Download Lie Groups and Lie Algebras Presentation Fall 2014 Chiahui

Lie Groups and Lie Algebras
Presentation
Fall 2014
Chiahui (Wendy) Cheng
Contents
I. Lie Groups
II. Vector Fields
III. The Lie Algebra of a Lie Group
IV. The Lie Algebra of a Lie Subgroup
1
I Lie Groups
Definition 1: A Lie group is a smooth manifold G that is also a group in the algebraic sense,
with the property that the multiplication map m : G × G → G and inversion map i : G → G,
given by
m(g, h) = gh, i(g) = g −1
are both smooth. A Lie group is, in particular, a topological group (a topological space with
a group structure such that the multiplication and inversion maps are continuous).
(Alternative definition: If G is a smooth manifold with a group structure such that the
map G × G → G given by (g, h) 7→ gh−1 is smooth, then G is a Lie group.)
If G is a Lie group, any element g ∈ G defines maps Lg , Rg : G → G, called left translation
and right translation respectively by
Lg (h) = gh,
Rg (h) = hg.
Since Lg can be expressed as the composition of smooth maps
ιg
m
G−
→ G × G −→ G,
where ιg (h) = (g, h) and m is multiplication, it follows that Lg is smooth. It is actually a
diffeomorphism of G, because Lg−1 is a smooth inverse for it.
Example 2
(a) The general linear group GL(n, R) is the set of invertible n × n is the set of invertible
n × n matrices with real entries. It is an open submanifold of vector space M (n, R) (since it is
an open subset of the n2 -dimensional vector space M (n, R), i.e. the set where the determinant
function is nonzero). Multiplication is smooth because the matrix entries of a product matrix
AB are polynomials in the entries of A and B. Inversion is smooth by Cramer’s rule.
(b) Let GL+ (n, R) denote the subset of GL(n, R) consisting of matrices with positive determinant. Because det(AB) = (det A)(det B) and det(A−1 ) = 1/ det(A), it is a subgroup of
GL(n, R); and because it is the preimage of (0, ∞) under the continuous determinant function,
it is an open subset of GL(n, R) and therefore an n2 -dimensional manifold. The group operations are the restrictions of those of GL(n, R), so they are smooth. Thus, GL+ (n, R) is a Lie
group.
(c) The complex general linear group GL(n, C) is the group of invertible complex n × n
matrices under matrix multiplication.
(d) Let V be any real or complex vector space, GL(V ) is the set of invertible linear maps from
V to itself. It is a group under composition.
(e) The real number R and the Euclidean space Rn are Lie groups under addition. Similarly,
C and Cn are Lie groups under addition.
(f) The circle S 1 ⊆ C∗ is a smooth manifold and a group under complex multiplication. With
appropriate angel functions as local coordinates on open subsets of S 1 (recall that an angle
function on a subset U ⊆ S 1 is a continuous function θ : U → R such that eiθ(z) for all z ∈ U ),
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multiplication and inversion have the smooth coordinates (θ1 , θ2 ) 7→ θ1 + θ2 and θ 7→ −θ, and
therefore S 1 is a Lie group.
Lie Group Homomorphisms
If G and H are Lie groups, a Lie group homomorphism from G to H is a smooth map
F : G → H that is also a group homomorphism. It is called a Lie group isomorphism
if it is also a diffeomorphism, which implies that it has an inverse that is also a Lie group
homomorphism. In this case we say that G and H are isomorphic Lie groups.
Example 3
(a) The inclusion map S 1 → C∗ is a Lie group homomorphism.
(b) Considering R as a Lie group under addition, and R∗ as a Lie group under multiplication,
the map exp: R → R∗ given by exp(t) = et is smooth, and is a Lie group homomorphism
because es+t = es et . The image of exp is the open subgroup R+ consisting of positive real
numbers, and exp : R → R+ is a Lie group isomorphism with inverse log : R+ → R.
