Lie Groups Tifr14
-
Upload
shekhardeodhar -
Category
Documents
-
view
221 -
download
0
Transcript of Lie Groups Tifr14
-
7/31/2019 Lie Groups Tifr14
1/140
Lectures on
Lie Groups and Representationsof Locally Compact Groups
By
F. Bruhat
Tata Institute of Fundamental Research, Bombay
1958
(Reissued 1968)
-
7/31/2019 Lie Groups Tifr14
2/140
Lectures on
Lie Groups and Representationsof Locally Compact Groups
By
F. Bruhat
Notes by
S. Ramanan
No part of this book may be reproduced in anyform by print, microfilm or any other means with-
out written permission from the Tata Institute of
Fundamental Research, Colaba, Bombay 5
Tata Institute of Fundamental Research, Bombay
1958
(Reissued 1968)
-
7/31/2019 Lie Groups Tifr14
3/140
Introduction
We shall consider some heterogeneous topics relating to Lie groups and
the general theory of representations of locally compact groups. The
first part exclusively deals with some elementary facts about Lie groupsand the last two parts are entirely independent of the material contained
in the first. We have rigidly adhered to the analytic approach in estab-
lishing the relations between Lie groups and Lie algebras. Thus we do
not need the theory of distributions on a manifold or the existence of
integral manifolds for an involutory distribution.
The second part concerns itself only with the general theory of mea-
sures on a locally compact group and representations in general. Only a
passing reference is made to distributions (in the sense of L. Schwartz),
and induced representations are not treated in detail.
In the third part, we first construct the continuous sum (the direct
integral) of Hilbert spaces and then decompose a unitary representa-tions into a continuous sum of irreducible representations. We derive
the Plancherel formula for a separable unimodular group in terms of
factorial representations and derive the classical formula in the abelian
case.
iii
-
7/31/2019 Lie Groups Tifr14
4/140
-
7/31/2019 Lie Groups Tifr14
5/140
Contents
Introduction iii
I Lie Groups 1
1 Topological groups 3
2 Local study of Lie groups 9
3 Relations between Lie groups and Lie algebras - I 19
4 Relation between Lie groups and Lie algebras - II 31
II General Theory of Representations 47
5 Measures on locally compact spaces 47
6 Convolution of measures 53
7 Invariant measures 61
8 Regular Representations 73
9 General theory of representations 83
v
-
7/31/2019 Lie Groups Tifr14
6/140
vi Contents
III Continuous sum of Hilbert Spaces 89
10 Continuous sum of Hilbert Spaces-I 91
11 Continuous sum of Hilbert spaces - II 103
12 The Plancherel formula 115
Bibliography 127
-
7/31/2019 Lie Groups Tifr14
7/140
Part I
Lie Groups
1
-
7/31/2019 Lie Groups Tifr14
8/140
-
7/31/2019 Lie Groups Tifr14
9/140
Chapter 1
Topological groups
1.1
We here assemble some results on topological groups which we need in 1
the sequel.
Definition. A topological group G is a topological space with a compo-
sition law G G G, (x,y) xy which is(a) a group law, and
(b) such that the map G G G defined by (x,y) x1y is contin-uous.
The condition (b) is clearly equivalent to the requirement that the
maps G G G defined by (x,y) xy and G G defined byx x1 be continuous.
It is obvious that the translations to the right: x xy are homeo-morphisms of G. A similar statement is true for left translations also.
We denote by A1, AB the subsets {a1 : a A}, {ab : a A, b B} re-spectively. It is immediate from the definition that the neighbourhoods
of the identity element e satisfy the following conditions:
(V1) For every neighbourhood V of e, there exists a neighbourhood
W of e such that W1
W V. (This follows from the fact that(x,y) x1y is continuous).
3
-
7/31/2019 Lie Groups Tifr14
10/140
4 1. Topological groups
(V2) For every neighbourhood V of e and for every y G, there existsa neighbourhood W of e such that yWy
1
V. (This is becausex yxy1 is continuous).
These conditions are also sufficient to determine the topology of the
group. More precisely,
Proposition 1. LetVbe a family of subsets of a groups G, such that2
(a) For every V V, e V.
(b) Any finite intersection of elements ofV is still in V.
(c) For every V V, there exists W Vsuch that W1W V.(d) For every V V, and for every y G there exists W V such
that yW y1 V.
Then G can be provided with a unique topology compatible with
the group structure such that the family V is a fundamental system ofneighbourhoods at e.
Supposing that such a topology exists, a fundamental system of
neighbourhoods at y is given by either Vy or yV. It is, therefore, anatural requirement that these two families generate the same filter. It is
this that necessitates the condition (d).We may now take the filter generated by yV and Vy as the neigh-
bourhood system at y. This can be verified to satisfy the neighbourhood
axioms for a topology. It remains to show that (xy) x1y is continu-ous.
Let V x10
y0 be any neighbourhood of x10
y0. By (c), (d), there exist
W, W1 V such that W1W V and x10 y0W1y10 x0 W. If we takex x0W, y y0W1, we have x1y W1x10 y0W1 W1W x10 y0 V x1
0y0. This shows that (x,y) x1y is continuous.
Examples of topological groups.
(1) Any group G with the discrete topology.
-
7/31/2019 Lie Groups Tifr14
11/140
Topological groups 5
(2) The additive group of real numbers R or the multiplicative group
of non-zero real numbers R with the usual topology.
(3) The direct product of two topological groups with the product 3
topology.
(4) The general linear group GL(n,R) with the topology induced by
that ofRn2
.
(5) Let X be a locally compact topological space, and G a group of
homeomorphisms of X onto itself. This group G is a topologi-
cal group with the compact open topology. (The compact-open
topology is one in which the fundamental system of neighbour-
hoods of the identity is given by finite intersections of the setsu(K, U) = {f G : f(x) U, f1(x) U for every x K}, Kbeing compact and U an open set containing K).
1.2 Topological subgroups.
Let G be a topological group and g a subgroup in the algebraic sense.
g with the induced topology is a topological group which we shall call
a topological subgroup of G. We see immediately that the closure g is
again a subgroup.
If g is a normal subgroup, so is g. Moreover, an open subgroup is
also closed. In fact, if g is open, G g = xgxg, which is open as lefttranslations are homeomorphisms. Hence g is closed.
Let now V be a neighbourhood of e and g the subgroup generated
by the elements of V. g is open containing as it does a neighbourhood
of every element belonging to it. Consequently it is also closed. So that
we have
Proposition 2. If G is a connected topological group, any neighbour-
hood of e generates G.
On the contrary, ifG is not connected, the connected component G0 4
ofe is a closed normal subgroup ofG. Since G0xG0 G1
0 G0 is contin-uous and G0xG0 connected, G
10
G0 is also connected, and hence G0.
-
7/31/2019 Lie Groups Tifr14
12/140
6 1. Topological groups
This shows that G0 is a subgroup. As x yxy1 is continuous, yG0y1
is a connected set containing e and consequently G0. Therefore, G0 isa normal subgroup.
Proposition 3. Every locally compact topological group is paracom-
pact.
In fact, let Vbe a relatively compact open symmetric neighbourhood
of e. G =
n=1
Vn is an open and hence closed subgroup of G. G is
countable at and therefore paracompact. Since G is the topologicalunion of left cosets modulo G, G is also paracompact.
1.3 Factor groups.
Let g be a subgroup of a topological group G. We shall denote by x
the right coset gx containing x. On this set, we already have the quotient
topology. Then the canonical map : G G/g is open and continuous.For, if (U) is the image of an open set U in G, it is also the image of
gU which is an open saturated set. Hence (U) is also open. But this
canonical map is not, in general, closed. The space G/g is called a
homogeneous space. Ifg is a normal subgroup, G/g is a group and is a
topological group with the above topology. This is the factor group of
G by g.
1.4 Separation axiom.
Theorem 1. The homogeneous space G/g is Hausdorffif and only if the
subgroup g is closed.
If G/g is Hausdorff, g = 1((e)) is closed, since (e) is closed.5Conversely, let g be closed. Let x, y G/g and x y. Since xy1 gand g is closed, there exists a symmetric neighbourhood V of the identity
such that xy1V g = . Hence xy1 gV. Now choose a neighbour-hood W of e such that WW1 y1Vy. We assert that gxW and gyWare disjoint. For, if they were not, 1, 2 g, w1, w2 W exist such that
1xw1 = 2yw2; i.e. 12 1x = yw2w
11 yWW1 Vy.
-
7/31/2019 Lie Groups Tifr14
13/140
Topological groups 7
Hence 12
1xy1 V, or xy1 1
12V gV.
This being contradictory to the choice of V, gxW, gyW are disjointor (gxW) (gyW) = . Hence G/g is Hausdorff.
In particular, ifg = {e}, G is Hausdorffif and only if{e} is closed. Onthe other hand, if{e} is not closed, {e} is a normal subgroup and G/{e} isa Hausdorff topological group. We shall hereafter restrict ourselves to
the consideration of groups which satisfy Hausdorffs axiom.
1.5 Representations and homomorphisms.
Definition . A representation h of a topological group G into a topo-
logical group H is a continuous map : G H which is an algebraic
representation. In other words, h(xy)=
h(x) h(y) for every x, y G.Obviously the image ofG by h is a subgroup of H and the kernel N
ofh is a closed normal subgroup ofG. The canonical map {h} : G/N H is a representation and is one-one.
Definition. A representation h is said to be a homomorphism if the in-
duced map h is a homeomorphism.
Proposition 4. Let G and H be two locally compact groups, the former 6
being countable at . Then every representation h of G onto H is ahomomorphism.
It is enough to show that for every neighbourhood V of e in G/N,
h(V) is a neighbourhood ofh(e) in H. Choose a relatively compact open
neighbourhood W of e such that WW1 V. G is a countable unionof compact sets and
xG W x = G. Therefore, one can find a sequence
{xj} of points such that G =
j W xj. Since h is onto, H =
j h(W xj) =j h(W)h(xj). H is a locally compact space and hence a Baire space
(Bourbaki, Topologie generale, Ch. 9). There exists, therefore, an in-
teger j such that h(W xj) has an interior point. h(W) being compact,
h(W) = h(W). Consequently, h(W)h(xj) and hence h(W) has an interior
point y. There exists a neighbourhood U of e such that h(W) Uy.Now h(V) h(WW1) = h( (W) h (W)1 Uy(y1U1) = UU1.h(V) is therefore a neighbourhood ofh(e), which completes the proof of
proposition 4.
