lie groups and lie algebras - kangnikinvi · 2019-04-18 · lie groups and lie algebras professor...

Lie groups and Lie algebras

Professor Kinvi KANGNI

Department of Mathematics and

computer sciences.

University FHB, Abidjan;

Ivory Coast

Lie groups and Lie algebras

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Kangni Kinvi

Université Felix Houphouet Boigny d�Abidjan

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UFR Mathématiques et informatique

Table des matières

1 Linear Lie groups 4

1.1 One parameter subgroups . . . . . . . . . . . . . . . . . . . . 4

1.2 Lie algebra of a linear Lie group . . . . . . . . . . . . . . . . . 5

1.3 Linear Lie groups are submanifolds . . . . . . . . . . . . . . . 8

1.4 Campbell�Hausdor¤ formula . . . . . . . . . . . . . . . . . . . 10

2 Lie algebras 13

2.1 Nilpotent and solvable Lie algebras . . . . . . . . . . . . . . . 20

2.2 Semi-simple Lie algebras. . . . . . . . . . . . . . . . . . . . . . 28

3 Lie groups 35

3.1 De�nitions and some proprieties. . . . . . . . . . . . . . . . . 35

3.2 Lie Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.3 The Lie Algebra of a Lie group . . . . . . . . . . . . . . . . . 43

4 Solvable Lie Groups and Algebras 51

4.1 Solvable Lie group . . . . . . . . . . . . . . . . . . . . . . . . 51

4.2 Solvable Lie Algebras and Radicals . . . . . . . . . . . . . . . 54

4.3 Lie�s Theorem on Solvability . . . . . . . . . . . . . . . . . . . 57

5 Nilpotent Lie groups and algebras 67

5.1 Nilpotent Lie Groups . . . . . . . . . . . . . . . . . . . . . . 67

5.2 Nilpotent Lie Algebras . . . . . . . . . . . . . . . . . . . . . 70

5.3 Nilpotent Lie Algebras of Endomorphisms . . . . . . . . . . . 76

2

TABLE DES MATIÈRES

6 Topoligical groups 80

6.1 BASICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

6.2 Subgroups and Homogeneous Spaces . . . . . . . . . . . . . . 82

6.3 Connected Groups . . . . . . . . . . . . . . . . . . . . . . . . 85

7 Representations of compact groups 87

7.1 Unitary representations . . . . . . . . . . . . . . . . . . . . . . 88

7.2 Compact self-adjoint operators . . . . . . . . . . . . . . . . . . 91

7.3 Schur orthogonality relations . . . . . . . . . . . . . . . . . . . 96

7.4 Peter�Weyl Theorem . . . . . . . . . . . . . . . . . . . . . . . 100

7.5 Characters and central functions . . . . . . . . . . . . . . . . . 106

8 Induced Representations 108

8.1 The De�nition . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

8.2 The carrier space of the induced representation . . . . . . . . 114

Kangni Kinvi 3 Analyse Harmonique

Chapitre 1

Linear Lie groups

A linear Lie group is a closed subgroup of GL (n;R). To a linear Liegroup one associates its Lie algebra. In this way the properties of the group

are translated in terms of the linear algebra properties of its Lie algebra. We

saw several examples in Section 1.3. Let us observe that is a line GL (n;C)is a Lie group since it can be seen as a closed subgroup of GL(n;R). In fact,to a matrix Z = X + iY in M(n;C) one associates the matrix

eZ =

X �YY X

!in M(2n;C), and the map Z 7�! eZ is an algebra

morphism which maps GL (n;C) onto a closed subgroup of GL (n;R) :

1.1 One parameter subgroups

Let G be a topological group. A one parameter subgroup of G is a conti-

nuous group morphism : R �! G, R being equipped with the additive

group structure.

Théorème 1.1.1 Let : : R �! GL (n;R) be a one parameter subgroup ofGL (n;R). Then is C1 and (t) = exp(tA), with A = (0). In fact is

even real analytic, as can be proved.

4

CHAPITRE 1. LINEAR LIE GROUPS

Preuve. Assume that is C1. Then

0(t) = lims!0

(t+ s)� (t)

s

= (t)lims!0

(s)� (0)

s= (t) 0(0) = 0(0) (t)

Put A = 0(0): Then 0(t) = A (t): This di¤erential equation has a unique

solution such that (0) = 1 , which is given by (t) = exp(tA) In fact, if

is such a solution

d

dt(exp(�tA) (t)) = exp(�tA)( 0(t)� A (t)) = 0

We will now show that is C1. Let � be a C1 function on R with compactsupport, and consider the regularised function f of :

f(t) =

Z +1

�1�(t� s) (s)ds

Then f : R �!M(n;R) is C1, and

f(t) =

Z +1

�1�(s) (t� s)ds =

Z +1

�1�(s) (�s)ds: (t):

We will choose the function � in such a way that the matrix

B =

Z +1

�1�(s) (�s)ds

is invertible. It will followthat is C1. If kB � Ik < 1 then it holds. Let � �0, with integral equal to one. Then kB � Ik � f(t) =

R +1�1 �(s) k (�s)� Ik ds:

Since is continuous at 0, for every

" > 0 there exists � > 0 such that, if jsj � �, then k (�s)� Ik � ": If

the support of � is contained in [��; �], then kB � Ik � ":

1.2 Lie algebra of a linear Lie group

Let G be a linear Lie group, that is a closed subgroup of GL(n;R).We associate to the group G the set g = Lie(G) = fX 2 M(n;R)j8t 2R; exp(tX) 2 Gg:



Théorème 1.2.1 (i) The set g is a vector subspace of M(n;R).

(ii) If X; Y 2 g, then [X; Y ] := XY � Y X 2 G:Preuve. (a) If X; Y 2 G, then

�exp t

kX exp t

kY�k 2 G, and, since G is closed,

as k �!1, exp(t(X + Y )) 2 G by Corollary 2.2.4, hence X + Y 2 G:

(b) Similarly, for t > 0, limk!1

�exp

ptkX exp

ptkY exp�

ptkX � exp�

ptkY�k2

=

exp(t [X;Y ] 2 G, hence [X; Y ] 2 G:A real (respectively complex) Lie algebra is a vector space G over R

(respectively C) equipped with a linear map G�G �! G by (X; Y ) 7�! [X; Y ]

called the bracket or commutator of X and Y , such that

(1) [X;Y ] = �[Y;X];(2) [X; [Y; Z]] = [[X; Y ] ; Z] + [Y; [X;Z]] :

Relation (2) is called the Jacobi identity.

The space M(n;R) equipped with the product [X; Y ] = XY � Y X is a

Lie algebra. If G � GL(n;R) is a linear Lie group, then G = Lie(G) is a

subalgebra of M(n;R), it is the Lie algebra of G.

Exemple 1.2.2 LieGL(n;R) =M(n;R);

Lie [SL(n;R)] = fX 2M(n;R)jtrX = 0g ;Lie [SO(n)] =

�X 2M(n;R)jXT = �X

;

Lie [Sp(n;R)] =

( A B

C �AT

!nA 2M(n;R); B; C 2 Sym(n;R)

);

Lie [(U(n))] = fX 2M(n;R)jX� = �XgConsider G = SL(2;R) and let G = sl(2;R) be its Lie algebra. The

following matrices constitute a basis of G :

H =

1 0

0 �1

!; E =

0 1

0 0

!; F =

0 0

1 0

!and [H;E] = 2E;

[H;F ] = �2F; [E;F ] = H:

Let G be the group `ax+ b0, that is the group of a¢ ne linear transforma-tions of R. It is the set R� �R equipped with the product (a1; b1)(a2; b2) =(a1a2; a1b2+ b1): This is not a group of matrices, but it can be identi�ed with

the closed subgroup of GL(2;R) whose elements are the matrices

a b

0 1

!:



The matrices X1 =

1 0

0 0

!; X2 =

0 1

0 0

!constitute a basis of its

Lie algebra and [X1; X2] = X2: Let G be the motion group of R2, that is thegroup of a¢ ne linear transformations of the form (x; y) 7�! (x cos ��y sin �+a; x sin � + y cos � + b): The group G can be identi�ed with the subgroup of

GL(3;R) whose elements are the matrices

0BB@cos � � sin � a

sin � cos � b

0 0 1

1CCAIts Lie algebra G has dimension 3. The following matrices constitute a

basis

for g : X1 =

0BB@0 �1 0

1 0 0

0 0 0

1CCA ; X2 =

0BB@0 0 1

0 0 0

0 0 0

1CCA ; X1 =

0BB@0 0 0

0 0 1

0 0 0

1CCA ;

and [X1; X2] = X3; [X1; X3] = �X2; [X2; X3] = 0: Let G and H be two

Lie algebras over R (or C). A Lie algebra morphism of G into h is a linearmap A : G �! Hsatisfying[AX;AY ] = A[X; Y ]: The group of automorphisms of the Lie algebra g

is denoted by Aut(G):Let G be a linear Lie group, and G = Lie(G) its Lie algebra. By the

de�nition of the Lie algebra of G, the exponential map maps g into G : exp :

G �! G. For y 2 G;X 2 G; t 2 R;y exp(tX)y�1 = exp(tyXy�1): Hence yXy�1 2 G: The map Ad(y) : X !

Ad(y)X = yXy�1 is an automorphism of the Lie algebra G; Ad(y)[X; Y ] =[Ad(y)X;Ad(y)Y ], (X; Y 2 G):Furthermore Ad(y1y2) = Ad(y1) � Ad(y2); and this means that the map

Ad : G �! Aut(G) is a group morphism.

Proposition 1.2.3 (i) For X 2 G;�ddtAd(exp(tX)

�t=0= adX:

(ii) Let us denote by Exp the exponential map from End(G) into GL(G).Then Exp(adX) = Ad(expX); (X 2 G):

Preuve. (a)�ddtAd(exp(tX)

�t=0=�ddtexp(tX)Y exp(�tX)

�t=0= [X; Y ] :



(b) Put 1(t) = Exp(tadX); 2(t) = Ad(exp(tX));They are two one

parameter subgroups ofGL(G), and 01(0) = ad(X); 02(0) = ad(X) Therefore

1(t) = 2(t) (t 2 R).

1.3 Linear Lie groups are submanifolds

Let us recall �rst the de�nition of a submanifold in a �nite dimensional

real vector space. A submanifold of dimension m in RN is a subset M with

the following property : for every x 2 M there exists a neighbourhood U of

0 in RN , a neighbourhood W of x in RN and a di¤eomorphism from U onto

W such that �(U \ Rm) =W \M:

Théorème 1.3.1 Let G be a linear Lie group and G = Lie(G) be its Lie

algebra.There exists a neighbourhood U of 0 in G and a neighbourhood V of

I in G such that

exp : U �! V is a homeomorphism.

Preuve. LetG � GL(n;R) be a linear Lie group, and G �M(n;R) be its Lie

algebra. Let U0 be a neighbourhood of 0 inM(n;R) and V0 a neighbourhoodof I in GL(n;R) for which exp : U0 �! V0 is a di¤eomorphism. Then U0 \ Gis a neighbourhood of 0 in G, the restriction of the exponential map to U0\Gis injective and maps U0 \G into V0 \G, but one does not know yet whetherexp(U0 \ G) = V0 \G, even if one assumes that G is connected.

Lemme 1.3.2 Let (gk ) be a sequence of elements in G which converges to

I . One assumes that, for all k, gk = I . Then the accumulation points of the

sequence Xk =log gkklog gkk

belong to g.

Preuve. We may assume that limk!1

Xk = X 2M(n;R) Put Yk = log gk and,for t 2 R, �k = t

klog gkk ; then exp(tX) = limk!1

exp(�kYk):

Let us denote by [�k] the integer part of �k. We can write exp(�kYk) =

(exp(Yk))[�k] exp((�k � [�k])Yk); and k�k � [�k])Ykk � kYkk �! 0;



hence, since expYk = gk, exp(tX) = limk!1

(gk)[�k] 2 G; and this proves

that X belongs to G.

Lemme 1.3.3 LetM be a subspace of M(n;R), complementary to G. Thenthere exists a neighbourhood U of 0 in m such that expU \G = fIg:

Preuve. Let us assume the opposite. In this case there exists a sequence

Xk 2 m with limit 0 such that gk = expXk; gk 6= I; gk 2 G: Let Y be an

accumulation point of the sequence XkkXkk : By Lemma Y 2 G \M = f0g, and

this is impossible since kY k = 1:

Lemme 1.3.4 Let E and F be two complementary subspaces in M(n;R).Then the map � : E � F �! GL(n;R) by (X; Y ) 7�! expX expY: is di¤e-

rentiable, and D�(0;0) (X; Y ) = X + Y

The proof is left to the reader. We can now �nish the proof of Theorem .

Let m be a subspace of M(n;R) complementary to g, and consider the map� : g�m �! GL(n;R) by (X; Y ) 7�! expX expY There exists a neighbou-

rhood U of 0 in G; a neighbourhood V of 0 in m, and a neighbourhood W of I

in GL(n;R) such that the restriction of toU�V is a di¤eomorphism onto W.Observe that expU = �(U�f0g) � W \G: By Lemma the neighbourhood Vcan be chosen such that expV \G = fIg. Let us show that expU = W \G.Let g 2 W \ G. One can write g = expX expY (X 2 U; Y 2 V ); and then

expY = exp(�X)g 2 expV \G = fIg; hence g = expX:

Corollaire 1.3.5 Alinear Lie group G � GL(n;R) is a submanifold ofM(n;R)of dimension m = dimg.

Preuve. Let g 2 G and let L(g) be the map L(g) : GL(n;R)! GL(n;R); byh 7�! gh: Let U be a neighbourhood of 0 inM(n;R) andW0 a neighbourhood

of I in GL(n;R) such that the exponential map is a di¤eomorphism from U

onto W0 which maps U \G onto W0 \G. The composed map � = L(g) � expmaps U ontoW = gW0, and U\G ontoW \G. An important consequence of



Theorem is that the set exp g is a neighbourhood of I in G, hence generates

the identity component G0 of G by Proposition1[k=1

(exp g)k = G0:

Corollaire 1.3.6 If two closed subgroups G1 and G2 of GL(n;R) have thesame Lie algebra then the identity components of G1 and G2 are the same.

It also follows from Theorem that the group G is discrete if and only if its

Lie algebra reduces to f0g : Lie(G) = f0g: To every closed subgroup G of

GL(n;R) one associates its Lie algebra G = Lie(G) �M(n;R). However, notevery Lie subalgebra ofM(n;R) corresponds to a closed subgroup of GL(n;R).

1.4 Campbell�Hausdor¤ formula

LetG be a linear Lie group and G = Lie(G) its Lie algebra. The Campbell�

Hausdor¤ formula expresses log(expX expY ) (X; Y 2 G) in terms of a series,each term of which is a homogeneous polynomial in X and Y involving ite-

rated brackets. Let us introduce the functions

�(z) =1� e�z

z=

1Xk=0

(�1)k zk

(k + 1)!(z 2 C)

;

(z) =z log z

z � 1 = z1Xk=0

(�1)k (z � 1)k

(k + 1); (jz � 1j < 1):

If jzj < log 2, then jez � 1j � ejzj�1 < 1, and (ez)�(z) = ezzez�1

1�e�zz

= 1:

Therefore, if L is an endomorphism such that kLk < log 2, then(ExpL)�(L) =Id:With this notation the di¤erential of the exponential map can be written

(D exp)A = expA�(adA):

Théorème 1.4.1 If kXk ; kY k < r = 12log(2� 1

2

p2) then

log(expX expY ) = X +

Z 1

0

(Exp(adX)Exp(tadY ))Y dt

Lemme 1.4.2 If kXk ; kY k < �; then kexpX expY � Ik � e2� � 1:



Preuve. expX expY �I = (expX�I)(expY �I)+(expX�I)+(expY �I);and, since kexpX � Ik � ekXk � 1 � e� � 1 , kexpX expY � Ik � (e� �1)2 + 2(e� � 1) = e2� � 1:

Lemme 1.4.3 If kg � Ik � � < 1, then klog gk � log 11�� :

Preuve. klog gk �1Xk=0

k(g�I)kkk

�1Xk=0

�k

k= log 1

1�� : Let us now prove Theo-

rem. For kXk ; kY k < 12log 2, put F (t) = log(expX exp tY ): By Lemma,

the function F is de�ned for jtj � 1 1. If furthermore kXk ; kY k < r (ob-

serve that r < 12log 2; then, by Lemmas kF (t)k < 1

2log 2 From the in-

equality kXY � Y Xk � 2 kXk kY k it follows that kadXk � 2 kXk, hencekadF (t)k < log 2. Let us prove that the function F satis�es the di¤erential

equation F 0(t) = (Exp(adF (t))Y: One can write expF (t) = expX exp tY:

Taking the derivative at t :

(D exp p)F (t)(F0(t)) = (expX exp tY )Y: By Theorem, we obtain�(adF (t))F 0(t) =

Y . Since kadF (t)k < log 2 this can be written F 0(t) = (Exp(adF (t)))Y:

We can also write

F 0(t) = (Ad(expF (t))Y

= (Ad(expX)Ad(exp tY ))Y

= (Exp(adX)Exp(adtY ))Y:

Furthermore F (0) = log(expX) = X; and F (1) = F (0) +R 10F 0(t)dt; hence

log(expX expY ) = X +R 10(Exp(adX)Exp(tadY ))Y dt:

Théorème 1.4.4 (Campbell�Hausdor¤ formula) If kXk ; kY k < r = 12log(2�

12

p2), then

log(expX expY ) = X+

1Xk=0

(�1)kk + 1

X"(k)

1

(q1 + q2 + :::qk + 1)

(adX)p1(adY )q1 ::::(adX)pk(adY )qk(adX)m

p1!q1!::::::pk!qk!m!Y

where, for k � 1; "(k) = fp1; q1; :::; pk; qk;m 2 N jpi + qi > 0; i = 1; :::; kg;and "(0) = fm 2 Ng.



Preuve. If A and B are two endomorphisms

(expA expB � I)k expA =X"(k)

(A)p1(B)q1 ::::(A)pk(B)qk(A)m

p1!q1!::::::pk!qk!m!

Since

(z) =z log z

z � 1 = z

1Xk=0

(�1)k (z � 1)k

(k + 1); (jz � 1j < 1)

we have

(Exp(adX)Exp(adtY ))Y = Exp(adX)Exp(adtY )Y1Xk=0

(�1)k(k + 1)

(Exp(adX)Exp(adtY ))�I)k:

Observing that Exp(tadY )Y = Y; we obtain

(Exp(adX)Exp(adtY ))Y =1Xk=0

(�1)kk + 1

X"(k)

(adX)p1(adY )q1 ::::(adX)pk(adY )qk(adX)m

p1!q1!::::::pk!qk!m!Y

The convergence of the series is uniform for t in [0, 1]. The statement is

obtained by termwise integration sinceZ 1

0

tq1+q2+:::qkdt =1

(q1 + q2 + :::qk + 1)

Corollaire 1.4.5

log(expX expY ) = X+Y+1

2[X;Y ]+

1

12[X; [X; Y ]]+

1

12[Y; [Y;X]]+termsof deg ree � 4:


Chapitre 2

Lie algebras

In this chapter we consider Lie algebras from an algebraic point of view.

We willLie algebras In this chapter we consider Lie algebras from an alge-

braic point of view.We will see how some properties of linear Lie groups can

be deduced from the corresponding properties of their Lie algebras. Then

we present the basic properties of nilpotent, solvable, and semi-simple Lie

algebras.

Dé�nition 2.0.6 A Lie algebra over K = R or C is a vector space G equip-ped with a bilinear map G � G �! G; (X; Y ) 7�! [X; Y ] satisfying

(1) [Y;X] = � [X; Y ](2) [[X; Y ] ; Z] + [[Y; Z] ; X] + [[Z;X] ; Y ] = 0: The equality (2) is called

the Jacobi identity. Assume G is �nite dimensional, and let (X1; :::; Xn) be a

basis of G. One can write

[Xi; Xj] =

nXk=1

ckijXk

The numbers ckij are called the structure constants of the Lie algebra G: Pro-perty (1) can be written ckij = �ckji, and property (2) says that, for any m;nXl=1

(clijcmlk + cljkc

mli + clkic

mlj ) = 0An automorphism of a Lie algebra is a linear

automorphism y 2 GL(g) such that

[yX; yY ] = y[X; Y ]:

13

CHAPITRE 2. LIE ALGEBRAS

The group of all automorphisms of the Lie algebra G is denoted by Aut(G):If G

is �nite dimensional, it is a closed subgroup of GL(g): A derivation of Gis a linear endomorphism D 2 End(G) such that

D([X;Y ]) = [DX; Y ] + [X;DY ]:

For X 2 G let ad X denote the endomorphism of g de�ned by adX:Y =

[X; Y ]: The Jacobi identity (2) says that ad X is a derivation. The space

Der(G) of thederivations of G is a Lie algebra for the bracket de�ned by

[D1; D2] = D1D2 �D1D2;

and the map ad : g �! Der(G) is a Lie algebra morphism :

ad[X;Y ] = [adX; adY ]:

Proposition 2.0.7 Let G be a �nite dimensional Lie algebra. The Lie alge-bra of Aut(G) is equal to Der(G):

Preuve. Let D 2 Lie(Aut(G)): For every t 2 R; Exp(tD) is an automor-phism of G : for X; Y 2 G;

Exp(tD)[X;Y ] = [Exp(tD)X;Exp(tD)Y ]

. Taking derivatives of both sides at t = 0 we obtain

D[X; Y ] = [DX; Y ] + [X;DY ];

which means that D is a derivation :D 2 Der(G): Conversely, letD 2 Der(G)and put, for X; Y 2 G;

F1(t) = Exp(tD)[X;Y ];

F2(t) = [Exp(tD)X;Exp(tD)Y ]



We have

F 01(t) = DExp(tD)[X; Y ] = DF1(t);

F 02(t) = [DExp(tD)X;Exp(tD)Y ] + [Exp(tD)X;DExp(tD)Y ] ;

and, since D is a derivation of G;

F2(t) = D [Exp(tD)X;Exp(tD)Y ] = DF2(t):

Thus F1 and F2 are solutions of the same di¤erential equation with the same

initial data : F1(0) = F2(0) = [X; Y ]: Hence, for every t 2 R; F1(t) = F2(t):

This means that, for every t; Exp(tD) is an automorphism of G, and thatD 2Lie(Aut(g)): An ideal J of a Lie algebra G is a subalgebra which furthermoresatis�es 8X 2 J;8Y 2 g; [X; Y ] 2 J: Let G be a linear Lie group, and H a

closed subgroup. Then h = Lie(H) is a subalgebra of G = Lie(G) and, if H is

a normal subgroup of G, then h is an ideal of G: The converse holds if G andH are connected. Let G be a topological group and V a �nite dimensional

vector space over R or C. A representation of G on V is a continuous map

� : G �! GL(V );

which is a group morphism :

�(g1g2) = �(g1)�(g2) (g1; g2 2 G); �(e) = Id;

A vector subspaceW � V is said to be invariant if, for every g 2 G; �(g)W =

W: Let us denote by �0(g) the restriction of �(g) toW : �0(g) = �(g)W: Then

�0 is a representation of G onW, one says that �0 is a subrepresentation of �.

