# Differentiable Manifolds/Product manifolds and Lie groups

## The product manifold edit

**Definition 10.1**:

Let be sets and and be functions. Then the **cartesian product of and **, denoted by , is defined to be the function

**Definition 10.2**:

Let be a manifold with atlas . The **product manifold** of is defined to be the product space with the atlas

**Lemma 10.3**:

Let be open. Then for each , there exists such that

**Proof**:

Let be arbitrary. We choose such that . Then we define . Then, by the triangle inequality we have for every :

Therefore, , and since was arbitrary:

**Lemma 10.4**:

Let be open sets. Then is open in .

**Proof**:

Due to the openness of and for each point in , we find such that and . If we define , we have for all :

and

and since the function is monotonely increasing on , it follows

and

and thus:

**Lemma 10.5**: Let be sets and let and be functions. If and , then

**Proof**: See exercise 2.

**Theorem 10.6**: The product manifold of a manifold of class , where with atlas really is a manifold; i. e. really is an atlas of class .

**Proof**:

1. We show that for all the set is open.

We have:

This set is open in due to lemma 10.4, since it is the cartesian product of two open sets.

2. We prove that for all , the function is a homeomorphism.

2.1. For bijectivity, see exercise 1.

2.2. We prove continuity.

Let be open. Due to the definition of the subspace topology,

Lemma 10.4 implies that we have

(this equation can be proven by showing ' ' and ' ') and thus follows with lemma 10.5, that:

, which is open as the union of open sets (since we had equipped with the product topology).

2.3. We prove continuity of the inverse.

Let be open.

3. We prove that

## Lie groups edit

**Definition 10.7**:

Let be a -dimensional manifold of class , which is simultaneously a group, i. e. we have a group operation regarding to which there exist an identity, which we in the following shall denote by , and inverses in , and which is associative. We call a ** -dimensional Lie group of class ** iff

- the function is differentiable of class , where has the product manifold atlas, AND
- the function is also of class .

**Definition 10.8**:

Let be a manifold, and let be a subset of which is also a manifold. The **inclusion of into ** is defined to be the function

**Definition 10.9**:

Let be a Lie group of class , and let be a subset of . We call a **Lie subgroup of ** iff

- is a Lie group of class, but not necessarily with the subspace topology induced by
- the inclusion of into is differentiable of class and its differential is injective at every point.

**Definitions 10.10**:

Let be a Lie group with group operation , and let . The **left multiplication function with respect to **, denoted by , is defined to be the function

The **right multiplication function with respect to **, denoted by , is defined to be the function

**Theorem 10.11**:
Let be a -dimensional Lie group of class . Then for each the respective left multiplication function and the respective right multiplication are diffeomorphisms of class from to itself.

**Proof**:

We only prove the claim for the left multiplication. The proof for the right multiplication goes the same way.

In this proof, the group operation of is denoted by .

1. We show that is differentiable of class .

Let be arbitrary. Since is a Lie group, the function

is differentiable of class , where is equipped with the product manifold structure.

Let now and be two arbitrary elements in the atlas of . We choose in the atlas of such that . As is differentiable of class , the function

is contained in . Therefore, also the function

is contained in ; the partial derivatives exist and are equal to the partial derivatives of the last variables of the function

. But we have for all :

and therefore the function

is contained in , which means the definition of differentiability of class is fulfilled.

2. We show that is bijective.

We do so by noticing that an inverse function of is given by : For arbitrary , we have:

and

3. We note that the inverse function is differentiable of class :

We use 1. with ; also is an element of , and 1. proved that the left multiplication function is differentiable for every element of , including .

## Left invariant vector fields edit

**Definition 10.12**:

Let be a Lie group. We call a vector field **left invariant** iff

The set of all left invariant vector fields of a lie group is denoted by the corresponding lower case fraktur letter; for example, the set of all left invariant vector fields of a Lie group one would denote by , and the set of all left invariant vector fields of a Lie group would be denoted by .

Let us repeat the definition of a Lie subalgebra:

**Definition 6.2**:

Let with be a Lie algebra. A subset of which is a Lie algebra with the restriction of on that subset is called a **Lie subalgebra**.

**Theorem 10.13**:

Let be a Lie group. Then , together with pointwise addition and the vector field Lie bracket, is a Lie subalgebra of with pointwise addition and the vector field Lie bracket.

**Proof**:

Let . It suffices to show that , because then is a Lie algebra with the restriction of the vector field Lie bracket.

Indeed, we have for all and :

, where by the function

is meant and by the function

is meant (as both are equal to , both are differentiable of class ).

**Theorem 10.14**:

Let be a Lie group of class . Then there exists a vector space isomorphism between and .

**Proof**:

We choose the function

We will now show that this function is the desired isomorphism.

1. We prove linearity: Let and . We have:

2. We prove bijectivity.

2.1. We prove injectivity.

Let (i. e. ) for . Since are left invariant, it follows for all and , that

2.2. We prove surjectivity.

Let be arbitrary. We define

Due to theorem 2.19, this is a vector field. It is also left invariant because for all and , we have:

Further, we have for all :

3. We note that the inverse of is linear since the inverse of a linear bijective function is always linear.

The next theorem shows that in a Lie group, all left invariant vector fields are complete.

**Lemma 10.15**:

Let be a Lie group, let and let be the flow of . Then for all and all in the domain of , we have:

**Proof**:

Let be arbitrary, and let <mat>I_h</math> be the unique largest interval such that and there exists a unique integral curve such that

**Theorem 10.16**:

Let be a Lie group and let . Then is complete.

**Proof**:

Let be arbitrary and let be an integral curve at .

## The exponential function edit

**Definition 10.17**:

## The adjoint function edit

For the next definition, we recall that the automorphism group of a group was given by the set of group isomorphisms from the group to itself with composition as the group operation. Indeed, this is a group (see exercise 3).

We further recall that for a group , the automorphism group is denoted by .

**Definition 10.18**:

Let be a Lie group, whose group operation we shall denote by . If for all we define the function by , the **adjoint function of **, denoted by is given by

**Theorem 10.19**:

Let be a manifold of class , where . For each , is of class .

**Proof**:

We have:

Therefore, the claim follows from theorems 2.29 and 10.11.

**Theorem 10.20**:

Let be a Lie group and . Then

## Exercises edit

- Let be sets and be two bijective functions. Prove that is bijective.
- Prove lemma 10.5.
- Let be a group and be the set of group isomorphisms from to . Prove that together with the composition as operation is a group.