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(• —¦ ψ)(t1 , . . . , tn ) = (• exp(t1 X1 )) · · · (• exp(tn Xn )), so • —¦ ψ is smooth. Thus
• is smooth near e ∈ G and consequently everywhere on G.

4.22. Theorem. Let G and H be Lie groups (G separable is essential here),
and let • : G ’ H be a continuous bijective homomorphism. Then • is a
di¬eomorphism.

Proof. Our ¬rst aim is to show that • is a homeomorphism. Let V be an
open e-neighborhood in G, and let K be a compact e-neighborhood in G such
that K.K ’1 ‚ V . Since G is separable there is a sequence (ai )i∈N in G such

that G = i=1 ai .K. Since H is locally compact, it is a Baire space (Vi open
and dense implies Vi dense). The set •(ai )•(K) is compact, thus closed.
Since H = i •(ai ).•(K), there is some i such that •(ai )•(K) has non empty
interior, so •(K) has non empty interior. Choose b ∈ G such that •(b) is an
interior point of •(K) in H. Then eH = •(b)•(b’1 ) is an interior point of
•(K)•(K ’1 ) ‚ •(V ). So if U is open in G and a ∈ U , then eH is an interior
point of •(a’1 U ), so •(a) is in the interior of •(U ). Thus •(U ) is open in H,
and • is a homeomorphism.
Now by 4.21 • and •’1 are smooth.

4.23. Examples. The exponential mapping on GL(n, R). Let X ∈ gl(n, R) =
L(Rn , Rn ), then the left invariant vector ¬eld is given by LX (A) = (A, A.X) ∈
GL(n, R) — gl(n, R) and the one parameter group ±X (t) = FlLX (t, I) is given
d
by the di¬erential equation dt ±X (t) = LX (±X (t)) = ±X (t).X, with initial con-
dition ±X (0) = I. But the unique solution of this equation is ±X (t) = etX =
∞ tk k
k=0 k! X . So

expGL(n,R) (X) = eX = 1
Xk.
k=0 k!

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993
38 Chapter I. Manifolds and Lie groups


If n = 1 we get the usual exponential mapping of one real variable. For all Lie
subgroups of GL(n, R), the exponential mapping is given by the same formula
exp(X) = eX ; this follows from 4.20.
4.24. The adjoint representation. A representation of a Lie group G on a
¬nite dimensional vector space V (real or complex) is a homomorphism ρ : G ’
GL(V ) of Lie groups. Then by 4.13 ρ : g ’ gl(V ) = L(V, V ) is a Lie algebra
homomorphism.
For a ∈ G we de¬ne conja : G ’ G by conja (x) = axa’1 . It is called
the conjugation or the inner automorphism by a ∈ G. We have conja (xy) =
conja (x) conja (y), conjab = conja —¦ conjb , and conj is smooth in all variables.
Next we de¬ne for a ∈ G the mapping Ad(a) = (conja ) = Te (conja ) : g ’ g.
By 4.13 Ad(a) is a Lie algebra homomorphism, so we have Ad(a)[X, Y ] =
[Ad(a)X, Ad(a)Y ]. Furthermore Ad : G ’ GL(g) is a representation, called
the adjoint representation of G, since Ad(ab) = Te (conjab ) = Te (conja —¦ conjb ) =
Te (conja ) —¦ Te (conjb ) = Ad(a) —¦ Ad(b). We will use the relations Ad(a) =
Te (conja ) = Ta (ρa’1 ).Te (»a ) = Ta’1 (»a ).Te (ρa’1 ).
Finally we de¬ne the (lower case) adjoint representation of the Lie algebra g,
ad : g ’ gl(g) = L(g, g), by ad := Ad = Te Ad.
Lemma. (1) LX (a) = RAd(a)X (a) for X ∈ g and a ∈ G.
(2) ad(X)Y = [X, Y ] for X, Y ∈ g.
Proof. (1) LX (a) = Te (»a ).X = Te (ρa ).Te (ρa’1 —¦ »a ).X = RAd(a)X (a).
(2) Let X1 , . . . , Xn be a linear basis of g and ¬x X ∈ g. Then Ad(x)X =
n ∞
i=1 fi (x).Xi for fi ∈ C (G, R) and we have in turn

Ad (Y )X = Te (Ad( )X)Y = d(Ad( )X)|e Y = d( fi Xi )|e Y =
dfi |e (Y )Xi =
= LY (fi )(e).Xi .
LX (x) = RAd(x)X (x) = R( fi (x)Xi )(x) = fi (x).RXi (x) by (1).
[LY , LX ] = [LY , fi .RXi ] = 0 + LY (fi ).RXi by 3.4 and 4.12.
[Y, X] = [LY , LX ](e) = LY (fi )(e).RXi (e) = Ad (Y )X = ad(Y )X.