The representation �1 of G on the quotient space V=W is called a quotient

representation. The representation � is said to be irreducible if the only

invariant subspaces are f0g and V . Two representations (�1; V1) and (�2; V2)are said to be equivalent if there exists an isomorphism A : V 1 �! V 2 (A is

an invertible linear map) such that

A�1(x) = �2(x)A



, for every x 2 G. One says that A is an intertwinning operator or that A

intertwins the representations �1 and �2.

A representation of a Lie algebra G on a vector space V is a linear map

� : G �! End(V )

which is a Lie algebra morphism :

�([X; Y ]) = [�(X); �(Y )] = �(X)�(Y )� �(Y )�(X):

One also says that V is a module over G, or that V is a G-module.The map ad : G �! Der(G) � End(G) is a representation of G, which is

called the adjoint representation. Let G be a linear Lie group with Lie algebra

G and let � be a representation of G on a �nite dimensional vector space V.Then, for X 2 G; t ! (t) = �(exp tX) is a one parameter subgroup of

GL(V ); hence di¤erentiable by Theorem. Put

d�(X) =

�d

dt�(exp tX)

�t=0

(X 2 G);

then d� is a representation of the Lie algebra of G on V , which is called thederived representation of �: Let us prove this fact. By Theorem,

�(expX) = Expd�(X) (X 2 g):

From the de�nition of d� it follows at once that, for t 2 R; d�(tX) = td�(X):

By Corollary

�(exp t(X + Y )) = limk!1

(�(exptX

k)�(exp

tY

k))k

= limk!1

(Expd�(tX)

kExp

d�tY

k)k

= Exp(d�(tX) + d�(tY ))

= Exp(td�(X) + td�(Y ))

and, by taking the derivatives at t = 0, we get

d�(X + Y ) = d�(X) + d�(Y ):



Furthermore

� (exp(tAd(x)Y )) = �(x)�(exp tY )�(x�1):

By taking the derivatives at t = 0, we get

d�(Ad(g)Y ) = �(x)d�(Y )�(x�1):

Then put x = exp sX and take the derivatives at s = 0,

d�([X;Y ]) = d�(X)d�(Y )� d�(Y )d�(X):

The adjoint representation � = Ad of G on G is a special case for which thederived representation is the adjoint representation ad of G on G. If �1 and�2 are two equivalent representations, then the derived representations d�1and d�2 are also equivalent. The converse holds if G is connected. The kernel

of a representation of a Lie algebra is an ideal. The center of a Lie algebra

G, denoted by Z(G); is de�ned as

Z(G) = fX 2 Gj8Y 2 G; [X; Y ] = 0g:

It is an Abelian ideal. It is the kernel of the adjoint representation.

Remarque 2.0.8 One can show that every �nite dimensional Lie algebra

admits a faithful (i.e. injective) �nite dimensional representation. This is the

theorem of Ado. Hence every �nite dimensional Lie algebra can be seen as a

subalgebra of gl(N;|) =M(N; |); for some N:Let G and H be two linear Lie groups and ' a continuous morphism of

G into H. One puts, for X 2 G = Lie(G);

d'(X) =d

dt'(exp tX) �t=0 :

From what we have seen, d' is a Lie algebra morphism from G into h =Lie(H): Observe that d' is the di¤erential of ' at the identity element I of

G :

d' = (D')I:



Proposition 2.0.9 (i) The Lie algebra of the kernel of the morphism ' is

equal to the kernel of d' :

Lie(ker') = ker(d'):

Therefore the kernel of ' is discrete if and only if d' is injective.

(ii) If d' is surjective, then the image of ' contains the identity com-

ponent H0 of H.

(iii) If G and H are connected and if d' is an isomorphism, then (G;')

is a covering of H.

Let us recall the de�nition of a covering. Let X and Y be two connected

topological spaces and ' : X �! Y a continuous map. The pair (X;') is

called a covering of Y if ' is surjective and if, for every x 2 X, there exist

neighbourhoods V of x and W of y = '(x) such that the restriction of ' to

V is a homeomorphism from V onto W . Let (X;') be a covering of Y ; if,

for y0 2 Y , the pullback '�1(y0) � X is a �nite set, then the same holds

for every y 2 Y , and the pullbacks '�1(y) all have the same number of

elements. Let k be that number. Then one says that (X;') is a covering of

order k of Y (or a covering with k sheets).

Preuve. (a) From Theorem it follows that, for X 2 G; t 2 R;

'(exp tX) = exp(td'(X))

Hence

Lie(ker') = ker(d'):

In particular, d' is injective if and only if the Lie algebra of ker ' reduces

to f0g, that is if ker' is discrete.(b) Recall that G0 denotes the identity component of G. Assume that d'

is surjective. This means that the di¤erential of ' at the identity element e of

G is surjective. Then V = '(G0) is a neighbourhood of the identity element

e of H. We saw that

H0 =1[k=1

V k



(Proposition). Since ' is a group morphism, V = '(G0) is a subgroup of H,

and V k = V; hence H0 = '(G0):

(c) Assume that G and H are connected and that d' is an isomorphism.

Let us show that (G;') is a covering of H. From (ii) it follows that ' is

surjective. By using Theorem and the relation

'(expX) = exp(d'(X)) (X 2 g);

one can show that there is a neighbourhood V � G of the identity element

of G, and a neighbourhood W � H of the identity element of H such that

' is an isomorphism from V onto W: It follows that, for every g 2 G; ' is ahomeomorphism of the neighbourhood gV of g onto the neighbourhood hW

of h = '(g) since

'(gv) = h'(v) (v 2 V ):

If ker' is a �nite group, then (G;') is a covering of order k of H; where k

is the number of elements in ker':

Exemple 2.0.10 Let V be the vector space of 2 � 2 Hermitian matrices

with zero trace. Such a matrix can be written x =

x1 x2 + ix3

x2 � ix3 �x1

!(x1; x2; x3 2 R):Then V � R3: For g 2 G = SU(2) the transformation

x �! gxg�1 = gxg�;

is a linear map �(g) from V onto V . From the relation

detx = �x21 � x22 � x23;

it follows that the transformation �(g) is orthogonal. Then one gets a mor-

phism ' from SU(2) into O(3). For T 2 su(2); d�(T )x = Tx� xT: If,

T =

iu v + iw

�v + iw �iv

!



one can establish easily that the matrix of d�(T ) is0BB@0 2v 2w

�2v 0 �2u�2w 2u 0

1CCATherefore d' is a bijection from su(2) onto LieO(3) = Asym(3;R): Thegroup SU(2) is connected. It follows that the group '(G) is the identity com-

ponent of O(3); that is SO(3): The kernel of ' is discrete. In fact one can

check that ker' = fI;�Ig. This establishes that SO(3) ' SU(2)=f�Ig; andthat (SU(2); ') is a covering of order two of SO(3):

2.1 Nilpotent and solvable Lie algebras

Let us recall some de�nitions and notation in group theory. Let G be a

group. If feg andG are the only normal subgroups,G is said to be simple. IfGis commutative, every subgroup is normal. The commutator of two elements

x and y of G is de�ned as

[x; y] = x�1y�1xy:

The derived group D(G) is the subgroup of G which is generated by the com-

mutators. If H is a normal subgroup, then G=H is a group. It is commutative

if and only if H contains the derived group D(G): One de�nes the successive

derived groups : D0(G) = G and Di+1(G) = D(Di(G)): The group G is said

to be solvable if there exists an integer n � 0 such that Dn(G) = feg: (Theterminology comes from the fact that, in Galois theory, such groups make it

possible to characterise polynomial equations which are solvable by radicals.)

Let g be a �nite dimensional Lie algebra over | = R or C. If A and B are twovector subspaces of g, then [A;B] denotes the vector subspace of g generated

by the brackets[X; Y ] with X 2 A and Y 2 B: One puts D(g) = [g; g]: Thisis an ideal of g which is called the derived ideal. The descending central series

Ck(G) is de�ned recursively by :



C1(G) = G; Ck(G) = [Ck�1(G);G]:

It is also denoted by Ck(G) = Gk: Observe that C2(G) = D(G): The derivedseries is de�ned by

D1(G) = D(G); Dk(G) = D(Dk�1(G)) =�Dk�1(G); Dk�1(G)

�It is also denoted by Dk(G) = G(k): The subspaces Ck(g) and Dk(G) (k =1; 2; :::) are ideals. The sequence Ck(G) is decreasing, hence constant for klarge enough. The Lie algebra g is said to be nilpotent if there exists an

integer n � 1 such that Cn(G) = f0g: Similarly the sequence Dk(G) isdecreasing, hence constant for k large enough. The Lie algebra G is said tobe solvable if there exists n � 1 such that Dn(G) = f0g: Observe that anilpotent Lie algebra is solvable. A subalgebra of a nilpotent Lie algebra is

nilpotent. A subalgebra of a solvable Lie algebra is solvable. Let X be an

element in a nilpotent Lie algebra G; then adX is a nilpotent endomorphism.

(Recall that an endomorphism T is said to be nilpotent if there exists an

integer k � 1 such that T k = 0.) In fact adX maps Ck(G) into Ck+1(G) and,if Cn(G) = f0g; then (adX)n�1 = 0:

Exemple 2.1.1 (1) Let G be the group `ax + b0; that is the group of a¢ netransformations of R. The Lie algebra G = Lie(G) has dimension 2. It has

a basis fX1; X2g satisfying

[X1; X2] = X1:

Hence D(G) = RX1; C3(G) = C2(G) = RX1; D

2(G) = f0g: Therefore G issolvable, but not nilpotent.

(2) The Heisenberg Lie algebra G of dimension 3 has a basis fX1; X2; X3gsatisfying

[X1; X2] = X3; [X1; X3] = 0; [X2; X3] = 0:

Hence C2(G) = |X3; which is the centre of G; and C3(G) = f0g: ThereforeG is nilpotent.



(3) Let G = sl(2;|) be the Lie algebra of the group SL(2;|): It has a basisfX1; X2; X3g satisfying

[X1; X2] = 2X2; [X1; X3] = �2X3; [X2; X3] = X1

. Hence D(G) = G: Therefore G is neither nilpotent, nor solvable.(4) Let T0(n; |) be the group of upper triangular matrices with diagonal

entries equal to one. Its Lie algebra g = t0(n;|) consists of the upper trian-gular matrices with zero diagonal entries

t0(n; |) = fx 2M(n;|)j xij = 0 if i � jg:

For 1 � k � n � 1; Ck(G) = fx 2 gj xij = 0if i � j � k + 1g: In particularCn(G) = f0g, and g is nilpotent. This is the basic example of a nilpotent Liealgebra.

(5) Let T (n;|) be the group of upper triangular matrices with non-zerodiagonal entries. Its Lie algebra g = t(n;|) consists of the upper triangularmatrices,

t(n;|) = fx 2M(n; |)jxij = 0 if i > jg:

We have

C2(G) = C3(G) = ::: = t0(n; |);

Dk(G) = fx 2M(n;|)jxij = 0 if i > j � 2k�1g:

Hence

Dk(g) = f0g if 2k�1 � n� 1:

Therefore g is solvable, but is not nilpotent. This is the basic example of a

solvable Lie algebra. Let g be a Lie algebra and � a representation of g on a

�nite dimensional vector space V: The representation � is said to be nilpotent

if, for every X of g, the endomorphism �(X) is nilpotent.

Lemme 2.1.2 If X is a nilpotent endomorphism acting on a vector space

V; then adX is nilpotent.



Preuve. Let k � 1 be such that Xk = 0: We have

(adX)N = (LX �RX)N =

NXj=0

(�1)N�jCNj LXjRXN�j :

Hence, if N � 2k � 1; then (adX)N = 0.

Théorème 2.1.3 Let � be a nilpotent representation of a Lie algebra G on avector space V: There exists a vector v0 = 0 in V such that, for every X 2 G;�(X)v0 = 0:

Preuve. Let ker(�) be the kernel of �. It is an ideal of G: It is enough toprove the statement for the representation � of the quotient algebra G= ker(�):This representation is faithful (i.e. injective). Hence we may assume that Gis a subalgebra of gl(V ): We have to show the following statement : if Gis a Lie subalgebra of gl(V ) made of nilpotent endomorphisms, then there

exists v0 = 0 in V such that, for every X 2 G; Xv0 = 0: The statement

will be proved recursively with respect to the dimension of G: If dimG = 1;then G = |X; and X is nilpotent. Hence there exists v0 = 0 in V such

that Xv0 = 0: Assume that the property holds for every Lie algebra with

dimension � n� 1:(a) Let G be a subalgebra of dimension n of gl(V ) made of nilpotent

endomorphisms, and let h be a proper subalgebra of G with maximal dimen-sion.We will show that h is an ideal of dimension n� 1: Let us consider therepresentation � of h on W = G=h de�ned by

�(X) : Y + h �! [X; Y ] + h:

By Lemma it follows that the representation � is nilpotent.Bythe recursion

assumption it follows that there exists w0 = 0 in W such that, for every X

in h,

�(X)w0 = 0:

Let X0 2 G be a representative of w0: Then X0 does not belong to h and

[X0; h] � h. Hence |X0 + h is a subalgebra of g whose dimension is greater



than that of h, therefore G = |X0 + h, and dimh = n � 1: Furthermore[G; h] � h; and this means that h is an ideal.

(b) Let us use for a second time the recursion assumption : there exists

v1 = 0 in V such that, for every X in h;

Xv1 = 0:

Put

V0 = fv 2 V j8X 2 h;Xv = 0g:

Since v1 2 V0; V0 = f0g: The subspace V0 is invariant under G: In fact letX 2 G; v 2 V0; and show that Xv 2 V0: For

Y 2 h; Y Xv = XY v � [X; Y ]v = 0:

In particular, X0V0 � V0: Since X0 is nilpotent there exists in V0 a vector

v0 = 0 such that X0v0 = 0; and then, for every X in G; Xv0 = 0:

Théorème 2.1.4 Let � be a nilpotent representation of a Lie algebra G ona vector space V: There exists a basis of V such that, for every X in G; thematrix of �(X) is upper triangular with zero diagonal entries.

Preuve. Let us prove the statement recursively with respect to the dimension

of V . By Theorem there exists a vector v1 such that, for every X 2 g;

�(X)v1 = 0:

From the recursion assumption applied to the quotient W = V=Kv1 we get

the result.

Corollaire 2.1.5 (Engel�s Theorem) A Lie algebra is nilpotent if and only

if, for every X 2 G; adX is a nilpotent endomorphism

Preuve. (a) Assume that the Lie algebra g is nilpotent : there exists an

integer n such that Cn(G) = f0g: For every X in G; adX maps Ck(G) intoCk+1(G); hence (adX)n�1 = 0: (b) Assume that, for every X in G; adX is



nilpotent. By Theorem the Lie algebra adG is isomorphic to a subalgebraof t0(N;|); hence ad G is nilpotent. There exists an integer n such that

Cn(adG) = f0g; hence Cn(G) � Z(G); the centre of G: Therefore Cn+1(G) =f0g: Let I be an ideal of G: If g is solvable, then g=I is solvable too. In fact,

Dk(g=I) ' Dk(G)=I \Dk(G)

Proposition 2.1.6 If I and G=I are solvable, then G is solvable.

Preuve. There is an integerm such that Dm(G=I) = f0g; henceDm(g) � I:

There exists n such that Dn(I) = f0g: Therefore Dm+n(G) = f0g:

Proposition 2.1.7 If I1 and I2 are two solvable ideals then the ideal I1+ I2is also solvable.

Preuve. The Lie algebra (I1+I2)=I2 is isomorphic to I1=I1\I2: This followsfrom the preceding proposition. Hence, if G is �nite dimensional, there existsa largest solvable ideal : the sum of all solvable ideals. It is called the radical

of G, and is denoted by rad(G):

Théorème 2.1.8 (Lie�s Theorem) Let G be a solvable Lie algebra over C,and let � be a representation of G on a �nite dimensional complex vectorspace V: There exists a vector v0 = 0 in V; and a linear form � on G suchthat, for every X in G; �(X)v0 = �(X)v0:

Preuve.We will prove the statement recursively with respect to the dimen-

sion of G: If dimG = 1; then g = CX0; and �(X0) has an eigenvector. Assume

that the property holds for every solvable Lie algebra of dimension � n� 1:Let g be a solvable Lie algebra of dimension n; and let h be a subspace of Gof dimension n� 1 containing D(G): Such a subspace exists since, because Gis solvable, D(G) = G: The subspace h is an ideal because

[G; h] � [G;G] = D(G) � h:



By the recursion assumption there is a vector w0 = 0 in V and a linear form

� on h such that, for every Y in h;

�(Y )w0 = �(Y )w0:

Let X0 2 gnh; and put

wj = �(X0)jw0; j � 1:

Let k be the largest integer for which the vectors w0; :::; wk are linearly

independent, and let Wj be the subspace which is generated by w0; :::; wj(0 � j � k):Observe that wj 2 Wk for j � k: Hence �(X0) maps Wk into

Wk and, for 0 � j < k; Wj into Wj+1: We will show that, for Y 2 h; the

restriction of �(Y ) to Wk is equal to �(Y )I: In a �rst step we will show that

the matrix of �(Y ) with respect to the basis fw0; :::; wkg is upper triangularwith diagonal entries equal to �(Y ): Let us show recursively with respect to

j(0 � j � k) that

�(Y )wj = �(Y )wj

mod Wj�1: (One puts W�1 = f0g:) This holds clearly forj = 0: Assume thatit holds for j < k: Then, for Y 2 h;

�(Y )wj+1 = �(Y )�(X0)wj = �(X0)�(Y )wj + �([Y;X0])wj;

and, since [Y;X0] 2 h;

�([Y;X0])wj = �([Y;X0])wj

mod Wj�1 by the recursion assumption. Hence

�(Y )wj+1 = �(Y )wj+1

mod Wj: This shows that the subspace Wk is invariant under the represen-

tation �: For Y 2 h;Tr(�([Y;X0])Wk) = 0:

On the other hand, for Z 2 h;

Tr(�(Z) �Wk) = (k + 1)�(Z):



Hence, if Z = [Y;X0]; then �(Z) = 0: In a second step we will showthat, for

Y 2 h; and w 2 Wk; �(Y )w = �(Y )w: Let us show recursively with respect

to j(0 � j � k) that, for Y 2 h;

�(X)wj = �(Y )wj:

This holds for j = 0: Assume that �(Y )wj = �(Y )wj: Then

�(Y )wj+1 = �(X0)�(Y )wj + �([Y;X0])wj

= �(Y )�(X0)wj + �([Y;X0])wj = �(Y )wj+1:

Let v0 2 Wk be an eigenvector of �(X0);

�(X0)v0 = �v0;

and extend the linear form � to g by putting

�(X0) = �:

Then, for every X in g;

�(X)v0 = �(X)v0:

Corollaire 2.1.9 Let G be a solvable Lie algebra over C, and � be a repre-sentation of G on a �nite dimensional complex vector space V: There exists abasis of V such that, for every X in G; the matrix of �(X) is upper triangu-lar. The diagonal entries can be written �1(X); :::; �m(X); where �1; :::; �m;

are linear forms on G.

The statements of Theorem and Corollary do not hold if G is a solvableLie algebra over R. (In fact one knows that, if A is an endomorphism of a

�nite dimensional real vector space, in general there is no basis with respect

to which the matrix of A is upper triangular.) See Exercise.



2.2 Semi-simple Lie algebras.

A Lie algebra is said to be simple if it has no non-trivial ideal, and if it

is not commutative. In other words a Lie algebra is simple if its dimension

is greater than 1 and if the adjoint representation ad is irreducible. If G issimple then [G;G] = G; because [G;G] is an ideal. The Lie algebra sl(n; |) issimple (n � 2): Let us show that sl(2;C) is simple. (For n � 3, see Exercise.) For that consider the following basis of sl(2;C) :

H =

1 0

0 �1

!; E =

0 1

0 0

!; F =

0 0

1 0

!The commutation relations are :

[H;E] = 2E; [H;F ] = �2F; [E;F ] = H:

Let I be an ideal of sl(2;C) which does not reduce to f0g: If one of theelements H;E, or F belongs to J; then J = sl(2;C): The basis elementsH;E; F are eigenvectors of ad H for the eigenvalues 0; 2;�2; and J is invariantunder adH; hence one of the eigenvectors belongs to J . A Lie algebra G is saidto be semi-simple if the only commutative ideal is f0g: A simple Lie algebra issemi-simple. There is no semi-simple Lie agebra of dimension 1 or 2. But there

exist semi-simple Lie algebras of dimension 3 ; in fact sl(2;C); sl(2;R) andsu(2) are semi-simple Lie algebras. The centre of a semi-simple Lie algebra

reduces to f0g: Hence, if G is semisimple, then the adjoint representation isfaithful, ad(G) ' G: A direct sum of semi-simple Lie algebras is semi-simple.Let � be a representation of a Lie algebra G on a �nite dimensional vectorspace V . For X; Y 2 G one puts

B�(X; Y ) = Tr(�(X)�(Y ))

. This is a symmetric bilinear form on G which is associative :

B�([X; Y ]; Z) = B�(X; [Y; Z]):

This means that the transformations adX are skewsymmetric with respect

to the form B�. The orthogonal of an ideal with respect to the form B� is



an ideal. The Killing form is the symmetric bilinear form associated to the

adjoint representation (� = ad) :

B(X; Y ) = Tr(adXadY ):

Exemple 2.2.1 (1) Let G =M(n;|):

B(X; Y ) = 2nTr(XY )� 2TrXTrY:

In order to establish this formula let us consider the canonical basis fEijg of

G =M(n; |). IfX =nX

i;j=1

xijEij 2 G; then ad:Ekl = [X;Ek;l] =nX

i;j=1

(xikEil � xliEki)

(k;= 1; :::; n): Hence, if Y =nX

i;j=1

yijEij then

(adX � adY )Ekl =nX

i;j=1

(xikyjiEij + xliyijEkl)�nX

i;j=1

(xikylj + xljyij)Ekl)

Therefore

Tr(adXadY ) = nnX

i;j=1

(xijyij + xjiyij)� 2nX

i;j=1

(xiiyjj)

= 2nTr(XY )� 2TrXTrY:

(2) Let g = sl(n; |): If n � 2; B(X; Y ) = 2nTr(XY ): (This follows from

Proposition 4.3.1 below.)

