>0.

Proof.Takek N. By Theorem 3.4.2, there exists ξk (0,ξ0) such that for every x C and∈ every T>0, ∈ ∈ ξ0

1 (k+1) ( t [0,T]:u(t)x C ) > 1 2− . (3.16) T { ∈ ∈ ξk } − 1 1 Fix α>1. Let g NG(U) be such that gu(t)g− = u(α− t) for all t R. Due to Lemma 3.3.6,∈ there exists n N such that ∈ k ∈ gnk X (U) X (U). (3.17) k ⊂ ξk/2

Let ηk > 0 be such that

n g− k X (U) X (U). ξ0 ⊂ ηk

Replacing the sequence xi by a subsequence, we may assume that for every { } nk k N and every i k, xi Xηk (U). Therefore g xi Xξ0 (U) and hence by ∈ ≥ ∈ ∈ nk Proposition 3.3.7, there exists T0 = Tk,i 0 such that u(T0)g xi Cξ0 . Now, after the change of variable: t + T t in≥ Eq. 3.16, for every T>T∈ , we have 0 → 0 1 n (k+1) ( t [T ,T]:u(t)g k x C ) 1 2− . T T { ∈ 0 i ∈ ξk } ≥ − − 0 Therefore for every T T , ≥ 0 1 nk T ( t [T,2T ]:u(t)g xi Cξk ) (k+1) (1{ 2 ∈ )(2T T0) (T T0) ∈ } − − − − − (3.18) T ≥ k 1 2− . ≥ − For every t R, ∈ n n n g− k u(t)g k = u(α k t).

74 Therefore, after the change of variable (αnk t t) in Eq. 3.18, for every T nk → ≥ α T0, we have

1 n k t [T,2T ]:u(t)x g− k C 1 2− . T { ∈ i ∈ ξk } ≥ − Now take i N. Put ∈ n C(i)= g− k C X X (U) ξk ⊂ \ k 1 k i 1 k i ≤≤ ≤≤ and nk T (i) = max α Tk,i. 1 k i ≤ ≤ Then for every T T (i), ≥ i 1 k i ( t [T,2T ]:u(t)x C(i) ) 1 2− =2− . T { ∈ i ∈ } ≥ − k=1 Therefore there exists t [T (i), 2T (i)] such that u(t )x C(i). i ∈ i i ∈ Put C = i NC(i). For every i N, C(i) is compact, C(i +1) C(i), ∩ ∈ ∈ ⊂ and by Eq. 3.17, C(i) Xi(U)= . Hence the set u(ti)xi : i N has a limit point in C and C X∩(U)= . ∅ { ∈ } ∩ ∅ The next result is quite similar.

Corollary 3.4.7 (Cf. [DM2, Prop. A.3]) Let U be a nontrivial horospherical subgroup of G and u(t):t R be a nontrivial one-parameter subgroup { ∈ } of U. Let y X X(U) and let a sequence xi X be such that xi y as i . Then∈ given\ a sequence t R and{ a }λ⊂>1, there exists a sequence→ →∞ { i}⊂ αi [1,λ] such that a subsequence of u(αiti)xi converges to a point in X{ }X⊂(U). { } \ Proof. Given α>1, there exists g N (U) such that ∈ G n n U = u G : g ug− e as n { ∈ → → ∞} 1 1 and gu(t)g− = u(α− t) for all t R. ∈ Since y X(U), by Lemma 3.4.5, there exists ξ (0,ξ0) such that the set = n N∈ : gny C is unbounded. ∈ R { ∈ ∈ ξ} Take k N. By Theorem 3.4.2, there exists a ξk (0,ξ) such that for every x C∈ and every T>0, ∈ ∈ ξ

1 k ( t [0,T]:u(t)x C ) 1 2− (λ 1)/λ. T { ∈ ∈ ξk } ≥ − − 75 Then k 1 (1 2− (λ 1)/λ)λT T ( t [T,λT]:u(t)x C ) > − − − λT T { ∈ ∈ ξk } (λ 1)T − k − =1 2− . −

Since the set is unbounded, by Lemma 3.3.6, there exists nk N such nk R nk ∈ that g Xk(U) Xξk/2(U) and g y Cξ. Since Cξ is open and xi y, by passing to a⊂ subsequence we may assume∈ that for every k N and every→ i k, ∈ ≥ gnk x C . i ∈ ξ Hence for every k N,everyi k, and every T>0, ∈ ≥ 1 n k ( t [T,λT]:u(t)g k x C ) 1 2− . (3.19) λT T { ∈ i ∈ ξk } ≥ − − n n n Note that for every t R, we have g− k u(t)g k = u(α k t). Therefore after the change of variable: α∈nt t in Eq. 3.19, for every T>0, we have that → 1 n k t [T,λT]:u(t)x g− k C 1 2− . λT T { ∈ i ∈ ξk } ≥ − − Take i N. Put ∈ n C(i)= g− k C X X (U). ξk ⊂ \ k 1 k i 1 k i ≤≤ ≤≤ Then for any T>0,

i 1 k i ( t [T,λT]:u(t)x C(i) ) 1 2− =2− . λT T { ∈ i ∈ } ≥ − − k=1 Therefore there exists α [1,λ] such that u(α t )x C(i). i ∈ i i i ∈ Put C = i NC(i). For every i N, C(i) is compact and C(i +1) C(i). ∈ Therefore the∩ sequence u(α t )x ∈has a limit point in C and C X(U⊂)= . { i i i} ∩ ∅ The above two corollaries will be used in the next chapter to study the closures of orbits of unipotent subgroups acting on L-homogeneous spaces of G.

3.5 Description of noncompact L-homogeneous spaces of SL2(C) Let X be a noncompact L-homogeneous space of G. In this section we shall obtain detailed information about the structure of the set X C . \ ξ0 76 Let U be a nontrivial horospherical subgroup of G. There exists k K 1 1 1∈ such that U = kNk− .IfU = k1Nk1− for another k1 K, then k− k1 N (N) K = M. Therefore the groups ∈ ∈ G ∩ 1 A(U) := kAk− and 1 M(U) := kMk− =N (U) K G ∩ are well deﬁned. Now G = KA(U)U is an Iwasawa decomposition of G and NG(U)=M(U)A(U)U.Forη>0, deﬁne

NG(U)η = g NG(U): det(Ad g ) η { ∈ | |U |≤ } and A (U)=A(U) N (U) . η ∩ G η Then NG(U)η = Aη(U)M(U)U.

Remark.Foranyx X(U) the orbit Ux is compact, and hence the orbit ∈ M(U)Ux is also compact. For any sequence gn A(U), if det(Ad gn ) 0 U as n , then the sequence g x X diverges{ }⊂ as n . | → →∞ { n }⊂ →∞ Description of the set X(U)

Lemma 3.5.1 For any η>0 and x Xη(U), there exists a neighbourhood Ω of e in G such that ∈ Ωx X (U) (Ω N (U))x. ∩ η ⊂ ∩ G Proof. By Lemma 3.3.6, for k K, ∈ kX (U) X (U) = k N (U). (3.20) ξ0 ∩ ξ0 ∅⇔ ∈ G Let Ψ be a neighbourhood of e in N (U) such that Ψ X (U) X (U). 1 G 1 ξ0/2 ⊂ ξ0 Due to Iwasawa decomposition, Ω1 := KΨ1 is a neighbourhood of e in G. Now due to Eq. 3.20, for any y X (U), we have that ∈ ξ0/2 Ω y X (U) (N (U) Ω )y. (3.21) 1 ∩ ξ0 ⊂ G ∩ 1 Let g N (U) be such that ∈ G gX (U) X (U). η ⊂ ξ0/2 1 Put Ω= g− Ω g and y = gx X (U). Then due to Eq. 3.21, 1 ∈ ξ0/2 1 1 Ωx X (U) g− (Ω y X (U)) g− (N (U) Ω )gx =(N (U) Ω)x. ∩ η ⊂ 1 ∩ ξ0/2 ⊂ G ∩ 1 G ∩ First we record two corollaries of this lemma for future reference.

77 Corollary 3.5.2 For any three distinct nontrivial horospherical subgroups U1, U2, and U3 of G and any η1,η2,η3 > 0, the set S = X (U ) X (U ) X (U ) η1 1 ∩ η2 2 ∩ η3 3 is ﬁnite.

Proof.Letg N (U ) be such that gX (U ) X (U ). Then ∈ G 1 η1 1 ⊂ ξ0 1

gS Xξ (U1) Xη (U ), ⊂ 0 ∩ 2 2 1 where U2 = gU2g− and η2 > 0. Therefore without loss of generality we may assume that η ξ . By Lemma 3.3.6, X (U ) X (U )= . Therefore 1 ≤ 0 ξ0 1 ∩ ξ0 2 ∅ S X (U ) (X (U ) X (U )). ⊂ η1 1 ∩ η2 2 \ ξ0 2 The set on the right hand side is relatively compact. Therefore S is compact. Now take x S. By Lemma 3.5.1, there exists a neighbourhood Ωof e in ∈1 G such that ΩΩ− G = e ,Ω C(G)= e , and ∩ x { } ∩ { } Ωx X (U ) (Ω N (U ))x ∩ ηi i ⊂ ∩ G i for i =1, 2, 3. Since N (U ) N (U ) N (U )=C(G), we have Ωx S = x . G 1 ∩ G 2 ∩ G 3 ∩ { } Therefore S is a discrete set. Now since S is compact, it is ﬁnite. Corollary 3.5.3 Let U be a nontrivial horospherical subgroup of G. For a one-parameter subgroup ρ(t) t R G, a point x X(U), and a δ>0, { } ∈ ⊂ ∈ suppose that ρ(t)x X(U) for all t [ δ,δ]. Then ρ(t) t R NG(U). ∈ ∈ − { } ∈ ⊂ Proof. Take an increasing sequence η in R.Forn N deﬁne n →∞ ∈ S(n)= t [ δ,δ]:ρ(t)x X (U) . { ∈ − ∈ ηn }

Then [ δ,δ]= n N S(n). Since each Sn is closed, by Baire’s category theo- − ∈ rem there exists n0 N such that S(n0) contains an open interval. Thus there ∈ exists t0 S(n0)andδ > 0 such that u(t+t0)x Xη (U) for all t ( δ,δ). ∈ ∈ n0 ∈ − Therfore due Lemma 3.5.1 there exists (0,δ) such that ρ(t) N (U)for ∈ ∈ G all t ( ,). But then ρ(t) NG(U) for all t R. To∈ describe− the structure∈ of a noncompact∈ L-homogeneous space X of G, ﬁrst we need to understand the structure of the set X(U).