4.25. Corollary. From 4.20 and 4.23 we have
Ad —¦expG = expGL(g) —¦ ad

G
(ad X)k Y = ead X Y
1
Ad(exp X)Y = k!
k=0
1 1
]]] + · · ·
= Y + [X, Y ] + 2! [X, [X, Y ]] + 3! [X, [X, [X, Y

4.26. The right logarithmic derivative. Let M be a manifold and let f :
M ’ G be a smooth mapping into a Lie group G with Lie algebra g. We de¬ne
the mapping δf : T M ’ g by the formula δf (ξx ) := Tf (x) (ρf (x)’1 ).Tx f.ξx .
Then δf is a g-valued 1-form on M , δf ∈ „¦1 (M, g), as we will write later. We
call δf the right logarithmic derivative of f , since for f : R ’ (R+ , ·) we have
δf (x).1 = f (x) = (log —¦f ) (x).
(x)
f


Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993
4. Lie groups 39


Lemma. Let f, g : M ’ G be smooth. Then we have
δ(f.g)(x) = δf (x) + Ad(f (x)).δg(x).

Proof. We just compute:
δ(f.g)(x) = T (ρg(x)’1 .f (x)’1 ).Tx (f.g) =
= T (ρf (x)’1 ).T (ρg(x)’1 ).T(f (x),g(x)) µ.(Tx f, Tx g) =
= T (ρf (x)’1 ).T (ρg(x)’1 ). T (ρg(x) ).Tx f + T (»f (x) ).Tx g =
= δf (x) + Ad(f (x)).δg(x).

Remark. The left logarithmic derivative δ left f ∈ „¦1 (M, g) of a smooth mapping
f : M ’ G is given by δ left f.ξx = Tf (x) (»f (x)’1 ).Tx f.ξx . The corresponding
Leibnitz rule for it is uglier than that for the right logarithmic derivative:
δ left (f g)(x) = δ left g(x) + Ad(g(x)’1 )δ left f (x).
The form δ left (IdG ) ∈ „¦1 (G; g) is also called the Maurer-Cartan form of the Lie
group G.
ez ’ 1
4.27. Lemma. For exp : g ’ G and for g(z) := we have
z

(ad X)p = g(ad X).
1
δ(exp)(X) = T (ρexp(’X) ).TX exp = (p+1)!
p=0


Proof. We put M (X) = δ(exp)(X) : g ’ g. Then
(s + t)M ((s + t)X) = (s + t)δ(exp)((s + t)X)
= δ(exp((s + t) ))X by the chain rule,
= δ(exp(s ). exp(t )).X
= δ(exp(s )).X + Ad(exp(sX)).δ(exp(t )).X by 4.26,
= s.δ(exp)(sX) + Ad(exp(sX)).t.δ(exp)(tX)
= s.M (sX) + Ad(exp(sX)).t.M (tX).
Now we put N (t) := t.M (tX) ∈ L(g, g), then the above equation gives N (s+t) =
d
N (s) + Ad(exp(sX)).N (t). We ¬x t, apply ds |0 , and get N (t) = N (0) +
ad(X).N (t), where N (0) = M (0) + 0 = δ(exp)(0) = Idg . So we have the
di¬erential equation N (t) = Idg + ad(X).N (t) in L(g, g) with initial condition
N (0) = 0. The unique solution is

ad(X)p .sp+1 ,
1
N (s) = and so
(p+1)!
p=0

ad(X)p .
1
δ(exp)(X) = M (X) = N (1) = (p+1)!
p=0




Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993
40 Chapter I. Manifolds and Lie groups


4.28. Corollary. TX exp is bijective if and only if no eigenvalue of ad(X) :

g ’ g is of the form ’1 2kπ for k ∈ Z \ {0}.