(c) The map : R → S 1 defined by (t) = e2πit is a Lie group homomorphism whose kernel is
the set Z of integers.
(d) The determinant function det : GL(n, R) → R∗ is smooth because det A is a polynomial
in the matrix entries of A. It is a Lie group homomorphism because det(AB) = (det A)(det B).
Theorem 4 Every Lie group homomorphism has constant rank. (Proof omitted)
Lie Subgroups
Prerequisite: Given a smooth map F : M → N and a point p ∈ M , we define the rank of
F at p to be the rank of the linear map dFp : Tp M → TF (p) N , it is the rank of the Jacobian
matrix of F in any smooth chart. If F has the same rank r at every point, we say that it
has the constant rank. A smooth map F : M → N is called a smooth submersion if its
differential is surjective at each point (or rank F = dim N ). It is called a smooth immersion
if its differential is injective at each point. An immersed submanifold of M is a subset
S ⊆ M endowed with a topology (not necessarily the subspace topology) with respect to which
it is a topological manifold, and a smooth structure with respect to which the inclusion map
S ,→ M is a smooth immersion.
Definition 5: A Lie subgroups of G is a subgroup of G endowed with a topology and
smooth structure making it into a Lie group and and immersed submanifold of G.
Proposition 6 Let G be a Lie group, and suppose H ⊆ G is a subgroup that is also an
embedded submanifold. Then H is a Lie subgroup.
Proof
Only need to check that multiplication H × H → H and inversion H → H are smooth maps.
Because multiplication is a smooth map from G × G → G, its restriction is clearly smooth
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from H × H into G. Because H is a subgroup, multiplication takes H × H into H, and since
H is embedded, this is a smooth map into H. A similar argument applies to inversion. This
proves that H is a Lie subgroup.
Proposition 7 Let F : G → H be a Lie group homomorphism. Then kernel of F is a properly
embedded Lie subgroup of G, whose codimension is equal to the rank of F .
Proof
Because F has constant rank (Theorem 4), its kernel F −1 (e) is a properly embedded submanifold of codimension equal to rank F (by Constant-Rank Level Set Theorem: If Φ : M → N
is a smooth map with constant rank r, each level set of Φ is a properly embedded submanifold
of codimension r in M .). It is thus a Lie subgroup by Proposition 6.
Examples 8 (Embedded Lie subgroups)
(a) The set SL(n, R) of n × n real matrices with determinant equal to 1 is called the special
linear group of degree n. It is an embedded Lie subgroup of GL(n, R) as follows: because
SL(n, R) is the kernel of the Lie group homomorphism det : GL(n, R) → R∗ , it is a properly
embedded Lie subgroup by Proposition 7.
(b) The special orthogonal group of degree n is defined as SO(n) = O(n) ∩ SL(n, R). It
is an embedded Lie subgroup of GL(n, R).
(c) The unitary group of degree n is the subgroup U (n) ⊆ GL(n, C). It is a embedded Lie
subgroup of GL(n, C) of dimension n2 .
(d) The special unitary group of degree n, SU (n) = U (n) ∩ SL(n, C), is an embedding
Lie subgroup of GL(n, C).
(e) Let n be a positive integer, and define a map β : GL(n, C) → GL(2n, R) by replacing each
complex matrix entry a + ib with the 2 × 2 block
a −b
:
b a


a11 + ib11

..
β
.
···
a1n + ib1n · · ·
a11 −b11 · · ·
n
 b1 a1
a1 + ib1
1
 1
  ..
..
= .
.
 1
an −b1n · · ·
ann + ibnn
b1n a1n

n

an1 −bn1
bn1 an1 

.. 
. 

ann −bnn 
bnn ann
Then β is an injective Lie group homomorphism whose image is a properly embedded Lie
subgroup of GL(2n, R). Thus, GL(n, C) is isomorphic to this Lie subgroup of GL(2n, R).