-
7/31/2019 Lie Groups Tifr14
14/140
-
7/31/2019 Lie Groups Tifr14
15/140
Chapter 2
Local study of Lie groups
2.17
Definition. A Lie group G is a real analytic manifold with a composition
law (x,y) xy which is(a) a group law, and
(b) such that the map (x,y) x1y is analytic.(b) is equivalent to the analyticity of the maps (x,y) xy and x
x1.
Remarks. (1) A Lie group is trivially a topological group.
(2) We may replace real by complex and define the notion of a
complex Lie group. We shall not have occasion to study complex
Lie groups in what follows, though most of the theorems we prove
remain valid for them.
(3) It is natural to inquire whether every topological group with the
structure of a topological manifold is a Lie group. This problem
(Hilberts fifth problem) has been recently solved by Gleason [18]
who has proved that a topological group G which is locally com-
pact, locally connected, metrisable and of finite dimension, is a
Lie group.
9
-
7/31/2019 Lie Groups Tifr14
16/140
10 2. Local study of Lie groups
Examples of Lie groups.
(1) R - real numbers, C - complex numbers,
T - the one-dimensional torus and Rn, Cn and Tn in the usual
notation are all Lie groups.
(2) Product of Lie groups with the product manifold structure is a Lie8
group.
(3) GL(n,R) - the general linear group.
2.2 Local study of Lie groups.
We shall assume that V is a sufficiently small neighbourhood of e inwhich a suitably chosen coordinate system, which taken e into the ori-
gin, is defined.
The following notations will be adhered to throughout these lec-
tures:
If a V, (a1, . . . , an) will denote the coordinate of a. = (1, . . . ,n) is a multi-index with i, non-negative integers.
|| = 1 + 2 + + n[i] will stand for with i = 1 and j = 0 for j i.
! = 1! . . . n!
x = x11
. . . xnn
x=
1++n
x11
xnnLet x, y V be such that xy V. (xy)i are analytic functions of the
coordinate of x and y.
(1) (xy)i = i(x1, . . . , xn,y1, . . . ,yn) where the i are analytic func-
tions of the 2n variables x, y in a neighbourhood of 0 in R2n.
These i cannot be arbitrary functions, as they are connected by
the group relations. These are reflected in the following equa-
tions: (x e)i = (e x)i = xi, or
-
7/31/2019 Lie Groups Tifr14
17/140
Local study of Lie groups 11
(2) i(x, e) = i(e, x) = xi. 9
The i are analytic functions and so are of the form
i(x,y) =,
ixy, i = 1, 2, . . . , n
By equation (2), this can be written
(3) i(x,y) = xi +yi +
||1||1
i,
xy.
By associativity,
i(xy,z) = i(x,yz), or
i(1(x,y), . . . , n(x,y),z) = i(x, 1(y,z), . . . , n(y,z)).
These may also be written
(4) i((x,y),z) = i(x, (y,z)).
One is tempted to expect another equation in i, due to the exis-
tence of the inverse of every element. However, these two equations are
sufficient to characterise locally the Lie group, and the existence of the
inverse is, in a certain sense, a consequence of the associative law and
the existence of the identity. To be more precise,
Proposition 1. Let G be the semigroup with an identity element e. If
it can be provided with the structure of an analytic manifold such that
the map (x,y) xy of GxG G is analytic, then there exists an openneighbourhood of e which is a Lie group.
In fact, the existence of the inverse element of x depends upon the
existence of the solution for y of i(x,y) = 0, 1 = 1, . . . , n. Nowi
xj=
i j+ terms containing positive powers of the yi. If we put y = e, the
latter terms vanish and
i(x,y)xj
y=e
= i j.
-
7/31/2019 Lie Groups Tifr14
18/140
12 2. Local study of Lie groups
Hence 10
J = det i(x,y)xj y=e = 1J being a continuous function of x and y, J 0 in some neighbourhood
V ofe. Therefore there exists a neighbourhood V ofe every element ofwhich has an inverse. Then the neighbourhood W =
n=1(V V1)n is
a group compatible with the manifold structure.
2.3 Formal Lie groups.
Definition. A formal Lie group over a commutative ring A with unit el-
ements, is a system of n formal series i in 2n variables with coefficients
in A such that
i(x1, . . . , xn, 0, 0, . . .) = xi = i(0, 0, . . . , x1, . . . , xn)
and
i(1(x,y), . . . , n(x,y),z) = i(x, 1(y,z), . . . , n(y,z)).
Almost all that we prove in the next few lectures will be valid for
formal Lie groups over a field of characteristic zero also. For a study of
formal Lie groups over a field of characteristic p 0, one may see, for
instance, [9], [10].
2.4 Taylors formula.
Let f be a function on an open neighbourhood of e, and let y, z denote
respectively the right and left translates of f defined by y f(x) = f(xy);
zf(x) = f(z1x) for sufficiently small y and z. These two operators
commute,
i.e.
y(zf) = z(y f)
If f is analytic in V, y f is (for y W) analytic in W, where W is aneighbourhood ofe such that W
2
V.Now,11
-
7/31/2019 Lie Groups Tifr14
19/140
Local study of Lie groups 13
i f(x) = f(1(x,y), . . . , n(x,y))
with
i(x,y) = xi +yi +||1||1
i,xy.
If we set
ui = yi +||1||1
i,xy
yf(x) = f(x+u) in the usual notation. This can be expanded as a Taylor
series
y f(x) = 1
!
uf
x
(x).
We may now substitute for the ui in this convergent series.
u = u1 unn=
y1 +
||1||1
1,xy
1. . .
= y +
||||g (x)y
where the coefficient of powers of y are analytic functions of x and
g
(e) = 0. If here we take = 0, u = 1, and hence g0
= 0 for
every . Thus
yf(x) =
1
!
f
x(x)(y +
||||
g (x)y).
These are uniformly absolutely convergent in a suitable neighbour-
hood of e, on the explicit choice of which we shall not meticulously
insist. Hence the above formula can be written as
y f(x) =
y(1
!
f
x (x) + |||| g(x)
1
!
f
x (x)).
-
7/31/2019 Lie Groups Tifr14
20/140
14 2. Local study of Lie groups
This we shall denote by
y f(x) =
1
!y f(x)
where is a differential operator not depending on f. This formula is12
the generalised Taylors formula we sought establish.
2.5 Study of the operator .
Now, =
x+
||||
g(x)
x!
!with g
(e) = 0. Since at x = e,
=
x, the are linearly independent at the origin. At = 0, is
the identity operator and if 0, is without constant term as g0 = 0.
Let us denote [i] by Xi. Then Xi =
x1+
j ai j(x)
xj, with ai j(e) =
0. These are vector fields in a neighbourhood of e. Now we shall use
the fact that, for every y, z G, the operators z and y commute. Wehave
z(y f) = z( 1
!y f) =
1!
y(z )(f)
and, on the other hand,
y(zf) = 1
!y(zf).
Therefore, by the uniqueness of the expansion in power-series in y of
z(y f) = y(zf), we have z = z, for every z i n a s ufficientlysmall neighbourhood of e. Otherwise stated, is left invariant in this
neighbourhood. This enables us to define at every point z in the Lie
group by setting f(z) = z1 zf(z), so that the extended operator
remains left invariant.
Theorem 1. The linear differential operators form a basis for thealgebra of left invariant differential operators.
-
7/31/2019 Lie Groups Tifr14
21/140
-
7/31/2019 Lie Groups Tifr14
22/140
16 2. Local study of Lie groups
(3) Any associative algebra with the bracket operation
[X, Y] = XY Y X
is a Lie algebra.
In particular, the matrix algebra Mn(K) over a field K and the spaceof endomorphisms of a vector space Uare Lie algebras.Definition. A subspaceJ of a Lie algebra g is a subalgebra if for everyx, y J, [x,y] J.
A subspace Uof a Lie algebra g is an ideal, if for every x U,y
g, [x,y]
U.
Example. (1) The set of all matrices inMn(K) whose traces are zerois an ideal ofMn(K).
(2) The set of all elements z g such that [z, x] = 0 for every x g isan ideal ofg, called centre ofg.
IfUis an ideal in g, the quotient space g/U can be provided withthe structure of a Lie algebra by defining [(x +U), (y +U)] = [x,y] +U.This is called the factor algebra of g by U.
2.7 Representations of a Lie algebra.Definition . A representation of a Lie algebra g in another Lie algebra
g is a linear map f such that f([x,y]) = [f(x), f(y)] for every x, y g.
It can be verified that the image of g by f and the kernel of f are
subalgebras ofg and g respectively. The latter is, in fact, an ideal andg/ ker f is isomorphic to f(g).
In particular, g may be taken to the the space of endomorphisms of15a vector space V, leading us to the definition of a linear representation
ofg.
Definition. A linear representation of a Lie algebra g in a vector space
V is a representation ofg into the Lie algebra of endomorphisms of V.
-
7/31/2019 Lie Groups Tifr14
23/140
Local study of Lie groups 17
2.8 Adjoint representation.
Let g be a Lie algebra and x g. The map g g defined by y [x,y]is a linear map of g into itself. This map is denoted adx. Thus, adx
(y) = [x,y].
Remarks. (1) Ad x is a derivation of the Lie algebra. We recall here
the definition of a derivation in an algebra g, associative or not. A
linear map D ofg into itself is a derivation if for any two elements
x, y g, we have D(xy) = x(Dy) + (Dx)y. In a Lie algebraderivations of the type ad x are called inner derivations.
(2) x ad x is a linear map of g into the Lie algebra of endomor-phisms of the vector space g.
This is, moreover, a linear representation ofg in g. The verification
of the relation ad[x,y] = [adx, ady] is an immediate consequence of
Jacobis identity.
This linear representation will henceforth be referred to as the ad-
joint representation ofg.
-
7/31/2019 Lie Groups Tifr14
24/140
-
7/31/2019 Lie Groups Tifr14
25/140
Chapter 3
Relations between Lie groups
and Lie algebras - I
3.1 Differential of an analytic representation.16
Definition. An analytic representation of a Lie group G into a Lie group
G is an algebraic representation which is an analytic map.
Remark. It is true, as we shall see later (Cor. to Th. 4, Ch. 4.5) that
any representation of the underlying topological group G in G is itselfa representation in the above sense.
We now seek to establish a correspondence between analytic repre-sentations of Lie groups and algebraic representations of their Lie alge-
bras. As a first step, we prove the following
Proposition 1. To every analytic representation h : G G therecorresponds a map dh : U(G) U(G) which is a representation ofalgebras such that (f h) = (dh()f) h.
