Differentiable Manifolds/Bases of tangent and cotangent spaces and the differentials

Differentiable Manifolds
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In this section we shall

  • give one base for the tangent and cotangent space for each chart at a point of a manifold,
  • show how to convert representations in one base into another,
  • define the differentials of functions from a manifold to the real line, from an interval to a manifold and from a manifold to another manifold,
  • and prove the chain, product and quotient rules for those differentials.

Some bases of the tangent space edit

Definition 2.1:

Let   be a  -dimensional manifold of class   with   and atlas  , let   and let  . We define for every   and  ,  :

 

In the following, we will show that these functionals are a basis of the tangent space.

Theorem 2.2: Let   be a  -dimensional manifold of class   with   and atlas  , let   and let  . For all  :

 

i. e. the function   is contained in the tangent space  .

Proof:

Let  .

1. We show linearity.

 

From the second to the third line, we used the linearity of the derivative.

2. We show the product rule.

 

From the second to the third line, we used the product rule of the derivative.

3. It follows from the definition of  , that   if   is not defined at  . 

Lemma 2.3: Let   be a  -dimensional manifold of class   with atlas  , and let  . If we write  , then we have for each  , that  .

Proof:

Let  . Since   is an atlas,   and   are compatible. From this follows that the function

 

is of class  . But if we denote by   the function

 

, which is also called the projection to the  -th component, then we have:

 

It is not difficult to show that   is contained in  , and therefore the function

 

is contained in   as a composition of  -times continuously differentiable functions (or continuous functions if  ). 

Lemma 2.4: Let   be a  -dimensional manifold of class   with   and atlas  , let   and let  . If we write   we have:

 

Note that due to lemma 2.3,   for all  , which is why the above expression makes sense.

Proof:

We have:

 

Further,

 

and

 

Inserting this in the above limit gives the lemma. 

Theorem 2.5: Let   be a  -dimensional manifold of class   with   and atlas  , let   and let  . The tangent vectors

 

are linearly independent.

Proof:

We write again  .

Let  . Then we have for all  :

  

Lemma 2.6:

Let   be a manifold with atlas  ,  ,   be open, let   and   for a  ; i. e.   is a constant function. Then   and  .

Proof:

1. We show  .

By assumption,   is open. This means the first part of the definition of a   is fulfilled.

Further, for each   and  , we have:

 

This is contained in  .

2. We show that  .

We define  . Using the two rules linearity and product rule for tangent vectors, we obtain:

 

Substracting  , we obtain  . 

Theorem 2.7:

Let   be a  -dimensional manifold of class   with atlas  , let   and let  . For every   and every  , we have

 

Proof:

Let   be open, and let   be contained in  .

Case 1:  .

In this case,   and  , since   is not defined at   and both   and   are tangent vectors. From this follows the formula.

Case 2:  .

In this case, we obtain that the set   is open in   as follows: Since   is a homeomorphism by definition of charts, the set   is open in  . By definition of the subspace topology, we have   for a   open in  . But   is open in   as the intersection of two open sets; recall that   was required to be open in the definition of a chart.

Furthermore, from   and   it follows that  , and therefore  . Since   is open, we find an   such that the open ball   is contained in  . We define  . Since   is bijective,  , and since   is a homeomorphism, in particular continuous,   is open in   with respect to the subspace topology of  . From this also follows   open in  , because if   is open in  , then by definition of the subspace topology it is of the form   for an open set  , and hence it is open as the intersection of two open sets.

We have that  , is contained in  :   is an open subset of  , and if  , then

 ,

(check this by direct calculation!), which is contained in   as the restriction of an arbitrarily often continuously differentiable function.

We now define the function  ,  , and further for each  , we define

 

From the fundamental theorem of calculus, the multi-dimensional chain rule and the linearity of the integral follows for each  , that

 

If one sets   for  , one obtains, inserting the definition of  :

 

Now we define the functions

 

These are contained in   since they are defined on   which is open, and further, if  , then

 

, which is arbitrarily often differentiable by the Leibniz integral rule as the integral of a composition of arbitrarily often differentiable functions on a compact set.