z
’1
Proof. The zeros of g(z) = e z are exactly z = ’1 2kπ for k ∈ Z \ {0}. The
linear mapping TX exp is bijective if and only if no eigenvalue of g(ad(X)) =
T (ρexp(’X) ).TX exp is 0. But the eigenvalues of g(ad(X)) are the images under
g of the eigenvalues of ad(X).
4.29. Theorem. The Baker-Campbell-Hausdor¬ formula.
Let G be a Lie group with Lie algebra g. For complex z near 1 we consider the
n
function f (z) := log(z) = n≥0 (’1) (z ’ 1)n .
z’1 n+1
Then for X, Y near 0 in g we have exp X. exp Y = exp C(X, Y ), where
1
f (et. ad X .ead Y ).X dt
C(X, Y ) = Y +
0
n
1
(’1)n tk
(ad X)k (ad Y )
=X +Y + X dt
n+1 k! !
0 k, ≥0
n≥1
k+ ≥1

(’1)n (ad X)k1 (ad Y ) 1 . . . (ad X)kn (ad Y ) n
=X +Y + X
(k1 + · · · + kn + 1)k1 ! . . . kn ! 1 ! . . . n !
n+1
k1 ,...,kn ≥0
n≥1
1 ,... n ≥0
ki + i ≥1

= X + Y + 1 [X, Y ] + 1
]] ’ [Y, [Y, X]]) + · · ·
12 ([X, [X, Y
2


Proof. Let C(X, Y ) := exp’1 (exp X. exp Y ) for X, Y near 0 in g, and let C(t) :=
C(tX, Y ). Then

d
T (ρexp(’C(t)) ) dt (exp C(t)) = δ exp(C(t)).C(t)
™ ™
1
(ad C(t))k C(t) = g(ad C(t)).C(t),
= k≥0 (k+1)!
z k
’1
where g(z) := e z = z
k≥0 (k+1)! . We have exp C(t) = exp(tX) exp Y and
exp(’C(t)) = exp(C(t))’1 = exp(’Y ) exp(’tX), therefore
d d
T (ρexp(’C(t)) ) dt (exp C(t)) = T (ρexp(’Y ) exp(’tX) ) dt (exp(tX) exp Y )
d
= T (ρexp(’tX) )T (ρexp(’Y ) )T (ρexp Y ) dt exp(tX)
= T (ρexp(’tX) ).RX (exp(tX)) = X, by 4.18.4 and 4.11.

X = g(ad C(t)).C(t).
ead C(t) = Ad(exp C(t)) by 4.25
= Ad(exp(tX) exp Y ) = Ad(exp(tX)). Ad(exp Y )
= ead(tX) .ead Y = et. ad X .ead Y .

If X, Y , and t are small enough we get ad C(t) = log(et. ad X .ead Y ), where
n+1
log(z) = n≥1 (’1) (z ’ 1)n , thus we have
n

™ ™
X = g(ad C(t)).C(t) = g(log(et. ad X .ead Y )).C(t).

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993
5. Lie subgroups and homogeneous Spaces 41

(’1)n
log(z)
’ 1)n , which satis¬es
For z near 1 we put f (z) := = n≥0 n+1 (z
z’1
g(log(z)).f (z) = 1. So we have
™ ™
X = g(log(et. ad X .ead Y )).C(t) = f (et. ad X .ead Y )’1 .C(t),

C(t) = f (et. ad X .ead Y ).X,
C(0) = Y.
Passing to the de¬nite integral we get the desired formula
1

C(X, Y ) = C(1) = C(0) + C(t) dt
0
1
f (et. ad X .ead Y ).X dt
=Y +
0
n
1
(’1)n tk
(ad X)k (ad Y )
=X +Y + X dt.
n+1 k! !
0 k, ≥0
n≥1
k+ ≥1

Remark. If G is a Lie group of di¬erentiability class C 2 , then we may de¬ne
T G and the Lie bracket of vector ¬elds. The proof above then makes sense
and the theorem shows, that in the chart given by exp’1 the multiplication
µ : G — G ’ G is C ω near e, hence everywhere. So in this case G is a real
analytic Lie group. See also remark 5.6 below.
4.30. Convention. We will use the following convention for the rest of the
book. If we write a symbol of a classical group from this section without indi-
cating the ground ¬eld, then we always mean the ¬eld R ” except Sp(n). In
particular GL(n) = GL(n, R), and O(n) = O(n, R) from now on.