Let y G and y = h(y). If f is an analytic function on G, we have
y(f h)(x) = f(h(xy)) = f(h(x)h(y)) = (y f) h(x).
We may now write down the Taylor formula for both sides of theequation and equate the coefficients of powers of y (by the uniqueness
19
-
7/31/2019 Lie Groups Tifr14
26/140
20 3. Relations between Lie groups and Lie algebras - I
of development in power series). We have
1
!y(f h) = (
1
!y
f) h.
But h being an analytic map, (h(y))i =
iy
. There is no con-
stant term in this summation since (h(y))i = 0 at y = e. Hence y =
||| |
y, where
are constants, and on substitution in the above17
equation, we obtain
1
!y(f h) =
(1
!
||| |
y f) h
=
y( | || |
1! f) h.
the series being uniformly absolutely convergent. IfD denotes
| || |
!
! , which is a left invariant differential operator, then (fh) =
(D f) h. Moreover, this equation completely determines D since itsvalue at e given by Df(e
) = (f h)(e). As the form a basisfor U(G) in G, we may define a linear map dh : U(G) U(G) bysetting dh() = D
. It is obvious that ( f h) = (dh()f) h for any
(G). To complete the proof of proposition 1, one has only to showthat dh(12) = dh(1)dh(2). But this is obvious since
(dh(12)f) h = 12(f h)= 1(dh(2 )f h)= (dh(1dh(2)f) h.
Now, dh() =
| |||
!
! is of order less than or equal to
that of. By linearity, the same is also true of any operator U(G).Also, dh preserves constant terms. The image ofg is in g, and by Propo-
sition 1, dh restricted to g is a Lie algebra representation. This is said tobe the differential of the map h.
-
7/31/2019 Lie Groups Tifr14
27/140
Relations between Lie groups and Lie algebras - I 21
Remarks. (1) If we have another representation h : G G, it isobvious that d(h h) = dh dh.
(2) Ifh is an analytic map of a manifold V into a manifold W, then 18
we can define the differential of h at x, viz., dhx : Tx Th(x)where Tx, Th(x) are tangent spaces at the respective points. This
map makes correspond to a tangent vector X at x, the vector X
at h(x) such that Xf = X(f h) for every function f analytic ath(x). However, we cannot, in general, define the image a vector
field. As we have seen, in the case of Lie groups, as long as
one is considering only left invariant vector fields, one can talk
of an image vector field. Thus we now have, corresponding to
an analytic representation of a Lie group G into another Lie groupG, two notions of a differential map: the linear map of the tangentspace at a into that at h(e) = e, and the representation of the Liealgebra ofG in that ofG. These two notions are essentially thesame in the following sense. Let , be the canonical vectorspace isomorphisms of g, g with Te, Te respectively. Then thediagram
g
dh
// Te
dhe
g // Te
is commutative.
Proposition 2. Let G and G be two connected Lie groups. An analyticrepresentation of G G is surjective if and only if the differential mapis surjective.
Proposition 3. A representation h of a Lie group G in another Lie group
G is locally injective (i.e. there axises a neighbourhood of e on whichh is injective) if and only if dh is injective.
These two propositions are consequences of the corresponding prop- 19erties of manifolds, the proofs of which we omit.
-
7/31/2019 Lie Groups Tifr14
28/140
22 3. Relations between Lie groups and Lie algebras - I
3.2 Subgroups of a Lie group.
Definition . An analytic map f of a manifold U into another manifold
V is said to be regular at a point x in U if the differential map d fx is
injective.
Definition . A submanifold of an analytic manifold U is a pair (V, )
consisting of a manifold V which is countable at and an injectiveanalytic map of V into U which is everywhere regular.
Remarks. (1) The topology on (V) is not that induced from the
topology of U in general. For instance, if T2 is the two - di-
mensional torus, V the space of real numbers, and the map
t (t, t) ofV into T2, where is irrational, it is easy to see that is an injective, analytic, regular map. But cannot be a home-
omorphism of V into T2. For, every neighbourhood of (0, 0) in
T2 contains points (t, t) with arbitrarily large values of t. Hence,
the inverse image of this neighbourhood in (U) with the induced
topology can never be contained in a given neighbourhood of 0
in R.
(2) Nevertheless, it is true that locally, for every point x of V, there
exist neighbourhoods W in V and W1 in U which satisfy the fol-
lowing : A coordinate system (x1, . . . , xn) can be defined in W1
such that W is defined by the annihilation of certain coordinates.
Definition . A Lie subgroup of a Lie group G is a submanifold (H, ),20
(H) being a subgroup of G.
We define on H the group structure obtained by requiring that be
a monomorphism. Since the map of H in G is regular, locally the
analytic structure of H is induced form that of G. Hence the group op-
erations in H are analytic in H, as they are analytic in G. H is therefore
a Lie group.
Proposition 4. The Lie algebra of a Lie subgroup of a Lie group G canbe identified with a subalgebra of the Lie algebra of G.
-
7/31/2019 Lie Groups Tifr14
29/140
Relations between Lie groups and Lie algebras - I 23
In fact, if (H, ) is the subgroup, is a representation of H in G and
d is injective since is regular. We identify the Lie algebra S of Hwith the subalgebra d(S) or g.
3.3 One-parameter subgroups.
Definition. An analytic representation of R into G is said to be a one-
parameter subgroup of G.
We know that the representation gives rise to a differential map d
of the Lie algebra of R (spanned byd
dt) into the Lie algebra of G. Let
d(d
dt
) = X = jXj. We now form the differential equations satisfiedby the function .Let (x1, x2, . . . xn) be a coordinate system in a neighbourhood ofe in
G and Let i denote xi . Nowi(t+ t
) = 1((t), (t))
t(i(t+ t
)) =
k
i
y1((t), (t)).
dk
dt(t)
Putting t = 0, we get
di
dt(t) =
kdk
dt(0)
i
yk((t), e).
21
IfXi = [i], we have
Xi =
xi+
j
ai j(x)
xjwith ai j(e) = 0.
Since (X f) = ddt
(f ) for every function analytic at e, we getdi
dt= (X xi) by settingf = xi.
Hence
di
dt(0) = (X xi) (0)
-
7/31/2019 Lie Groups Tifr14
30/140
24 3. Relations between Lie groups and Lie algebras - I
= X xi(e)
= (jXj)xi(e)= i.
To sum up, satisfies the system of differential equations
(A)di
dt=
k ki
yk((t), e)
with the initial condition
(B) i(0) = 0.
(A) implies
di
dt (0) = i.
Now, conversely if are given the system of differential equations (A)
with the initial condition (B), then by Cauchys theorem on the exis-
tence and uniqueness of solutions of differential equations, there exists
one and only one solution t (t, ) which is analytic in t and in aneighbourhood of (0, ). We shall now show that (t + u) = (t) (u)for sufficiently small values of tand u.
Let22
i (t) = i((u), (t)). Then
didt
= l
i
yi((u), (t)) d
l
dt(t)
=k,j
ki
yl((u), (t))
l
yk((t), e)
since the i are solutions of (A). On the other hand, we have
i((u)(t),y) = i((u), (t)y)
= i((u), ((t),y))
i
yk((u)(t),y) =
n
l=1
i
yl ((u), (t)y)
l
yk((t),y).
-
7/31/2019 Lie Groups Tifr14
31/140
Relations between Lie groups and Lie algebras - I 25
Hence,
didt
= k
kiyk
((t), e).
i.e. i is a solution of (A) with the initial condition (C):
i (0) = i(e, (u)) = i(u).
Also, t (t) = (t + u) is a solution of (A) since the differentialequation (A) is a invariant for translations of it. Also (0) = (u).
Hence i
and i are two sets of solutions of (A) with the same initial
conditions, and therefore
i (t)=
(t+
u), i.e. i((t), (u))=
i(t+
u).
or (t)(u) = (t + u) for sufficiently small values of t and u. Also this
map t (t,X) is analytic. We assume the followingLemma 1. Let H be a connected, locally connected and simply con- 23
nected topological and f a local homomorphism of H G (i.e. a con-tinuous map of a neighbourhood of e into H such that f(xy) = f(x)f(y)
for all x, y such that x, y, xy V). Then there exists one and only onerepresentation f of H in G which coincides with f on V.
We immediately obtain (since R is simply connected), the
Theorem 1. For every X g, there exists one and only one one - pa-rameter subgroup (t,X) such that d
(d)
dt= X. The function (t,X) is
analytic in t and X.
One can assign to any finite dimensional vector space over the real
number field a manifold structure which is induced by that of the real
numbers. In particular, The Lie algebra of a Lie group also has an ana-
lytic structure. Whenever we talk of an analytic map into of from a Lie
algebra, it is to this analytic structure that we refer.
Proof of the Lemma Consider the Cartesian set product H = H G.We provide
H with a topology by defining the neighbourhood system at
each point (x,y) in the following way:
-
7/31/2019 Lie Groups Tifr14
32/140
26 3. Relations between Lie groups and Lie algebras - I
Let W be a neighbourhood ofe in H V, where V can be assumedto be connected since H is locally connected. The fundamental sys-tem of neighbourhoods at (x,y) is given by N(W, x,y) = {(x,y) : x xW,y = y f(x1x)}. It is easily verified that this satisfies the neigh-bourhood axioms for a topology, and that H with the usual projection
: H G H is a covering space of H. Let H1 be the connectedcomponent of (e, e) in H. Then H1 is a connected, covering space of H
and since H is simply connected, is a homeomorphism of H1 onto H.
Let be its inverse. Define f(x) = 2 (x) for every x in H where242 : H G G is the second projection. N(W, x,y) is mapped home-omorphically by onto xW. Hence N(W, x,y) H1 if W is connectedand (x,y) H1. It follows that f is a representation which extends f.
3.4 The exponential map.
We shall denote (t,X) by exp(tX).
But such a notation involves the tacit assumption that (t,X) de-
pends only on tX. In other words, one has to make sure that (1, sX) =
(s,X) before such a notation becomes permissible. But this is obvious
in as much as t (st,X) is a one -parameter subgroup with d (d)d(st)
=
X or d(d)
dt= sX. The one-parameter subgroup such that d
(d)
dt= sX
is, by definition, t
(t, sX). By uniqueness of the one-parameter sub-
groups, (st,X) = (t, sX), or in particular, (s,X) = (1, sX). It is easyto see that exp(tX) exp(tX) = exp(t + t)X and exp(X) = (expX)1.But, in general, exp Y exp Y exp(Y + Y).Theorem 2. The map h : X expX of g into G is an analytic iso-morphism of a neighbourhood of 0 in g onto a neighbourhood of e in
G.
