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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
BY
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 theoryon homogeneous spaces. This was achieved by proving a particular case ofa conjecture of Raghunathan about orbit closures for unipotent actions onhomogeneous spaces. In 1988, the National Board of Higher Mathematics or-ganized an instructional conference in ergodic theory, which provided me agood background to appreciate the significance of Raghunathan’s conjecture.After the conference S.G. Dani suggested me to read the work of Margulis andgive 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 developedby him and Margulis, and write my Master’s thesis on it.
While the group structure of SL2(C) is easy to understand, the dynamicsof unipotent flows on its homogeneous spaces illustrate many of the interest-ing ergodic theoretic phenomena. The study of this case provided me a goodinsight into Raghunathan’s conjecture in the general case. Some of the impor-tant results used in the proof are based on the work of Garland and Raghu-nathan about fundamental domains of noncocompact lattices in semisimplegroups of real rank 1. In turn, their work depends on the theorem of Kazhdanand Margulis about existence of unipotent elements in noncompact lattices insemisimple groups. The main endeavor in writing this thesis is to present thewhole theory for the group SL2(C), using only the language of linear algebraand topological groups.
I wish to thank Professor S.G. Dani for introducing me the subject of er-godic theory on homogeneous spaces. Each discussion with him gave me newinsights into the problems I was studying. I also thank him for suggestingseveral corrections in the thesis. Many of the papers about flows on homo-geneous spaces use the machinery of algebraic groups and discrete subgroups.I was always helped by Professor Gopal Prasad in understanding the resultsused in these papers and learning the proofs. I express my thanks to him. Iwould 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 wasmy most helpful source.
Nimish A. Shah
i
Contents
Introduction 1
1 Topological groups and homogeneous spaces 81.1 Definitions and immediate consequences . . . . . . . . . . . . 81.2 Invariant measures on topological groups . . . . . . . . . . . . 131.3 Invariant measures on homogeneous spaces . . . . . . . . . . . 171.4 Closed subgroups of Rn and Rn/Zn . . . . . . . . . . . . . . . 221.5 Covolumes of discrete subgroups in euclidean spaces . . . . . . 26
2 The group SL2(C) and its homogeneous spaces 302.1 Exponential map, Lie algebras, and Unipotent subgroups . . . 302.2 Subgroups of SL2(C) . . . . . . . . . . . . . . . . . . . . . . . 372.3 The invariant measure on SL2(C)/SL2(Z[i]) . . . . . . . . . . 442.4 Ergodic properties of transformations on homogeneous spaces 47
3 L-homogeneous spaces of SL2(C) 553.1 A result of Kazhdan and Margulis . . . . . . . . . . . . . . . . 553.2 Non-divergence of unipotent trajectories . . . . . . . . . . . . 623.3 Volumes of compact orbits of horospherical subgroups . . . . . 653.4 A lower bound for the relative measures of a large compact set
on unipotent trajectories . . . . . . . . . . . . . . . . . . . . . 703.5 Description of noncompact L-homogeneous spaces of SL2(C) . 76
4 Proof of Raghunathan’s conjecture for SL2(C) 814.1 Minimal invariant sets . . . . . . . . . . . . . . . . . . . . . . 814.2 Topological limits and inclusion of orbits-I . . . . . . . . . . . 844.3 Specialization to the case of SL2(C) . . . . . . . . . . . . . . . 874.4 Topological limits and inclusion of orbits-II . . . . . . . . . . . 914.5 Closures of orbits of N1 . . . . . . . . . . . . . . . . . . . . . . 944.6 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 inergodic theory. In this thesis we shall study in detail the homogeneous spaces ofG = SL2(C) admitting finite G-invariant measures, and verify Raghunathan’sconjecture about closures of orbits of unipotent subgroups of G acting on thesespaces.
A subgroup Γ of a locally compact group G is called a lattice in G, if Γ isdiscrete and the homogeneous space G/Γ admits a finite G-invariant measure.
A linear transformation T of a finite dimensional vector space V is calledunipotent, if (T − 1V )n = 0 for some n ∈ N.
Let G be a Lie group and let G denote its Lie algebra. An element u ∈ Gis called an Ad-unipotent element of G, if the linear automorphism Ad u of Gis unipotent. A subgroup U ⊂ G is called a unipotent subgroup of G, if all ofelements 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] aboutunitary representations of nilpotent groups. For a solvable group G, a resultdue to Mostow [R, Sect.3.3] says that if N is the maximal connected normalunipotent subgroup of G, then every N -orbit in G/Γ is compact. In this caseevery connected unipotent subgroup of G is contained in N , and hence theverification 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/Γ is either denseor periodic; the periodic orbits exist if and only if G/Γ is noncompact. Infact, for a cocompact lattice Γ, Furstenberg [F] proved that any U -invariantprobability measure on G/Γ is G-invariant, and hence it is unique.
1
Let G be a Lie group. For g ∈ G, define
U(g) = u ∈ G : gnug−n →∞ as n →∞.
Then U(g) is called the horospherical subgroup of G associated to g. Note thatU(g) is a closed connected unipotent subgroup of G.
Now let G be a semisimple group (with trivial center and no compactfactors) and Γ be a cocompact lattice in G. It was proved by Veech [V] andBowen [B] that every orbit of a nontrivial horospherical subgroup of G is densein 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, thenfor any x ∈ G/Γ, there exists a compact set C ⊂ G/Γ such that the set R(C)of return times, defined as
R(C) = n ∈ N : u(n)x ∈ C,
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 R(C) of return times haspositive density in the set of positive integers. Using the strenghened verson,in [D6] he verified Raghunathan’s conjecture when Γ is any lattice in a Liegroup G and U is a horospherical subgroup of G. This generalizes Veech’sresult for noncocompact lattices.
A major source of interest in Raghunathan’s conjecture is the fact, observedby Raghunathan, that a very special case of the conjecture implies the validityof 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. Let Q be the quadraticform 2x1x3 − x2
2, and define
H = SO(Q) = g ∈ SL3(R) : Q(gx) = Q(x) for all x ∈ R3.
In [M2], Margulis proved that every relatively compact orbit of H in the spaceSL3(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 techniquesof 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 unipotentelements 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 Gcontained in U . Then for any x ∈ G, there exists a closed subgroup L of Gcontaining 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 closuresof unipotent subgroups. Generalizing these techinques and the results aboutreturning 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)/Γ iseither dense or closed, and (ii) Raghunathan’s conjecture holds for the orbits ofunipotent subgroups of H acting on SL3(R)/Γ, where Γ is a lattice in SL3(R).
In this thesis we shall verify the generalized conjecture for the Lie groupG = SL2(C), using the methods of [M2], [DM1] and [DM2]. In this case itis possible to give a proof illustrating the ideas without having to go throughthe kind of technical work involved in the case of SL3(R).
It would be appropriate to mention here that recently Ratner [Ra2] hassettled Raghunathan’s conjecture by proving the following much stronger re-sult.
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 Lxis 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,
limT→∞
1
T
T
0
φ(u(t)x) dt =
Lx
φ dν.
It is hoped that this thesis will motivate the reader to study the recent workdone in this area. For various related developements, the reader is referred tothe 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 groupsand their homogeneous spaces. We give Fubini’s formula relating the invari-ant measures on a unimodular group G, a unimodular subgroup F , and thehomogeneous space G/F .
We begin Chapter 2 by introducing very basic concepts about linear Liegroups, their Lie algebras, and the exponential maps. We need to introducesome notations at this stage to decribe the furter contents.
3
Notation. Let G = SL2(C), Γ be any lattice in G, and µ be the G-invariantprobability measure on the homogeneous space X = G/Γ. Let
K = g ∈ G : ggt = 1,
D =
d(α)
def=
α 00 α−1
: α ∈ C∗
,
M = d(eti) : t ∈ R = D ∩K,
A = d(a) : a > 0,Aη = d(a) : a > 0, a2 < η, where η > 0,
N =
u(z)
def=
1 z0 1
: z ∈ C
,
H = SL2(R) and N1 = N ∩H.
We study properties of these subgroups, and obtain Iwasawa and Bruhatdecompositions of G. In the last section of this chapter, we study ergodicproperties of the subgroup actions on X. Using Mautner phenomenon weprove 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 followingimportant result.
Theorem 2 (Cf. [DM2, Sect. 1.7]) Let an ⊂ C∗be a sequence such that
|an|→∞ as n →∞. Then for every x ∈ X, the set
n∈N
d(an)Nx
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 homo-geneous space X of G admitting a finite G-invariant measure. We prove thefollowing result about the returning of unipotent trajectories to compact setswith 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 (t ∈ [0, T ] | v(t)x ∈ C ) > 1−
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 ad-
mitting a finite G-invariant measure. Then there exists a finite collection Cconsisting of compact N-orbits in X such that the following holds.
1. The set
X(N) :=
O∈C
MAO
is the union of all compact N-orbits in X. (Note that NG(N) = MAN .)
2. There exists η0 > 0 such that
X =
O∈C
KAη0O.
(Note that G = KAN .)
3. There exists 0 < η1 < η0 such that for O, O ∈ C, a, a ∈ Aη1, and
k, k ∈ K, if
kaO ∩ kaO = ∅then O = O
, a = a, kM = kM , and
kaO = kaO.
For a subgroup F ⊂ G and a closed F -invariant subset Z ⊂ X, if everyF -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 ele-
ment. 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 .
(2) Y is a closed orbit of vHv−1for 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, orbitsof larger subgroups in closed N1-invariant subset of X. The following resultabout 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 requiresbasically all the machinary developed before.
Lemma 2 Let Y be a closed N1-invariant subset of X. Suppose that Y con-
tains a closed AN1-minimal subset, say Z. Let z0 ∈ Z \X(N). Define
M = g ∈ G \NH : gz0 ∈ Y .
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 Xand prove the following result, which verifies the generalized Raghunathanconjecture.
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 Gsuch 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
F = gd(ekθi)k∈Z · u(z) : z ∈ Z[e2θi] · w
g−1,
the orbit (gNg−1)x is compact, (F ∩ gNg−1)x = Lx, and dim(L) = 1.
7
Chapter 1
Topological groups and homogeneousspaces
1.1 Definitions and immediate consequences
A group G which is also a topological space is called a topological group if themap m : G × G → G, which is given by m(x, y) = x−1y for all x, y ∈ G, iscontinuous.
For any g ∈ G, let Lg : G → G denote a left translation on G and Rg : G →G denote a right translation on G, defined by Lg(x) = gx and Rg(x) = xg−1
for all x ∈ G. The continuity of the map m is equivalent to the followingconditions:
(a) Lg and Rg are homeomorphisms of G for every g ∈ G.
(b) Given a neighbourhood Ω of the identity e in G, there exists a neigh-bourhood Ω1 of e such that Ω−1
1 Ω1 ⊂ Ω.
To avoid pathological situations, we shall always work with the topologicalgroups which are locally compact, hausdorff, and second countable as topo-logical spaces.
Some examples of topological groups.
1. A countable group with discrete topology is a topological group, calleda discrete group.
2. A finite dimensional real or complex vector space as an additive groupis a topological group.
3. The multiplicative group of n× n invertible real or complex matrices isa topological group, denoted by GLn(k), where k = R or C. Note thatthe space of all n × n matrices is naturally homeomorphic to kn
2and
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 isa topological group. The following are some closed subgroups of GLn(k):
SLn(k) = A ∈ GLn(k) : det A = 1U(n) = A ∈ GLn(C) : AA∗ = 1 (where A∗ = At)
O(n) = GLn(R) ∩ U(n)
SO(n) = SLn(R) ∩ U(n)
5. If G and H are topological groups then the direct product G × H is atopological group with respect to the product topology.
The multiplicative group of elements of unit norm in C∗ is a topologicalgroup, called the circle group and denoted by T1. For n > 1, define Tn
inductively as Tn = T1 ×Tn−1. The topological group Tn is called then-torus.
6. Let G and V be topological groups. Suppose that there exists a con-tinuous map φ : G × V → V with the following property: For everyg ∈ G, the map φg : V → V defined by φg(v) = φ(g, v) (∀v ∈ V ) is anautomorphism. Then we can give a group structure to the the productspace G× V as follows: Identify G with G× e, V with e× V . Forany g ∈ G and v ∈ V , define
gvg−1 = φ(g, v).
Then for all g1, g2 ∈ G and v1, v2 ∈ V , we obtain
(g1v1)(g2v2) = (g1g2)((g−12 v1g2)v2).
This defines a topological group called the semidirect product of G andV , and it is denoted by G φ V . Note that V is a normal subgroup ofG φ V .
Let G = GLn(k) and V = kn, where k = R or k = C, and defineφ(g, v) = gv for all (g, v) ∈ G×V . Then Gφ V is the group of all affinetransformations of kn. Its subgroup O(n) Rn is the group of all rigidmotions of Rn.
The following properties of topological groups are direct consequences ofthe 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 Ω be a symmetric neighbourhood of the identity in G; that is, Ω =Ω−1. Then the set ∪n∈NΩn is an open subgroup of G. In particular, ifG is connected then
G = ∪n∈NΩn.
(For subsets A and B of G, we define AB = ab ∈ G : a ∈ A, b ∈ Band A−1 = a−1 ∈ G : a ∈ A. Note that for any neighbourhood Ω of e,the set ΩΩ−1 is a symmetric neighbourhood of e.)
4. Let N be a discrete normal subgroup of G. If G is connected then N iscontained in the center of G. (To see this, fix n ∈ N and consider themap G g → gng−1 ∈ N .)
5. Let H be a topological group and let φ : G → H be a group homo-morphism. Then φ is continuous if and only if φ is continuous at theidentity.
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 pre-cisely, there exits a map ρ : G × X → X such that the following conditionsare satisfied:
1. ρ is a group action; that is, for any g1, g2 ∈ G, and x ∈ X, we thave that
ρ(g1, ρ(g2, x)) = ρ(g1g2, x).
When the action ρ is understood, we denote ρ(g, x) by simply g · x orgx.
2. ρ is a continuous map.
Some examples of G-spaces:
1. Let G = GLn(C). Then Cn is a G-space with respect to the standardaction of G via linear transformations.
2. Let X be a G-space and F be a closed subgroup of G. If a locally compactsubset Y ⊂ X is invariant under the action of F , then Y becomes anF -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 bardenotes the closure.
3. Let F be a closed subgroup of G. Consider the left action of G on thequotient space G/F . Let q : G → G/F be the natural quotient map.Equip the space G/F with the finest topology with respect to which qis continuous. For this topology a subset S ⊂ G/F is open if and onlyif q−1(S) is open in G. With this topology, the quotient space G/Fbecomes a G-space. Note that G/F is hausdorff because F is closed.
10
Definition. A G-space X is called a homogeneous space of G if the followingholds: G acts transitively on X and for every x ∈ X, the map ρx : G → X,defined by ρx(g) = gx for all g ∈ G, is open.
As a consequence of Baire’s category theorem, the condition in this defini-tion can be weakned. Recall that the category theorem says that any locallycompact, hausdorff, second countable topological space is of second category;that is, the space cannot be expressed as a countable union of closed subsetswith 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 Ω of e in G,the set Ωx is a neighbourhood of x in X. Let Ω1 be a compact neighbourhoodof e such that Ω−1
1 Ω1 ⊂ Ω. As G is second countable, there exists a sequencegi ⊂ G such that G = ∪i∈NgiΩ1. By transitivity of the G-action,
X =
i∈N
giΩ1x.
As Ω1 is compact, so is Ω1x. Since X is of second category, the interiorof giΩ1x is non-empty for some i ∈ N. Since the elements of G act on Xas homeomorphisms, the interior of Ω1x is nonempty. Therefore there existsω ∈ Ω1 such that Ω1x contains a neighbourhood of ωx. But then ω−1Ω1x ⊂ Ωxand ω−1Ω1x contains a neighbourhood of x. This completes the proof.
Notation. Let X 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
Fx = g ∈ F : gx = x.
Then Fx is closed subgroup of G. Let π : F → X be the map defined byπ(g) = gx, for all g ∈ F . Let q : F → F/Fx be the natural quotient map.Then there exists a map π : F/Fx → X such that π = π q. Clearly, π isan injective map. Now π is continuous and the topology of F/Fx is the finesttopology such that q is continuous. Therefore π 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 onFx, by Lemma 1.1.1, Fx is a homogeneous space of F . Therefore π mapsopen subsets of F onto open subset of Fx. Therefore π maps open subsets ofF/Fx onto open subsets of Fx. Thus π is a homeomorphism onto Fx.
We know that F/Fx is locally compact, therefore if π is a homeomorphismonto Fx then Fx is locally compact. This completes the proof of the first part.
Next suppose that Fx is closed. Then the intersection of any compactsubset of X with Fx is compact. In particular, Fx is locally compact. Nowfor a compact set C ⊂ X, π−1(C) = π−1(C∩Fx). Since π is a homeomorphismonto Fx, the set π−1(C ∩Fx) is compact. This shows that π 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 setK ⊂ G such that C ⊂ Kx.
2. Let C be a compact subset of X. Then given any open subset Ψ of Xcontaining C, there exits a neighbourhood Ω of 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 KYis closed in X.
5. Let F be a close subgroup of G and x ∈ X. Suppose if the orbit Fx isclosed then the orbit F 0x is a component of Fx. In particular, F 0x isclosed. (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 Ω of e such that for all x ∈ Xand all g ∈ Ω, we have that
|φ(gx)− φ(x)| < .
(Here Cc(X) denotes the space of all complex valued continuous functionson X with compact support; that is, thoes vanishing outside a compactsubset.)
12
The following applications of Corollary 1.1.2 are recorded here for futurereference.
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. Let z ∈ 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 Ω of the identity e in G such thatΩΩ−1 ∩ Gz = e, (Fz ∩ Ωz) = (F ∩ Ω)z, and (Hz ∩ Ωz) = (H ∩ Ω)z. Thisimplies that (Fz ∩ Hz ∩ Ωz) = (F ∩ H ∩ Ω)z. Hence (F ∩ H)z is open inFz∩Hz = Z for every z ∈ Z. Now (F ∩H)x is closed, because its complementin 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 resultis 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 measureon Y such that for every borel set E ⊂ Y , we have that
T∗µ(E) = µ(T−1E).
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 finiteon compact subsets. It is a fact that any locally finite measure on a locallycompact, hausdorff, second countable topological space is regular (see [Ru2,Sect. 2.8]). To avoid pathological situations, we always work with the measureswhich are locally finite and regular. Note that if µ is locally finite and T is aproper 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 righttranslations on Rn preserve the measure m. For any invertible lineartransformation T on Rn,
T∗m = | det T |−1m
(see [Ru2, Sect. 8.28]). Thus if T ∈ SLn(R) then T preserves m.
2. Let µ be the standard ‘arc length measure’ on the circle group T1. Thenthe rotation on T1 by any angle preserves µ. For every n ∈ N, thehomomorphism z → zn of T1 also preserves µ.
Definition. Let G be a topological group. A (nontrivial) locally finite measureµ on G is called a left haar measure if it is preserved by all the left translationson G. That is, for every g ∈ G, µ(gE) = µ(E) for all borel measurable subsetsE ⊂ G. Equivalently, for all φ ∈ Cc(G) and all g ∈ G,
G
φ(gx) dµ(x) =
G
φ(x) dµ(x).
Similarly one defines a right haar measure on G. Note that if T is thetransformation on G defined by T (g) = g−1 for all g ∈ G, and if µ is a left (orright) 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) topolog-
ical 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 onG. For the applications we shall explicitly construct left haar measures on thegroups we shall work with. For ‘uniqueness’ of the left haar measure, we givea 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 mea-sures on G. Then the product measure ν×µ on the group G×G is also a lefthaar 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 thetransformations S and T preserve the measure µ×ν. For every φ ∈ Cc(G×G),
G×G
φ S d(µ× ν) =
G
G
φ(x, xy) dν(y)
dµ(x)
=
G
G
φ(x, y) dν(y)
dµ(x)
=
G×G
φ d(µ× ν),
14
where the first and the third equalities hold due to Fubini’s theorem, andthe second equality holds because ν is a left haar measure on G. This showsthat the measure µ × ν is preserved by S. Similarly by changing the orderof integration and using the left invariance of µ one proves that T preservesµ× ν.
Let Ω be an open relatively compact subset of G. Now the measure µ× νis preserved by S−1 T . Note that S−1 T (x, y) = (yx, x−1) 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)
G
ψ dν =
G
G
χΩ(x)ψ(y) dµ(x)dν(y)
=
G×G
χΩ(yx)ψ(x−1) dµ× ν(x, y)
=
G
ν(Ωx−1)ψ(x−1) dµ(x). (1.1)
By Fubini’s theorem, x → ν(Ωx−1) is a measurable function on (G, µ) (see [Ru2,Sect.7.8]). Since µ and ν are arbitrary left haar measures on G, by puttingµ = ν, we obtain that y → ν(Ωy) is a measurable function on (G, ν). SinceΩy is an open and a relatively compact subset of G, 0 < ν(Ωy) < ∞ for everyy ∈ G. For any positive borel measurable function φ on G, define
ψ(y) = φ(y−1)/ν(Ωy)
for all y ∈ G. Then ψ is a measurable function on (G, ν). Now by Eq. 1.1,
G
ψ dν =1
µ(Ω)
G
φ dµ. (1.2)
By putting µ = ν in Eq. 1.2, we get
G
ψ dν =1
ν(Ω)
G
φ dν. (1.3)
Thus by Eq. 1.2 and Eq. 1.3, for every positive borel measurable functionφ on G,
1
µ(Ω)
G
φ dµ =1
ν(Ω)
G
φ dν.