(3) Let g = so(n;|): If n � 2;

B(X;Y ) = (n� 2)Tr(XY ):

The proof is left as an exercise.

Proposition 2.2.2 Let J be an ideal in a Lie algebra G: The Killing formof the Lie algebra J is the restriction to J of the Killing form of G:

Preuve.



Théorème 2.2.3 Let X; Y 2 J: The endomorphisms S = adX; T = adY

map G into J . Let us consider a basis of G obtained by completing a ba-sis of J . With respect to this basis the matrices of S and T have the follo-

wing shape Mat(S) =

S1 �0 0

!; Mat(T ) =

T1 �0 0

!and Mat(ST ) =

S1T1 �0 0

!Therefore Tr(ST ) = Tr(S1T1): We will see that a Lie algebra

is semi-simple if and only if the Killing form is non-degenerate. To prove

this we will use Cartan�s criterion for solvable Lie algebras. We will need the

properties of the decomposition of an endomorphism into semi-simple and

nilpotent parts. Let V be a �nite dimensional vector space over C. Recall that

an endomorphism T of V decomposes as

T = Ts + Tn;

where Ts is semi-simple (i.e. diagonalisable), and Tn is nilpotent, in such a

way that Ts and Tn are polynomials in T . The endomorphisms Ts and Tncommute. This decomposition is unique in the following sense : if

T = D +N;

with D semi-simple, N nilpotent, and DN = ND; then D = Ts; N = Tn: Ts

is called the semi-simple part of T; and Tn the nilpotent part We have

adT = adTs + adTn;

adTs is semi-simple, adTn is nilpotent (Lemma ). In order to show that adTsand adTn are the semi-simple and nilpotent parts of ad T it is enough to show

that adTs and adTn commute. But

[ad(Ts); ad(Tn)] = ad[Ts; Tn] = 0:

It follows that ad Ts and ad Tn are polynomials in ad T .



Théorème 2.2.4 (Cartan�s criterion) Let G be a Lie subalgebra ofM(m;C).Assume that Tr(XY ) = 0 for every X;Y 2 G: Then G is solvable.

Preuve. We will show that every X 2 [G;G] is a nilpotent endomorphism.(a) Let X = Xs + Xn be the decomposition of X 2 G into semi-simple

and nilpotent parts. We may assume that

Xs =

0BB@�1

:

�m

1CCA ;

the numbers �j being the eigenvalues of X. Let p be a polynomial in one

variable, and put

U = p(Xs) =

0BB@p(�1)

:

p(�m)

1CCA ;

Since Xs is a polynomial in X; one can write U = p0(X); and since Xn is

also a polynomial in X; U and Xn commute. Then

(UXn)k = UkXk

n

hence UXn is nilpotent, therefore Tr(UXn) = 0; or Tr(UXs) = Tr(UX):

(b) The eigenvalues of ad Xs are the numbers �i � �j , and the corres-

ponding eigenvectors are the matrices Eij,

adXsEij = (�i � �j)Eij:

Let us now choose a polynomial p in one variable with complex coe¢ cients

such that p(�i) = �i (i = 1; :::;m): Hence, if �i � �j = �k � �l, then

p(�i)� p(�j) = p(�k)� p(�l):

Therefore there exists a polynomial P such that, if U = p(Xs); then adU =

P (adXs) and, since adXs is a polynomial in adX; there exists a polynomial

P0 such that adU = P0(adX): Therefore adU(G) � G:



(c) Let us now take X 2 [g; g]; and show that

Tr(UX) = 0:

We can write X =nX;j=1

[Yj; Zj] ; Yj; Zj 2 G; and then

Tr(UX) =

nX;j=1

Tr(U [Yj; Zj]) =

nX;j=1

Tr([U; Yj]Zj) = 0

by assumption, since [U; Yj] = adUYj 2 G: But, by (a)

Tr(UX) = Tr(UXs);

hence

Tr(UX) =nX;j=1

p(�j)�j =nX;j=1

j�jj2 :

Therefore the eigenvalues �j of X vanish, and X is nilpotent. By Engel�s

Theorem (Corollary) it follows that [G;G] is nilpotent, hence G is solvable

Corollaire 2.2.5 If the Killing form of G vanishes identically, then G is

solvable.

Théorème 2.2.6 Let G be a Lie algebra. The following properties are equi-valent :

(i) G is semi-simple,(ii) the radical of g reduces to f0g;(iii) the Killing form of g is non-degenerate.

Preuve. (i) ) (ii). Assume that there exists a solvable ideal I 6= f0g ing. Let Dk�1(I) be the last non-zero derived ideal of I: Then Dk�1(I) is a

non-zero commutative ideal in G; and this contradicts (i).(ii))(iii). Put I = G?,

I = fX 2 Gj8Y 2 G; B(X; Y ) = 0g



This is an ideal and the restriction ofB to I vanishes identically. By Corollary,

I is a solvable ideal, and I = f0g by (ii).(iii) ) (i). Let I be a commutative ideal in G. For X 2 I; Y 2 G; the

endomorphism adXadY maps g into I; and (adXadY )2 maps g into [I; I] =

f0g; hence adXadY is nilpotent. Therefore

B(X; Y ) = Tr(adXadY ) = 0:

Since B is non-degenerate it follows that I = f0g:

Proposition 2.2.7 A semi-simple Lie algebra G is a direct sum of simple

subalgebras. Furthermore, [G;G] = G:

Preuve. Let I be an ideal of G; and let I? be its orthogonal complementwith respect to the Killing form,

I? = fX 2 Gj8Y 2 I; B(X;Y ) = 0g:

Since the Killing form is associative it follows that I? is an ideal and, by Co-

rollary, that the ideal I\I? is solvable, hence reduces to f0g since the radicalof G reduces to f0g . Therefore G = I � I?. To get the stated decompositionone starts from a minimal non-zero ideal I1 in g, which is necessarily simple,

then one obtains recursively a decomposition

G = I1 � I2 � �� Im;

where I1; :::; Im are simple ideals. It follows furthermore that

[G;G] =mMi=1

[Ii; Ii] =

mMi=1

Ii = G:

From this theorem it follows that, if I is a solvable ideal in G; then I =rad(g) if and only if G=I is semi-simple. Finally let us state without proofthe theorem of Levi�Malcev. Let G be a Lie algebra. A Levi subalgebra ofG is a Lie subalgebra which is a complement to rad(G). It is a semi-simplealgebra since it is isomorphic to G=rad(G): The theorem of Levi�Malcev says



that, in every Lie algebra G; there is a Levi subalgebra s. Therefore every Liealgebra decomposes as

G = s+ rad(g);

the sum of a semi-simple Lie algebra, and a solvable Lie algebra. This is the

so-called Levi decomposition.

Exemple 2.2.8 Let G be the Lie subalgebra of M(n+1;R) consisting of thematrices

x y

0 0

!(x 2 so(n); y 2 Rn)

(g is isomorphic to the Lie algebra of the motion group of Rn). Let I be theideal of g consisting of the matrices

0 y

0 0

!; (y 2 Rn)

It is Abelian, hence solvable. Let s be the subalgebra of g consisting of the

matrices x 0

0 0

!; x 2 so(n)

If

n � 3;

the subalgebra s, which is isomorphic to so(n), is semi-simple (even simple

for n = 4). Therefore, since G=I ' s; Iis the radical of G; and G = s + I is

a Levi decomposition of G.


Chapitre 3

Lie groups

3.1 De�nitions and some proprieties.

The topological space locally looks like Euclidean space via suitable choice

of coordinates,

ui : Rn ! R is called a coordinate function.

(a1; ::::an) 7! ai

Let M be a topological space and p 2M

Dé�nition 3.1.1 An m-dimensional chart at p 2 M is a pair (U;') where

U is an open neighborhood of p and ' a homomorphism of U onto an open

set in Rn .

The coordinates of the chart (U;') is 'i = uio'

Rn ui! R"U

% 'i U is called a coordinate neighborhood and the pair (U;') a

coordinate system, n is dimension of the chart (U;') :

Dé�nition 3.1.2 An m-dimensional topological manifold is a Hausdor¤ space

with a countable basis such that 8p 2 M;there exist an m-dimensional chart

at p.

Remarque 3.1.3 We can �nd a covering of M by open sets and each open

set U in the covering is homeomorphic to the open m-ball Bn = fa 2 Rn; kak � 1g

35

CHAPITRE 3. LIE GROUPS

A set A of charts of an m-dimensional manifold M is called a C1�atlasif A satisti�es the following conditions.a) 8p 2M; 9 (U;') 2 A and p2 U i.e M = [

(U;')2AU

b) (U;') and (V; ) 2 A then U \ V = ? or the map 'o �1 and o'�1

are of class C1

We say that (U;') and (V; ) are compatible.

Dé�nition 3.1.4 Let A be a C1�atlas on an m-dimensional manifold M ,then the chart (U;') is admissible to A or compatible with A if (U;') is

compatible with every chart in A.

Given any atlas A, one can adjoin all chart which are admissible to Aand obtain a collection

�A which is again an atlas on M .

�A is maximal relative to properties a) and b).

An m-dimensional topological manifold M has C1�di¤erentiable struc-ture or just a C1�structure if one give M a maximal C1�atlas.A di¤erentable manifold of class C1 or just C1�manifold is an m-

dimensional topological manifoldM to which is assigned a C1�di¤erentiablestructure.

Remarque 3.1.5 One obtain di¤erentable manifold of class C1 or real ana-

lytic manifold if the change of coordinate 'o �1 and o'�1 is of class C1

or analytic.

Basic Structures

Dé�nition 3.1.6 A Lie group is a set such that :

(a) G is a group ;

(b) G is an analytic maniforld ;

(c) the group multiplication in (a) of the product manifold � : G�G �!G : (x; y) 7�! xy and the group inversion operation in (a) ; � : G �! G :

x 7�! x�1 are analytic functions relative to the structure in (b).



The dimension of a Lie group is the common dimension of all connec-

ted components of G.

Every Lie group is a topological group where the topology comes from its

analytic structure because a manifold is a separate space and the map (x, y)

xy 1iscontinuous:

On the other hand, every Lie group is locally compact (a locally Euclidean

manifold), locally connected, metrizable and complete.

A connected Lie group is countable at in�nity and therefore separable.

1) The question of when a topological group is a Lie group is the �fth

problem of Hilbert (posited in 1900).

The solution was given in 1956 by Montgomery and Zippin.

A locally Euclidean topological group is isomorphic to a Lie group. An

interesting class of topological groups which are not Lie groups is the class

of topological groups in in�nite dimension which intervenes in quantum and

classical physics. For example, the abelian group of Gauge transformations

in electrodynamics, is not a Lie group because non locally Euclidean.

Let G be a locally compact group, and let F be the family of compact

compact subgroups k of G such that the quotientG / k is a Lie group. Bruhat

calls such a subgroup of G "a good subgroup".

The family F ordered by inclusion is decreasing �ltering.

In fact, if k1 and k2 2 F , the group k1=k1\k2 (' G=k1 � k1=k1\k2) is an ex-

tension of the Lie group G=k1by the subgroup k1=k1\k1 which is topologically

isomorphic k1k2=k2 because ki sont compacts.

Since k1k2=k2 is a Lie group, as a compact subgroup of the Lie group G=k2 .

.

We deduce that G=k1\k2 is a Lie group, so k1 \ k2 is a good subgroup.Suppose \fkg

k2F = feg. This is the case, according to Montgomery andZippin, if the quotient G=G0 of G by the connected connected component

G0 is compact. The group G is then canonically isomorphic to the projective

limit of the Lie groups G=k for k 2 F . If, moreover, G is metrizable, there



exists a decreasing sequence (kn) of good subgroups such that \kn = feg.The group G is then canonically isomorphic to the projective limit of G=kn.

1) The additive groups R and the groups of matricesGL (n;R) ; S0 (n;R),etc. are Lie groups.

2) If G is a discrete topological group, the neutral element e has an open

neighborhood {e} homeomorphic to R = {0}. A discrete topological group

can be considered as a Lie group of dimension 0 and vice versa.

3) All closed subgroups of GL (n, R) are Lie groups. For example, the

linear special group SL (n), othogonal groups (non-degenerate quadratic

forms), symplectic groups (orthogonal group of non-degenerate alternating

bilinear forms) and standard unipotent groups U_ {n}, ...

The map f induces an injective immersion ef : G1=H �! G2 which is

homomorphism Lie group .

G1f�! G2

p #G1=H %ef

Let G be a connected Lie group. The underlying manifold admits a simply

connected covering f : M G

Soit ' :M �M �! G

(m1; m2) �! f (m1) f (m2) � 8m1; m2 2M:

SinceM�M is simply connected, it is raised to a continuous mappinge' : M �M �! M such that e' (e; e) = e et f � e' = ' where e is chosen in

M such that f (e) = 1.

Me'% # f

M �M'�! G

where f is C1; F continuous and i an immersion .



For all p 2 P , there exist an neigborhood U of p, an neigborhood V of

i (p) 2 N and a map g of class C1 such that g � i = id=U . Since f = i � F ,we have

F = id � F = g � (i � F ) = g � f . Then F is C1 as a composition of C1

functions.

In particular, let put F = f and i = id we have the result.

Let G be a Lie group and H a submanifold of G which is also a subgroup

of G. If H is a topological group (the topology is deduced from the analytic

structure). Then H is a Lie subgroup

Since f is a covering, the kernel of f is necessarily a discrete invariant

subgroup of M. Then there exists on M a Lie group structure for which f

is a Lie group morphism. Thus, any connected Lie group G has a simply

connected cluster. In other words, there exists a simply connected Lie group

M and a surjective (discrete kernel) morphism f : M! G and therefore G is

identi�ed with the Lie group quotient M / D.

It is called the universal covering of G and is denoted by G.

1) The simply connected abelian Lie groups are the vector groups.

2) The Heisenberg groups, the unipotent groups,

The group ax + b are simply connected Lie groups ; In each case, the

underlying topological space is an R ^{m}, where m is the dimension of the

group.

Let G be a Lie group. A subset H G is an analytic subgroup of G if

1) H is a subgroup of G

2) H is an analytic submanifold of G

Any analytic subgroup H of a Lie group G is a Lie group.

In fact :



The map f induces uniquely an injective immersion ef : G1=H �! G2

which is a morphisme of Lie group.

G1f�! G2

p #G1=H %ef

Where f is C1, F is continuous and i is an immersion (P is a submanifold).

For any p P, there exists a neighborhood U of p, a neighborhood V of i

(p) N and an application g of class C1 such that g i = id / {U}.

Since H is a topological group, the map fH : H � H �! H is

continuous. fH is analytic then H is a Lie group.

Let G be a Lie group and G0 the identity component of G. Then

a) G0 is an open invariant Lie subgroup of G

b) - TeG = TeG0 and dimG = dimG0

c) G=G0 is a discret Lie group.

? If G is connected Lie group and H a proper Lie subgroup then

dimH < dimG:

? We will show later that if G is a Lie group and if H is closed

subgroup of G ; then H is a Lie subgroup of G.

Hence O (n) ; SL (n) and Sp (n) are closed Lie subgroup of GL (n; R).

3.2 Lie Subgroups

We shall di�ne the concept of a Lie subgroup of a Lie group G and

note that this concept di¤ers from a topological subgroup because a Lie

subgroup need not have the induced topolo

Dé�nition 3.2.1 Let G be a Lie group and let H be a Lie group. Then

H is a Lie subgroup of G if H is an analytic submanifod of G and if H is a

subgroup of G.



Exemple 3.2.2 (1) The torus T 2 is a Lie group when regarded as a

product group T 1�T 1 and has as a submanifold the �irrational wrap around�curve as discussed in section 2.3 (see Fig. 4.1). This curve is given by f(t) =

(exp 2�iat; exp 2�ibt) where a=b = � is irrational. As we saw, this curve is

a one-dimensional submanifold which is dense in T 2 and since f(s + t) =

f(s)f(t) in T 2, it is therefore a Lie subgoup which is not closed. This Lie

subgroup does not have the induced topology since there points on the curve

which are arbitrarily close in the topology induced from T 2 but are arbitrarily

far apart in the topology of the curve.

(2) The integrs Z are a zero-dimensional Lie subgroup of R.We shall now give various criterion for a subgroup to be a Lie sub-

group and we need the following result [Helgason, 1962, P. 78] :

Lemme 3.2.3 LetM an N be C1 (analytic) manifolds and let f :M �!N a C1 (analytic) mapping such that f(M) is contained in a submanifold

P . If the map f :M �! P is continous, then this map is also C1 (analytic).

PROOF We shall schow this as follows. Let the accompanying diagram

be commutative when f is C1, F continous and i an immersion (since P

is a submanifold). Then by the remark following Proposition 2.23, for each

p 2 P , there is neighborhood U of p and a neighborhood V of i(p) 2 N, anda C1-map g : V �! U so that g � i = identity j U . Thus, since f = i � F ,we have locally that

F = identity � F = g � (i � F ) = g � fand since the rigth side is a composition of C1-functions, F is C1. In par-

ticular, let F = f and i be the identity ; that is, letting P be a submanifold,

we obtain the result.

Proposition 3.2.4 Let G be a Lie group and let H be a submanifold of

G which is also an abstract subgroup of G. If H is also a topological (relative

to the topology induced from its analytic structure), then H is a Lie subgroup

of G.



PROOF It su¢ ces to schow H is a Lie group and this follows from the

preceding lemma. The mappingf : G�G �! G

(x; y) 7�! xy�1

is analytic and its restriction fH : H � H �! G is continous. Thus by

lemma 4.9, fH is analytic so that H is a Lie group.

The next result follows from previous facts.

Proposition 3.2.5 Let G be a Lie group and let H be a connected

topological subgroup of G. Then there is at most one analytic structure A(H)

on H which makes H into a Lie subgroup of G.

We now give some computional results which determine Lie subgroups.

Proposition 3.2.6 Let H be a Lie group which is an abstract subgroup of

the Lie group G. Assume at the identity e 2 G there exist an analytic chart

(U; x) in G and an analytic chart (V; y) at e in H such that xi j H = yi and

(@yi@yj

(e)) has rank equal to the dimension of H. Then H is a Lie subgroup of

G.

PROOF We �rst translate the Charts at e to any point a by the analytic

di¤eomorphism L(a) (of H and G) so that we can now apply corollary 2.12

to obtainH is a submanifold.

This result can also stated in term of local Lie groups which generate a

subgroup.

Corollaire 3.2.7 Let G be a Lie group and let B be a local Lie group

relative to the group operations in G. If there exist charts (U; x) at e in G

and (V; y) at e in B such that xi j B = yi and rank (@yi@yj

(e)) = dimB, then

the subgroup H generated by B is a connected Lie subgroup of G.

We now note that the topological and manifold structure is mostly in

the identity component. This will also become more evident when the Lie

algebras are also taken into consideration.



Proposition 3.2.8 Let G be a Lie group and G0 be the identity component

of G (as a topological group). Then :

(a) G0 is an open normal Lie subgroup of G ;

(b) T (G; e) = T (G0; e), and therefore dimG = dimG0 ;

(c) G=G0 is a Lie group which is discrete.

PROOF Since G is localy Euclidian, it has a connected open neigh-

borhood U of e in G and from Proposition 3.23, U generates a connected

subgroup H of G. Since G0 is the identity component, we have by maxima-

lity that H � G. However H contains the neighborhood e of G0 and G0 is

connected. Therefore G0 = H. Now G0 contains the neighborhood U of e

and U is open in G so that any a 2 G0 is in the open neighborhood aU .

Thus G0 is open in G. This means G0 is an open submanifold of G so that

dimG = dimG0 and T (G; e) = T (G0; e). Also by Theorem 3.22, G=G0 is

discrete.

Remarque 3.2.9 (1) If G is a connected Lie group and H a proper Lie

subgroup, them dimH < dimG for otherewise H contains an open nucleus

of G which generates G ; that is, G = H.

(2) We shall show later that if G is a Lie group and if H is a closed

subgroup of G, then H is a Lie subgroup of G. Thus the previously discussed

subgoups O(n), SL(n) and Sp(n) are all closed Lie subgroups of GL(n;R).(3) We shall consider later normal Lie subgroups when we discuss

homomorphisms.

3.3 The Lie Algebra of a Lie group

We have seen from previous examples that the tangent space T (G; e) of

a Lie group G can be used to give local informations about G. In this chap-

ter, we formalize this situation by introducing the G-invariant vector �elds

L(G) and seeing that it is a vector space which is isomorphic to T (G; e).Also L(G) is a Lie algebra over R and induces a Lie algebra structure on



T (G; e). Using this we de�ne the exponential map exp : L(G) �! G in

terms of homomorphisms of R into G. The exponential map is a local di¤eo-morphism exp : U0 �! Ue of a suitable neighborhood U0 of 0 in L(G) ontoa neighborhood Ue of e in G. Using the inverse function log : Ue �! U0

we de�ne canonical coordinates (Ue; log) at e in G. Thus by the action

L(a) : G �! G : x 7�! ax, we obtain coordinates at any point a 2 G.