In fact, since h is an analytic map, it is enough to show that the
Jacobian of the map h 0 in a neighbourhood of the origin. (X1, . . . ,Xn)
form a basis for g, where Xi = [i].
h i
yiXi = exp(i
yiXi)
-
7/31/2019 Lie Groups Tifr14
33/140
Relations between Lie groups and Lie algebras - I 27
hj
yk
(0) =d
dt
(exp tXk)j(t = 0) = (Xkxj)x=e = jk.
i.e. the jacobian = 1 at e. By continuity, the Jacobian does not vanish in 25
a neighbourhood ofe.
Now, let X1, . . . ,Xn be an arbitrary basis of g. This can be trans-
ported into a system of coordinates in G by means of the above map.
For every x G sufficiently near e, there exists one and only one system(x1, . . . , xn) near 0 such that x = exp(
i xiXi).
This system of coordinates is called the canonical system of coordi-
nates with respect to any given basis. Hereafter, we will almost always
operate only with a canonical system of coordinates.
Remark. Let x V, V being a neighbourhood ofe in which a canonicalcoordinate system exists and x is sufficiently near e. Now, if
x = exp(
i
xiXi),
xp = exp(
i
(pxi)Xi),
i.e. the coordinates of xp are (px1, . . . , pxn).
Proposition 5. Let h be a representation of G in H, and dh : g Fitsdifferential. Then the diagram
g
exp
dh// S
exp
Gh
// H
is commutative.
Consider t h(exp tX).
This is obviously a one-parameter subgroup, and d = dh d.Therefore
h(exp X) = exp(d( ddt
))
-
7/31/2019 Lie Groups Tifr14
34/140
28 3. Relations between Lie groups and Lie algebras - I
= exp(dh(X)).
It follows, therefore, that ifh(G) = (e), dh is the map g (0).Conversely, if dh = C, and G connected, h(exp X) = e for every26
element in a neighbourhood of e, and h = e. Again, ifG is connected
and two representations h1, h2 ofG in H are such that dh1 = dh2, then
h1 = h2.
Proposition 6. For every analytic function f on a neighbourhood of e,
we have
f(exp tX) =
n=0
tn
n!(Xnf)(e).
In fact, d
dtf(exp tX) = (X f)(exp tX).
By induction on n, we have
dn
dtnf(exp tX) = (Xnf)(exp tX)
or dn
dtnf(exp tX)
t=0
= Xnf(e).
Now, f(exp tX) is an analytic function of t and by Taylors formula,
we have
f(exp tX) =
n=0
tn
n!(Xnf)(e)
Theorem 3. In canonical coordinates, we have =!
||! S where Sis the coefficient of t in the expansion of(
ni=1 tiXi)
|| and S U(G).
In fact, it is enough to prove the equality of and!
||! S at e sinceboth and S are invariant. Now,
f(y) = y f(e) =
1!
y f(e) with f(e) = y
f(y)y=e
-
7/31/2019 Lie Groups Tifr14
35/140
Relations between Lie groups and Lie algebras - I 29
y = exp(
yiXi) where yi are the canonical coordinates of y.
Therefore27
f(y) = f(exp(
yiXi))
=
p=0
1
p!
(
yiXi)pf
(e)
by Prop. 6, chapter 3, 5.
Taking partial derivatives at y = e, we have
yf(y) at y = e is
!
|
|!
(S f)(e).
Hence, Delta f(e) =!
||! Sf(e) which is what we wanted to prove.
-
7/31/2019 Lie Groups Tifr14
36/140
-
7/31/2019 Lie Groups Tifr14
37/140
Chapter 4
Relation between Lie groups
and Lie algebras - II
4.1 The enveloping algebras.28
Let g be a Lie algebra, and T the tensor algebra of the underlying vector
space ofg. Consider the two-sided ideal I generated in T By the ele-
ments of the form x y y x [x,y]. Then the associative algebra T/Iis said to be the universal enveloping algebra of the Lie algebra g.
Definition. A linear map h of al Lie algebra g into as associative algebra
A is said to be a linearisation if h([x,y])=
h(x)h(y) h(y)h(x) for everyx, y g.
We have obviously a canonical map ofg in to T/I, which we shall
denote by j.
Proposition 1. To any linearisation f ofg in an associative algebra A,
there corresponds one and only one representation f of T/I in such thatf j = f .
In fact, f being a linear map, it can be lifted uniquely into a rep-
resentation f of the tensor algebra T in A. Obviously, the kernel of f
contains elements of the form x y y x [x,y] and hence contains I.Hence this gives rise to a map f with the required property.
31
-
7/31/2019 Lie Groups Tifr14
38/140
32 4. Relation between Lie groups and Lie algebras - II
4.2 The Birkhoff-Witt theorem.
With reference to the universal enveloping algebra of a Lie algebra of a
Lie group, we have the following
Theorem 1. The algebra of left invariant differential operators Uis29canonically isomorphic to the universal enveloping algebra of the Lie
algebra.
In fact, the inclusion map of the Lie algebra g into Ucan be liftedto a representation of the enveloping algebra U of g in Uby propo-sition 1. It only remains to show that this map h : U Uis anisomorphic. IfS is the coefficient oft
in the expansion of (n
i=1tiXi)
||,
it has been proved (Th. 3, Ch. 3.4) that S form a basis for U. S is anoperators of the form
(. . .)Xi1 Xi|| , where Xi1 Xi|| are obtained
by certain permutations of xi1
Xnn . Let S denote the element of theform
(. . .)Xi1 . . . Xi|| , where S =
(. . .)Xi1 . . . xi|| . By definition
of h, we have h(S) = S. To prove that h is an isomorphism, it istherefore sufficient to show that the S generate U. We shall do thisshowing that the S for || rgenerate the space Tr of tensors of order r modulo I. The statement being trivially true for r = 0, we shallassume it verified for (r 1) and prove it for r. Again, it is enough toprove that S for || = r generate the space Tr of tensors of order = r,modulo I + Tr1. Let Xi1 Xir be an element Tr. ThenXi1 Xi2 . . . Xir Xi2 Xi1 Xi3 . . . + Xi1 ,Xi2Xi3 . . . mod I.
Hence, if is a permutation of (1, 2, . . . , r),
Xi1 Xi2 . . . Xir Xi(1) . . .Xi(r) mod (Tr1 +I)by successive transpositions. Now, S =
UXi (1) . . . xi(r), where30
U are positive integers. Therefore(
U)Xi1 . . . Xir S mod (Tr1 + I).
Since
(
U) 0,Xi1 . . . Xir kS mod (Tr1 + I),
-
7/31/2019 Lie Groups Tifr14
39/140
Relation between Lie groups and Lie algebras - II 33
and hence the theorem is completely proved.
Incidentally we have proved that the Lie algebra g can be embeddedin its universal enveloping algebra by the natural map h. This is known
as the Birkhoff-Witt theorem, and it true in the more general case when
the Lie algebra is over a principal ideal ring.
4.3 Group law in terms of structural constants.
We now show that the Lie algebra of a Lie group completely charac-
terises the group locally. In other words, the group laws of the Lie group
can be expressed in terms of the structural constants of its Lie algebra.
(If (X)A be a basis of the Lie algebra, and [Xi,Xj] =
kC
ki,j
Xk, Cki,j
are called the structural constants of the Lie algebra).Theorem 2. Lie groups having isomorphic Lie algebras are locally iso-
morphic. If they are connected and simply connected, they are isomor-
phic.
Choose a basis X1, . . . ,Xn of the Lie algebra g. If i(x) be the ith
coordinates ofx in the canonical of coordinates with respect to the above
basis we have.
i(x,y) = i(xy) = yi(x) = 1
!yi(x)
But i(x) = x(i)(e)
=
1!
x(i)(e).
Hence 31
i(x,y) =,
1
!
1
!yx(i)(e).
If
=
d
, = d
,
are completely known, once the Lie algebra g is given, because =!||! S. Since
ri(e) = (r
xrxi)x=e =
1 if r=[i]
0 if r[i]
-
7/31/2019 Lie Groups Tifr14
40/140
34 4. Relation between Lie groups and Lie algebras - II
we have i(x,y)= ,
d[i]
,
!!x
y
.
Thus the group law is completely determined by the constants d[i]
,.
If two Lie groups have isomorphic Lie algebras, the constants d[i]
,are
the same for both, and the group operation is given locally by the above
formula, which is to say the groups are locally isomorphic. By Lemma
1, Ch. 3.3, if the groups are connected and simply connected, they are
isomorphic.
We can compute the constants d[i]
,in terms of the structural con-
stants of the Lie algebra and obtain a universal formula (i.e. a for-
mula which is the same for all Lie groups - the CampbellHausdor ff
formula). For instance, if is a multi-index of order 2 with 1 in the jth
and kth indices and 0 elsewhere, it can easily be seen that =12
S =12
(XjXk +XkXj) and ifCij,k
are the structural constant of the Lie algebra,
XjXk =1
2
Xj,Xk
+
1
2(XjXk +XkXj)
=1
2
i
Cij,kXi +1
2(XjXk +XkXj)
and hence d[i]
[j],[k]=
1
2Cij,k.
We have therefore
i(x,y) = xi +yi +1
2
Cij,kxjxk + terms of order 3.
Again, if x = expX, y = exp Y (xi, yi begin canonical coordinates)32
and xy = expZ, we have
Z = X + Y +1
2[X, Y] + terms of order 3.
4.4
We have proved (Prop. 4, Ch. 3.2) that the Lie algebra of a Lie subgroup
can be identified with a subalgebra of the Lie algebra. We now establishthe converse by proving the following
-
7/31/2019 Lie Groups Tifr14
41/140
Relation between Lie groups and Lie algebras - II 35
Theorem 3. To every subalgebraJ of a Lie algebra g of a Lie group G,
there corresponds one and only one connected Lie subgroup H, havingit for its Lie algebra.