Now for any borel measurable subset F of G, by putting φ = χF in thisequality, we get
µ(F )
µ(Ω)=
ν(F )
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 righttranslation on G commute with each other, (Rg)∗µ is a left haar measure onG for every g ∈ G. Therefore by the ‘uniqueness’ of left haar measures thereexists a positive constant, say ∆(g), such that (Rg)∗µ = ∆(g)µ. Again due tothe ‘uniqueness’, the function ∆(g) does not depend on the choice of µ. Thusfor any measurable subset E ⊂ G,
(Rg)∗µ(E) = µ(Eg) = ∆(g)µ(E).
Equivalently, for every φ ∈ Cc(G),
G
φ Rg dµ = ∆(g)
G
φ dµ.
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 grouphomomorphism.
To prove the continuity, let > 0 be given. Let C be a compact subset ofG such that µ(C) > 0. By the regularity of µ, there exists an open subset Uof G containing C such that µ(U \ C) ≤ · µ(C). Let Ω be a neighbourhoodof e in G such that C(Ω ∪ Ω−1) ⊂ U . Then for every g ∈ Ω,
∆(g) = µ(Cg)/µ(C) ≤ µ(U)/µ(C) ≤ 1 +
and∆(g) = µ(C)/µ(Cg−1) ≥ 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 groupsis 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 ∈ Cc(G),
G
φ(g) dµ(g) =
G
φ(g−1)∆(g−1) dµ(g).
If ∆ is a trivial then G is called a unimodular group. Thus G is unimodularif and only if G admits a measure which is invariant under all the left and theright translations.
Some examples:
16
1. Any discrete group Γ is unimodular. Because the ‘counting measure’ onΓ is 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,where [G, G] is the subgroup generated by xyx−1y−1 : x, y ∈ G. Thegroup G = SLn(k) has this property, where k = R or k = C.
5. The following is a typical example of a nonunimodular group: Let
B = ρ(a, b) =
a b0 a−1
∈ SL2(R) : a ∈ R∗, b ∈ R.
Let m be the labesgue measure on R2, restricted to (R \ 0)×R, andµ = ρ∗(m) be its projection on B. Using Fubini’s theorem one can verifythe 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, letµn be the product measure on Tn. Observe that any invertible n×nmatrix A with integral entries acts on Tn ∼= Rn/Zn (see Sect. 1.4)as follows: for every x ∈ Rn, A(x + Zn) = Ax + Zn. This action ofA 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. Letx0 ∈ X and π : G → X be the map defined by π(g) = gx0 for all g ∈ G. PutF = Gx0 . Then F is a closed subgroup of G. Define the map π : G/F → Xas π(gF ) = gx0 for all g ∈ G. Then π is a homeomorphism preserving theG-actions.
17
Fix a left haar measure λ0 on F = π−1(x0). Then for every x ∈ X there ex-ists a unique measure λx supported on the fiber π−1(x) such that the followingconditions 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) andevery x ∈ X,
I(φ)(x) =
π−1(x)
φ dλx. (1.5)
Note that supp I(φ) ⊂ π(supp φ) = (supp φ)x0. Also since φ is uniformlycontinuous on G, it follows that the function I(φ) is uniformly continuous onX. Thus I(φ) ∈ Cc(X). Due to Eqs. 1.5 and 1.4, for every g ∈ G,
I(φ)(gx0) =
F
φ(gh) dλ0(h)
andI(φ Lg) = I(φ) Lg.
Lemma 1.3.1 ([R, Sect. 1.1]) I is surjective. In fact,
I(C+c
(G)) = C+c
(X),
where C+c
(X) denotes the set of all nonnegative functions in Cc(X).
Proof. Given ψ ∈ C+c
(X), put ψ1 = ψ π ∈ C+(G). Then (supp ψ1)x0 =supp ψ. Let K be a compact subset of G such that supp ψ ⊂ Kx0. Letθ1 ∈ C+
c(G) be such that θ1 ≡ 1 on K. Now for every x ∈ supp ψ, we have
π−1(x) ∩K = ∅, and hence I(θ1)(x) > 0. Define a function θ on G as follows:
θ(g) =
ψ1(g)θ1(g)/I(θ1)(gx0) if g ∈ supp ψ1
0 otherwise.
Then θ is continuous and supp θ ⊂ supp θ1. Thus θ ∈ C+c
(G). It is straightforward 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 : Cc(G) → R be the positive linear functional
Λ(φ) =
G
φ dµ.
Then
ker I ⊂ ker Λ.
18
Proof. Let φ ∈ ker I; i.e. I(φ) = 0. By Lemma 1.3.1, there exists θ ∈ C+c
(G)such that I(θ) ≡ 1 on π(supp φ). Now
0 =
G
I(φ)(gx0)θ(g) dµ(g)
=
G
F
φ(gh) dλ0(h)
θ(g) dµ(g)
=
F
G
φ(gh)θ(g) dµ(g)
dλ0(h)
=
F
G
φ(g)θ(gh−1) dµ(g)
dλ0(h)
=
G
φ(g)
F
θ(gh−1) dλ0(h)
dµ(g)
=
G
φ(g)I(θ)(gx0) dµ(g)
=
G
φ(g) dµ(g) = Λ(φ),
where the fourth equality holds because µ is right invariant, the sixth equalityholds because F is unimodular (see Lemma 1.2.3), and the last equality holdsbecause I(θ) ≡ 1 on (supp φ)x0. Thus we have shown that ker I ⊂ ker Λ.
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 ∈ Cc(G),
G
φ dµ =
X
I(φ) dν. (1.6)
Moreover, any locally finite G-invariant measure on X is a constant mul-
tiple of ν.
Proof. Let Λ : Cc(G) → R be the positive linear functional defined by Λ(φ) =G
φ dµ for all φ ∈ Cc(G). Since I : C+c
(G) → C+c
(X) is surjective andker I ⊂ ker Λ, there exists a unique positive linear functional Ψ : Cc(X) → Rsuch that
Λ = Ψ I. (1.7)
By Riesz representation theorem ([Ru2, Sect. 2.14]), there exists a uniquelocally finite measure ν on X such that for all φ ∈ Cc(X),
Ψ(φ) =
X
φ dν.
19
By Eqs. 1.4 and 1.7, for every φ ∈ Cc(G) and every g ∈ G,
Ψ(I(φ) Lg) = Ψ(I(φ Lg)) = Λ(φ Lg) = Λ(φ) = Ψ(I(φ)).
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 : Cc(G) → R such that for all φ ∈ Cc(G),
Λ(φ) =
X
I(φ) dν . (1.8)
By Riesz representation theorem, there exists a unique measure µ on G suchthat for all φ ∈ Cc(G),
G
φ dµ = Λ(φ). (1.9)
Now for any g ∈ G,
Λ(φ Lg) =
X
I(φ Lg) dν =
X
I(φ) Lg dν =
X
I(φ) dν = Λ(φ).
Thus µ is a left haar measure on G. By the ‘uniqueness’ of left haar mea-sures 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 groupand the stabilizer of every point of the homogeneous space X is a discretesubgroup of G. Fix a point x0 ∈ X, then Γ = Gx0 is a discrete subgroup of Gand X ∼= G/Γ. The counting measure on Γ is a natural choice for a left and aright haar measure on Γ. Now for every φ ∈ Cc(G) and x = gx0 ∈ X,
I(φ)(x) =
δ∈G:δx0=x
φ(δ)
=
γ∈Γ
φ(gγ).
Fix a haar measure µ on G. Then by Theorem 1.3.3, there exists a uniqueG-invariant measure ν on G/Γ such that for all φ ∈ Cc(G),
G
φ dµ =
G/Γ
I(φ)(y) dν(y)
=
G/Γ
γ∈Γ
φ(gγ) dν(gΓ).
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 = gx0 for some g ∈ G. Since G is unimodular, replacing Eby Eg there is no loss of generality in proving the theorem for x0 in place ofx.
Take > 0. By regularity of µ and ν, there exists a compact set C ⊂ Esuch that µ(E \ C) < and ν(Ex0 \ Cx0) < . Also there exists an open setD ⊃ C such that µ(D \ C) < and ν(Dx0 \ Cx0) < .
Let φ ∈ Cc(G) be such that φ ≡ 1 on C and supp φ ⊂ D. Then
I(φ)(y) ≥ 1 if y ∈ Cx0 and I(φ)(y) = 0 if y ∈ X \Dx0.
Therefore
ν(Cx0) ≤
X
I(φ) dν =
G
φ dµ ≤ µ(D).
Thus,ν(Ex0) ≤ ν(Cx0) + ≤ µ(D) + ≤ µ(E) + 3.
Since is arbitrary,ν(Ex0) ≤ µ(E).
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) ≤ 1 fory ∈ Dx0. Therefore
µ(C) ≤
G
φ dµ =
X
I(φ) dν ≤ ν(Dx0).
Hence
µ(E) ≤ µ(C) + ≤ ν(Dx0) + ≤ ν(Cx0) + 2 ≤ ν(Ex0) + 2.
Since is arbitrary,µ(E) ≤ ν(Ex0).
Therefore by the first part, ν(Ex0) = µ(E).
Notation. For a neighbourhood Ω of e in G, define
XΩ = x ∈ X : Gx ∩ Ω = e.
21
Lemma 1.3.5 For any compact set Y ⊂ X, there exists a neighbourhood Ωof e in G such that Y ∩XΩ = ∅. Thus if Ωn is a sequence of neighbourhoods
of e in G such that Ωn ↓ e as n →∞, then X \XΩn ↑ X as n →∞.
Proof. Fix x0 ∈ X. There exists a compact set C ⊂ G such that Y ⊂ Cx0.Let Ω0 be a neighbourhood of e in G such that Gx0 ∩Ω0 = e. Note that forany x ∈ Cx0, Gx = cGx0c
−1 for some c ∈ C. Since C is compact and Ω0 isopen, there exists a neighbourhood Ω of e in G such that c−1Ωc ⊂ Ω0 for allc ∈ C. Therefore, Gx ∩ Ω = e for all x ∈ Cx0. Hence Y ∩XΩ = ∅.
Definition. A discrete subgroup Γ of G is called a lattice in G if the homoge-neous space X = G/Γ admits 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 \XΩ is relatively compact.
Proof. Without loss of generality we may assume that Ω is relatively compact.Now suppose that X \ XΩ is not relatively compact. Then there exists asequence xi ⊂ X \ XΩ such that Ωxi ∩ Ωxj = ∅ for all i = j. Let Ω1 be arelatively compact neighbourhood of e in G such that Ω−1
1 Ω1 ⊂ Ω. Then forevery x ∈ X\XΩ, the map Ω1 g → gx ∈ X is injective. Due to Lemma 1.3.4,ν(Ω1xi) = µ(Ω1) for all i ∈ N. Therefore
ν(X) ≥ ν(∪∞i=1Ω1xi) =
∞
i=1
ν(Ω1xi) =∞
i=1
µ(Ω1) = ∞.
This is a contradiction to our assumption.
Definition. Let Γ be a discrete subgroup of G and X = G/Γ. We say that Γis an L-subgroup of G if for every neighbourhood Ω of e in G, the set X \XΩ
is relatively compact in X. In this case we call X an L-homogeneous space ofG.
Thus by Lemma 1.3.6, every lattice is an L-subgroup. In the next chapterwe 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 ofinterest 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
∆ = Zx1 ⊕ Zx2 ⊕ · · ·⊕ Zxk.
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 . Takelinearly independent vectors y1, . . . , yk ∈ ∆ such that V = spany1, · · · , yk.Let
∆ = Zy1 ⊕ · · ·⊕ Zyk ⊂ ∆.
And let π : V → V/∆ be the quotient map. If
I =
k
i=1
aiyi ∈ V : 0 ≤ ai < 1 for 1 ≤ i ≤ k
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 finitelygenerated, so is ∆. Since ∆ is abelian and torsion free, there exist x1, . . . , xn ∈∆ such that
∆ = Zx1 ⊕ · · ·⊕ Zxn
(see [L, Chap. 1, Sect. 10, Theorem 7]).By a rearrangement, we can assume that x1, . . . , xk are linearly inde-
pendent. Let ∆ be the free abelian group generated by x1, . . . , xk. Thenspan(∆) = V . Repeating the above argument for ∆ in place of ∆, weobtain 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 0is a subspace of V and ∆ is a
discrete subgroup of V .
Proof. If F is discrete, there is nothing to be proved. Otherwise there existsa sequence vi ⊂ F \ 0 such that vi → 0 as i → ∞. For each i ∈ N,put xi = vi/vi ∈ S(1), where S(1) is the unit sphere in V . Since S(1) iscompact, by passing to a subsequence we may assume that as i →∞, xi → xfor some x ∈ S(1). Let W1 be the one dimensional subspace passing throughx.
We claim that W1 ⊂ F . To prove the claim, let λ ∈ R and let > 0 begiven. For each i ∈ N, there exists ni ∈ Z such that
nivi − λxi ≤ vi.
23
Choose i0 ∈ N such that for all i ≥ i0,
|λ| · xi − x < /2 and vi ≤ /2.
Thereforenivi − λx ≤ nivi − λxi+ |λ| · xi − x ≤ .
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 W1 ⊂ F , we have F = W1 ⊕ F1. Now since F1 is a closed subgroup ofV1, 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 condi-
tions are equivalent:
(a) span(F ∩Qn) = span(F ).
(b) The group F/(F ∩ Zn) is compact.
(c) F = F 0⊕∆, where F 0∩Qnis dense in F 0
and ∆ is a discrete subgroup
contained in Qn.
(d) The subgroup F ∩Qnis dense in F .
Proof. Put FZ = F ∩ Zn and FQ = F ∩Qn. Then
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
π(F 0) is an open, and hence a closed subgroup of F/FZ.Now suppose that the condition (b) holds. Then the group F 0/(F0∩Zn) ∼=
π(F 0) is compact. By Theorem 1.4.2, the group F 0 is a subspace of Rn. Nowby 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
F1 := FQ = F 0 ∩Qn ⊕ (∆ ∩Qn).
Since FZ ⊂ F1, the quotient group F/F1 is compact. Therefore the group∆/(∆ ∩ Qn) ∼= F/F1 is finite. Now for any δ ∈ ∆, there exists n ∈ N such
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 thecondition (d) implies the condition (a). This completes the proof.
Definition. A closed subgroup F of Rn is called rationally defined, if it satisfiesany one of the equivalent conditions of Lemma 1.4.3.
Let T1 denote the circle group and Tn denote the n-torus as before. Thereis a homomprphism φ : R → T1 given by φ(t) = e2πit for all t ∈ R. Thenker φ = Z and we have a topological group isomorphism φ : R/Z → T1.Clearly, Rn/Zn ∼= (R/Z)n. Therefore Rn/Zn ∼= Tn 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 Rncontaining F . Then the closure of
any orbit of F on Rn/Zn ∼= Tnis a compact orbit of L.
Proof. Let π : Rn → Rn/Zn be the quotient homomorphism. Since Rn isabelian, 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, thereforeπ(F ) ⊂ π(L).
Let F = π−1(π(F )). Then F is a closed subgroup of Rn containing Fand π(F ) = π(F ) is compact. Therefore by Lemma 1.4.3, F is a rationallydefined closed sugroup of Rn. By minimality, L ⊂ F , and hence π(L) ⊂ π(F ).Therefore π(L) = π(F ).
Lemma 1.4.5 Let v = (a1, . . . , an) ∈ Rnand L be the smallest closed ratio-
nally defined subgroup of Rncontaining v. Think of R as a vector space over
Q. Then
dimQ Q- span1, a1, . . . , an = 1 + dim L0.
Proof. Let m = dim L0 and k = dimQ Q- span1, a1, . . . , an. There exists abasis e1, . . . , em ⊂ Qn of L0 and a discrete subgroup ∆ ⊂ Qn such thatL = L0 ⊕∆. Since v ∈ L, there exist x1, . . . , xm ∈ R and δ ∈ ∆ such that
v =m
i=1
xiei + δ.
Therefore1, a1, . . . , an ⊂ Q- span1, x1, . . . , xm.
Hence k ≤ m + 1.
25
Without loss of generality we may assume that the reals 1, a1, . . . , ak−1 arelinearly independent over Q, and for every 1 ≤ j ≤ n and 0 ≤ i ≤ k− 1, thereexists rij ∈ Qn, such that
aj = r0j +k−1
i=1
rijai. (1.10)
For 0 ≤ i ≤ k − 1, put fi = (ri1, . . . , rin). Let
E = R- spanfi : 1 ≤ i ≤ k − 1⊕ Z · f0.
Then E is a rationally defined closed subgroup Rn, and by Eq. 1.10, we havethat 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 theoremdue to Kronecker.
Corollary 1.4.6 Let v = (a1, . . . , an) ∈ Rn. If the elements 1, a1, . . . , an are
linearly independent over Q then every orbit of the cyclic subgroup generated
by v is dense in Rn/Zn ∼= Tn.
1.5 Covolumes of discrete subgroups in eu-clidean 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 uniquefinite measure m on W/∆. By Lemma 1.3.4, if the map π : W → W/∆ isinjective on a measurable set E ⊂ W then m(π(E)) = m(E). We define
Vol(W/∆) = m(W/∆),
which can be computed as follows.Let x1, . . . , xk ∈ ∆ be a Z-basis of ∆. Then ∆ = Zx1 ⊕ · · · ⊕ Zxk and
W = Rx1 ⊕ · · ·⊕Rxk. If we define
I =
k
i=1
aixi ∈ W : 0 ≤ ai < 1 for 1 ≤ i ≤ k
,
then W = I + ∆ and I ∩∆ = 0; that is π is injective on I. Therefore
m(W/∆) = m(I).
26
To compute m(I), let e1, . . . , ek be the standard orthonormal basis ofW , and let T : W → W be the linear transformation such that T (ei) = xi forall 1 ≤ i ≤ k.
If
J =
k
i=1
aiei : 0 ≤ ai < 1 for 1 ≤ i ≤ k
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 euclideanmetric. Let ·, · denote the associated positive definite inner product.
Definition. Given a discrete subgroup ∆ of rank k in V , we associate a quantitycalled the covolume to ∆ in the following way: Let W = span ∆. Then Winherites 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-thexterior power ∧kV of V . Then the set
B = ei1 ∧ · · · ∧ eik: 1 ≤ i1 < i2 < . . . < ik ≤ n
is a basis of ∧kV . We equip ∧kV with a unique inner product, again denotedby ·, ·, so that B becomes an orthonormal basis of ∧kV . Let · denote theassociated norm on ∧kV .
Theorem 1.5.1 Let ∆ be a discrete subgroup of V generated by k linearly
independent vectors, say x1, . . . , xk. Then
covol(∆) = x1 ∧ · · · ∧ xk.
27
We recall some standard facts used for the proof.A set of vectors x1, . . . , xk ⊂ V is linearly independent if and only if
x1 ∧ · · · ∧ xk is a nonzero vector in ∧kV . Also if y1, . . . , yk is another set oflinearly independent vectors in V then
spany1, . . . , yk = spanx1, . . . , xk⇔ y1 ∧ . . . ∧ yk = λ(x1 ∧ . . . ∧ xk)
for some λ ∈ R∗. In this case, let W = spanx1, . . . , xk and T be the lineartransformation on W such that Txi = yi for 1 ≤ i ≤ k then
y1 ∧ . . . ∧ yk = det T (x1 ∧ . . . ∧ xk).
If A is a linear transformation on V then there exists a unique lineartransformation ∧kA on ∧kV such that for any k vectors x1, . . . , xk ∈ V ,
∧kA(x1 ∧ . . . ∧ xk) = (Ax1) ∧ · · · ∧ (Axk).
If B is another linear transformation on V then
∧k(A B) = ∧k(A) ∧k(B).
Definition. Let E be an inner product space with the inner product ·, ·. Forany linear transformation A on E, there exists a unique linear transformationA∗ 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 innerprod-uct on E. Thus A is orthogonal ⇔ AA∗ = 1 ⇔ A transforms one orthonormalbasis into another orthonormal basis.
Lemma 1.5.2 For any linear transformation A on V ,
(∧kA)∗ = ∧kA∗
on ∧kV . In particular, if A is an orthogonal transformation of V then ∧kA is
an orhogonal transformation of ∧kV .
Proof. Let 1 ≤ i1 < i2 < . . . < ik ≤ n and 1 ≤ j1 < j2 < . . . < jk ≤ n. Then
∧kA∗(ei1 ∧ · · · ∧ eik), ej1 ∧ · · · ∧ ejk
= (A∗ei1) ∧ · · · ∧ (A∗eik
), ej1 ∧ · · · ∧ ejk
= det X,
28
where X is the k × k matrix whose (p, q)-th entry is
A∗eip , ejq = eip , Aejq.