The exponential map is used to obtain a local representation of the mul-

tiplication in G analogous to the results of the Section 1.6. Thus for X and

Y su¢ ciently near 0 in L(G) we can write expX expY = expF (X; Y ) whereF : L(G)�L(G) �! L(G) is analytic at (0; 0) 2 L(G)�L(G). We show thatthe terms F (k)(0; 0)(X; Y )(k) of the Taylors series for F are in the subalgebra

of L(G) generated by X and Y . We brie�y discuss the actual formula for

F (X;Y ) which is known as the Campbel-Hausdor¤ formula. Finaly we show

that a continous homomorphism of a Lie group is analytic. This yields the

fact that the analytic structure of a Lie group is uniquely determined by its

topology.

We now introduce the Lie algebra of a Lie group G in terms of invariant

vector �elds. Thus the Lie algebra will be determined by the tangent space

T (G; e) and the action of G detemines the values of the vector �elds at any

other point in G.

Dé�nition 3.3.1 An analytic vector �eld X 2 D(G) de�ned on a Lie

group G is called invariant if for all a 2 G we have [(TL(a))(e)]X(e) = X(a).

Thus as in Section 2.7 we have since (TL(a)(e)) : T (G; e) �! T (G; a),

then the value [(TL(a))(e)]X(e) actually equals X(a):

Next we note that if X is invariant, then X is L(a)-invariant for all

a 2 G ; that is X is actually G-invariant or left invariant according to

Section 2.7. For let p 2 G, then



X(L(a)p) = X(ap) = [(TL(ap)(e)]X(e)

= [T (L(a) � L(p)(e)]X(e)= [TL(a)(p)] � (TL(p)(e))(X(e))= TL(a)(p) �X(p)

which gives the results.

Proposition 3.3.2 Let G be a Lie group, let X 2 T (G; e) and let

eX : G �! T (G) : p 7�! eX(p),where T (G) is the tangent bundle of G with projection map � and eX(p)

is given by f(Xf)(p) = X(f � L(p))where f is any real-valued analytic function onG. Then eX is aG-invariant

analytic vector �eld on G such that eX(e) = X. Furtheremore eX is the unique

G-invariant vector �eld on G such that eX(e) = X. Thus any G-invariant

vector �eld on G is of the form eX.PROOF Letting TL(p) = TL(p)(e) we �rst note that ( eXf)(p) =

(TL(p)X)(f) so that eX(p) 2 T (G; p) and therefore (� � eX)(p) = p. ThuseX is a vector �eld on G. Since eX(p) = TL(p)X, we have eX(e) = X and

the above computations show eX is G-invariant. For the uniqueness, we use

Proposition 2.34 with f(a) = L(a) for any a 2 G or directly as follows. Let Zbe a G-invariant vector �eld with Z(e) = X. Then Z(p) = TL(p)(e)Z(e) =

TL(p)(e)X = eX(p). Finally we shall show eX is analytic and derive an other

formula for it. Thus let � : I �! G : t 7�! �(t) be an analytic curve on an

interval I containing 0 2 R so that ��(0) = X

h= eX(e)i and �(0) = e. Then

analogous to the results in section 2.7 we use the sesults on curves in section

2.5 to obtain( eXf)(p) = X(f � L(p))

=d

dt(0)(f � L(p) � �) = d

dt[f(p�(t)]t=0 (�)



where p�(t) is the analytic product on G. Thus since f ,� and the multi-

plication on G are analytic we have eXf is an analytic function ; that is, eXis analytic.

Let L(G) denote the set of G-invariant vector �elds on G. Then from theabove result we see that L(G) consists of all vector �elds of the form eX for

X 2 T (G; e).From eX(p) = TL(p)(e) and TL(p)(e) being injective we obtain the follo-

wing.

Corollaire 3.3.3 The map � : L(G) �! T (G; e) : eX 7�! X is a vector

space isomorphism. In particular the dimension of L(G) over R equals the

dimension of G and is �nite.

Corollaire 3.3.4 L(G) is a Lie algebra relative the bracket operationh eX; eY i= eX eY�eY eX.Soient eX et eY deux champs de vecteurs invariants.

Si eX et eY 2 L (G) alorsh eX; eY i 2 L (G).

Preuve :

Il su¢ t de montrer queh eX; eY i est invariant à gauche.

Soient a2 G et ' 2 A (a). On a

[dL (a)]h eX; eY i

e' =

h eX; eY ie(' � La)

= eXe

� eY (' � La)�� eYe � eX (' � La)�= eXe

��eY '� � La�� eYe �� eX'� � La�= dL (a) eXe

�eY '�� dL (a) eYe (X')= Xa

�eY '�� eYa � eX'� = h eX; eY ia':

d�où dLah eX; eY i

e= [X; Y ]a :

L�ensemble L (G) des champs de vecteurs invariants à gauche estdonc une sous-algèbre de Lie de l�agèbre des champs de vecteurs C1. L (G)



est appelé algèbre de Lie de G. On la note G.

TeG est aussi muni d�une structure d�algèbre de Lie en posant pour

tous

X; Y 2 Te (G) ; [X; Y ] =h eX; eY i

e

Par conséquent l�application � : L (G) �! TeGeX �! X

est un isomorphisme d�algèbre de Lie.

L�algèbre de Lie TeG est aussi appelé l�algèbre de Lie de G.

Exemple 3.3.5 :

Dans les paragraphes suivants, nous allons voir comment déterminer l�al-

gèbre de Lie des groupes de Lie linéaires.

Soit G = GL (n;R) :L�application � : L (G) �! TI (G) est un isomorphisme d�algèbre de Lie où

le produit est [X; Y ] =h eX; eY i

g(I) dans TI (G). On montre que L (G) est

isomorphe à gl (n;R) :

Dé�nition 3.3.6 (a) The Lie algebra of a Lie group G is the Lie algebra

L(G) of G-invariant vector �elds of G.

(b) The Lie algebra G with product [ ]G is homomorphic to the Lie algebra

h with product [ ]h if there is a vector space homomorphism � : G �! h

such that � [X; Y ]G = [�X; �Y ]h for all X, Y 2 G. If � is a vector space

isomorphism, then G and h are isomorphic Lie algebras.

By means of corollary 5.4 we can make T (G; e) into a Lie algebra as

follows. Let X,Y 2 T (G; e) and eX, eY as above. Then de�ne the product

[XY ] =h eX; eY i (e) which is in T (G; e) and makes t(G,e) into a Lie alge-

bra. This yields the following.

Corollaire 3.3.7 The map � : L(G) �! T (G; e) : eX 7�! X is a Lie

algebra isomorphism.



Frequently the lie algebra T (G; e) is also called the �Lie algebra of G�.

Exemple 3.3.8 (1) Let G = GL(V ) Then from corollary 5.6 we have the

map � : L(G) �! T (G; I) is a Lie algebra isomorphism using the product

[XY ] =h eX; eY i (I) in T (G; I). However, we have the Lie algebra gl(V ) atta-

ched to G and we now show that L(G) is isomorphic to gl(V ) as Lie algebras.Recall from example (3), Section 2.5 that for each A 2 gl(V ) we de�ned anelement A 2 T (G; I) by

(Ah) = [Dh(I)]A,

h analytic at I. The map gl(V ) �! T (G; I) : A 7�! A is a vector space

isomorphism. Thus we obtain a vector �eld eA in L(G) and consequently avector space isomorphism gl(V ) �! T (G; I) : A 7�! eA. We now show thisis a Lie algebra isomorphism ; that is

heA; eBi = [̂A;B]. Usually T (G; I) andgl(V ) are considered the same and the overbar is omited as done before,

but we shall not do this now. Let p 2 G and A, B 2 gl(V ): Then using

L(p)A = pA, the product in End(V ), we have for f analytic in G



(eA eB) = eA(eB(f))(p)= A(eB(f) � L(p), definition of X

=hD(eB(f) � L(p))(I)iA, definition of A

=hD(eB(f)(p) �D(L(p))(I)iA, chain rule

=hD(eB(f)(p)i (pA), L(p) linear

= limt�!0

1t

h(eBf)(p+ tpA)� (eBf)(p)i

= limt�!0

1t

�B(f � L(p+ tpA)�B(f � L(p)

�= lim

t�!01tf[D(f � L(p+ tpA))(I)]B � [D(f � L(p))(I)]Bg

= limt�!0

1tf[Df(p+ tpA)] (pB + tpAB)� [Df(p)] pBg , chain rule

= limt�!0

1t[Df(p+ tpA)(pB)�Df(p)(pB)] + lim

t�!01t[Df(p+ tpA)] (tpAB)

= D2f(p)(pB; pA) +Df(p)(pAB)

Interchangin A and B, subtracting the equations and using the fact that

D2f(p) is symmetric we obtain



(eA eB)(f)(p) = Df(p)(pAB � pA)

= Df(p)([A;B])

= [D(f � L(p))(I)] [A;B]

= [A;B](f � L(p))

= [̂A;B](f)(p)which proves the result.


Chapitre 4

Solvable Lie Groups and

Algebras

We now start the structural development of Lie groups and algebras.

First we de�ne a Lie group to be solvable if it is solvable as an abstract goup.

Then the "derivative" of these results we discuss solvable Lie algebras. Thus

we show that a connected Lie group is solvable if and only if its Lie algebra

is solvable.Finillly we discuss Lie�s theorem which involves �nding a common

characteristic vector for a solvable Lie algebra of endomorphisms acting on a

complex vector space. This eventually yields that the matrices representing

a solvable Lie algebra of endomorphisms acting on a complex vector space

can be put into triangular form by using a suitable basis of the vector space.

Once again, all �elds in this chapter will be assumed to be of characteristic

zero.

4.1 Solvable Lie group

Let G be an abstract group and let A and B be subgroups of G. Then

we have the following notation.

(1) We denote by (A;B) the subgroup of G generated by all elements

xyx�1y�1 for x 2 A, y 2 B.

51

CHAPITRE 4. SOLVABLE LIE GROUPS AND ALGEBRAS

(2) If A is a normal subgroup of G, we writr A C G or G B A.

Note that if A and B are nomal subgroups of G, then (A;B) is a normal

subgroup of G.

Dé�nition 4.1.1 Let G(1) = (G;G) and de�ne by induction G(k+1) =

(G(k); G(k)).Then we have the sequence of normal subgroups

G B G1 B G2 B ::: .

Thus G is solvable if this sequence is and terminates at feg ; that is thereexists n so that G(n) = feg and G is called solvable of length n.

From results of Lang [1965] we have the following theorem :

Théorème 4.1.2 Let G be an abstract goup. Then the following are

equivalent.

(a) The group G is solvable.

(b) There is a �nite sequence of subgroups G = G0 B G1 B G2::: B Gn =

feg such that Gk=Gk+1 is commutative for k = 0; 1; :::; n� 1.PROOF First we observe that by induction each G(k) is a normal

subgroup of G. Now assume (a). Then note from the de�nition of (G;G) that

G=G(1) is commutative and by induction and de�nition, G(i)=G(i+1) is also

commutative. Thus we have (b) by taking Gk = G(k). Conversly, assume we

have a descending sequence G = G0 B G1 B G2::: B Gn = feg with Gi=Gi+1commutative. Then G=G1 being commutative implies xyx�1y�1G1 = eG1

which yields G1 � G(1). Now assume Gk � G(k) ; Then since Gk=Gk+1 is

commutative we see

Gk+1 � (Gk)(1) � (G(k); G(k)) = G(k+1).

However, since Gn = feg we see G(n) = feg which gives (a).

Corollaire 4.1.3 (a) A subgroup H of a solvable group is solvable.

(b) If G is a solvable group of length n and H a normal subgroupn, then

G=H is a sovable group of length less than or equal to n.



(c) If G is a group and H is a normal solvable subgroup of length n such

that G=H is solvable of length m, then G is solvable less than or equal to

n+m.

PROOF (a) We just note that by induction H(k) � G(k).

(b) Let G = G=H.Then by induction we see that (G)(k) = G(k) using

� : G �! G=H : x 7�! x = xH is a homomorphism. Thus the series for G

yields the series G B G1 B G

2 B ::: B G(n)= feHg.

(c) Note that from the series G B G1 B G

2 B ::: B G(n)= feHg we

obtain the series G B G1 B G2 B ::: B G(n) and G(n) � H. However, since

H is solvable we have H B H1 B H2 B ::: B feg and we put these two seriestogether to see that G is solvable.

Dé�nition 4.1.4 Let G be a Lie group. Then G is a solvable Lie group

if G is solvable as an abstract group.

Théorème 4.1.5 Let G be a Lie group. Then the folowing are equiva-

lent.

(a) The Lie group is solvable.

(b) There exists a �nite sequence G = G0 B G1 B G2::: B Gn = fegsuch that each Gk is a closed subgroup of G with Gk=Gk+1 commutative for

k = 0; 1; :::; n� 1.(c) There exists a �nite sequence G = G0 B G1 B G2::: B Gr = feg such

that for k = 0; 1; :::; r� 1, we have Gk=Gk+1 is a connected one dimensionalgroup or a discrete group.

PROOF To show (a) implies (b), we recall that if H is a normal sub-

group of G, then its closure H is a closed normal subgroup of G. Next assume

G is solvable so we obtain the series G B G1 B G2 B ::: B G(m) = feg and letG0 = G and Gk = Gk. Then Gk are closed normal subgroups (and, therefore,

Lie groups) such that G0 B G1 B G2::: B Gm = feg. Furthermore Gk=Gk+1is a Lie group whitch is commutative, for let � : G �! G=Gk+1. Then since

�(G(k)) is commutative, we have �(G(k)) is commutative. However, since �



is continuous �(Gk) = �(G(k)) � �(G(k)), so that Gk=Gk+1 = �(Gk) is com-

mutative.

The converse (b) implies (a) is clear. Also (c) implies (b) is clear, so it

remains to show (b) implies (c). Thus le Hk be the connected component

of the commutative Lie group Gk=Gk+1. Then by the result outlined in the

exercice (1), section 6.5, we have for the Gk in (b) Gk=Gk+1 = Hk�Dk where

Dk is a discret group, and Hk�= Rq(k) � T p(k) by theorem 6.20.

Next let � : G �! G=Gk+1. Then in the series for (b) we replace each of

the terms Gk by the series [with p = p(k); q = q(k)]

��1(Rp�T q�Dk) B ��1(Rp�1�T q�Dk) B ::: B ��1(T q�Dk) B ��1(T q�1�Dk) B ::: B ��1(Dk)

Then we obtain the series in (c)

4.2 Solvable Lie Algebras and Radicals

Let G be a �nite dimensional Lie algebra over a �eld | and let h, k be

subspaces of G. Then we shall use the following notation.

(1) We denote by [h,k] the subspace of G generated by all products [x,y]for x 2 h and y 2 k. In particular G(1) = [G,G] is a subalgebra of G.(2) If h is an ideal of G, then we write G Bh or h C G. In paricular, note

G B G(1). A Lie algebra G is abelian or commutative if G(1) = f0g.

Dé�nition 4.2.1 Let G be a fnite-dimensional Lie algebra over |, setG(1) = [GG], and de�ne by induction G(k+1) =

�G 0k);G(k)

�.

From the Jacobi identity for G we obtain G B G(1) B G(2) B ::: and we call

G solvable if there exists n with G(n) = f0g. The smallest such n is calledthe length of the solvable algebra G.

Théorème 4.2.2 Let G be a �nite-dimensional Lie algebra over |.Then the folowing are equivalent.



(a) The algebra G is solvable.(b) There exists a sequence G = G0 B G1 B G2::: B Gr = f0g so that the

quotient algebra Gk=Gk+1 is commutative. Each Gk can be taken to be an idealin G.(c) There exists a �nite sequence of subalgebras G = G0 B G1 B G2::: B

Gs = f0g such that dimGk=Gk+1 = 1. In general Gk is not an ideal in G butonly in Gk�1.PROOF The equivalence of (a) an (b) is similar to those for groups

in section 10.1. Thus for example, if G is solvable, then take Gk = G(k) fork = 1; :::; r to obtain the sequence in (b) and also note

�G 0k)G(k)

�= G(k+1) so

the desireted quotient algebra is commutative.

Next assume (c) where we have Gk=Gk+1 = |X = |X+Gk+1 since Gk=Gk+1is one dimensional. Then, since

�X X

�= 0 we have Gk=Gk+1 is a commutative

Lie algebra. Thus (c) implies (b). Conversly, if the sequence in (b) is such

that Gk=Gk+1 = |X1 + ::: + |Xr + Gk+1 is commutative, then each suspacehi = |X1 + ::: + |Xi + Gk+1 is an ideal in Gk=Gk+1 for i = 1; :::; r. Thus

the corresponding subspaces hi generated by fX1; :::; Xig [ Gk+1 where Xi +

Gk+1 = Xi is an ideal in Gk. Thus we obtain a sequence Gk = hr B hr�1 B::: B h1 B Gk+1, so that the quoteint ideals are one dimensional and thisyields (c).

The proof of the following is similar to corollary 10.3.

Corollaire 4.2.3 Let G be a Lie algebra containig the Lie algebra h.

(a) If G is solvable, then h is solvable:(b) If G is solvable and h an ideal of G, then G=h is solvable of length

less than or equal to the length of G.(c) If h is a solvable ideal of G such that G=h is solvable, then G is

solvable.

Théorème 4.2.4 Let G be a Lie group with Lie algebra G.

(a) If G is solvable, then G is solvable.



(b) If G is connected and G is solvable, then G is solvable.

PROOF (a) If G is a solvable lie gruop, then we have a sequence

G = G0 B G1 B G2::: B Gn = feg with each GK is a closed normal Lie sub-group so that GK=GK+1 is commutative. Then we obtain the corresponding

sequence G = G0 B G1 B G2::: B Gr = f0g of ideals of G so that Gk=Gk+1 is acommutative Lie algebra.

(b) If G is connected and G is solvable, then we shall show G is solvable byinduction on the length of G. Thus let G B G(1) B ::: B G(n�1) B G(n) = f0gbe the sequence for G and let K be the Lie subgroup of G generated by

exp(G(n�1)). Then K is a commutative normal subgroup of G (since G(n�1) isa commutative ideal of G) and its closueK = H is also a commutative normal

Lie subgroup ofG. Now let h be the Lie algebra ofH. Then h is a commutative

lie algebra of G and G(n�1) � h. From this we have [exercicev (2) above],

G=h �= (G=G(n�1))=(h=G(n�1)) and since G=G(n�1) is solvable of length lessthan or equal to n � 1 we have G=h is solvable of length less than or equalto n� 1.(corollary 10.8). Thus by the induction hypotheses G=H is solvable

and since H is solvable, we have G is solvable using corollary 10.3.

Lemme 4.2.5 Let G be a �nite-dimensional Lie algebra over |. Thenthere exists a unique maximal sovable ideal of G/ namely the sum of the

solvable ideals of G. This solvable maximal ideal is called the radical of Gand is denoted by r. Moreever G=r is

�0or contains no proper solvable

ideals ; that is the radical of G=r is�0.

PROOF Let h and k be solvable ideals of G. Then the vector subspaceh + k is an ideal of G. Now by the above exercicise (2) we see (h + k)=k �=h=(h\ k) and since h\ k � h is solvable we have h=(h[ k) is solvable ; Thguswe have (h+ k)=k is solvable and k is solvable so that by corollary 10.8 h+ k

is solvable Thus since G is �nite dimensional, the solvable ideal of maximimdimension is unique and by the above, contains every solvable ideal of G ;denote this maximal solvable ideal by r.

Next let h = h/r be solvable ideal of G = G=r where h is some ideal



of G with h � r. Then since h/r is solvable and r is solvable we have by

corollary 10.8 that h is solvable. Thus h � r, so that h =�0.

Dé�nition 4.2.6 Let G be a Lie group with lie algebra G and let r bethe radical of G. Then we de�ne the radical of G, R = rad G, to be the

connected Lie subgroup of G whose Lie algebra is r = rad G.

Proposition 4.2.7 Let G be a Lie group with radical R. Then R is

closed and R is the maximal solvable normal connected Lie subgroup of G.

PROOF Let R denote the closure of R. Then R is a normal, solvable

Lie subgroup (since it is closed) ; Thus its Lie algebra r is solvable (Theorem

10.9) so that r = r and consequently R = R ; that is R is closed, normal,

solvable Lie subgroup of G. The fact that R is maximal among connected

Lie subgroups with those proprerties also uses the maximally of r.

Corollaire 4.2.8 The radical of G=R = feRg.

Dé�nition 4.2.9 (a) A �nite-dimensional Lie algebra is called semi-

simple if it has no proper solvable ideal. Thus G is semi-simple if r = f0g.Similary a Lie group G is semisimple if its radical R = feg.

(b) A Lie group G is simple if its Lie algebra G is simple. That is[GG] 6= f0g G has no proper ideals.

We shall eventually show that a semisimple Lie algebra over a �eld of

characteristic 0 is a direct sum of simple lie algebras which are ideals. Conse-

quently many problems involving semisimple Lie groups can be done in terms

of simple Lie algebras.

4.3 Lie�s Theorem on Solvability

We now describe how a solvable Lie group or Lie algebra of endomor-

phisms can be represented by triangular matricies. To do this we compute

characteristic foots so we consider real Lie groups or algebras acting on com-

plex vector spaces.



Dé�nition 4.3.1 (a) Let | be a �eld of characteristic 0 and let V be

a �nite-dimensional vector space over |. Let T 2 End|(V ) and � 2 |. ThenV� = fX 2 V : TX = �Xg and V (�) = fX 2 V : (T � �I)nX = 0 for some n 2 Ngwhere N is the set of natural numbers (which are greater than 0). If V� 6= f0gthen � is called a characteristic value or eigenvalue of T and 0 6= X 2 V�is called an eigen vector or characteristic vector of T with characterisc-

tic value �. If V (�) 6= f0g, then � is called a weight of T and V (�) a weightspace and 0 6= X 2 V (�) is called a weight vector of T

A characteristic value or a weight � of T is a solution of the equation

det(Ix � T ) = 0 and if the solutions of this (characteristic) equation are in

|, then we say that the characteristic values or weights are in | ; recallthe de�nition of a split endomorphism in section 9.2.