Let X1, . . . ,Xn be a basis of the Lie algebra g such that X1, . . . ,Xris a basis ofJ. Let be the subalgebra generated by Xi with i r inU(G). We assert that the subspace ofU(G) generated by S with =(1, . . ., n ) is the same as . By definition ofS, it is evident thatV V. It is enough to show that elements of the form Xi1 Xis Uif 1 ik r. We prove this by induction on the length of the product.Now,
Xi1Xi2 . . . = Xi2Xi1 . . . +
Xi1 ,Xi2
Xi3 . . .
and since Xi1 ,Xi2 , J being a subalgebra, we have, by inductionassumption
Xi1Xi2 . . . Xi2Xi1 . . . mod .If is any permutation of (1, 2, . . . , s), we have
Xi1 . . .Xis Xi(1) . . . xi(s) ( mod ).
It follows (as in Th. 1), that = .Now, let U be a symmetric neighbourhood ofe in which the system 33
of canonical coordinates with respect to X1, . . . ,Xn is valid. Let N de-
note the subset ofU consisting of points for which xr+1 =
= xn = 0.
N is obviously a closed submanifold ofU. Let x, y N sufficiently neare.
(xy)i =,
1
!
1
!xyd
[i],
We now show that (xy)i = 0 for i > r. In the summation, unless
both and are off the form (1, . . . , r, 0 . . .)xy = 0. If both and
are of the above form, , and being generated by , ofthe same form d
[i],
= 0 for i > r. Hence (xy)i = 0 for i > r. Thus,
x,y N xy N and x x1 N for x,y sufficiently near e.Finally, let H be the subgroup algebraically generated by the con-
nected component N1
of e in N. Then H can be provided with an an-alytic structure such that the map H G is everywhere regular. We
-
7/31/2019 Lie Groups Tifr14
42/140
36 4. Relation between Lie groups and Lie algebras - II
define neighbourhoods of e in H by intersecting neighbourhood of e in
G with N1
. This system can easily be seen to satisfy the neighbourhoodaxioms for a topological group (Prop 1, Ch. 1.1). For every x H,the neighbourhood xN1 of x can be provided with an analytic structure
induced by that of G, since xN is a closed submanifold of xU. For
x,y H, these analytic structure agree an xN1 yN1 because those ofxU and yU agree on xUyU. H of course has J as its Lie algebra.
We now prove the uniqueness of such a group. Let H1 be another
connected Lie subgroup having J for its Lie algebra. Exp J is openin H1, as the map h exp h is open (Th. 2, Ch. 3.4). But expJ H.34Hence H is open in H1. As H is open, it is also closed (Ch. 1.2) and
therefore =H1. This completes the demonstration of Theorem 3.
Remark. We have incidentally proved that if a Lie subgroup hasJ forits Lie algebra, it contains H as an open subgroup.
It has already been proved (Prop. 1, Ch. 3.1) that if f : G H is arepresentation of Lie groups, there exists a representation d f : g Jof Lie algebras. Now, we establish the converse in the form of a
Corollary. Let G and H be two Lie groups having g andJ as their Liealgebras. If G is connected and simply connected, to every representa-
tion ofg in J, there corresponds one and only one representation fof G
H such that d f = .
If there exists one such representation, by Prop. 5, Ch. 3.4, it is
unique. We shall now prove the existence of such an f.
We first remark that if f is a representation ofG in H, K the graph
of f in GxHviz. the set {(x, f(x)), x G}, and the restriction to K ofthe projection of GxH G, then is an analytic isomorphism. Con-versely, to every subgroup ofGxH the first projection from which is an
isomorphism to G, there corresponds one and only one representation
ofG in H. gxJ is evidently the Lie algebra ofGxH.Now, Let be a representation of g in J. Let K be the subset
{(x, (x)), x g} ofg J. It can easily be seen that Kis a subalgebra.Then there exists (Th. 3) a connected Lie subgroup K of GxH whoseLie algebra is isomorphic to K. Let be the restriction to K of the35
-
7/31/2019 Lie Groups Tifr14
43/140
Relation between Lie groups and Lie algebras - II 37
projection of G H G. Then d is the map K g defined byd(x, (x)) = x. Obviously d is an isomorphism. Hence is a localisomorphism of K in G. K is therefore a covering space of G and G
being simply connected, is actually an isomorphism. To this there
corresponds (by our remark above) a representation f of G in H,the
graph of whose differential is K, i.e. d f = .
4.5
Theorem 4 (E. Cartan). Every closed subgroup of a Lie group is a Lie
subgroup.
For proving this, we require the following
Lemma 1. Let G be a Lie group with Lie algebra g. Letg be the direct
sum of vector subspaces , . Then the map f : (A,B) expA exp B ofg G is a local isomorphism.
It is obvious that f is an analytic map. To prove that it is a local
isomorphism, it is enough to show that the Jacobian of the map 0
in a neighbourhood of (0, 0). Let (X1, . . . ,Xr) be a basis of U andXr+1, . . . ,Xn, a basis of . Let (y1, . . . ,yn) be the canonical coordinate
system with respect to (X1 . . . ,Xn) and yi = fi. Then, ifX =n
i=1
yiXi,
f{X} = exp (r
i=1
yiXi) exp (
nj=r+1
yjXj)
fk
yl(0) =
d
dt( exp t Xl)k (t = 0) = (Xl.xk)x=e = k,l.
Hence Jacobian 0 at (0, 0) and by continuity, 0 in a neighbour-
hood of the origin. This completes the proof of the lemma.
Let H be a closed subgroup of a Lie group G. We first construct a 36
subgroup J ofG and prove that the Lie subgroup H1 ofG with J asits Lie algebra is relatively open in H. Then H is the topological union
of cosets of H modulo H1 and is hence a closed submanifold of G.
Let J be the set {X g : exp tX H for every t R}. We assertthatJ is a Lie subalgebra ofg. To prove this, we have to verify
-
7/31/2019 Lie Groups Tifr14
44/140
38 4. Relation between Lie groups and Lie algebras - II
(i) X J X J for every R(ii) X, Y J X+ Y J
(iii) X, Y J [X, Y] J.(a) is a trivial consequence of the definition.
(b) Let now X, Y J. We have seen that (Ch. 4.3)
expXexp Y = exp(X + Y +1
2[X, Y] + )
Hence (exptX
nexp
tY
n)n = (exp{tX + Y
n+
1
2
t2
n2[X, Y] + 0(
1
n2})n
= exp t(X + Y) + t22n
[X, Y] + 0( 1n
).But exp
tX
n, exp
tY
n H and H is a subgroup. Therefore (exp tX
n
exptY
n)n H and since H is closed, lim
n(exptX
nexp
tY
n)n =
exp t(X + Y) (by the above formula) H. Hence X + Y J.(iv) As before,
X, Y J limn(exp
tX
nexp
tY
nexp
tYn
exptY
n)n
2 H.
The right hand side in this case tends to exp t2[X, Y] as n .Hence exp t[X, Y] H for positive values of t, and since exp(t[X, Y]) = (exp t[X, Y])1 for all values oft, i.e. [x,y] J.
Let K be the connected Lie subgroup of G having J for its Lie37algebra (Th. 3, Ch. 4.4). We now show that K is open in H. It is
obviously sufficient to prove that K contains a neighbourhood of e in
H. IfUis a vector subspace ofg supplementary to J, by Lemma 1,there exists a neighbourhood V of 0 in g such that the map : (X,A) expXexpA, X J, A Uis an isomorphism of V onto (V) = W.Suppose that K does not contain any neighborhood of e in H. Then,we can find a sequence of points an H W which are not in K and
-
7/31/2019 Lie Groups Tifr14
45/140
Relation between Lie groups and Lie algebras - II 39
which tend to e. There is no loss of generality in assuming an to be of
the form exp An,An U V, An 0. For, If an = expXn expAn, then(expXn)
1an = expAn K. Let V1 be a compact neighborhood of 0 ing V/2. For sufficiently large n, An V1. Let rn be the largest integerfor which rnAn V1. i.e. (rn + 1)An V1.
But
(rn + 1)An = rn(An) + An V/2 + V/2 = V arnn W1 = (V1)and a
rn+1n W
1 but W. Since W1 is compact, we may assume (bytaking a suitable subsequence) that a
rnn converges to an a W1. Now,
we assert that a e. For, ifa = e, ar
n+1
n = anrn an e. But arnn cannot
tend to e. Hence a e W1. Therefore a = lim anrn = expA withA 0 and A U V
1
.We shall now show that A J, which will imply that JU (0)
and hence will give the contradiction we were seeking. It is enough
to show that exp p/qA H for every rational number p/q. Now letprn/q = sn + tn/q, sn an integer and 0 tn q.
exppA
q= lim
n exp(prn
qAn)
= limn
exp snAn exp(tnAn
q)
Now, exptn
q
An
e as n
, and lim
n
exp snAn = limn
ansn
H 38
as H is closed. Hence A J, and Theorem 4 is completely proved.Remark. The theorem is not true in the case of complex Lie groups. For
instance, the space of real numbers is a closed subgroup of the complex
plane, but is not a complex Lie group.
Corollary 1. Every continuous representation f of the underlying topo-
logical group of a Lie group G into that of another Lie group H is an
analytic representation.
In fact, the graph K of f is a closed subgroup of the Lie group GxH,
and hence is a Lie subgroup. Then f is the composite of the maps G K, and K H, and both of them can be seen to be analytic. As animmediate consequence, we have the following
-
7/31/2019 Lie Groups Tifr14
46/140
40 4. Relation between Lie groups and Lie algebras - II
Corollary 2. Lie groups with isomorphic underlying topological group
structures are analytically isomorphic.
4.6 Some examples.
We have seen that the general linear group GL(n,R) is a Lie group, and
by Theorem 4, every closed subgroup, and in particular, the orthogonal
and symmetric groups are Lie groups. A group matrices defined by
some polynomial identities in the coefficients of the matrices is a Lie
group.
We proceed to study GL(n,R) in greater detail. If x GL(n,R) is39the matrix (ai j), xi,j = ai,j i,j is a coordinate system which takes theunit matrix to origin in the space R
n2
. Now,
i,j(x,y) = xi,j +yi,j +
k
xi,kyk,j
Setting ui,j = yi,j +
k xi,kyk,j, we have
y f(x) = f(x + u)
= f(x) +
i,j
yi,j(f
xi,j+
xk,ik
f
xk,j)
The left invariant differential operators of order 1 are therefore gen-
erated by xi,j =n
k=1ak.i
xk,j. The Xi,j form a basis of the Lie algebra
of GL(n,R).Y =
i,j Xi,j is a generic element of the Lie algebra. We
associate the matrix Y = (i,j) with this element Y. we now have the
Proposition 2. The map Y Y of the Lie algebra of GL(n,R) into thealgebra of all n-square matrices Mn(R) is a Lie algebra isomorphism.