Therefore
det X = ei1 ∧ · · · ∧ eik, Aej1 ∧ · · · ∧ Aejk
= ei1 ∧ · · · ∧ eik
, ∧kA(ej1 ∧ · · · ∧ ejk).
Since B is a basis of ∧kV , we get
∧kA∗ = (∧kA)∗.
Proof of Theorem 1.5.1. Let W = span ∆ and let w1, . . . , wk be an or-thonormal basis of W . Extend this basis to an orthonormal basis of V givenby w1, . . . , wn.
Let T : W → W be the linear transformation given by Twi = xi for1 ≤ i ≤ k. Then by Eq. 1.11 and 1.12,
covol(∆) = | det T |.
Nowx1 ∧ · · · ∧ xk = det T (w1 ∧ · · · ∧ wk) ∈ ∧kW ⊂ ∧kV.
Thereforex1 ∧ · · · ∧ xk = | det T | · w1 ∧ · · · ∧ wk.
Note that there exits a unique orthogonal transformation A of V which takese1, . . . , en to the orthonormal basis w1, . . . , wn. Therefore by Lemma 1.5.2,w1 ∧ · · · ∧ wk = ∧kA(e1 ∧ · · · ∧ ek) is an element of an orthonormal basis of∧kV . Hence w1 ∧ · · · ∧ wk = 1. Thus
covol(∆) = | det T | = x1 ∧ · · · ∧ xk.
29
Chapter 2
The group SL2(C) and its homogeneousspaces
2.1 Exponential map, Lie algebras, and Unipo-tent subgroups
In this section we shall introduce some very basic concepts regarding LinearLie groups. Part of the material in this section is taken from an excellentarticle 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 ) denotethe group of invertible linear maps from V to itself. The usual name forGL(V ) is the general linear group for V . If V = kn, where k = R or k = C,then End(V ) ∼= Mn(k), the space of n × n matrices with entries from k, andGL(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 = supAv/v : v ∈ V \ 0.
The norm on End(V ) makes End(V ) into a matric space. Thus GL(V ) alsobecomes a metric space with respect to the restricted metric.
For A ∈ End(V ) and r > 0, define
Br(A) = A ∈ End(V ) : A − A < r.
The following basic formulae about the norm are staightforward to verify: Forall A, B ∈ End(V ),
AB ≤ A · B
30
andA + B ≤ A+ B.
The formula(1V − A)−1 = 1V + A + A2 + · · ·
is valid for A < 1, where 1V is the identity transformation on V . Therefore aneighbourhood of 1V in End(V ) is contained in GL(V ). Since the left and theright multiplications by an element of GL(V ) are homeomorphisms of End(V ),the group GL(V ) is an open subset of End(V ). Using the formula
(1V + A)(1V + B) = 1V + A + B + AB.
we obtain that for r < 1/2,
Br(1V )−1Br(1V ) ⊂ B4r(1V ).
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 iscalled a one-parameter subgroup of G. Note that ρ(0) is the identity elementof 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 expo-nential map on matrices.
For A ∈ End(V ), define
exp(A) =∞
n=0
An/n!.
Since An ≤ An, we see by the standard estimates in the exponential series,that the series defining exp A converges absolutely for all A and uniformly onBr(0) for every r > 0. Hence exp defines a smoodh, in fact an analytic, mapfrom End(V ) to itself.
For s, t ∈ R,
exp(sA) exp(tA) =∞
m=0
(sm/m!)Am ·∞
n=0
(tn/n!)An
=∞
m,n=0
(smtn/m!n!)Am+n
=∞
l=0
m+n=l
((m + n)!/m!n!)smtn
Al/l!
=∞
l=0
((s + t)A)l/l! = exp((s + t)A).
31
In particular exp(A) exp(−A) = 1V . Hence exp(A) ⊂ GL(V ) for every A ∈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) = exp tA (∀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 someA ∈ End(V ) (see [Ho, Thm. 10]). We will not use this fact later in thesenotes.
We shall need the following result.
Lemma 2.1.1 There exists r0 > 0 such that the function exp maps Br0(0) ⊂End(V ) homeomorphically onto an open neighbourhood of 1V in GL(V ).
Proof. Let D expA
be the differential of exp at A. It is a linear map fromEnd(V ) to End(V ) defined by
D expA(B) =
d
dtexp(A + tB)
t=0
= limt→0
(exp(A + tB)− exp A)/t.
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 in-vertible. Therefore the lemma follows from the Inverse Function Theorem(see [Ru1, Chap. 9]).
Remark. If one defines,
log(1V − A) = −∞
n=1
An/n,
then just as in the case of real numbers, this series converges absolutely forA < 1. Now for all B ∈ B1(1V ),
exp(log B) = B. (2.1)
The formula is known in the scalar case, and this implies that in fact Eq. 2.1is an identity in the absolutely convergent power series, whence it follows inthe 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 H1 and H2 be linear subspaces of End(V ) such that
End(V ) = H1 ⊕H2.
Let pi be the projection of End(V ) on Hi with respect to the above decomposi-
tion. Define a map φ : End(V ) = H1 ⊕H2 → GL(V ) by
φ(A) = exp(p1(A)) exp(p2(A)).
Then φ maps a neighbourhood of 0 in End(V ) homeomorphically onto a neigh-
bourhood of 1V in GL(V ).
Proof. From the formula of the exp map one computes that
d
dt
t=0
exp(p1(tA)) exp(p2(tA)) = p1(A) + p2(A) = A
for every A ∈ End(V ). Therefore the differential of φ at 0 is the identity mapon End(V ). Now the conclusion of the lemma follows by the Inverse FunctionTheorem.
The Lie algebra associated to a matrix group
Definition. A real Lie algebra G is a real vector space equipped with a skewsymmetric bilinear product [·, ·] : G × G → G satisfying the Jacobi Identity:
[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 theproduct 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 Liealgebra in its own right.
Remark. Let A ∈ End(V ). If the one-parameter group exp tA : t ∈ R isregarded as a curve inside the vector space End(V ), then this curve passesthrough the identity 1V at time t = 0. By differentiating the formula forexp tA, we see that the the tangent vector at the point 1V to this curve is justA. Thus A is called an infinitesimal generator of the one-parameter groupexp tA.
To any closed subgroup G ⊂ GL(V ) one associates a set
G = A ∈ End(V ) : exp tA ∈ G for all t ∈ R,which is a collection of the infinitesimal generators of all one-parameter sub-groups of G.
33
Theorem 2.1.3 The set G is a Lie subalgebra of End(V ). In particular, G is
a vector subspace of End(V ).The map exp : G → G maps a neighbourhood of 0 in G bijectively onto a
neighbourhood of 1V ∈ G.
Due to this theorem the set G is called the Lie algebra associated to G.The theorem asserts that the set of tangent vectors at 1V to the curves de-
fined by one-parameter subgroups of G actually fills out some linear subspace,namely G, of End(V ), and further, if we make a smooth change of co-ordinatesgiven by A → exp A, then this linear subspace G is bent in such a way that itlies entirely in G, and fills up G around 1V . In other words, G is shown to bea multidimensional surface inside End(V ), and G is simply its tangent spaceat the point 1V .
See [Ho, Thm. 17] for a proof of this theorem. We will directly verifythis theorem for G = SLn(k) and the ‘horospherical subgroups’. These areessentially 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 G associated to G is Mn(R) ⊂ Mn(C).
The following formula is useful in describing Lie algebras of certain closedsubgroups 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 ∈ Mn(C). Note that if B is anupper triangular matrix with the diagonal (b11, b22, . . . , bnn) then the diagonalof exp B is (exp b11, . . . , exp bnn). Therefore
exp(tr B) = exp(b11 + · · ·+ bnn)
= (exp b11) · · · (exp bnn)
= det(exp B).
Now given A ∈ Mn(C) there exists g ∈ GLn(C) such that gAg−1 is uppertriangular. Therefore,
exp(tr A) = exp(tr(gAg−1))
= det(exp(gAg−1))
= det(g(exp A)g−1) = det(exp A).
The group G = SL(V ) = g ∈ GL(V ) : det g = 1 is a closed subgroup of
GL(V ). If G denotes the Lie algebra associated to G then
G = A ∈ End(V ) : det(exp tA) = 1 ∀t ∈ R.
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
G = A ∈ End(V ) : tr A = 0.Clearly G is a vector subspace of End(V ). Since tr(AB) = tr(BA), we havetr([A, B]) = 0 for all A, B ∈ End(V ). Thus G is a Lie subalgebra of End(V ).Now let r0 > 0 be such that | tr A| < 1 for all A ∈ Br0(0). Then due toLemma 2.1.1, there exists r ∈ (0, r0) such that the function exp maps the setBr(0) ∩ G homeomorphically onto an open neighbourhood of 1V in G. Thuswe have verified Theorem 2.1.3 in this case.
The Adjoint representation
For g ∈ GL(V ) and A ∈ End(V ), we can form the conjugate
Ad g(A) = gAg−1.
Let A, B ∈ End(V ), a, b ∈ R, and g, g1, g2 ∈ GL(V ). Then the followingidentities 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) = Ad g1(Ad g2(A)),
(v) exp(Ad g(A)) = g(exp A)g−1.
The identities (i) and (iii) say that Ad g is a Lie algebra automorphism ofEnd(V ), and formula (iv) says that the map Ad : g → Ad g is a group homo-morphism 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 G be the corresponding Liealgebra. Then for any g ∈ GL(V ), the Lie algebra associated to the groupgGg−1 is
Ad g(G) = Ad g(A) : A ∈ G.In particular, for all g ∈ G, we have that Ad g(G) = G. Note that Ad : G →Aut(G) is a continuous homomorphism. Thus Ad provides a natural linearaction 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. Let G be a closed subgroup of GL(V ). For g ∈ G, define
U(g) = u ∈ G : gnug−n → 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 G denote the Lie algebra of G. Then the Lie algebra associated to U(g)can be given by
U(g) = A ∈ G : Ad gn(A) → 0 as n →∞.
Since Ad g is a linear transformation on G, the set U(g) is a linear subspaceof G. The other part of Theorem 2.1.3 is also straight forward to verify forU(g). Moreover, in this case the map exp : U(g) → U(g) is a homeomorphismof U(g) on to U(g).
Definition. An element u ∈ GL(V ) is called unipotent if every eigenvalue of uis 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 Nn = 0 for somen ≤ 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 thecharacteristic polynomial P (λ) = det(u−λ1V ). Therefore if N = u−1V
then Nn = 0 for some n ≤ dim V . Define
log(u) = −∞
k=1
Nk/k = −(N + N/2 + · · ·+ Nn−1/(n− 1)). (2.2)
Then log(u) is nilpotent and u = exp(log u). This shows that everyunipotent 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(t0)v = v for some t0 = 0 then u(t)v = v for all t ∈ R.
36
Proof. Let N be a nilpotent element such that u(t) = exp tN for all t ∈ R.Since Nn = 0 for some n ≤ dim V , we have
c(t) = (exp tN)v = v + tNv + (t2/2)N2v + · · ·+ (tn−1/(n− 1)!)Nn−1v.
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 constantpolynomial. Therefore c(t) = v for all t ∈ R.
The polymomial behaviour of the orbits of a one-parameter group of unipo-tent transformations on a vector space turns out to be a key property forstudying 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 conjugateare same. Also given an > 0, there exists an r > 0 such that the eigenvaluesof every A ∈ Br(0) are of absolute value less than . Therefore the Lie algebraof any horospherical subgroup consists of nilpotent elements. And hence everyhorospherical subgroup consists of unipotent elements.
Let g = diag(a1, . . . , an) be a diagonal matrix, where 0 < a1 < a2 < · · · <an. Let X = (xij) ∈ Mn(C). Then for all 1 ≤ i, j ≤ n, the (i, j)-th entry ofgXg−1 is (ai/aj)xij. Therefore the horospherical subgroup U(g) of GLn(C)consists of all the upper triangular matrices with diagonal entries 1.
Any unipotent element of GLn(C) has a conjugate which is an uppertrian-gular matrix with all diagonal entries 1. Therefore every unipotent element ofGLn(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 horo-
spherical subgroup of G associated to an element g ∈ G. Let ρ be a con-
tinuous 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 itssubgroups on the homogeneous spaces of G. In this section we shall obtainsome useful properties of this group.
Consider the standard action of G on E = C2. Fix a point p = (1, 0) ∈ E.
For any g =
a bc d
∈ G, gp = (a, c). Thus Gp = E \ 0 and the stabilizer
37
of p in G is the closed subgroup
N =
u(z)
def=
1 z0 1
: z ∈ C
.
By Lemma 1.1.2, G/N ∼= E \ 0 and the isomorphism preserves the standardactions of G on both the spaces.
Note that N is the horospherical subgroup of G associated to
a 00 a−1
,
for every a ∈ C∗ with |a| < 1. The Lie subalgebra of G associated to N is
N =
0 z0 0
: z ∈ C
and the exponential map exp : N → N is an isomorphism of topologicalgroups. Thus N is a complex one-parameter subgroup of G via the map
C z → exp
0 z0 0
∈ N.
Define L = v ∈ E : Nv = v. Then L = Cp = (x, 0) ∈ E : x ∈ C. Forany g ∈ G,
gp ∈ L ⇔ N(gp) = gp
⇔ (g−1Ng)p = p
⇔ g−1Ng ⊂ N
⇔ g ∈ NG(N).
Thus
NG(N) = g ∈ G : gp ∈ L
=
a b0 a−1
: a ∈ C∗, b ∈ C
.
Define
D =
d(a) :=
a 00 a−1
: a ∈ C∗
.
Then NG(N) = DN . Also NG(N)p = L \ 0 = Dp.Let · denote the euclidean norm on C2. Define
K = g ∈ G : gv = v for all v ∈ C2.
ThenK = U(2) := g ∈ G : gg∗ = g∗g = 1.
38
By definition, K acts on the unit sphere S(1) in E. Let (z1, z2) ∈ S(1). Then
|z1|2 + |z2|2 = 1. If g =
z1 −z2
z2 z1
then g ∈ K and gp = (z1, z2). Thus S(1)
is a homogeneous space of K. Now K∩N = e. Therfore the map K → S(1)given by k → kp is a homeomorphism and it preserves the standard actions ofK 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 Mp is a circle, which is theintersection 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 thatkp = v/v. Then (ka)p = v. Thus the map ψ : K × A → C2 \ 0 given byψ(k, a) = (ka)p (∀(k, a) ∈ K × A) is a homeomorphism.
Now given g ∈ G, let (k, a) = ψ−1(gp). Then gp = (ka)p. Hence n =g−1ka ∈ N . Thus the map G g → (k, a, n) ∈ K ×A×N is continuous, andit is the inverse of φ. Therefore φ is a homeomorphism.
Lemma 2.2.2 (1) Given g ∈ G, there exits k ∈ K such that kgk−1 ∈ DN .
(2) Given h ∈ DN \C(G)N , there exits v ∈ N such that vhv−1 ∈ D. (Here
C(G) = d(1), d(−1) is the center of G.)
(3) Let w =
0 −11 0
. Then for every α ∈ C∗
, wd(α)w−1 = d(α−1).
(4) For any z ∈ C and α ∈ C∗, d(α)u(z)d(α−1) = u(α2z).
In particular, given g ∈ G there exists x ∈ G such that either xgx−1 = d(α)with |α| ≤ 1 or xgx−1 = u(1).
39
Proof. (1) Let v ∈ C2 be an eigen vector of g with unit norm. Let k ∈ K besuch that kv = p. Therefore p is an eigen vector of kgk−1. Hence kgk−1 ∈NG(N) = DN .
(2) Any h ∈ DN is of the form d(a)u(z) for some a ∈ C∗ and z ∈ C. Forany y ∈ C,
u(y)hu(−y) = d(a)u(a−2y + z − y).
If h ∈ C(G)N then a = ±1. If we choose y = −(a−2−1)−1z and v = u(y) ∈ N ,then vhv−1 = d(a) ∈ D.
The statements (3) and (4) are evident.
Corollary 2.2.3 Any nontrivial horospherical subgroup of G is of the form
kNk−1for some k ∈ K.
Proof. Let g ∈ G be such that U(g) is the given horospherical subgroup. ByLemma 2.2.2, there exists x ∈ G such that if we put h = xgx−1 then eitherh = u(1) or h = d(α) with |α| ≤ 1. Now U(g) = xU(h)x−1.
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 .Due to Lemma 2.2.1, we can express x−1 = kb, where k ∈ K and b ∈ AN ⊂NG(N). Then U(g) = kNk−1.
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∗.
(3) L = Z[eθi] · w and B = ekθik∈Z, where θ ∈ π/3, π/2, 2π/3 and
w ∈ C \ 0.
Proof. If L0 = C then (1) holds. If L0 is one-dimensional then B ⊂ R∗. SinceB ⊂ 1,−1, for any z ∈ C, the additive group generated by Bz is dense inRz. Hence in particular L is connected and (2) holds.
Now suppose that L0 = 0. Then L is a discrete subgroup of C. Put
r = infz∈L\0
|z| > 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| < r, and hence αw ∈ L. ThereforeB ⊂ eθi : θ ∈ R.
For any θ ∈ (2π/3, π) and v ∈ L ∩ S(r), we have that
0 < |eθiw + w| = |eθi + 1|r < r.
Hence eθiv ∈ L. In particular, B contains no element of the arc eθi : θ ∈(2π/3, π). This readily implies that B = ekθi : k ∈ Z for some θ ∈π/2, π/3, 2π/3 and that S(r) ∩ L = B · w. Define L1 = Z[eθi] · w. Then L1
is a subgroup of L. If L1 = L then there exists x ∈ L such that |x| < r, acontradiction. Therefore L1 = L and (3) holds.
Proposition 2.2.5 Let F be a closed subgroup of DN . Then there exists
v ∈ N such that for F v := vFv−1, one of the following holds.
(1) F v ⊂ d(1), d(−1) ·N .
(2) F v ⊂ D.
(3) F v =(d(ik)k∈Z · A) ∩ F v
(F ∩N).
(4) F v = (D ∩ F v)N .
(5) F v = d(ekθi)k∈Z · u(z) : z ∈ Z[e2iθ] · w, where θ ∈ π/6, π/4, π/3and 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. Let π : 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
hu(y)h−1 = u(a2y) = π(h)u(y)π(h)−1 ∈ F.
Therefore if we defineL = z ∈ C : u(z) ∈ F
andB = a2 ∈ 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:
(i) For θ ∈ π/2, π/3, if a ∈ ekθik∈Z, then B = akk∈Z.
41
(ii) If a ∈ R∗, then B ⊂ R∗.
If B = 1 then (1) holds. Now suppose that B = 1. Then byLemma 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 existsz ∈ C \ 0 and b ∈ C∗ such that h1 = d(b)u(z) ∈ F v. Now
h−11 d(a)h1d(a)−1 = u(z1) ∈ F v,
where z1 = (a2 − 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 =k ·Cp. Then U(L) = kNk−1, and it is the horospherical subgroup associatedto kd(a)k−1 for any a ∈ C∗ with |a| < 1. Note that
NG(U(L)) = g ∈ G : gL = L.
Let L1 and L2 be two distinct complex lines in C2; that is,
L1 ⊕ L2 = C2.
For i = 1, 2, if we put Ui = U(Li) then for any v ∈ C2 \ Li, we have
Uiv = v + Li. (2.3)
Lemma 2.2.6 The group G is generated by unipotent elements. In fact
G = U2U1U2U1.
Proof. Let q ∈ L1 \ 0. Since the map φ : G/U1 → C2 \ 0 defined byφ(gU1) = gq (∀g ∈ G), is a homeomorphism, it is enough to prove that
U2U1U2q = C2 \ 0.
Now due to Eq. 2.3,
U2q = q + L2,
U1(U2q) = q + (L2 \ 0) + L1, and
U2(U1U2q) = C2 \ 0.
This completes the proof.
42
Lemma 2.2.7 (Cf. [R, Sect. 12.6]) Let U1 and U2 be two distinct nontriv-
ial horospherical subgroups of G then U1 ∩ U2 = e.
Proof. Let u ∈ U1 ∩ U2. Then u fixes the lines L1 and L2 pointwise. SinceL1 ⊕ L2 = C2, the linear transformation u = 1C2 .
Lemma 2.2.8 Let U1, U2, and U3 be three distinct nontrivial horospherical
subgroups of G. Then
NG(U1) ∩ NG(U2) ∩ NG(U3) = C(G).
Proof. For i ∈ 1, 2, 3, suppose Ui fixes the complex line Li pointwise. Ifg ∈ NG(U1)∩NG(U2)∩NG(U3) then gLi = Li for all i. Since the three lines arepairwise independent and any two of them span C2, the linear transformationg acts as a scaler on C2. Hence g ∈ C(G).
Notation. Let w =
0 −11 0
∈ K. Then w2 = d(−1). Put L = wL and
N− = U(L) = wNw−1. Then N− is the horospherical subgroup associated tod(a) for any a ∈ C∗ with |a| > 1. Note that
N− =
1 0z 1
: z ∈ C
.
