Lie Groups
Contents
Chapter 0: Topological Groups .......................................................................................................... 2
Chapter 1: Manifolds ........................................................................................................................... 4
Chapter 2: Lie Groups ......................................................................................................................... 7
Chapter 3: Lie Algebras ..................................................................................................................... 14
Chapter 4: Tangent Space ................................................................................................................. 16
Chapter 5: The Lie Algebra of a Lie Group .................................................................................... 22
Chapter 6: Integrating Vector Fields and the Exponential Map .................................................. 27
Chapter 7: Lie Subgroups ................................................................................................................. 34
Chapter 8: Continuity implies Smoothness .................................................................................... 42
Chapter 0: Topological Groups
Topological Group
A Topological Group is a set with group structure and a topology. We write the group
)
structure as maps
with (
and
with ( )
and
require these maps to be continuous. Usually we require that the topology be nice, for
example Hausdorff.
Lemma 0.1
If is a topological group then
( )
defined by
is a homeomorphism for
each .
Proof
( )
is continuous since
( ) and
map
( )
(
). Now, (
that is
. Therefore
)( )
, so for any
has continuous inverse
Notation
If
are subsets let
define
{
(
is invertible with inverse
and thus a homeomorphism.
} i.e.
{
|
|
etc and
( ))
(
}
). Then we can recursively
( ) so
(
)
Lemma 0.2
a. If
then
)
b. (
( )
c. If
and
generated by .
, then ⋃
is a subgroup of , known as the subgroup
Proof
a. Take any element
then
and as
by definition,
(
)
b.
(
for
. Then
.
)
(
) .
c. We need to check that if
then
and
then
⋃
⋃
⋃
.
⋃
As and are products of elements of , clearly
is a product of elements of . If
⋃
and
Lemma 0.3
If is a topological group with
and
Proof
Clearly
so
so
( ) (
)
. As
inverse map so open implies
we have
⋃
open and
then
. Moreover,
is continuous,
open.
.
⋃
is an open set with
(
)
( )
(
)
( ) as is a self-
Theorem 0.4
i.
ii.
If
If
is a topological group and
is connected then
⋃
is open and
then ⋃
is open.
Proof
i.
(By induction)
is a homeomorphism and
is open we
( ) and as
⋃
have
open. As is open, by the principle of induction ⋃
is open.
By Lemma 0.3,
is an open set with
and
therefore by
Lemma 0.2
is a subgroup of and by part i. is open.
⋃
( )
Consider the cosets
is open for all
and
. Therefore is the disjoint union of open sets hence disconnected,
unless
for all
; that is
.
Clearly
ii.
Examples
1. Any group with the discrete topology is a topological group.
)
2.
with the usual topology is a group under addition; (
3.
{ } is a topological group under multiplication
(
)
4.
5. The set of
6. The set of
rule.
(
( )
)
( )
is a topological subgroup.
are all similar.
invertible real matrices is a topological group.
invertible real matrices is also a topological group using Cramer’s
Chapter 1: Manifolds
Definition 1.1
) on a topological space of dimension
A chart (
a homeomorphism of into an open set in
.
is a non-empty open set
with
Examples
) is a chart of dimension 1. Similarly, taking
1. Take
then (
and
(
) gives a chart of dimension on
or
where is a nonempty open set and the inclusion map
. In these cases all the charts
are known as the ‘standard’ or ‘canonical’ charts.
{(
}
)
|
2. The Circle
Charts can be defined in a number of ways, including using angle variables,
stereographic projection, orthogonal projection etc.
Definition 1.2
Two charts (
) and (
) of dimension are compatible of class
(for
(
)
) if either
or
and
(
) is a map between two open sets of
which is
along with its inverse (that is a
-diffeomorphism).
Example
Take a chart given by stereographic projection and work out
or take two stereographic projections and check compatibility:
{ } then
Suppose
which is
(
)
(
)
(
(
)
and
explicitly
)
(
)
(
)
(
)( )
.
Definition 1.3
An atlas of class
and dimension on a topological space is a collection of charts on
of dimension which are pairwise compatible of class
such that the open sets cover .
Maximal Atlas
An atlas is maximal if every charts compatible with all the charts of the atlas is already in
the atlas. Every Atlas extends to a unique maximal atlas.
Definition 1.4
A manifold structure (or a smooth) structure on a topological space of class
dimension is a choice of maximal atlas of class
and dimension .
and
A smooth manifold of dimension is a topological space with a manifold structure of class
and dimension . Usually we assume the topology is Hausdorff and paracompact
(partition of unity).
From the remark before Definition 1.4, we can specify a manifold by giving a topological
space and an atlas. For many spaces
we have a standard topology and a standard
atlas.
Example
)
For
the standard topology is the Euclidean topology and the atlas contains (
For
give it the subspace topology and charts compatible with stereographic, angle
variable and projections onto axes.
Examples
The following are all examples of smooth manifolds. If no atlas is given, it is the canonical
atlas.
1.
2.
{ }(
)
{
(
)
From this we define charts
( )
respective maps
(
}
{
)
}
(
( )
{
} and
)
Note that each chart is homeomorphic with
and there is an explicit formula for
(
) which is easily seen to be
)(
)} is a
on
. {(
atlas of dimension .
( ) to be an 3. Let
be a
map and
then we would like
dimensional manifold. We say that is a regular value on if for every which
satisfies the equation ( )
the derivative of at (denoted
) is surjective.
(
) is -dimensional. To verify this we apply:
I.e.
Theorem (Implicit Function Theorem)
( ) with the subspace
If
is a regular value of a
map
then
(relative) topology is a
manifold of dimension with a covering of open sets .
, define (
For
) to be ( )
( )
(
( )) where (
) are the
Euclidean coordinates for
(Solutions defined locally therefore we use these to construct charts which are
automatically smoothly compatible.)
Examples:
1. For
defined as (
, take
(
)(
)
(
(
)
. Then
)|
)|
)
) ( ). Therefore by the Regular
Then observe that (
when (
)(
( ) is a 1 dimensional
Value Theorem,
manifold for
.
( )
2. More generally Take
defined by ( ) | | then
. Applying
the Regular value theorem we observe that
is an -dimensional smooth manifold.
The charts given by this theorem are obviously compatible with stereographic charts.
Definition 1.5
A continuous map
) a chart on
pair (
1.
2.
between two smooth manifolds is said to be smooth if for any
) a chart on either:
and (
( )
( )
( )), the composition
and on (
between open sets in Euclidean Space.
is
as a map
We say that is a diffeomorphism if is a homeomorphism and both and
are smooth.
Two manifolds are diffeomorphic if there exists a diffeomorphism between them.
Remark
For a function
on ( ).
,
is smooth if for every chart (
) on
,
Denote by
(
)( )
( ) the set of smooth functions on
( )
( )
and ( )( )
Moreover,
(
)( )
( ) is a ring using addition and pointwise multiplication
( ) ( )
with identity the constant function ( )
. Note that
( ) for all
is a
function
( ) is a vector space:
and
.
.
To get every ingredient for the definition of a Lie group, we need the notion of a product of
two manifolds.
Definition 1.6
) with
Let and be smooth manifolds of dimension and . A chart on is a pair (
) for
open and
. Likewise a chart on is a pair (
open and
. Then
is open in
and
is a homeomorphism
( ) which is open in
) is a chart on
onto ( )
. Hence (
of
dimension
. This construction respects compatibility of charts. So an atlas on and
an atlas on give rise to an atlas
on
which we call the product manifold. It
has dimension
.
Example
⏟
, the torus of
dimensions is a smooth product manifold.
Chapter 2: Lie Groups
For a group , it has two associated maps
taking ( )
taking
(
)
and
Definition 2.1
A Lie group is a set with two structures:
)
1. A group structure with maps
,
where (
and
( )
2. A
manifold structure with a 2nd countable Hausdorff topology such that the maps
and are smooth maps of manifolds.
Note that it is sufficient to require that the map ̃ (
)
is smooth.