(b) Let N � End|(V ), let f : N �! | be a function, and setVf = fX 2 V : for all T 2 N; TX = f(T )Xg andV (f) = fX 2 V : for all T 2 N; there exits nm 0 with (T � f(T )I)n = 0g.

If Vf 6= f0g, then f is called characteristic function on N and 0 6= X 2Vf is called a characteistic vector of N for the characteristic fucnction f .

Similary one de�nes a weght function, weight space and weight vector

in case V (f) 6= f0g.Thus these functions on N asign to each T in N a characteristic root

f(T ) of T . Of course, in actual computations, the characteristic roots dis-

cussed above might be in the algebraic closure of |.with these de�nitions and results on canonical forms of endomorphisms

[Jacobson,1953,Vol.II ;Lang,1965] we state the following :

Proposition 4.3.2 Let V be a �nite-dimensional vector space over K

and let T 2 End|(V ) have its (distinct) weights �1; �2; :::; �m in |. Then theweight spaces V (�i) are T -invariant and V = V (�1) + V (�2) + ::: + V (�m)

(direct sum).

Remarque 4.3.3 This direct sum decomposition will be generalized in

the next chapter to a direct sum decompostion of weight spaces of a nilpotent



Lie group or Lie algebra.

Proposition 4.3.4 Let V be a �nite-dimensional vector space over R,let G be a Lie subgroup of GL(V ), and let f : G �! R be a characteristicfunction with ; f(G) � R� = R� f0g. Then regarding R� as a multiplicativeLie group, the map f : G �! R� is an analytic homomorphism of Lie groups.f is frequently called a character of G.

PROOF Let S,T 2 G. Then for 0 6= X 2 Vf we have SX = f(S)X

and T (X) = f(T )X. Thus STX = Sf(T )X = f(T )SX = f(T )f(S)X.

However since (ST )X = f(ST )X this gives f(ST ) = f(S)f(T ) so that

f : G �! R� is a homomorphism. To see that f is analytic, let X1; X2; :::; Xm

be a basis of V so X1 is a charateristic vector of G for the characteristic

function f . Noting the mappings r : G �! V : S 7�! S(X1) and s : V �!R :

nPi=1

�iXi 7�! �1 are analytic, so is the map f = s � r : G �! R�.

Analogous to the lemma 7.15 we are the following result :

Lemme 4.3.5 Let V be a �nite-dimensional vector space over R andlet G be a real connected solvable Lie group which is a subgroup of GL(V )

and has real Lie algebra. Let W be a subspace of V .

(a) W is invariant under the action of G if and only if W is invariant

under the action of G.(b) For A 2 G, the vector X 2 V is a characteristic vector of A with

characteristic value � if and only if X is characteristic vector of the group

fexp tA : t 2 Rg for the characteristic function f : exp tA 7�! et�.

The following result or some of its equivalent conseqences is known as "

Lie�s theorem on sovability". We follow the work of Tits [1965] for the group

proof.

Théorème 4.3.6 (Lie�s theorem). Le V be �nie-dimensioal vector space

over C and let G be a real connected solvable Lie group which is a subgroup

of GL(V;C). Then there exists a nonzero characteristic vector of G for somecharacteristic function.



PROOF We shall proof the results by induction on the dimension of

G.

First, if G is one dimensional and 0 6= A 2 G which is the Lie algebra ofG, then since G � gl(V ) we see that A has a nonzero characteristic vector

X 2 V . However by lemma 10.18 and exercise (2), X is also a characteristic

vector of G. Net assume G is of dimension n and assume as an induction

hypothesis that we have shone the result for all such groups of smaller di-

mension. Now since G is connected and solvable, G has a connected solvable

normal subgroup H of dimension n� 1 [Exercice (5), section 10.2]. Thus bythe induction hypothesis we can conclude there is a characteristic function

f : H �! C� = C � f0g and analogous as Proposition 10.17 we have f iscontinous.

We shall now show that the subspace Vf = fX 2 V : SX = f(S)X for all S 2 Hgis invariant under G. Thus let X 2 Vf , S 2 H and T 2 G. Then

S(T (X)) = (ST )(X) = T (T�1ST )(X) = f(T�1ST )T (X) (?)

using T�1ST 2 H. Thus the number f(T�1ST ) is a characteristic valueof S with characteristic vector T (X) and also the fonction k : G �! C� :T 7�! f(T�1ST ) is continous ; However, since G is connected and the set of

characteristic values of S is discrete, the image k(G) consists of a single point.

( This uses the characterization : the topological spaceM is connected if and

only ifM is mapped continously into a discrete space implies the image ofM

consists on a single point.) Thus we have k(T ) = k(I) = f(S) Using this in

(?) we have for any X 2 Vf , T 2 G, and S 2 H that S(T (X)) = f(S)T (X)

which shows by the de�nition of Vf that Vf is invariant under the action of

G.

Next by lemma 10.8 we have that Vf is invariant under the action

of G � gl(V ). Therefore if A 2 G and A =2 h which is the Lie algebra of H,then, since the subspace Vf is invariant under the linear transformation A,

there exists a characteristic vector 0 6= Vf for A. Thus since G=h = RA + h(using the hypothesis that dimH is n� 1) we see that X is a characteristic

vector for G. For let B = aA + bC 2 G with C 2 h and for AX = �X 2 Vf



and CX = �X [Using lemma 10.8 (b) applied to h and H] we have BX =

(aA+bC)X = (a�+b�)X. Thus by lemma 10.18, X is a characteristic vector

of G.

Dé�nition 4.3.7 Let V be an m�dimensional vector space other the�eld |. Then a sequence of subspaces f0g � V1 � V2 � ::: � Vm = V such

that dimVi = i for i = 1; ::;m is called a �ag of V . Let G � GL(V ) be a Lie

group. Then the �ag is G-invariant if for every T 2 G we have T (Vi) � Vi

for i = 1; :::;m. Similarly for a Lie algebra G of endomorphisms, we de�ne

a G-invariant �ag.

Proposition 4.3.8 Let V be an n-dimensional vector space over C andlet G be a real connected Lie group which is a subgroup of GL(V;C). Thenthe following are equivalent.

(a) The group G is solvable.

(b) There exists a �ag which is G-invariant.

(c) There exists a �ag of V such that the matrices for the elements

in G can be put simultaneously into triangular form. (The matrices might

have compex entries).

PROOF Assume G is solvable. Then to show (b) we use induction on

the dimension of V . From Lie�s theorem there is a one-dimensional subspace

W of V which is invariant under G. Therefore an element T 2 G induces a

nonsingular linear map

T : V=W �! V=W : x+W 7�! Tx+W and the map

G �! GL(V=W;C) : T 7�! T is an analytic homomorphism. Thus

the image G =�T 2 GL(V=W;C) : T 2 G

is a real connected solvable Lie

group which is a subgroup of GL(V=W;C) and by the induction hypoyhesisthere exists a �ag in V=W which is invariant under G

�0� V 2 � V 3 �

::: � V = V=W . Now let � : V �! V=W , let Vi = ��1(V i), and set

V1 = W = ��1(�0). Then dimVi = i and fOg � V1 � V2 � ::: � Vn = V is

a �ag which is invariant under G.



Next to show (b) implies (c) we choose a basis of V from the cor-

responding �ag as follows. Let V1 = fX1g. Then since TV1 � V1 for T 2 G

we have TX1 = a11(T )X1. Next let V2 = fX1; X2g where X1 and X2 are

independent using dim(V2=V1) = 1. Then since TV2 � V2 we have TX2 =

a12(T )X1+a22(T )X2 for all T 2 G. Continuing in this manner we can choosea basis of V so that any T 2 G has a matrix of the form266666666664

a11(T ) a12(T ) : : : a1n(T )

a22(T )

:

:

:

0 ann(T )

377777777775with 0 6= a11(T ):::ann(T ) = detT .

Finaly to show (c) implies (a) let G be represented by the group

triangular matrices as above ; Let G1 be the normal subgroup of triangular

matrces of the form 266666666664

1 �0 1

:

:

:

0 ::: 0 1

377777777775with 10s on the diagonal. Let G2 be the normal subgroup of G1 of the

form 266666666664

1 0 �1 0

1

:

: 0

0 1

377777777775Kangni Kinvi 62 Analyse Harmonique


with 10s on the diagonal and 00s on the superdiagonal. Continuing this

way we obtain the sequence G B G1 B G2::: B Gn = fIg with Gi=Gi+1commutative.

The preceding results on Lie groups can be translated into results on Lie

algebras via the exp mapping or directly as follows. This proof involves some

computations we shall see again in chapter 11.

Théorème 4.3.9 Let P be the algebraic closure of the �eld | and letV be a nonzero �nite-dimensional vector space of P . Let G be a solvable Liealgebra over | and let � be a homomorphism of G into gl(V; P ). Then thereexists a vector 0 6= X 2 V which is a characteristic vector for all the membersof �(G) for some characteristic funtion.

PROOF We prove this by induction on the dimensional of G. FordimG = 1;the theorem follows from the results on canonical form (Proposi-

tion 10.16). We are assume the results hold for all Lie algebras of dimension

less than dimG. From theorem 10.7 (d) we can �nd an ideal h in G so thatdimG=h = 1. By corollary 10 we have h is solvable so that by the inductionassumption there exists a characteristic function f : h �! P so that for all

S 2 h, '(S)X = f(S)X. From dimG=h = 1 we can �nd T 2 G so that T =2 h.Thus G = |T + h. Let W be the subspace of V spanned by all the vectors

X1 = X and Xk+1 = '(T )kX. for k = 1; 2:::. Now note W is �(T )-invariant

subspace of V .

We shall now show : For all S 2 h, '(S)X = f(S)X for all Y 2 W ;

that is, W is �(T )-invariant and furthermore �(S) = f(S)I on W .

We �rst prove by induction that for all S 2 h and k = 1; 2; :::

�(S)Xk = f(S)Xk+ak�1Xk�1+:::+a1XI (�)

where aj = aj(S) are in P . By the choice for X1 = X the result holds for

k = 1. Assuming (�) for k, we have



�(S)Xk+1 = �(S)�(T )Xk, de�nition of Xk+1

= �([ST ])Xk + �(T )�(S)Xk

= �([ST ])Xk + �(T )(f(S)Xk + ak�1Xk�1 + :::+ a1XI)

= f(S)�(T )Xk + bkXk + :::+ b1X1

= f(S)Xk+1 + bkXk + :::+ b1X1

using ST 2 h and the induction assumption.We next prove '(S)X = f(S)X for all Y 2 W . From (�) and the

de�nitiin of Xk we �rst observe that W is �(T )-invariant. Next note that

from the above, the restriction �(S) j W has matrix266666664

f(S) �:

:

:

0 f(S)

377777775so that tr(�(S) j W ) = f(S) dimW for S 2 h. Next note that �(S)

and �(T ) map W into W so that �([ST ]) = �(S)�(T ) � �(T )�(S) as en-

domorphisms of W . Since tr(AB)) = tr(BA) for endomorphisms, we see

0 = �([ST ]) = f([ST ]) dimW . Since dimW m 1 this gives f([ST ]) = 0. Thus

�(S)Xk+1 = �(S)�(T )Xk

= �([ST ])Xk + �(T )�(S)Xk

= f([ST ])Xk + f(S)�(T )Xk

= f(S)Xk+1



that is '(S)X = f(S)X for all Y 2 W .Since W is �(T )-invariant and P is algebraically closed we se that

�(T ) has a characteristic vector A 2 W : �(T )A = tA. Also �(S)A = f(S)A

for all S 2 h and since G = |T + h we have for any Z = aT + S that

�(Z)A = a�(T )A+ �(S)A = (at+ f(S))A. Thus A is a characteristic vector

of �(G) and F : aT +S 7�! at+f(S) de�nes the corresponding characteristic

funtion.

The formalities in the proof of propositioin 10.21 yield the following :

Proposition 4.3.10 Let P be the algebraic closure of the �eld | andlet V a nonzero �nite-dimensional vectoe space over P . Let G be a Lie algebraover | and let � be a homomorphism of G into gl(V; P ). Then the followingare equivalent.

(a) The Lie algebra �(G) is solvable.(b) There is a �ag in G which is invariant under �(G).(c) There is a basis of V such that the matrices for the endomorphisms in

�(G) can be put simultaneously in the triangular form. (The matrices mighthave entries from P .)

These results apply when we take the �el | to be algebraically closeditself. Thus | = P and we obtain the following :

Proposition 4.3.11 Let G be a Lie algebra over the albebraically closed�eld |. Then G is solvable if and only if thre is a �ag in G

f0g � G1 � G2 � ::: � Gn = G such that each Gi is an ideal of G.PROOF Assume G is solvable. Then since G �! ad(G) : X 7�! adX

is a homomorphism of Lie algebras over |, we see that ad(G) is a solvableLie algebra of endomorphisms acting on the vector space G. By proposition10.23 (b) there is a �ag f0g � G1 � G2 � ::: � Gn = G which is invariantunder ad(G) ; that is, each Gi is an ideal of G.



Conversely, assuming such a �ag exists we see that ad(G) is solvable ;using (b) implies (a) in proposition 10.23. However, ad : G �! ad(G) is ahomomorphism so that ad(G) u G= ker(ad). Since ker(ad) is the center of Gwhich is solvable and since G= ker(ad) is solvable, we have by corollary 10.8that G is solvable.


Chapitre 5

Nilpotent Lie groups and

algebras

We continue the concepts given in the preceding chapter and call a Lie

group nilpotent if it is nilpotent as an abstract group. Then we discuss nil-

potent Lie algebras and obtain the result that a connected Lie group is nil-

potent if and only if its Lie algebra is nilpotent. In the last section we consider

the vector space decomposition which yiels the Jordan canonical form for an

endomorphism and extend this compostion to a nilpotent group of automor-

phisms.

5.1 Nilpotent Lie Groups

We now give a variation of the results on solvable groups using some of

the notation of the preceding chapter.

Dé�nition 5.1.1 Let G be an abstract group.

(a) Let C0G = G and Cn+1G = (G;CnG). Then CnGB Cn+1G and we

have the descending central series G = C0GB C1GB C2G:::

67

CHAPITRE 5. NILPOTENT LIE GROUPS AND ALGEBRAS

(b) Let C0G = feg and let CnG = ��1(Z(G=Cn�1G) where Z(G=Cn�1G

is the center of G=Cn�1G noting by inductioin

CnG C Cn+1G C G and where � : G �! G=Cn�1G is the corresponding

projection map. Thus we have the ascending central series

feg = C0G C C1GC C2GC :::.

Théorème 5.1.2 Let G be an abstrct group. Then the following are

equivalent.

(a) There exists a series of normal subgroups of G

G = G0 BG1 B :::BGs = feg such that(G;Gn) � Gn+1 for n = 0; :::; s� 1.

(b) There exists a positive integer p such that

GB C1GB :::B CpG = feg.(c) There exists a psitive integer q such that

fegC C1GC :::C CqG = G.

PROOF Assume there is a series as given in (a). Then by induction

we have Gn � CnG. Thus CsG = feg. Conversely if (b) holds, then weautomatically have a series satisfaying (a).

Next we have (a) implies (c), for if we have a series as in (a), then

we shall show by induction Gs�n � CnG so that for n = s we obtain

CsG � G. Thus feg = Gs � C0G = feg and assume Gs�1 � CiG. Then

(G=CiG;Gs�i�1=CiG) � Gs�i=CiG � CiG=CiG = feg using the inductionhypothesis for the second inclusion ; that is, (G;Gs�i�1) � CiG. Thus if

� : G �! G=CiG is the projection, we se that

Gs�i�1 � ��1(Z(G=CiG)) = Ci+1G, using the de�nition of Ci+1G.

Conversely, to see (c) impliies (a), we �rst note that

(G;CiG)=Ci�1G � (G=Ci�1G;CiG=Ci�1G) = feg usingCiG = ��1(Z(G=Ci�1G)),

where � : G �! G=Ci�1G. Thus (G;CiG) � Ci�1G. So that for

CqG = G = G0; Cq�1G = G1; :::; Cm�1G = Gm;etc, we see that the series

in (c) yields the series in (a).



Dé�nition 5.1.3 An abstract group G is nilpotent if it satisfaties any

one of the conditions of theorem 11.2.

Remarks (1) Note that nilpotency involves a descending series com-

mutators of the terms of the series with the group, whereas solvability involves

descending series using commutators of the terms of the series with itself.

(2) Subgroups, quotient groups, and �nite direct products of nilpotent

groups are nilpotent. The proofs run as expected. For example, if Gi are

groups with CniG = fejg for i = 1; :::;m and if G = G1 � ::: � Gm, then

CnG = feg, where n = max fn1; :::; nmg.

Théorème 5.1.4 Let G be a Lie group. The following are equivalent.

(a) As an abstract group G is nilpotent.

(b) There is a series G = G0 B G1 B ::: B Gs = feg where each Gn is aclosed normal Lie subgroup of G and (G;Gn) � Gn+1.

(c) If C0G = G and C

n+1G = (G;C

nG), then there exists a positive

integer p such that GB C1GB :::B C

pG = feg.

PROOF Showing (c) if and if (b) is similar to theorem 11.2 ; (b) im-

plies (a) is also clear. Next assume (a). Then there is a series of normal

subgroups G B G1 B ::: B Gs = feg with (G;Gn) � Gn+1. Consequently we

obtain Gs = feg, using G as Hausdor¤, and G B G1 B ::: B Gs = feg with(G;Gn) � Gn+1 which proves (b).

Dé�nition 5.1.5 A Lie group G is a nilpotent Lie group if it is

nilpotent as an abstract group.



5.2 Nilpotent Lie Algebras

Let G be a Lie group with Lie algebra G. We shall now de�ne the notionof nilpotent Lie algebra so that if G is connected, then G is a nilpotent Lie

group if and only if G is a nilpotent Lie algebra.

Dé�nition 5.2.1 (a) Let G be an algebra over a �eld | and let

C0G = G ; Cn+1G = [G; CnG]. Thus we see thatC1G = [G;G], ..., CkG =(adG)k(G), ...

are ideals of G and we obtain the descending central series G = C0GBC1G B :::.

(b) Set C0G = f0g and Cn+1G = ��1(Z(G=CnG), where by induction CnG C Gand � : G �! G=CnG is the Lie algebra homomorphism and Z(G=CnG) thecenter of the Lie algebra G=CnG. Thus we see that C0G = f0g, C1G =Z(G),etc. are ideals of G and we obtain the ascending central series f0g =C0G CC1G C:::.

Théorème 5.2.2 Let G be a Lie algebra over |. Then the followindare equivalent.

(a) There exists a sequence G = G0BG1B :::BGs = f0g where all the Gnare ideals of G such that [G;Gn] = Gn+1.(b) There exists a positive integer p such that

G = C0G B C1G B :::B CpG = f0g.(c) There exists a positive integer q such that f0g = C0G CC1G C:::CCqG =C G.(d) There rexists an integer r such that for all X1; X2; :::; Xr 2 G we have

adX1 � adX2 � ::: � adXr = 0.

PROOF The equivalence (a)-(c) are similar to Theorem 11.2. To see

that (b) if and only (d) use the fact that CkG is generated by the elements(adX1 � adX2 � ::: � adXk)Y for any X1; X2; :::; Xr and Y 2 G ; see de�nition11.6.



Dé�nition 5.2.3 A Lie algebra G is called nilpotent if it sati�es anyone of the conditions of theorem 11.7

Remarks (1) Subalgebras, quotient algebras and �nite direct sums

of nilpotent Le algebras are also nilpotent.

(2) A nilpotent Lie algebra is solvable, for by induction we obtain

CnG � G(n+1):

Proposition 5.2.4 Let G be a Lie algebra over |. G is solvable if andonly if [G;G] is nilpotent.

PROOF Suppose [G;G] is nilpotent, then [G;G] is solvable. Also sinceG= [G;G] is a commutative Lie algebra, it is solvablle. Thus by corollary 10.8,G is solvable.Conversly, �rst let P be the algebraic closure of | and let G be a solvable

Lie algebra over P contained in gl(V; P ) where V is a �nite-dimensional

vector space over P . Then by the results following Lie�s theorem (Proposition

10.23) there exists a basis of V so that =the matrices of G have triangularform 266666664

a11 �a22

:

:

0 amm

377777775and consequently the matrices for elements of [G;G] have the form266666664

0 �0

:

:

0 0



Thus by exercice (2), [G;G] is a nilpotent Lie algebra.Next if G is an arbritrary Lie algebra over P , then ad(G) is solvable, and

therefore [ad(G); ] ad(G) = ad([G;G]) is nilpotent. However sincead : G �! ad(G) is a Lie algebra homomorphism with Ker(ad) = Z(G)

we see that G = G=Z(G) u ad(G). Therefore G(2) =�G;G

�u ad([G;G]) is

nilpotent. Consequently, there exists a positive integer p such that�0= Cp+1G(2) = Cp+1G(2)=Z(G). ThusCp+1G(2) � Z(G) so thatCp+1G(2) =

f0g ; that is, G(2) = [G;G] is nilpotent.Finally, if G is a Lie algebra over |, we let h = P | G be the tensor

product of the algebras P and G over | as in 9.1. Then h is a Lie algebraover P and a straightforward computation shows that G is solvable, thenh is solvable. Thus since [G;G] � [h,h] we use the results of the preceding

paragraph to conclude [G;G] is nilpotent.

Théorème 5.2.5 (Engel�s theorem) Let V be a nonzero �nite-dimensional

vector space over the �eld | and le G be a Lie subalgebra of gl(V ) whichconsists of nilpotent linear transformations (that is, An = 0 for some n).

Then there is a nonzero vector X 2 V such that for all A 2 G, we haveAX = 0.

PROOF First we shall show that A being a linear nilpotent transfor-

mation implies that adGA is a nilpotent linear transformation acting on G.Thus since gl(V ) = End(V ) as sets, we can de�ne the endomorphisms :

R(A) : gl(V ) �! gl(V )

Z 7�! ZAand

L(A) : gl(V ) �! gl(V )

Z 7�! AZ

and see (adA)Z = AZ � ZA = (L(A) � R(A))Z in gl(V ). Also noting

that L(A)R(A) = R(A)L(A) we have by the binomial theorem of any integer

k � 0, (adA)kZ = [L(A)�R(A)]k Z =kPi=0

(�1)i k

i

!Ak�iZAi. However,

since A 2 G is nilpotent, all the factors Ak�i or Ai are 0 for suitably larger kor i. Thus adGA is nilpotent.



