Ta ν = ’Te (ρa’1 ).Ta (»a’1 ) = ’Te (»a’1 ).Ta (ρa’1 ).

Proof. The equation µ(x, ν(x)) = e determines ν implicitly. Since we have

Te (µ(e, )) = Te (»e ) = Id, the mapping ν is smooth in a neighborhood of e by

the implicit function theorem. From (ν —¦ »a )(x) = x’1 .a’1 = (ρa’1 —¦ ν)(x) we

may conclude that ν is everywhere smooth. Now we di¬erentiate the equation

µ(a, ν(a)) = e; this gives in turn

0e = T(a,a’1 ) µ.(Xa , Ta ν.Xa ) = Ta (ρa’1 ).Xa + Ta’1 (»a ).Ta ν.Xa ,

Ta ν.Xa = ’Te (»a )’1 .Ta (ρa’1 ).Xa = ’Te (»a’1 ).Ta (ρa’1 ).Xa .

4.4. Example. The general linear group GL(n, R) is the group of all invertible

real n — n-matrices. It is an open subset of L(Rn , Rn ), given by det = 0 and a

Lie group.

Similarly GL(n, C), the group of invertible complex n — n-matrices, is a Lie

group; also GL(n, H), the group of all invertible quaternionic n — n-matrices, is

a Lie group, but the quaternionic determinant is a more subtle instrument here.

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4. Lie groups 31

4.5. Example. The orthogonal group O(n, R) is the group of all linear isome-

tries of (Rn , , ), where , is the standard positive de¬nite inner prod-

n

uct on R . The special orthogonal group SO(n, R) := {A ∈ O(n, R) : det A = 1}

is open in O(n, R), since

’1 0

O(n, R) = SO(n, R) SO(n, R),

0 In’1

where Ik is short for the identity matrix IdRk . We claim that O(n, R) and

SO(n, R) are submanifolds of L(Rn , Rn ). For that we consider the mapping

f : L(Rn , Rn ) ’ L(Rn , Rn ), given by f (A) = A.At . Then O(n, R) = f ’1 (In );

so O(n, R) is closed. Since it is also bounded, O(n, R) is compact. We have

df (A).X = X.At + A.X t , so ker df (In ) = {X : X + X t = 0} is the space o(n, R)

of all skew symmetric n — n-matrices. Note that dim o(n, R) = 1 (n ’ 1)n. If

2

A is invertible, we get ker df (A) = {Y : Y.A + A.Y = 0} = {Y : Y.At ∈

t t

o(n, R)} = o(n, R).(A’1 )t . The mapping f takes values in Lsym (Rn , Rn ), the

space of all symmetric n — n-matrices, and dim ker df (A) + dim Lsym (Rn , Rn ) =

1 1 2 n n n n

2 (n ’ 1)n + 2 n(n + 1) = n = dim L(R , R ), so f : GL(n, R) ’ Lsym (R , R )

is a submersion. Since obviously f ’1 (In ) ‚ GL(n, R), we conclude from 1.10

that O(n, R) is a submanifold of GL(n, R). It is also a Lie group, since the group

operations are obviously smooth.

4.6. Example. The special linear group SL(n, R) is the group of all n — n-

matrices of determinant 1. The function det : L(Rn , Rn ) ’ R is smooth and

d det(A)X = trace(C(A).X), where C(A)i , the cofactor of Aj , is the determinant

j i

j

of the matrix, which results from putting 1 instead of Ai into A and 0 in the rest

of the j-th row and the i-th column of A. We recall Cramer™s rule C(A).A =

A.C(A) = det(A).In . So if C(A) = 0 (i.e. rank(A) ≥ n ’ 1) then the linear

functional df (A) is non zero. So det : GL(n, R) ’ R is a submersion and

SL(n, R) = (det)’1 (1) is a manifold and a Lie group of dimension n2 ’ 1. Note

¬nally that TIn SL(n, R) = ker d det(In ) = {X : trace(X) = 0}. This space of

traceless matrices is usually called sl(n, R).

