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

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Differentiable Manifolds
 ← What is a manifold? Bases of tangent and cotangent spaces and the differentials Maximal atlases, second-countable spaces and partitions of unity → 

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 | edit source]

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 | edit source]

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 | edit source]

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 | edit source]

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 | edit source]

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:

Now, let's go on to proving the chain rule for functions from manifolds to manifolds. But to do so, we first need another theorem about the pullback.

Theorem 2.28: Let be three manifolds, and let and be two functions differentiable of class . Then

Proof: Let . Then we have:

Theorem 2.29:

Let be manifolds of class , and let and be two functions which are differentiable of class . Then is differentiable of class and

Proof:

1. We prove that is differentiable of class .

Let be contained in the atlas of and let be contained in the atlas of such that , and let be arbitrary. We choose in the atlas of such that .

We have ; indeed, due to the choice of and because . Further, we choose . Then the function

is contained in as the composition of two functions.

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

2. We prove that .

For all and , we have:

Theorem 2.30:

Let be a manifold of class , (!) an open interval, a curve in which is differentiable of class and . Then and

Proof:

1. Among another thing, theorem 2.22 states that is contained in .

2. We show that :

Intuition behind the tangent space[edit | edit source]

In this section, we want to prove that what we defined as the tangent space is isomorphic to a space whose elements are in analogy to tangent vectors to, say, tangent vectors of a function .

We start by proving the following lemma from linear algebra:

Lemma 2.33:

Let be vector spaces and let , and be linear functions such that

, and .

Then , and are all vector space isomorphisms.

Proof:

We only prove that is a vector space isomorphism; that and are also vector space isomorphisms will follow in exactly the same way.

From and follows that is the inverse function of .

Definition 2.34:

Let be a manifold of class , , and let . Then the space of velocities of at is defined to be the vector space consisting of the equivalence classes of the equivalence relation of the set of curves

.

Sources[edit | edit source]

  • Torres del Castillo, Gerardo (2012). Differentiable Manifolds. Boston: Birkhäuser. ISBN 978-0-8176-8271-2.
Differentiable Manifolds
 ← What is a manifold? Bases of tangent and cotangent spaces and the differentials Maximal atlases, second-countable spaces and partitions of unity →