5. Lie subgroups and homogeneous spaces

5.1. De¬nition. Let G be a Lie group. A subgroup H of G is called a Lie
subgroup, if H is itself a Lie group (so it is separable) and the inclusion i : H ’ G
is smooth.
In this case the inclusion is even an immersion. For that it su¬ces to check
that Te i is injective: If X ∈ h is in the kernel of Te i, then i —¦ expH (tX) =
expG (t.Te i.X) = e. Since i is injective, X = 0.
From the next result it follows that H ‚ G is then an initial submanifold in
the sense of 2.14: If H0 is the connected component of H, then i(H0 ) is the Lie
subgroup of G generated by i (h) ‚ g, which is an initial submanifold, and this
is true for all components of H.
5.2. Theorem. Let G be a Lie group with Lie algebra g. If h ‚ g is a Lie
subalgebra, then there is a unique connected Lie subgroup H of G with Lie
algebra h. H is an initial submanifold.
Proof. Put Ex := {Te (»x ).X : X ∈ h} ‚ Tx G. Then E := x∈G Ex is a
distribution of constant rank on G, in the sense of 3.18. The set {LX : X ∈ h}

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993
42 Chapter I. Manifolds and Lie groups


is an involutive set in the sense of 3.23 which spans E. So by theorem 3.25 the
distribution E is integrable and by theorem 3.22 the leaf H through e is an initial
submanifold. It is even a subgroup, since for x ∈ H the initial submanifold »x H
is again a leaf (since E is left invariant) and intersects H (in x), so »x (H) = H.
Thus H.H = H and consequently H ’1 = H. The multiplication µ : H — H ’ G
is smooth by restriction, and smooth as a mapping H — H ’ H, since H is an
initial submanifold, by lemma 2.17.
5.3. Theorem. Let g be a ¬nite dimensional real Lie algebra. Then there
exists a connected Lie group G whose Lie algebra is g.
Sketch of Proof. By the theorem of Ado (see [Jacobson, 62] or [Varadarajan, 74,
p. 237]) g has a faithful (i.e. injective) representation on a ¬nite dimensional
vector space V , i.e. g can be viewed as a Lie subalgebra of gl(V ) = L(V, V ).
By theorem 5.2 above there is a Lie subgroup G of GL(V ) with g as its Lie
algebra.
This is a rather involved proof, since the theorem of Ado needs the struc-
ture theory of Lie algebras for its proof. There are simpler proofs available,
starting from a neighborhood of e in G (a neighborhood of 0 in g with the
Baker-Campbell-Hausdor¬ formula 4.29 as multiplication) and extending it.
5.4. Theorem. Let G and H be Lie groups with Lie algebras g and h, re-
spectively. Let f : g ’ h be a homomorphism of Lie algebras. Then there
is a Lie group homomorphism •, locally de¬ned near e, from G to H, such
that • = Te • = f . If G is simply connected, then there is a globally de¬ned
homomorphism of Lie groups • : G ’ H with this property.
Proof. Let k := graph(f ) ‚ g — h. Then k is a Lie subalgebra of g — h, since f is a
homomorphism of Lie algebras. g — h is the Lie algebra of G — H, so by theorem
5.2 there is a connected Lie subgroup K ‚ G — H with algebra k. We consider
the homomorphism g := pr1 —¦ incl : K ’ G — H ’ G, whose tangent mapping
satis¬es Te g(X, f (X)) = T(e,e) pr1 .Te incl.(X, f (X)) = X, so is invertible. Thus
g is a local di¬eomorphism, so g : K ’ G0 is a covering of the connected
component G0 of e in G. If G is simply connected, g is an isomorphism. Now we
consider the homomorphism ψ := pr2 —¦ incl : K ’ G — H ’ H, whose tangent
mapping satis¬es Te ψ.(X, f (X)) = f (X). We see that • := ψ —¦ (g|U )’1 : G ⊃
U ’ H solves the problem, where U is an e-neighborhood in K such that g|U is a
di¬eomorphism. If G is simply connected, • = ψ —¦ g ’1 is the global solution.
5.5. Theorem. Let H be a closed subgroup of a Lie group G. Then H is a Lie
subgroup and a submanifold of G.
Proof. Let g be the Lie algebra of G. We consider the subset h := {c (0) : c ∈
C ∞ (R, G), c(R) ‚ H, c(0) = e}.
Claim 1. h is a linear subspace.
If ci (0) ∈ h and ti ∈ R, we de¬ne c(t) := c1 (t1 .t).c2 (t2 .t). Then c (0) =
T(e,e) µ.(t1 .c1 (0), t2 .c2 (0)) = t1 .c1 (0) + t2 .c2 (0) ∈ h.
Claim 2. h = {X ∈ g : exp(tX) ∈ H for all t ∈ R}.
Clearly we have ˜⊇™. To check the other inclusion, let X = c (0) ∈ h and consider