4.7. Example. The symplectic group Sp(n, R) is the group of all 2n — 2n-

matrices A such that ω(Ax, Ay) = ω(x, y) for all x, y ∈ R2n , where ω is the

standard non degenerate skew symmetric bilinear form on R2n .

Such a form exists on a vector space if and only if the dimension is even, and

on Rn —(Rn )— the standard form is given by ω((x, x— ), (y, y — )) = x, y — ’ y, x— ,

n

i.e. in coordinates ω((xi )2n , (y j )2n ) = i=1 (xi y n+i ’ xn+i y i ). Any symplectic

i=1 j=1

form on R2n looks like that after choosing a suitable basis. Let (ei )2n be the

i=1

standard basis in R2n . Then we have

0 In

(ω(ei , ej )i ) = =: J,

j

’In 0

and the matrix J satis¬es J t = ’J, J 2 = ’I2n , J x = ’x in Rn — Rn , and

y

y

ω(x, y) = x, Jy in terms of the standard inner product on R2n .

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32 Chapter I. Manifolds and Lie groups

For A ∈ L(R2n , R2n ) we have ω(Ax, Ay) = Ax, JAy = x, At JAy . Thus

A ∈ Sp(n, R) if and only if At JA = J.

We consider now the mapping f : L(R2n , R2n ) ’ L(R2n , R2n ) given by

f (A) = At JA. Then f (A)t = (At JA)t = ’At JA = ’f (A), so f takes val-

ues in the space o(2n, R) of skew symmetric matrices. We have df (A)X =

X t JA + At JX, and therefore

ker df (I2n ) = {X ∈ L(R2n , R2n ) : X t J + JX = 0}

= {X : JX is symmetric} =: sp(n, R).

We see that dim sp(n, R) = 2n(2n+1) = 2n+1 . Furthermore we have ker df (A) =

2 2

{X : X t JA + At JX = 0} and X ’ At JX is an isomorphism ker df (A) ’

Lsym (R2n , R2n ), if A is invertible. Thus dim ker df (A) = 2n+1 for all A ∈

2

GL(2n, R). If f (A) = J, then At JA = J, so A has rank 2n and is invertible, and

dim ker df (A) + dim o(2n, R) = 2n+1 + 2n(2n’1) = 4n2 = dim L(R2n , R2n ). So

2 2

f : GL(2n, R) ’ o(2n, R) is a submersion and f ’1 (J) = Sp(n, R) is a manifold

and a Lie group. It is the symmetry group of ˜classical mechanics™.

4.8. Example. The complex general linear group GL(n, C) of all invertible

complex n — n-matrices is open in LC (Cn , Cn ), so it is a real Lie group of real

dimension 2n2 ; it is also a complex Lie group of complex dimension n2 . The

complex special linear group SL(n, C) of all matrices of determinant 1 is a sub-

manifold of GL(n, C) of complex codimension 1 (or real codimension 2).

The complex orthogonal group O(n, C) is the set

{A ∈ L(Cn , Cn ) : g(Az, Aw) = g(z, w) for all z, w},

n

where g(z, w) = i=1 z i wi . This is a complex Lie group of complex dimension

(n’1)n

, and it is not compact. Since O(n, C) = {A : At A = In }, we have

2

1 = detC (In ) = detC (At A) = detC (A)2 , so detC (A) = ±1. Thus SO(n, C) :=

{A ∈ O(n, C) : detC (A) = 1} is an open subgroup of index 2 in O(n, C).

The group Sp(n, C) = {A ∈ LC (C2n , C2n ) : At JA = J} is also a complex Lie

group of complex dimension n(2n + 1).

These groups here are the classical complex Lie groups. The groups SL(n, C)

for n ≥ 2, SO(n, C) for n ≥ 3, Sp(n, C) for n ≥ 4, and ¬ve more exceptional

groups exhaust all simple complex Lie groups up to coverings.

4.9. Example. Let Cn be equipped with the standard hermitian inner product

n ii

i=1 z w . The unitary group U (n) consists of all complex n — n-

(z, w) =

matrices A such that (Az, Aw) = (z, w) for all z, w holds, or equivalently U (n) =

t

{A : A— A = In }, where A— = A .