Further, again denoting  , the functions  ,   are contained in   due to lemma 2.3.

Since  ,   is defined. We apply the rules (linearity and product rule) for tangent vectors and lemma 2.6 (we are allowed to do so because all the relevant functions are contained in  ), and obtain:

 

, since due to our notation it's clear that  .

But

 

Thus we have successfully shown

 

But due to the definition of subtraction on  , due to lemma 2.6, and due to the fact that the constant zero function is a constant function:

 

Due to linearity of   follows  , i. e.  . Now, inserting in the above equation gives the theorem. 

Together with theorem 2.5, this theorem shows that

 

is a basis of  , because a basis is a linearly independent generating set. And since the dimension of a vector space was defined to be the number of elements in a basis, this implies that the dimension of   is equal to  .

Some bases of the cotangent space edit

Definition 2.8:

Let   be a  -dimensional manifold of class   and atlas  , let   and let  . We write  . Then we define for  :

 

Note that   is well-defined because of lemma 2.3.

Theorem 2.9: Let   be a  -dimensional manifold of class   and atlas  , let   and let  . For all  ,   is contained in  .

Proof:

By definition,   maps from   to  . Thus, linearity is the only thing left to show. Indeed, for   and  , we have, since addition and scalar multiplication in   are defined pointwise:

  

Lemma 2.10: Let   be a  -dimensional manifold of class   and atlas  , let   and let  . For  , the following equation holds:

 

Proof:

We have:

  

Theorem 2.11: Let   be a  -dimensional manifold of class   and atlas  , let   and let  . The cotangent vectors   are linearly independent.

Proof:

Let  , where by   we mean the zero of  . Then we have for all  :

  

Theorem 2.12:

Let   be a  -dimensional manifold of class   and atlas  , let   and let  . If  , then for all  :

 

Proof:

Let   and  . Due to theorem 2.7, we have

 

Therefore, and due to the linearity of   (because   was the space of linear functions to  ):

 

Since   was arbitrary, the theorem is proven. 

From theorems 2.11 and 2.12 follows, as in the last subsection, that

 

is a basis for  , and that the dimension of   is equal to  , like the dimension of  .

Expressing elements of the tangent and cotangent spaces in different bases edit

If   is a manifold,   and   are two charts in  's atlas such that   and  . Then follows from the last two subsections, that

  •   and   are bases for  , and
  •   and   are bases for  .

One could now ask the questions:

If we have an element   in   given by  , then how can we represent   as linear combination of the basis  ?

Or if we have an element   in   given by  , then how can we represent   as linear combination of the basis  ?

The following two theorems answer these questions:

Theorem 2.13:

Let   be a manifold,   and   are two charts in  's atlas such that   and  . If   is given by  , then we have:

 

Proof:

Due to theorem 2.7, we have for  :

 

From this follows:

  

Theorem 2.14:

Let   be a manifold,   and   are two charts in  's atlas such that   and  . If   is given by  , then we have:

 

Proof:

Due to theorem 2.12, we have for  :

 

Thus we obtain:

  

The pullback and the differentials edit

In this subsection, we will define the pullback and the differential. For the differential, we need three definitions, one for each of the following types of functions:

  • functions from a manifold to another manifold
  • functions from a manifold to  
  • functions from an interval   to a manifold (i. e. curves)

For the first of these, the differential of functions from a manifold to another manifold, we need to define what the pullback is:

Definition 2.15:

Let   be two manifolds of class   and   be differentiable of class  . We define the pullback with respect to   of  , where   as

 ,

where   is the open set on which   is defined.

Lemma 2.16: Let   be a  -dimensional and   be a  -dimensional manifold, let   and let   be differentiable of class  . Then   is continuous.

Proof:

We show that for an arbitrary  ,   is continuous on an open neighbourhood of  . There is a theorem in topology which states that from this follows continuity.