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,
NwDp = N(L \ 0) = L \ 0+ L = C2 \ 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
unipotent element. Then there exists g ∈ G such that either gFg−1 ⊂ NG(N)or gFg−1 ∩N = e and gFg−1 ∩N− = e.
Moreover, in the second case, F ∩ U = e for infinitely many distinct
horospherical subgroups U of G.
43
Proof. By Lemma 2.2.2, there exists k ∈ k such that kFk−1 ∩ N = e.Put F1 = kFk−1. If F1 ⊂ NG(N), then by the Bruhat decomposition, thereexits g1 ∈ F1 such that g1 = nwh for some n ∈ N and h ∈ DN . Thereforeg1Ng−1
1 ∩F1 = e. But g1Ng−11 = nN−n−1. Put g = n−1k then gFg−1∩N =
e and gFg−1 ∩N− = e.Now every any v ∈ gFg−1∩N−, if we put N v = vNv−1 then N v∩gFg−1 =
e. For any v1 = v2 in N−, we have N v1 = N v2 . Now since gFg−1 ∩ N− isan 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 ahomogeneous space of G and for the point p = (1, 0) ∈ X, the stabilizerGp = N . The lebesgue meausure on C2 restricted to X, say m, is G-invariant.Since N is isomorphic to C as a topological group, the lebesgue measure, sayλ, on C corresponds to a haar measure on N under the isomorphism. Now asin Section 1.3, there exists a linear map I : Cc(G) → Cc(X) with the followingproperties:
(i) I(C+c
(G)) = C+c
(X).
(ii) For any g ∈ G and ξ ∈ Cc(G),
I(ξ Lg) = I(ξ) Lg. (2.5)
Therefore there exists a unique locally finite measure µ on G such that forevery ξ ∈ Cc(G),
G
ξ dµ =
X
I(ξ) dm.
Due to Eq. 2.5, the measure µ is left G-invariant. Thus we have explicitlyconstructed a haar measure on G.
Note that due to Lemma 2.2.6 and Lemma 2.1.6, there does not exist anynontrivial continuous homomorphism of G into GL1(C) ∼= C∗. Therefore themodular character of G must be trivial. Hence G is a unimodular group andµ is right invariant.
Due to Lemma 2.2.1, there is a continuous function ψ : X → KA ⊂ Gsuch that for any x ∈ X, k ∈ K, and a ∈ A if ka · p = x then ψ(x) = ka.Then for any ξ ∈ Cc(G) and x ∈ X,
I(ξ)(x) =
C
ξ(ψ(x)u(z)) dλ(z).
44
Hence for all ξ ∈ Cc(G),
G
ξ dµ =
C2\0
C
ξ(ψ(x)u(z)) dm(x)dλ(z).
As a consequence we obtain the following.
Lemma 2.3.1 For measurable subsets F1 ⊂ K, F2 ⊂ A and F3 ⊂ N , if
F = F1F2F3 then,
µ(F ) = m(F1F2 · p)λ(F3).
For η > 0, defineAη = d(a) : a > 0, a2 < η
and for t > 0, define
N(t) = u(z) : z ∈ C, |(z)| ≤ t, |(z)| ≤ t..
Definition. For η > 0 and a compact set C ⊂ N , define
S(η, C) = KAηC.
A subset of G of the form S(η, C) is called a Siegel set.
Lemma 2.3.2 The haar measure of a Siegel set is finite.
Proof.
µ(S(η, C)) = m(KAη · p)λ(C)
= 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. LetZ[i] denote the ring of Gaussian integers in C. Then ∆ = Z[i]×Z[i] is a latticein C2. If we define
Γ = SL2(Z) =
a bc d
∈ G : a, b, c, d ∈ Z[i]
then Γ is 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-
gorithm, given any a, b ∈ Z[i], there exists an element
a xb y
∈ Γ if and
only if gcd(a, b) = 1 ([J, Sect.2.15,Exercise 11]). Also for every g ∈ G, thediscrete subgroup g∆ is a Z[i]-module with respect to the action of Z[i] on C2
by complex multiplication. Moreover, by Eq. 1.11,
covol(g∆) = | det g| · covol(∆) = 1. (2.6)
Therefore G-acts transitively on the set L of all discrete Z[i]-submodules ofC2 with covolume one. Since elements L are lattices, we can topologize L bydeclaring that two lattices are ‘nearby’ if they admit basises which are ‘nearby’.With this topology, L becomes a homogeneous space of G and it is isomorphicto G/Γ, via the map gΓ → g∆ (∀g ∈ G).
We claim that L admits a finite G-invariant meausre; in other words, Γ isa lattice in G. Due to Lemma 1.3.4 and Lemma 2.3.2, it is enough to showthe following:
Proposition 2.3.3 ([R, Sect. 10.4]) G = S(η, N(1/2)) · Γ for some η > 0.
Proof. Since N ∩ Γ = u(z) : z ∈ Z[i], we have N = N(1/2) · (N ∩ Γ), andhence NΓ = N(1/2)Γ. Therefore to prove the proposition, it is enough toshow that
G = (KAηN)Γ
for some η > 0. By Iwasawa decomposition, G = KAN . Therefore givenh ∈ G, h ∈ KAηN if and only if hp ∈ B(η). Therefore it is enough to provethat 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 δ1, δ2 ∈ ∆ withδ1 = δ2,
(B(η) + gδ1) ∩ (B(η) + gδ2) = ∅.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 allg ∈ G. This completes the proof.
46
2.4 Ergodic properties of transformations onhomogeneous spaces
Let G be a (locally compact, hausdorff, second countable) topological groupand 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 measurespace (X, ν) if the following condition is satisfied: For any measurable subsetE ⊂ X, if ν(gE∆E) = 0 for every g ∈ F , then either ν(E) = 0 or ν(X\E) = 0,where A∆B = (A \B) ∪ (B \ A).
We say that an element g ∈ G acts ergodically on (X, ν) if the subgroupgenerated by g acts ergodically on (X, ν).
Note that ergodicity is the condition of irreducibility of an action in mea-sure 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 B be a countable basis of open subsets of X. Replacing X by supp ν,we can assume that supp ν = X. Then for every E ∈ B \ ∅, ν(E) > 0. Nowν(FE) > 0 and the set FE is F -invariant. Therefore by the ergodicity of theF -action, ν(X \ FE) = 0. Now if
Y =
E∈B\∅
FE
then ν(Y ) = ν(X). Now take any y ∈ Y . Then Fy ∩ E = ∅ for everyE ∈ B \ ∅. Thus Fy is dense in X.
In the second part, since we consider only the positive trajectory of gnx :n ∈ Z, we need to argue as follows: For any E ∈ B \ ∅, define
X(E) =∞
n=0
g−nE.
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 =
E∈B\∅
X(E)
47
then ν(X \ Y ). Now take any y ∈ Y . Then for every nonempty open subsetE of X, there exists n ≥ 0 such that gny ∈ E. Therefore the trajectorygny : n ≥ 0 is dense in X.
Ergodicity and unitary representations
Now we reformulate the condition of ergodicity in terms of unitary representa-tions of G. Let L2(X, ν) denote the space of complex valued square integrablefunctions on the measure space (X, ν). It is well known that L2(X, ν) is ahilbert space with respect to the L2-norm, which is denoted by · L2 . Thegroup G acts on L2(X, ν) by the following formula: For every g ∈ G andξ ∈ L2(X, ν),
(g · ξ)(x) = ξ(g−1x)
for almost all x ∈ X. Since the measure ν is G-invariant, g · ξL2 = ξL2
for every g ∈ G and every ξ ∈ L2(X, ν). Thus G acts on L2(X, ν) by unitaryautomorphisms.
Lemma 2.4.2 The above unitary representation of G on L2(X, ν) is contin-
uous; that is, the map G× L2(X, ν) (g, ξ) → g · ξ ∈ L2(X, ν) is continuous.
Proof. Since Cc(X) is a dense subset of L2(X, ν), it is enough to show that forevery ψ ∈ Cc(X), the map (g, ξ) → g · ξ, ψ is continuous.
Since g acts by a unitary automorphism on L2(X, ν),
g · ξ, ψ = ξ, g−1 · ψ =
X
ξ(x)ψ(gx) dν(x).
Since ψ ∈ Cc(X), given ∈ (0, 1) there exists a neighbourhood Ω of the identityin G such that for all h ∈ Ω and x ∈ X, |ψ(hx)ψ(x)| < . In particular,h−1 · ψ − ψL2 ≤ . Now for any ξ ∈ L2(x, ν) such that ξ − ξL2 < ,
g · ξ, ψ − hg · ξ, ψ = ξ, g−1 · ψ − ξ, g−1h−1 · ψ= ξ − ξ, g−1 · ψ+ ξ, g−1 · ψ − g−1h−1 · ψ≤ ξ − ξL2ψL2 + ξL2ψ − h−1 · ψL2
≤ (ψL2 + ξL2 + 1).
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 sub-
group F of G acts ergodically on (X, ν) if and only if every F -invariant func-
tion 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) bean F -invariant function. Then for any measurable subset D ⊂ C,
ν(g(ξ−1(D))∆ξ−1(D)) = 0
for every g ∈ F , where ξ−1(D) = x ∈ X : ξ(x) ∈ D. By ergodicity of theF -action, either ν(ξ−1(D)) = 0 or ν(ξ−1(C \D)) = 0. Since D is an arbitrarymeasurable subset of C, there exists z0 ∈ C such that ν(X \ ξ−1(z0)) = 0.Thus ξ(x) = z0 for almost all x ∈ (X, ν).
Next suppose that every F -invariant function in L2(X, ν) is constant almosteverywhere. Let E be a measurable subset of G such that ν(gE∆E) = 0 forevery g ∈ F . Let χE denote the charasteristic function of E on X. ThenχE ∈ L2(X, ν) and χE is F -invariant. Therefore it is constant ν-a.e. andhence either ν(E) = 0 or ν(X \ E) = 0.
Mautner phenomenon - a criterion for ergodicity
Definition. Let G 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 followingcondition is satisfied: For any continuous unitary representation of G on ahilbert space H and an element ξ ∈ H, if F ξ = ξ then Lξ = ξ.
Remark. Thus if the triple (G, F, G) has the Mautner property and if G actsergodically on the probability space (X, ν), then due to Lemma 2.4.3, F actsergodically on (X, ν).
For a subgroup F ⊂ G, define
L(F ) = g ∈ G : there exist sequences fi ⊂ F , f i ⊂ F ,
and mi → e in G such that fimif i → g as i →∞.
Let < L(F ) > denote the closed subgroup generated by L(F ).
Example. Let G be a closed subgroup of GLn(C). For g ∈ G, let U(g) denotethe horospherical subgroup of G associated to g. If F = gk : k ∈ Z thenU(g) ⊂ L(F ) and U(g−1) ⊂ 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 Hand an element ξ ∈ H, suppose that F ξ = ξ. Then for any f, f ∈ F andm ∈ G,
fmf · ξ − ξ = mf · ξ − f−1 · ξ = m · ξ − ξ.Now since the representation of G on H is continuous, the lemma follows
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 theMautner property if and only if the triple (G, hgh−1, G) has the Mautnerproperty for any h ∈ G. For g · ξ = ξ ⇔ (hgh−1)(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 . Let F = gk : k ∈ Z.
First suppose that g = d(a), 0 < |a| < 1. Then U(g) = N and U(g−1) =N−. Therefore N ⊂ L(F ) and N− ⊂ L(F ). Since G is generated by N andN− (see Lemma 2.2.6), G =< L(F ) >. 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,
put mk =
1 0
1/(2k) 1
. Then
g2kmkg−k =
1 2k0 1
1 0
1/(2k) 1
1 −k0 1
=
2 0
1/(2k) 1/2
→ d(2) as k →∞.
Hence < L(F ) > d(2). This completes the proof.
Notation. Till the end of this chapter, let X denote a homogeneous space ofG = SL2(C) admitting a (unique) G-invariant borel probability measure, sayν. 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 |an|→∞ as n →∞. Then for every x ∈ X, the set
n∈N
d(an)Nx
50
is dense in X.
In particular, every AN-orbit in X is dense. (Cf. [DR, Sect. 1.5])
Proof. Let
Y =
n∈N
d(an)Nx.
Then Y is N -invariant.For r1, r2 ≥ 0, define
U(r1, r2) =
a bc d
∈ G : |c| ≤ r1, |a− 1| ≤ r2
.
Take r1 > 0 and r2 ∈ (0, 1/2). Then the set U(r1, r2)x contains a neighbour-hood of x in X, and hence
ν(U(r1, r2)x) = α > 0. (2.8)
Now for any an ∈ C∗,
d(an)U(r1, r2)d(a−1n
) = U(|an|−2r1, r2).
Therefore,
ν(U(r1, r2)x) = ν(d(an)U(r1, r2)x)
= ν(U(|an|−2r1, r2)d(an)x)
≤ ν(U(|an|−2r1, r2)Y ). (2.9)
Since |an|→∞, for every n ∈ N there exists m ≥ n such that
n
k=0
U(|ak|−2r1, r2)Y
⊃ U(|am|−2r1, r2)Y. (2.10)
Note the following statements.
(i) Y is closed and N -invariant.
(ii) U(r1, r2)/N is a compact subset of G/N .
(iii) U(|ak|−2r1, r2)/N ↓ U(0, r2)/N in G/N as k →∞.
Therefore∞
k=0
U(|ak|−2r1, r2)Y
= U(0, r2)Y. (2.11)
Now since ν is finite, we conclude that
0 < α ≤ ν(U(0, r2)Y ) ≤ ν(DY ). (2.12)
51
Since N acts ergodically on (X, ν), by Lemma 2.4.1 there exists ω ∈ D andy ∈ Y such that X = N(ωy) = ωNy. Hence Y ⊃ Ny = X. This completesthe proof.
The last part can also be argued alternatively as follows: Since U(0, r2) ⊂NG(N), the set U(0, r2)Y is N -invariant. From Eq. 2.12 and the ergodicity ofthe 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 andν(X \ Y ) = 0. Therefore X = Y .
Now we prove a version of the Borel density theorem for SL2(C). Althoughit is not essential for the proof of our main theorem, it is of interest in its ownright.
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 ∈ Xand let Gx denote its stabilizer in G. Then any Gx-invariant subspace W of
V is G-invariant.
Proof. Let k = dim W and define a homomorphism ∧kρ : G → GL(∧kV ) asfollows: For every g ∈ G and every subset v1, . . . , vk ⊂ V ,
∧kρ(g)(v1 ∧ · · · ∧ vk) = (ρ(g)v1) ∧ · · · ∧ (ρ(g)vk).
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, wecan assume that W is one dimensional.
Let P1(V ) denote the set of all one-dimensional subspaces of V . Let π :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 ) withthe finest topology for which the map π is continuous.
The linear G action ρ on V induces a ‘projective linear’ action ρ of G onP1(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 action ρ : G × P1(V ) → P1(V ) iscontinuous.
By Theorem 2.4.6, the element u(1) ∈ N acts ergodically on X. ByLemma 2.4.1, there exists y ∈ X such that the set u(n)y : n ∈ N is dense
52
in X. Let h ∈ G be such that y = hx. Put W = ρ(h)W . Then W is Gy-invariant. Now if we show that W is G-invariant then W = ρ(h−1)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 ∈ Gx, we have
ρ(δ)π(w) = π(w),
where π and ρ 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 groupof unipotent transformations of V . Therefore by Lemma 2.1.5 there exist realpolynomials φ1, . . . ,φn such that for every t ∈ R,
φ(t) = ρ(u(t))w =n
i=1
φi(t)ei.
Now for every 1 ≤ i ≤ n, the rational function φ2i(t)/
n
j=1 φ2j(t) converges as
t →∞. Therefore there exists p ∈ V \ 0 such that
limt→∞
φ(t)/φ(t) = p.
Since G = u(n) : n ∈ NGx, given any g ∈ G there exist sequences nk ⊂N and δk ⊂ Gx such that nk →∞ and u(nk)δk → g as k →∞. Now
ρ(g)π(w) = limk→∞
ρ(u(nk)δk)π(w)
= limk→∞
ρ(u(nk))π(w)
= limk→∞
π(φ(nk))
= π(p).
Putting g = e, we get π(p) = π(w). Hence ρ(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 Gsuch that X ∼= G/Γ.
Proof. Fix x ∈ X. Let F be the connected component of e in Gx. Thus F isa closed connected subgroup of G.
53
Let F denote the Lie algebra of F . Then F is also the Lie algebra of Gx. ByTheorem 2.1.3, F is a Vector subspace of G and there exists r0 > 0 such thatthe map exp : F ∩ Br0(0) → Gx is a homeomorphism onto a neighbourhoodof 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(G). Then Ad(Gx)F = F .
By Lemma 2.4.8, Ad(g)F = F for every g ∈ G. Therefore by Lemma 2.2.2and Corollary 2.2.10, either F = 0 or F = G. Since F = G and F is connected,we have F = e. Hence Gx is discrete.
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. Let π : G/Γ → G/F be the natural quotient map which preserves theG-actions on both the spaces. Let µ be the unique G-invariant probabilitymeasure 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 adiscrete subgroup of G. Hence F is a lattice in G.
Applying the arguments of this section to the group SL2(R) one can provethe 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 ai →∞ in R, the set
i∈N
d(ai)N1x
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 Γ be a discrete subgroup of G, and denote by X thehomogeneous space G/Γ. Our main objective is to study closures of orbits ofone-parameter unipotent subgroups of G acting on X, when X admits a finiteG-invariant measure; i.e. when Γ is a lattice in G. In our method we oftenneed to construct convergent sequences in X with some ‘control’ over theirlimit 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 justifiedby general topological arguments. But when X is noncompact, one needs touse the fact that X admits a finite G-invariant measure in order to make theconstructions. We shall first study in detail the geometric structure of thecomplement of a large compact subset in X. The construction of convergentsequences will then be made using the structural properties of the complementof 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 otherwords, Γ is an L-subgroup of G. Being an L-homogeneous space is a topologicalcondition on a homogeneous space which is, a priori, a weaker condition thanthe measure-theoretic condition of admitting a finite invariant measure. Forthe results of this chapter it is enough to assume that X is an L-homogeneousspace 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 betweenthe 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. Let U be a nontrivial unipotent subgroup of G. Suppose that for somex ∈ X, the orbit Ux is compact. Since Ux = Ux = Ux, replacing U by U , wemay assume that U is closed. Then by Corollary 1.1.2, the map φ : U/Ux → Xgiven by φ(gUx) = gx (∀ g ∈ U) is proper. Hence U/Ux is compact. Since U isnoncompact, Ux contains a nontrivial element, say u. Therefore there exists anelement g ∈ G such that gnug−n → e as n → ∞. But gnug−n ∈ Ggnx and Xis 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 ofa more general theorem due to Kazhdan and Margulis, by specializing theirmethod for G = SL2(C). The result, in particular, shows that every pointof a noncompact L-homogeneous space X is fixed by a nontrivial unipotentelement 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 Gx ∩ Ω ⊂ U .
First we introduce the concept of Zassenhaus neighbourhood, which playsa crucial role in the proof.
Notation. For s1, s2 ∈ GLn(C), we denote by [s1, s2], the commutator of s1
and s2, namely s1s2s−11 s−1
2 . For subsets S1, S2 ⊂ GLn(C), define
[S1, S2] = [s1, s2] : si ∈ Si, i = 1, 2.
For S ⊂ GLn(C) and k ≥ 0, define S(k) inductively as: S(0) = S and S(k) =[S, S(k−1)] for k ≥ 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
[g, h]− 1 ≤ 2A · B · (1− A) ·−1 (1− B)−1. (3.1)
Proof.
ghg−1 − h = g(1−B)g−1 − (1−B)
= B − gBg−1
= B − gB(1 + Ag−1)
= B − (1− A)B − gBAg−1
= AB − gBAg−1.
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Therefore
[g, h]− 1 ≤ A · B · (1 + g · g−1) · h−1≤ A · B · (1 + (1 + A)(1− A)−1) · (1− B)−1
= 2A · B · (1− A)−1 · (1− B)−1.
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. DefineΩ = g ∈ GLn(C) : g − 1 < 1/8.
It is enough to prove that for any k ≥ 0 and g ∈ Ω(k),
g − 1 ≤ (1/8)2−k. (3.2)
For k = 0 the Eq. 3.2 obviously holds. Now suppose it holds for somek ≥ 0. By Lemma 3.1.2, for any g ∈ Ω and h ∈ Ω(k),
[g, h]− 1 ≤ (1/8)2−k−1.
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. Let G be a closed subgroup of GLn(C). A neighbourhood Ω ofthe identity e in G is called a Zassenhaus neighbourhood if for every discretesubgroup Γ of G, the set Ω ∩ Γ is contained in a closed connected nilpotentsubgroup of G.
It is a fact that every closed subgroup of GLn(C) admits a Zassenhausneighbourhood; (see [R, Sect. 8.16] for a proof). Since we shall deal only withG = SL2(C), here we give a proof just for this case.
