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Example 1. Any Lie group left-acts on itself by left multiplication. I.e., we set
M = G and de¬ne »: G — M ’ M to simply be µ. This action is both free and transitive.
Example 2. Given a homomorphism of Lie groups φ: H ’ G, de¬ne a smooth left
action »: H —G ’ G by the rule »(h, g) = φ(h)g. Then He = ker(φ) and H ·e = φ(H) ‚ G.
In particular, Theorem 1 implies that the kernel of a Lie group homomorphism
is a (closed, normal) Lie subgroup of the domain group and the image of a Lie group
homomorphism is a Lie subgroup of the range group.
Example 3. Any Lie group acts on itself by conjugation: g · g0 = gg0g ’1 . This action
is neither free nor transitive (unless G = {e}). Note that Ge = G and, in general, Gg is
the centralizer of g ∈ G. This action is e¬ective (respectively, almost e¬ective) if and only
if the center of G is trivial (respectively, discrete). The orbits are the conjugacy classes
of G.
Example 4. GL(n, R) acts on Rn as usual by A · v = Av. This action is e¬ective but
is neither free nor transitive since GL(n, R) ¬xes 0 ∈ Rn and acts transitively on Rn \{0}.
Thus, there are exactly two orbits of this action, one closed and the other not.

L.3.2 39
Example 5. SO(n + 1) acts on S n = {x ∈ Rn+1 | x · x = 1} by the usual action
A · x = Ax. This action is transitive and e¬ective, but not free (unless n = 1) since, for
example, the stabilizer of en+1 is clearly isomorphic to SO(n).
Example 6. Let Sn be the n(n + 1)/2-dimensional vector space of n-by-n real sym-
metric matrices. Then GL(n, R) acts on Sn by A · S = A S tA. The orbit of the identity
matrix In is S+ (n), the set of all positive-de¬nite n-by-n real symmetric matrices (Why?).
In fact, it is known that, if we de¬ne Ip,q ∈ Sn to be the matrix
« 
Ip 0 0
= 0 0,
0 0 0

(where the “0” entries have the appropriate dimensions) then Sn is the (disjoint) union of
the orbits of the matrices Ip,q where 0 ¤ p, q and p + q ¤ n (see the Exercises).
The orbit of Ip,q is open in Sn i¬ p + q = n. The stabilizer of Ip,q in this case is de¬ned
to be O(p, q) ‚ GL(n, R).
Note that the action is merely almost e¬ective since {±In} ‚ GL(n, R) ¬xes every
S ∈ Sn .
Example 7. Let J = J ∈ GL(2n, R) | J 2 = ’I2n . Then GL(2n, R) acts on J on
the left by the formula A · J = AJ A’1 . I leave as exercises for the reader to prove that J
is a smooth manifold and that this action of GL(2n, R) is transitive and almost e¬ective.
The stabilizer of J0 = multiplication by i in Cn ( = R2n ) is simply GL(n, C) ‚ GL(2n, R).
Example 8. Let M = RP1 , denote the projective line, whose elements are the lines
through the origin in R2 . We will use the notation x to denote the line in R2 spanned
by the non-zero vector .

Let G = SL(2, R) act on RP1 on the left by the formula

ab x ax + by
· = .
cd y cx + dy

This action is easily seen to be almost e¬ective, with only ±I2 ∈ SL(2, R) acting trivially.
Actually, it is more common to write this action more informally by using the iden-
ti¬cation RP1 = R∪{∞} which identi¬es x when y = 0 with x/y ∈ R and 1 with ∞.
y 0
With this convention, the action takes on the more familiar “linear fractional” form

ax + b
·x= .
cd cx + d

Note that this form of the action makes it clear that the so-called “linear fractional”
action or “M¨bius” action on the real line is just the projectivization of the usual linear
representation of SL(2, R) on R2 .