Group Actions and Equivariant Maps
The most important applications of Lie groups to smooth manifold theory involve actions by
Lie groups on other manifolds. If G is a group and M is a set, a left action of G on M is a
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map G × M → M , often written as (g, p) 7→ g · p, that satisfies
g1 · (g2 · p) = (g1 g2 ) · p
e·p=p
for all g1 , g2 ∈ G andp ∈ M ;
for all p ∈ M.
Example 9 (Lie Group Actions)
(a) If G is any Lie group and M any smooth manifold, the trivial action of G on M is
defined by g · p = p for all g ∈ G and p ∈ M . It is a smooth action, for which each orbit is a
single point and each isotropy group is all of G.
(b) The natural action of GL(n, R) on Rn is the left action given by matrix multiplication
: (A, x) 7→ Ax, considering x ∈ Rn as a column matrix. This is an action because In x = x
and matrix multiplication is associative: (AB)x = A(Bx). Because any nonzero vector can be
taken to any other by some invertible linear transformation, there are exactly two orbits: {0}
and Rn \{0}.
(c) Every Lie group G acts smoothly on itself by left translation. Given any two point
g1 , g2 ∈ G, there is a unique left translation of G taking g1 to g2 , namely left translation
by g2 g1−1 ; thus the action is both free and transitive.
(d) Every Lie group acts smoothly on itself by conjugation: g · h = gh−1 g.
Equivariant Maps
A map F : M → N is said to be equivariant with respect to the given G-action if for each
g ∈ G,
F (g · p) = g · F (p) (for left actions)
F (p · g) = F (p) · g
(for right actions)
Theorem 10 (Equivariant Rank Theorem) Let M and N be smooth manifolds and let
G be a Lie group. Suppose F : M → N is a smooth map that is equivariant with respect to a
transitive smooth G-action on M and any smooth G-action on N . Then F has constant rank. (e) The Orthogonal Group O(n) is a Lie subgroup of GL(n, R) A real n × n matrix A
is said to be orthogonal if as a linear map A : R → R it preserves the Euclidean dot product:
(Ax) · (Ay) = x · y
for all x, y ∈ Rn .
The set O(n) of all orthogonal n×n matrices is a subgroup of GL(n, R), called the orthogonal
group of degree n. A matrix A is orthogonal if and only if it takes the standard basis of Rn
to an orthonormal basis, which is equivalent to the columns of A being orthonormal. Since
the (i, j)-entry of the matrix AT A is the dot product of the ith and j th columns of A, this
condition is also equivalent to the requirement that AT A = In . i.e.
O(n) = {A ∈ M (n, R) | AT A = In }.
Define a smooth map Φ : GL(n, R) → M (n, R) by Φ(A) = AT A. Then O(n) is equal to the
level set Φ−1 (In ). To show that Φ has a constant rank and therefore that O(n) is an embedded
Lie subgroup, we show that Φ is equivariant with respect to suitable right action of GL(n, R).
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Let GL(n, R) act on itself by right multiplication, and define a right action of GL(n, R) on
M (n, R) by
X · B = B T XB for X ∈ M (n, R), B ∈ GL(n, R).
It is easy to check that this is a smooth action, and Φ is equivariant because
Φ(AB) = (AB)T (AB) = B T AT AB = B T Φ(A)B = Φ(A) · B.
Thus by Theorem 10 O(n) has a constant rank. Since O(n) = ker Φ, by Proposition 7, O(n)
is a properly embedded Lie subgroup of GL(n, R). It is compact because it is closed (since
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O(n) = Φ−1 (In ) is closed) and bounded in M (n, R) ∼
= Rn : closed because it is a level set of√Φ,
and bounded because every A ∈ O(n) has columns of norm 1, and therefore satisfies |A| = n.
To determine the dimension of O(n), we need to compute the rank of Φ. Because the
rank is constant, it suffices to compute it at the identity In ∈ GL(n, R). Thus for any
B ∈ TIn GL(n, R) = M (n, R), let γ : (−, ) → GL(n, R) be the curve γ(t) = In + tB,
and compute
d d dΦIn (B) = Φ ◦ γ(t) = (In + tB)T (Tn + tB) = B T + B
dt
dt
t=0
t=0
So it is clear that the image of dΦIn is contained in the vector space of symmetric matrices.