Next we shall use induction on m = dimG to prove the result. For m = 1

we have G = |A where A is a nonzero nilpotent linear transformation. Thussince there exists X 2 V with AX = 0, the same results holds for every

B = bA 2 G. Now assume as an induction hypothesis that the result holdsfor Lie algebras of dimension less than m and let h be a proper subalgebra of

G of maximum dimension. Then by the results of the above paragraph adGAis a nilpotent endomorphism on G for all A 2 h. Thus since adA : h �! h we

see adA induces a nilpotent endomorphism A on the vector space G = G=h.Furtheremore the set h =

�A : A 2 h

is a subalgebra of gl(G) which consists

of nilpotent endomorphisms and dim h < m.

By the induction hypothesis with V = G we can conclude that thereexists B 6=

�0in G such that for all A 2 h we have AB = 0 ; that is, there

exists B 2 G with B =2 h and [h; B] � h. Thus the subspace h+ |B of G is asubalgebra which contains h. However, by the maximal choice of h we have

h+ |B = G.Finally let W = fZ 2 V : AZ = 0, for all A 2 hg. Then by the above

induction hypothesis W 6= f0g. Furtheremore for A 2 h and B 2 G as abovewe have, since [A;B] 2 h, that for any Z 2 W

A(BZ) = (AB)Z = [A;B]Z + (BA)Z:

Thus by the de�nition of W we obtain BW � W . However, since B

is nilpotent on V we have B is nilpotent on W . Consequently there exists

0 6= X 2 W with BX = 0 and since G = h + |B we see this X has the

desired property.

Corollaire 5.2.6 Let V be a �nite dimensional vector space over | andG � gl(V ) be a Lie algebra of nilpotents endomorphisms of V .

(a) There exists a basis of V such that the matices of the endomorphisms

in G relative to this basis have the form



266666664

0 �0

:

:

0 0

377777775(b) G is a nilpotent Lie algebra of endomorphisms.(c) The associative algebra G� generated by the endomorphisms of G is a

nilpotent associative algebra ; that is there is an integer r such that for any

endomorphisms, A1; A2; :::; Ar 2 G� we have A1A2:::Ar = 0.

PROOF (a) Let X1 2 V such that AX1 = 0 for all A 2 G. If thesubspace V1 = |X1 6= V , then each A 2 G induces a nilpotent endomorphismA on the nonzero vector space V = V=V1. Thus we can �nd X2 = X2+V1 6= 0in V such that A X2 = 0 for all A 2 G ; that is, there exists X2 2 V and

X2 =2 V1 with AX2 = a21(A)X1 + 0X2 for all A 2 G. Continuing we obtain abasis X1; X2; :::; Xn of V such that for all A 2 G,

AX1 = 0 and AXn � 0 mod(X1; X2; :::; Xn�1) where (X1; X2; :::; Xn�1)

denotes the subspace spanned by these vectors. Thus the matrix of A has 00s

on and below the diagonal.

Part(b) follows from (a) and exercise 2, while (c) follows from (a) and

matrix multiplication.

Corollaire 5.2.7 Let G be an abstract Lie algebra over |.Then G is anilpotent Lie algebra if and only if for X 2 G adGX is a nilpotent endomor-

phism of G.

PROOF If G is nilpotent, them from theorem 11.7(d) we have adX is

nilpotent. Conversly, if each adX is nilpotent, then by corollary 11.11, (adG)�

is a nilpotent associative algebra. Thus there exists a positive integer p with

f0g = (adG)p = CpG ; that is, G is nilpotent.



Théorème 5.2.8 Let G be a connected real Lie group with Lie algebra

G. Then G is a nilpotent Lie group if and only if G is a nilpotent Lie algebra.

PROOF Thist assume that G is nilpotent and let

G = G0 BG1 B :::BGn = feg a series of closed normal Lie subgroups ofG such that (G;Gn) � Gn+1. Consequently we have the corresponding series

G = G0 B G1 B ::: B Gn = f0g of ideals of G. Next (G;Gk) � Gk+1 implies

[G;Gk] � Gk+1, for let X 2 G, Y 2 Gk. Then for t near 0 in R we have fromtheorem 5.16(c) that

(exp(tX); exp(tY ) = exp(t2 [X; Y ] + o(t3)) is in Gk+1.

However from the charactezation of the Lie algebra of Gk+1 from the

theorem 6.9, this implies t2 [X;Y ] + o(t3) 2 Gk+1 ; that is, [G;Gk] � Gk+1.We now sketch the main parts of the proof of the converse and leave the

details as exercises. First, since G is a nilpotent Lie algebra we see that adG isa nilpotent Lie algebra of endomorphisms (with index of nilpotency N). Thus

for anyZ 2 G, adZ is nilpotent. Consequently in the expansion of theorem

5.18 expX: expY = expF (X;Y ) for X, Y in G near the origin 0 2 G, wehave that the Campbell-Hausdor¤ formula

F (X; Y ) = X + Y + 12[X; Y ] + ::: is of �nite lenght since

(adX)N = (adY )N = 0.

Secondly from the chain of ideals G = G0 B G1 B ::: B Gn = f0g, where[G;Gk] � Gk+1, we obtain for the connected subgroup Gk generated by expGkthe chain G = G0 BG1 B :::BGn = feg :Finally, for X 2 G, Y 2 Gk near enough the origin 0, we have for x =

expY , y = expY : xyx�1y�1 = exp([X; Y ] + :::) = expP (X;Y ) where

P (X; Y ) is a �nite sum of commutators, using the �rst part of the proof.Now

each commutator term in P (X; Y ) contains Y 2 Gk. However Gk is an idealof G so that [G;Gk] � Gk+1 and therefore P (X; Y ) 2 Gk+1. Thus

xyx�1y�1 2 expGk+1 � Gk+1 and by induction on the lenght of products

of elements G and Gk we obtain (G;Gk) � Gk+1.



Exercice. Show that each commutatorterm in P (X; Y ) contains Y 2 Gk.Also complete the induction.

5.3 Nilpotent Lie Algebras of Endomorphisms

We shall now generalize the process of �nding the Jordan canonical form

matrix of an endomorphism to the process of decomposing a vector space

into weight spaces relative to a nilpotent Lie algebra of endomorphisms ;

that is, �nding simultaneously "Jordan forms" for nilpotent Lie algebra of

endomorphisms. Recall frol Section 9.2 that a Lie algebra over | is split if allthe characteristic roots of adX are in | for all X 2 G.

Theorem 11.14 Let V be a �nite-dimensional vector space over |and let G be a split nilpotent Lie subalgebra of gl(V ).(a) There exists a direct sum decomposition V = V (�1) + ::: + V (�m),

where V (�k) = fX 2 V : for all T 2 G, (T � �k(T )I)pX = 0g are G-invariant

weight spaces for G for k = 1; :::;m.(b) There exists a basis of V so that the matrices of the endomorphisms

in G relative to this basis all have the block form

2666666666666666666664

266666664

�1(T ) �:

:

:

0 �1(T )

3777777750

0

266666664

�2(T ) �:

:

:

0 �2(T )

377777775



(c) The functions �k : G �! | are linear ; that is, �k 2 G�. Furtheremore�k([G;G]) = fOg.

PROOF We break the proof into several parts. First we have the

following formulla for an associative algenra A. Let s, t 2 A, and let s(0) = s,

s(1) = ts� st = (adt)s, and s(k) = (adt)ks. Then we obtain by induction for

k = 1; 2; :::

tks =kPi=0

k

i

!s(i)tk�i

= stk +

k

1

!s(1)tk�1 + :::+ s(k) �

Next we have the following result.

Lemma 11.15 Let V be a �nite-dimensional vector space over |,and let G be a split nilpotent Le subalgebra of gl(V ). Let T 2 G, � 2 | andlet V (�) = fX 2 V : (T � �I)nX = 0 for some ng be a weight space for T .Then V (�) is a G-invariant subspace of V .

PROOF We noted in section 10.3 that V (�) is a subspace. Since G isa nilpotent Lie algebra, the algebra h = G + |I is also nilpotent where I isthe identity endomorphism. Therefore by corollary 11.12, [ad(T � �I)]N = 0

for some �xed N . Now for S 2 G � End(V ) let S(1) = [ad(T � �I)]S,

S(2) = [ad(T � �I)]2 S, etc. Then for X 2 V (�) with (T � �I)mX = 0, we

see by choosing k = N +m and using (�) that

(T��I)k(SX) =kPi=0

k

i

!S(i)(T��I)k�i = 0 noting S(N) = [ad(T � �I)]N S =

0. Thus by de�nition of V (�) we see SX 2 V (�) ; that is, V (�) is G-invariant.

PROOF OF THEOREM 11.14 (continued) To prove part (a), we use

induction on the dimension m of V . If m = 1, then every T 2 G has acharacteristic root so that TX = �(T )X for V = |X. This yields the resultin this case. For m � 1 we let T 2 G, and by Proposition 10.6 we have



the direct sum V = V (�1) + ::: + V (�n) where V (�i) are weight spaces for

T . By Lemma 11.15 the V (�i) are G-invariant, and consequently G restrictsto a nilpotent Lie algebra of endomorphisms on each V (�i). Thus conclude

the proof by induction, because we can assume T has at least two distinct

characteristic roots (why?) so that the dimension of V (�i) is less than the

dimension of V . Now since V is �nite-dimensional, we see that there are only

�nitely many distinct weights �i of G.To construct the basis of V whitch gives the matrix in (b), it su¢ cies

to �nd a basis for each V (�i) which gives the corresponding block ma-

trix. Thus let V (�) be a typical weight space as in (a). Then there is a

nonzero X 2 V (�) such that TX = �(T )X. To see this, we use Lie�s

theorem (Theorem 10.23) replacing algebraic closure by "split" as follows.

Since G is nilpotent on V (�), it is solvable on V (�). Thus there is a non-zero X 2 V (�) and a characteristic function F so that for all T 2 G,TX = F (T )X. Therefore [F (T )� �(T )]X = [T � �(T )I]X, and by induc-

tion [F (T )� �(T )]kX = [T � �(T )I]kX = 0 for k large enough, remembe-

ring X 2 V (�). Thus F (T ) = �(T ) ; that si, �(T ) is the only characteristic

root.

Thus, for X1 = X as above, the one-dimensional subspace W = |X1 is

G-invariant. Set V = V (�)=W . Then G induces a nilpotent Lie algebra ofendomorphisms G by T X = TX From this, the characteristic roots of T are

�(T ), and V is a weight space of dimension less than the dimension of V . By

induction we can �nd a basis X2; :::; Xm of V so that

T X2 = �(T )X2

T X3 = a23(T )X2 + �(T )X3

:

:

:

T Xm =m�1Pj=2

ajm(T )Xj + �(T )Xm.

Thus for X i = Xi +W and W = |X1, we can �nd a basis X1; :::; Xm of



V (�) so that

TX1 = �(T )X1

TX2 = a12(T )X1 + �(T )X2

:

:

:

TXm =m�1Pj=1

ajm(T )Xj + �(T )Xm.

Thus we have the desired basis of V (�):

For part (c), we show a weight � is a linear funtional as follows. As in

(b), let 0 6= X 2 V (�) be such that for all T 2 G, TX = �(T )X. Then for

S, T 2 G we have S + T 2 G and�(S+T )X = (S+T )X = SX+TX = [�(S) + �(T )]X so that �(S+T ) =

�(S) + �(T ). Similarly, �(aS) = a�(S) for a 2 |, and also �([S; T ])X =

[S; T ]X = (ST )X � (TS)X = �(S)�(T )X � �(T )�(S)X = 0.

Thus since the elements of [G;G] are of the formP[S; T ], this implies

�([G;G]) = 0.


Chapitre 6

Topoligical groups

In our previous discussion of some matrix groups it was observed that

we were studying not only the group operations but also the continuity of

these operation. Thus in this chapter we abstract the situation and consider

groups which are topological spaces so that the group operations are conti-

nuous relative to the topology of the space. We then prove facts for these

topological groups which indicate that much information can be obtained

from a neighborhood of the identity element ; this leads to local groups and

local isomorphisms. Next we consider topological subgroups, coset spaces,

and normal subgroups. Finally, for connected topological groups, we show

that any neighborhood of the identity actually generates the group as an

abstract group.

6.1 BASICS

In the next, we shall apply the results of the preceding chapters to obtain

elementary results on Lie groups. However, since a Lie proup is a topological

group, we shall briefty discuss this more general situation.

Dé�nition 6.1.1 A topological group is a set G such that :

(a) G is a Hausdor¤ topological space ;

(b) G is a group ;

80

CHAPITRE 6. TOPOLIGICAL GROUPS

(c) the mappings G � G �! G : (x; y) 7�! xy and G �! G : x 7�! x�1

are continuous, where G�G has the product topology.

Thus the set G has two structures-topological and algebraic-and they are

related by property (c) that is, the group structure is compatible with the

topological structure. The compatibility conditions in (c) are equivalent to

the following single condition : (c0) the mapping G � G �! G : (x; y) 7�!xy�1 is continuous.

This condition holds for if (c) holds, then we have that G�G �! G�G :(x; y) 7�! (x; y�1) is continuous. Consequently the map G � G �! G :

(x; y) 7�! (x; y�1) 7�! xy�1 is continuous. Conversely if (c0) holds, then set

x = e (the identity) to obtain (e; y) 7�! (e; y�1) 7�! y�1 is a continuous

map. Also from xy = x(y�1)�1 the map (x; y) 7�! xy is continuous. We

can express (c) in terms of neighborhoods as follows. For any x; y 2 G and

for any neighborhoodsW of xy in G, there exist neighborhoods U of x and

V of y with UV � W . Also for any neighborhoods U of x�1, we have if

U�1 = fa�1 : a 2 Ug is a neighborhoods of x. Thus replacing x by x�1, wehave if V is a neighborhoods of x, then V �1 is a neighborhoods of x�1:

Dé�nition 6.1.2 Let G be a topological group and let a 2 G. Then the mapL(a) : G �! G : x 7�! ax is called a left tranlation. Similarty the map

R(a) : G �! G : x 7�! xa is called a right translation.

It should be noted that the maps L(a) and R(a) for a 2 G are ho-

meomorphisms of G. Furthermore given any two points x; y 2 G, then the

homeomorphism L(xy�1) maps x into y: In particular, there always exists a

homeomorphism which maps e 2 G onto any other element a 2 G and usingthis, we shall see many of the local properties of a 2 G are determined by

those of e: Thus, for example, U is a neighborhoods of a 2 G if and only if

U = L(a)V = aV where V is a neighborhoods of e 2 G:

Proposition 6.1.3 Let G be a topological space which is also a group. Then

G is a topological group relative to these two structures if only if :



(a) the set feg is closed ;(b) for all a 2 G the translations R(a) and L(a) are continuous ;

(c) the mappings G�G �! G : (x; y) 7�! xy�1 is continuous at the point

(e; e):

6.2 Subgroups and Homogeneous Spaces

We shall consider subgroups H of a topological group G and the cor-

responding space of left cosets G=H = faH : a 2 Gg : Then we eventuallyconsider the case when H is a normal subgroup so that G=H becomes a

topological group.

Dé�nition 6.2.1 Let G be a topological group and let H be a subset of G

such that HH�1 � H: Then H is a subgroup of G (in the abstract sense).

The topology of G induces a topology on the subgroup H by requiring U � H

to be open if only if U = H \ V where V is open in G. If with this induced

topology, the subgroup H becomes a topological group we call H a topological

subgroup.

Remarque 6.2.2 We shall be interested in closed subgroups H of G ; that

is, H is closed as a subset of G. However, we note that if H is an open

subgroup of G, then H is closed. For, since H is open, so is aH for all a 2 G:Therefore K = [faH : a =2 Hg is open so that the complement of K, whichis H, is closed.

Théorème 6.2.3 Let H be a topological subgroup of the topological group G

and let � : G �! G=H : a 7�! aH be the natural projection. Then

(a) G=H can be made into a topological space such that :

(i) the projection � : G �! G=H is continuous, and

(ii) if N is a topological space and if f : G=N �! N is such that f � � :G �! N is continuous, then f is continuous.

The topology de�ned on G=H is uniquely determined by (i) and (ii) and

is called the quotient topology.



(b) G=H with the quotient topology is such that � is an open map ; that

is U is open in G implies �(U) is open in G/H.

(c) The quotient topology is Hausdor¤ if and only if H is a closed subset

of G.

Dé�nition 6.2.4 A subgroup H is a normal subgroup of G if aHa�1 � H

all a 2 G and then G=H is a group relative to aH � bH = abH which is called

the quotient ggroup.

Corollaire 6.2.5 Let H be a clesed normal subgroup of the topological group

G and let G=H be the quotient group. Then relative to the quotient topology,

G=H becomes a topological group such that the projection � : G �! G=H is

an open continuous homomorphism.

Preuve. It su¢ ces to show that G=H � G=H �! G=H : (aH; bH) 7�!ab�1H is continuous. Let U be a neighborhood of ab�1H = �(ab�1) in G=H:

Then ��1(U) is a neighborhood of ab�1 in G. Now there exist neighborhood

V of a and W of b in G such that VW�1 � ��1(U): However, since � is

open, �(V ) and �(W ) are neighborhoods of aH = �(a) and bH = �(b),

respectively. Thus �(V )�(W )�1 = �(VW�1) � U which proves continuity.

Corollaire 6.2.6 If H is an open normal subgroup of the topological group

G, then G=H is discrete.

Preuve. Since H is open, the cosets aH for a 2 G are open in G. Thus

since � is an open map, the sets faHg in G=H are open. Therefore G=H is

discrete.

Corollaire 6.2.7 Let f : G �! G be a homomorphism of topological groups.

Then f is continuous if only if f is continuous at the identity e 2 G:

Preuve. Assume f is continuous at the identity. Let a 2 G and let f(a)Ube a neighborhood of f(a) in G where U is a neighborhood of e in G: Since



f is continuous at e 2 G and since e = f(e), there exists a neighborhood

U of e in G such that f(U) � U which proves continuity at a since aU is a

neighborhood of a in G with f(aU) � f(a)U .

Théorème 6.2.8 Let f : G �! G be a continuous homomorphism of the

topological groups G and G and let H = fx 2 G : f(x) = eg be the kernel off where e is the identity of G. Then : (a) H is a closed normal subgroup of

G and � : G �! G=H is a continuous homomorphism ;

(b) there is a continuous monomorphism g : G=H �! G such that f =

g � �;(c) let H and N be clased normal topological subgroups of G such that

N � H: Then G=H is topologically isomorphic to (G=N) (H=N) :

Dé�nition 6.2.9 Let M be a Hausdor¤ topological space and let G be a

topological group. Then :

(a) G operates onM if there is a surjection G�M �!M : (g; p) 7�! g�psuch that (xy) � p = x � (y � p) and e � p = p for all x; y 2 G; and p 2 M

where e is the identity of G:

(b) G operates transitively on M if for every p; q 2M , there exists x 2 Gsuch that x � p = q

(c) G operates continuously on M if the map G �M �! M : (g; p) 7�!g � p is continuous.(d) G is called a topological transformation group on M if G operates

continuously on M: Note that for each x 2 G, the map �(x) : M �! M :

p 7�! x � p is a homeomorphism.(e) G is e¤ective if a � p = p for all p 2M implies a = e:

(f) Let p be �xed in M: Then G(p) = fx 2 G : x � p = pg is a groupcalled the isotropy subgroup of G at p or �xed point subgroup at p: The set

G � p = fx � p 2M : x 2 Gg is called an orbit under G:

Théorème 6.2.10 Let M be a Hausdor¤ space and let G be a transitive

topological transformation group operating on M: Let p be some (�xed) point

in M and let G(p) be the isotropy group at p: Then G(p) is a closed subgroup



of G; and the map f : G �! G�p : a 7�! a�p induces a continuous bijectionf : G=G(p) �!M such that f � � = f ; that is, the accompanying diagram is

commutative.

Corollaire 6.2.11 If f : G=G(p) �! M is open or if G=G(p) is compact,

then f is a homeomorphism ; that is, M is a homogenous space.

If G is a locally compact group with countable basic and if M is a locally

compact Hausdor¤ space, then f is a homeomorphism of G=G(p) onto M .

6.3 Connected Groups

Dé�nition 6.3.1 Let M be a topological space and let p 2 M: Then p is

contained in a unique maximal connected subset C(p) is closed and is called

the connected component of p. For M = G a topological group, the connected

component of the identity e 2 G is called the identity component of G and is

denote by G0:

Théorème 6.3.2 Let G be a topological group, and let G0 be the identity

component. Then :(a) G0 is a closed normal topological subgroup of G and

the connected component C(a) of a 2 G equals aG0;

(b) If G is loccally connected (that is ; if every point a 2 G has a connectedneighborhood), then G=G0 is dixrete.

Proposition 6.3.3 Let G be a topological group, let G0 be the identity com-

ponent, and let U be any open neighborhood of e 2 G:

(a) If U is a symmetric neighborhood, then H =

1[k=1

Uk is an open and

closed subgroup of G. If U is connected, so is H.

(b) G0 =

1[k=1

Uk

!\G0:

(c) If G is connected, then1[k=1

Uk = G: Thus any open neighborhood, of e

is a set of generators of a connected topological group as an abstract group.



Dé�nition 6.3.4 Let G be a topological group. Then the center C of G equals

fx 2 G : xa = ax for all a 2 Gg : The center is a normal subgroup of G andis also denote by Z(G):

Proposition 6.3.5 Let G be a connected topological group and let H be a

discrete normal topological subgroup of G: Then H � G; the center of G.

Let G be a topological group and let H be a closed topological subgroup

such that H is connected and G=H is connected. Then G is connected.