Lemma 3.5.4 Let U be a nontrivial horospherical subgroup of G. Then X(U) is the union of ﬁnitely many orbits of NG(U). More precisely, there exists a ﬁnite set Σ X(U) such that ⊂

X(U)= NG(U)x x Σ ∈ 78 and for any η>0,

Xη(U)= NG(U)ηx. x Σ ∈ (Note that due to Theorem 2.4.7, each NG(U)-orbit in X is dense.) Proof. Put C = X (U) X (U). ξ0 \ ξ0/2

Then C is a compact subset of Xξ0 (U). By Lemma 3.5.1, NG(U)x Xξ0 (U) is an open subset of X (U)foreveryx X(U). Therefore there exists∩ a finite ξ0 ∈ set Σ C such that ⊂ C N (U)x. ⊂ G x Σ ∈ Now for any z X(U), there exists g N (U) such that ∈ ∈ G Vol(Ugz)= det(Ad g ) Vol(z) (ξ0/2,ξ0). | |U |· ∈ Therefore X(U)=N (U)C N (U)x X(U). G ⊂ G ⊂ x Σ ∈ Thus X(U) is a union of finitely many orbits of NG(U). Replace each x Σ by gx for some g N (U) such that ∈ ∈ G Vol(Ugx)= det(Ad g ) Vol(Uy)=1 | |U | and denote the new set by Σ. Then Σsatisfies the conclusion of the lemma.

A function V on X

Take any x X. Then its stabilizer-Gx contains a nontrivial element of a horospherical∈ subgroup of G,sayW . Now by Corollary 3.2.4, the orbit Wx is compact. Therefore we can deﬁne a function V : X (0, ) such that for → ∞ every x X, V(x) is the inﬁmum of Vol(Wx) over all nontrivial horospherical ∈ subgroup W of G such that the orbit Wx is compact. It is straight forward to verify the following properties of V. 1. V(kx)=V(x) for all k K and x X. ∈ ∈ 2. is a continuous function. V

3. V(x)=V(x) for all x X Cξ . In particular, V vanishes at the point 0 at inﬁnity. ∈ \ Thus def η0 = max V(x) < . x X ∈ ∞ 79 Fix a nontrivial horospherical subgroup U of G.Foraz X, let W be a ∈ horospherical subgroup of G such that Wz is compact and V(z) = Vol(Wz). 1 1 Let k K be such that W = kUk− . Put x = k− z. Then ∈ 1 V(x)=V(z) = Vol((kUk− )z) = Vol(kUx) = Vol(Ux).

Therefore x X (U). Thus we have proved that ∈ η0 

X = KXη0 (U)= KNG(U)η0 x. x Σ ∈ From the above discussion we obtain the following structure theorem.

Theorem 3.5.5 (Cf. [GR, Main theorem]) Let X be a noncompact L-homogeneous space of G and U be a nontrivial horospherical subgroup of G. Then there exists a ﬁnite collection consisting of compact U-orbits in X such that the following holds: C

1. There exists η0 > 0 such that

X = KA (U) . η0 O O∈C

2. There exists an ξ > 0 such that for , , a, a A (U), and 0 O O ∈C ∈ ξ0 k, k K, if ∈ ka ka = O∩ O ∅ then = , a = a, kM(U)=kM(U), and O O

ka = ka . O O

Remark. Put U = N in the above theorem and apply Lemma 2.3.2 to see that X admits a ﬁnite G-invariant measure. This provides another proof of the fact that an L-homogeneous space of G admits a ﬁnite G-invariant measure (see Corollary 3.4.3).

80 Chapter 4

Proof of Raghunathan’s conjecture for SL2(C)

In this chapter we shall study closures of orbits of unipotent one-parameter subgroups of G acting on an L-homogeneous space of G. To analyse the closures we use the techniques introduced by G.A. Margulis in [M2] and developed further in [DM1, DM2]. The strategy is to show that under certain conditions, the closures of orbits of smaller subgroups contain orbits of larger subgroups. In this method the minimal closed invariant subsets for the subgroup actions play an important role.

4.1 Minimal invariant sets

Deﬁnition.LetF be a (locally compact) topological group and Z be an F - space. We say that Z is F -minimal, if no (nonempty) proper closed subset of Z is invariant under the action of F , or equivalently if every F -orbit in Z is dense.

Lemma 4.1.1 Any compact F -space contains a compact F -minimal subset.

Proof.LetY be a compact F -space. Let = Yα α I be the family of all ∈ nonempty closed F -invariant subsets of Y .C Then{ } is partially ordered by set theoretic inclusion. We want to show that this familyC contains a minimal element. Due to Zorn’s lemma, it is enough to show that contains a lower bound for every totally ordered subfamily. C Let J = Yα α J be a totally ordered subfamily of , where J I.Now C { } ∈ C ⊂ J has the ﬁnite intersection property. Since Yα is compact for each α J, theC set ∈ YJ = Yα α J ∈ is nonempty. Also YJ is closed and F -invariant. Therefore YJ is a lower bound for the subfamily in . This completes the proof. CJ C 81 In general a noncompact F -space need not contain a closed F -minimal subset. While for the unipotent actions on L-homogeneous spaces, we have the following aﬃrmative results. Proposition 4.1.2 (Cf. [DM1, Cor.1.3]) Let X be an L-homogeneous space of G = SL2(C) and F be a closed subgroup of G containing a nontrivial unipotent element. Then there exists a compact set C in X such that given any nonempty closed F -invariant subset Y of X, at least one of the following holds: (i) Y contains a compact F -invariant subset. (ii) Y C = . ∩ ∅

Proof. There exists a horospherical subgroup U1 of G such that F U1 = e . Due to Corollary 2.2.10, there are the following cases: ∩ { } (a) F (K N (U ))U . ⊂ ∩ G 1 1

(b) There exists g F NG(U1) such that det(Ad g 1 ) > 1. ∈ ∩ | |U | (c) There exists a horospherical subgroup U2 of G such that U1 = U2 and F U = e . ∩ 2 { } For notational convenience, put I = 1 for the cases (a) and (b), and put { } I = 1, 2 for the case (c). For i I, let ui(t):t R be a nontrivial one-parameter{ } subgroup of U such that∈ u (1){ F . Deﬁne∈ a} compact set i i ∈

C = ui([0, 1])Cξ0 , i I ∈ where Cξ0 is the set as deﬁned following Lemma 3.3.5.

Then by Lemma 3.3.7, for each i I, and any given x X Xξ0 (Ui), there exists n N such that u (n)x C. Therefore∈ ∈ \ ∈ i ∈ if Y X (U ) then Y C = . ⊂ ξ0 i ∩ ∅ i I ∈

In the Case (c), due to Lemma 3.3.6, Xξ0 (U1) Xξ0 (U2)= . Hence Y C = . ∩ ∅ ∩ ∅ In the Case (b), for any x Y , if x Xξ0 (U1) then there exists m N such that ∈ ∈ ∈ m m Vol(U1(g x)) = det(Ad g 1 ) Vol(U1x) >ξ0. | |U | · m Since g F ,wehaveg x Y Xξ0 (U1). Hence Z C = . In the∈ Case (a), if there∈ exists\ x Y X (U )∩ then ∅ ∈ ∩ ξ0 1 Fx (K N (U ))U x, ⊂ ∩ G 1 1 which is a compact F -invariant subset of Y . This completes the proof.

82 Corollary 4.1.3 (Cf. [DM1, Cor. 1.5]) Let X be an L-homogeneous space of G = SL2(C) and F be a closed subgroup of G containing a nontrivial unipotent element of G. Then any closed F -invariant subset of X contains a closed F -minimal subset.

Proof.IfY contains a compact F -invariant subset, then we are through by applying Lemma 4.1.1 to that compact subset. Otherwise due to Proposition 4.1.2, there exists a compact set C in X such that any closed F -invariant subset of Y intersects C. Let the notation be as in the proof of Lemma 4.1.1. To apply Zorn’s lemma, we only need to show that the set YJ is non-empty. Now

Y C = Y C. J ∩ α ∩ α J ∈

The family Yα C α J consists of (non-empty) compact sets and it has the ∈ ﬁnite intersection{ ∩ property.} Therefore Y C is non-empty. This completes J ∩ the proof. Proposition 4.1.4 (Cf. [DM1, Cor. 1.7]) Let X be an L-homogeneous space of G = SL2(C). Let U be a horospherical subgroup of G and V be a subgroup of U. Then every closed V -minimal subset of X is compact.

Proof. (Cf. [M4, Sect. 2]) Let Z be a V -minimal subset of X.IfZ intersects a compact U-orbit, say Y , then Z Y is V -invariant. By minimality, Z Y , and hence Z is compact. ∩ ⊂ Therefore we are left with the case when Z X(U)= . We can assume that V = e .Let u(t):t R be a nontrivial∩ one-parameter∅ subgroup of U such that{ }u(1) {V . Put ∈ } ∈

C = u([0, 1])Cξ0 . Then by Proposition 3.3.7, for every x X X(U), the sets ∈ \ n>0:u(n)x C and n<0:u(n)x C (4.1) { ∈ } { ∈ } are unbounded. Suppose Z is not compact. Then there exists a sequence zi Z C such that z has no convergent subsequence. Let a sequence{ n}⊂ N\be { i} { i}⊂ such that yi := u(ni)zi C and u(k)zi C for all 0 k ni 1. Since C is relatively compact and∈ z , we have∈ n ≤as i ≤ −. Since C is i →∞ i →∞ →∞ relatively compact, by passing to a subsequence we may assume that yi y for some y C Z. Now for every k N, → ∈ ∩ ∈ u( k)y = u(n k)z C for all large i N. − i i − i ∈ ∈ 83 Since C is open, u( k)y C for every k N. This contradicts Eq. 4.1, − ∈ ∈ because y Z X X(U). We note∈ some⊂ simple\ useful observations about minimal invariant sets.

