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 .
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 .
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.