Conversely, if B ∈ M (n, R) is an arbitrary symmetric n × n matrix, then dΦIn ( 12 B) = B. It
follows that the image of dΦIn is exactly the space of symmetric matrices. This is the linear
subspace of M (n, R) of dimension n(n + 1)/2, because each symmetric matrix is uniquely determined by its values on and above the main diagonal. It follows that O(n) is an embedded
Lie subgroup of dimension n2 − n(n + 1)/2 = n(n − 1)/2.
II Vector Fields
Definition 11 A vector field on M is a section of the map π : T M → M . i.e. a vector
field is a continuous map X : M → T M , p 7→ Xp , with the property that π ◦ X = IdM . Or
equivalently, Xp ∈ Tp M for each p ∈ M . If (U, (xi )) is any smooth coordinate chart for M , we
can write the value of X at any point p ∈ U in terms of the coordinate basis vectors:
∂ i
Xp = X (p) i .
∂x
p
Or simply
X = Xi
∂
∂xi
where X i is the ith component function of X.
Vector Fields as Derivations of C ∞ (M )
Let X(M ) denotes the set of all smooth vector fields on M . An essential property of vector
fields is that they define operators on the space of smooth real-valued functions. If X ∈ X(M )
and f is a smooth real-valued function defined on an open subset U ⊆ M , we obtained a new
function Xf : U → R, defined by
(Xf )(p) = Xp f
In another word, a smooth vector field X ∈ X(M ) defines a map from C ∞ (M ) to itself by
f 7→ Xf . This map is clearly linear over R. Moreover, the product rule for tangent vectors
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translations into the following product rule for vector fields:
X(f g) = f Xg + gXf
In general, a map X : C ∞ (M ) → C ∞ (M ) is called derivation if it is linear over R and satisfies
above equation for all f, g ∈ C ∞ (M ).
Conversely, the derivations of C ∞ (M ) can be identified with smooth vector fields: If a map
D : C ∞ (M ) → C ∞ (M ) is a derivation, then it is of the form Df = Xf for some smooth
vector field X ∈ X(M ). As a result, we sometimes identify smooth vector fields on M with
derivations of C ∞ (M ), using the same letter for both the vector fields and the derivation.
Example
12: Coordinate vector fields If (U, (xi )) is any smooth chart on M , the map
∂ th coordinate vector field and denote by
p 7→ ∂x
i p determines a vector field on U , called the i
∂/∂xi .
Definition 13 Suppose F : M → N is smooth and X is a vector field on M , and suppose
there is a vector field Y on N with the property that for each p ∈ M , dFp (Xp ) = YF (p) . In
this case, we say the vector field X and Y are F -related.
III The Lie Algebra of a Lie Group
Lie bracket
Motivation Let X and Y be smooth vector fields on a smooth manifold M . Given a smooth
function f : M → R, we can apply X to f and obtain another smooth function Xf . In turn,
we can apply Y to this function, and obtain yet another smooth function Y Xf = Y (Xf ). The
operation f 7→ Y Xf , however, does not in general satisfy the product rule and thus cannot be
a vector field, as the following example shows.
Example 14: Define vector fields X = ∂/∂x and Y = x∂/∂y on R2 , and let f (x, y) = x,
g(x, y) = y. Then direct computation shows that XY (f g) = 2x, while f XY g + gXY f = x, so
XY is not derivation of C ∞ (R2 ).
