Differentiable Manifolds/Product manifolds and Lie groups

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The product manifold

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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

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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

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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

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Definition 10.17:

The adjoint function

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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

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

Sources

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