L.3.3 40
We now turn to the proof of Theorem 1.
Proof of Theorem 1: Fix m ∈ M and de¬ne φ: G ’ M by φ(g) = »(g, m) as in the
theorem. Since Gm = φ’1 (m), it follows that Gm is a closed subset of G. The axioms for
a left action clearly imply that Gm is closed under multiplication and inverse, so it is a
I claim that Gm is a submanifold of G. To see this, let gm ‚ g = Te G be the kernel
of the mapping φ (e): Te G ’ Tm M. Since φ —¦ Lg = »g —¦ φ for all g ∈ G, the Chain Rule
yields a commutative diagram:
Lg (e)
’’ Tg G
¦ ¦
¦ ¦
φ (e) φ (g)
»g (m)
Tm M Tg·m M

Since both Lg (e) and »g (m) are isomorphisms, it follows that ker(φ (g)) = Lg (e)(g m ) for
all g ∈ G. In particular, the rank of φ (g) is independent of g ∈ G. By the Implicit
Function Theorem (see Exercise 2), it follows that φ’1 (m) = Gm is a smooth submanifold
of G.
It remains to show that the orbit G · m can be given the structure of a smooth
submanifold of M with the stated properties. That is, that G · m can be given a second
countable, Hausdor¬, locally Euclidean topology and a smooth structure for which the
inclusion map G · m ’ M is a smooth immersion and for which the map φ: G ’ G · m is
a submersion.
Before embarking on this task, it is useful to remark on the nature of the ¬bers of
the map φ. By the axioms for left actions, φ(h) = h · m = g · m = φ(g) if and only if
g ’1 h · m = m, i.e., if and only if g ’1 h lies in Gm . This is equivalent to the condition that
h lie in the left Gm -coset gGm . Thus, the ¬bers of the map φ are the left Gm -cosets in G.
In particular, the map φ establishes a bijection φ: G/Gm ’ G · m.
First, I specify the topology on G· m to be quotient topology induced by the surjective
map φ: G ’ G · m. Thus, a set U in G · m is open if and only if φ’1 (U) is open in G. Since
φ: G ’ M is continuous, the quotient topology on the image G · m is at least as ¬ne as
the subspace topology G · m inherits via inclusion into M. Since the subspace topology is
Hausdor¬, the quotient topology must be also. Moreover, the quotient topology on G · m
is also second countable since the topology of G is. For the rest of the proof, “the topology
on G · m” means the quotient topology.
I will both establish the locally Euclidean nature of this topology and construct a
smooth structure on G · m at the same time by ¬nding the required neighborhood charts
and proving that they are smooth on overlaps. First, however, I need a lemma establishing
the existence of a “tubular neighborhood” of the submanifold Gm ‚ G. Let d = dim(G) ’
dim(Gm ). Then there exists a smooth mapping ψ: B d ’ G (where B d is an open ball
about 0 in Rd ) so that ψ(0) = e and so that g is the direct sum of the subspaces gm and
V = ψ (0)(Rd ). By the Chain Rule and the de¬nition of gm , it follows that (φ—¦ψ) (0): Rd ’

L.3.4 41
Tm M is injective. Thus, by restricting to a smaller ball in Rd if necessary, I may assume
henceforth that φ —¦ ψ: B d ’ M is a smooth embedding.
Consider the mapping Ψ: B d — Gm ’ G de¬ned by Ψ(x, g) = ψ(x)g. I claim that Ψ
is a di¬eomorphism onto its image (which is an open set), say U = Ψ(B d — Gm ) ‚ G.
(Thus, U forms a sort of “tubular neighborhood” of the submanifold Gm in G.)
To see this, ¬rst I show that Ψ is one-to-one: If Ψ(x1 , g1 ) = Ψ(x2 , g2 ), then

(φ —¦ ψ)(x1 ) = ψ(x1 ) · m = (ψ(x1 )g1 ) · m = (ψ(x2 )g2 ) · m = ψ(x2 ) · m = (φ —¦ ψ)(x2 ),

so the injectivity of φ —¦ ψ implies x1 = x2 . Since ψ(x1 )g1 = ψ(x2 )g2 , this in turn implies
that g1 = g2 .
Second, I must show that the derivative