Dé�nition 6.3.6 A topological spaceM is locally Euclidean of dimension m

if each point p 2 M has a neighborhood, which is homeomorphic to an open

set in Rm. Note that an open subset of Rm cannot be homeomorphic to an

open subset of Rn if m 6= n:


Chapitre 7

Representations of compact

groups

In the mathematical �eld of representation theory, group representations

describe abstract groups in terms of linear transformations of vector spaces ;

in particular, they can be used to represent group elements as matrices so

that the group operation can be represented by matrix multiplication. Repre-

sentations of groups are important because they allow many group-theoretic

problems to be reduced to problems in linear algebra, which is well unders-

tood. They are also important in physics because, for example, they describe

how the symmetry group of a physical system a¤ects the solutions of equa-

tions describing that system.

The term representation of a group is also used in a more general sense to

mean any "description" of a group as a group of transformations of some ma-

thematical object. More formally, a "representation" means a homomorphism

from the group to the automorphism group of an object. If the object is a

vector space we have a linear representation. Some people use realization for

the general notion and reserve the term representation for the special case of

linear representations. The bulk of this article describes linear representation

theory ; see the last section for generalizations.

In this chapter we present the Peter�Weyl theory for compact groups. By

87

CHAPITRE 7. REPRESENTATIONS OF COMPACT GROUPS

using spectral theory for compact operators we will see that an irreducible

representation of a compact group is �nite dimensional. Using the Peter�Weyl

theory, classical Fourier analysis is extended to compact groups.

7.1 Unitary representations

Let G be a topological group and V a normed vector space over R or

C (V 6= f0g). Let L(V ) denote the algebra of bounded operators on V . Arepresentation of G on V is a map

� : G �! L(V )g 7�! �(g)

;

such that

1. �(xy) = �(x)�(y), �(e) = I,

2. for every v 2 V; the map� : G �! V

g 7�! �(g)v; is continuous.

The de�nition, we give here, di¤ers slightly from that given in Section,

where we only considered the case of a �nite dimensional vector space V: A

subspace W � V is said to be invariant if, for every g 2 G; �(g)W = W:

Putting �0(g) = �(g)W; the restriction of �(g) to W; we get a representation

of G on W . One says that �0 is a subrepresentation of �. Assume W to be

closed. The representation �1 of G on the quotient space V=W is called the

quotient representation. The representation � is said to be irreducible if the

only invariant closed subspaces are f0g and V: Observe that, by de�nition,a one dimensional representation is irreducible. Let (�1; V1) and (�2; V2) be

two representations of G: If a continuous linear map A from V1 into V2 sa-

tis�es the relation A�1(g) = �2(g)A; for every g 2 G; one says that A is anintertwinning operator or that A intertwins the representations �1 and �2.

The representations (�1; V1) and (�2; V2) are said to be equivalent if there

exists an isomorphism A : V1 �! V2 which intertwins the representations �1and �2. Let H be a Hilbert space. Recall that an operator A onHis said to

be unitary if it is invertible and A�1 = A�: A representation � of G on H is



said to be unitary if, for every g 2 G; �(g) is a unitary operator ; this can bewritten

8g 2 G; 8v 2 H; k�(g)vk = kvk :

If the representation � is unitary, and if W is an invariant subspace, then

the orthogonal subspace W? is invariant as well. If W is closed, the quotient

representation on H=W is equivalent to the subrepresentation on W?.

Proposition 7.1.1 Let � be a representation of a compact group G on a �-

nite dimensional vector space V: There exists on V a Euclidean inner product

for which � is unitary.

Preuve. Let us choose arbitrarily on V a Euclidean inner product (�j�)0 andput

(u; v) =

ZG

(�(g)u; �(g)v)0 �(dg);

where � is a Haar measure on G: One can check easily that (�j�) is a Euclideaninner product on V; and that the representation � is unitary with respect to

this Euclidean inner porduct.

Corollaire 7.1.2 Let � be a representation of a compact group G on a �nite

dimensional vector space V:

(i) For every invariant subspace there is an invariant complementary sub-

space.

(ii) The vector space V can be decomposed into a direct sum of irreducible

invariant subspaces :

V = V1 � �� VN :

Preuve. By Proposition there exists on V a Euclidean inner product for

which the representation � is unitary. IfW is an invariant subspace, then the

orthogonal subspace W? is invariant and complementary to W: Let V1 be a

non-zero invariant subspace with minimal dimension. Then

V = V1 � V ?1



If V ?1 6= f0g; let V2 be a non-zero invariant subspace in V ?

1 with minimal

dimension. One continues the process as long as the subspace V ?

is not zero. Since the dimension of V is �nite, the process stops necessarily.

Théorème 7.1.3 (Schur�s Lemma) (i) Let (�1; V1) and (�2; V2) be two �-

nite dimensional irreducible representations of a topological group G. Let

A : V1 �! V2 be a linear map which intertwins the representations �1 and

�2 : A�1(g) = �2(g)A for every g 2 G: Then either A = 0; or A is an

isomorphism.

(ii) Let � be an irreducible C-linear representation of a topological groupG on a �nite dimensional complex vector space V: Let A : V ! V be a

C-linear map which commutes to the representation � :

A�(g) = �(g)A

for every g 2 G: Then there exists � 2 C such that A = �I:

Preuve. (i) In fact, the kernel ker(A) and the image Im(A) are two invariant

subspaces. The statement follows immediately.

(ii) There exists � 2 C such that A� �I is not invertible. It follows from(i) that A � �I = 0: If the group G is commutative, by Schur�s Lemma an

irreducible C-linear representation is one dimensional. It is a character of G.In this setting a character is de�ned as a continuous function � : G ! C

satisfying

�(xy) = �(x)�(y):

For instance the characters of the group G = SO(2) ' U(1) ' R=2�Z arethe functions

�m(�) = eim� (m 2 Z):

In part (ii) of Theorem, the assumption that the representation � is C-linear cannot be dropped. For the R-linear representations the situation is



quite di¤erent. Consider for instance the representation � of the group G =

SO(2) ' R=2�Z on R2 de�ned by

�(�) =

cos � � sin �sin � cos �

!

This representation is irreducible. But the matrices :

A =

a b

�b a

!

(a; b 2 R); commute with the matrices �(�). (See Exercise 4 about irredu-cible R-linear representations.) In the same way one can establish similarstatements for representations of Lie algebras.

Proposition 7.1.4 (i) Let (�1; V1) and (�2; V2) be two �nite dimensional

irreducible representations of a Lie algebra g: Let A : V1 �! V2 be a linear

map which intertwins the representations �1 and �2 :

A�1(X) = �2(X)A

for every X 2 g. Then either A = 0; or A is an isomorphism.(ii) Let � be a C-linear representation of a complex Lie algebra g on a

�nite dimensional complex vector space V . Let A : V �! V be a C-linearmap which commutes with the representation � :

A�(X) = �(X)A

for every X 2 g: Then there exists � 2 C such that A = �I:

7.2 Compact self-adjoint operators

Let A be a bounded operator on a Hilbert space H: Its norm kAk isde�ned by kAk = sup

kuk�1kAuk. For v �xed, the map

u �! (Aujv)



is a continuous linear form on H. By the Riesz representation theorem there

exists a unique w 2 H such that

(Aujv) = (ujw)

for every u 2 H: The map v �! w is linear, it is denoted by A� and is

called the adjoint operator of A: It is de�ned by the relation

(Aujv) = (ujA�v):

One can show that A� = A and that (A�)� = A: If A� = A; one says that

the operator A is self-adjoint, that is (Aujv) = (ujAv) for every u; v 2 H:

Proposition 7.2.1 Let A be a self-adjoint operator.

(i) The eigenvalues of A are real.

(ii) If � and � are distinct eigenvalues of A; the corresponding eigenspaces

are orthogonal.

Preuve. (a) Let � be an eigenvalue of A; and u an associated eigenvector :

Au = �u; u 6= 0:

Then (Auju) = (ujAu) and � kuk2 = � kuk2.(b) Let � and � be two eigenvalues of A; � = �; u and v associated

eigenvectors :

Au = �u;Av = �v:

Then

(Aujv) = (ujAv)

and

(�� )(ujv) = 0:

Proposition 7.2.2 Let A be a self-adjoint operator. Then kAk = sup j(Au; ujkuk�1



Preuve. Put

M = sup j(Au; ujkuk�1

Observe �rst that, by the Schwarz inequality, M � A: On the other hand,

kAk = sup jRe(Au; vjkuk�1

In fact, for w 2 H;kwk = sup j(w; v)j

kuk�1;

and, by de�nition of the norm of an operator, and, by de�nition of the norm

of an operator,

kAk = sup kAukkuk�1

From the identity

4Re(Aujv) = (A(u+ v)ju+ v)� (A(u� v)ju� v)

it follows that

jRe(Aujv)j � M

4(ku+ vk2 + ku� vk2) = M

2(kuk2 + kvk2):

Hence, if u � 1; v � 1;jRe(Aujv)j �M;

therefore A � M: Let A be an operator acting on H: The operator A is

said to be compact if the following property holds : the image under A of a

bounded set is relatively compact. This property is equivalent to each of the

two following : the image underA of the unit ball is relatively compact ; if (un)

is a bounded sequence, there is a subsequence (unk) such that the sequence

(unk) converges. A�nite rank operator is compact. If A is a compact operator

and B a bounded operator, then AB and BA are compact operators.

Proposition 7.2.3 If (An) is a sequence of compact operators with limit A;

limn!1

kAn � Ak = 0 then the operator A is compact.



Preuve. Let (uk) be a sequence in H such that uk � 1. Since the operatorA1 is compact, there is a subsequence (u

(1)k ) such that the sequence (A1u

(1)k )

converges. Since the operator A2 is compact, one can extract from the sub-

sequence (u(1)k ) a subsequence (u(2)k ) such that (A2u

(2)k ) converges, and so on.

Then one considers the sequence

(u0k) = (u

(2)k )For every n the sequence k 7�! (Anu

0k)converges. Let us show

that (Au0k) is a Cauchy sequence. Let " > 0: There exists n such that

kAn � Ak � "

3

Since (Anu0k) is a Cauchy sequence, there exists K > 0 such that, if k; l �

K; Anu0k � Anu

0l

� "3; Hence, if k; l � K; Anu0k � Anu

0

l

� Au0k � Anu0

l

+ Anu0k � Anu0

l

+ Anu0l � Au0

l

� "

Finally we can state that the set of compact operators is a closed two-sided

ideal in L(H):

Exemple 7.2.4 Let H = l2(N): Let (�n) be a sequence of complex numbers

with limit 0; and let A 2 L(H) be de�ned by

A(un) = (�nun):

The operator A is compact. In fact, let AN be the operator de�ned as follows :

if v = ANu; vn = �nun if n � N; vn = 0 if n > N: The operator AN has

�nite rank and kA� ANk = supn>N

j�nj

Théorème 7.2.5 Let A be a compact self-adjoint operator. Then, either A

or �A is an eigenvalue of A: Hence, a non-zero compact self-adjoint operatorhas a non-zero eigenvalue.

Preuve. Since the operator A is self-adjoint, kAk = sup j(Au; v)jkuk�1

(Proposi-

tion). Observe that the numbers (Auju) are real ; one may assume, by taking-A instead of A if necessary, that kAk = sup j(Au; u)j

kuk�1: Let us then show that

� = kAk is an eigenvalue of A. There is a sequence (un): such that kunk = 1;

limn!1

(Aun; un) = �



Since the operator A is compact, there is a subsequence (unk) such that the

sequence (Aunk) converges : limk!1

Aunk = v: From the expansion

kAunk � �unkk2 = kAunkk

2 � 2� (Aunk ; unk) + �2

it follows that

limk!1

kAunk � �unkk2 = kvk2 � �2;

On the other hand, since kAk = �; limk!1

kAunkk = kvk � �; hence limn!1

kAunk � �unkk =0 Therefore the sequence (unk) converges : lim

k!1unk = u: Furthermore Au = v

and Au = �u:

Théorème 7.2.6 (Spectral theorem) Let A be a compact self-adjoint opera-

tor. The non-zero eigenvalues of A form a sequence (�n) which is �nite or

converges to 0: Let Hn be the eigenspace associated to �n and let Pn be the

orthogonal projection onto Hn: The dimension of Hn is �nite and

A =NXn=1

�nPn

if the number N of non-zero eigenvalues is �nite, otherwise

A =1Xn=0

�nPn

as a convergent series in the norm topology.

Lemme 7.2.7 Let H be a Hilbert space. If the unit ball in H is compact,

then H is �nite dimensional.

Preuve. If H were not �nite dimensional, there would be in H an in�nite

orthonormal sequence (en): Since

kep � eqk =p2

for p = q; there cannot be a converging subsequence. Let A be a self-adjoint

operator, and � a non-zero eigenvalue of A: From this lemma it follows that



the associated eigenspace is �nite dimensional. By Theorem there exists an

eigenvalue �1 of A such that j�1j = kAk : LetH1 be the associated eigenspace.

From Lemma it follows that H1 is �nite dimensional. Put A1 = A � �1P1:

The operator A is self-adjoint and compact, and kA1k � kAk : By continuing,either one �nds an integer N such that AN = 0; and then

A =

NXn=1

�nPn;

or the sequence (�n) is in�nite. Observe that the sequence (j�nj) is decreasingby construction. Let us show that, when in�nite, the sequence (�n) goes to

0: Let us assume the opposite, that j�nj � � > 0: For every n let us take

vn 2 Hn; vn = 1: Since A is compact, one can extract from the sequence

(Avn) a converging subsequence. But this is impossible since

kAvp � Avqk2 = k�pvp � �qvqk2 = �2p + �2q � 2�2:

It follows that

A =1Xn=1

�nPn

In fact,

A =NXn=1

�nPn + AN+1;

and kAN+1k = j�N+1j : Finally, the dimension of Hn is �nite since the unitball of Hn is compact.

7.3 Schur orthogonality relations

Let G be a compact group, and � the normalised Haar measure of G:

Let (�;H) be a unitary representation of G: For v 2 H one considers the

operator Kv of H de�ned by

Kvw =

ZG

(w; �(g)v)�(g)v�(dg):



This can also be written

(Kvw;w0) =

ZG

(w; �(g)v) (w0; �(g)v)�(dg):

Proposition 7.3.1 (i) Kv is bounded, kKvk � kvk2 :(ii) Kv is self-adjoint : K�

v = Kv:

(iii) Kv commutes with the representation � : for every g 2 G; Kv�(g) =

�(g)Kv:

(iv) Kv is a compact operator.

Preuve. (i)

kKvwk � kvk2 kwk :

(ii)

(K�vwjw0) = (wjKvw

0) = (Kvwjw0):

(iii) Let g0 2 G;

Kv�(g0)w =

ZG

(wj�(g�10 g)v)�(g)v�(dg);

and, by the invariance of the measure �,

Kv�(g0)w =

ZG

(wj�(g)v)�(g0g)v�(dg) = �(g0)Kvw:

(iv) For v 2 H let Pv be the rank one operator de�ned by

Pvw = (wjv)v:

It is a compact operator and, for v �xed, the mapG �! L(H)

g 7�! P�(g)vis continuous

for the norm topology. The operator Kv can be written

Kv =

ZG

P�(g)v�(dg)

Since the space of compact operators is closed for the norm topology (Pro-

position), the operator Kv is compact. Observe that

(Kvwjw) =ZG

j(�(g)v; w)j2 �(dg);

and that, if v = 0; (Kvvjv) > 0; hence Kv 6= 0:



Théorème 7.3.2 (i) Every unitary representation of a compact group contains

a �nite dimensional subrepresentation.

(ii) Every irreducible unitary representation of a compact group is �nite

dimensional.

Preuve. Let (�;H) be a unitary representation of a compact group. The

operator Kv is self-adjoint, compact (Proposition), and non-zero if v 6= 0: ByTheorem it has a non-zero eigenvalue, and the corresponding eigenspace is

�nite dimensional. This subspace is invariant under the representation �.

Théorème 7.3.3 Let � be an irreducible unitary C-linear representation ofa compact group G on a complex Euclidean vector space H with dimension

d�. Then, for u; v 2 H;ZG

j(�(g)u; v)j2 �(dg) = 1

d�kuk2 kvk2 ;

and, by polarisation, for u; v; u0; v0 2 H;ZG

(w; �(g)v) (w0; �(g)v)�(dg) =1

d�(u; u0) (v; v0)

Preuve. For v 2 H; the operator Kv commutes with the representation �.

By Schur�s Lemma (Theorem) there is �(v) 2 C such that

Kv = �(v)I:

Hence, ZG

j(�(g)u; v)j2 �(dg) = �(v) kuk2 ;

By permuting u and v we get

�(u) kvk2 = �(v) kuk2 ;

hence

�(u) = �0 kuk2

, where �0 is a constant. Let fe1; :::; eng be an orthonormal basis of

H(n = d�) :

nXi=1

j�(g)u; eij2 = kuk ;



By integration over G we get

kuk2 =nXi=1

ZG

j(�(g)u; ei)j2 �(dg) = n�0 kuk2 ;

hence �0 = 1=n: FinallyZG

j(�(g)u; v)j2 �(dg) = 1

nkuk2 kvk2 :

Let �ij(g) denote the entries of the matrix �(g) with respect to the basis

feig;�ij(g) = (�(g)ejjei)

From the preceding theorem one obtains Schur�s orthogonality relations :ZG

�ij(g)�kl(g)�(dg) =1

d��ik�jl:

This can be written in the following alternative form : if A and B are two

endomorphisms of H; thenRGtr(A�(g))tr(B�(g)�(dg) = 1

d�tr(AB�): In fact

one can check that, if A and B are two rank one endomorphisms, the above

formula is precisely the second formula of the preceding theorem. Let M�

denote the subspace of L2(G) generated by the entries of the representation

�, that is by the functions of the following form : g �! (�(g)ujv) (u; v 2 H).

Théorème 7.3.4 Let (�;H) and (�0; H 0) be two irreducible unitary repre-

sentations of a compact group G which are not equivalent. Then M� and M�0

are two orthogonal subspaces of L2(G) :ZG

(v; �(g)u) (v0; �(g)u0)�(dg) = 0; (u; v 2 H; u0; v0 2 H 0):

Preuve. Let A be a linear map from H into H 0 and put

eA = ZG

�0(g�1)A�(g)�(dg):

Then eA is a linear map from H into H 0 which intertwins the representations

� and �0 eA � �(g) = �0(g) �fA:Kangni Kinvi 99 Analyse Harmonique


By Schur�s Lemma (Theorem ), eA. Hence� eAu; u0� = ZG

(A�(g)u; �0(g)u0)�(dg) = 0:

Take for A the rank one operator de�ned by

Au = (u; v) v0 (v 2 H; v0 2 H 0);

then A�(g)u = (�(g)ujv)v, andZG

(v; �(g)u) (v0; �(g)u0)�(dg) = 0; (u; v 2 H; u0; v0 2 H 0):

It follows that two irreducible representations �1 and �2 of a compact group

G are equivalent if and only if the spaces M�1 and M�2 agree.

7.4 Peter�Weyl Theorem

Let G be a compact group, and let R denote the right regular represen-

tation of G on L2(G) :

R(g)f(x) = f(xg):

Let (�;H) be an irreducible representation of G; and let fe1; :::; eng be anorthonormal basis of H (n = d�): One puts

�ij(x) = (�(x)ejjei):

Let M (1)� be the subspace of M� generated by the entries of the �rst row,

that is by the functions x �! �1j(x); for j = 1; :::; n. Observe that

�1j(xy) =nXk=1

�1k(x)�kj(y):

This shows that the subspace M (1)� is invariant under R: Furthermore, the

map

A :nXj=1

cjej 7�!nXj=1

cj�1j(x)



from H into M (1)� is an isomorphism, and intertwins the representations �

and R: In fact, if u =nXj=1

cjej, then

A�(g)u = A

nXj=1

cj�(g)ej = A

nXj=1

cj

nXi=1

�ij(g)ej

!

=

nXi=1

nXj=1

�ij(g)cj

!�1i(x) =

nXj=1

�1j(xg)ej

!= R(g)Au

Furthermore kAuk2 = 1nkuk2 : LetM (i)

� denote the subspace ofM� generated

by the coe¢ cients of the ithline. Then G

M� =M (1)� � �� M (n)

� ;

and the restriction toM� of the representation R is equivalent to �� =n�: By considering the columns instead of the rows one gets the same sta-

tement with, instead of the representation R; the regular left representation

L :

L(g)f(x) = f(g�1x):

Théorème 7.4.1 (Peter�Weyl Theorem)

Let bG be the set of equivalence classes of irreducible unitary representa-

tions of the compact group G and, for each � 2 bG; let M� be the space ge-

nerated by the coe¢ cients of a representation in the class �. Then L2(G) =

[��2 bGM�: Recall that [��2 bGM� denotes the closure in L2(G) of M = [��2 bGM�

which is the space of �nite linear combinations of coe¢ cients of �nite dimen-

sional representations of G:

Preuve. We saw that the subspaces M� are two by two orthogonal (Theo-

rem). Put

[��2 bGM� = H

and

H0 = H?:



We will show that H0 = f0g: Let us assume the opposite, that H0 6= f0g:The spaceH0 is invariant under the representation R and closed. By Theo-

rem it contains a closed subspace Y 6= f0g which is invariant under R and

irreducible. The restriction of R to Y belongs to one of the classes �. Let

f 2 Y; f 6= 0; and put

F (g) =

ZG

f(xg)f(x)�(dx) = (R(g)f; f) :

The function F belongs to M�. We will see that it is orthogonal to M�. Let

(�; V ) be a representation of the class �, and u; v 2 V: ThenZG

F (g)(�(g)u; v)�(dx) =

ZG

f(xg)f(x)(�(g)u; v)�(dx)

and, by putting xg = g0;ZG

F (g)(�(g)u; v)�(dx) = 0

Therefore F = 0; and, since

F (e) =

ZG

f(x)f(x)�(dx) =

ZG

jf(x)j2 �(dx) = (R(e)f; f) :

it follows that f = 0: This yields a contradiction. Let H be a �nite dimensio-

nal Hilbert space and A 2 L(H): The Hilbert�Schmidt norm of A is de�ned

by

kAk2 = tr(AA�)

If fe1; :::; eng is an orthonormal basis of H; and if (aij) is the matrix of A

with respect to this basis, kAk =nX

i;j=1

jaijj2 : For every � 2 bG one chooses

a representative (��; H�): Let d� denote the dimension of H�. If f is an

integrable function on G; its Fourier coe¢ cient bf(�) is the operator actingon the space H� de�ned by

bf(�) = ZG

f(g)��(g�1)�(dg):

The following theorem follows directly from the Peter�Weyl Theorem and

from Schur�s orthogonality relations.