Examples:
1. Any finite or countable group with zero dimensional manifold Structure is a Lie
Group.
)
( )
2.
under addition and inverses given by negatives (
for
.
{
} under
| |
3.
can be viewed as a subset of the complex numbers
)
complex multiplication and the inverse (
4. If and are Lie groups then so is
with the product manifold structure.
5.
is a Lie group of dimension . This is compact.
{ } under multiplication (
{ }
) under multiplication
6.
under complex multiplication
(
), the general linear group which is the group of invertible real
7.
matrices. ( ( )
( ) since we have a polynomial map
(
(
) is an open set in
)
(
and (
) then
( )
) has a chart given by the inclusion map
( )
. In this case
).
(
) is
)
a quadratic polynomial as a map
so restricts to (
(
)
(
) as a
map. ( ) is given by a matrix whose entries are
polynomials divided by
Lie Group of dimension
8. Similarly we can define
and
.
(
which is open in
9. Denote
. Then
then ̅ (
invertible in
and
is
on
) to be the set of
(
) therefore
(
) is a
invertible complex matrixes
with
)
̅
| |
then ̅
| | . This implies that if
. Then we have |
|
then
is
| || |. The quaternions are an
associative, non-commutative skew-field or division algebra.
}. This is a group under multiplication and
| |
Define ( ) {
̅.
is a quadratic polynomial map and ( ) ̅ which are obviously smooth functions
on .
Moreover, | |
so ( ) is the unit sphere in
(as
vector spaces) and is a regular value of ( )
hence ( ) is a smooth
manifold and so a Lie group.
As a manifold, ( ) is
so
and
both have group structures making them into
Lie Groups. Although not proved here, these are the only spheres with smooth
group multiplication.
) to be the invertible
10. Similarly to Example 7 and 8, define (
quaternion
( )
matrices. As before, this is an open set in
and during matrix
multiplication the real components of the quaternion entries are smooth functions so
is smooth. is smooth by writing quaternion matrices of twice the size in terms of
complex numbers in a way that respects multiplication. (See the Pauli-Spin matrices)
). It follows that (
) is a Lie group of
and then use inverses defined in (
dimension
.
∑
11. Define ( ), the orthogonal group have an inner product in
if
(
) and
(
) then define ( ) {
( ) ( ) ( )
}. Rewrite
the product of matrices where
are column
vectors and
is transpose.
( )
Since ( )
, we have ( ) ( )
so
. That is
}
( ) {
( )
( )
First this implies that ( ) is a group; ( ) ( )
so
( ). As
( ) we have
( )
( ).
( )
( ) by ( )
Define
; we need to prove that is a regular value.
(
)
( )
( )
First note that if ( )
then
so
}.
( ) {
( )
Take
( ) and compute (
)( )
(
)
(
)(
)
( ) can we find
Given
Then (
)
(
)
so that
(
Try
(
)
(
)
)
so the map
.
is
surjective.
Therefore ( ) has a manifold structure by the implicit function theorem, of
dimension
of ( ) {
)
As (
In particular
( )
( )
and ( )
( )
(
(
)
)
(
)
which is the dimension
}.
are smooth maps it follows that ( ) is a Lie Group.
( )
( )
( ) { }
(
)
12. We denote by
the special linear group in dimensions. This is the set
{
}. Clearly this is a subgroup of
( )
( ). To show
( ) is a
( )
manifold, we need to prove that 1 is a regular value of the map
:
)( )
(
(
)
( (
(
))
)
Therefore (
)( ) is surjective since given a real value we can find a matrix
( ) so that the sum of that matrix’s eigenvalues is .
}
( ) {
( )
( )
( ) is called the special
13.
orthogonal group.
( ) { }
If (
( ) then (
)
)(
)
(
) with
. We have to solve
a set of simultaneous equations. In particular as
we note
that ( ) ( ) are unit vectors which are perpendicular in
has basis (– ) so ( )
(
)
(
(
( ) but
) where
and
. The line perpendicular
)
. Therefore
(
. Observe
)
so we
conclude the following:
( )
)
{(
( )
{(
}
)
}
Therefore we conclude that ( ) is as a manifold and as a group structure. ( )
is a disjoint union of 2 circles as a manifold and ( ) is the connected component of
( ) containing the identity .
Definition 2.2
If is a Lie Group then the connected component of which contains (as a topological
space) is also a Lie group of the same dimension, denoted . This Lie subgroup is
sometimes called the component of the identity.
Example
Example 13 above proves that
( )
More generally, ( )
by induction on :
( )
(
(
)(
( ) .
( ) so they have the same dimension. This can be proved
) as a row of column vectors in
)
(
). Then if
. Therefore
(
) so
we have
{
The idea is then to rotate
to line up with the standard vector ( ) and apply these
to give a matrix of the form (
rotations to
(
) (
) component belongs in (
( ).
two components and ( )
( ) in turn gives
Taking each column of
). Moreover the
). Then an induction shows ( ) has
vectors in
of unit length in
and hence get a map ( )
whose image is a closed
and bounded set since it is defined by polynomial equations
for
.
( ) is homeomorphic to this image set so it is compact.
( ) is an open subgroup of ( ) so closed and hence compact and connected.
{ } and
. Then define
) and (
) as follows, for
14. Now we define (
( ) ∑
∑
For
, define
(
( )
(
)
( )
{
We want to first rewrite this definition so it does not involve
)
(
)
(
⏟
}
and .
)
(
)
(
) {
( )
So
From here, the argument can proceed analogously to ( ):
( )
( ) by ( )
Define
and now prove that
}
is a regular
) is a manifold of the same dimension as ( ) which is
(
value. So
.
(
)
.
We need to check that (
) is a group. Multiplication and inverses are smooth
(( )
) So (
) is a Lie Group of dimension
)
)
(
We define
(
(
(
)
.
)
Examples
Consider the case
(
. Then for (
)
(
)(
) (
) we have
)
( ) is a non-zero vector. As
we have ( )
so for some
(
)
In particular if (
)
Now (
)
) is perpendicular to ( )
this implies that
(
)
(
)
(
)
}. We can now find a unique
)
)
{(
)(
)
so
) as a manifold is two copies of
Observe that (
we have (
( ). As
(
(
)
)
{(
so
)
so we can rewrite
(
(
(
so that
and
}
(
{
)
}).
(
(
(
)
)
(
(
)
)
)
So
(
) is a homomorphism of groups from
map given by (
)
(
function on
(
). Hence if (
(
)
(
to
)
). It has an inverse
(
) is a smooth
).
Definition 2.3
A homomorphism from a Lie Group to a Lie Group is a map
which is both a
group homomorphism and a smooth map of manifolds. is an isomorphism if it is a
bijection and
is a homomorphism of Lie Groups. In particular then is a
diffeomorphism.
So the above explanation shows that
(
and
) are isomorphic as Lie Groups.
( )
We now briefly consider the case
. Then ( ) preserves
which is the basis of special relativity. ( ) is called the Lorentz Group.
( )
( )
(
) has two components and ( ) is the subgroup preserving
time orientations.
In particular, it is clear that (
)
(
)
( ) and (
)
(
).
) the Symplectic Group in
15. In this example we consider (
dimensions. It is
the group which preserves a non-degenerate antisymmetric bilinear form on
.
(
Take
where
Define
(
)
( ). Observe that
)
(
. Observe that
{
( )
(
)
)
. Then take
(
(
)( )
(
}
)
{
(
)
(
)
( )
(
( )
)
(
( )
).
( ) so
)
( ) can be written as
( ).
for some
( ) is surjective for
(
) and so is a regular
) has a manifold structure of dimension
( )
Check that every
( )
Therefore
value. Therefore (
( )
)
)
).