Ψ (x, g): Tx Rd • Tg Gm ’ Tψ(x)g G

is an isomorphism for all (x, g) ∈ B d — Gm . However, from the beginning of the proof,
ker(φ (ψ(x)g)) = Lψ(x)g (e)(gm ) and this latter space is clearly Ψ (x, g)(0 • Tg Gm ). On
the other hand, since φ(Ψ(x, g)) = φ —¦ ψ(x), it follows that

φ (Ψ(x, g)) Ψ (x, g)(Tx Rd • 0) = (φ —¦ ψ) (x)(Tx Rd )

and this latter space has dimension d by construction. Hence, Ψ (x, g)(Tx Rd • 0) is
a d-dimensional subspace of Tψ(x)g G which is transverse to Ψ (x, g)(0 • Tg Gm ). Thus,
Ψ (x, g): Tx Rd • Tg Gm ’ Tψ(x)g G is surjective and hence an isomorphism, as desired.
This completes the proof that Ψ is a di¬eomorphism onto U. It follows that the inverse
of Ψ is smooth and can be written in the form Ψ’1 = π1 — π2 where π1 : U ’ B d and
π2 : U ’ Gm are smooth submersions.
Now, for each g ∈ G, de¬ne ρg : B d ’ M by the formula ρg (x) = φ gψ(x) . Then
ρg = »g —¦ φ —¦ ψ, so ρg is a smooth embedding of B d into M. By construction, U =
φ’1 (φ —¦ ψ(B d )) = φ’1 (ρe (B d )) is an open set in G, so it follows that ρe (B d ) is an open
neighborhood of e · m = m in G · m. By the axioms for left actions, it follows that
φ’1 ρg (B d ) = Lg (U) (which is open in G) for all g ∈ G. Thus, ρg (B d ) is an open
neighborhood of g · m in G · m (in the quotient topology). Moreover, contemplating the
commutative square Lg
U ’’ Lg (U)
¦ ¦
¦ ¦φ

Bd ρg (B d )
whose upper horizontal arrow is a di¬eomorphism which identi¬es the ¬bers of the vertical
arrows (each of which is a topological identi¬cation map) implies that ρg is, in fact, a
homeomorphism onto its image. Thus, the quotient topology is locally Euclidean.
Finally, I show that the “patches” ρg overlap smoothly. Suppose that

ρg (B d ) © ρh (B d ) = ….

L.3.5 42
Then, because the maps ρg and ρh are homeomorphisms,

ρg (B d ) © ρh (B d ) = ρg (W1 ) = ρh (W2 )

where Wi = … are open subsets of B d . It follows that

Lg Ψ(W1 — Gm ) = Lh Ψ(W2 — Gm ) .

Thus, if „ : W1 ’ W2 is de¬ned by the rule „ = π1 —¦ Lh’1 —¦ Lg —¦ ψ, then „ is a smooth map
with smooth inverse „ ’1 = π1 —¦ Lg’1 —¦ Lh —¦ ψ and hence is a di¬eomorphism. Moreover, we
have ρg = ρh —¦ „ , thus establishing that the patches ρg overlap smoothly and hence that
the patches de¬ne the structure of a smooth manifold on G · m.
That the map φ: G ’ G· m is a smooth submersion and that the inclusion G· m ’ M
is a smooth one-to-one immersion are now clear.

It is worth remarking that the proof of Theorem 1 shows that the Lie algebra of Gm
is the subspace gm . In particular, if Gm = {e}, then the map φ: G ’ M is a one-to-one

The proof also brings out the fact that the orbit G · m can be identi¬ed with the left
coset space G/Gm , which thereby inherits the structure of a smooth manifold. It is natural
to wonder which subgroups H of G have the property that the coset space G/H can be
given the structure of a smooth manifold for which the coset projection π: G ’ G/H is a
smooth map. This question is answered by the following result. The proof is quite similar
to that of Theorem 1, so I will only provide an outline, leaving the details as exercises for
the reader.