Proof
Since
(XY )(f g) = (XY )(xy) = X(x
∂ 2
∂
)(xy)) = X(x · x) = X(x2 ) =
x = 2x
∂y
∂x
and
f XY g + gXY f = x · X(x
∂
∂
∂
)y + y · X(x )x = x · Xx + y · X0 = x ·
x=x
∂y
∂y
∂x
So XY is not a derivation of C ∞ (R2 ). We can apply the same two vector fields in the opposite order, obtaining a function XY f . Applying both of these operators to f and subtracting, we obtain an operator [X, Y ] : C ∞ (M ) →
C ∞ (M ), called the Lie bracket of X and Y , defined by
[X, Y ]f = XY f − Y Xf
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The key fact is that this operator is a vector field.
Lemma 15 The Lie bracket of any pair of smooth vector fields is a smooth vector field. Proposition 16 (Coordinate Formula for the Lie bracket) Let X, Y be smooth vector
fields on a smooth manifold M with or without boundary, and let X = X i ∂/∂xi and Y =
Y j ∂/∂xj be the coordinate expressions for X and Y in terms of some smooth local coordinates
(xi ) for M . Then [X, Y ] has the following coordinate expression:
j
j
∂
iX ∂
iY ∂
−Y
,
[X, Y ] = X
i
i
∂x
∂x
∂xi
or more precisely,
[X, Y ] = (XY j − Y X j )
∂
∂xj
Proof
We know already that [X, Y ] is smooth vector field, its action on a function is determined
locally: ([X, Y ]f )|U = [X, Y ](f |U ). Thus it suffices to compute in a single smooth chart, where
we have
∂f
∂
∂f
∂
[X, Y ]f = X i i Y j j − Y j j X i i
∂x
∂x
∂x
∂x
2
2
j
i
∂Y ∂f
i j ∂ f
j ∂X ∂f
j i ∂ f
= Xi i
+
X
Y
−
Y
−
Y
X
∂x ∂xj
∂xi ∂xj
∂xj ∂xi
∂xj xi
j
i
i
∂Y ∂f
∂X ∂X
= Xi i
−Yj j
,
j
∂x ∂x
∂x ∂xi
where in the last step we have used the fact that mixed partial derivatives of a smooth function
can be taken in any order. Interchanging the roles of the dummy indices i and j in the second
term. Example 17 Define smooth vector fields X, Y ∈ X(R3 ) by
X=x
∂
∂
∂
+
+ x(y + 1) ,
∂x ∂y
∂z
Y =
∂
∂
+y .
∂x
∂z
Then we get
∂
∂
∂
∂
∂
+ X(y)
− Y (x)
− Y (1)
− Y (x(y + 1))
∂x
∂z
∂x
∂y
∂z
∂
∂
∂
∂
∂
= 0
+1
−1
−0
− (y + 1)
∂x
∂z
∂x
∂y
∂z
∂
∂
= −
−y
∂x
∂z
[X, Y ] = X(1)
Properties of the Lie Bracket The Lie bracket satisfies Bilinearity, Antisymmetry, Jacobi
Identity and
[f X, gY ] = f g[X, Y ] + (f Xg)Y − (gY f )X.
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Lie Algebra of a Lie Group
One of the most important applications of Lie brackets occurs in the context of Lie groups.
Suppose G is a Lie group. Recall that G acts smoothly and transitively on itself by left
translation: Lg (h) = gh, in the sense that it is Lg -related to itself for every g ∈ G. A vector
field X on G is said to be left-invariant if it is invariant under all left translations. More
explicitly, this means
d(Lg )g0 (Xg0 ) = Xgg0 for all g, g 0 ∈ G.
by Definition 13.
Proposition 18 (Left-invariant is closed under Lie brackets) Let G be a Lie group,
and suppose X and Y are smooth left-invariant vector fields on G. Then [X, Y ] is also leftinvariant. Definition 19 A Lie algebra (over R) is a real vector space g endowed with a map from g × g
to g, denoted by (X, Y ) 7→ [X, Y ], that satisfying properties for all X, Y, Z ∈ g:
(i) Biliearity: for a, b ∈ R,
[aX + bY, Z] = a[X, Z] + b[Y, Z]
[Z, aX + bY ] = a[Z, X] + b[Z, Y ].