Proposition 3.1.4 The group G = SL2(C) admits a Zassenhaus neighbour-
hood. 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. Let Ω be as in the proof of Lemma 3.1.3. Given a discrete subgroupΓ ⊂ G, put ∆ = Γ∩Ω. Then ∆(k) = Ω(k)∩Γ. If ∆ = e then there is nothingto be proved. Otherwise, since Ω(k) ↓ e as k ↑ ∞ and since Γ is discrete,there exists a k ≥ 1 such that ∆(k−1) = e and ∆(k) = e.
57
Let h ∈ ∆(k−1) \ e. Then h commutes with every element of ∆. ByLemma 2.2.2 there exists g ∈ G such that ghg−1 ∈ D or ghg−1 ∈ C(G)N .Since h ∈ Ω, h is not in the center of G. Therefore the centralizer of ghg−1
is contained in either D or C(G)N . Note that Ω ∩ C(G)U = Ω ∩ U for anyhorospherical subgroup U of G. Hence either g∆g−1 ⊂ D or g∆g−1 ⊂ N . Thiscompletes the proof.
We need to setup some notations and definitions before we proceed to theproof of the main theorem of this section.
Notation. Let G denote the Lie algebra associated to G. Then
G = A ∈ M2(C) : tr(A) = 0.
The space G inherits the operator norm, denoted by · , from M2(C) =End(C2) (see §2.1). For r > 0, let Br denote the open ball of radius r inG centered at the origin. Put Br = expBr ⊂ G. By Lemma 2.1.1, thereexists r0 > 0 such that the map exp : Br0 → Br0 is a homeomorphism. Letlog : Br0 → Br0 denote its inverse. For g ∈ Br0 , define
|g| = log(g).
Now let X be an L-homogeneous space of G. Recall that for any neigh-bourhood Ω of e in G,
XΩ = x ∈ X : Gx ∩ Ω = e.
Define a function R : XBr0→ (0, r0) in the following way: For every x ∈ XBr0
,
R(x) = inf|δ| : δ ∈ Gx ∩Br0 \ e.
Note the following properties of the map R:
(i) R is continuous.
(ii) For every k ∈ K and x ∈ XBr0, we have R(kx) = R(x).
(iii) R vanishes at the infinity.
For convenience we choose r0 > 0 small enough so that Br0 is a Zassenhausneighbourhood in G. Now by Proposition 3.1.4, for any x ∈ XBr0
, the setGx ∩ Br0 is contained in a conjugate of either D or N . Therefore Gx ∩ Br0 ⊂exp(L), where L = Ad g(D) or L = Ad g(N ) for some g ∈ G; (in fact everymaximal abelian Lie subalgebra L ⊂ G is of this form).
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 L ⊂ G, there exists g ∈ E such that for every x ∈ L,
Ad(g)x ≥ cx.
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Proof. Given any abelian Lie subalgebra L, there exists k ∈ K such that eitherAd k(L) ⊂ N or Ad k(L) ⊂ Ad v(D) for some v ∈ N .
Since Ad K preserves the norm on G and since K is compact, without lossof generality we may assume that either L = N or L = Ad v(D).
If L = N then for any x ∈ L and a > 0,
Ad(d(a))x = a2x. (3.3)
Now suppose L = Ad v(D), where v =
1 z0 1
and z ∈ C. Let ξ ∈ C
be such that |ξ| = 1 and |z + ξ| = |z| + 1; that is ξ = z/|z| if z = 0. For anya > 0, put g = d(a)u(ξ) ∈ NG(N). Now any x ∈ L is of the form
x = Ad v
y 00 −y
= y
1 −2z0 −1
for some y ∈ C. Therefore
Ad g(x) = y Ad(d(a)) Ad(u(z + ξ))
1 00 −1
= y
1 −2a2(z + ξ)0 −1
.
Note that for any t ∈ C,
|(|t|− 1)| ≤
1 t0 −1
≤ |t|+ 1.
Therefore if a > 1 then
Ad g(x)x ≥ 2a2(|z|+ 1)− 1
2|z|+ 1≥ a2. (3.4)
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(G), the operator norm of Ad g is bounded forall g ∈ E. Therefore there exist 0 < r3 < r2 < r1 < r0 such that for i = 1, 2, 3and every g ∈ E ∪ E−1, Ad g(Bri) ⊂ Bri−1 or equivalently gBrig
−1 ⊂ Bri−1 .
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Proposition 3.1.6 Let x ∈ X be such that R(x) < r3, i.e. Gx ∩ Br3 = e.Then there exists g ∈ E such that the following conditions are satisfied.
(1) For every γ ∈ Gx ∩Br1,
|gγg−1| ≥ c|γ|.
(2)
cR(x) < R(gx) < r2.
(3) For any t ∈ (0, r2],
Ggx ∩Bt ⊂ g(Gx ∩Bc−1t)g−1.
Proof. (cf. [R, Sect. 11.7]) There exists an abelian subalgebra L ⊂ G suchthat Gx ∩ Br0 ⊂ expL. Now by Lemma 3.1.5 there exists a g ∈ E such thatfor every γ ∈ expL ∩Br1 , gγg−1 ⊂ Br0 and
|gγg−1| ≥ c|γ|.
This verifies (1).Note that g(Gx ∩ Br3)g
−1 ⊂ Ggx ∩ Br2 . Therefore R(gx) < r2. Now letγ ∈ Ggx ∩Br2 . Then γ := g−1γg ∈ Gx ∩Br1 . Hence
|γ| = |gγg−1| ≥ c|γ|.
Thus γ = gγg−1 and |γ| ≤ c−1|γ|. This verifies (2) and (3).
Notation. For n ∈ N ∪ 0, define En inductively as: E0 = E and En =E · E(n−1) if n > 0.
Proposition 3.1.7 For any x ∈ X with R(x) < r2, there exist n ∈ N ∪ 0and g ∈ En
such that
r3 < R(gx) < r2
and
Ggx ∩Br2 ⊂ g(Gx ∩Bc−nr2)g−1.
Proof. For x ∈ X with R(x) < r2, define
n(x) = infn ∈ N ∪ 0 : cnR(x) > r3.
We prove the proposition by induction on n(x).If n(x) = 0 then R(x) > r3 and the proposition holds for n = 0 and g = e.
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Now suppose n(x) ≥ 1. Then R(x) ≤ r3. By Proposition 3.1.6, there existsh ∈ E such that if we put x1 = hx then
cR(x) < R(x1) < r2
and for every t ∈ (0, r2),
Gx1 ∩Bt ⊂ h(Gx ∩Bc−1t)h−1. (3.5)
Hence n(x1) < n(x). By induction there exists an integer n1 ≥ 0 and anelement g1 ∈ En1 such that
r3 < R(g1x1) < r2
andGg1x1 ∩Br2 ⊂ g1(Gx1 ∩Bc−n1r2
)g−11 . (3.6)
Now from Eq. 3.5 and Eq. 3.6, the conclusion of the proposition holds forn = 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 Gsuch that for every discrete subgroup Γ ⊂ G, there exists g ∈ G such that
gΓg−1 ∩ Ω = e.
Proof. Put X = G/Γ and Ω = Br3 . Let x = eΓ ∈ X. If Γ ∩ Ω = ethen by Proposition 3.1.7, there exists g ∈ G such that R(gx) > r3. ThusgΓg−1 ∩ Ω = Ggx ∩ Br3 = e.
Theorem 3.1.9 Let X be an L-homomogeneous space of G. Then there ex-
ists δ0 > 0 such that for any x ∈ X with R(x) < δ0, there exists a unique
horospherical subgroup U of G such that
Gx ∩Br0 ⊂ U.
Proof. (cf. [R, Sect. 11.10]) Fix a point x0 ∈ X. Since X is an L-homogeneousspace, there exists a compact set C ⊂ G such that X \XBr3
⊂ Cx0. For anyy ∈ X \XBr2
take h ∈ C such that y = hx0. Then
h−1(Gy ∩Br2)h ⊂ Gx0 ∩ C−1Br2C =: S. (3.7)
Now S is a finite set, because Gx0 is discrete and C−1Br2C is relatively compactin G. Let S1 be the set of all unipotent elements in S and put S2 = S \ S1.
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Since every non-unipotent element of G has an eigenvalue different from 1 andsince eigenvalues are preserved under conjugation, there exists δ ∈ (0, r2) suchthat no element of S2 has a conjugate in Bδ. Let p ∈ N be such that
cpδ > r2.
Since E is compact, there exists δ0 ∈ (0, r3) such that for all g ∈ Ep,
gBδ0g−1 ⊂ Br3 . (3.8)
Take x ∈ X such that R(x) < δ0. From Proposition 3.1.7, it follows thatthere exists n ∈ N and g ∈ En such that
r3 < R(gx) < r2 (3.9)
and there exists γ ∈ Ggx ∩Br2 \ e such that
g−1γg ∈ Gx ∩Bc−nr2 . (3.10)
Because R(x) < δ0, due to Eq. 3.8 and Eq. 3.9, we have n > p. Thereforec−nr2 < δ. By Eq. 3.7, the element γ has a conjugate in S, say γ. Then byEq. 3.10, γ has a conjugate g−1γg ∈ Bδ. Therefore γ ∈ S2. Therefore γ ∈ S1.Hence γ ∈ Ggx and g−1γg ∈ Gx are nontrivial unipotent elements of G.
Since a conjugate of D cannot contain a nontrivial unipotent element, byProposition 3.1.4, there exists a horospherical subgroup U of G such thatGx ∩Br0 ⊂ U . The uniqueness of U follows from Corollary 2.2.7.
3.2 Non-divergence of unipotent trajectories
In this section we shall show that unipotent trajectories on L-homogeneousspaces are nondivergent. As a consequence we show that every horosphericalsubgroup 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,
Ggnx ∩Br0 ⊂ gn(Gx ∩ U)g−n.
Proof. ([GR, Lemma 3.5]) Let δ ∈ (0, δ0) be such that gBδg−1 ⊂ Br0 , where δ0
is defined as in Theorem 3.1.9. Since X is an L-homogeneous space, R(gnx) →0 as n →∞. Without loss of generality we may assume that R(gnx) < δ for all
62
n ≥ 0. By Theorem 3.1.9, for each n ≥ 0 there exists a unique horosphericalsubgroup Un such that
Ggnx ∩Br0 ⊂ Un.
Since R(gnx) < δ, we have that Ggnx ∩Bδ = e. Also
g(Ggnx ∩Bδ)g−1 ⊂ Ggn+1x ∩Br0 .
Therefore gUng−1 ∩ Un+1 = e. By Corollay 2.2.7, gUng−1 = Un+1. Thus forall n ≥ 0,
Un = gnU0g−n
and henceGgnx ∩Br0 ⊂ gn(Gx ∩ U0)g
−n.
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
gnung−n → 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, ifwe replace g with its conjugate by an element of G. Due to Lemma 2.2.2, theelement g has a conjugate in either D or C(G)N . Therefore we may assumethat only one of the following is possible: Either g = d(a) for some a ∈ C∗
with |a| ≤ 1 or g = u(z) for some z ∈ C. Since g is not contained in a compactsubgroup of G, in first case we have |a| < 1.
Define
B =
x z0 −x
: x, z ∈ C
and
N− =
0 0z 0
: z ∈ C
.
Then G = N− ⊕ B.Let U be the Lie algebra associated to U . Since U is a horospherical
subgroup of G, the map exp : U → U is a homeomorphism. Now put xn =exp−1(un) ∈ U for every n ≥ 0. Since un ∈ Ω, there exits c > 0 such thatxn ≥ c for all n ≥ 0. Since gnung−n → e as n →∞, we have that
limn→∞
Ad gn(xn) = 0. (3.11)
Suppose that U∩B = 0. Since every element of U is nilpotent, U∩B ⊂ N .Therefore by Corollary 2.2.7, U = N . Now due to Eq. 3.11, g ∈ N . Therefore
63
g = d(a) for some a ∈ C∗, |a| < 1 and hence U = N = U(g). This completesthe proof for the case at hand.
Now suppose that U ∩ B = 0. Let p : G = N− ⊕ B → N− be theprojection on the first factor. According to the hypothesis, p is injective whenrestricted to U . Therefore there exists µ > 0 such that for all x ∈ U ,
p(x) ≥ µx.
For every y ∈ N−, Ad(d(a))y = a−2y and p(Ad(u(z))y) = y. Therefore
p(Ad g(x)) = p(Ad g(p(x)))
for all x ∈ G, andp(Ad g(y)) ≥ y
for all y ∈ N−. Thus for every x ∈ U and n ≥ 1,
p(Ad gn(x)) = p(Ad gn(p(x))) ≥ p(x) ≥ µx ≥ 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
gnγg−n → e as n →∞.
In particular, a unipotent element u ∈ G does not have a divergent trajec-
tory on X; that is, for any x ∈ X, there exists a compact set C = C(x) ⊂ Xsuch that unx ∈ C for infinitely many n ∈ N.
Proof. By Lemma 3.2.1, there exists a unique nontrivial horospherical sub-group U such that for all large n ∈ N,
Ggnx ∩Bδ0 ⊂ gn(Gx ∩ U)g−n.
Since R(gnx) → 0 as n → ∞, there exist un ∈ Gx ∩ U \ e such thatgnung−n → e as n →∞. Therefore by Lemma 3.2.2, g is a semisimple elementand U = U(g). Hence for any γ ∈ Gx ∩ U , gnγg−n → e as n →∞.
Compact orbits of horospherical subgroups
Corollary 3.2.4 (Cf. [GR, Lemma 3.16]) Let X be a noncompact L-homo-
geneous 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.
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Proof. Let u ∈ Gx∩U\e. Because U is a horospherical subgroup, there existsg ∈ NG(U) such that |gug−1| < δ0. Also gug−1 ∈ Ggx∩U . Since g(Ux) = Ugx,the orbit Ux is compact if and only if the orbit Ugx is compact. Thereforereplacing x by gx, we may assume that there exists
γ ∈ Gx ∩Bδ0 \ e ⊂ U.
Let Ω be an open neighbourhood of e in G such that for every ω ∈ Ω,
ωBδ0ω−1 ⊂ Br0
andΩγΩ−1 ∩Gx = γ.
Note that for any y ∈ Ux, Gy ∩ U = Gx ∩ U . Since the set of fixed pointsof γ in X is a closed subset, γ ∈ Gz for every z ∈ Ux. Take z ∈ Ux. Thereexists an ω ∈ Ω such that y = ωz ∈ Ux. Thus
ωγω−1 ∈ Gy ∩Br0 ⊂ Gy ∩ U ⊂ Gx.
Therefore ωγω−1 = γ, and hence ω ∈ U and in turn z = w−1y ∈ Ux. Thisproves that Ux is closed. (Notice that to prove that Ux is closed, we only usethe fact that Gx is discrete; the L-homogeneous condition is not used).
Now by Lemma 1.1.2, the map π : U/Ux → X defined by π(uUx) =ux (∀u ∈ U) is proper. Recall that U/Ux is a topological group and it isisomorphic to R2/∆, where ∆ is a discrete subgroup. Therefore if U/Ux is notcompact, there exists u ∈ U such that the sequence unUx : n ∈ N ⊂ U/Ux
has no convergent subsequence. Since the map π is proper, π(unUx) = unx →∞ in X as n →∞. This contradicts Theorem 3.2.3. Hence U/Ux is compact.
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 Gx ∩ Ω ⊂ U ; in this case the orbit Ux is compact.
3.3 Volumes of compact orbits of horospheri-cal subgroups
Define a positive definite innerproduct ·, · on the Lie algebra G as follows:For x,y ∈ G, x,y = tr(xy∗). Let · 1 denote the associated norm on G.
65
Then for any x = (xij) ∈ G,
x1 =
2
i,j=1
|xij|2.
Note that for every k ∈ K, Ad k(x)1 = x1.Recall that for any nontrivial horospherical subgroup U of G and its asso-
ciated Lie subalgebra U ⊂ G, the map exp : U → U is an isomorphism of thetopological groups. Now U inherits a euclidean metric from G, and hence itadmits the standard lebesgue measure associated to the metric. This measureon U induces a unique haar measure on U , say σU .
Note that for any other nontrivial horospherical subgroup U of G, thereexists k ∈ K such that U = kUk−1. It is straight forward to verify that forany measurable subset E ⊂ U ,
σU (kEk−1) = 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). Nowif Ux = exp−1(Ux) then Ux is a lattice in U , and Vol(Ux) = covol(Ux) (see§1.5 and Lemma 1.3.4). Let x1,x2 ⊂ U form a basis of Ux as a free abeliangroup. Then by Theorem 1.5.1,
Vol(Ux) = x1 ∧ x21,
where · 1 denotes the euclidean norm on ∧2G induced by the norm · 1 onG.
Lemma 3.3.1 The volumes of compact orbits of horospherical subgroups have
the following properties.
1. For every g ∈ G,
Vol(gUx)def= Vol((gUg−1)gx) = ∧2 Ad g(x1 ∧ x2)1.
2. For every k ∈ K,
Vol(kUx) = Vol(Ux).
3. For every h ∈ NG(U),
Vol(Uhx) = Vol(hUx) = | det(Ad h|U)| · Vol(Ux).
(Recall that by Iwasawa decomposition G = KNG(U).)
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Proof. Statement 1 follows from the above discussion. Statement 2 followsfrom Statement 1 and the equality
∧2 Ad k(x1 ∧ x2)1 = x1 ∧ x21.
Statement 3 follows from Statement 1 and the equality
∧2 Ad h(x1 ∧ x2) = det(Ad h|U) · (x1 ∧ x2).
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 thatGx ∩ Br0 ⊂ U(x). Let U(x) denote the Lie subalgebra of G associated toU(x). Recall that due to Corollary 3.2.4, the orbit U(x)x is compact. Definea function V : XBδ0
→ (0,∞) such for every x ∈ XBδ0,
V(x) = Vol(U(x)x).
Lemma 3.3.2 (1) V is uniformly continuous. (2) V is bounded on XBδ0\XBδ
for any δ ∈ (0, δ0).
Proof. Given α > 1, let Ω be a neighbourhood of e ∈ G such that Ω =Ω−1 and for every g ∈ Ω, (i) gBδ0g
−1 ⊂ Br0 and (ii) the operator norm ∧2 Ad g1 ∈ (α−1, α). Then for any x ∈ XBδ0
and g ∈ Ω, if gx ∈ XBδ0 thenGgx ∩ Br0 ⊃ g(Gx ∩ Bδ0)g
−1 = e. Therefore U(gx) = gU(x)g−1 and henceby Lemma 3.3.1,
V(gx) = Vol((gU(x)g−1)gx) = Vol(gU(x)x) ∈ (α−1 V(x), α V(x)).
This equation implies (1). Since the set XBδ0\XBδ
is relatively compact, theuniform continuity implies (2).
Lemma 3.3.3 V vanishes at the point at infinity (for X).
Proof. Take a sequence xi : i ∈ N ⊂ X which has no convergent subse-quence. Since X is an L-homogeneous space, R(xi) → 0 as i → ∞. We mayassume that R(xi) < δ0 for every i ∈ N.
Let λi ≥ 1 be such that λiR(xi) ∈ (δ0/2, δ0) and let gi ∈ NG(U(xi)) besuch that Ad gi(x) = λix for every x ∈ U(xi). Then
R(gixi) = λiR(xi) ∈ (δ0/2, δ0),
67
U(gixi) = giU(xi)g−1i
,
andV(gixi) = | det(Ad gi|U(xi))| · V(xi) = λ2
iV (xi).
Since R(xi) → 0 as i → ∞, we have that λi → ∞ as i → ∞. Now V isbounded on XBδ0
\XBδ0/2. Therefore V(xi) → 0 as i →∞.
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) < δ0 for all t ∈ I. Then
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)Bδ0u(−t) ⊂ Br0 for all |t| < . ExpressI as a union of countably many open intervals of length at most . It isenough to prove the result for each interval separately. Also there is no loss ofgenerality 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. ThusGu(t)x ∩Br0 ⊃ u(t)(Gx ∩Bδ0)u(−t), and hence
U(u(t)x) = u(t)U(x)u(−t) (3.12)
for all t ∈ I.Let U = U(x), U = U(x) and Ux = exp−1(Ux) ⊂ U . Let x1,x2 ⊂ U be a
basis of the lattice Ux. Let the function ψ : R → ∧2G be defined by
ψ(t) = ∧2 Ad(u(t))(x1 ∧ x2) (∀t ∈ R). (3.13)
Since ∧2 Ad(u(t)) : t ∈ R is a one-parameter group of unipotent lineartransformations of ∧2G, the co-ordinates of ψ with respect to a basis of ∧2Gare polynomials of degree at most 2 (see Lemma 2.1.5 and note that takingthe 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(∧2G). Thus V2(u(t)x) = φ(t) for all t ∈ I. This proves the first part.Now suppose 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))(x1 ∧ x2) = (x1 ∧ x2) for all t ∈ R.
68
Hence Ad(u(t))U = U for all t ∈ R. Therefore U(u(t)x) = U(x) for all t ∈ Rand by Corollary 2.2.7, u(t) : t ∈ R ⊂ U(x). This completes the proof.