Let Y be an element of the Lie algebra of GL(n,R). We show that
the map t exp tY assigns to t the usual exponential matrix exp tY. Ithas been proved (Ch. 3.3) that x = exp tY satisfies
xi.j
t = k,l
k,li,j
yk,l(x(t), e)
-
7/31/2019 Lie Groups Tifr14
47/140
Relation between Lie groups and Lie algebras - II 41
=
k,lk.lj,l(i,k + xi,k)
=
k
k,j(i,k + xi,k)
i.e. x is a matrix satisfyingd
dtx(t) = x(t)Y with x(O) = I. These two
conditions can easily be seen to be satisfied by exp tY. By the unique-
ness theorem on differential equations, exp tY = exp tY, where the latter 40
exponential is in the sense of the exponential matrix. Let Y, Z g theLie algebra of GL(n,R). The map X X is trivially a vector spaceisomorphism ofg onto Mn(R). Now,
exp YexpZ = exp(Y +Z + 12
[Y,Z] + ) by Ch. 4.3.
But exp YexpZ =
Ynn!
Zmm!
= 1 + Y + Z +
Y2
2!+
and
exp(Y +Z +1
2[Y,Z] + ) = 1 + Y + Z + 1
2(Y2 + y Z + ZY + Z2) +
Comparing the coefficients, we get
[ Y,Z] = [Y, Z].
Remarks. (1) Y =d
dt(exp tY)t=0.
(2) The Lie algebra of a closed subgroup H ofGL(n,R) is simply the
Lie algebra of matrices Y such that exp tY Hfor every t R (byTheorem 4, Ch. 4.5).
(3) Let B(a, b) be a bilinear form on Rn.
Then, the set of all regular matrices x which leave B(a, b) invariantis a Lie subgroup H ofGL(n,R). Then the Lie algebra of this Lie Group
-
7/31/2019 Lie Groups Tifr14
48/140
42 4. Relation between Lie groups and Lie algebras - II
consists of matrices Y such that B(Ya, b) +B(a, Yb) = O for every a, b R
n
. In fact, ifY is in the Lie algebra, exp tY H,B being invariant underH,B (exp tY a, exp tY b) = B(a, b). But
B(exp tY a, exp tY b) =p,Q
tp+q
p!q!B(Ypa, Yqb).
Hence
B(Ya, b) + B(a, Yb) = 0.
Conversely, if Y satisfies this condition, by induction it can be seen41
that
p+q=ntp+q
p!q!B(YPa Yqb) = 0, which proves that B(exp tY a, exp
tY b) = B(a, b), i.e. exp tY H for every t R. This proves that Yis in the Lie algebra of H.
4.7 Group of automorphisms.
Let G be a connected Lie group. Then the set of automorphisms of G
(continuous representations of G onto itself), form a group. We shall
denote this group by Aut G.
Let Aut G. This gives rise to a map d : g g where d is anautomorphism of the Lie algebra. We thus have a map; Aut G Aut g.This map is one-one, and, if G is simply connected, onto. We have, in
this case, an isomorphism of Aut G Aut g, for d(1, 2) = d1 d2and d1 = (d)1. Now, Autg GL(g). Aut g is actually a closedsubgroup of GL(g). For, ifCk
i,jbe the structural constants of g, A
Aut g A(xi),A(xj) = k
Cki,j
A(xk) for every i, j and A GL(g), {xi}being a basis ofg. Since Aut g is determined by these n2 equations, it is
a closed subgroup ofGL(g). Hence, Aut g is a Lie group.
Proposition 3. Let be the Lie algebra of Aut g. Then X Hom(g, g)(which is the Lie algebra of GL(g)) is in if and only ifexp tX Aut g
for every t in R. This is obvious from the proof of Theorem 4, Ch. 4.5.
Proposition 4. X Hom(g, g) is in if and only if X is a derivation.
-
7/31/2019 Lie Groups Tifr14
49/140
Relation between Lie groups and Lie algebras - II 43
By Proposition 3,42
(exp tX)[Y,Z] = [exp tX Y, exp tX Z]
i.e., tnXn
n!
[Y,Z]
=p,q
tp+q
p!q![xpY,XqZ]
Xn[Y,Z] =
p+q=n
n!
p!q![XpY,XqZ]
for every n. In particular
X[Y,Z] = [XY,Z] + [Y,XZ]
X is therefore a derivation. Conversely,
X[Y,Z] = [XY,Z] + [Y,XZ] Xn[Y,Z] =
p+q=n
n!
p!q![XpY,XqZ]
by induction on n exp tX[Y,Z] = [exp tX Y, exp tX Z].Hence X .In other words, the Lie algebra of Aut g is only the Lie algebra of
derivations ofg (it is a trivial verification to see that the derivations of g
form a Lie algebra and the set of inner derivations (Remark 1, Ch. 2.8)
form an ideal in that algebra).
Now, corresponding to every y G, there exists an inner automor-phism y : x yxy1 ofG. Obviously y y is an algebraic represen-tation ofG in Aut G. y induces an automorphism dy ofg. We denote
this by ady.
y ady is an algebraic representation ofG in Aut g. We now showthat this is an analytic representation. By Corollary to Theorem 4, Chap-
ter 4.5, it is enough to show that this is continuous, i.e. if y e thenadyX X for every X g. Since G and g are locally isomorphic, itsuffices to prove that as y e, y exp Xy1 exp X but this is obvious. 43This analytic representation of G in Aut G = Aut g is called the ad-
joint representation ofG. Let be the differential of the representation
y ady. We now show that this is actually the adjoint representation(Ch. 2.8) of the Lie algebra g.
-
7/31/2019 Lie Groups Tifr14
50/140
44 4. Relation between Lie groups and Lie algebras - II
Theorem 5. (X) = ad X for every X g.By Remark 1, Prop. 2, Ch. 4.6,
(X) = { ddt
(exp t (X))}t=0
=d
dt(ad exp t x)t=0
by definition of exponential. We have now to show that (X)Y = [X, Y]
for every Y g. Let x = exp tX (X)Y = ddt
(ad Y)t=0. But (adx Y f ) x = Y(f x) where f is any analytic function on G. It follows that
x1 x1 adx Y f = Yx1 x1 for (adx Y) x = x xYx1
= xY
since Y is left invariant
adx Y = xYx1 .
But
x
f(e) = f(exp tx) = ntnXn
n!f(e) (Prop. 6, Ch. 3.4)
Hence adx Y =m,n
tmXm
m!
(t)nXnn!
=m,n
(1)n tm+n
m!n!XmY Xn
Therefore
(X)Y = { ddt
(adx Y)}t=0 = XY Y X = [X, Y].
Corollary . Let H be a connected Lie subgroup of G. H is a normalsubgroup if and only if its Lie algebra J is an ideal in g.
-
7/31/2019 Lie Groups Tifr14
51/140
Relation between Lie groups and Lie algebras - II 45
In fact, if H is a normal subgroup, y(H) H for every y G, i.e.44adyJ J for every y G. Let X g. Then ad exp tXJ J.
d
dt(ad exp tX)t=0J = ad XJ J
i.e. J is an ideal.
Reciprocally, let J be an ideal. ad XJ J ( tnadXnn!
)J J. But exp t ad X = ad exp tX (adexp tX)H H for every X in aneighborhood ofe. Since G is connected, H is normal.
4.8 Factor groups.
Theorem 6. Let H be a closed subgroup of a Lie group G. The homo-
geneous space G/H is an analytic manifold in a canonical way. The
operations by G on G/H are isomorphisms. If H is a normal subgroup,
is a Lie group, and its Lie algebra is isomorphic to g/J.
Let J be the Lie algebra of H, and let Ube a vector subspaceof g supplementary to J. We have seen in the proof of Theorem 4,Ch. 4.5, that there exists a neighborhood V1 of (0, 0) in J Usuchthat the map : (X,A) expXexpA is an isomorphism of V1 ontoa neighborhood W1 of e in G. Let U and V be neighborhoods of 0 in
J and Urespectively such that U V V1
and WW1
W1
withW = (U V). We now show that L = exp V is a cross-section of thecanonical map : G G/H in the neighborhood W = (W) of (e).In other words, L H x contains one and only one element for everyx W. For, we have x = expXexpA with X U and A V andexpA L H x. On the other hand, if expA1 and expA2 belong to H x(with A1, A2 V), then expA1(expA2)1 H W1; hence there exists 45an X V1 J such that expA1 = expXexpA2 and this implies X = 0,A1 = A2 because is an isomorphism from V
1 onto W1.
We can, therefore, provide W with a manifold structure induced
from that of U. This can be extended globally by translating that onW. It is easily seen that on the overlaps x
W, y
W, the analytic structures
agree because the analytic structure on W is induced from that ofU. By
-
7/31/2019 Lie Groups Tifr14
52/140
46 4. Relation between Lie groups and Lie algebras - II
the definition of the manifold G/H, it is obvious that the operations by
G on G/H are analytic isomorphisms.IfHis normal subgroup, G/Hhas also a group structure and is a Lie
group with the above manifold. By Theorem 6, Ch. 4.8, J ia an idealofJ and ifKis the Lie algebra ofG/H, the map : G/H gives riseto a representation d : J K. The kernal of this map is J sinceKis isomorphic as a vector space to the tangent space at e of L whichis U. Hence H/J is isomorphic to Kas a Lie algebra also, i.e. G/Hhas its Lie algebra isomorphic to g/J.
Corollary. Let f be a representation of a Lie group G which is count-
able at in another Lie group H. Then the image f(G) is a Lie sub-
group. If N is the kernel of f , then f can be factored into G
G/Nf
H where is the canonical map and f an injective regular map.
The proof is an immediate consequence of the isomorphism theorem
on Lie algebras and Theorem 6.
-
7/31/2019 Lie Groups Tifr14
53/140
Part II
General Theory of
Representations
47
-
7/31/2019 Lie Groups Tifr14
54/140
-
7/31/2019 Lie Groups Tifr14
55/140
-
7/31/2019 Lie Groups Tifr14
56/140
Chapter 5
Measures on locally compact
spaces
1.1 Definition of a measure.
In this chapter and the following, we shall give a brief summary of cer- 46
tain results on measure theory, a knowledge of which is essential in what
follows.