Lemma 4.1.5 Let G be a (locally compact) topological group and X be a G- space. Let F be a closed subgroup of G and Z be a closed F -minimal subset of X. Then the following holds:

1 (1) For any g G, gZ is a closed gFg− -minimal subset of X. In particular, if g N (∈F ) then gZ is F -minimal. ∈ G

(2) Let Y be any closed F -invariant subset of X.Ifg NG(F ) and gZ Y = , then gZ Y . ∈ ∩ ∅ ⊂ 1 Proof. The statement (1) holds because if Y is a closed gFg− -invariant subset 1 of gZ then g− Y is a closed F -invariant subset of Z. The statement (2) holds because gZ Y is a closed F -invariant subset of ∩ gZ, which is F -minimal.

4.2 Topological limits and inclusion of orbits-I

Our method of obtaining orbits of larger subgroups in the closure of an orbit of a given subgroup is based on the following observation in t¸eM-qforms.

Lemma 4.2.1 (Cf. [M2, Sect. 2.1]) Let G be any locally compact group, Γ be a discrete subgroup of G, and let X = G/Γ. Let F be a closed subgroup of G and Y be the closure of an F -orbit in X. Suppose Y contains a compact F -minimal subset, say Z. Deﬁne

M = g G N (F ):gZ Y = . { ∈ \ G ∩ ∅}

If e M then Y = Z and Y is an orbit of a closed subgroup of NG(F ) containing∈ F . If e M then hZ Y for every h NG(F ) FMF. In particular, if Y is F -minimal∈ (i.e. Y⊂= Z) then Y is∈ invariant∩ under the closed subgroup generated by N (F ) FMF. G ∩ Proof. Suppose that e M. Then there exists a neighbourhood Ωof e in G such that ΩZ Y (N∈ (F ) Ω)Z. Therefore (N (F ) Ω)Z contains an ∩ ⊂ G ∩ G ∩ open subset of Y . Since Y is the closure of an F -orbit, there exists g NG(F ) and z Z such that F (gz)=Y . Now gZ = gFz = Fgz = Y Z.∈ Now by Lemma∈ 4.1.5, gZ is F -minimal. Therefoere Z = gZ and hence Y⊃ = Z.Let

L = g N (F ):gY Y = . { ∈ G ∩ ∅} 84 Again by Lemma 4.1.5, gY = Y for every g L. Since Y is compact, L is a closed subgroup of N (F ) containing F . Also∈ for every x Y , G ∈ Ωx Y (Ω N (F ))x Y Lx. ∩ ⊂ ∩ G ∩ ⊂ Therefore every L-orbit in Y is open. Since Y is F -minimal, Y is itself an orbit of L. This completes the proof of the ﬁrst part. Next suppose that e M. First take h FMF.Letf,f F and ∈ ∈ ∈ m M be such that h = f mf.Letz Z be such that mz Y . Then ∈ 1 ∈ ∈ z = f − z Z and f (mfz ) Y . Thus hZ Y = . Now take h FMF 1 ∈ 1 ∈ ∩ ∅ ∈ and let hi FMF be a sequence such that hi h as i . Then there exists a{ sequence}⊂ z Z such that h z Y →for all i →∞N. Since Z is { i}⊂ i i ∈ ∈ compact, by passing to subsequences we may assume that zi z for some z Z as i . Therefore h z hz as i . Since Y is closed,→ we have ∈ →∞ i i → →∞ hz Y . Further if we assume that h NG(F ) then since Z is F -minimal, by Lemma∈ 4.1.5, hZ Y . This completes∈ the proof. ⊂ The second part of this lemma can be generalized as follows.

Lemma 4.2.2 ([DM1, Lemma 2.1]) Let G be a locally compact group, let X be a homogeneous space of G, and let P and Q be closed subgroups of G. Let Y be a closed P -invariant subset of X and Z be a compact Q-invariant subset of X. Put M = g G : gZ Y = . { ∈ ∩ ∅} Then M = PMQ and M is closed. Let F be a closed subgroup of P Q, and further assume that Z is F - minimal. Then for every g N (F ) ∩M, we have gZ Y . ∈ G ∩ ⊂ The conclusion of Lemma 4.2.1 is of signiﬁcance mainly when the set S := NG(F ) FMF is considerably larger than F . In a linearised formulation, the next lemma∩ shows that if F is a one-parameter unipotent subgroup of G then the set S/F NG(F )/F contains an unbounded curve passing through the identity coset.⊂

Lemma 4.2.3 ([M2, Lemma 13],[DM1, Lemma 2.2]) Let E be a finite dimensional real vector space and V = u(t):t R be a one-parameter group of unipotent linear transformations{ of E. Define∈ } a subspace L = x E : V x = x and fix a point p L. Let Y be a subset of E L such{ that∈ p Y L. Then} there exists a nonconstant∈ polynomial function\ φ : R L such∈ that∩ φ(0) = p and φ(R) VY L. → ⊂ ∩ More precisely, there exists a sequence t and a sequence p Y i →∞ { i}⊂ with pi p as i , such that for any sequence si s in R, we have u(s t )p → φ(s) as→∞i . → i i i → →∞ 85 Proof. (Cf. [Si, pp. 312]) Let N be a nilpotent linear transformation on E such that u(t) = exp(tN) for all t R. Since N = 0, there exists r 0 such that N r = 0 and N r+1 = 0. Consider∈ the following filteration of E: ≥ 0 L =kerN ... ker N r ker N r+1 = E. { }⊂ ⊂ ⊂ ⊂ j j+1 For 0 j r, let Ej be a complementary subspace of ker N in ker N . Then ≤ ≤ E = r E . ⊕j=0 j For x E, write x = r x , where x E for 0 j r. Then for t R, ∈ j=0 j j ∈ j ≤ ≤ ∈ r r r k k u(t)x = exp(tN)xj = (t /k!)N xj =: A(x,t)+B(x,t), j=0 j=0 k=0 where r r j j A(x,t)= (t /j!)N xj =: Aj(x,t) j=0 j=0 and r j 1 − k k B(x,t)= (t /k!)N xj . j=1 k=0 Put ∆(x,t) = max A (x,t) :1 j r , { j ≤ ≤ } where denotes a euclidean norm on E. Then there exists a constant c>0 such that· for all t>1,

1 B(x,t) ct− ∆(x,t). ≤

If x L then ∆(x,t) > 0. Take a sequence pi Y E L such that p p ∈as i . Since p L = E ,forany1{ }j⊂ r⊂, the\ projection of i → →∞ ∈ 0 ≤ ≤ pi on Ej converges to 0 as i . Therefore for any t>0, ∆(pi,t) 0as i .Moreover,foreveryi,→∞∆(p ,t) as t . Therefore there→ exists →∞ i →∞ →∞ a sequence ti R such that ∆(pi,ti)=1foreveryi N and ti as i . { }⊂ ∈ →∞ →∞By passing to a subsequence, we may assume that for every 1 j r, there exists w L such that A (p ,t) w as i . Also w ≤=1for≤ j ∈ j i i → j →∞ j0 some j0 1,...,r . ∈{ } 1 Moreover, B(pi,ti) ctk− ∆(pi,ti) 0asi . Deﬁne a polynomial function φ : R L such≤ that for every s→ R, →∞ → ∈ r j φ(s)=p + s wj. j=1 86 Now for every s R and i N, ∈ ∈ r j u(sti)pi = p + s Aj(pi,ti)+B(pi,ti). j=1

Hence for any sequence si s in R, u(siti)pi φ(s)asi . In particular, VY L φ(R). → → →∞ ∩ ⊃ 

4.3 Specialization to the case of SL2(C)

The following notations will be used throughout the remaining part.

Notation.LetG = SL2(C) and X be an L-homogeneous space of G. Recall that a homogeneous space of G admits a ﬁnite G-invariant measure if and only if it is an L-homogeneous space. Let

K = g G : gg¯t =1 , { ∈ } def α 0 D = d(α) = 1 : α C∗ , 0 α− ∈ M = d(eti):t R = D K, { ∈ } ∩ A = d(a):a>0 , { } def 1 z N = u(z) = : z C , 01 ∈ H = SL2(R) K = H K 1 ∩ N = u(t):t R = N H, and 1 { ∈ } ∩ N = v(t) := u(it):t R N. 2 { ∈ }⊂

Our main interest is to study the closures of N1-orbits in X. Using the linear formulation in Lemma 4.2.3, we shall obtain some group theoretic information about the set FMF for F = N1 or F = N and M G NG(F ) such that e M. ⊂ \ ∈

Proposition 4.3.1 Let V = N or N1. Then there exists a ﬁnite dimensional vector space E, a point p E, and a continuous representation of G on E such that the following conditions∈ are satisﬁed:

(i) V = g G : g p = p ; and { ∈ · } (ii) the orbit G p is open in its closure; that is, the set G p G p is closed. · · \ ·

87 Proof. First let V = N. Consider the standard action of G = SL2(C)on E = C2 and take p =(1, 0). Then N = g G : g p = p and G p = C2 0 . Thus (i) and (ii) are satisﬁed. { ∈ · } · \{ } 2 2 2 2 Now let V = N1.LetE = C (C C ), where C is treated as a four dimensional real vector space. Consider⊕ the∧ G-action on E, where an element g G acts on E as a linear transformation such that for all v , v , v C2, ∈ 1 2 3 ∈ g (v +(v v )) = gv +(gv gv ). · 1 2 ∧ 3 1 2 ∧ 3 Take e =(1, 0) and f =(0, 1) in C2. Put p = e + e f E. Then for any g G, ∧ ∈ ∈ g p = p ge = e and ge gf = e f · ⇔ ∧ ∧ g N and e gf = e f ⇔ ∈ ∧ ∧ g N and gf = λe + f for some λ R ⇔ ∈ ∈ g N SL (R)=N . ⇔ ∈ ∩ 2 1 Thus condition (i) is satisﬁed. Due to Iwasawa decomposition of G (see Lemma 2.2.1),

G p = KAN p = K(AN p); · · · the second equality holds because K is compact. Therefore

G p G p = K(AN p G p). · \ · · \ · Now to prove that G p G p is closed, it is enough to show that AN p G p is closed. · \ · · \ · Deﬁne L = x E : g x = x for all g N . { ∈ · ∈ 1} Due to condition (i),

g G : g p L = g G : N1(g p)=g p { ∈ · ∈ } { ∈ 1 · · } = g G : g− N g N { ∈ 1 ⊂ 1} =NG(N1). Therefore AN p L and · ⊂ AN p G p = AN p N (N ) p. · \ · · \ G 1 · Now

AN p = AN p = d(a)v(t)(e +(e f)) : a>0,t R · 2 · { ∧ ∈ } = d(a)(e + t(e ie)+(e f)) : a>0,t R { ∧ ∧ ∈ } = ae + t(e ie)+(e f):a>0,t R . { ∧ ∧ ∈ } 88 k Recall that NG(N1)= d(i ) k ZAN. Therefore { } ∈ AN p N (N ) p = t(e ie)+(e f):t R · \ G 1 · { ∧ ∧ ∈ } is a closed set. Thus condition (ii) is satisﬁed.

Proposition 4.3.2 (Cf. [DM1, Prop. 3.6]) Let V = N or V = N1, and let M G N (V ) be such that e M. Then the following holds: ⊂ \ G ∈ 1. If V = N then there exists a nonconstant complex polynomial σ such that σ(0) = 1 and for any t R, if σ(t) =0then ∈ d(σ(t)) N MV. ∈ 1

2. If V = N1 then there exist two real polynomials σ and ν such that atleast one of them is non-constant, σ(0) = 1, ν(0) = 0, and for any t R, if σ(t) =0then ∈ v(ν(t))d(σ(t)) N MV. ∈ 1 Proof. Let the linear representation of G on E and the point p E be as in the proof of Proposition 4.3.1. Let ∈ L = x E : g x = x for all g N . { ∈ · ∈ 1} For any g G, ∈ g p L N1(g p)=g p · ∈ ⇔ 1 · · (g− N1g) p = p ⇔ 1 · g− N g V ⇔ 1 ⊂ g N (V ), ⇔ ∈ G 1 where the last implication is due to the fact that N gNg− = if and only if g N (N) (see Lemma 2.2.7). Thus ∩ ∅ ∈ G N (V )= g G : g p L . (4.2) G { ∈ · ∈ } Put Y = M p. Then Y L = , because M N (V )= . Also · ∩ ∅ ∩ G ∅ p Y L, because e M. By Lemma 2.1.6, N1 acts on E by unipotent linear∈ transformations.∩ Therefore∈ by Lemma 4.2.3, there exists a nonconstant polynomial function φ : R L such that φ(0) = p and for every t R, → ∈ φ(t) N Y = N M p. (4.3) ∈ 1 1 · Let ψ : G/V G p be the map deﬁned by ψ(gV )=g p for all g G. Since G p is→ locally· compact, the map ψ is a homeomorphism· due to∈ Lemma 1.1.2.· Therefor ψ(N MV/V) G p = ψ N MV/V (4.4) 1 ∩ · 1 89 From Eqs. 4.2, 4.3, and 4.4, we conclude that g G : φ(t)=g p for some t R N MV N (V ). (4.5) { ∈ · ∈ }⊂ 1 ∩ G First suppose that V = N. In this case E = C2 and p =(1, 0). Since L = Cp, there exists a complex polynomial σ such that for all t R, φ(t)= σ(t) p. Now for any t R such that σ(t) = 0, we have ∈ · ∈ d(σ(t)) p = σ(t) p = φ(t). · · Therefore d(σ(t)) N1MN. This completes the proof of part 1. ∈ 2 2 2 Now suppose that V = N1. In this case E = C (C C ) and p = e +(e f), where e =(1, 0) and f =(0, 1). Since G p⊕is open∧ in its closure and φ(0)∧ = p, there exists δ>0 such that φ(t) G· p for all t ( δ,δ). Therefore by Eq. 4.5, we have that ∈ · ∈ − φ(t) N (N )0 p = AN p = N A p for all t ( δ,δ). (4.6) ∈ G 1 · · 2 · ∈ − For any a>0 and s R, we have ∈ v(s)d(a) p = ae + s(ie f)+(e f). (4.7) · ∧ ∧ From Eqs. 4.6 and 4.7 we deduce the following: There exist real polynomials σ and ν such that σ(0) = 1, ν(0) = 0, and for any t ( δ,δ), we have that σ(t) = 0 and ∈ − φ(t)=σ(t)e + ν(t)(ie f)+(e f)=v(ν(t))d(σ(t)) p. (4.8) ∧ ∧ · Since ν, σ, and φ are polynomial functions, the Eq. 4.8 holds for all t R Z(σ), where Z(σ)= t R : σ(t)=0 . Therefore v(ν(t))d(σ(t)) N∈MN\ { ∈ } ∈ 1 1 for all t R Z(σ). This completes the proof of part 2. Now∈ we are\ ready to prove the homogeneity of the closures of N-orbits. Theorem 4.3.3 (Cf. [M2, Lemma 9]) Any N-orbit in X is either compact or dense. Proof.LetY be the closure of an orbit of N. By Corollary 4.1.3 and Propo- sition 4.1.4, there exists a compact N-minimal subset Z Y . Deﬁne ⊂ M = g G N (N):gZ Y = . { ∈ \ G ∩ ∅} If e M then by Lemma 4.2.1, Y = Z and there exists a closed subgroup L N ∈(N)=DN containing N such that Y is a compact orbit of L. Fix ⊂ G z Y . Then L/Lz is compact. Therefore by Corollary 2.2.5, the orbit Nz is compact.∈ Now due to minimality, Y is a compact N-orbit. Now suppose that e M. Then by Lemma 4.2.1 and Proposition 4.3.2, there exists a nonconstant∈ complex polynomial σ such that for all large t>0, d(σ(t))Z Y . Since σ(t) as t , by Theorem 2.4.7, we have ⊂ | |→∞ →∞ Y = X. This completes the proof.

90 4.4 Topological limits and inclusion of orbits- II

The next result is very useful in applying Lemma 4.2.1. Lemma 4.4.1 (Cf. [DM1, Sect. 3.6]) Let F be a closed subgorup of G and let Y and Z be closed F -invariant subsets of X. Suppose that there exists a continuous map Φ:(t0, ) NG(F ) such that Φ(t)Z Y for all t>t0, where t R. Deﬁne ∞ → ⊂ 0 ∈ Y = y Y : there exist sequences z Z and t in R 1 { ∈ { i}⊂ i →∞ such that Φ(t )z y as i , and i i → → ∞} S = g NG(F ):there exists a function f :(0, ) R such that { ∈ 1 ∞ → t + f(t) and Φ(t + f(t))Φ(t)− g as t . →∞ → → ∞} Then Y1 is closed and it is invariant under the closed subgroup of NG(F ) generated by SF.

Proof. It is straight forward to verify that Y1 is a closed F -invariant subset of Y . Now for any y Y and g S, with the notations above, as i , ∈ 1 ∈ →∞ 1 Φ(t + f(t ))z =Φ(t + f(t ))Φ(t )− Φ(t )z gy. i i i i i i · i i → Therefore gy Y . This completes the proof. ∈ 1 Corollary 4.4.2 (Cf. [DM1, Lemma 2.3]) Let X be an L-homogeneous space of G = SL2(C) and W = w(t):t R be a one-parameter unipotent subgroup of G. Then for any{ x X,∈ the} set Y = w(t)x : t>0 contains a W -orbit. ∈ { } Proof. Let the notations be as in Lemma 4.4.1. Take F = e , Z = Y , and Φ(t)=w(t) for all t>0. Then S = W and due to Proposition{ } 3.2.3, the set Y is nonempty. Therefore by Lemma 4.4.1, WY = SY Y . 1 1 1 ⊂ To study the closures of orbits of N1, in view of Proposition 4.3.2 and the above lemma, the next result is very useful. Proposition 4.4.3 ([DM1, Prop. 2.4]) Let σ and ν be real polynomials such that at least one of them is nonconstant. Let t 0 be such that σ(t) =0 0 ≥ for all t>t0. Deﬁne a function Φ:(t0, ) AN2 as Φ(t)=v(ν(t))d(σ(t)) for all t>t. Then there exists a nontrivial∞ → one-parameter subgroup ρ : R 0 → AN2, and for every s R there exists a function fs :(0, ) R such that as t , ∈ ∞ → →∞ 1 t + f (t) and Φ(t + f (t))Φ(t)− ρ(s). s →∞ s → (Note that due to Lemma 2.2.2, any nontrivial one-parameter subgroup of AN2 1 is either N or vAv− for some v N .) 2 ∈ 2 91 Proof.Fort>t0 and ξ>0, we have

1 1 Φ(t + ξ)Φ(t)− = v(ν(t + ξ))d(σ(t + ξ)) d(σ(t)− )v( ν(t)) · 1 − = v(ψ(t, ξ)) d(σ(t + ξ)σ(t)− ), · where ψ(t, ξ)=ν(t + ξ) σ2(t + ξ)/σ2(t) ν(t). − · Suppose that ψ 0. Since ψ(t, 0) = 0 for all t R, there exist k N and ≡ ∈ ∈ rational functions ψi, where i =1,...,k, such that

k i ψ(t, ξ)= ψi(t)ξ . i=1 For a rational function R(t)=P (t)/Q(t), we deﬁne deg(R) = deg(P ) deg(Q), where P (t) and Q(t) are polynomials in t. Now put −

q = max (1/i) deg ψi. 1 i k ≤ ≤ Then there exists a polynomial P without a constant term and with deg P 1, such that for any sequence s s in R and for any sequence t , ≥ i → i →∞ q ψ(t ,st− ) P (s). i i i → In the case when ψ 0, put q = . ≡ −∞ Note that for any p>q,asi , →∞ p ψ(t ,st− ) 0. i i i → Also as i , →∞ 1 deg σ σ(t + s t )σ(t )− (1 + s) i i i i → and q 1 σ(t + s t− )σ(t )− 1 if q> 1. i i i i → − For s R and t>0, deﬁne ∈ st q if q> 1 f (t)= − s (es 1)t if q −1. − ≤−

Then for any s R, there exists ρ(s) AN2, such that for any sequences s s and t ∈ in R, ∈ i → i →∞ 1 Φ(t + f (t ))Φ(t )− ρ(s); (4.9) i si i i → 92 in fact one can verify that, for every s R, ∈ v(P (s)) if q> 1 − ρ(s)= v(P (es 1))d(es deg(σ)) if q = 1  − − d(es deg(σ)) if q< 1.  − To show that ρ is a one-parameter subgroup of AN ,takes ,s R.For  2 1 2 ∈ t>t0, 1 Φ(t + fs1+s2 (t))Φ(t)− 1 1 (4.10) =Φ(t + f (t))Φ(t + f (t))− Φ(t + f (t))Φ(t)− . s1+s2 s2 · s2

For t R, put yt = t + fs2 (t). Then yt as t . First∈ suppose that q 1. Then t +→∞f (t)=e→∞st. Therefore ≤− s s1+s2 s1 t + fs1+s2 (t)=e t = e yt = yt + fs1 (yt). (4.11)

For notational convenience put rt = s1, for all t>t0, in this case. Now suppose that q> 1. Then − q t + fs1+s2 (t)=t +(s1 + s2)t− q q = yt + s1(t/yt)− yt− q · = yt + rtyt−

= yt + frt (yt), (4.12) q where rt = s1(t/yt)− . Thus due to Eqs. 4.9, 4.11 and 4.12, as t , we have that yt , r s , and →∞ →∞ t → 1 Φ(t + f (t))Φ(t + f (t)) = Φ(y + f (y )) ρ(s ). s1+s2 2 t rt t → 1 Now by Eq. 4.10, we obtain that ρ(s1 + s2)=ρ(s1)ρ(s2). Hence ρ is a one- parameter subgroup of AN2. This completes the proof. In order to put Proposition 4.4.3 in a proper perspective, we give a more general result without a proof. A possibility of such a generalization was suggested to the author by Professor G.A. Margulis. Lemma 4.4.4 Let B denote the group of all upper-triangular matrices in GLn(C). Let Φ:(t0, ) B be a function satisfying the following conditions: (i) for all 1 i,∞ j →n, the map t Φ (t) is a rational function in ≤ ≤ → i,j the variable t, where Φi,j(t) denotes the (i, j)-th co-ordinate of Φ(t); and (ii) Φ(t) as t . Then there exists a nontrivial one-parameter subgroup →∞ →∞ ρ : R B, and for every s R there exists a function fs :(0, ) R such that as→t , we have ∈ ∞ → →∞ 1 t + f (t) and Φ(t + f (t))Φ(t)− ρ(s). s →∞ s → Moreover, all the eigenvalues of ρ(s) are real for every s R. ∈

93 4.5 Closures of orbits of N1

Using the techniques developed in the earlier sections, now we give a proof of the homogeniety of closures of orbits of N1.

Proposition 4.5.1 Let Y be the closure of an N1-orbit in X. Then one of the following possibilities occurs.

1. Y is a compact orbit of N1 or N.

1 2. Y contains an orbit of vAv− for some v N . ∈ 2 3. Y = X.

4. Every compact N1-minimal subset of Y is contained in a compact N- orbit, which is also contained in Y .

Proof. (Cf. [DM2, Prop. 2.3]) Let Z is a given compact N1-minimal subset of Y . It is enough to prove that either (1), (2), or (3) occurs or NZ is a compact N-orbit contained in Y .Let

M = g G N (N ):gZ Y = . { ∈ \ G 1 ∩ ∅} If e M then by Lemma 4.2.1, Y = Z and Y is a compact orbit of a closed ∈ 0 subgroup of NG(N1) = AN containing N1. Therefore by Corollary 2.2.5, Y is a compact orbit of either N1 or N. Now suppose e M. Then by Lemma 4.2.1 and Proposition 4.3.2, there exist polynomials σ∈and ν such that atleast one of them is nonconstant, σ(0) = 1, ν(0) = 0, and for any t R, if σ(t) = 0 then v(ν(t))d(σ(t))Z Y . Put Φ(t)=v(ν(t))d(σ(t)) for every∈ t R such that σ(t) =0. ⊂ ∈ For a while, put Y = Z in the above discussion. Then either (i) Z is a compact orbit of N1 or N or (ii) Z is invariant under the closed subgroup L generated by Φ(t):t R,σ(t) =0 . { ∈ } Comming back to the original situation: In the case (i), if Z is a compact N-orbit then we are through. In the case (ii), by Proposition 4.4.3, either 1 1 L N or L vAv− for some v N .IfL vAv− then possibility (2) ⊃ 2 ⊃ ∈ 2 ⊃ occurs. If L N2 and if Z is not a compact N-orbit then by Theorem 4.3.3, Z = X and hence⊃ possibility (3) occurs. Thus we only need to consider the case when Z is a compact N1-orbit. First suppose that σ 1. In which case ν is a nonconstant polynomial and v(ν(t))Z Y for all t ≡R. By Corollary 3.2.4, NZ is a compact N-orbit. By ⊂ ∈ Corollary 4.4.2, v(ν(R))Z contains an N2-orbit. Therefore NZ Y , and we are through. ⊂

94 Now assume that σ is not a constant polynomial. Deﬁne Y1 Y and S AN as in Lemma 4.4.1. Note that for all large t>0 and any z⊂ Z, ⊂ 2 ∈ Vol(N(Φ(t)z)) = det(Ad Φ(t) ) Vol(Nz)= σ(t) 4 Vol(Nz). | |N |· | | · Since σ(t) as t , by Corollary 3.4.6, Y1 X(N). By| Proposition|→∞ 4.4.3,→∞ the closed subgroup generated⊂ by S contains either 1 N2 or vAv− for some v N2. Therefore by Lemma 4.4.1, either NY1 Y 1 ∈ ⊂ or vAN1v− Y1 Y . Since Y1 X(N), by Theorem 4.3.3, NY1 is dense in X. Thus if σ is nonconstant⊂ then⊂ possibilities (2) or (3) occur. This completes the proof. Proposition 4.5.2 If possibility (4) of Proposition 4.5.1 occurs, then either Y is a compact N-orbit or Y = X.

Proof. By Corollary 4.1.3 and Proposition 4.1.4, Y contains a compact N1- minimal subset. Therefore by the hypothesis, there exists a compact N-orbit, say Z, contained in Y .Let M = g G N (N):gZ Y = . { ∈ \ G ∩ ∅} If e M then there exists z Z and g N (N) such that gz Y and ∈ ∈ ∈ G ∈ Y = N1(gz). Thus Y is contained in the compact N-orbit gZ. Hence Y is a compact N-orbit. Now suppose that e M. Then by Lemma 4.2.2 and Proposition 4.3.2, there exists a nonconstant∈ complex polynomial σ such that d(σ(t))Z Y = for all large t>0. Now d(σ(t))Z is a compact N-orbit. Therefore for∩ every ∅ y d(σ(t))Z, the set N1y is a compact N1-minimal subset. Hence by the hypothesis∈ of this proposition, d(σ(t))Z Y for all large t>0. Now by ⊂ Theorem 2.4.7, Y = X. This completes the proof. In view of the above propositions, to prove that the closure of any N1- orbit is an orbit of a closed subgroup of G, we are left to show this under an 1 additional assumption that Y contains an orbit of vAv− for some v N2. Observe that if a set S X is the closure of an orbit of a subgroup∈ F of ⊂ 1 G then for any g G, the set gS is the closure of an orbit of gFg− . ∈ 1 Since v N (N ), the set Y := v− Y is the closure of an orbit of N and ∈ G 1 1 Y contains an orbit of A. Hence for further investigation we can assume that the closure of the given N1-orbit contains an orbit of AN1, and we should show that Y is an orbit of a closed subgroup of G.

Orbit closures containing AN1-orbits Now in view of Lemma 4.2.1, we need to see how large is the set

N M(AN ) N (N ) /N, 1 1 ∩ G 1 95 where M G N (N ) and e M. The next lemma is useful in this regard. ⊂ \ G 1 ∈ Lemma 4.5.3 Given sequences h H(= SL (R)) and t in R, there { i}⊂ 2 i →∞ exists a sequence ik k N such that the following condition is satisﬁed: There { } ∈ exists s∗ R such that for any s R s∗ and a sequence sk s in R, we∈ have∪{±∞} ∈ \{ } → π(u(s t )h ) π(e) k ik ik → as i , where π : H H/AN is the natural quotient map. →∞ → 1 Further if h is a constant sequence then s∗ can be chosen to be . { i} ±∞

Proof. (Cf. [DM1, Sect. 3.5]) The group H = SL2(R) acts on the unit circle S1 R2 as follows: For all ξ S1 and g H, ⊂ ∈ ∈ g ξ = gξ/ gξ . · 

This action is transitive and the stabilizer of p =(1, 0) is AN1. Therefore by Lemma 1.1.2, 1 H/AN1 ∼= S (4.13) as H-spaces. Let i N and put p = h p =: (a ,b) S1. Then for s R, ∈ i i · i i ∈ ∈

2 2 u(sti)pi =(ai + stibi,bi)/ (ai + stibi) + bi . Note that

1 b b t− | i| | i| = i . 2 1 (a + s t b )2 + b ≤ ai + stibi ai(tibi)− + s i i i i i | | | | Now there exists a sequence i in N such that as k , k →∞ →∞ (i) a a, b b, where a, b [0, 1]; and ik → ik → ∈ 1 (ii) a (t b )− s∗, where s∗ R . − ik ik ik → ∈ ∪{±∞} Then for every s R s∗ and a sequence s s, ∈ \{ } k → b /(a + s t b ) 0. ik ik k ik ik → Therefore u(s t )p p,ask . Now the conclusion of the lemma k ik ik → →∞ follows from the equivariant isomorphism given by Eq. 4.13. Notation. Recall that = A M (C):trA =0 is the Lie algebra as- G { ∈ 2 } sociated to G, and = A M2(R):trA =0 is the the Lie subalgebra of associatedH to H.{ Put∈ = i . Then } = and is in- variantG under the action of Ad(HP)on H. Now theG one-parameterP⊕H subgroupP G 96 Ad N1 = Ad u(t) t R acts on by unipotent linear transformations. Let 2 { } ∈ P N denote the Lie subalgebra associated to N2. Then = q : (Ad N ) q = q . N2 { ∈P 1 · }

Remark. By Lemma 2.1.2, there exists a neighbourhood Ψ0 of the origin in and a neighbourhood Ω0 of the identity in G, such that the map (q, y) (expG q)(exp y) is a homeomorphism from ( Ψ ) ( Ψ )ontoΩ. → P∩ 0 × H∩ 0 0 Now using Lemma 4.2.3 we deduce the following. Proposition 4.5.4 (Cf. [M2, Lemma 7], [DM1, Cor. 3.5]) Let M be a subset of G (N Ω0)H such that e M. Then there exist a sequence mi M, m \e, a∩ sequence t ∈in R, and a nonconstant polynomial{ }⊂ν i → i →∞ such that the follwing holds: There exists s∗ R such that for any ∈ ∪{±∞} s R s∗ , and a sequence s s, we have ∈ \{ } i → π(u(s t )m ) π(v(ν(s))), i i i → as i , where π : G G/AN1 is the natural quotient map. In→∞ particular, → N MAN v(ν(R)). 1 1 ⊃ Proof.Let m M Ω be a sequence such that m e as i . Then { i}⊂ ∩ 0 i → →∞ for every i N, there exist qi Ψand hi exp( Ψ) H such that m =(expq∈)h . Then q 0 and∈Ph∩ e as i ∈ . NowH∩ for all⊂ large i N, i i i i → i → →∞ ∈ we have that qi 2,becausemi (N Ω0)H. By Lemma 4.2.3,∈ N if we replace∈ m ∩by a subsequence then there exist a { i} sequence ti in R and a nonconstant polynomial ν such that for any sequence s →∞s in R, i → 0 iν(s) Ad u(s t )q as i i i i → 00 ∈N2 →∞ and hence as i , →∞ u(s t )(exp q )u( s t ) v(ν(s)). (4.14) i i i − i i → By Lemma 4.5.3, there exist a sequence i in N and an element s∗ k →∞ ∈ R such that given a sequence s s R s∗ , we have ∪{±∞} k → ∈ \{ } π(u(s t )h ) π(e) (4.15) k ik ik → as k . Thus by Eq. 4.14 and Eq. 4.15, →∞ π(u(s t )m )=(u(s t )expq u( s t ))π(u(s t )h ) v(ν(s))π(e) k ik ik k ik ik − k ik k ik ik → As k . This completes the proof. It→∞ will be convenient to note the following simple observation.

97 Lemma 4.5.5 1. The set X(N) does not contain a closed AN1-invariant subset.

2. Let Z be a closed AN1-invariant subset of X.IfZ contains an orbit of 1 vAv− for some v N e , then Z = X. ∈ 2 \{ }

3. Let Z be a proper, closed AN1-minimal subset of X. Then for every z Z X(N), we have that N z = Z. ∈ \ 1

Proof. By Corollary 3.4.6, any closed N1-invariant subset of X(N) is contained in X (N) for some η>0. For any z X(N) and a A, η ∈ ∈ Vol(N(az)) = det(Ad a ) Vol(Nz). | |N |

But then Xη(N) does not contain an orbit of A. Hence X(N) does not contain a closed AN1-invariant subset. This proves Part 1). For t = 0 and v = v(t) N , ∈ 2 1 1 2 AvAv− d(a)v(t)d(a− )v( t)=v((a 1)t):a>0 . ⊃{ − − } Therefore 1 AvAv− v([0, )) or v(( , 0]). (4.16) ⊃ ∞ −∞ 1 Suppose that Z is AN1-invariant and contains an orbit of vAv− . Then 1 AvAv− z Z for some z Z. Now by Eq. 4.16 and Corollary 4.4.2, we have ⊂ ∈ that Z contains an orbit of N2 and hence it contains an orbit of AN. Therefore by Theorem 2.4.7, Z = X. This proves Part 2). Now suppose that Z is AN1-minimal. Let z Z X(N). Then by Proposi- ∈ \ 1 tions 4.5.1 and 4.5.2, the set N z contains an orbit of vAv− for some v N . 1 ∈ 2 If v = e then for some y N1z, we have N1z AN1y and hence AN1y = Z by the minimality of Z.Ifv∈= e then by Part 2),⊃Z = X, which is a contradiction. This proves Part 3). The next conditional result is quite important.

Lemma 4.5.6 Let Y be a closed N1-invariant subset of X. Suppose that Y contains a closed AN -minimal subset, say Z, and let z Z X(N). Deﬁne 1 0 ∈ \ M = g G (N Ω )H : gz Y { ∈ \ ∩ 0 0 ∈ }

and suppose that if e M, (where Ω0 is the set as deﬁned in the remark before Proposition 4.5.4). Then∈ Y = X.

Proof. (Cf. [DM1, Prop. 4.3]) Let the notations be as in the statement of Proposition 4.5.4. Put yi = miz0 for every i N. Now yi z0, z0 X(N), and t . Therefore by Corollary 3.4.7, for∈ every s R→and every∈ λ>1 i →∞ ∈ 98 there exists a sequence α [1,λ] such that a subsequence of u(α st )y i ∈ { i i i} converges to a point y Y X(N). By passing to subsequences we may assume that α α [1,∈λ] and\ i → ∈

u(α st )y y. (4.17) i i i →

Given s R s∗ , let λ>1 be such that s∗ [s, λs]. Now by Propo- ∈ \{ } ∈ sition 4.5.4, there exist an increasing sequence ik N and a sequence g AN such that as k , { }⊂ { k}⊂ 1 →∞ u(α st )m g v(ν(αs)). (4.18) ik ik ik k → Therefore by Eq. 4.17 and Eq. 4.18, as k , →∞

1 1 gk− z0 =(u(αik stik )mik gk)− (u(αik stik ))yi 1 v(ν(αs))− y =: z Z. → ∈

Since z Ny, we have z X(N). As we can assume that Z = X, due to ∈ ∈ Lemma 4.5.5, Z = N1z. Therefore

Y N y = v(ν(αs))N z = v(ν(αs))Z. ⊃ 1 1 Now λ>1 can be chosen arbitrarily close to 1, α [1,λ], and Y is closed. ∈ Therefore v(ν(s))Z Y for every s R s∗ and hence for every s R. Deﬁne N (+) = ⊂v(t):t 0 and∈ N\{( )=} v(t):t 0 . Then ∈ν(R) 2 { ≥ } 2 − { ≤ } ⊃ N2( ), and hence N2( )AN1Z Y . Note that N2( )AN1 = AN1N2( ). By Corollary± 4.4.2, the set±N ( )Z⊂contains an N -orbit.± Therefore Y contains± 2 ± 2 an orbit of AN. Hence by Theorem 2.4.7, Y = X. In the next proposition we remove the extra conditions involved in the previous proposition.

Proposition 4.5.7 Let Y be the closure of an orbit of N1. Suppose Y contains an orbit of AN1. Then either Y is a closed orbit of H or Y = X.

Proof. (Cf. [DM1, Prop. 4.1]) Let y Y be such that Y = N y . Due 0 ∈ 1 0 to Corollary 4.1.3, Y contains a closed AN1-minimal subset, say Z.By Lemma 4.5.5, there exists z Z X(N) such that N z = Z. Deﬁne 0 ∈ \ 1 0 M = g G : gz N y . { ∈ 0 ∈ 1 0} There are three cases:

1. There exists an open relatively compact neighbourhood Ωof the identity in G such that M Ω H. ∩ ⊂ 99 2. There exist v N and h H such that v = e and vh M. ∈ 2 ∈ ∈ 3. e M N H. ∈ \ 2 In Case 3, by Lemma 4.5.6, Y = X. In Case 1, Ωz N y (H Ω)z . Therefore Ωz N y (H Ω)z . 0 ∩ 1 0 ⊂ ∩ 0 0 ∩ 1 0 ⊂ ∩ 0 Hence Hz0 contains a neighbourhood of z0 in Y . Therefore Z Hz0 z0 and it is AN -invariant. Since Z is closed and AN -minimal, we have\ Hz Z and 1 1 0 ⊃ Z = AN1z0. By Iwasawa decomposition H = K1AN1.ThereforeK1Z = Hz0. Since K1 is compact, the orbit Hz0 is closed. By Corollary 3.4.4, Hz0 admits a ﬁnite H-invariant measure. Hence by Theorem 2.4.11, Hz0 = AN1z0 = Z. Since Hz0 N1y0 = , we have Hz0 = Y . Thus in the Case 1, Y is a closed H-orbit. ∩ ∅ Now consider the Case 2. Take any z Z. Then there exists a sequence t such that u(t )y z as i . Because∈ otherwise az N y for all i →∞ i 0 → →∞ ∈ 1 0 a in a neighbourhood of the identity in AN1, which is not possible as Gy0 is discrete. Now since vh M, there exists t0 R such that u(t0)y = vhz0.By Lemma 4.5.3, if we replace∈ t by a subsequence,∈ then there exists a sequence { i} gi AN1 such that u(ti)hgi e as i . Since u(ti + t0)y0 u(t0)z,we have{ }⊂ as i , → →∞ → →∞

1 u(t + t )y = u(t )vhz = v(u(t )hg )(g− z ) vz i 0 0 i 0 i i i 0 →

1 for some z Z such that u(t0)z = vz. Hence v− z = u( t0)z Z. Thus we ∈ 1 − ∈ have shown that v− Z Z. Now AZ = Z, therefore Z contains an orbit of 1 ⊂ v− Av, and hence by Lemma 4.5.5, Z = X. Thus in the Case 2, Y = X. In view of the remark after the proof of Proposition 4.5.2, we can summerise the Propositions 4.5.1, 4.5.2, and 4.5.7 in the following result.