}. Observe that
and so is invertible and (
(
)
(
)
. Therefore
(
)
(
)
Put ( )
. Then ( )
( )
( ) is
. Differentiating we get
(
(
(
)
(
)
) is dimensional, the same dimension as (
)
In particular (
16. There are complex matrix versions of the previous few examples and are denoted
) (
)
(
)
(
) (
)
(
)
similarly as in (
17. On
For
we can define a sesquilinear form as follows:
, the inner product ⟨
⟩
( ) {
( ) ⟨
⟩ ⟨
It is easy to see that ( ) is a group.
∑
⟩
. Then define
}
{
( )
}
Put ( )
then ( )
( ) so
maps to the
complex
matrices such that
. These are known as Hermitian or Self-Adjoint Matrices.
Let ( ) denote the set of self-adjoint matrices. ( ) is a real vector space of
(
)
( )
( ) we need a regular
dimension
. As a map
( ) we need to check if
( )
( ) is surjective as a
value, so for each
linear map.
( )
( ), choose
If
(
so
)|
; that is
( ) is surjective
. Therefore
so is a regular value so ( ) has manifold structure with charts given by choosing
some of the real and imaginary parts of the matrix entries. Hence ( ) has
( )
( )
dimension
.
[
( ) as a row of columns:
Write
⟨
So
⟩
⟨
⟩
]
⟩
⟨
[
⟨
⟩
⟨
if ⟨
⟩
then the vectors
Writing the vector
] then
where
requirements ⟨
⟩
form an orthonormal basis of
⟨ ⟩
| |
| | so
( )
Similarly,
( )
(
.
which identifies the
. Hence ( ) is a
to a closed subset of
compact Lie Group of dimension
] then
⟩
. This gives a map ( )
( )
[
.
) is called the special unitary group. For any
( ), |
|
so we already have a real
(
)
condition on
. Then setting
imposes one further condition on any
( )
so we drop one dimension. Hence
. If necessary, one can
look at ( )
Case
( )
.
{
}
as a Lie Group. In particular,
( )
{ }.
Case
Take (
(
| |
( ) for some
)
| |
| |
| |
)
(
. Then (
) so | |
| |
The vector ( )
so ( ) is a complex multiple of (
and
| |
Therefore
.
| |
| |
so | |
.
)(
| |
)
(
| | and
) i.e. there exists
)
.
so that
(
(| |
)
| | )
so
( )
) | |
{(
| |
}
as a
manifold, while ( ) as a manifold is diffeomorphic to ( )
.
Is it an isomorphism of Lie Groups? No.
}
} so the
| |
The centre of ( ) is the set {
. The centre of ( ) is {
centre of ( ) is connected, but the centre of ( )
has two components.
( ) are diffeomorphic but not isomorphic.
Therefore ( ) and ( )
( )
( )
( ) mapping ( )
(There is a homomorphism of Lie Groups
but is not injective.)
We have shown that
and
are Lie Groups. No other spheres are Lie Groups. Both
and ( ) are Lie Groups diffeomorphic to .
18. Similarly, we can define (
) and
⟨
(
⟩
We can write it in matrix form as ⟨
( )
) using the form
∑
∑
⟩
where
(
). It
) {
( ) ⟨
⟩
⟨
⟩
follows analogously that (
) is a Lie Group of dimension (
) .
and that (
(
) then |
|
) is a Lie Group of dimension (
If
and so (
.
}
)
In particular, ( ) is of dimension and used by Roger Penrose in “twister
theory” to convert some zero-mass field equations in Relativity into CauchyRiemann Equations. This group is closely related to ( ) (the conformal group)
which is locally isomorphic to ( ).
) are the quaternion versions of the unitary groups:
19. ( ) (
( )
(
{
( )
}
)
( )
{
These are both Lie Groups, of dimensions (
These matrix groups
exceptional Lie Groups
}
).
etc are called classical Lie Groups. There are 5
.
Chapter 3: Lie Algebras
Definition 3.1
A vector space with a bilinear map [
we have:
1. [
2. [[
]
]
[
]
] [[
In this course, usually
]
]
[[
]
is called a Lie Algebra if for all
]
]
(Jacobi Identity)
will be a real vector space of finite dimension.
Examples:
]
1. Any vector space with the bracket [
is a Lie Algebra (Usually called the
Trivial Lie Algebra).
]
2. Take any associative Algebra and define [
(The Commutator
]
bracket). If is commutative, then [
so we get the trivial bracket.
It is easy to check that [ ] satisfies the properties above, making into a Lie
Algebra.
Special Example:
{
} using composition of maps, denoted by
( ). For
Let
( )
]
3. If is a Lie Algebra and
is a subspace we call a Lie Subalgebra if [
for all
. is itself then a Lie Algebra.
Special Example:
( ) where is a commutative associative Algebra,
If we take the Lie Algebra
( ) of derivations
( ) defined by
( )
then the subspace of
{
( )
( )
( )} can be shown to be a Lie
( ).
Subalgebra of
Important Example:
( ) for a manifold . Then
( ) is the set of linear maps
Take
( )
( ) satisfying ( )
( )
( ). We can show that
( ) is a
( ).
Lie-Subalgebra of
( )
( ) is a linear map with (
)
In the case that
, suppose
( )
( ). We want to find all such maps:
( )
( )
We first show that ( )
for all constant functions : ( )
( )
( )
( )
. Then by -linearity, ( )
.
Now use Taylor’s Formula: For
( )
( ) (
) ( ) (
) ( ) for some
( ) as a function of .
( ) ( )
((
) ) then
Then ( )
((
) ) (
) ( )
((
) ) (
) ( )
(
) ( )
( ) ( )( )
( )( ) then
In particular, ( )( )
. If we define ( )
( )( )
( ) ( )
( )
so
is of the form
where
( ).
Remark
In the important example, we can show for
( )
For
∑
( ) if
[
(
with
]( )
and
) and
(
) has the form
( ).
then
( ( ))
(
So
that any derivation of
( ( ))
(
)
( ) becomes a Lie Algebra with Bracket [
)
(
(
)
)
]
(
)
Chapter 4: Tangent Space
Assume
in
(
each
(
is a smooth map
)
) we have a tangent vector
from some open interval (
)
. Then at
.
If we have a curve in a manifold can we make sense of its derivative and have it lie in a
vector space? We can use charts to transport the curve to
and differentiate it there.
(
)
(
)
So suppose
is a smooth map and
so ( )
. Take a chart
(
) with ( )
. Then ( ( )) is a curve in
defined on an open interval around
( ( ))
|
We can then form
Taking a second chart (
.
) also with ( )
( ( ))|
.
(
( ( ))|
we can form
)( ( ))|
(
.
) ( ( ( )))|
Then by chain rule:
( ( )) (
Call ( )
(
) (
Space of
(
at .
( ( ))|
( ( )) (
)
) by
then we have a relation defined on the triples (
) if
) . Clearly is an equivalence relation.
( )(
Definition 4.1
Using the equivalence relation
{
)
)
defined above, let
(
)
}. Then
is called the Tangent
) with
We make
into a vector space by picking a chart (
and defining the sum
)(
).
of two classes by addition of the vectors when the representatives are (
]
]
) with respect to
If the class of (
is denoted by [
then [
[
]
[
] and [
]
[
] for
.
Since the maps
choice of chart.
( )(
) are linear isomorphisms this notion is independent of the
Definition 4.2
If
(
)
also write
is a smooth curve then we define
( ) or ̇ ( ) for
( )
( ( ))]
[
( )
( )
as [
( ( ))
]
( )
. We may
( )
Theorem 4.3
For a manifold of dimension and a point
,
is a vector space of dimension . If
is a smooth map then there is a corresponding linear map
( ) called
the derivative of at which satisfies the chain rule: if
is another smooth map then
)
(
)
is smooth and (
. In particular,
( )
( ) ( )
(
) ( ). This is a special case of the chain rule thinking of as a smooth map between two
manifolds.
Proof
]
) with
Given a point
we pick a chart (
then each [
so
[
] is a linear map
which is a bijection hence a linear isomorphism. We
[
] by picking (
) around ( ) then
define
is a map between open
[
]
[
(
)( )] ( ) .
sets in Euclidean Spaces and set
) and (
). Then
Observe that the equivalence class is independent of the choice of (
)
) (
))
( )(
( ) ((
)
)
( )(
( ( )) (
This gives the Chain Rule for (
) .