Notation. Let δ1 ∈ (0, δ0) be such that B−1δ1
Bδ1 ⊂ Bδ0 . Let U be a nontrivialhorospherical subgroup of G and σU denote the haar measure on U as definedearlier. Put
ξ0 = σU(Bδ1 ∩ U).
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.Then there exists k ∈ K such that U = kUk−1. Since kBδ1k
−1 = Bδ1 , wehave
σU (Bδ1 ∩ U ) = σU (k(Bδ1 ∩ U)k−1) = σU(Bδ1 ∩ U) = ξ0.
For (2), by Lemma 1.3.4 and the definition of δ1, we have Gx∩Bδ0∩U = e.Therefore (2) holds by the definitions of U(x) and V(x).
Notation. For ξ ∈ (0, ξ0], define
Xξ = x ∈ X : V(x) ≤ ξ
andCξ = X \Xξ.
Then Xξ is closed and Cξ is open. By Lemma 3.3.3, Cξ is relatively compactand Cξ ↑ X as ξ ↓ 0.
For a nontrivial horospherical subgroup U of G, define
X(U) = x ∈ X : the orbit Ux is compact
and for η > 0, define
Xη(U) = x ∈ X(U) : Vol(Ux) < 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,
kXη(U) = Xη(kUk−1).
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2. For any g ∈ NG(U) and η > 0,
gXη(U) = Xη(U),
where η = | det (Ad g|U) · η.
3. For any nontrivial horospherical subgroup U = U ,
Xξ0(U) ∩Xξ0(U) = ∅.
4. For any ξ ∈ (0, ξ0],Xξ = K ·Xξ(U).
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 ∈ Xξ. Then thereexists a nontrivial horospherical subgroup U such that the orbit U x is com-pact and Vol(U x) = V(x) < ξ. Thus x ∈ Xξ(U ). Now take k ∈ K such thatU = kUk−1. Then by Statement 1, x ∈ Xξ(U ) = kXξ(U).
From the above notations and Lemma 3.3.4, the following result is imme-diate.
Proposition 3.3.7 For any x ∈ X, a nontrivial horospherical subgroup Uof G, and any ξ ∈ (0, ξ0], either (a) x ∈ Xξ(U) or (b) for every nontrivial
one-parameter subgroup u(t) : t ∈ R ⊂ U and every T0 ∈ R, there exists
T > T0, such that u(T )x ∈ Cξ.
3.4 A lower bound for the relative measuresof a large compact set on unipotent tra-jectories
To study asymptotic behaviour of trajectories of one-parameter unipotent ub-group of G, the following observation about the growths of polynomials ofbounded degree is very useful.
Notation. For a polynomial φ, a subset I ⊂ R, and α > 0, define
Iα(φ) = t ∈ I : |φ(t)| ≤ 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
sups∈I
|φ(s)| ≥ α,
70
then
(Iβ(φ)) ≤ (Iα(φ)),
where denotes the lebesgue measure on R.
Proof. Put Iα = Iα(φ) and Iβ = Iβ(φ). Note that for any component [a, b] ofIα either |φ(a)| = α or |φ(b)| = α. Since it is enough to prove the lemma foreach component of Iα, without loss of generality we can assume that Iα = I.We may also assume that (Iβ) > 0.
Claim. There exist t0 < t1 < . . . < tm in Iβ such that
ti − ti−1 ≥ (1/m)(Iβ) (3.14)
and
((0, ti) ∩ Iβ) ≤ (i/m)(Iβ) (3.15)
for all i ∈ 1, . . . ,m.
We shall prove the claim by induction. Put t0 = inf Iβ. Suppose that forsome i ∈ 0, . . . ,m− 1, we have chosen t0 < . . . < ti in Iβ satisfying Eq. 3.14and Eq. 3.15. Then
((0, ti + (1/m)(Iβ)) ∩ Iβ) ≤ ((0, ti) ∩ Iβ) + (1/m)(Iβ)
≤ ((i + 1)/m)(Iβ)
≤ (Iβ).
Now putti+1 = inf (Iβ \ (0, ti + (I)/m)) .
Then ti+1 ∈ Iβ, |ti+1 − ti| ≥ (1/m)(Iβ), and
((0, ti+1) ∩ Iβ) ≤ ((0, ti + (1/m)(Iβ)) ∩ Iβ) ≤ ((i + 1)/m)(Iβ).
Thus ti+1 satisfies Eq. 3.14 and Eq. 3.15 and the induction is complete. Thisproves the claim.
Now since deg φ ≤ m, by Lagrange’s interpolation formula, for any t ∈ R,
φ(t) =m
i=0
φ(ti)
j∈0,...,m\i
t− tjti − tj
.
Take t ∈ I such that |φ(t)| ≥ α. Since |φ(ti)| ≤ β and |tj − ti| ≥ (1/m)(Iβ),we have
α ≤ |φ(t)| ≤ (m + 1)β
(I)
(1/m)(Iβ)
m
.
Since β = (m/(m + 1)mm)α, we have (Iβ) ≤ (Iα).
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Theorem 3.4.2 ([D3, Thm. 0.2]) Given ξ ∈ (0, ξ0) and > 0, there exists
ξ ∈ (0, ξ) such that for every one-parameter unipotent subgroup u(t) : t ∈ Rof G, every point x ∈ Cξ, and every T > 0,
1
T (t ∈ [0, T ] : u(t)x ∈ Cξ) > 1− .
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 componentof E. Since x ∈ Cξ, we have a = 0. Hence by continuity of V , we haveV(u(a)x) = ξ. By Lemma 3.3.4, there exists a polynomial φ of degree at mostm = 4 such that V2(u(t)x) = φ(t) for all t ∈ I.
Put ξ = (m/(m + 1)mm)1/2ξ. Then by Lemma 3.4.1,
(Iξ2(φ)) ≤ (Iξ2(φ)).
Hence
(t ∈ I : u(t)x ∈ Cξ) ≤ (I).
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 ofG and u(t) : t ∈ R be a nontrivial one-parameter subgroup of U . For anyξ ∈ (0, ξ0), define
f(x) = lim infT→∞
(1/T )
T
0
χCξ (u(t)x) dt
for every x ∈ X. Then using Proposition 3.3.7 and Theorem 3.4.2, we canchoose ξ ∈ (0, ξ0) such that
f(x) ≥ 1/2
for all x ∈ X \Xξ0(U).By Theorem 1.3.3, there exists a locally finite G-invariant measure µ on
X. Now
X
f dµ ≥ (1/2)µ(X \Xξ0(U)).
72
By Fatou’s Lemma ([Ru2, Sect. 1.28]),
X
f dµ ≤ lim infT→∞
(1/T )
X
T
0
χCξ (u(t)x) dt
dµ(x)
= lim infT→∞
(1/T )
T
0
X
χCξ (u(t)x) dµ(x)
dt
= lim infT→∞
(1/T )
T
0
µ(Cξ) dt
= µ(Cξ) < ∞.
Thereforeµ(X \Xξ0(U)) < ∞.
Let U be a nontrivial horospherical subgroup of G such that U = U . Thensimilarly we have µ(X\Xξ0(U
)) < ∞. By Lemma 3.3.6, Xξ0(U)∩Xξ0(U) = ∅.
Thereforeµ(X) ≤ µ(X \Xξ0(U)) + µ(X \Xξ0(U
)) < ∞.
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 byLemma 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 containsatleast two nontrivial one-parameter unipotent subgroups which are not con-tained in the same horospherical subgroup of G. Now arguing as in the proofof Corollary 3.4.3, with µH in place of µ, we obtain that
µH(X) = µH(Hx) < ∞.
Remark. Given an L-homogeneous space Y of H, all the results of this chaptergo 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).
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Proof. Observe that g ∈ NG(U) and | det(Ad g|U)| < 1. Let x ∈ D(g). Thenby 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 followthe 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 ti ⊂ R>0 such that a subsequence of
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. Take k ∈ N. By Theorem 3.4.2, there exists ξk ∈ (0, ξ0) such that forevery x ∈ Cξ0 and every T > 0,
1
T (t ∈ [0, T ] : u(t)x ∈ Cξk
) > 1− 2−(k+1). (3.16)
Fix α > 1. Let g ∈ NG(U) be such that gu(t)g−1 = u(α−1t) for all t ∈ R.Due to Lemma 3.3.6, there exists nk ∈ N such that
gnkXk(U) ⊂ Xξk/2(U). (3.17)
Let ηk > 0 be such that
g−nkXξ0(U) ⊂ Xηk(U).
Replacing the sequence xi by a subsequence, we may assume that for everyk ∈ N and every i ≥ k, xi ∈ Xηk
(U). Therefore gnkxi ∈ Xξ0(U) and hence byProposition 3.3.7, there exists T0 = Tk,i ≥ 0 such that u(T0)gnkxi ∈ Cξ0 . Now,after the change of variable: t + T0 → t in Eq. 3.16, for every T > T0, we have
1
T − T0 (t ∈ [T0, T ] : u(t)gnkxi ∈ Cξk
) ≥ 1− 2−(k+1).
Therefore for every T ≥ T0,
1T (t ∈ [T, 2T ] : u(t)gnkxi ∈ Cξk
)≥ (1−2−(k+1))(2T−T0)−(T−T0)
T
≥ 1− 2−k.(3.18)
For every t ∈ R,g−nku(t)gnk = u(αnkt).
74
Therefore, after the change of variable (αnkt → t) in Eq. 3.18, for every T ≥αnkT0, we have
1
Tt ∈ [T, 2T ] : u(t)xi ∈ g−nkCξk
≥ 1− 2−k.
Now take i ∈ N. Put
C(i) =
1≤k≤i
g−nkCξk⊂ X \
1≤k≤i
Xk(U)
andT (i) = max
1≤k≤i
αnkTk,i.
Then for every T ≥ T (i),
1
T (t ∈ [T, 2T ] : u(t)xi ∈ C(i)) ≥ 1−
i
k=1
2−k = 2−i.
Therefore there exists ti ∈ [T (i), 2T (i)] such that u(ti)xi ∈ C(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 limitpoint 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 horospher-
ical 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 ti ⊂ R and a λ > 1, there exists a sequence
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 ∈ NG(U) such that
U = u ∈ G : gnug−n → e as n →∞
and gu(t)g−1 = u(α−1t) for all t ∈ R.Since y ∈ X(U), by Lemma 3.4.5, there exists ξ ∈ (0, ξ0) such that the set
R = n ∈ N : gny ∈ Cξ is unbounded.Take k ∈ N. By Theorem 3.4.2, there exists a ξk ∈ (0, ξ) such that for
every x ∈ Cξ and every T > 0,
1
T (t ∈ [0, T ] : u(t)x ∈ Cξk
) ≥ 1− 2−k(λ− 1)/λ.
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Then
1
λT − T (t ∈ [T,λT ] : u(t)x ∈ Cξk
) >(1− 2−k(λ− 1)/λ)λT − T
(λ− 1)T
= 1− 2−k.
Since the set R is unbounded, by Lemma 3.3.6, there exists nk ∈ N suchthat gnkXk(U) ⊂ Xξk/2(U) and gnky ∈ Cξ. Since Cξ is open and xi → y,by passing to a subsequence we may assume that for every k ∈ N and everyi ≥ k,
gnkxi ∈ Cξ.
Hence for every k ∈ N, every i ≥ k, and every T > 0,
1
λT − T (t ∈ [T,λT ] : u(t)gnkxi ∈ Cξk
) ≥ 1− 2−k. (3.19)
Note that for every t ∈ R, we have g−nku(t)gnk = u(αnkt). Therefore after thechange of variable: αnt → t in Eq. 3.19, for every T > 0, we have that
1
λT − T
t ∈ [T,λT ] : u(t)xi ∈ g−nkCξk
≥ 1− 2−k.
Take i ∈ N. Put
C(i) =
1≤k≤i
g−nkCξk⊂ X \
1≤k≤i
Xk(U).
Then for any T > 0,
1
λT − T(t ∈ [T,λT ] : u(t)xi ∈ C(i)) ≥ 1−
i
k=1
2−k = 2−i.
Therefore there exists αi ∈ [1, λ] such that u(αiti)xi ∈ C(i).Put C = ∩i∈NC(i). For every i ∈ N, C(i) is compact and C(i+1) ⊂ C(i).
Therefore the sequence u(αiti)xi has a limit point in C and C ∩X(U) = ∅.
The above two corollaries will be used in the next chapter to study theclosures of orbits of unipotent subgroups acting on L-homogeneous spaces ofG.
3.5 Description of noncompact L-homogeneousspaces of SL2(C)
Let X be a noncompact L-homogeneous space of G. In this section we shallobtain detailed information about the structure of the set X \ Cξ0 .
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Let U be a nontrivial horospherical subgroup of G. There exists k ∈ Ksuch that U = kNk−1. If U = k1Nk−1
1 for another k1 ∈ K, then k−1k1 ∈NG(N) ∩K = M . Therefore the groups
A(U) := kAk−1
andM(U) := kMk−1 = NG(U) ∩K
are well defined. Now G = KA(U)U is an Iwasawa decomposition of G andNG(U) = M(U)A(U)U . For η > 0, define
NG(U)η = g ∈ NG(U) : | det(Ad g|U)| ≤ η
andAη(U) = A(U) ∩ NG(U)η.
Then NG(U)η = Aη(U)M(U)U .
Remark. For any x ∈ X(U) the orbit Ux is compact, and hence the orbitM(U)Ux is also compact. For any sequence gn ⊂ A(U), if det(Ad gn|U) → 0as n →∞, then the sequence gnx ⊂ X diverges as n →∞.
Description of the set X(U)
Lemma 3.5.1 For any η > 0 and x ∈ Xη(U), there exists a neighbourhood Ωof e in G such that
Ωx ∩Xη(U) ⊂ (Ω ∩ NG(U))x.
Proof. By Lemma 3.3.6, for k ∈ K,
kXξ0(U) ∩Xξ0(U) = ∅ ⇔ k ∈ NG(U). (3.20)
Let Ψ1 be a neighbourhood of e in NG(U) such that Ψ1Xξ0/2(U) ⊂ Xξ0(U).Due to Iwasawa decomposition, Ω1 := KΨ1 is a neighbourhood of e in G. Nowdue to Eq. 3.20, for any y ∈ Xξ0/2(U), we have that
Ω1y ∩Xξ0(U) ⊂ (NG(U) ∩ Ω1)y. (3.21)
Let g ∈ NG(U) be such that
gXη(U) ⊂ Xξ0/2(U).
Put Ω = g−1Ω1g and y = gx ∈ Xξ0/2(U). Then due to Eq. 3.21,
Ωx ∩Xη(U) ⊂ g−1(Ω1y ∩Xξ0/2(U)) ⊂ g−1(NG(U) ∩ Ω1)gx = (NG(U) ∩ Ω)x.
First we record two corollaries of this lemma for future reference.
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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 = Xη1(U1) ∩Xη2(U2) ∩Xη3(U3)
is finite.
Proof. Let g ∈ NG(U1) be such that gXη1(U1) ⊂ Xξ0(U1). Then
gS ⊂ Xξ0(U1) ∩Xη2(U
2),
where U
2 = gU2g−1 and η2 > 0. Therefore without loss of generality we mayassume that η1 ≤ ξ0. By Lemma 3.3.6, Xξ0(U1) ∩Xξ0(U2) = ∅. Therefore
S ⊂ Xη1(U1) ∩ (Xη2(U2) \Xξ0(U2)).
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 Ω of e in
G such that ΩΩ−1 ∩Gx = e, Ω ∩ C(G) = e, and
Ωx ∩Xηi(Ui) ⊂ (Ω ∩ NG(Ui))x
for i = 1, 2, 3. Since NG(U1)∩NG(U2)∩NG(U3) = C(G), we have Ωx∩S = x.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 ηn →∞ in R. For n ∈ N define
S(n) = t ∈ [−δ, δ] : ρ(t)x ∈ Xηn(U).
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 thereexists t0 ∈ S(n0) and δ > 0 such that u(t+ t0)x ∈ Xηn0
(U) for all t ∈ (−δ, δ).Therfore due Lemma 3.5.1 there exists ∈ (0, δ) such that ρ(t) ∈ NG(U) forall 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) =
x∈Σ
NG(U)x
78
and for any η > 0,Xη(U) =
x∈Σ
NG(U)ηx.
(Note that due to Theorem 2.4.7, each NG(U)-orbit in X is dense.)
Proof. PutC = Xξ0(U) \Xξ0/2(U).
Then C is a compact subset of Xξ0(U). By Lemma 3.5.1, NG(U)x∩Xξ0(U) isan open subset of Xξ0(U) for every x ∈ X(U). Therefore there exists a finiteset Σ ⊂ C such that
C ⊂
x∈Σ
NG(U)x.
Now for any z ∈ X(U), there exists g ∈ NG(U) such that
Vol(Ugz) = | det(Ad g|U)| · Vol(z) ∈ (ξ0/2, ξ0).
ThereforeX(U) = NG(U)C ⊂
x∈Σ
NG(U)x ⊂ X(U).
Thus X(U) is a union of finitely many orbits of NG(U).Replace each x ∈ Σ by gx for some g ∈ NG(U) such that
Vol(Ugx) = | det(Ad g|U)|Vol(Uy) = 1
and denote the new set by Σ. Then Σ satisfies the conclusion of the lemma.
A function V on X
Take any x ∈ X. Then its stabilizer-Gx contains a nontrivial element of ahorospherical subgroup of G, say W . Now by Corollary 3.2.4, the orbit Wxis compact. Therefore we can define a function V : X → (0,∞) such that forevery x ∈ X, V(x) is the infimum of Vol(Wx) over all nontrivial horosphericalsubgroup 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. V is a continuous function.
3. V(x) = V(x) for all x ∈ X \ Cξ0 . In particular, V vanishes at the pointat infinity.
Thusη0
def= max
x∈X
V(x) < ∞.
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Fix a nontrivial horospherical subgroup U of G. For a z ∈ X, let W be ahorospherical subgroup of G such that Wz is compact and V(z) = Vol(Wz).Let k ∈ K be such that W = kUk−1. Put x = k−1z. Then
V(x) = V(z) = Vol((kUk−1)z) = Vol(kUx) = Vol(Ux).
Therefore x ∈ Xη0(U). Thus we have proved that
X = KXη0(U) =
x∈Σ
KNG(U)η0x.
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 ex-
ists a finite collection C consisting of compact U-orbits in X such that the
following holds:
1. There exists η0 > 0 such that
X =
O∈C
KAη0(U)O.
2. There exists an ξ0 > 0 such that for O, O ∈ C, a, a ∈ Aξ0(U), and
k, k ∈ K, if
kaO ∩ kaO = ∅then O = O
, a = a, kM(U) = kM(U), and
kaO = kaO.
Remark. Put U = N in the above theorem and apply Lemma 2.3.2 to see thatX admits a finite G-invariant measure. This provides another proof of the factthat an L-homogeneous space of G admits a finite G-invariant measure (seeCorollary 3.4.3).
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Chapter 4
Proof of Raghunathan’s conjecture forSL2(C)
In this chapter we shall study closures of orbits of unipotent one-parametersubgroups of G acting on an L-homogeneous space of G. To analyse the clo-sures we use the techniques introduced by G.A. Margulis in [M2] and developedfurther 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 actionsplay an important role.
4.1 Minimal invariant sets
Definition. Let F 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 ofZ is invariant under the action of F , or equivalently if every F -orbit in Z isdense.
Lemma 4.1.1 Any compact F -space contains a compact F -minimal subset.
Proof. Let Y be a compact F -space. Let C = Yαα∈I be the family of allnonempty closed F -invariant subsets of Y . Then C is partially ordered byset theoretic inclusion. We want to show that this family contains a minimalelement. Due to Zorn’s lemma, it is enough to show that C contains a lowerbound for every totally ordered subfamily.
Let CJ = Yαα∈J be a totally ordered subfamily of C, where J ⊂ I. NowCJ has the finite intersection property. Since Yα is compact for each α ∈ J ,the set
YJ =
α∈J
Yα
is nonempty. Also YJ is closed and F -invariant. Therefore YJ is a lower boundfor the subfamily CJ in C. This completes the proof.
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In general a noncompact F -space need not contain a closed F -minimalsubset. While for the unipotent actions on L-homogeneous spaces, we havethe 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 unipo-
tent 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 ∩ NG(U1))U1.
(b) There exists g ∈ F ∩ NG(U1) such that | det(Ad g|U1)| > 1.
(c) There exists a horospherical subgroup U2 of G such that U1 = U2 andF ∩ U2 = e.
For notational convenience, put I = 1 for the cases (a) and (b), and putI = 1, 2 for the case (c). For i ∈ I, let ui(t) : t ∈ R be a nontrivialone-parameter subgroup of Ui such that ui(1) ∈ F . Define a compact set
C =
i∈I
ui([0, 1])Cξ0 ,
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 ui(n)x ∈ C. Therefore
if Y ⊂
i∈I
Xξ0(Ui) then Y ∩ C = ∅.
In the Case (c), due to Lemma 3.3.6, Xξ0(U1) ∩ Xξ0(U2) = ∅. HenceY ∩ C = ∅.