Theorem 4.5.8 Let Y be the closure of an orbit of N1. Then one of the following holds:

1. Y is a compact orbit of N1 or a compact orbit of N.

1 2. Y is a closed orbit of vHv− for some v N. ∈ 3. Y = X.

This theorem settles Raghunathan’s closure conjecture for unipotent ﬂows in L-homogeneous spaces of G = SL2(C).

100 4.6 Closures of orbits of subgroups containing a nontrivial unipotent element

Closed AN1-invariant subsets

Theorem 4.6.1 Let Y be the closure of an AN1-orbit in X. Then either Y is a closed H-orbit or Y = X. In particular, any H-orbit in X is either closed or dense.

Proof.Lety Y be such that Y = AN y . If the orbit N y is not compact 0 ∈ 1 0 1 0 then due to Theorem 4.5.8, Theorem 2.4.7, and Lemma 4.5.5, either Y = Hy0 or Y = X. Therefore we can assume that the orbit Y0 := N1y0 is compact. Let Z be a closed AN -minimal subset of Y . Since we can assume that Z = X, 1 by Lemma 4.5.5, there exists z0 Z such that Z = N1z0. Now by Proposi- tion 4.5.7, Z is a closed H-orbit. ∈ Deﬁne M = g G : gz AY . { ∈ 0 ∈ 0} If e M H, the orbit Hz is open in Y and hence AY Hz = . ∈ \ 0 0 ∩ 0 ∅ Therefore Y = Hz0. If e M (N Ω )H then by Lemma 4.5.6, Y = X. ∈ \ 2 ∩ 0 Now suppose that e M H and e M (N2 Ω0)H. Note that d(a)Y as a 0∈ (see\ the proof of∈ Lemma\ 3.1.1).∩ Therefore the 0 →{∞} → following holds: There exist sequences ai and si 0 in R, and yi y in Y such that for every i N, →∞ → → 0 ∈ d(a )y v(s )Hz . i i ∈ i 0 Thus for every i N, ∈ 1 2 y d(a− )v(s )Hz = v(a− s )Hz . i ∈ i i 0 i i 0

Therefore y Hz0 and hence y0 N1y Hz0. Thus Y = Hz0, which is a contradiction∈ to the assumption that∈ e ⊂M H. This completes the proof. ∈ \ 

Proposition 4.6.2 Let Yi i N be a sequence of closed H-orbits in X and { } ∈ let vi i N be a sequence in N2. Suppose that any one of the following two ∈ conditions{ } is satisﬁed:

(i) Y = Y for all i = j. i j (ii) v as i . i →∞ →∞ 101 Put ∞ Y = viYi i=1 Then Y = X.

1 Proof. (Cf. [M5, Thm. 4]) First note that viYi is a closed orbit of viHvi− for each i N. In particular, Y is N1-invariant. Due to Corollary 3.4.6, there exists∈y v Y X(N)foreveryi N such that a subsequence of y i ∈ i i \ ∈ { i} converges to a point y0 X X(N). ∈ \ 1 We can assume that N1y0 = X. Then by Theorem 4.5.8, N1y0 = vHv− y0 1 1 for some v N2.Foreveryi N, replace vi by v− vi and yi by v− yi. Replace 1 ∈ 1 ∈ y0 by v− y0 and Y by v− Y . Now without loss of generality by passing to a subsequence we may assume that y y and N y = Hy . i → 0 1 0 0 For every i N, let mi G be such that yi = miy0 and mi e as i . Put M = m :∈i N and∈Z = Hy . → →∞ { i ∈ } 0 If e M (N Ω )H then by Lemma 4.5.6, Y = X and we are through. ∈ \ 2 ∩ 0 Now suppose that mi (N2 Ω0)H for all large i N. For any such large i N, express m = w h,∈ where∩w N and h H.∈ Then by Theorem 4.5.8, ∈ i i i ∈ 2 ∈

viYi = N1yi = N1miy0 = N1wi(hy0)=wiN1(hy0).

Since N1hy0 cannot be a compact N1-orbit, we have N1(hy0)=Z. Thus viYi = 1 1 wiZ. Now viYi is a closed orbit of viHvi− and wiZ is a closed orbit of wiHwi− . 1 1 Therefore by Lemma 1.1.3, v Hv− = w Hw− . Since N N (H)= e ,we i i i i 2 ∩ G { } get vi = wi and hence Yi = Z. As i : mi e and hence vi = wi e. Thus the conditions (i) and (ii) are violated,→∞ if e→ M (N Ω )H. This→ completes the proof. ∈ \ 2 ∩ 0

Corollary 4.6.3 Any closed AN1-invariant subset is either dense or it is a ﬁnite union of closed H-orbits.

Proof. By Theorem 4.6.1, any closed AN1-invariant subset is H-invariant. Now the conclusion follows from Proposition 4.6.2. 

Corollary 4.6.4 Let Z be a closed H-orbit and let g G be a sequence { i}⊂ such that the sequence π(gi) is not contained in any compact subset of G/H, where π : G G/H denotes{ } the natural quotient map. Then →

X = giZ. i N ∈

102 Proof. By Iwasawa decomposition (see Lemma 2.2.1), for each i N,wecan express g = k v b , where k K, v N , and b AN . Since AN∈ H,by i i i i i ∈ i ∈ 2 i ∈ 1 1 ⊂ passing to a subsequence, we may assume that ki k K and vi as i . Now by Proposition 4.6.1, for any m N, → ∈ →∞ →∞ ∈

∞ 1 ki− giZ = X. i=m Hence 1 giZ = k− X = X. i N ∈ Corollary 4.6.5 For any x X and a sequence a in R, let ∈ i →∞

Y = d(ai)N1x. i N ∈ Then either Y is a closed H-orbit or Y = X. (Compare this result with Theorem 2.4.7 and Theorem 2.4.11.)

Proof. Follows from Theorem 4.5.8, Proposition 4.6.2, and an argument as in the proof of Theorem 4.6.1. We omit the details. 

A description of closed N1-invariant subsets of X

Let Y be a proper closed N1-invariant subset of X.Foranyy Y X(N), 1 ∈ \1 there exists v N such that N y =(vHv− )y. If we put z = v− y then ∈ 2 1 N1z = Hz and vHz Y . For any closed H⊂-orbit Z X, deﬁne ⊂ L = v N : vZ Y . Z { ∈ 2 ⊂ }

Then LZ is closed. Let

Σ= Z X : Z is a closed H-orbit and L = . { ⊂ Z ∅} By Proposition 4.6.2, Σis ﬁnite and L is compact for every Z Σ. Put Z ∈ Y = L Z. H Z · Z Σ ∈ Then Y is closed subset of Y . Also Y X(N) Y . Therefore Y X(N)= H ⊂ ∪ H \ YH .

103 Note that due to Lemma 1.1.3, for any Z = Z Σ, ∈

L Z L Z X(N). Z ∩ Z ⊂

Also note that for any t>0, the set N(t)YH is closed and N1-invariant, where N(t)= u(z):z C, (z) t, (z) t . { ∈ | |≤ | |≤ }

Theorem 4.6.6 Let Y be a proper, closed N1-invariant subset of X. Then for every t>0, there exists η>0 such that

Y N(t)Y X (N). ⊂ H ∪ η Proof. Suppose Y N(t)Y X (N)foreveryη>0. Then by Corol- \ H ⊂ η lary 3.4.6, there exists a sequence yi Y N(t)YH such that yi y for { }⊂ \ 1 → some y Y X(N)=YH X(N). Now N1y = vHv− y for some v N2. Put ∈1 \ 1 \ 1 1 ∈ Y = v− Y , Y = v− Y , y = v− y for all i N, and y = v− y. Note that H H i i ∈ Y is a proper closed N -invariant subset of X. Also y y and N y = Hy. 1 i → 1 Since Hy Y , we have that y N(t)Hy for all i N. Therefore by ⊂ H i ∈ ∈ Lemma 4.5.6, Y = X, a contradiction to the assumption that Y , and hence Y is a proper subset of X. We shall see that by group theoretic manipulations using the above result, one can derive a good amount of information about the dynamics of unipotent ﬂows on X.

Generalized Raghunathan conjecture for SL2(C) Proposition 4.6.7 Let Y be the closure of an orbit of the discrete unipotent subgroup u(n):n Z N . Then one of the following holds. { ∈ }⊂ 1 1. Y is contained in a compact N-orbit. In which case there are the following possibilities:

(a) Y is a ﬁnite set.

(b) Y is a compact N1-orbit.

(c) Y is a union of ﬁnitely many compact orbits of N3, where N3 is a one-parameter subgroup of N and N1N3 = N. (d) Y is a compact N-orbit.

1 2. Y is a closed orbit of vHv− for some v N . ∈ 2 3. Y = X.

In particular for any x X X(N), the set u(n)x : n N is u(t) t R- ∈ invariant. ∈ \ { ∈ } { }

104 Proof. Put Y1 = u(t):t [0, 1] Y . Then Y1 is the closure of an N1-orbit. Therefore by Proposition{ ∈ 4.5.8 and}· Proposition 4.5.7, one of the following holds:

(1) Y1 is contained in a compact N-orbit.

1 (2) Y is a closed vHv− -orbit for some v N . 1 ∈ 2

(3) Y1 = X.

2 In the Case (1), Y is contained in a compact N-orbit. Since N ∼= R , the further deductions follow from Theorem 1.4.4. In the Cases (2) and (3), the group u(n):n Z acts ergodically on Y1 (see Theorem 2.4.6). Therefore by Lemma{ 2.4.1, there∈ } exists y Y such that 1 ∈ 1 the set u(n)y1 : n Z is dense in Y1. Now y1 = u(t)y for some t [0, 1] and y Y .{ Therefore ∈ } ∈ ∈ Y u(n)y : n Z = u( t)Y = Y . ⊃ { ∈ } − 1 1

Thus Y = Y1 in the Cases (2) and (3), and the proof is complete. 