Examples:
) so we can use this to make an
1. Take
; this has standard chart (
identification between
and
as a vector space:
[
]
( )
)
Under this identification, if we have a curve (
then the abstract tangent
(
vector is [
In
2. If
( ) as a vector valued function.
( )
we can write down a curve which has tangent vector at for any fixed
; namely the straight line
.
) is a chart with
is a manifold,
and (
. Then take
then
[
] for some
)
and we get a curve in on some interval (
containing
[
)]
( )
by setting ( )
|
]
( ( )
[
]
) then
( )
[
( ( ))|
]
.
Definition 4.4
( ) and
Let
. We define ( )
as a directional derivative at by choosing
[
] and taking
) with
a chart (
, transporting to ( ) as
, writing
the directional derivative of a function of variables at ( ) in the direction ; that is
( )
(
( ( )
))|
Remark
This expression of ( ) is independent of the choice of chart.
Proposition 4.5
1.
2.
3.
(
( ) is linear as a map
( ) is linear as a map
)
( )
( )
( )
Proof
All these statements are easy consequences of the directional derivatives of functions on
open sets in
.
Remark
This can be used to give another definition of tangent space at
( ) to satisfying the product rule (
)
( )
from
form a vector space with
a linear subspace.
In these particular identifications,
( )
; namely all linear maps
( ). They clearly
( ( )); the derivative of a real-valued
function of 1 variable.
Examples
1. Let
be an open set in a real vector space . We can get a chart on by
identifying with
where
and defining a map from as follows:
(
)
∑
Pick a basis
for and let (
if
for
)( )
)(
. Then (
. It is clear that changing the basis gives a
) is a chart on
compatible chart since they are related by invertible linear maps of
which are
∑
diffeomorphisms. We can identify
with by [
(
) ]
and check that this does not depend on the choice of basis.
Given
and
we denote the corresponding tangent vector by
[
Then if
( ),
( )
(
( (
)
]
))|
;
∑
( )( ) with ( )
.
( ) for a regular value of . If
2. For a smooth map
let
)
and
let ( )
be a curve (
with ( )
.
Then ( )
means ( ( ))
for all . Differentiating this expression with
respect to ;
then put
to get ( ( ))
so ( )
( ) ( ( ))
This gives a map
subspace of
.
E.g. Take
(
)
so
orthogonal to .
Lie Group Examples:
which allows us to identify
( ) with ( )
as
{
. Then
}. That is
with a
, the set of vectors
(
1.
) is an open set in
( ),
For
(
(
(
( ) so
(
)
( )
)|
( ) so
(
)
( )
( ) so
(
)
( )
3. Similarly,
(
)
( ) where
( )
4.
is a smooth map. We showed
∑
( ). We can introduce the function
for any
(
)
( )
( ) ∑
( )
( )
by
. Observe that
(
)
( ) so by Jordan Normal Form
( )
(
) and for
( )
Therefore
,
}
(
) {
( )
( )
Therefore
}
) we can have
(
) {
( )
( )
5. Similarly, for (
}. Let
( ) {
( )
( )
( ) by ( )
6.
then
2. Similarly,
) is an open set in
) is an open set in
)( )
(
{
Therefore
( )(
(
) (
}
( )
( ) and so
( ( ))
( )
{
( )
{
} we can see
{
}
}
( )
8. In particular,
is a smooth map, we have defined
( ) for some curve ( ) with ( )
, then if
and
( ( ))( )
( ( ))|
( )
then (
)( )
( ) then (
we think of
as a directional derivative on
( ): if
( ) then take a curve
a directional derivative on
( )
. Then
(
)( )( )
(
)( ( ))( )
( )(
That is, we get
If
( ( ))
( ))
7. Using the same method as above, for ( )
Remarks
If
)|
(
(
) ( )( )
)
(
)
)( )( )
(
)
((
Proposition 4.8
For a Lie Group , the functions
Proof
By definition of , the multiplication map
Therefore by fixing one of these entries, ( )
(
) is smooth.
)( ) can be computed as
so that ( )
and
)( ))|
is a Lie group, the left and right translations are defined by
. If
( )
(
and
)( )|
( )
are smooth and satisfy
)
defined by (
(
) is smooth and
is smooth.
( )
Observe that for any
we have
(
( ))
(
)
(
( ))
(
)
( )
( )
( )
Corollary 4.9
For any
,
and
( )
are diffeomorphisms of
(
)
(
with
)
Lemma 4.10
(
)
(
)
Definition 4.11
A vector field on a smooth manifold
The tangent vector at is defined by
is the choice of a tangent vector to
at each point.
Definition 4.12



( ) is smooth for all smooth functions
A vector field is said to be smooth if
( ). Then we set ( )( )
( ) for all
.
( )
( ) which is a linear map. A
A smooth vector field defines a map
)
( ) ( )
( ) ( )
tangent vector at satisfies (
)
( )
( ) ; that is smooth vector fields are derivations of
Hence (
( ).
We denote the set of smooth vector fields on by ( ). This is a vector space and
has a Lie Algebra bracket given by the commutator:
[
]( )
( ( ))
( ( ))
which is called the Lie Bracket of Vector Fields.
Definition 4.13
Let
be a smooth map, then two vector fields
( )
-related if
( ) for all
( ) and
( ) are said to be
then
. Then
Examples:
1. If is constant ( )
for all
for fixed
related to if and only if vanishes at .
2. Suppose is a diffeomorphism of onto . Then
(
to ( ) so
( ) (
( ) ) and
is -
is a linear isomorphism from
) ( ( ) ) so and
determine each other completely.
3. If is not surjective then outside ( ), the closure of the image, the values of
unknown.
are
Theorem 4.14
If
and
[
].
( ) is -related to
Proof
( ) then for
( ) we have (
First, take
related to then the LHS becomes ( ) ( ) so
( ( )
)( )
Therefore ( )
( ).
Now suppose
[
](
( ( ))( ( ))
(
). That is
is -related to
)
(
(
( )(
( ( ))
So by the above condition, we have [
(
)
(
( ( ))
] is -related to
(
)
( (
) and if
is -
))
iff ( )
(
) for all
( ) then using the above condition;
and
(
))( )
(
is -related to
for
))
then [
( ) for
))
( ( )
([
] is -related to [
]( ))
].
)
( ( )
)
Chapter 5: The Lie Algebra of a Lie Group
Definition 5.1
A Vector Field
all
is said to be left-invariant if (
on a Lie Group
)
( )
for
.
(There is a similar notion of Right-Invariance using
Remark
is left-invariant means that
Lemma 5.2
If is left-invariant, then
is
).
-related to itself for all
.
is smooth.
Proof
( )( )
( )
( ) (
( )
)( )
(
). Then note that (
)( )
(
)(
) is jointly smooth in and and
differentiates . Therefore ( )( ), being
the derivative of a smooth map, is smooth as a function of so is smooth by definition.
Theorem 5.4
The Lie Bracket of two invariant vector fields is again left-invariant. Left-invariant vector
fields form a linear subspace of ( ) and the evaluation map
from ( )
is a
linear isomorphism on the subspace of Left-invariant vector fields. Hence the left-invariant
vector fields on form a finite dimensional Lie subalgebra of ( ) whose dimension is the
same as the dimension of as a manifold.
Proof
If and are left-invariant then is -related to and is -related to for all
.
] is -related to [
] by Theorem 4.14 so [
] is left-invariant.
Therefore [
The fact that left-invariant vector fields form a subspace of ( ) is a consequence of the
defining equation (
)
map. The evaluation map
fields.
If
is
involving the derivative of , which is a linear
is a linear map by the definition of addition of vector
( )
for some left-invariant vector field then taking
so
gives
for all
is injective.