In the Case (b), for any x ∈ Y , if x ∈ Xξ0(U1) then there exists m ∈ Nsuch that
Vol(U1(gmx)) = | det(Ad g|U1)|m · Vol(U1x) > ξ0.
Since g ∈ F , we have gmx ∈ Y \Xξ0(U1). Hence Z ∩ C = ∅.In the Case (a), if there exists x ∈ Y ∩Xξ0(U1) then
Fx ⊂ (K ∩ NG(U1))U1x,
which is a compact F -invariant subset of Y . This completes the proof.
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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 unipo-
tent element of G. Then any closed F -invariant subset of X contains a closed
F -minimal subset.
Proof. If Y contains a compact F -invariant subset, then we are through byapplying Lemma 4.1.1 to that compact subset.
Otherwise due to Proposition 4.1.2, there exists a compact set C in X suchthat any closed F -invariant subset of Y intersects C. Let the notation be asin the proof of Lemma 4.1.1. To apply Zorn’s lemma, we only need to showthat the set YJ is non-empty. Now
YJ ∩ C =
α∈J
Yα ∩ C.
The family Yα ∩ Cα∈J consists of (non-empty) compact sets and it has thefinite intersection property. Therefore YJ ∩ C is non-empty. This completesthe 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. If Z intersectsa 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 assumethat V = e. Let u(t) : t ∈ R be a nontrivial one-parameter subgroup ofU 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 zi has no convergent subsequence. Let a sequence ni ⊂ N besuch that yi := u(ni)zi ∈ C and u(k)zi ∈ C for all 0 ≤ k ≤ ni − 1. Since Cis relatively compact and zi → ∞, we have ni → ∞ as i → ∞. Since C isrelatively compact, by passing to a subsequence we may assume that yi → yfor some y ∈ C ∩ Z. Now for every k ∈ N,
u(−k)yi = u(ni − k)zi ∈ C for all large i ∈ N.
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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) For any g ∈ G, gZ is a closed gFg−1-minimal subset of X. In particular,
if g ∈ NG(F ) then gZ is F -minimal.
(2) Let Y be any closed F -invariant subset of X. If g ∈ NG(F ) and gZ∩Y =∅, then gZ ⊂ Y .
Proof. The statement (1) holds because if Y is a closed gFg−1-invariant subsetof gZ then g−1Y is a closed F -invariant subset of Z.
The statement (2) holds because gZ ∩ Y is a closed F -invariant subset ofgZ, 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 orbitof a given subgroup is based on the following observation in teM-qforms.
Lemma 4.2.1 (Cf. [M2, Sect. 2.1]) Let G be any locally compact group, Γbe 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 \ NG(F ) : gZ ∩ Y = ∅.
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 NG(F ) ∩ FMF .
Proof. Suppose that e ∈ M . Then there exists a neighbourhood Ω of e in Gsuch that ΩZ ∩ Y ⊂ (NG(F ) ∩ Ω)Z. Therefore (NG(F ) ∩ Ω)Z contains anopen 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 byLemma 4.1.5, gZ is F -minimal. Therefoere Z = gZ and hence Y = Z. Let
L = g ∈ NG(F ) : gY ∩ Y = ∅.
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Again by Lemma 4.1.5, gY = Y for every g ∈ L. Since Y is compact, L is aclosed subgroup of NG(F ) containing F . Also for every x ∈ Y ,
Ωx ∩ Y ⊂ (Ω ∩ NG(F ))x ∩ Y ⊂ Lx.
Therefore every L-orbit in Y is open. Since Y is F -minimal, Y is itself anorbit of L. This completes the proof of the first part.
Next suppose that e ∈ M . First take h ∈ FMF . Let f, f ∈ F andm ∈ M be such that h = f mf . Let z ∈ Z be such that mz ∈ Y . Thenz1 = f−1z ∈ Z and f (mfz1) ∈ Y . Thus hZ ∩ Y = ∅. Now take h ∈ FMFand let hi ⊂ FMF be a sequence such that hi → h as i → ∞. Then thereexists a sequence zi ⊂ Z such that hizi ∈ Y for all i ∈ N. Since Z iscompact, by passing to subsequences we may assume that zi → z for somez ∈ Z as i → ∞. Therefore hizi → hz as i → ∞. Since Y is closed, we havehz ∈ Y . Further if we assume that h ∈ NG(F ) then since Z is F -minimal, byLemma 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 ∈ NG(F ) ∩M , we have gZ ⊂ Y .
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, thenext lemma shows that if F is a one-parameter unipotent subgroup of G thenthe set S/F ⊂ NG(F )/F contains an unbounded curve passing through theidentity 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 → Lsuch that φ(0) = p and φ(R) ⊂ V Y ∩ L.
More precisely, there exists a sequence ti → ∞ and a sequence pi ⊂ Ywith pi → p as i → ∞, such that for any sequence si → s in R, we have
u(siti)pi → φ(s) as i →∞.
85
Proof. (Cf. [Si, pp. 312]) Let N be a nilpotent linear transformation on Esuch that u(t) = exp(tN) for all t ∈ R. Since N = 0, there exists r ≥ 0 suchthat N r = 0 and N r+1 = 0. Consider the following filteration of E:
0 ⊂ L = ker N ⊂ . . . ⊂ ker N r ⊂ ker N r+1 = E.
For 0 ≤ j ≤ r, let Ej be a complementary subspace of ker N j in ker N j+1.Then
E = ⊕r
j=0Ej.
For x ∈ E, write x =
r
j=0 xj, where xj ∈ Ej for 0 ≤ j ≤ r. Then for t ∈ R,
u(t)x =r
j=0
exp(tN)xj =r
j=0
r
k=0
(tk/k!)Nkxj
=: A(x, t) + B(x, t),
where
A(x, t) =r
j=0
(tj/j!)N jxj =:r
j=0
Aj(x, t)
and
B(x, t) =r
j=1
j−1
k=0
(tk/k!)Nkxj
.
Put∆(x, t) = maxAj(x, t) : 1 ≤ j ≤ r,
where · denotes a euclidean norm on E. Then there exists a constant c > 0such that for all t > 1,
B(x, t) ≤ ct−1∆(x, t).
If x ∈ L then ∆(x, t) > 0. Take a sequence pi ⊂ Y ⊂ E \ L such thatpi → p as i → ∞. Since p ∈ L = E0, for any 1 ≤ j ≤ r, the projection ofpi on Ej converges to 0 as i → ∞. Therefore for any t > 0, ∆(pi, t) → 0 asi →∞. Moreover, for every i, ∆(pi, t) →∞ as t →∞. Therefore there existsa sequence ti ⊂ R such that ∆(pi, ti) = 1 for every i ∈ N and ti → ∞ asi →∞.
By passing to a subsequence, we may assume that for every 1 ≤ j ≤ r,there exists wj ∈ L such that Aj(pi, ti) → wj as i → ∞. Also wj0 = 1 forsome j0 ∈ 1, . . . , r.
Moreover, B(pi, ti) ≤ ct−1k
∆(pi, ti) → 0 as i →∞. Define a polynomialfunction φ : R → L such that for every s ∈ R,
φ(s) = p +r
j=1
sjwj.
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Now for every s ∈ R and i ∈ N,
u(sti)pi = p +r
j=1
sjAj(pi, ti) + B(pi, ti).
Hence for any sequence si → s in R, u(siti)pi → φ(s) as i →∞. In particular,V Y ∩ L ⊃ φ(R).
4.3 Specialization to the case of SL2(C)
The following notations will be used throughout the remaining part.
Notation. Let G = SL2(C) and X be an L-homogeneous space of G. Recallthat a homogeneous space of G admits a finite G-invariant measure if and onlyif it is an L-homogeneous space. Let
K = g ∈ G : ggt = 1,
D =
d(α)
def=
α 00 α−1
: α ∈ C∗
,
M = d(eti) : t ∈ R = D ∩K,
A = d(a) : a > 0,
N =
u(z)
def=
1 z0 1
: z ∈ C
,
H = SL2(R)
K1 = H ∩K
N1 = u(t) : t ∈ R = N ∩H, and
N2 = v(t) := u(it) : t ∈ R ⊂ N.
Our main interest is to study the closures of N1-orbits in X. Using thelinear formulation in Lemma 4.2.3, we shall obtain some group theoretic in-formation 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 Esuch 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) onE = C2 and take p = (1, 0). Then N = g ∈ G : g·p = p and G·p = C2\0.Thus (i) and (ii) are satisfied.
Now let V = N1. Let E = C2 ⊕ (C2 ∧C2), where C2 is treated as a fourdimensional real vector space. Consider the G-action on E, where an elementg ∈ G acts on E as a linear transformation such that for all v1,v2,v3 ∈ C2,
g · (v1 + (v2 ∧ v3)) = gv1 + (gv2 ∧ gv3).
Take e = (1, 0) and f = (0, 1) in C2. Put p = e + e ∧ f ∈ E. Then for anyg ∈ 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 ∩ SL2(R) = N1.
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 ·pis closed.
DefineL = x ∈ E : g · x = x for all g ∈ N1.
Due to condition (i),
g ∈ G : g · p ∈ L = g ∈ G : N1(g · p) = g · p= g ∈ G : g−1N1g ⊂ N1= NG(N1).
Therefore AN · p ⊂ L and
AN · p \G · p = AN · p \ NG(N1) · p.
Now
AN · p = AN2 · p = d(a)v(t)(e + (e ∧ f)) : a > 0, t ∈ R= d(a)(e + t(e ∧ ie) + (e ∧ f)) : a > 0, t ∈ R= ae + t(e ∧ ie) + (e ∧ f) : a > 0, t ∈ R.
88
Recall that NG(N1) = d(ik)k∈ZAN . Therefore
AN · p \ NG(N1) · p = t(e ∧ ie) + (e ∧ f) : t ∈ R
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 \ NG(V ) be such that e ∈ M . Then the following holds:
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) = 0 then
d(σ(t)) ∈ N1MV .
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) = 0 then
v(ν(t))d(σ(t)) ∈ N1MV .
Proof. Let the linear representation of G on E and the point p ∈ E be as inthe proof of Proposition 4.3.1. Let
L = x ∈ E : g · x = x for all g ∈ N1.
For any g ∈ G,
g · p ∈ L ⇔ N1(g · p) = g · p⇔ (g−1N1g) · p = p
⇔ g−1N1g ⊂ V
⇔ g ∈ NG(V ),
where the last implication is due to the fact that N ∩ gNg−1 = ∅ if and onlyif g ∈ NG(N) (see Lemma 2.2.7). Thus
NG(V ) = g ∈ G : g · p ∈ L. (4.2)
Put Y = M · p. Then Y ∩ L = ∅, because M ∩ NG(V ) = ∅. Alsop ∈ Y ∩ L, because e ∈ M . By Lemma 2.1.6, N1 acts on E by unipotentlinear transformations. Therefore by Lemma 4.2.3, there exists a nonconstantpolynomial function φ : R → L such that φ(0) = p and for every t ∈ R,
φ(t) ∈ N1Y = N1M · p. (4.3)
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 toLemma 1.1.2. Therefor
ψ(N1MV/V ) ∩G · p = ψN1MV/V
(4.4)
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From Eqs. 4.2, 4.3, and 4.4, we conclude that
g ∈ G : φ(t) = g · p for some t ∈ R ⊂ N1MV ∩ NG(V ). (4.5)
First suppose that V = N . In this case E = C2 and p = (1, 0). SinceL = 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.Now suppose that V = N1. In this case E = C2 ⊕ (C2 ∧ C2) and p =
e + (e ∧ f), where e = (1, 0) and f = (0, 1). Since G · p is open in its closureand 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) ∈ NG(N1)0 · p = AN · p = N2A · p for all t ∈ (−δ, δ). (4.6)
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 polynomialsσ and ν such that σ(0) = 1, ν(0) = 0, and for any t ∈ (−δ, δ), we have thatσ(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)) ∈ N1MN1
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. Let Y 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 \ NG(N) : gZ ∩ Y = ∅.
If e ∈ M then by Lemma 4.2.1, Y = Z and there exists a closed subgroupL ⊂ NG(N) = DN containing N such that Y is a compact orbit of L. Fixz ∈ Y . Then L/Lz is compact. Therefore by Corollary 2.2.5, the orbit Nz iscompact. 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 haveY = X. This completes the proof.
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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 t0 ∈ R. Define
Y1 = y ∈ Y : there exist sequences zi ⊂ Z and ti →∞ in R
such that Φ(ti)zi → y as i →∞, and
S = g ∈ NG(F ) : there exists a function f : (0,∞) → R such that
t + f(t) →∞ and Φ(t + f(t))Φ(t)−1 → 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 ofY . Now for any y ∈ Y1 and g ∈ S, with the notations above, as i →∞,
Φ(ti + f(ti))zi = Φ(ti + f(ti))Φ(ti)−1 · Φ(ti)zi → gy.
Therefore gy ∈ Y1. This completes the proof. 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 sub-
group 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 , andΦ(t) = w(t) for all t > 0. Then S = W and due to Proposition 3.2.3, the setY1 is nonempty. Therefore by Lemma 4.4.1, WY1 = SY1 ⊂ Y .
To study the closures of orbits of N1, in view of Proposition 4.3.2 and theabove 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 t0 ≥ 0 be such that σ(t) = 0for all t > t0. Define a function Φ : (t0,∞) → AN2 as Φ(t) = v(ν(t))d(σ(t))for all t > t0. Then there exists a nontrivial one-parameter subgroup ρ : R →AN2, and for every s ∈ R there exists a function fs : (0,∞) → R such that
as t →∞,
t + fs(t) →∞ and Φ(t + fs(t))Φ(t)−1 → ρ(s).
(Note that due to Lemma 2.2.2, any nontrivial one-parameter subgroup of AN2
is either N2 or vAv−1for some v ∈ N2.)
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Proof. For t > t0 and ξ > 0, we have
Φ(t + ξ)Φ(t)−1 = v(ν(t + ξ))d(σ(t + ξ)) · d(σ(t)−1)v(−ν(t))
= v(ψ(t, ξ)) · d(σ(t + ξ)σ(t)−1),
whereψ(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 andrational functions ψi, where i = 1, . . . , k, such that
ψ(t, ξ) =k
i=1
ψi(t)ξi.
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 = max1≤i≤k
(1/i) deg ψi.
Then there exists a polynomial P without a constant term and with deg P ≥ 1,such that for any sequence si → s in R and for any sequence ti →∞,
ψ(ti, sit−q
i) → P (s).
In the case when ψ ≡ 0, put q = −∞.Note that for any p > q, as i →∞,
ψ(ti, sit−p
i) → 0.
Also as i →∞,σ(ti + siti)σ(ti)
−1 → (1 + s)deg σ
andσ(ti + sit
−q
i)σ(ti)
−1 → 1 if q > −1.
For s ∈ R and t > 0, define
fs(t) =
st−q if q > −1(es − 1)t if q ≤ −1.
Then for any s ∈ R, there exists ρ(s) ∈ AN2, such that for any sequencessi → s and ti →∞ in R,
Φ(ti + fsi(ti))Φ(ti)−1 → ρ(s); (4.9)
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in fact one can verify that, for every s ∈ R,
ρ(s) =
v(P (s)) if q > −1v(P (es − 1))d(es deg(σ)) if q = −1d(es deg(σ)) if q < −1.
To show that ρ is a one-parameter subgroup of AN2, take s1, s2 ∈ R. Fort > t0,
Φ(t + fs1+s2(t))Φ(t)−1
= Φ(t + fs1+s2(t))Φ(t + fs2(t))−1 · Φ(t + fs2(t))Φ(t)−1.
(4.10)
For t ∈ R, put yt = t + fs2(t). Then yt →∞ as t →∞.First suppose that q ≤ −1. Then t + fs(t) = est. Therefore
t + fs1+s2(t) = es1+s2t = es1yt = yt + fs1(yt). (4.11)
For notational convenience put rt = s1, for all t > t0, in this case.Now suppose that q > −1. Then
t + fs1+s2(t) = t + (s1 + s2)t−q
= yt + s1(t/yt)−q · y−q
t
= yt + rty−q
t
= yt + frt(yt), (4.12)
where rt = s1(t/yt)−q.Thus due to Eqs. 4.9, 4.11 and 4.12, as t → ∞, we have that yt → ∞,
rt → s1, and
Φ(t + fs1+s2(t))Φ(t + f2(t)) = Φ(yt + frt(yt)) → ρ(s1).
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 moregeneral result without a proof. A possibility of such a generalization wassuggested 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 condi-
tions: (i) for all 1 ≤ i, j ≤ n, the map t → Φi,j(t) is a rational function in
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
t + fs(t) →∞ and Φ(t + fs(t))Φ(t)−1 → ρ(s).
Moreover, all the eigenvalues of ρ(s) are real for every s ∈ R.
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4.5 Closures of orbits of N1
Using the techniques developed in the earlier sections, now we give a proof ofthe 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 .
2. Y contains an orbit of vAv−1for some v ∈ N2.
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 ofY . It is enough to prove that either (1), (2), or (3) occurs or NZ is a compactN -orbit contained in Y . Let
M = g ∈ G \ NG(N1) : gZ ∩ Y = ∅.
If e ∈ M then by Lemma 4.2.1, Y = Z and Y is a compact orbit of a closedsubgroup of NG(N1)0 = AN containing N1. Therefore by Corollary 2.2.5, Yis a compact orbit of either N1 or N .
Now suppose e ∈ M . Then by Lemma 4.2.1 and Proposition 4.3.2, thereexist 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 . PutΦ(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 acompact orbit of N1 or N or (ii) Z is invariant under the closed subgroup Lgenerated by Φ(t) : t ∈ R, σ(t) = 0.
Comming back to the original situation: In the case (i), if Z is a compactN -orbit then we are through. In the case (ii), by Proposition 4.4.3, eitherL ⊃ N2 or L ⊃ vAv−1 for some v ∈ N2. If L ⊃ vAv−1 then possibility (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. ByCorollary 4.4.2, v(ν(R))Z contains an N2-orbit. Therefore NZ ⊂ Y , and weare through.
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Now assume that σ is not a constant polynomial. Define Y1 ⊂ Y andS ⊂ AN2 as in Lemma 4.4.1. Note that for all large t > 0 and any z ∈ Z,
Vol(N(Φ(t)z)) = | det(Ad Φ(t) |N )| · Vol(Nz) = |σ(t)|4 · Vol(Nz).
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
N2 or vAv−1 for some v ∈ N2. Therefore by Lemma 4.4.1, either NY1 ⊂ Yor vAN1v−1Y1 ⊂ 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 completesthe 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 \ NG(N) : gZ ∩ Y = ∅.
If e ∈ M then there exists z ∈ Z and g ∈ NG(N) such that gz ∈ Y andY = N1(gz). Thus Y is contained in the compact N -orbit gZ. Hence Y is acompact 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 everyy ∈ d(σ(t))Z, the set N1y is a compact N1-minimal subset. Hence by thehypothesis of this proposition, d(σ(t))Z ⊂ Y for all large t > 0. Now byTheorem 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 anadditional assumption that Y contains an orbit of vAv−1 for some v ∈ N2.
Observe that if a set S ⊂ X is the closure of an orbit of a subgroup F ofG then for any g ∈ G, the set gS is the closure of an orbit of gFg−1.
Since v ∈ NG(N1), the set Y := v−1Y is the closure of an orbit of N1 andY contains an orbit of A. Hence for further investigation we can assume thatthe closure of the given N1-orbit contains an orbit of AN1, and we should showthat 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 setN1M(AN1) ∩ NG(N1)
/N,
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where M ⊂ G \ NG(N1) and e ∈ M . The next lemma is useful in this regard.
Lemma 4.5.3 Given sequences hi ⊂ H(= SL2(R)) and ti →∞ in R, there
exists a sequence ikk∈N such that the following condition is satisfied: There
exists s∗ ∈ R ∪ ±∞ such that for any s ∈ R \ s∗ and a sequence sk → sin R, we have
π(u(sktik)hik) → π(e)
as i →∞, where π : H → H/AN1 is the natural quotient map.
Further if hi is a constant sequence then s∗ can be chosen to be ±∞.
Proof. (Cf. [DM1, Sect. 3.5]) The group H = SL2(R) acts on the unit circleS1 ⊂ 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 byLemma 1.1.2,
H/AN1∼= S1 (4.13)
as H-spaces.Let i ∈ N and put pi = hi · p =: (ai, bi) ∈ S1. Then for s ∈ R,
u(sti)pi = (ai + stibi, bi)/
(ai + stibi)2 + b2i.
Note that
|bi|(ai + sitibi)2 + b2
i
≤ |bi||ai + stibi|
=t−1i
|ai(tibi)−1 + s| .
Now there exists a sequence ik →∞ in N such that as k →∞,
(i) aik→ a, bik
→ b, where a, b ∈ [0, 1]; and
(ii) −aik(tikbik
)−1 → s∗, where s∗ ∈ R ∪ ±∞.
Then for every s ∈ R \ s∗ and a sequence sk → s,
bik/(aik
+ sktikbik) → 0.
Therefore u(sktik)pik→ p, as k → ∞. Now the conclusion of the lemma
follows from the equivariant isomorphism given by Eq. 4.13.