Theorem 2: If H is a closed subgroup of a Lie group G, then the left coset space G/H
can be given the structure of a smooth manifold in a unique way so that the coset mapping
π: G ’ G/H is a smooth submersion. Moreover, with this smooth structure, the left action
»: G — G/H ’ G/H de¬ned by »(g, hH) = ghH is a transitive smooth left action.

Proof: (Outline.) If the coset mapping π: G ’ G/H is to be a smooth submersion,
elementary linear algebra tells us that the dimension of G/H will have to be d = dim(G) ’
dim(H). Moreover, for every g ∈ G, there will have to exist a smooth mapping ψg : B d ’ G
with ψg (0) = g which is transverse to the submanifold gH at g and so that the composition
π —¦ ψ: B d ’ G/H is a di¬eomorphism onto a neighborhood of gH ∈ G/H. It is not
di¬cult to see that this is only possible if G/H is endowed with the quotient topology. The
hypothesis that H be closed implies that the quotient topology is Hausdor¬. It is automatic
that the quotient topology is second countable. The proof that the quotient topology is
locally Euclidean depends on being able to construct the “tubular neighborhood” U of H
as constructed for the case of a stabilizer subgroup in the proof of Theorem 1. Once this
is done, the rest of the construction of charts with smooth overlaps follows the end of the
proof of Theorem 1 almost verbatim.

L.3.6 43
Group Actions and Vector Fields. A left action »: R — M ’ M (where R has
its usual additive Lie group structure) is, of course, the same thing as a ¬‚ow. Associated
to each ¬‚ow on M is a vector ¬eld which generates this ¬‚ow. The generalization of this
association to more general Lie group actions is the subject of this section.
Let »: G — M ’ M be a left action. Then, for each v ∈ g, there is a ¬‚ow Ψ» on M
de¬ned by the formula
Ψ» (t, m) = etv · m.

This ¬‚ow is associated to a vector ¬eld on M which we shall denote by Yv» , or simply
Yv if the action » is clear from context. This de¬nes a mapping »— : g ’ X(M), where
»— (v) = Yv» .

Proposition 1: For each left action »: G — M ’ M, the mapping »— is a linear anti-
homomorphism from g to X(M). In other words, »— is linear and
»— ([x, y]) = ’[»— (x), »— (y)].

Proof: For each v ∈ g, let Yv denote the right invariant vector ¬eld on G whose value
at e is v. Then, according to Lecture 2, the ¬‚ow of Yv on G is given by the formula
Ψv (t, g) = exp(tv)g. As usual, let ¦v denote the ¬‚ow of the left invariant vector ¬eld Xv .
Then the formula
Ψv (t, g) = ¦’v (t, g ’1 )
is immediate. If ι— : X(G) ’ X(G) is the map induced by the di¬eomorphism ι(g) = g ’1 ,
then the above formula implies
ι— (X’v ) = Yv .
In particular, since ι— commutes with Lie bracket, it follows that
[Yx , Yy ] = ’Y[x,y]
for all x, y ∈ g.
Now, regard Yv and Ψv as being de¬ned on G— M in the obvious way, i.e., Ψv (g, m) =
(etv g, m). Then » intertwines this ¬‚ow with that of Ψ» :

» —¦ Ψv = Ψ» —¦ ».

It follows that the vector ¬elds Yv and Yv» are »-related. Thus, [Yx , Yy» ] is »-related to

[Yx , Yy ] = ’Y[x,y] and hence must be equal to ’Y[x,y] . Finally, since the map v ’ Yv is

clearly linear, it follows that »— is also linear.
The appearance of the minus sign in the above formula is something of an annoyance
and has led some authors (cf. [A]) to introduce a non-classical minus sign into either the
de¬nition of the Lie bracket of vector ¬elds or the de¬nition of the Lie bracket on g in order
to get rid of the minus sign in this theorem. Unfortunately, as logical as this revisionism
is, it has not been particularly popular. However, let the reader of other sources beware
when comparing formulas.


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