To see the map is surjective, take a tangent vector at the identity of
( )
. Then
(
(
So
)
(
)
)(
)
)(
(
(
is left-invariant and hence smooth. Evaluating at
(
)(
)
Hence the map defined by evaluation at
(
)(
. Then set
)
is surjective.
))
(
gives:
(
)
)
that
Summarising, the evaluation map is a linear isomorphism from left-invariant vector fields to
.
is a vector space of the same dimension as .
Definition 5.5
Let
then we denote by ̃ the left-invariant vector field on
That is, ̃
whose value at
is .
( ). ̃ is called the left-invariant extension of .
Definition 5.6
If is a Lie Group then its Lie Algebra is the vector space
with bracket [
]
[ ̃ ̃] .
[ ]) is denoted by the lower case fraktur letter . As a vectorspace,
The pair (
.
Remark 5.7
We could use right-invariant vector fields as each result above has a right-invariant
].
counterpart. The resulting bracket on
is [
Corollary 5.8
If is a Lie Group and Abelian as a group, then [
]
for all
.
Examples
1.
is a Lie Group under addition and is abelian, so its Lie Algebra is
with zero
bracket.
The same holds for any finite dimensional real vector space and for , so
and
have the same Lie Algebra,
with the zero bracket.
is also an -dimensional abelian Lie Group. We will show later that these
account for all connected Abelian Lie Groups.
(
)
( ) is an open set so its tangent space at any
(
) is just
( )
2.
( ) and
(
)
with tangent vectors being directional derivatives: if
(
then we define
except for a 1 in entry (
Then for
(
(
(
)|
) then the curve
)),
) which we can call
These give a basis for
By taking ( )
. In particular if we take
(
( )
to have all zeroes
only adds to the (
( ) so as
)th entry of .
varies we get a vector field on
.
) at every point. In fact,
then
Claim
This means that at each point
(
) the corresponding
| are
linearly independent.
Proof
Suppose ∑
Then
|
. Then apply this expression to the function
.
∑
This holds for all
If
so all
(
is any vector field on
are zero.
) then
∑
( )
( ). As varies then we get
functions
on (
( ), so if is a smooth vector field, then all
are
∑
| for some coefficients
). In particular,
(
)
functions. Furthermore,
.
( )
If
̃(
), let ̃ be the left-invariant extension. Then
(
)|
(
)
)(
(
)
( (
Since
(
))|
)|
)(
(
)|
of a matrix is a linear map so:
̃(
)|
(
( )
))|
) so ̃ (
(
This is true for all
Moreover, if ̃
(
∑
(
)
)
∑
(∑
and hence ̃
∑
)( )
∑
.
then
[ ̃ ̃ ](
)
(̃̃
Finally we get a Lie Algebra called
]
Commutator Bracket [
̃ ̃ )(
(
( ̃
)
)(
)
) which is the vector space
( ) with
3. All the other Matrix Lie Groups:
) is a smooth map say which identifies ( ( )) as
The inclusion of ( ) in (
(
) namely ( )
( ) and if
( ) then we can
a subspace of
) so we get two left-invariant extensions ̃ ( ) and
extend it to ( ) or to (
̃
[̃
(
)
( )
̃
which are -related. Hence
( )
] is -related to [ ̃ ( ) ̃
( ) so [ ̃
( )
̃
( )
]|
( ̃
(
)
)
(
( ̃
]
)
|
)
(
)
and
.
So the Lie Algebra of ( ) is ( ) with Commutator bracket. It is denoted by ( ).
[Note that this argument used no information about ( ) and works for any matrix
(
) (
) (
) all have respective Lie
Lie Group. For example ( )
) (
) (
), which are the Tangent Spaces at the
Algebras ( ) (
Identity equipped with the Commutator Bracket.
Definition 5.10
A homomorphism of Lie Algebras and is a linear map
preserving brackets:
([ ] ) [ ( ) ( )]
If
is a homomorphism of Lie Groups then a homomorphism implies ( )
and so the differential of at
sends
to
. Denote this linear map by ; that is
.
Theorem 5.11
If
is a homomorphism of Lie Groups then
is a homomorphism of
Lie Algebras.
Proof
If
then
( )
( ) .
and we can find the left-invariant extension ̃ and ̃
Claim
̃ and ̃
( ) are -related.
Proof
Since ̃ is left-invariant by definition,
(
̃
(̃
)( )
(
(
(̃ )
(
)
( ))
then [ ̃
([
] )
([
)
(
)
(
)
( ) ( )
( )(
( ))
⏞
)
)( )
( )
)
(
( )(
Hence, for
we have
(
)( )
( )
([ ̃
( )(
̃
( )
( ))
̃ ] is -related to [ ̃
] )
⏞
̃ ]. Evaluating at
̃ ] )
[̃
̃ ]
and (
[ ( )
)
( )]
If
and
are homomorphisms of Lie Groups then this induces
homomorphisms of Lie Algebras
The Chain rule implies that
)
has derivative at
and (
that is compositions are preserved.
Corollary 5.12
Isomorphic Lie Groups have Isomorphic Lie Algebras.
Proof
If
Therefore
is an isomorphism, then
and differentiating this gives:
is an isomorphism and ( )
is a homomorphism and
(
and
)
However, the converse to this is not true. It is possible to have a homomorphism
is an isomorphism:
so that
Examples
( ). ( ) has two connected components, while
is the inclusion map ( )
( ) is connected, so is certainly not an isomorphism. However the tangent space
( )
( )
( ). More generally,
at the identity is the same:
,
not connected. Therefore only gives information about .
2. It can happen that for and not connected,
can be a homomorphism but
not an isomorphism, while
is an isomorphism. (In particular, any two
Abelian Lie Groups have isomorphic Lie Algebras if they have the same dimension
such as
( ) ( ). We know
For an example of a non-trivial Lie Algebra consider ( )
1.
( )
( ). Computing Centres: (
( ))
}. Meanwhile, ( ) has centre {
( ) and ( )
( ).
and thus ( )
( ) ( ) is
( ) and
However, ( )
[ ]
[ ]
where [ ]
{
{
( )
So has basis
(
)
}
}
(
}
so (
( )
.
)
{(
{
)
( ( ))
( ))
{ }
( ) has basis
}
(
) and [
[
]
[
]
therefore identifying
isomorphism between ( ) and ( ).
[There is a homomorphism inducing this called the spin covering.]
]
gives an
Chapter 6: Integrating Vector Fields and the Exponential Map
Definition 6.1
Let be a smooth vector field on the manifold. A smooth curve
integral curve of the vector field through at time if ( )
(
(
)
is called an
and ( )
( ) for all
).
Theorem 6.2 (Existence and Uniqueness of Integral Curves)
( ),
)
(
) and a
If
and
there exists an interval (
with
)
smooth curve (
which is an integral curve of passing through at time .
Moreover, if we have two integral curves of passing through at time then they agree
on the intersection of their domains.
(
)
(
)
( )
( ) for
(
) (
)
Note
That we have a “larger” solution on (
Sketch of Proof
) and
Take a chart around , say (
vector valued function on ( ). For
( ( ))
)
(
)
and transfer the problem to
. becomes a
,
[
( ( ))] and we look for so that
( ( ( )))
( ( ))
( )
Then we solve the above first order autonomous linear ODE. The existence and uniqueness
Theorem for ODEs guarantees a solution:
We find ( ) satisfying
( ( )) and ( )
( ) then set ( )
( ( ))
Example
Take (
(
)
,
( )
)
( )
(
(
(
)
( )
( )
) with
Take
or
(
{ } and
[
](
( )
)
where
( ). The ODE we solve is
so the integral curve through ( ) at time
is
).
then the integral curve through (
) at
is (
) with
).
Definition 6.3
An integral curve is called maximal if any integral curve through one of the points on at
the same time is the restriction of to a subinterval. A vector field is called complete if
). Otherwise, is called incomplete.
every maximal integral curve is defined on (
Theorem 6.4
If is a Lie Group and
is a left-invariant vector field on
then
is complete.
Lemma 6.5
)
If is a smooth vector field on a manifold and (
(
) is an integral curve on (
for any
then ̅ ( )
is an integral curve of
).
then
Proof
Since the ODE is autonomous.
Lemma 6.6
)
If is a left-invariant vector field on a Lie Group and (
( ) for all
(
) is an integral curve.
for any
, ( )
is an integral curve then
Proof
( )
( )
( ( ))
( )(
)(
( ))
(
( )
)
( )
⏞
( )
( )
So ( ) is an integral curve.
Remark
These two Lemmas tell us that a left-invariant vector field on a Lie Group has essentially
one integral curve from which all others can be obtained. Let
be the maximal integral
(
) in the maximal integral curve
curve for passing through at
. Then
passing through at
. So all maximal integral curves are defined on the interval
around .
Lemma 6.7
Let be a left-invariant vector field on the Lie Group and
its maximal integral curve
(
) for all
with ( )
then ( ) ( )
and ( ) is defined for all
.
Proof
Let
be such that they are all in the interval where
is defined. Fix and let vary
( ) ( ) and
(
) which are both
in some open neighbourhood and consider
integral curves of as a consequence of Lemmas 6.6 and 6.5. Then the uniqueness of
Theorem 6.2 says that they agree for all where defined.
We claim this formula allows us to extend
arbitrarily:
) with
If
were only defined on some interval (
finite then ( ) ( ) as a
) but (
) is defined on (
). Taking
function on is defined on (
positive contradicts the minimality of and taking negative contradicts the
maximality of .
Hence
are infinite so (
)
(
).
Proof of Theorem 6.4
The Three Lemmas 6.5, 6.6, 6.7 together prove 6.4.
The property
Groups.
(
)
( )
( ) says the map
is a homomorphism of Lie
Definition 6.8
A Lie Group homomorphism
of or 1-PSG.
with
a Lie Group is called a 1-parameter subgroup
Remark
A 1-PSG is a smooth curve in which is defined for all and a homomorphism of groups.
By the observation above, a maximal integral curve is also a 1-PSG.
Theorem 6.9
There are natural bijection between the following sets:
1. The set of 1-PSG of
2. The set of maximal integral curves of left invariant vector fields on
3. The Lie Algebra
Proof
[
] Given a 1-PSG
( ). Then
let
[
] Take the Left-Invariant Extension ̃
[
] Take the integral curve
(
)|
̃
̃(
and observe that
̃
( ) ( )|
( )
̃(
Example
For
, suppose
( )
then
is a 1-PSG so ( )
( ) for
(
( ) if
for all
Note that
)
. Moreover (
( )
( ) therefore
( ) then
that is
( ):
( )|
( )(
( ))
( )(
( )
Then by uniqueness of integral curves we have
( )
)
is an integral curve of ̃, then use uniqueness of integral
This can be seen by checking that
curves:
( )
.
( )
.
)
)
( ) for all .
(
). Then an induction shows
( ) so ( )
(
( ). Then for
)
is dense in
( ) for all
and
we have
. If
( )
is continuous therefore ( )
is a straight line through .
( ), so this gives the Lie Algebra of
Definition 6.10
If is the Lie Algebra of the Lie Group
map by setting
̃( )
we get a map
Proposition 6.11
is a smooth map of manifolds.
as
.
called the exponential
)
Proof
̃ is a family of solutions to a family of ODEs parameterised by . Smoothness depends on
the families version of existence and uniqueness theorem for ODEs
Proposition 6.12
Let
then for
( )
Proof
Consider the map
̃(
̃(
) with
)
fixed which is a 1-PSG so
) ̃(
)
) )
̃ ((
Therefore the map above is the integral curve
̃(
Hence
̃(
̃(
)
̃(
( )
for some
)|
) for all . Putting
Corollary 6.13
( ) is a 1-PSG; that is
̃
̃(
)|
̃(
)
yields
( )
(
and
̃(
)
( )
)
Proof
By Theorem 6.12 and Lemma 6.7, using the fact that the integral curves of left-invariant
vector fields are defined on ;
(
)
)
( )
( )
̃(
̃( ) ̃( )
Corollary 6.14
The integral curve of ̃ through
at time
(
is
) .
Examples
( )
1.
( ) for 1-PSG of
tells us that
is the identity map.
{ } under multiplication. If
2.
then ̃
̃(
)
so
(
̃(
That is
(
) If
3.
)|
)
(
(
then
) so
̃
is the solution of
̃(
)|
)
̃(
) and
.
(
)
( ) then
satisfies
̃
̃(
)
̃(
) with
̃(
)
the solution is
̃(
Therefore
(
. For
)
)
∑
∑
(
( )
)
4. Same for all our matrix groups (but need to check that
Theorem 6.13
Let
be a homomorphism of Lie Groups and
for
the derivative at
then
and
(
)
( )
Proof
we compare 1-PSG
(
( ))
tangent vectors at
( ̃ ( )) and
( )
̃
( )(
) are 1-PSG and have the same
.
Example
(
)
The determinant map
polynomial map hence smooth and
given by the trace so
is a homomorphism of Lie Groups (it is a
(
)
( )
( ) we have
(
)
( )
. Therefore by Theorem 6.13,
is smooth so has derivative and in particular
. Since
we have
⏟
Theorem 6.14
under the identifications
and
.
Proof
If
it defines a tangent vector at 0 as the tangent to the cuve
since
̃( )
( )
(
(
Corollary 6.15
There is a neighbourhood of
diffeomorphism of with .
)|
)
⏞
(
in and a neighbourhood
at
)|
of
. Therefore
̃(
in
)|
such that
is a
Proof
Apply inverse function theorem after picking a basis of to identify it with
for some
) on with
and a chart (
. Then
becomes a map
and
( )
(
(∑
)) if
is a basis for . Then
is an isomorphism of
with
itself, so by IFT we have a diffeomorphism near
.
Corollary 6.16
has an atlas where the open sets are translates of obtained in the previous Corollary, and
the chart maps are a left-translation to bring it into a neighbourhood of followed by
followed by picking a basis for to get to . These charts are compatible.
Remark
Such charts are called canonical and the atlas also canonical.
Corollary 6.17
If is connected, every element of
is a finite product of exponentials of elements of .
Proof
For a connected Group, any open set containing
from Corollary 6.15
Proposition 6.18
If is connected and
then
.
Proof
If
generates the group so we can take
is a homomorphism of Lie Groups for
and
then by the previous corollary it is a product of exponentials
( )
( )
( ) so
( )
(
( )) (
(
(
( ))
(
)
(
( )) (
(
)
( ))
))
(
(
(
)
(
)
(
)
)
(
))
( )
Remark
Notice the strength of this condition: is one derivative of a homomorphism at one point;
by the Theorem above this completely determines on . General smooth maps do not
have this property.
Proposition 6.19
Let be a connected Lie Group then
only if is abelian.
is a homomorphism of Lie Groups if and
Proof
If
is a homomorphism then
Therefore
(
)
(
)
connected implies every element is a product of exponentials so for any
∏
If
is abelian, then
( )∏
∏
( )
( )
( )∏
is a 1-PSG therefore
(
)
(
)
Lemma 6.20
{
( )
Let be a Lie Group. Then
integer span of a finite set of linearly independent vectors in .
Theorem 6.21
Let be a connected Abelian Lie Group. Then
Proof
By Corollary 6.17
above,
} is either 0 or an
is the product of tori and Euclidean Space.
is surjective since is connected and abelian. By the Lemma
{ } or the integer span of
is a discrete set and so either
,
:
linearly independent in . In the first case, by the first isomorphism theorem we have
{ }
. The consider
⏟
{
. In the second case let
and
Group as
. Hence can see
homomorphism) and so
diffeomorphism.
and
is a bijection from
} and
. Since
to
be a supplement so
so
identifies
so we can make
with
into a Lie
is in bijection with
(
is a surjective
using the group structure and the map is a
Chapter 7: Lie Subgroups
There are “bad” subgroups which ought to be Lie Groups but are not embedded
submanifolds.
Example
Take
, then
with the zero bracket so every subspace of is a Lie subalgebra; so
the subspaces are { } and lines through the origin.
These exponentiate to subgroups of the form
(
) if ( ) is a non-zero vector in
.
These can either be closed curves in
in
(The condition for a circle is
lines with irrational slope in
(so embedded circles) or copies of
(assuming
)). Copies of
which are dense
are projections of
.
Definition 7.1
1. A smooth map
is an immersion if
( ) is injective for all
.
2. A smooth map
is called a submersion if
( ) is surjective for
all
.
3. A smooth map is a diffeomorphism if it is a bijection and both an immersion and a
submersion.
Examples



The inclusion
is an immersion. The exponential map
an immersion.
The projection
is a submersion.
are diffeomorphisms

is an immersion and a submersion but not a diffeomorphism.
Definition 7.2
A submanifold of a manifold is a pair (
which is injective and immersive.
Definition 7.3
We say a submanifold
the subspace topology.
Example
is
) where
is embedded if
is a manifold with a smooth map
is a homeomorphism from
( ) with
1.
is embedded
2. Lines of irrational slope on a torus are not embedded, but are immersed.
Definition 7.4
If is a Lie Group, a Lie Subgroup is a submanifold
where is a Lie Group and
) is an immersed submanifold with a homomorphism of groups.
the pair (
This implies:
1.
is a homomorphsm of Groups and is smooth so is a homomorphism of Lie
Groups
2.
is injective and a homomorphism of groups, so is an isomorphism onto its image,
which is a subgroup of
3.
is an immersion, so
,
( ) is injective and in particular at
is injective and a homomorphism of Lie Algebras and hence is an
isomorphism of with a Lie subalgebra of .
Examples
(
1.
) defined by (
)
(
) is a Lie Subgroup. The image is
( )
( )
(
) is a Lie Subgroup and any matrix group defined earlier are Lie
) or (
) or (
) for some .
Subgroups of (
3.
defined by ( ) (
are independent over is a Lie
) where
Subgroup.
2.
Proposition 7.6
A Lie Subgroup of a Lie Group
both injective.
Proof
Suppose
(
)( ̃ )
is a Lie Group
with a homomorphism
)( ̃ ) where
is a Lie Subgroup, then consider (
(
)(
(
Therefore
So in particular,
⏟
( ))
(
)
( )
(
)( )
)( )
( )
(
( )(
)( )
with
.
⏞
(
( )
)( )
( ))
⏟
so
By definition , since
is a Lie Subgroup, is an injective immersion so and
injective for all
. But from the above this occurs if and only if
is injective.
is
Often we have either the image subgroup ( )
(just as a group) or the image subspace
( )
as a Lie Subalgebra. We want, when it is possible, to find the Lie Group structure of
.
Theorem 7.7
If
is a Lie subgroup then
( ) {
( )
Proof
If
put
(
)
( )}
( ) then
( )
}
is a homomorphism of Lie Groups, so
for all by Theorem 6.13. Then ( )
For the other inclusion, we start with
(
. Then we have a curve ( ) for
Step 1
We will prove that
(
) and consider
( ) ( )
)
(
( ( )) for all
) for some
(
).
( )
; that is
. Then
, mapping
( )
We want to show that the derivative
Take
(
is smooth.
Proof
Pick a subspace
so supplementing ( )
is a manifold of the same dimension as .
Pick
(
( ) for all
such that
) such that
{
is an invertible linear map:
then
(
)
)
(
)|
(
)(
( )
(
) (
)
( )|
)
( )
Therefore ( )
is an isomorphism
so by the Inverse
) in
Function Theorem
is a diffeomorphism of a neighbourhood
of (
(
). Then let
to a neighbourhood of
in , say
on
(where
So
is projection onto the first component, which is smooth).
is smooth near
Step 2
We finish the proof; Take
( )
Proof
.
, and
(
arbitrary implies ( ) is smooth on (
) and set
( )
( )
( ( ))
).
. Want to show
( )
(
( )
( )
( ( )))
( )(
( ( ))
( )
( ( )
(
)( ( ))
( )
)
)
(
From the above, we note finally that ( )
⏞
( )(
( ( ))
( ( ))
)
( ( )
( ( )
(
)
) ( ( ))
)
)(
( ( )))
(
(
)
( ))
.
Proposition 7.8
Let be a Lie Group,
a subgroup (in the usual algebraic sense) and
a subspace
(not necessarily a Lie Subalgebra). Suppose there is a neighbourhood of in and of
(
)
in such that
is a diffeomorphism with
. Then with the
relative topology has an atlas formed by left-translates around of the chart on
constructed by inverting
and picking a basis for .
Proof
Left translates of
by elements of are open sets in the relative topology and the charts
constructed as in the statement of the proposition are restrictions of canonical charts of ,
hence compatible and so given an atlas.
are restrictions of
hence smooth.
Therefore is a Lie Group with this atlas and the inclusion map is smooth and an
immersion, so is a Lie Subgroup of with
.
Note
We didn’t need to know that is a Lie Subalgebra of .
Lemma 7.9
Let be a Lie Group and
]
( )) and
(
smooth in with
[
function ( ) defined near
then there is a
and
bounded. [In fact, ( )
( )]
Proof
Pick
with
for some
and
with
a diffeomorphism. Then
(
)
and | |
. For | |
put ( )
( )) for | |
and apply Taylor’s Theorem on
(
(
) then
( ) around
:
( )
( )
( )
( ) with ( ) bounded as
( )
(
( )
( ))
(
)
Then by Theorem 6.14 ( )
. Then
(
Corollary 7.10
Take
( )
. Then
(
)
(
)
) (
)
Proof
Pick
as in the proof of 7.9. Take
large enough so that
(
. Then
( ))
Therefore
(
And
( )
Therefore
)
( ))
(
( )
( )
(⏟
)
(
)
(
(
( ))
)
Theorem 7.11 (Closed Subgroup Theorem)
Let be a closed subgroup of a Lie Group. Then has a unique structure as a Lie Group
with the relative topology making the inclusion map
into a Lie subgroup.
Proof
Uniqueness follows from Proposition 7.8, since there will be an atlas of charts formed by
intersecting exponential charts of with .
The main idea of the proof is to build a Lie Algebra inside and show it is of the correct
{
}. We need to show that is
form. Guided by Theorem 7.7, set
a vector space, and closed under brackets:
1. Since
2. If
3. If
so (
, and
then
then
)
it is clear that
.
since
( ( ))
for all
(
)
(
( )
) but
for all . By Corollary 7.10, (
(
); since is closed, the limit is in , so
(
Since
for all , the above argument shows that
Therefore
.
, for all .
so
and
) converges in
)
(
to
.
)
for all .
Hence is a subspace of .
We will show is a Lie Algebra of a Lie Group structure on by using Proposition 7.8. We
need to find
with
and
with
such that
is a
(
)
( ) we look at open subsets in
diffeomorphism and
. Since
with
on which
is a diffeomorphism onto its image and
(
) (
)
We argue by contradiction; suppose there is no open subset around 0 which satisfies the last
equation and on which
is a diffeomorphism. Fix a Euclidean Structure on and look at
balls around of smaller and smaller radius.
Using Corollary 6.15, fix some open set
balls inside
of radius for
is a diffeomorphism on
and hence
is a diffeomorphism, and consider
sufficiently large for the ball
By hypothesis, there must be an element
Since
on which
( ))
(
,
( )
.
(
and not in
( ) so
with
( )
).
in at
.
Claim
(
)
Proof
If
( )
(
and
(
(
) then
is bijective on
( )
( )
with
so
but
. Hence
( )
). But this is a contradiction, since
(
). Therefore
(
Therefore we have a sequence
( ))
with
and so
was chosen so that
).
(
and
) and
as
.
Pick a supplement for in so
and consider the map
mapping
( )
. Then has derivative at ( ) given by the identification of
with
so in particular, this identification is an isomorphism (from the choice of ) so
by the inverse function theorem, is a diffeomorphism near ( ). That is, there exist open
sets
in and respectively with
and
and
is a
diffeomorphism onto its image in .
Claim
We can shrink
{ }.
so that
Proof
Suppose not. Then we can find a sequence
for all , but
with
so that
||
. In the Euclidean norm on , put
||
so
is a
sequence in the unit sphere in the metric on . Compactness of the sphere gives a
convergent subsequence which (for convenience) we relabel as
Take
and put
( )
so ( )
. Now
.
( )
||
as
therefore
.
( )
⌊ ⌋, the integer part. Then
|| so
. Let
( )
as
and
( )
(
(
)
for all
and
. Therefore by definition of ,
Therefore
closed implies
for all
)
(
.
( )
but
)
{ } (it is the limit point of a sequence in the unit sphere on
But
a contradiction (
by definition of ).
Therefore we can shrink
and
implies
) so we have
{ }.
so that
So the sequence
in and
. Therefore eventually,
(
), since this is
an open set in containing . Therefore for large ,
for some
and
. Note
, since otherwise
which contradicts the first claim.
But then,
Claim,
so (⏟
)
therefore
{ } so
. Contradiction.
Therefore there has to be an open set in with
and
Proposition 7.8, we have a Lie Subgroup structure for
topology.
Corollary 7.12
If is a closed subgroup of a Lie Group
{
Moreover,
.
. By the Previous
(
with
)
. By
having the subspace
then its Lie Algebra is
}
Proof
From the proof of Theorem 7.11, we showed
then
is a 1-PSG of so therefore
derivatives at
gives
.
{
Theorem 7.13
If
is a homomorphism of Lie Groups then
hence a Lie Subgroup with Lie Algebra
.
}. Furthermore if
for some and taking
is a closed subgroup of ,
Proof
{ }. is continuous and
By definition,
are Hausdorff so { } is closed and thus
is closed in . is a subgroup of and so a closed subgroup of . Therefore is a Lie
{
}. Therefore
Subgroup and its Lie Algebra by Corollary 7.12 is
(
)
( )
( )
(
) so
. But
( )
So is the kernel of .
Example
(
)
) as its kernel. Therefore we have a
The determinant map
has (
) is a Lie Group and has Lie Algebra the kernel of
different proof that (
(
)
Similarly, we can apply the same result to
.
.
Theorem 7.14
If
is an automorphism of Lie Groups; that is an isomorphism from to itself, then
{
} is a Lie Subgroup with Lie Algebra
( )
the fixed point set
{
}
( )
Proof
closed in , since a given a convergent sequence of fixed points, its limit is a fixed point
because is continuous. Moreover,
is a subgroup of so it is a closed subgroup. By
Theorem 7.11,
is a Lie Subgroup and by Corollary 7.12, the Lie Algebra of
is the
}. But
set {
(
)
( )
( )
Examples
1.
2.
( ) is the set
(
) with
( ) which is a fixed point
condition for the map ( ) ( ) (this is an automorphism) and
. Hence
( )
(
) so it is a Lie Subgroup and ( )
therefore its Lie Algebra is
}
( ) {
( )
is the fixed points of
3. Let
[
] then ( )
(
)
(
) given by ( )
is an automorphism of
(
)
(
) and the fixed
}. Therefore the fixed point set is a Lie Group
(
)
points are {
{
}. The Lie Group obtained is the centraliser of
( )
with Lie Algebra
)
, which is isomorphic to (
) given by ( )
( ) . Then the fixed
4. Let be the automorphism of (
(
) with ( )
(
)
points are all
Chapter 8: Continuity implies Smoothness
The Motivating Idea behind this chapter is the following:
)
If
is a continuous map and (
( ) ( )
homomorphism then ( )
. Then
(
continuity and
)
( ) so
( )
( ) which implies ( )
( ) is a smooth function of so
Lemma 8.1
If is a Lie Group and
closed subgroup of .
( )
( ) for all
and it is a continuous
( ) for all
( ) for all
so ( )
( ) for all
using
is smooth.
is a subgroup then the closure ̅ in the topology of
Proof
Obviously ̅ is closed. We need to prove it is a subgroup. For
(
and
with
. Then as is smooth,
( ) ̅
( )
is a
̅ take sequences
)
(
) ̅ . Similarly,
Lemma 8.2
If is a continuous 1-PSG of a Lie Group (That is
is continuous and (
̅̅̅̅̅̅̅
( ) ( )) then ( ) is a closed connected abelian subgroup of .
)
Proof
̅̅̅̅̅̅̅
( ) is a closed subgroup already by the previous Lemma. ( ) is the continuous image of
( ) is connected, and by the previous argument, for any
a connected set, hence ̅̅̅̅̅̅̅
(
)
( ) ( )
(
)
(
)
( ) ( )
(
)
̅̅̅̅̅̅̅
)
)
( ) and so ̅̅̅̅̅̅̅
( ) is a closed
And (
, (
so
for all
connected abelian subgroup of .
Corollary 8.3
( ) is a Lie Group
Let
be a continuous homomorphism into a Lie Group . Then ̅̅̅̅̅̅̅
̅̅̅̅̅̅̅
( ) is a continuous homomorphism.
and is connected, abelian and
Theorem 8.4
If
is a continuous homomorphism into a Lie Group then it is smooth, hence a 1PSG so ( )
for
.
Proof
We only need to prove
.
is smooth. Theorem 6.9 then shows that ( )
for some
By Corollary 8.3, we only need consider the case
is a homomorphism which is
continuous and is a connected, abelian Lie Group. We know all connected Abelian Lie
Groups of dimension are isomorphic to
for some and so a homomorphism
is given by a pair of homomorphisms
and
. Moreover, is
continuous precisely when and are continuous.
( )
We already know that is smooth since it has the form
Therefore it is left to study continuous homomorphism
decomposes as
homomorphisms
and only if all are continuous.
Given
)
for
, and
(
)
)
)
(
(
) )
.
so
is continuous if
and it satisfies (
a continuous homomorphism, then we view
( ) ( ). It follows that (
(
for some
⏟
( ( )) and (
)
( ) . Then we need to take a th root. To do so we use the following
theorem:
{ } is continuous, there exists a continuous
If is a simply connected domain and
function
with
. Then
(
) ( )
( (
( )
)
( )
( ))
)
( )
( )
( ) for ( )
Therefore (
and the LHS is continuous
implies that ( ) is constant. Since
is connected, setting
implies ( )
for
)
( )
( ) and is continuous. Therefore ( )
( ) and so is
all
so (
smooth. Therefore
is smooth and so
is smooth.
Remark
This means that the Lie Algebra of is bijective with the set of continuous
homomorphisms of with . (Hence only one Lie Group structure). There is a Theorem due
to Mostow and others which says all continuous Lie Groups are smooth Lie Groups.
Theorem 8.5
If
is a continuous homomorphism between Lie Groups then
of Lie Groups.
Proof
It is enough to show that
If
then
( )
As
varies in
is smooth on some open set
will be an open set containing
(
(
then
( )))
( )(
(
( ) varies in =V and
containing
and if
( )))
is a homomorphism
then
(
( )
is smooth on
(
so on
)) ( )
is
( )
which is a composition of three smooth maps so smooth.
Thus
has an open covering by open sets on which
is smooth. Therefore
To find such a , we pick a basis
for . Consider
)
continuous 1-PSG of and hence smooth so (
Consider (
derivative (
the basis {
)⏞(
)
}. Hence
)(
)
(
(
is smooth.
) which is a
for some
.
) mapping from
. has
which is the identification of
using
has an invertible derivative at the origin so by the Inverse
)
(
) for some
Function Theorem, is a diffeomorphism of (
) is a canonical chart and
open set
containing . (
))
(
)
(
) (
) (
( (
Which is smooth in
hence
is smooth on .
End of Course
onto an
)