Notation. Recall that G = A ∈ M2(C) : tr A = 0 is the Lie algebra as-sociated to G, and H = A ∈ M2(R) : tr A = 0 is the the Lie subalge-bra of G associated to H. Put P = iH. Then G = P ⊕ H and P is in-variant under the action of Ad(H) on G. Now the one-parameter subgroup
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Ad N1 = Ad u(t)t∈R acts on P by unipotent linear transformations. Let N2
denote the Lie subalgebra associated to N2. Then
N2 = q ∈ P : (Ad N1) · q = q.
Remark. By Lemma 2.1.2, there exists a neighbourhood Ψ0 of the origin inG and a neighbourhood Ω0 of the identity in G, such that the map (q,y) →(expq)(expy) is a homeomorphism from (P ∩Ψ0)× (H ∩Ψ0) onto Ω0.
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 , mi → e, a sequence ti → ∞ in R, and a nonconstant polynomial νsuch that the follwing holds: There exists s∗ ∈ R ∪ ±∞ such that for any
s ∈ R \ s∗, and a sequence si → s, we have
π(u(siti)mi) → π(v(ν(s))),
as i →∞, where π : G → G/AN1 is the natural quotient map.
In particular,
N1MAN1 ⊃ v(ν(R)).
Proof. Let mi ⊂ M ∩ Ω0 be a sequence such that mi → e as i →∞. Thenfor every i ∈ N, there exist qi ∈ P ∩ Ψ and hi ∈ exp(H ∩ Ψ) ⊂ H such thatmi = (expqi)hi. Then qi → 0 and hi → e as i →∞. Now for all large i ∈ N,we have that qi ∈ N2, because mi ∈ (N ∩ Ω0)H.
By Lemma 4.2.3, if we replace mi by a subsequence then there exist asequence ti → ∞ in R and a nonconstant polynomial ν such that for anysequence si → s in R,
Ad u(siti)qi →
0 iν(s)0 0
∈ N2 as i →∞
and hence as i →∞,
u(siti)(expqi)u(−siti) → v(ν(s)). (4.14)
By Lemma 4.5.3, there exist a sequence ik → ∞ in N and an element s∗ ∈R ∪ ±∞ such that given a sequence sk → s ∈ R \ s∗, we have
π(u(sktik)hik) → π(e) (4.15)
as k →∞. Thus by Eq. 4.14 and Eq. 4.15,
π(u(sktik)mik) = (u(sktik) expqik
u(−sktik))π(u(sktik)hik) → v(ν(s))π(e)
As k →∞. This completes the proof. It will be convenient to note the following simple observation.
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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. If Z contains an orbit of
vAv−1for some v ∈ N2 \ e, then Z = X.
3. Let Z be a proper, closed AN1-minimal subset of X. Then for every
z ∈ Z \X(N), we have that N1z = Z.
Proof. By Corollary 3.4.6, any closed N1-invariant subset of X(N) is containedin Xη(N) for some η > 0. For any z ∈ X(N) and a ∈ A,
Vol(N(az)) = | det(Ad a|N )|Vol(Nz).
But then Xη(N) does not contain an orbit of A. Hence X(N) does not containa closed AN1-invariant subset. This proves Part 1).
For t = 0 and v = v(t) ∈ N2,
AvAv−1 ⊃ d(a)v(t)d(a−1)v(−t) = v((a2 − 1)t) : a > 0.
ThereforeAvAv−1 ⊃ v([0,∞)) or v((−∞, 0]). (4.16)
Suppose that Z is AN1-invariant and contains an orbit of vAv−1. ThenAvAv−1z ⊂ Z for some z ∈ Z. Now by Eq. 4.16 and Corollary 4.4.2, we havethat Z contains an orbit of N2 and hence it contains an orbit of AN . Thereforeby 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-tions 4.5.1 and 4.5.2, the set N1z contains an orbit of vAv−1 for some v ∈ N2.If v = e then for some y ∈ N1z, we have N1z ⊃ AN1y and hence AN1y = Z bythe minimality of Z. If v = 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 Ycontains a closed AN1-minimal subset, say Z, and let z0 ∈ Z \X(N). Define
M = g ∈ G \ (N ∩ Ω0)H : gz0 ∈ Y
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 ofProposition 4.5.4. Put yi = miz0 for every i ∈ N. Now yi → z0, z0 ∈ X(N),and ti → ∞. Therefore by Corollary 3.4.7, for every s ∈ R and every λ > 1
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there exists a sequence αi ∈ [1, λ] such that a subsequence of u(αisti)yiconverges to a point y ∈ Y \ X(N). By passing to subsequences we mayassume that αi → α ∈ [1, λ] and
u(αisti)yi → y. (4.17)
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 sequencegk ⊂ AN1 such that as k →∞,
u(αikstik)mik
gk → v(ν(αs)). (4.18)
Therefore by Eq. 4.17 and Eq. 4.18, as k →∞,
g−1k
z0 = (u(αikstik)mik
gk)−1(u(αik
stik))yi
→ v(ν(αs))−1y =: z ∈ Z.
Since z ∈ Ny, we have z ∈ X(N). As we can assume that Z = X, due toLemma 4.5.5, Z = N1z. Therefore
Y ⊃ N1y = v(ν(αs))N1z = v(ν(αs))Z.
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 N2(+) = v(t) : t ≥ 0 and N2(−) = v(t) : t ≤ 0. Then ν(R) ⊃N2(±), and hence N2(±)AN1Z ⊂ Y . Note that N2(±)AN1 = AN1N2(±). ByCorollary 4.4.2, the set N2(±)Z contains an N2-orbit. Therefore Y containsan orbit of AN . Hence by Theorem 2.4.7, Y = X.
In the next proposition we remove the extra conditions involved in theprevious 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 y0 ∈ Y be such that Y = N1y0. Dueto Corollary 4.1.3, Y contains a closed AN1-minimal subset, say Z. ByLemma 4.5.5, there exists z0 ∈ Z \X(N) such that N1z0 = Z. Define
M = g ∈ G : gz0 ∈ N1y0.
There are three cases:
1. There exists an open relatively compact neighbourhood Ω of the identityin G such that M ∩ Ω ⊂ H.
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2. There exist v ∈ N2 and h ∈ H such that v = e and vh ∈ M .
3. e ∈ M \N2H.
In Case 3, by Lemma 4.5.6, Y = X.In Case 1, Ωz0 ∩ N1y0 ⊂ (H ∩ Ω)z0. Therefore Ωz0 ∩ N1y0 ⊂ (H ∩ Ω)z0.
Hence Hz0 contains a neighbourhood of z0 in Y . Therefore Z \Hz0 z0 andit is AN1-invariant. Since Z is closed and AN1-minimal, we have Hz0 ⊃ Z andZ = AN1z0. By Iwasawa decomposition H = K1AN1. Therefore K1Z = Hz0.Since K1 is compact, the orbit Hz0 is closed. By Corollary 3.4.4, Hz0 admitsa 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 closedH-orbit.
Now consider the Case 2. Take any z ∈ Z. Then there exists a sequenceti →∞ such that u(ti)y0 → z as i →∞. Because otherwise az ∈ N1y0 for alla in a neighbourhood of the identity in AN1, which is not possible as Gy0 isdiscrete. Now since vh ∈ M , there exists t0 ∈ R such that u(t0)y = vhz0. ByLemma 4.5.3, if we replace ti by a subsequence, then there exists a sequencegi ⊂ AN1 such that u(ti)hgi → e as i →∞. Since u(ti + t0)y0 → u(t0)z, wehave as i →∞,
u(ti + t0)y0 = u(ti)vhz0 = v(u(ti)hgi)(g−1i
z0) → vz
for some z ∈ Z such that u(t0)z = vz. Hence v−1z = u(−t0)z ∈ Z. Thus wehave shown that v−1Z ⊂ Z. Now AZ = Z, therefore Z contains an orbit ofv−1Av, 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 summerisethe 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 .
2. Y is a closed orbit of vHv−1for some v ∈ N .
3. Y = X.
This theorem settles Raghunathan’s closure conjecture for unipotent flowsin L-homogeneous spaces of G = SL2(C).
100
4.6 Closures of orbits of subgroups containinga nontrivial unipotent element
Closed AN1-invariant subsets
Theorem 4.6.1 Let Y be the closure of an AN1-orbit in X. Then either Yis a closed H-orbit or Y = X.
In particular, any H-orbit in X is either closed or dense.
Proof. Let y0 ∈ Y be such that Y = AN1y0. If the orbit N1y0 is not compactthen 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 AN1-minimal subset of Y . Since we can assume that Z = X,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.
DefineM = g ∈ G : gz0 ∈ AY0.
If e ∈ M \H, the orbit Hz0 is open in Y and hence AY0 ∩ Hz0 = ∅.Therefore Y = Hz0.
If e ∈ M \ (N2 ∩ Ω0)H then by Lemma 4.5.6, Y = X.Now suppose that e ∈ M \H and e ∈ M \ (N2 ∩ Ω0)H. Note that
d(a)Y0 → ∞ as a → 0 (see the proof of Lemma 3.1.1). Therefore thefollowing holds: There exist sequences ai → ∞ and si → 0 in R, and yi → yin Y0 such that for every i ∈ N,
d(ai)yi ∈ v(si)Hz0.
Thus for every i ∈ N,
yi ∈ d(a−1i
)v(si)Hz0 = v(a−2i
si)Hz0.
Therefore y ∈ Hz0 and hence y0 ∈ N1y ⊂ Hz0. Thus Y = Hz0, which is acontradiction to the assumption that e ∈ M \H. This completes the proof.
Proposition 4.6.2 Let Yii∈N be a sequence of closed H-orbits in X and
let vii∈N be a sequence in N2. Suppose that any one of the following two
conditions is satisfied:
(i) Yi = Yj for all i = j.
(ii) vi →∞ as i →∞.
101
Put
Y =∞
i=1
viYi
Then Y = X.
Proof. (Cf. [M5, Thm. 4]) First note that viYi is a closed orbit of viHv−1i
for each i ∈ N. In particular, Y is N1-invariant. Due to Corollary 3.4.6,there exists yi ∈ viYi \X(N) for every i ∈ N such that a subsequence of yiconverges to a point y0 ∈ X \X(N).
We can assume that N1y0 = X. Then by Theorem 4.5.8, N1y0 = vHv−1y0
for some v ∈ N2. For every i ∈ N, replace vi by v−1vi and yi by v−1yi. Replacey0 by v−1y0 and Y by v−1Y . Now without loss of generality by passing to asubsequence we may assume that yi → y0 and N1y0 = Hy0.
For every i ∈ N, let mi ∈ G be such that yi = miy0 and mi → e as i →∞.Put M = mi : i ∈ N and Z = Hy0.
If e ∈ M \ (N2 ∩ Ω0)H then by Lemma 4.5.6, Y = X and we are through.Now suppose that mi ∈ (N2 ∩Ω0)H for all large i ∈ N. For any such large
i ∈ N, express mi = wih, where wi ∈ N2 and h ∈ H. Then by Theorem 4.5.8,
viYi = N1yi = N1miy0 = N1wi(hy0) = wiN1(hy0).
Since N1hy0 cannot be a compact N1-orbit, we have N1(hy0) = Z. Thus viYi =wiZ. Now viYi is a closed orbit of viHv−1
iand wiZ is a closed orbit of wiHw−1
i.
Therefore by Lemma 1.1.3, viHv−1i
= wiHw−1i
. Since N2 ∩ NG(H) = e, weget 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 \ (N2 ∩ Ω0)H. This completes the proof.
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. Nowthe conclusion follows from Proposition 4.6.2.
Corollary 4.6.4 Let Z be a closed H-orbit and let gi ⊂ G be a sequence
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 =
i∈N
giZ.
102
Proof. By Iwasawa decomposition (see Lemma 2.2.1), for each i ∈ N, we canexpress gi = kivibi, where ki ∈ K, vi ∈ N2, and bi ∈ AN1. Since AN1 ⊂ H, bypassing to a subsequence, we may assume that ki → k ∈ K and vi → ∞ asi →∞. Now by Proposition 4.6.1, for any m ∈ N,
∞
i=m
k−1i
giZ = X.
Hence
i∈N
giZ = k−1X = X.
Corollary 4.6.5 For any x ∈ X and a sequence ai →∞ in R, let
Y =
i∈N
d(ai)N1x.
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 inthe 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. For any y ∈ Y \ X(N),there exists v ∈ N2 such that N1y = (vHv−1)y. If we put z = v−1y thenN1z = Hz and vHz ⊂ Y .
For any closed H-orbit Z ⊂ X, define
LZ = v ∈ N2 : vZ ⊂ Y .
Then LZ is closed. Let
Σ = Z ⊂ X : Z is a closed H-orbit and LZ = ∅.
By Proposition 4.6.2, Σ is finite and LZ is compact for every Z ∈ Σ. Put
YH =
Z∈Σ
LZ · Z.
Then YH is closed subset of Y . Also Y ⊂ X(N)∪ YH . Therefore Y \X(N) =YH .
103
Note that due to Lemma 1.1.3, for any Z = Z ∈ Σ,
LZZ ∩ LZZ ⊂ X(N).
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)YH ∪Xη(N).
Proof. Suppose Y \N(t)YH ⊂ Xη(N) for every η > 0. Then by Corol-lary 3.4.6, there exists a sequence yi ⊂ Y \ N(t)YH such that yi → y forsome y ∈ Y \X(N) = YH \X(N). Now N1y = vHv−1y for some v ∈ N2. PutY = v−1Y , Y
H= v−1YH , y
i= v−1yi for all i ∈ N, and y = v−1y. Note that
Y is a proper closed N1-invariant subset of X. Also yi→ y and N1y = Hy.
Since Hy ⊂ Y
H, we have that y
i∈ N(t)Hy for all i ∈ N. Therefore by
Lemma 4.5.6, Y = X, a contradiction to the assumption that Y , and henceY 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 unipotentflows 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 ⊂ N1. Then one of the following holds.
1. Y is contained in a compact N-orbit. In which case there are the follow-
ing 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.
2. Y is a closed orbit of vHv−1for some v ∈ N2.
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 followingholds:
(1) Y1 is contained in a compact N -orbit.
(2) Y1 is a closed vHv−1-orbit for some v ∈ N2.
(3) Y1 = X.
In the Case (1), Y is contained in a compact N -orbit. Since N ∼= R2, thefurther 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 y1 ∈ Y1 such thatthe set u(n)y1 : n ∈ Z is dense in Y1. Now y1 = u(t)y for some t ∈ [0, 1] andy ∈ Y . Therefore
Y ⊃ u(n)y : n ∈ Z = u(−t)Y1 = Y1.
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) F ⊂ SN1 and Y is a closed orbit of vS Hv−1, where v ∈ N2 and S is a
subgroup of S = d(ik)k∈Z.
(2) F ⊂ MN , Y is contained in a compact MN -orbit and one of the follow-
ing 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
F = vd(ekθi)k∈Zv−1 · u(z) : z ∈ Z[e2θi] · w
and
Y = vd(ekθi)k∈Z · Lx,
where x ∈ Y , L is a closed subgroup of N containing F ∩ N , and
L0is one-dimensional.
(3) F ⊂ AN1, F ⊂ N1, and Y is a closed H-orbit.
(4) Y = X.
Proof. Due to Proposition 2.2.5, there are four cases:
105
(i) F ⊂ SN1.
(ii) F ⊂ SN1 and F ⊂ MN .
(iii) d(an) : n ∈ ZN1 ⊂ F ⊂ AN1 for some a > 1.
(iv) d(αn) : n ∈ ZN ⊂ 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 theu(1)-action and arguing as in the proof of Proposition 4.6.7, we obtain thatY = X. If Y1 = X then by Theorem 4.3.3, there exists x ∈ Y such that Nxis compact. Therefore Y ⊂ MNx. There exists a closed subgroup of L ofN containing F ∩ N , such that (F ∩N)x = Lx. Since F/(F ∩ N) is a finitegroup, Y = FLx. Now the conclusions (a) and (b) of the Statement (2) followfrom 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 Hand 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 existsh ∈ H such that hw(t) : t ∈ Rh−1 = N1 and hUh−1 = N . Therefore hY1 isN1-invariant, Z ⊂ hY1, and
hMh−1 = g ∈ G \NH : gZ ∩ hY1 = ∅.
Hence by Lemma 4.5.6, hY1 = X. Thus Y1 = X. Now using ergodicity of theaction of u(1) on X, and arguing as in the proof of Proposition 4.6.7, we getthat 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 horo-
spherical 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
(2) F ⊂ NG(gHg−1) and Y = Y0 ∪E, where g ∈ G, Y0 is a union of finitely
many closed gHg−1-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 subgroupUk of G, where k ∈ I = 1, 2, 3, and nontrivial one-parameter subgroupsuk(t)t∈R ⊂ Uk such that uk(1) ∈ F , where k ∈ I.
Let
Y0 = y ∈ Y : uk(t)y ∈ Y for some k ∈ I, and all t ∈ R.
If Y0 = ∅ then by Proposition 4.6.7 and Corollary 3.4.6, there exists η > 0such that
Y ⊂
k∈I
Xη(Uk).
Therefore by Corollary 3.5.2, Y is finite.Now we assume that Y0 = ∅. Clearly Y0 is a closed subset of X. Take
y ∈ Y0 and suppose that u1(t)y ⊂ Y0 for all t ∈ R. By Corollary 3.5.3, forevery δ > 0 there exists s ∈ (0, δ) such that u1(s)y ∈ Y \X(U2). Therefore byProposition 4.6.7, u2(t)(u1(s)y) ∈ Y for all t ∈ R. Since δ > 0 is arbitrary, wehave u2(t)y ∈ Y for all t ∈ R. From this argument one concludes that Y0 isinvariant under the closed subgroup, say L, generated by all uk(t)t∈R, k ∈ I.
Now either L = G or L = gHg−1 for some g ∈ G (see Corollary 2.2.10).If L = G then Y = X and we are through. If L = gHg−1 then replacingY by g−1Y and F by g−1Fg we may assume that L = H. Therefore byTheorem 4.6.1, either Y0 = X or Y0 is a finite union of closed H-orbits. In theformer case we are through. Now put E = Y \ Y0. Then E is F -invariant.
If E ⊂
k∈IX(Uk) then there exists η > 0 such that
E ⊂
k∈I
Xη(Uk).
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, uk(t)y : t ∈ R = Hy. Threfore thereexist y0 ∈ Hy \∪k∈IX(Uk) and a sequence yi ⊂ Y \ Y0 such that yi → y0 asi →∞. Let mi → e be a sequence in G such that yi = miy0 for all i ∈ N. Thenthere exist sequences qi → 0 in P and hi → e in H such that mi = (expqi)hi
for all i ∈ N (notations as in Proposition 4.5.4). Since U1 ∩ U2 = e, bypassing to subsequences we may assume that expqi ∈ Ul for all large i ∈ N,where l ∈ 1, 2. Now by Lemma 4.6.9, Y = X. In which case Y0 = Y , acontradiction. This completes the proof.
107
The Proposition 4.6.8 and Proposition 4.6.10 combined together give thefollowing 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
F = gd(ekθi)k∈Z · u(z) : z ∈ Z[e2θi] · w
g−1,
the orbit (gNg−1)x is compact, (F ∩ gNg−1)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 sub-
group 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 lat-
tice Γ in G, either [Γ : Γ ∩ Γ] < ∞ or the set ΓΓ is dense in G.
Proof. Since G/Γ is noncompact, by Theorem 3.1.9, Γ contains a nontrivialunipotent element. Consider the Γ action on G/Γ. By Corollary 2.4.10 andTheorem 4.6.11, either ΓΓ = G or ΓΓ/Γ is a finite subset of G/Γ. Thiscompletes the proof.
It is natural to ask what happens if both Γ and Γ in the above lemma arecocompact lattices. It was pointed out by Dave Witte that the question can beanswered, using the next lemma, if one assumes the generalized Raghunathanconjecture for the the diagonal action of G on G/Γ × G/Γ (note that thegeneralized Raghunathan conjecture has been proved by Ratner [Ra2]).
Notation. Let Γ and 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 con-
taining ∆(G) such that
∆(G) · (x0, y0) = F · (x0, y0).
Then either [Γ : Γ ∩ Γ] < ∞ or ΓΓ is dense in G.
Proof. Let C be a compact subset of G such that G = CΓ. Then,
∆(C) ·∆(Γ) · (x0, y0) = ∆(G) · (x0, y0) = F · (x0, y0).
Since Gx0 = Γ, we have ∆(Γ) · (x0, y0) = (x0, Γy0). Therefore
(x0, Γy0) = ∆(C) · (x0, Γy0) ∩ (x0× Y )
= (F · (x0, y0)) ∩ ((Γ×G) · (x0, y0))
= (x0, p2(F ∩ (Γ×G))y0)
= (x0, Ly0),
where p2 : G × G → G is the projection on the second factor, and L =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 Γ as a subgroup offinite index. In the first case ΓΓ is dense in G. In the second case, ΓΓ/Γ isa discrete and hence a finite subset of G/Γ. This completes the proof.
109
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