Proposition 4.6.8 Let F be a closed subgroup of NG(N)=DN containing u(1). Let Y be the closure of an orbit of F . Then one of the following holds:

1 (1) F SN1 and Y is a closed orbit of vSHv− , where v N2 and S is a ⊂ k ∈ subgroup of S = d(i ) k Z. { } ∈ (2) F MN, Y is contained in a compact MN-orbit and one of the following⊂ holds:

(a) Y is an orbit of a closed subgroup of MN. (b) There exist v N, θ π/4,π/6,π/3 , and w C 0 such that ∈ ∈{ } ∈ \{ } kθi 1 2θi F = v d(e ) k Zv− u(z):z Z[e ] w { } ∈ ·{ ∈ · } and kθi Y = v d(e ) k Z Lx, { } ∈ · where x Y , L is a closed subgroup of N containing F N, and L0 is one-dimensional.∈ ∩

(3) F AN , F N , and Y is a closed H-orbit. ⊂ 1 ⊂ 1 (4) Y = X.

Proof. Due to Proposition 2.2.5, there are four cases:

105 (i) F SN . ⊂ 1 (ii) F SN and F MN. ⊂ 1 ⊂ (iii) d(an):n Z N F AN for some a>1. { ∈ } 1 ⊂ ⊂ 1 n (iv) d(α ):n Z N F for some α C∗ with α > 1. { ∈ } ⊂ ∈ | | The case (i) was essentially considered in Proposition 4.6.7. In the case (ii), there exists t>0 such that if we deﬁne Y1 = N(t)Y then Y1 is closed and N-invariant. If Y1 = X then using ergodicity of the u(1)-action and arguing as in the proof of Proposition 4.6.7, we obtain that Y = X.IfY1 = X then by Theorem 4.3.3, there exists x Y such that Nx is compact. Therefore Y MNx. There exists a closed∈ subgroup of L of N containing F N, such⊂ that (F N)x = Lx. Since F/(F N) is a ﬁnite group, Y = FLx∩. Now the conclusions∩ (a) and (b) of the Statement∩ (2) follow from the description of closed subgroups of MN given by Proposition 2.2.5. In the case (iii) by Corollary 4.6.5, Y is either a closed H-orbit or Y = X. And in the case (iv) by Theorem 2.4.7, Y = X.

Lemma 4.6.9 Let w(t):t R be a nontrivial unipotent subgroup of H { ∈ } and let U be the horospherical subgroup of G containing w(t) t R. Let Y be ∈ a closed w(n):n Z -invariant subset of X, and suppose{ that} Y contains a closed H{-orbit, say∈Z.} Deﬁne

M = g G (UH):gZ Y = . { ∈ \ ∩ ∅} Now if e M, then Y = X. ∈

Proof. Put Y1 = w([0, 1])Y . Then Y1 is w(t) t R-invariant. Now there exists 1 { } ∈ 1 h H such that h w(t):t R h− = N1 and hUh− = N. Therefore hY1 is N∈-invariant, Z {hY , and∈ } 1 ⊂ 1 1 hMh− = g G NH : gZ hY = . { ∈ \ ∩ 1 ∅}

Hence by Lemma 4.5.6, hY1 = X. Thus Y1 = X. Now using ergodicity of the action of u(1) on X, and arguing as in the proof of Proposition 4.6.7, we get that Y = X. 

Proposition 4.6.10 Let F be a closed subgroup of G containing a nontrivial unipotent element of G. Suppose that F NG(U) for any nontrivial horospherical subgroup U of G. Let Y be a closed⊂ F -invariant subset of X. Then one of the following holds:

(1) Y is a ﬁnite set and F is a discrete group.

106 1 (2) F NG(gHg− ) and Y = Y0 E, where g G, Y0 is a union of ﬁnitely ⊂ 1 ∪ ∈ many closed gHg− -orbits, and E is a ﬁnite F -invariant subset. Further, if E = then F is a discrete group. ∅ (3) Y = X.

Proof. By Lemma 2.2.10, there exist three distinct horospherical subgroup U of G, where k I = 1, 2, 3 , and nontrivial one-parameter subgroups k ∈ { } uk(t) t R Uk such that uk(1) F , where k I. ∈ { Let} ⊂ ∈ ∈

Y = y Y : u (t)y Y for some k I, and all t R . 0 { ∈ k ∈ ∈ ∈ }

If Y0 = then by Proposition 4.6.7 and Corollary 3.4.6, there exists η>0 such that ∅ Y X (U ). ⊂ η k k I ∈ Therefore by Corollary 3.5.2, Y is finite. Now we assume that Y = . Clearly Y is a closed subset of X.Take 0 ∅ 0 y Y0 and suppose that u1(t)y Y0 for all t R. By Corollary 3.5.3, for every∈ δ>0 there exists s (0,δ)⊂ such that u (s∈)y Y X(U ). Therefore by ∈ 1 ∈ \ 2 Proposition 4.6.7, u2(t)(u1(s)y) Y for all t R. Since δ>0 is arbitrary, we have u (t)y Y for all t R.∈ From this argument∈ one concludes that Y is 2 ∈ ∈ 0 invariant under the closed subgroup, say L, generated by all uk(t) t R, k I. 1 { } ∈ ∈ Now either L = G or L = gHg− for some g G (see Corollary 2.2.10). ∈ 1 If L = G then Y = X and we are through. If L = gHg− then replacing 1 1 Y by g− Y and F by g− Fg we may assume that L = H. Therefore by Theorem 4.6.1, either Y0 = X or Y0 is a finite union of closed H-orbits. In the former case we are through. Now put E = Y Y . Then E is F -invariant. \ 0 If E k I X(Uk) then there exists η>0 such that ⊂ ∈ E X (U ). ⊂ η k k I ∈ Therefore by Corollary 3.5.2, E is finite and we are through. Otherwise there exists y Y Y0 X(Uk) for some k I. By Proposi- tion 4.6.7 and the above discussion,∈ \ \u (t)y : t R = Hy∈. Threfore there { k ∈ } exist y0 Hy k I X(Uk) and a sequence yi Y Y0 such that yi y0 as ∈ i .Let∈ m\∪ e be a sequence in G such{ that}⊂y =\m y for all i N→. Then →∞ i → i i 0 ∈ there exist sequences qi 0 in and hi e in H such that mi =(expqi)hi for all i N (notations→ as in PropositionP → 4.5.4). Since U U = e ,by ∈ 1 ∩ 2 { } passing to subsequences we may assume that exp qi Ul for all large i N, where l 1, 2 . Now by Lemma 4.6.9, Y = X. In∈ which case Y = ∈Y ,a ∈{ } 0 contradiction. This completes the proof.

107 The Proposition 4.6.8 and Proposition 4.6.10 combined together give the following result. Theorem 4.6.11 Let X be an L-homogeneous space of G and F be a closed subgroup of G containing a nontrivial unipotent element of G. Then for every x X, there exists a closed subgroup L of G such that the following holds: ∈ (a) The orbit Lx is closed and it admits a finite L-invariant measure; (b) the subgroup L F has finite index in F ; and ∩ (c) (the main property) (F L)x = Lx. ∩ In fact, the group L can be chosen to contain F , except in the following case: there exist g G, θ π/4,π/6,π/3 , and w C 0 such that ∈ ∈{ } ∈ \{ } kθi 2θi 1 F = g d(e ) k Z u(z):z Z[e ] w g− , { } ∈ ·{ ∈ · } 1 1 the orbit (gNg− )x is compact, (F gNg )x = Lx, and dim(L)=1. ∩ − This theorem verifies the generalized Raghunathan conjecture, which is stated as a result below. Corollary 4.6.12 Let X be an L-homogeneous space of G and F be a subgroup of G generated by unipotent elements of G contained in F . Then for every x X, there exits a closed subgroup L of G containing F such that ∈ Fx = Lx and Lx admits a finite L-invariant measure. The following is an interesting consequence of Theorem 4.6.11. Corollary 4.6.13 Let Γ be a non-co-compact lattice in G. Then for any lattice Γ in G, either [Γ:Γ Γ] < or the set ΓΓ is dense in G. ∩ ∞ Proof. Since G/Γis noncompact, by Theorem 3.1.9, Γcontains a nontrivial unipotent element. Consider the Γaction on G/Γ. By Corollary 2.4.10 and Theorem 4.6.11, either ΓΓ = G or ΓΓ/Γ is a finite subset of G/Γ. This completes the proof. It is natural to ask what happens if both Γand Γ in the above lemma are cocompact lattices. It was pointed out by Dave Witte that the question can be answered, using the next lemma, if one assumes the generalized Raghunathan conjecture for the the diagonal action of G on G/Γ G/Γ (note that the generalized Raghunathan conjecture has been proved by× Ratner [Ra2]).

Notation. LetΓand Γ be cocompact lattices in G. Put X = G/Γ, Y = G/Γ, x0 = eΓ X, and y0 = eΓ. Then X Y is a homogeneous space of G G. Let ∆: G∈ G G denote the diagonal× inclusion. × → × 108 Lemma 4.6.14 Suppose that there exists a closed subgroup F of G G containing ∆(G) such that ×

∆(G) (x ,y )=F (x ,y ). · 0 0 · 0 0

Then either [Γ:Γ Γ] < or ΓΓ is dense in G. ∩ ∞ Proof.LetC be a compact subset of G such that G = CΓ. Then,

∆(C) ∆(Γ) (x ,y )=∆(G) (x ,y )=F (x ,y ). · · 0 0 · 0 0 · 0 0 Since G =Γ, we have ∆(Γ) (x ,y )=(x , Γy ). Therefore x0 · 0 0 0 0 (x , Γy )=∆(C) (x , Γy ) ( x Y ) 0 0 · 0 0 ∩ { 0}× =(F (x ,y )) ((Γ G) (x ,y )) · 0 0 ∩ × · 0 0 =(x ,p (F (Γ G))y ) 0 2 ∩ × 0 =(x0,Ly0), where p : G G G is the projection on the second factor, and L = 2 × → p2(F (Γ G)) is a closed subgroup of G. Thus we conclude that ΓΓ = LΓ. Since∩ ∆×(G) F , we have that Γ L. Therefore by Corollary 2.4.10, either (i) L = G or (ii)⊂L is a discrete subgroup⊂ of G containing Γas a subgroup of finite index. In the first case ΓΓ is dense in G. In the second case, ΓΓ/Γ is a discrete and hence a finite subset of G/Γ. This completes the proof.

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