Let X be a locally compact topological space and CX the algebra ofcontinuous complex-valued functions on X with compact support. Let
K be a compact subset of X and CK the subset {f: support of f K} ofCX. Then CK is a Banach space under the norm ||f|| = supxK |f(x)|.Definition. A measure on X is a linear form on CX such that the restric-tion to CK is continuous for every compact subset K of X. A measure is said to be positive if(f) 0 for every f 0.Proposition 1. Every positive linear form on CX is a measure on X.
In fact, if K is any compact subset of X, there exists a continuous
function f on X which = 1 in K, = 0 outside a compact neighbourhood
of K and 0 f 1. If g is a function belonging to CK, obviously
||g
||f
g
||g
||f and hence
||g||(f) (g) ||g||(f).
47
-
7/31/2019 Lie Groups Tifr14
57/140
48 5. Measures on locally compact spaces
i.e.,
|(g)| ||g||(f),which shows that is continuous when restricted to CK.
On the other hand, if is any real measure (i.e. a measure such that
(f) is real whenever f is real) it can be expressed as the difference of
two positive measures. Moreover, a complex measure can be uniquely47
decomposed into + i where and are real. Hence a measure canalternatively be defined as a linear combination of positive linear forms
on CX. For a real measure , there exists a unique minimal decompo-sition 1 2 (1, 2 positive) in the sense that if = 1 2 be anyother decomposition, we have
1= 1 + ,
2
= 2 + with positive.
1.2 Topology on CX.
The space CX =KCK where K runs through all the compact subsets of
X can be provided with the topology obtained by taking the direct limit
of the topologies on CK. This topology makes ofCX a locally convextopological vector space. The fundamental property of this space is that
a linear map ofCX in a locally convex space is continuous if and onlyif its restriction to each CK is continuous. (Bourbaki, Espaces vectorielstopologiques, Chapter 6). A measure is, by definition, a continuous
linear form on CX with its topology of direct limit. The space MX ofmeasures is none other than the dual ofCX. One can provide MX withseveral topologies, as for instance, the weak topology in which 0 for every f CX, (f) 0.
1.3 Support of a measure.
Definition. The support of a measure is the smallest closed set S such
that for every function f CX whose support is contained in X S ,(f) = 0.
Let Mc be the space of measures with compact support. If f is acontinuous function on X, for every Mc, we can define (f) =
(f) where is a function 1 on a neighbourhood of the support K of48, and 0 outside a compact neighbourhood of K. It is obvious that the
-
7/31/2019 Lie Groups Tifr14
58/140
Measures on locally compact spaces 49
value of(f) does not depend on . We shall denote by EX the space of
continuous function on X. EX with the topology of compact convergenceis a locally convex topological vector space. defined on E
Xin the
above manner is continuous with respect to this topology. Conversely,
let be a continuous form on EX
. The topology on CX is finer thanthat induced from the topology ofE
X. Hence restricted to CX is again
continuous and is consequently a measure. We now show that this has
compact support. Since is a continuous function on EX
, we can find
a neighbourhood V of 0 such that |(f)| < 1 for every f V. V maybe taken to be of the form {f : |f| < on K} as the topology on E
Xis
the topology of compact convergence. Let g CX be a function 0 onK. Then |(g)| < 1. If is any complex number, (g) = (g), and|(g)| < 1. Hence (g) = 0, i.e. the support of is contained in K.
1.4 Bounded measures.
Let ba a measure MX. We define a positive measure || in thefollowing way:
||f = sup0|g|f
|(g)| for every positive function f and extend it bylinearity to all functions CX. If is a real measure with the minimaldecomposition (Ch. 1.1) = 1 2, then || = 1 +2.
Definition . A measure is bounded if and only if there exists a real
number k such that |(f)| k||f|| with ||f|| = supxX
|f(x)|.
Obviously is bounded if and only if|| is bounded, is continuous 49for this norm and can be extended to the completion CX (which is onlythe space of continuous functions tending to zero at ). CX is actuallythe adherence ofCX in the space of all continuous bounded functions.The space of bounded measures is a Banach space under the norm |||| =sup
fCX
|(f)|||f|| . |||| is the smallest number k such that |(f)| k||f||. It can
proved that every bounded continuous function is integrable with respect
to a bounded measure, and we have still the inequality |(f)| |||| ||f||for bounded continuous functions f.
-
7/31/2019 Lie Groups Tifr14
59/140
50 5. Measures on locally compact spaces
1.5 Integration of vector valued functions.
We introduce here the notion of integration of a vector valued function
with respect to a scalar measure, a use of which we will have frequent
occasions to resort to in the sequel. Let Kbe a compact space and Ea lo-
cally convex quasi-complete topological vector space (i.e. every closed
bounded subset is complete). We shall provide the space C(K,E) ofcontinuous functions of K into E with the topology of uniform conver-
gence.
Theorem 1. Corresponding to every measure on K, there exists one
and only one continuous linear map ofC(K,E) in E such that (fa) =(f)
a for every continuous complex valued function f on K and a
E.
can obviously be lifted to a linear map ofC E in E by setting(ce) = (c)e and extending by linearity. Also, CEcan be identifiedwith a subset of the space C(K,E) of continuous functions of K into50E. We will now show that is continuous with respect to the induced
topology on C E and that C E is dense in C(K,E). We will in factprove more generally that every function f C(K,E) is adherent to abounded subset ofC E.
Let V be a convex neighbourhood of 0 in E. Then there exists a
neighbourhood Ax of each point x Ksuch that f(y)f(x) Vfor everyy Ax. Now the Ax cover the compact space K and let Ax1 , . . . ,Axn bea finite cover extracted from it. Let i be positive continuous functionson K such that
ni=1
i = 1 and the support of Axi . Ifg =
i f(xi),
then g C E, and we have
g(y) f(y) =
i(y)f(xi)
i(y)f(y)
=
i(y)[f(xi) f(y)] V since V is convex.
By allowing V to describe fundamental system of neighbourhoods
of 0, we see that f is adherent to the set of such functions g. This set is a
bounded subset ofC(K,E). For, i(y)f(xi) is in the convex envelopeof f(K) for every y K and the convex envelope of a compact set is
-
7/31/2019 Lie Groups Tifr14
60/140
Measures on locally compact spaces 51
bounded. It follows that
if(xi) are uniformly bounded and hence
form a bounded subset ofC(K,E). This proves, in particular,that CEis dense in C(K,E).
Now let g =
giai be a function C E tending to zero in thetopology ofC(K,E). Then giai, a 0 uniformly on the compactset K and on any equicontinuous subset H of the dual of E.
g, a =
giai, a
= (
giai, a) =
(gi)ai, a=
(gi)ai, a
= g, a
51
Since g, a 0 uniformly on K H, g, a 0 and henceg, a 0 uniformly on any equicontinuous subset H of E. Conse-quently g 0. This shows that is continuous on C E.
Therefore can be extended uniquely to a continuous linear map of
C(K,E) in the completion EofE. But if f C(K,E), it is adherent to abounded set B and(B) is also bounded in E. By the quasi-completeness
of E, the closure of (B) in E and E are the same. Hence (f) (B) (B) E. Thus we have extended to a continuous linear map ofC(K,E) E and it is obvious this is unique. Now by Theorem1, if G be any locally compact space and a measure on G, we can
define f(x)d = (f) for every continuous function f from G to Ewith compact support.Remark. The measure with this extended meaning is factorial in char-
acter in the following sense: Let E and F be two locally convex spaces
and f a continuous map of a compact space K into E. IfA is a continuous
map ofE in F, we have A f C(K, F) and satisfies (A f) = A(f).
-
7/31/2019 Lie Groups Tifr14
61/140
-
7/31/2019 Lie Groups Tifr14
62/140
Chapter 6
Convolution of measures
2.1 Image of a measure52
Definition. Let X, Y be two locally compact topological spaces and a
map X Y. Let be a positive measure on X. Then is said to be proper if for every function f CY, f is integrable with respectto . The value (f ) depends linearly on f and is therefore a linear
form on CY. In other words, (f ) defines a positive measure onY, which we denote by (). We have, by definition,
Y
f(y)d()(y) =X
f (x)d(x).If is not positive, but is equal to (1
2) + i(3
4), 1, 2, 3,
4 positive, and if is || - proper, we can define the image measure
() = (1) (2) + i(3) i(4).
Examples.
(1) A continuous proper map of X Y (i.e. a map such that inverseimage of every compact set is compact) is - proper for every .
In fact, f CK f C1(K) and 1(K) is compact.(2) Let be a continuous map G Hand let have compact support
K. Then is -proper; the support of () (K) and is hencecompact.
53
-
7/31/2019 Lie Groups Tifr14
63/140
54 6. Convolution of measures
If f is a continuous function on H with compact support K, f is continuous and hence -integrable (Ch. 1.3). This shows that is -proper and if f = 0 on (K), then f = 0 on Kand therefore
(f ) = 0, i.e. support of() (K).
(3) More generally, when is continuous and bounded, is -53
proper. Also () is bounded and ||()|| ||||.
In fact, f is bounded and in view of the remark in Ch. 1.4, f is integrable with respect to . Moreover,
||()|| = supg
CY
|(g)|
||g
||= sup
g
CY
|(g )|
||g
|| ||||
2.2 Convolution of two measures.
Let G, H be two locally compact topological spaces and , measures
on G, Hrespectively. Then there exists one and only measure on GHsuch that if f, g be functions with compact support respectively on G,
H we havef(x)g(y)d(x,y) = (
f(x)d(x))(
g(y)d(y)).
shall be called the product measure of and .
If and are two measures on a locally compact group G, we denote
the product measure by and, if the group operation : G G Gdefined by (x,y) xy is - proper, its image in G by . The latteris said to be the convolution productof and . The most general class
of measures for which convolution product can be defined are those for
which f(xy) is integrable with respect to the product measure for every
function f CG. The following cases are the particular interest to us:
(1) If and are bounded, the convolution product exists and is
bounded.
This is almost obvious, being continuous and bounded(Example 3, Ch. 2.1).
-
7/31/2019 Lie Groups Tifr14
64/140
Convolution of measures 55
(2) If and are measures on G with compacts K, K respectively,54
exists and has compact support. In fact, has support K K. Hence, convolution product exists, and has support KK.