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993
5. Lie subgroups and homogeneous Spaces 43


v(t) := (expG )’1 c(t) for small t. Then we get X = c (0) = dt |0 exp(v(t)) =
d
1 1 1
v (0) = limn’∞ n.v( n ). We put tn = n and Xn = n.v( n ), so that exp(tn .Xn ) =
1 1
exp(v( n )) = c( n ) ∈ H. By claim 3 below we then get exp(tX) ∈ H for all t.
Claim 3. Let Xn ’ X in g, 0 < tn ’ 0 in R with exp(tn Xn ) ∈ H. Then
exp(tX) ∈ H for all t ∈ R.
Let t ∈ R and take mn ∈ ( tt ’ 1, tt ] © Z. Then tn .mn ’ t and mn .tn .Xn ’ tX,
n n
and since H is closed we may conclude that exp(tX) = limn exp(mn .tn .Xn ) =
limn exp(tn .Xn )mn ∈ H.
Claim 4. Let k be a complementary linear subspace for h in g. Then there is
an open 0-neighborhood W in k such that exp(W ) © H = {e}.
If not there are 0 = Yk ∈ k with Yk ’ 0 such that exp(Yk ) ∈ H. Choose a
norm | | on g and let Xn = Yn /|Yn |. Passing to a subsequence we may assume
that Xn ’ X in k, then |X| = 1. But exp(|Yn |.Xn ) = exp(Yn ) ∈ H and
0 < |Yn | ’ 0, so by claim 3 we have exp(tX) ∈ H for all t ∈ R. So by claim 2
X ∈ h, a contradiction.
Claim 5. Put • : h — k ’ G, •(X, Y ) = exp X. exp Y . Then there are 0-
neighborhoods V in h, W in k, and an e-neighborhood U in G such that • :
V — W ’ U is a di¬eomorphism and U © H = exp(V ).
Choose V , W , and U so small that • becomes a di¬eomorphism. By claim
4 W may be chosen so small that exp(W ) © H = {e}. By claim 2 we have
exp(V ) ⊆ H © U . Let x ∈ H © U . Since x ∈ U we have x = exp X. exp Y for
unique (X, Y ) ∈ V — W . Then x and exp X ∈ H, so exp Y ∈ H © exp(W ), thus
Y = 0. So x = exp X ∈ exp(V ).
Claim 6. H is a submanifold and a Lie subgroup.
(U, (•|V — W )’1 =: u) is a submanifold chart for H centered at e by claim 5.
For x ∈ H the pair (»x (U ), u —¦ »x’1 ) is a submanifold chart for H centered at
x. So H is a closed submanifold of G, and the multiplication is smooth since it
is a restriction.

5.6. Remark. The following stronger results on subgroups and the relation
between topological groups and Lie groups in general are available.
Any arc wise connected subgroup of a Lie group is a connected Lie subgroup,
[Yamabe, 50].
Let G be a separable locally compact topological group. If it has an e-
neighborhood which does not contain a proper subgroup, then G is a Lie group.
This is the solution of the 5-th problem of Hilbert, see the book [Montgomery-
Zippin, 55, p. 107].
Any subgroup H of a Lie group G has a coarsest Lie group structure, but
it might be non separable. To indicate a proof of this statement, consider all
continuous curves c : R ’ G with c(R) ‚ H, and equip H with the ¬nal topology
with respect to them. Then the component of the identity satis¬es the conditions
of the Gleason-Yamabe theorem cited above.

5.7. Let g be a Lie algebra. An ideal k in g is a linear subspace k such that
[k, g] ‚ k. Then the quotient space g/k carries a unique Lie algebra structure
such that g ’ g/k is a Lie algebra homomorphism.

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993
44 Chapter I. Manifolds and Lie groups


Lemma. A connected Lie subgroup H of a connected Lie group G is a normal
subgroup if and only if its Lie algebra h is an ideal in g.

Proof. H normal in G means xHx’1 = conjx (H) ‚ H for all x ∈ G. By remark
4.20 this is equivalent to Te (conjx )(h) ‚ h, i.e. Ad(x)h ‚ h, for all x ∈ G. But
this in turn is equivalent to ad(X)h ‚ h for all X ∈ g, so to the fact that h is an
ideal in g.

5.8. Let G be a connected Lie group. If A ‚ G is an arbitrary subset, the
centralizer of A in G is the closed subgroup ZA := {x ∈ G : xa = ax for all a ∈

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