We consider the mapping f : LC (Cn , Cn ) ’ LC (Cn , Cn ), given by f (A) =

A— A. Then f is smooth but not holomorphic. Its derivative is df (A)X =

X — A + A— X, so ker df (In ) = {X : X — + X = 0} =: u(n), the space of all skew

hermitian matrices. We have dimR u(n) = n2 . As above we may check that

f : GL(n, C) ’ Lherm (Cn , Cn ) is a submersion, so U (n) = f ’1 (In ) is a compact

real Lie group of dimension n2 .

The special unitary group is SU (n) = U (n) © SL(n, C). For A ∈ U (n) we

have | detC (A)| = 1, thus dimR SU (n) = n2 ’ 1.

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4. Lie groups 33

4.10. Example. The group Sp(n). Let H be the division algebra of quater-

nions. Then Sp(1) := S 3 ‚ H ∼ R4 is the group of unit quaternions, obviously

=

a Lie group.

Now let V be a right vector space over H. Since H is not commutative, we

have to distinguish between left and right vector spaces and we choose right ones

as basic, so that matrices can multiply from the left. By choosing a basis we get

n

V = Rn —R H = Hn . For u = (ui ), v = (v i ) ∈ Hn we put u, v := i=1 ui v i .

is R-bilinear and ua, vb = a u, v b for a, b ∈ H.

Then ,

An R linear mapping A : V ’ V is called H-linear or quaternionically linear

if A(ua) = A(u)a holds. The space of all such mappings shall be denoted by

LH (V, V ). It is real isomorphic to the space of all quaternionic n — n-matrices

with the usual multiplication, since for the standard basis (ei )n in V = Hn we

i=1

have A(u) = A( i ei ui ) = i A(ei )ui = i,j ej Aj ui . Note that LH (V, V ) is

i

only a real vector space, if V is a right quaternionic vector space - any further

structure must come from a second (left) quaternionic vector space structure on

V.

GL(n, H), the group of invertible H-linear mappings of Hn , is a Lie group,

because it is GL(4n, R) © LH (Hn , Hn ), open in LH (Hn , Hn ).

A quaternionically linear mapping A is called isometric or quaternionically

unitary, if A(u), A(v) = u, v for all u, v ∈ Hn . We denote by Sp(n) the

group of all quaternionic isometries of Hn , the quaternionic unitary group. The

reason for its name is that Sp(n) = Sp(2n, C) © U (2n), since we can decompose

the quaternionic hermitian form , into a complex hermitian one and a

complex symplectic one. Also we have Sp(n) ‚ O(4n, R), since the real part of

is a positive de¬nite real inner product. For A ∈ LH (Hn , Hn ) we put

,

t

A— := A . Then we have u, A(v) = A— (u), v , so A(u), A(v) = A— A(u), v .

Thus A ∈ Sp(n) if and only if A— A = Id.

Again f : LH (Hn , Hn ) ’ LH,herm (Hn , Hn ) = {A : A— = A}, given by f (A) =

A— A, is a smooth mapping with df (A)X = X — A+A— X. So we have ker df (Id) =

{X : X — = ’X} =: sp(n), the space of quaternionic skew hermitian matrices.

The usual proof shows that f has maximal rank on GL(n, H), so Sp(n) = f ’1 (Id)

is a compact real Lie group of dimension 2n(n ’ 1) + 3n.

The groups SO(n, R) for n ≥ 3, SU (n) for n ≥ 2, Sp(n) for n ≥ 2 and

real forms of the exceptional complex Lie groups exhaust all simple compact Lie

groups up to coverings.

4.11. Invariant vector ¬elds and Lie algebras. Let G be a (real) Lie group.

A vector ¬eld ξ on G is called left invariant, if »— ξ = ξ for all a ∈ G, where

a

»— ξ = T (»a’1 )—¦ξ —¦»a as in section 3. Since by 3.11 we have »— [ξ, ·] = [»— ξ, »— ·],

a a a a

the space XL (G) of all left invariant vector ¬elds on G is closed under the Lie

bracket, so it is a sub Lie algebra of X(G). Any left invariant vector ¬eld ξ

is uniquely determined by ξ(e) ∈ Te G, since ξ(a) = Te (»a ).ξ(e). Thus the Lie

algebra XL (G) of left invariant vector ¬elds is linearly isomorphic to Te G, and

on Te G the Lie bracket on XL (G) induces a Lie algebra structure, whose bracket

is again denoted by [ , ]. This Lie algebra will be denoted as usual by g,

sometimes by Lie(G).

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34 Chapter I. Manifolds and Lie groups

We will also give a name to the isomorphism with the space of left invariant

vector ¬elds: L : g ’ XL (G), X ’ LX , where LX (a) = Te »a .X. Thus [X, Y ] =

[LX , LY ](e).

A vector ¬eld · on G is called right invariant, if ρ— · = · for all a ∈ G. If

a

—

ξ is left invariant, then ν ξ is right invariant, since ν —¦ ρa = »a’1 —¦ ν implies

that ρ— ν — ξ = (ν —¦ ρa )— ξ = (»a’1 —¦ ν)— ξ = ν — (»a’1 )— ξ = ν — ξ. The right invariant

a

vector ¬elds form a sub Lie algebra XR (G) of X(G), which is again linearly

isomorphic to Te G and induces also a Lie algebra structure on Te G. Since

ν — : XL (G) ’ XR (G) is an isomorphism of Lie algebras by 3.11, Te ν = ’ Id :

Te G ’ Te G is an isomorphism between the two Lie algebra structures. We will

denote by R : g = Te G ’ XR (G) the isomorphism discussed, which is given by

RX (a) = Te (ρa ).X.

4.12. Lemma. If LX is a left invariant vector ¬eld and RY is a right invariant

one, then [LX , RY ] = 0. Thus the ¬‚ows of LX and RY commute.

Proof. We consider 0 — LX ∈ X(G — G), given by (0 — LX )(a, b) = (0a , LX (b)).

Then T(a,b) µ.(0a , LX (b)) = Ta ρb .0a + Tb »a .LX (b) = LX (ab), so 0 — LX is µ-

related to LX . Likewise RY —0 is µ-related to RY . But then 0 = [0—LX , RY —0]

is µ-related to [LX , RY ] by 3.10. Since µ is surjective, [LX , RY ] = 0 follows.

4.13. Let • : G ’ H be a homomorphism of Lie groups, so for the time being

we require • to be smooth.

Lemma. Then • := Te • : g = Te G ’ h = Te H is a Lie algebra homomor-

phism.

Proof. For X ∈ g and x ∈ G we have

Tx •.LX (x) = Tx •.Te »x .X = Te (• —¦ »x ).X =

Te (»•(x) —¦ •).X = Te (»•(x) ).Te •.X = L• (X) (•(x)).

So LX is •-related to L• (X) . By 3.10 the ¬eld [LX , LY ] = L[X,Y ] is •-related

to [L• (X) , L• (Y ) ] = L[• (X),• (Y )] . So we have T • —¦ L[X,Y ] = L[• (X),• (Y )] —¦ •.

If we evaluate this at e the result follows.

Now we will determine the Lie algebras of all the examples given above.

4.14. For the Lie group GL(n, R) we have Te GL(n, R) = L(Rn , Rn ) =: gl(n, R)

and T GL(n, R) = GL(n, R) — L(Rn , Rn ) by the a¬ne structure of the sur-

rounding vector space. For A ∈ GL(n, R) we have »A (B) = A.B, so »A

extends to a linear isomorphism of L(Rn , Rn ), and for (B, X) ∈ T GL(n, R)

we get TB (»A ).(B, X) = (A.B, A.X). So the left invariant vector ¬eld LX ∈

XL (GL(n, R)) is given by LX (A) = Te (»A ).X = (A, A.X).

Let f : GL(n, R) ’ R be the restriction of a linear functional on L(Rn , Rn ).

Then we have LX (f )(A) = df (A)(LX (A)) = df (A)(A.X) = f (A.X), which we

may write as LX (f ) = f ( .X). Therefore

L[X,Y ] (f ) = [LX , LY ](f ) = LX (LY (f )) ’ LY (LX (f )) =

= LX (f ( .Y )) ’ LY (f ( .X)) =

= f ( .X.Y ) ’ f ( .Y.X) = LXY ’Y X (f ).

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4. Lie groups 35

So the Lie bracket on gl(n, R) = L(Rn , Rn ) is given by [X, Y ] = XY ’ Y X, the

usual commutator.

4.15. Example. Let V be a vector space. Then (V, +) is a Lie group, T0 V = V

is its Lie algebra, T V = V —V , left translation is »v (w) = v+w, Tw (»v ).(w, X) =

(v + w, X). So LX (v) = (v, X), a constant vector ¬eld. Thus the Lie bracket is

0.

4.16. Example. The special linear group is SL(n, R) = det’1 (1) and its Lie

algebra is given by Te SL(n, R) = ker d det(I) = {X ∈ L(Rn , Rn ) : trace X =

0} = sl(n, R) by 4.6. The injection i : SL(n, R) ’ GL(n, R) is a smooth

homomorphism of Lie groups, so Te i = i : sl(n, R) ’ gl(n, R) is an injective

homomorphism of Lie algebras. Thus the Lie bracket is given by [X, Y ] =

XY ’ Y X.

The same argument gives the commutator as the Lie bracket in all other

examples we have treated. We have already determined the Lie algebras as Te G.

4.17. One parameter subgroups. Let G be a Lie group with Lie algebra g.

A one parameter subgroup of G is a Lie group homomorphism ± : (R, +) ’ G,

i.e. a smooth curve ± in G with ±(0) = e and ±(s + t) = ±(s).±(t).

Lemma. Let ± : R ’ G be a smooth curve with ±(0) = e. Let X = ±(0) ∈ g.

™

Then the following assertions are equivalent.

(1) ± is a one parameter subgroup.

±(t) = FlLX (t, e) for all t.

(2)

±(t) = FlRX (t, e) for all t.

(3)

x.±(t) = FlLX (t, x), or FlLX = ρ±(t) , for all t.

(4) t

±(t).x = FlRX (t, x), or FlRX = »±(t) , for all t.

(5) t

Proof. (1) =’ (4).

d d d d

ds |0 x.±(t ds |0 x.±(t).±(s) ds |0 »x.±(t) ±(s)

dt x.±(t) = + s) = =

d

= Te (»x.±(t) ). ds |0 ±(s) = LX (x.±(t)).

By uniqueness of solutions we get x.±(t) = FlLX (t, x).

(4) =’ (2). This is clear.

d d d

(2) =’ (1). We have ds ±(t)±(s) = ds (»±(t) ±(s)) = T (»±(t) ) ds ±(s) =

T (»±(t) )LX (±(s)) = LX (±(t)±(s)) and ±(t)±(0) = ±(t). So we get ±(t)±(s) =

FlLX (s, ±(t)) = FlLX FlLX (e) = FlLX (t + s, e) = ±(t + s).

s t —

(4) ⇐’ (5). We have Flν ξ = ν ’1 —¦ Flξ —¦ν by 3.14. Therefore we have by 4.11

t t

—

(FlRX (x’1 ))’1 = (ν —¦ FlRX —¦ν)(x) = Flν RX

(x)

t t t

= FlLX (x) = x.±(’t).

’t

So FlRX (x’1 ) = ±(t).x’1 , and FlRX (y) = ±(t).y.

t t

(5) =’ (3) =’ (1) can be shown in a similar way.

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993

36 Chapter I. Manifolds and Lie groups

An immediate consequence of the foregoing lemma is that left invariant and

the right invariant vector ¬elds on a Lie group are always complete, so they

have global ¬‚ows, because a locally de¬ned one parameter group can always be

extended to a globally de¬ned one by multiplying it up.

4.18. De¬nition. The exponential mapping exp : g ’ G of a Lie group is

de¬ned by

exp X = FlLX (1, e) = FlRX (1, e) = ±X (1),

where ±X is the one parameter subgroup of G with ±X (0) = X.

™

Theorem.

exp : g ’ G is smooth.

(1)

exp(tX) = FlLX (t, e).

(2)

FlLX (t, x) = x. exp(tX).

(3)

FlRX (t, x) = exp(tX).x.

(4)

exp(0) = e and T0 exp = Id : T0 g = g ’ Te G = g, thus exp is a

(5)

di¬eomorphism from a neighborhood of 0 in g onto a neighborhood of e

in G.

Proof. (1) Let 0 — L ∈ X(g — G) be given by (0 — L)(X, x) = (0X , LX (x)). Then

pr2 Fl0—L (t, (X, e)) = ±X (t) is smooth in (t, X).

(2) exp(tX) = Flt.LX (1, e) = FlLX (t, e) = ±X (t).

(3) and (4) follow from lemma 4.17.

(5) T0 exp .X = dt |0 exp(0 + t.X) = dt |0 FlLX (t, e) = X.

d d

4.19. Remark. If G is connected and U ‚ g is open with 0 ∈ U , then the

group generated by exp(U ) equals G.

For this group is a subgroup of G containing some open neighborhood of e,

so it is open. The complement in G is also open (as union of the other cosets),

so this subgroup is open and closed. Since G is connected, it coincides with G.

If G is not connected, then the subgroup generated by exp(U ) is the connected

component of e in G.

4.20. Remark. Let • : G ’ H be a smooth homomorphism of Lie groups.

Then the diagram

w

•

g h

u u

expG expH

wH

•

G

commutes, since t ’ •(expG (tX)) is a one parameter subgroup of H and

d G G H

dt |0 •(exp tX) = • (X), so •(exp tX) = exp (t• (X)).

If G is connected and •, ψ : G ’ H are homomorphisms of Lie groups with

• = ψ : g ’ h, then • = ψ. For • = ψ on the subgroup generated by expG g

which equals G by 4.19.

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4. Lie groups 37

4.21. Theorem. A continuous homomorphism • : G ’ H between Lie groups

is smooth. In particular a topological group can carry at most one compatible

Lie group structure.

Proof. Let ¬rst • = ± : (R, +) ’ G be a continuous one parameter subgroup.

Then ±(’µ, µ) ‚ exp(U ), where U is an absolutely convex open neighbor-

hood of 0 in g such that exp |2U is a di¬eomorphism, for some µ > 0. Put

β := (exp |2U )’1 —¦ ± : (’µ, µ) ’ g. Then for |t| < 1 we have exp(2β(t)) =

µ

exp(β(t))2 = ±(t)2 = ±(2t) = exp(β(2t)), so 2β(t) = β(2t); thus β( 2 ) = 1 β(s) s

2

for |s| < µ. So we have ±( 2 ) = exp(β( 2 )) = exp( 1 β(s)) for all |s| < µ and by

s s

2

recursion we get ±( 2s ) = exp( 21 β(s)) for n ∈ N and in turn ±( 2k s) = ±( 2s )k =

n n n n

1 k k

k

exp( 2n β(s)) = exp( 2n β(s)) for k ∈ Z. Since the 2n for k ∈ Z and n ∈ N are

dense in R and since ± is continuous we get ±(ts) = exp(tβ(s)) for all t ∈ R. So

± is smooth.

Now let • : G ’ H be a continuous homomorphism. Let X1 , . . . , Xn be a lin-

ear basis of g. We de¬ne ψ : Rn ’ G as ψ(t1 , . . . , tn ) = exp(t1 X1 ) · · · exp(tn Xn ).

Then T0 ψ is invertible, so ψ is a di¬eomorphism near 0. Sometimes ψ ’1 is called

a coordinate system of the second kind. t ’ •(expG tXi ) is a continuous one

parameter subgroup of H, so it is smooth by the ¬rst part of the proof. We have