(3) If either or has compact support, exists. Let f be acontinuous function on G with compact support K and let have
compact support K. Obviously
f(xy)d(x)d(y) =
K
d(x)
KK1
f(xy)d(y)
Hence f(xy) is integrable with respect to
. Consequently,
exists.We denote as usual by M1, Mc, E0
Gthe spaces of bounded mea-
sures, the space of measures with compact support and the space of all
continuous functions on G respectively. Let , , be three measures
on G such that either all three are bounded or two of them have compact
support. In any case the function (x,y,z) f(xyz) is integrable withrespect to and hence Fubinis theorem can be applied.
f(xyz)d(x)d(y)d(z) =
d(z)
f(xyz)d(x)d(y)
= d(z) f(tz)d( )(t)=
f(tz)d( )(t)d(z)
= {( ) }f= { ( )}f
by a similar computation. This shows that m1 with the convolution prod-
uct is an associative algebra and that Mc acts on M on both sides and 55makes it a two-sided module. Moreover, M1 is actually a Banach alge-bra under the usual norm, since we have || || ||| | | |||.
Remarks. (1) It is good to point out here that the associativity doesnot hold in general. Take,for instance, R to be the locally compact
-
7/31/2019 Lie Groups Tifr14
65/140
56 6. Convolution of measures
group and the Lebesgue measure. Let be 1 0 (a begin theDirac measure at a - see Ch. 2.5 and = (x)dx where is theHeaviside function viz. = 0 for x < 0 and =1 for x 1. Then( ) = 0 and ( ) = dx. Again ( ) may existwithout begin well defined. Let R be the locally compactgroup, and Lebesgue measures on R and = 1 0. Then ( ) = 0 but is not defined. However, when f(xyz) isintegrable with respect to then the convolution productis associative.
(2) The formula for the integration of functions with respect to the
convolution of two measures is valid also for vector-valued func-
tion. Thus we haveG
f(x)d( ) =
f(xy)d(x)d(y).
2.3 Continuity of the convolution product.
That the convolution product is continuous in M1 is trivial in virtue ofour remark that it is a Banach algebra. Regarding the continuity of the
convolution product in the other cases, we have the following
Lemma 1. Let f be a continuous function and a measure on G, one of
them having compact support. Then (i) the function g(x) = f(xy)d(y)is continuous; (ii) the map f g is a continuous linear map ofCG in56E0
G(with the usual topologies); (iii) if has compact support, then the
above map f
f(xy)d(y) is also continuous from EG
EG
and
from CG CG.
(1) Let H and K be two compact subsets of G. Then H K is alsocompact and f(xy) is uniformly continuous on H K. For every > 0, and for every x H, there exists a neighbourhood U of xsuch that |f(xy) f(xy)| < for every x U H and y K.If f has compact support S, we choose a compact neighbourhoodH of x and K such that HS1 K. If y K, then xy, xy S.
-
7/31/2019 Lie Groups Tifr14
66/140
Convolution of measures 57
Hence
GK
f(xy) f(xy)d(y) = 0.So, we have
|g(x) g(x)|
K
f(xy) f(xy)
d||(y) ||(K).
This shows that g is continuous. If however has compact support
C, we take K = C and the same inequality as above results.
(2) Again, as in (i) if we assume that f has compact support S ans
HS1
K, it is immediate that
|g(x)|
K
|f(xy)|d||(y) for every x H
K
|f(y)|d||(x1y)
sup |f|||(H1K).
Hence we have supxH
|g(x)| sup |f|||(H1 K).
It follows that whenever f 0 on CS, g(x) 0 uniformly onthe compact set H. A similar proof holds when has compact
support.
(3) Let now C be the support of , and f has compact port K; obvi- 57
ously g CKC1. Since the map CG EG is continuous, so alsois the map CK CKC1 and by the properly of the direct limittopology, CG CG is continuous. An analogous proof holds forthe other part.
2.4 Duality and convolution products
Let E be a locally convex topological vector space an E its dual. ThenE can be provided with several interesting topologies (Bourbaki, Es-
paces vectoriels topologiques, Chapter 8). The following three are offundamental importance:
-
7/31/2019 Lie Groups Tifr14
67/140
58 6. Convolution of measures
(i) The weak topology, in which x E 0 if and only ifx, x 0 for every x E.
(ii) The convex compact topology, in which x E 0 if and onlyifx, x 0 uniformly on every convex compact subset, and
(iii) The strong topology, where x E 0 if and only ifx, x 0uniformly on every bounded set.
In general, these topologies are distinct. If E is a Banach space, the
usual dual is the E with the strong topology. However, the convex com-pact topology is often the most useful, in as much as it shares the good
properties of both the weak and the strong topologies. To mention but
one such, (E) = E is true for the weak, but not for the strong, topology.The convex compact topology possesses this property. We shall almost
always restrict ourselves to the consideration of this topology.
In particular, the spaces M, Mc being duals ofCG and EG respec-58tively, they can be provided with the convex compact topology. With
reference to the convolution map we have the
Proposition 1. The convolution map (, v) is continuous in eachvariable separately in the following situations:
Mc
Mc
Mc;Mc
M
M;M
Mc
M.
In fact, let be fixed in Mc and 0 in M. Then
(f) =
f(xy)d(x)d(y)
=
d(y)
f(xy)d(x).
Denoting by f0(y) the function
f(xy)d(x) the map f f0 is con-tinuous from CG CG (Lemma 1, Ch. 2.3). The image of a convexcompact subset being again a convex compact subset, 0 uni-formly on a convex compact subset. All other assertions in the proposi-tion can be demonstrated in an exactly similar manner.
-
7/31/2019 Lie Groups Tifr14
68/140
Convolution of measures 59
2.5 Convolution with the Dirac measure
If x is a point of G, the Dirac measure x is defined by x(f) = f(x).
This is trivially a measure with compact support. Let be any arbitrary
measure. Then
x (f) =
f(yz)dx (y)d(z)
=
f(xz)d(z)
= (x 1f).
In a similar manner,
x(f) = (x f). We may define left and right
translation of a measure by setting d(x)(y) = d(yx) and d(x)(y) =d(x1y). It requires a trivial verification to establish that x(f) = 59(x1 f) and x(f) = (x1 f). Hence we have
x = x, and x = 1x .
In particular, x y = xy = xy. In other words, the map x xis a representation in the algebraic sense of the group G into the algebra
Mc or M1. As a matter of fact, this can be proved to be a topologicalisomorphism (Bourbaki, Interation, Chapter 7).
-
7/31/2019 Lie Groups Tifr14
69/140
-
7/31/2019 Lie Groups Tifr14
70/140
Chapter 7
Invariant measures
3.1 Modular function on a group.60
We assume the fundamental theorem relating to measures on locally
compact groups, namely the existence and uniqueness (upto a positive
constant factor) of a right invariant positive measure. If is such a
measure, we have
(y ) x = y ( x) = y .
Hence y is also a right invariant positive measure. By our remarkabove, y
= k where, of course, kdepends on y. We shall denote kby
(y)1 where (y) is a positive real number. It is immediate that (yz) =(y)(z). is therefore a representation ofG in the multiplicative group
ofR+. In fact, the continuity of(y) =
f(y1x)d(x)
f(x)d(x)follows at once
from that of
f(y1x)d(x) (Lemma 1, Ch. 2.3). This representation of a locally compact group is said to be its modular function.
Proposition 1. If a right invariant positive measure on G is denoted by
dx, then the following identity holds: dx1 = (x1)dx
In fact, ifd stands for (x1)dx, we have
d(yx) = (x1y1)d(yx) = (x1)dx = d.
61
-
7/31/2019 Lie Groups Tifr14
71/140
62 7. Invariant measures
Hence d is left invariant. So also is dx1 for,
f(yx)dx1 =
f(x)d(y1x)1 =
f(x)d(x1).
So kd x1 = (x1)dx where k is a constant. We now prove thatk = 1. When x is near e, (x1) is arbitrarily near 1 and if we take g(x) =f(x) + f(x1), f being a positive continuous function with sufficiently61small support, we have
g(x)dx =
g(x)dx1 =
1
k
g(x)(x1 )dx
k being a fixed number and (x1
) arbitrarily near 1; it follows thatk = 1.
Definition . A locally compact group G is said to be unimodular if its
modular function is a trivial map which maps G onto the unit element
of R.
The group of triangular matrices of the type
a11...0
.........ann
can be proved
to be non-unimodular.
Examples of unimodular groups.
(1) A trivial example of unimodular groups is that of commutativegroups.
(2) Compact groups are unimodular. This is due to the fact that (G)
is a compact subgroup ofR+ which cannot but be (1).
(3) If in a group the commutator subgroup is everywhere dense, then
the group is unimodular. This again is trivial as maps the com-
mutator subgroup and consequently the whole group onto 1.
(4) A connected semi-simple Lie group is unimodular. (A Lie group
G is said to be semi-simple if its Lie algebra g has no proper
abelian ideals. Consequently, it dose not have proper ideals suchthat the quotient is abelian). The kernel N of the representation
-
7/31/2019 Lie Groups Tifr14
72/140
Invariant measures 63
d of the Lie algebra into the real number is an ideal in g such that
g/N is abelian and is therefore the whole Lie algebra. It followsthat the map d maps the Lie algebra onto (0). This shows that
group is unimodular.
3.2 Haar measure on a Lie group.62
Let G be a Lie group with a coordinate system (x1, . . . , xn) in a neigh-
bourhood of e. We now investigate the form of the right invariant Haar
measure on the Lie group. By invariance of a measure here we mean
that
f(xy1)d(x) =
f(x)d(x) for y which are sufficiently near
e and for f whose supports are sufficiently small. We set d(x) =
(x)dx1 . . .dxn and enquire if integration with respect to this measureis invariant under right translations. For invariance, we requiref(x)(xy)
det i(x,y)xj
x1 . . . dxn=
f(x)(x)dx1 . . . dxn,
or still (x) = (xy)J(x,y) with J(x,y) =
det ixj (x,y).
For this it is obviously necessary and sufficient to take (x) = J1(e,x). This gives an explicit construction of the Haar measure in the case
of Lie groups.
3.3 Measure on homogeneous spaces.
IfG/His the quotient homogeneous space of a locally compact group G
by a closed subgroup H, we denote the elements ofG by letters x,y, . . .
those of H by , , . . . the respective Haar measure by dx, dy, . . . d,
d . . . and the respective modular functions by , . Also is the canon-
ical map G G/H. Let f be a continuous function on G with compactsupport K. Then fo(x) =
H
f(x)dis a continuous function on G (as in
lemma 1, Ch. 2.3) and we have fo(x) = fo(x) for every H. There-fore f
o
may be considered as a continuou