Théorème 7.4.2 (Plancherel�s Theorem) Let f 2 L2(G): Then f is equal tothe sum of its Fourier series :

(i)

f(g) =X�2 bG

d�tr( bf(�)��(g):This holds in the L2 sense.

(ii) ZG

jf(x)j2 �(dx) =X�2 bG

d�

( bf(�) 2And, if f1; f2 2 L2(G);Z

G

f1(x)f2(x)�(dx) =X�2 bG

d�tr(bf1(�)(f2(�))�:(iii) The map f 7�! bf is a unitary isomorphism from L2(G) onto the space

of sequences of operators A = (A�) (A� 2 L(H�)), for which

kAk2 =X�2 bG

d� kA�k2 ;

and equipped with this norm. If the compact group G is commutative then a

C-linear irreducible representation is one dimensional, and bG is the set of

continuous characters. Recall that, in this setting, a continuous character is

a continuous function

� : G �! C�

satisfying

�(xy) = �(x)�(y):

Since G is compact, the set �(G) is a compact subgroup of C�, hence consistsof modulus one complex numbers. Therefore � : G! fz 2 C tel que jzj = 1g:The set bG is a commutative group for the ordinary product of the characters

which is called the dual group of G; and the continuous characters form a

Hilbert basis of L2(G):The Fourier coe¢ cient bf(�) of a square integrableKangni Kinvi 103 Analyse Harmonique


function f on G is given by bf(�) = RGf(x)�(x)�(dx) The Fourier series of

f is written as : X�2 bG

bf(�)�(x)and the Plancherel formula :Z

G

jf(x)j2 �(dx) =X�2 bG

�� bf(�)��2 :For instance, if G = SO(2) ' U(1) ' R=2�Z; then a character � has theform

�(�) = eim�;

where m 2 Z: Hence bG = Z. In this case one obtains the classical formu-lae. If f is an integrable function on R=2�Z; the Fourier coe¢ cients of f aregiven by bf(m) = 1

2�

ZG

f(�)e�im��(d�):

The Fourier series of f is written as ;Xm2Z

bf(m)eim�and the Plancherel formula, if f is square integrable, 1

2�

RGjf(�)j2 d� =

Xm2Z

�� bf(m)��2 :Recall thatMdenotes the space of �nite linear combinations of coe¢ cients of

�nite dimensional representations of G; M =M�2 bG

M� We will show that M

is dense in the space of continuous complex valued functions on G: For that

we will apply the Stone�Weierstrass Theorem which we recall below.

Théorème 7.4.3 (Stone�Weierstrass Theorem) Let X be a compact topolo-

gical space, and C(X) the space of complex valued continuous functions on

X; equipped with the topology of uniform convergence. Let A be a subspace of

C(X) with the following properties :

(i) A is an algebra (for the ordinary product of functions),

(ii) A separates the points of X; and constant functions belong to A;



(iii) if f belongs to A; then ¯f also belongs to A.

Then A is dense in C(X): See for instance : K. Yosida (1968): Functional

Analysis. Springer (Corollary). Let (�1; H1) and (�2; H2) be two �nite dimen-

sional representations of G. The tensor product �1�2 is the representationof G on H1 H2 such that

(�1 �2)(g)(u1 u2) = �1(g)u1 �2(g)u2:

If H1 and H2 are �nite dimensional Hilbert spaces, then H1H2 is equipped

with an inner product such that

(u1 u2jv1 v2) = (u1jv1)(u2jv2);

and

(�1 �2)(g)(u1 u2)j(v1 v2) = (�1(g)u1jv1)(�2(g)u2jv2):

Therefore the product of a coe¢ cient of �1 and a coe¢ cient of �2 is a coe¢ -

cient of �1 �2: For �; � 2 bG the representation �� can be decomposedinto a sum of irreducible representations :

�� =M

v2E(�;�)

c(�; �; v)�v;

, where E(�; �) is a �nite subset of bG. The numbers c(�; �; v), which arepositive integers, are called Clebsch�Gordan coe¢ cients. This shows that the

space A of �nite linear combinations of coe¢ cients of �nite dimensional re-

presentations of G is an algebra. Let V be a normed vector space, and V

its topological dual. Let � be a representation of G on V: The contragredient

representation of � is the representation � of G on V de�ned by

(�0(g)f; u) = (f; �(g � 1)u) (f 2 V 0; u 2 V ):

Assume now that V = H is a Hilbert space and that � is unitary. There is

an antilinear isomorphism T from H onto H 0 de�ned by

(Tv; u) = (ujv):

Lemme 7.4.4 Let G be a compact group. If g 6= e, there exists a �nite

dimensional representation � of G such that �(g) 6= I:



7.5 Characters and central functions

Let G be a compact group. A function f which is de�ned on G is said to

be central if

f(gxg�1) = f(x) (g; x 2 G):

Let � be a representation of G on a �nite dimensional complex vector space

V . The character of � is the function �� de�ned on G by

��(g) = tr�(g):

It is a central function which only depends on the equivalence class of �. One

can establish easily the following properties :

��(e) = dimV;

��1��2(g) = ��1(g) + ��2(g);

��1��2(g) = ��1(g)��2(g);

��(g) = ��(g�1) = ��(g):

Let us denote by V G the subspace of invariant vectors :

V G = fv 2 V j8g 2 G; �(g)v = vg:

The operator P; de�ned by

Pv =

ZG

�(g)v�(dg);

where � is the normalised Haar measure of G; is a projection onto V G: Since

trP = dimV G;

it follows that ZG

��(g)d�(g) = dimV G

. If (�1; V1) and (�2; V2) are two �nite dimensional representations of G one

puts

E(�1; �2) = fA 2 L(V1; V2)j8g 2 G;A�1(g) = �2(g)Ag:



This is the space of operators which intertwin the representations �1 and �2.

The group G acts on the space L(V1; V2) by the representation T de�ned by

T (g)A = �2(g)A�1(g�1):

This representation is equivalent to �2 �1. Observe that

E(�1; �2) = L(V1; V2)G;

Proposition 7.5.1 Let �1 and �2 be two �nite dimensional representations

of G: Then ZG

��1(g)��2(g)�(dg) = dimE(�1; �2):

Assume that �1 and �2 are irreducible. They are equivalent if and only if they

have the same character :

��1(g) = ��2(g) (g 2 G)

A �nite dimensional representation � of G is irreducible if and only ifZG

j��(g)j2�(dg) = 1:

If � is an irreducible representation of the compact group G, thenZG

��(xgyg�1)�(dg) =

1

d��(x)��(y):


Chapitre 8

Induced Representations

8.1 The De�nition

Let L be a unitary representation of H on a Hilbert space E.

Dé�nition 8.1.1 Let EL be the linear space of all functions f from G to E

such that :

1. f is dG-measurable ;

2. f (x�) = (�H(�)=�G(�))1=2 L

��1�f(x) whenever � 2 H, x 2 G ;

3. kf(�)k2 is locally summable on G.

It is clear that x 7! kf(x)k2 is dG-measurable ( f 2 EL ) ; moreover,the polarization identity together with (3) imply that x 7! (f(x); g(x)) is

dG-measurable and locally summable ( f; g 2 EL).

Given � 2 Cc(G), set _� ( _x) =RH�(x�)dH(�) _x ( _x = xH)� then the

assignment � 7! _� is a continuous surjection of Cc(G) onto Cc(G=H) (cf.

infra).

Lemme 8.1.2 Let f; g 2 EL� then we may de�ne a Radon measure �f;g onG=H by the rule :Z

G

(f(x); g(x))�(x)dG(x) =

ZG=H

_� ( _x) d�f;g ( _x) (� 2 Cc(G)):

108

CHAPITRE 8. INDUCED REPRESENTATIONS

Preuve. To verify the lemma, one need only prove that

_� = 0)ZG

(f(x); g(x))�(x)dG(x) = 0:

For this purpose, �x an arbitrary 2 Cc(G).Noting that _� = 0)Z

H

�(x�)dH(�) =

ZH

��x��1

��H��1�dH(�) = 0

we have

0 =

ZG

ZH

(f(x); g(x)) (x)��x��1

��H��1�dH(�)dG(x)

=

ZH

ZG

(f(x); g(x)) (x)��x��1

��H��1�dG(x)dH(�)

=

ZH

ZG

(f(x�); g(x�)) (x�)�(x)�G(�)�H��1�dG(x)dH(�)

=

ZG

(f(x); g(x))�(x)

�ZH

(x�)dH(�)

�dG(x):

All that remains to do is choose so that the function x 7!RH (x�) dH(�)

is 1 on the compact support of �. �

Dé�nition 8.1.3 Let EL denote the subset of EL consisting of those f suchthat �f;f (G=H) <1.Due to the fact that

kf(�) + g(�)k2 6 kf(�)k2 + kg(�)k2�f; g 2 EL

�;

it is clear that EL is a linear subspace of EL. Evidently f; g 2 EL ) �f;g(G=H) <

1. Set (f; g) = �f;g(G=H)�f; g 2 EL

�; plainly (�; �) is a positive semi-

de�nite Hermitian form on EL. Agreeing to identify functions which are

equal almost everywhere, we then see that the form (�; �) equips EL with thestructure of a pre-Hilbert space. In order to show completeness, we need the

following estimate. Write kfk for (f; f)1=2�f 2 EL

�.

Lemme 8.1.4 For each compact subset ! of G, there exists a constant r!such that for all f 2 EL, Z

!

kf(x)kdG(x) 6 r!kfk:



Preuve. Choose � 2 C+c (G) such that � = 1 on !� thenZ!

kf(x)k2dG(x) 6ZG

�(x)kf(x)k2dG(x) = �f;f

�_��6 _�

1kfk2:

Therefore, thanks to Schwarz inequality, we may take

r! =

� _� 1

Z!

dG(x)

�1=2:�

Proposition 8.1.5 The space EL is complete.

Preuve. Let ffng be a Cauchy sequence in EL ; in order to show that ffnghas a limit in EL we may, by passing to a subsequence if necessary, assume

that kfn � fn+1k < 2�n.(1) We begin by proving that lim

n!1fn(x) exists in E for almost all x 2 G.

Thus let ! be a compact subset of G� then, owing to Lemma 8.1.4, we haveZ!

kfn(x)�fn+1(x)kdG(x) < 2�nr! )Z!

1X1

kfn(x)� fn+1(x)k!dG(x) < r!:

Therefore, for almost all x 2 !, ffn(x)g is Cauchy in E. Put f(x) =

limn!1

fn(x) or 0 according to whether the limit exists or not. Of course f

is dG�measurable and veri�es, almost everywhere,

f (x�) =

��H(�)

�G(�)

�1=2L��1�f(x) (� 2 H; x 2 G):

(2) We still have to show that kf(�)k2 is locally summable, that kfk <1,and that kfn � fk ! 0. Let � 2 C+c (G)� then, iterating the parallelogram

identity for E, we �nd thatZG

kfn(x)� fn+mk2�(x)dG(x) 61Xi=1

2iZG

kfn+i�1(x)� fn+i(x)k2�(x)dG(x)

61Xi=1

2ikfn+i�1 � fn+ik2 _�

1

< 2�2n+2 _�

1:

By Fatou�s Lemma,RGkfn(x) � f(x)k2�(x)dG(x) 6 2�2n+2

_� 1. Letting

! be compact in G and taking � = 1 on !, we see that kfn(�) � f(�)k2 is



summable on !, whence fn � f 2 EL ) f 2 EL. On the other hand, as �was arbitrary in G+c (G), our computation also tells us that

kfn � fk 6 2�2n+2 ) f 2 EL;

�nally, it is now obvious that fn ! f in EL.

The proposition is thereby established. �

A priori, our Hilbert space EL might consist of zero alone ; to show that

this is not the case, we shall now discuss a procedure, introduced by Mackey

and elaborated upon by Bruhat, which serve to establish among other things,

that EL contains plenty of functions.

Convention Let E and F be two Hausdor¤ topological vector spaces

over C, f a linear map of E onto F . We shall say that f is a strict morphismif the canonical bijection of E=f�1(0) onto F which is associated with f is

an isomorphism (of topological vector spaces). For this, it is necessary and

su¢ cient that f be continuous and open.

It will be convenient to make a temporary change in our hypotheses ;

thus, let E be a Fréchet space, L a di¤erentiable representation of H on

E�consider the space LC1(G;E) of all functions f on G with values in E

such that :

1. [(a)]

2. f(x�) = �H(�)1=2L

��1�f(x) for all � 2 H and x 2 G ;

3. The canonical image in G=H of the support of f is compact (brie�y f

has compact support mod H) ;

4. f 2 C1(G;E).[Here �H is to be taken as in appendix 1 ; in particular, then, �H(�) = (�H(�)=�G(�))

(� 2 H):]The space LC1(G;E)may be topologized in the following way. Let ! be

a compact subset of G ; let LC1! (G;E) be the subspace ofLC1(G;E)



comprised of those functions with support contained in !H. Place onLC1! (G;E) the ralative topology inherited from C1(G;E) ; in this to-

pology LC1! (G;E) is a Fréchet space.

Equip LC1! (G;E) with the strict inductive limit of the topologies of

the LC1! (G;E)� then LC1(G;E) is an LF�space.Given f 2 C1c (G;E), put

fL(x) =

ZH

�H(�)�1=2L(�)f(x�)dH(�) (x 2 G; dH left Haar measure on H):

[The integral appearing on the right here exist in the Bochner sense, since

the integrand is continuous and has compact support.]

Lemme 8.1.6 (Bruhat) The map �L, f 7! fL, is a continuous surjection

of C1c (G;E) ontoLC1(G;E)� in fact, it is a strict morphism.

Preuve. (1) Plainly fL satis�es (a) and (b) above. As for (c) observe that

for each X 2 g (viewed as a right invariant di¤erential operator on G),

XfL(x) = limt!0

ZH

�H(�)�1=2L(�)

�f (exp(�tX)x�)� f(x�)

t

�dH(�) (x 2 G):

Since spt(f) is compact and since the operators L(�) constitute an equi-

continuous set when � runs through a compact subset of H, we see that the

limit on the right hand side, as t! 0, exists. Hence fL 2 C1(G;E) and foreach right invariant di¤erential operator D 2 G we have

DfL(x) =

ZH

�H(�)�1=2L(�)Df(x�)dH(�) (x 2 G):

In addition note that f 2 C1! (G;E)) fL 2L C1! (G;E) (! a compact subset of G):

(2) Let us now establish the continuity of the map f 7! fL. Il su¢ ces to

show that the map f 7! fL of C1! (G;E) intoLC1! (G;E) is continuous (!

a compact subset of G). So suppose fn ! 0 in C1! (G;E). We must prove

that DfLn ! 0 on each compact subset !1 � G (D 2 G). Let P1 be a closedconvex balanced neighborhood of zero in E ; per P1, select a neighborhoodP2 of zero in E such that a 2 P2; � 2 !�11 ! \H ) L(�)a 2 P1.



�Evidently the function � 7! f(x�) (f 2 C1! (G;E)) has its support in 2 !�11 ! \H if x 2 !1:

�Now we may �nd an index n0 so that n > n0 ) Dfn(x) 2 P2 for all x 2 G ;therefore, if n > n0, then

DfLn (x) 2�Z

!2

�H(�)�1=2dH(�)

�P1 (!2 = !�11 ! \H)

for all x 2 !1 and so the continuity of the map f 7! fL follows.

(3) We have yet to prove that f 7! fL is surjective ; in so doing it will be

seen that our mapping is actually a strict morphism. First of all observe that

if � 2 C1c (G=H) and h 2L C1(G;E), then the function x 7! � ( _x)h(x) is

still in LC1(G;E) ( the map x 7! �(x) = _x denoting, as usual, the canonical

projection of G onto G=H). Hence LC1(G;E) is a module over C1c (G=H).

Viewing G in the usual way as a principal �ber bundle over G=H, select

a covering fOig of G=H by open sets in each of which there exists a C1

section _x 7! si( _x) of ��1(Oi) �bered by H. Let f�ig be a C1 partition of

unity subordinate to this covering and let be a function in C1c (H) such

thatRH�H(�)

�1=2 (�)dH(�) = 1. For h 2L C1(G;E), put

Lh(x) =Xi

�i ( _x) �si ( _x)

�1 x�L�x�1si ( _x)

�h (si ( _x)) (x 2 G):

Then Lh 2 C1c (G;E); h 7!L h is continuous map from LC1(G;E) into

C1c (G;E) and LhL = h, which shows that h 7!L h is a continuous right

inverse to f 7! fL. Therefore the map f 7! fL of C1c (G;E) intoLC1(G;E)

is indeed a strict morphism.

Hence the lemma. �Remarks (1) The proof of lemma 8.1.6 shows that the map f 7! fL of

Cpc (G;E) into the spaceLCpc (G;E) of p times (0 6 p 6 1) di¤erentiable

functions on G with values in E verifying (a) and (b) above, equipped with

the evident topology, is a strict morphism. Take in particular for L the repre-

sentation � 7! �H(�)1=2 (dim(E) = 1) - then LCpc (G) is canonically identi�ed

with the space Cpc (G=H), and thus the map f 7! fL of Cpc (G) into Cpc (G=H)

is a strict morphism, a well-known result due to Weil[1].



(2) Let us suppose that L is merely a continuous representation of H on

our Fréchet space E ; then, of course, the natural counterpart to the spaceLC1(G;E) (in the di¤erentiable case) is the space LC0(G;E) =L C(G;E) of

continuous E-valued functions f on G, having compact support modulo H

and verifying the relation

f(x�) = �H(�)1=2L

��1�f(x) (� 2 H; x 2 G):

Equipping the spaces Cc(G;E), LC(G;E) with the evident inductive limit

topologies, one veri�es without di¢ culty that the map

�L; f 7! fL; fL(x) =

ZH

�H(�)�1=2L(�)f(x�)dH(�) (x 2 G);

is a strict morphism of Cc(G;E) onto LC(G;E) (cf. Remark 1).

We shall now return to the situation initially under consideration ; thus,

in particular, we are once again working with a unitary representation L

of H on a Hilbert space E. As in number 4.4.1, let L1 be the di¤erentiable

representation of H on E1 canonically associated with L. As we know, E1 is

a Fréchet space ; hence it makes sense to consider the space L1C1(G;E1)�plainly L1C1(G;E1) �L C(G;E) � EL. [Let us recall that the topology

on E1 is not the relative topology induced by E but rather the (in general

�ner) topology which E1 inherits from C1(H;E) via the map a 7! ~a; ~a(�) =

L(�)a (a 2 E1; � 2 H) .]

8.2 The carrier space of the induced repre-

sentation

Lemme 8.2.1 The injection of L1C1(G;E1) into EL is continuous and, in

fact, admits a continuous extension to LC(G;E)� furthermore, L1C1(G;E1)is dense in EL.

Preuve. (1) Let us begin by estimating fL1 (f 2 C1c (G;E1)). Choose

� 2 C+c (G) such that _� = 1 on _spt(f). Let g 2 EL. Since fL1 vanishes



outside of !H(! = spt(f)), the measure �fL1;g must be carried by_spt(f).

Hence, using Fubini�s Theorem, we �nd that

(fL1 ; g) = �fL1 ;g(G=H)

=

ZG

�(x)�fL1(x); g(x)

�dG(x)

=

ZG

ZH

�(x)�G(�)1=2�H(�)

�1=2(f(x�); L(��1)g(x))dH(�)dG(x)

=

ZG

ZH

�(x)�G(�)�H(�)�1(f(x�); g(x�))dH(�)dG(x)

=

ZG

ZH

�(x)�G��1� �f�x��1

�; g�x��1

��dH(�)dG(x)

=

ZH

ZG

�(x�)(f(x); g(x))dG(x)dH(�)

=

ZG

(f(x); g(x))dG(x):

Conclusion :��(fL1 ; g)�� 6 rspt(f)kfk1kgk ) fL1 6 rspt(f)kfk1

(cf. Lemma 8.1.4).

(2) The estimate in (1), together whith Lemma 8.1.6, implies that the

injection of L1C1(G;E1) into EL is continuous ; the argument also shows

that this injection admits a continuous extension to LC(G;E). [As was shown

in part (3) of the proof of Lemma 8.1.6, the map f 7! fL1 of C1c (G;E1) ontoL1C1(G;E1) admits a continuous right inverse h 7!L1 h

�h 2L1 C1(G;E1)

�.

So suppose, for instance, that hn ! 0 in L1C1(G;E1) ; to verify that hn ! 0

in EL, �rst note that L1hn ! 0 in C1c (G;E1). This being the case, select

a compact set ! � G such that spt (L1hn) � ! (all n) ; in view of preceding

estimate, we then have that

kL1hL1n k = khnk 6 r!kL1hnk1 ! 0

as n!1: ]

(3) Let f 2 C1c (G); a 2 E1� then�fO

a�L1

(x) =

ZH

�H(�)�1=2f(x�)L1(�)adH(�) (x 2 G):



To prove that L1C1(G;E1)) is dense in EL, it su¢ ces to show that the

(fN

a)L1

(f 2 C1c (G); a 2 E1) span a dense subspace of EL (since EL is complete) ;in turn, for this, it will be enough to show that only zero is orthogonal to all

the (fN

a)L1.

If�(fN

a)L1 ; g�= 0 for some g 2 EL (all f 2 C1c (G); a 2 E1), then, by

(1) above, ZG

f(x)(a; g(x))dG(x) = 0 (f 2 C1c (G); a 2 E1):

Hence (a; g(x)) = 0 almost everywhere onG (all a 2 E1). Therefore g(x) = 0almost everywhere on G.

The proof of Lemma 8.2.1 is now complete. �With these preliminaries out of the way, let us proceed to the de�nition

of the unitary representation UL of G induced by the unitary representation

L of H on E.

The reprentation space for UL will be the Hilbert space EL, given f 2EL,UL (x) f = f � Lx�1 i.e the left translation of f by x�1 (x2 G):


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