(ii) Antisymmetry:
[X, Y ] = −[Y, X].
(iii) Jacobi identity:
[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]] = 0.
If g is a Lie algebra, a linear subspace h ⊆ g is called a Lie subalgebra of g if it is closed
under brackets. If g and h are Lie algebras, a linear map A : g → h is called a Lie algebra
homomorphism if it preserves brackets: A[X, Y ] = [AX, AY ]. An invertible Lie algebra
homomorphism is called a Lie algebra isomorphism.
Examples 20 (Lie Algebras)
(a) The space of X(M ) of all smooth vector fields on a smooth manifold M is a Lie algebra
under the Lie bracket.
(b) If G is a Lie group, the set of all smooth left-invariant vector fields on G is a Lie subalgebra
of X(G) and is therefore a Lie algebra. - The most important one! called the Lie algebra of
G and is denoted by Lie(G).
(c) The vector space M (n, R) of n × n real matrices becomes an n2 -dimensional Lie algebra
under the commutator bracket:
[A, B] = AB − BA
Bilinearity and antisymmetry are obvious from the definition, and the Jacobi identity follows
from a straightforward calculation. When are regarding M (n, R) as a Lie algebra with this
bracket, we denote it by gl(n, R).
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(d) If V is a vector space, the vector space of all linear maps from V to itself becomes a Lie
algebra, which we denote by gl(V ), with the commutator bracket:
[A, B] = A ◦ B − B ◦ A.
gl(Rn )
is the same as gl(n, R).
(e) Any vector space V becomes a Lie algebra if we define all brackets to be zero. Such a Lie
algebra is said to be abelian.
Examples of Lie algebras of some Lie groups 21
(a) Euclidean space Rn , with Lie(Rn ) ∼
= Rn . Consider Rn as a Lie group under addition,
n
left translation by an element b ∈ R is given by the affine map Lb (x) = b + x, whose differential d(Lb ) is represented by the identity matrix in standard coordinates. Thus a vector
field X i ∂/∂xi is left-invariant if and only if its coefficients X i are constants. Because the Lie
bracket of two constant-coeffcient vector fields is zero by Proposition 16, the Lie algebra of Rn
is abelian (Example 20(e): [X, Y ] = XY − Y X = 0), and is isomorphic to Rn itself with the
trivial bracket. In brief, Lie (Rn ) ∼
= Rn .
(b) S 1 with Lie(S 1 ) ∼
= R. Recall that an angle function on a subset U ⊆ S 1 is a continuous
function θ : U → R such that eiθ(z) for all z ∈ U . Let d/dθ denote the corresponding coordinate
vector field. Each left translation has a local coordinate representation of the form θ 7→ θ + c.
Since the differential of this map is the 1 × 1 identity matrix, it follows that the vector field
d/dθ is left-invariant, and is therefore a basis for the Lie algebra of S 1 . This Lie algebra is
1-dimensional and abelian. Hence Lie(S 1 ) ∼
= R.
(c) T n = S 1 × · · · × S 1 . An analysis similar to the one above show that ( ∂θ∂ 1 , · · · , ∂θ∂n ) is a
basis for Lie(T n ), where ∂θ∂ i is the angle coordinate vector field on the ith S 1 factor. Since the
Lie brackets of these coordinate vector fields are all zero, Lie (T n ) = Rn .
(d) Lie algebra of the GL(n, R). The composition of the natural maps
Lie(GL(n, R))
/ TI GL(n, R)
n
/ gl(n, R)
gives a Lie algebra isomorphism between Lie(GL(n, R)) and the matrix algebra gl(n, R). IV The Lie Algebra of a Lie Subgroups
If G is a Lie group and H ⊆ G is a Lie subgroup, we might hope that the Lie algebra of H
would be a Lie subalgebra of that of G. However, elements of Lie(H) are vector fields on H,
but not on G, so are not elements of Lie(G). The following proposition gives us a way to view
Lie(H) as a subalgebra of Lie(G).
Theorem 22 (The Lie algebra of a Lie subgroup). Suppose H ⊆ G is a Lie subgroup,
and ι : H ,→ G is the inclusion map. There is a Lie subalgebra h ⊆ Lie(G) that is canonically
isomorphic to Lie(H), characterized by either of the following descriptions:
h = ι∗ (Lie(H))
= {X ∈ Lie(G) : Xe ∈ Te H}
10
Lemma Suppose S ⊆ M is a level set of a smooth map φ : M → N with constant rank. Then
Tp S = ker dφp for each p ∈ S.
(a) The Lie algebra of O(n)
The orthogonal group O(n) is a Lie subgroup of GL(n, R). Define a smooth map φ : GL(n, R) →
M (n, R) by φ(A) = AT A. Then O(n) is equal to the level set φ−1 (In ), where Φ : GL(n, R) →
M (n, R) is the map Φ(A) = AT A. By Lemma, TIn O(n) is equal to the kernel of dΦIn :
TIn GL(n, R) → TIn M (n, R). By computation in Example 9(e), this differential is dΦIn (B) =
B T + B, so
TIn O(n) = {B ∈ gl(n, R) : B T + B = 0}
= {skew-symmetric n × n matrices}
We denote this subspace of gl(n, R) by o(n). Theorem 22 implies that o(n) is a Lie subalgebra
of gl(n, R) that is canonically isomorphic to Lie(O(n)).
(b) The Lie algebra of GL(n, C)
As in the real case, our usual identification of GL(n, C) as an open subset of gl(n, C) yields a
sequence of vector space isomorphisms
Lie(GL(n, C))
/ TI GL(n, C) ϕ
n
/ gl(n, C)
where is the evaluation map and ϕ is the usual identification between the tangent space to
an open subset of a vector space and the vector space itself.
The composition of the maps above yields a Lie algebra isomorphism between Lie(GL(n, C))
and the matrix algebra gl(n, C) as below: The Lie group homomorphism constructed in Example 8(e) induces a Lie algebra homomorphism
β∗ : Lie(GL(n, C)) → Lie(GL(2n, R)).
Composing β∗ with our canonical isomorphisms yields a commutative diagram
Lie(GL(n, C))
/ TI GL(n, C)
n
β∗
Lie(GL(2n, R))
ϕ
dβIn
/ TI GL(2n, R) ϕ
2n
/ gl(n, C)
α
/ gl(2n, R)
in which α = ϕ ◦ dβIn ◦ ϕ−1 . Example 21(d) showed that the composition of the isomorphisms
in the bottom row is a Lie algebra isomorphism, we need to show the same thing for the top row.
It is easy to see that from the formula in Example 8(e) that β is (the restriction of) a linear
map. It follows that dβIn : TIn GL(n, C) → TI2n GL(2n, R) is given by exactly the same formula
as β, as in α : gl(n, C) → gl(2n, R). Because β(AB) = β(A)β(B), it follows that α preserves
matrix commutators:
α[A, B] = α(AB − BA) = α(A)α(B) − α(B)α(A) = [α(A), α(B)]
Thus α is an injective Lie algebra homomorphism from gl(n, C) to gl(2n, R) (considering both
as matrix algebras). Replacing the bottom row in the diagram above by the image of the
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vertical maps, we obtain a commutative diagram of vector space isomorphisms
∼
=
Lie(GL(n, C))
β∗
β∗ (Lie(GL(n, C)))
∼
=
/ gl(n, C)
α
/ α(gl(n, C))
in which the bottom map and the two vertical maps are Lie algebra isomorphisms. It follows
that the top map is also a Lie algebra isomorphism. References
M. Lee, John Introduction to Smooth Manifolds, Second Edition 2013
Karin Erdmann and Mark J. Wildon, Introduction to Lie Algebras, Springer 2006
Spivak, A Comprehensive Introduction to Differential Geometry, Volume One, 2005
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