We choose   in the atlas of   such that  , and   in the atlas of   such that  . Due to the differentiability of  , the function

 

is contained in  , and therefore continuous. But   and   are charts and therefore homeomorphisms, and thus the function

 

is continuous as the composition of continuous functions. 

Lemma 2.17: Let   be two manifolds, let   be differentiable of class  , and let   be defined on the open set  . In this case, the function   is contained in  ; i. e. the pullback with respect to   really maps to  .

Proof:

Since   is continuous due to lemma 2.16,   is open in  . Thus   is defined on an open set.

Let   be an arbitrary element of the atlas of   and let   be arbitrary. We choose   in the atlas of   such that  . The function

 

is  -times continuously differentiable (or continuous if  ) at   as the composition of two   times continuously differentiable (or continuous if  ) functions. Thus, the function

 

is  -times continuously differentiable (or continuous if  ) at every point, and therefore contained in  . 

Definition 2.18:

Let   be two manifolds of class  , let   be differentiable of class   and let  . The differential of   at   shall be defined as the function

 

Theorem 2.19:

Let   be two manifolds of class  , let   be differentiable of class   and let  . We have  ; i. e. the differential of   at   really maps to  .

Proof:

Let   be open,   and   be arbitrary. In the proof of the following, we will use that for all open subsets  ,   (which follows from the linearity of  ).

1. We prove linearity.

 

2. We prove the product rule.

  

Definition 2.20:

Let   be a manifold of class  , let   and let  . The differential of  , denoted by  , is defined as the function

 

Definition 2.21:

Let   be a manifold of class  ,  , let   be an interval, let   and let   be a differentiable curve of class  . The differential of   at   shall be defined as the function

 

Theorem 2.22: Let   be a manifold of class  ,  , let   be an interval, let   and let   be a differentiable curve of class  . Then   is contained in   for every   and   is a tangent vector of   at  .

Proof:

1. We show  

Let   be arbitrary, and let   be the set where   is defined (  is open by the definition of   functions. We choose   in the atlas of   such that  . Then the function

 

is contained in   as the composition of two   times continuously differentiable (or continuous if  ) functions.

Thus,   is   times continuously differentiable (or continuous if  ) at every point, and hence   times continuously differentiable (or continuous if  ).

2. We show that   in three steps:

Let   and  .

2.1 We show linearity.

We have:

 

2.2 We prove the product rule.

 

2.3 It follows from the definition of   that   is equal to zero if   is not defined at  . 

Linearity of the differential for Ck(M), product, quotient and chain rules edit

In this subsection, we will first prove linearity and product rule for functions from a manifold to  .

Theorem 2.23:

Let   be a manifold,  ,   and  . Then   and

 

Proof:

1. We show that  .

Let   be the (open as intersection of two open sets) set on which   is defined, and let   be contained in the atlas of  . The function

 

is contained in   as the linear combination of two   functions.

2. We show that  .

For all   and  , we have:

  

Remark 2.24: This also shows that for all  ,  .

Theorem 2.25:

Let   be a manifold,   and  . Then   and

 

Proof:

1. We show that  .

Let   be the (open as intersection of two open sets) set on which   is defined, and let   be contained in the atlas of  . The function

 

is contained in   as the product of two   functions.

2. We show that  .

For all   and  , we have:

  

Theorem 2.26:

Let   be a  -dimensional manifold of class   and let   such that   is zero at no point. Then   and

 

Proof:

1. We show that  :

Let   be the (open as the intersection of two open set) set on which   is defined, and let   be in the atlas of   such that  . The function

 

is contained in   as the quotient of two   from which the function in the denominator vanishes nowhere.

2. We show that  :

Choosing   as the constant one function, we obtain from 1. that the function   is in  . Hence follows from the product rule:

 

which, through equivalent transformations, can be transformed to

 

From this and from the product rule we obtain:

  

Theorem 2.27:

Let   be manifolds of class  , let   be differentiable of class   and let   be differentiable of class  . Then   is differentiable of class   and for all   we have the equation:

 

Proof:

1. We already know that   is differentiable of class  ; this is what lemma 2.17 says.

2. We prove that  .

Let  . Then we have: