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

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Differentiable Manifolds
 β† What is a manifold? Bases of tangent and cotangent spaces and the differentials Maximal atlases, second-countable spaces and partitions of unity β†’ 

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

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Definition 2.1:

Let M be a d-dimensional manifold of class π’žn with nβ‰₯1 and atlas {(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, let (O,Ο•)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯} and let p∈O. We define for every j∈{1,…,d} and Ο†βˆˆπ’žn(M), Ο†:O→ℝ:

(βˆ‚βˆ‚Ο•j)p:π’žn(M)→ℝ,(βˆ‚βˆ‚Ο•j)p(Ο†)={(βˆ‚xj(Ο†βˆ˜Ο•βˆ’1))(Ο•(p))p∈O0pβˆ‰O

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

Theorem 2.2: Let M be a d-dimensional manifold of class π’žn with nβ‰₯1 and atlas {(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, let (O,Ο•)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯} and let p∈O. For all j∈{1,…,d}:

(βˆ‚βˆ‚Ο•j)p∈TpM

i. e. the function (βˆ‚βˆ‚Ο•j)p:π’žn(M)→ℝ is contained in the tangent space TpM.

Proof:

Let Ο†,Ο‘βˆˆπ’žn(M).

1. We show linearity.

(βˆ‚βˆ‚Ο•j)p(Ο†+cΟ‘)=(βˆ‚xj((Ο†+cΟ‘)βˆ˜Ο•βˆ’1))(Ο•(p))=(βˆ‚xj(Ο†βˆ˜Ο•βˆ’1+cΟ‘βˆ˜Ο•βˆ’1))(Ο•(p))=(βˆ‚xj(Ο†βˆ˜Ο•βˆ’1)+cβˆ‚xj(Ο‘βˆ˜Ο•βˆ’1)))(Ο•(p))=(βˆ‚βˆ‚Ο•j)p(Ο†)+c(βˆ‚βˆ‚Ο•j)p(Ο‘)

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

2. We show the product rule.

(βˆ‚βˆ‚Ο•j)p(φϑ)=(βˆ‚xj((φϑ)βˆ˜Ο•βˆ’1))(Ο•(p))=(βˆ‚xj((Ο†βˆ˜Ο•βˆ’1)(Ο‘βˆ˜Ο•βˆ’1)))(Ο•(p))=(Ο†βˆ˜Ο•βˆ’1)(Ο•(p))(βˆ‚xj(Ο‘βˆ˜Ο•βˆ’1))(Ο•(p))+(Ο‘βˆ˜Ο•βˆ’1)(Ο•(p))(βˆ‚xj(Ο†βˆ˜Ο•βˆ’1))(Ο•(p))=Ο†(p)(βˆ‚βˆ‚Ο•j)p(Ο‘)+Ο‘(p)(βˆ‚βˆ‚Ο•j)p(Ο†)

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

3. It follows from the definition of (βˆ‚βˆ‚Ο•j)p, that (βˆ‚βˆ‚Ο•j)p(Ο†)=0 if Ο† is not defined at p.β—»

Lemma 2.3: Let M be a d-dimensional manifold of class π’žn with atlas {(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, and let (O,Ο•)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯}. If we write Ο•=(Ο•1,…,Ο•d), then we have for each k∈{1,…,d}, that Ο•kβˆˆπ’žn(M).

Proof:

Let (U,ΞΈ)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯}. Since {(OΟ…,ϕυ)|Ο…βˆˆΞ₯} is an atlas, ΞΈ and Ο• are compatible. From this follows that the function

Ο•|U∩O∘θ|O∩Uβˆ’1

is of class π’žn. But if we denote by Ο€k the function

Ο€k:ℝd→ℝ,Ο€k(x1,…,xd)=xk

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

Ο•k|U∩O∘θ|O∩Uβˆ’1=Ο€kβˆ˜Ο•|U∩O∘θ|O∩Uβˆ’1

It is not difficult to show that Ο€k is contained in π’žβˆž(ℝd,ℝ), and therefore the function

Ο€kβˆ˜Ο•|U∩O∘θ|O∩Uβˆ’1

is contained in π’žn(ℝd,ℝ) as a composition of n-times continuously differentiable functions (or continuous functions if n=0).β—»

Lemma 2.4: Let M be a d-dimensional manifold of class π’žn with nβ‰₯1 and atlas {(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, let (O,Ο•)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯} and let p∈O. If we write Ο•=(Ο•1,…,Ο•d) we have:

(βˆ‚βˆ‚Ο•j)p(Ο•k)={1j=k0jβ‰ k

Note that due to lemma 2.3, Ο•kβˆˆπ’žn(M) for all k∈{1,…,d}, which is why the above expression makes sense.

Proof:

We have:

(βˆ‚βˆ‚Ο•j)p(Ο•k)=(βˆ‚xj(Ο•kβˆ˜Ο•βˆ’1))(Ο•(p))=limyβ†’0(Ο•kβˆ˜Ο•βˆ’1)(x1,…,xjβˆ’1,xj+y,xj+1,…,xd)βˆ’(Ο•kβˆ˜Ο•βˆ’1)(x1,…,xd)y

Further,

(Ο•kβˆ˜Ο•βˆ’1)(x1,…,xd)=xk

and

(Ο•kβˆ˜Ο•βˆ’1)(x1,…,xjβˆ’1,xj+y,xj+1,…,xd)={xk+yk=jxkkβ‰ j

Inserting this in the above limit gives the lemma.β—»

Theorem 2.5: Let M be a d-dimensional manifold of class π’žn with nβ‰₯1 and atlas {(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, let (O,Ο•)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯} and let p∈O. The tangent vectors

(βˆ‚βˆ‚Ο•j)p∈TpM,j∈{1,…,d}

are linearly independent.

Proof:

We write again Ο•=(Ο•1,…,Ο•d).

Let βˆ‘j=1daj(βˆ‚βˆ‚Ο•j)p=0p. Then we have for all k∈{1,…,d}:

0=0p(Ο•k)=βˆ‘j=1daj(βˆ‚βˆ‚Ο•j)p(Ο•k)=akβ—»

Lemma 2.6:

Let M be a manifold with atlas {(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, p∈M, VβŠ†M be open, let 𝐕p∈TpM and Ο†:V→ℝ,Ο†(q)=c for a cβˆˆβ„; i. e. Ο† is a constant function. Then Ο†βˆˆπ’žβˆž(M) and 𝐕p(Ο†)=0.

Proof:

1. We show Ο†βˆˆπ’žβˆž(M).

By assumption, VβŠ†M is open. This means the first part of the definition of a π’žβˆž(M) is fulfilled.

Further, for each (U,ΞΈ)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯} and x∈θ(V∩U), we have:

Ο†βˆ˜ΞΈ|U∩V(x)=c

This is contained in π’žβˆž(ℝd,ℝ).

2. We show that 𝐕p(Ο†)=0.

We define Ο‘:V→ℝ,Ο‘(q)=1. Using the two rules linearity and product rule for tangent vectors, we obtain:

𝐕p(Ο†)=𝐕p(ϑφ)=1𝐕p(Ο†)+Ο†(p)𝐕p(Ο‘)=𝐕p(Ο†)+𝐕p(ϑφ(p))=𝐕p(Ο†)+𝐕p(Ο†)

Subtracting 𝐕p(Ο†), we obtain 𝐕p(Ο†)=0.β—»

Theorem 2.7:

Let M be a d-dimensional manifold of class π’žβˆž with atlas {(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, let (O,Ο•)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯} and let p∈O. For every 𝐕p∈TpM and every Ο†βˆˆπ’žn(M), we have

𝐕p(Ο†)=βˆ‘j=1d𝐕p(Ο•j)(βˆ‚βˆ‚Ο•j)p(Ο†)

Proof:

Let UβŠ†M be open, and let Ο†:U→ℝ be contained in π’žn(M).

Case 1: pβˆ‰U.

In this case, 𝐕p(Ο†)=0 and (βˆ‚βˆ‚Ο•j)p(Ο†)=0, since Ο† is not defined at p and both 𝐕p and (βˆ‚βˆ‚Ο•j)p are tangent vectors. From this follows the formula.

Case 2: p∈U.

In this case, we obtain that the set Ο•(U∩O) is open in ℝd as follows: Since Ο•:Oβ†’Ο•(O) is a homeomorphism by definition of charts, the set Ο•(U∩O) is open in Ο•(O). By definition of the subspace topology, we have Ο•(U∩O)=Vβˆ©Ο•(O) for a V open in ℝd. But Vβˆ©Ο•(O) is open in ℝd as the intersection of two open sets; recall that Ο•(O) was required to be open in the definition of a chart.

Furthermore, from p∈U and p∈O it follows that p∈U∩O, and therefore Ο•(p)βˆˆΟ•(O∩U). Since Ο•(O∩U) is open, we find an Ο΅>0 such that the open ball BΟ΅(Ο•(p)) is contained in Ο•(O∩U). We define W:=Ο•βˆ’1(BΟ΅(Ο•(p))). Since Ο• is bijective, WβŠ†U∩O, and since Ο• is a homeomorphism, in particular continuous, W is open in O with respect to the subspace topology of O. From this also follows O open in M, because if W is open in O, then by definition of the subspace topology it is of the form V∩O for an open set VβŠ†M, and hence it is open as the intersection of two open sets.

We have that Ο†|W:W→ℝ, is contained in π’žβˆž(M): W is an open subset of M, and if (V,ΞΈ)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, then

Ο†|W∩V∘θ|W∩Vβˆ’1=(Ο†|U∩V∘θ|U∩Vβˆ’1)|ΞΈ(W∩V),

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

We now define the function F:BΟ΅(Ο•(p))→ℝ, F(x)=(Ο†βˆ˜Ο•βˆ’1)(x), and further for each x∈BΟ΅(Ο•(p)), we define

ΞΌx(ΞΎ):=F(ΞΎx+(1βˆ’ΞΎ)Ο•(p))

From the fundamental theorem of calculus, the multi-dimensional chain rule and the linearity of the integral follows for each x∈BΟ΅(Ο•(p)), that

F(x)=ΞΌx(1)=ΞΌx(0)+∫01ΞΌx′(ΞΎ)dΞΎ=F(Ο•(p))+βˆ‘j=1d(xjβˆ’Ο•(p)j)∫01βˆ‚xjF(ΞΎΟ•(p)+(1βˆ’ΞΎ)x)dΞΎ

If one sets x=Ο•(q) for q∈W, one obtains, inserting the definition of F:

Ο†(q)=Ο†(p)+βˆ‘j=1d(Ο•(q)jβˆ’Ο•(p)j)∫01βˆ‚xj(Ο†βˆ˜Ο•βˆ’1)(ΞΎΟ•(p)+(1βˆ’ΞΎ)Ο•(q))dΞΎ

Now we define the functions

fj:W→ℝ,fj(q):=∫01βˆ‚xj(Ο†βˆ˜Ο•βˆ’1)(ΞΎΟ•(p)+(1βˆ’ΞΎ)Ο•(q))dΞΎ

These are contained in π’žβˆž(M) since they are defined on W which is open, and further, if (V,ΞΈ)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, then

fj|V∩W∘θ|V∩Wβˆ’1=∫01βˆ‚xj(Ο†βˆ˜Ο•βˆ’1)(ΞΎΟ•|V∩W∘θ|V∩Wβˆ’1+(1βˆ’ΞΎ)Ο•|V∩W∘θ|V∩Wβˆ’1)dΞΎ

, 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 Ο•=(Ο•1,…,Ο•d), the functions Ο•k, k∈{1,…,d} are contained in π’žβˆž(M) due to lemma 2.3.

Since Ο†|Wβˆˆπ’žβˆž(M), 𝐕p(Ο†|W) 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 π’žβˆž(M)), and obtain:

𝐕p(Ο†|W)=𝐕p(Ο†(p)+βˆ‘j=1d(Ο•jβˆ’Ο•(p)j)fj)=βˆ‘j=1d(Ο•j(p)𝐕p(fj)+fj(p)𝐕p(Ο•j)βˆ’Ο•(p)j𝐕p(fj))=βˆ‘j=1dfj(p)𝐕p(Ο•j)

, since due to our notation it's clear that Ο•j(p)=Ο•(p)j.

But

fj(p)=∫01βˆ‚xj(Ο†βˆ˜Ο•βˆ’1)(ΞΎΟ•(p)+(1βˆ’ΞΎ)Ο•(p))dΞΎ=∫01βˆ‚xj(Ο†βˆ˜Ο•βˆ’1)(Ο•(p))dΞΎ=βˆ‚xj(Ο†βˆ˜Ο•βˆ’1)(Ο•(p))=(βˆ‚βˆ‚Ο•j)p(Ο†)

Thus we have successfully shown

𝐕p(Ο†|W)=βˆ‘j=1d𝐕p(Ο•j)(βˆ‚βˆ‚Ο•j)p(Ο†)

But due to the definition of subtraction on π’žβˆž(M), due to lemma 2.6, and due to the fact that the constant zero function is a constant function:

𝐕p(Ο†|Wβˆ’Ο†)=𝐕p(0)=0

Due to linearity of 𝐕p follows 0=𝐕p(Ο†|W)βˆ’π•p(Ο†), i. e. 𝐕p(Ο†|W)=𝐕p(Ο†). Now, inserting in the above equation gives the theorem.β—»

Together with theorem 2.5, this theorem shows that

{(βˆ‚βˆ‚Ο•j)p|j∈{1,…,d}}

is a basis of TpM, 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 TpM is equal to d.

Some bases of the cotangent space

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Definition 2.8:

Let M be a d-dimensional manifold of class π’žβˆž and atlas {(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, let (O,Ο•)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯} and let p∈O. We write Ο•=(Ο•1,…,Ο•d). Then we define for j∈{1,…,d}:

(dΟ•j)p:TpM→ℝ,(dΟ•j)p(𝐕p):=𝐕p(Ο•j)

Note that (dΟ•j)p is well-defined because of lemma 2.3.

Theorem 2.9: Let M be a d-dimensional manifold of class π’žβˆž and atlas {(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, let (O,Ο•)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯} and let p∈O. For all j∈{1,…,d}, (dΟ•j)p is contained in TpMβˆ—.

Proof:

By definition, dΟ•k maps from TpM to ℝ. Thus, linearity is the only thing left to show. Indeed, for 𝐕p,𝐖p∈TpM and bβˆˆβ„, we have, since addition and scalar multiplication in TpM are defined pointwise:

(dΟ•j)p(𝐕p+b𝐖p)=(𝐕p+b𝐖p)(Ο•k)=𝐕p(Ο•k)+b𝐖p(Ο•k)=(dΟ•j)p(𝐕p)+b(dΟ•j)p(𝐖p)β—»

Lemma 2.10: Let M be a d-dimensional manifold of class π’žβˆž and atlas {(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, let (O,Ο•)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯} and let p∈O. For j,k∈{1,…,d}, the following equation holds:

(dΟ•j)p((βˆ‚βˆ‚Ο•k)p)={1k=j0kβ‰ j

Proof:

We have:

(dΟ•j)p((βˆ‚βˆ‚Ο•k)p)=(βˆ‚βˆ‚Ο•k)p(Ο•j)=lemma 2.4{1k=j0kβ‰ jβ—»

Theorem 2.11: Let M be a d-dimensional manifold of class π’žβˆž and atlas {(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, let (O,Ο•)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯} and let p∈O. The cotangent vectors (dΟ•j)p,j∈{1,…,d} are linearly independent.

Proof:

Let 0=βˆ‘j=1daj(dΟ•j)p, where by 0 we mean the zero of TpMβˆ—. Then we have for all k∈{1,…,d}:

0=βˆ‘j=1daj(dΟ•j)p((βˆ‚βˆ‚Ο•k)p)=lemma 2.10akβ—»

Theorem 2.12:

Let M be a d-dimensional manifold of class π’žβˆž and atlas {(OΟ…,ϕυ)|Ο…βˆˆΞ₯}, let (O,Ο•)∈{(OΟ…,ϕυ)|Ο…βˆˆΞ₯} and let p∈O. If Ξ±p∈TpMβˆ—, then for all 𝐕p∈TpM:

Ξ±p(𝐕p)=βˆ‘j=1dΞ±p((βˆ‚βˆ‚Ο•j)p)(dΟ•j)p(𝐕p)

Proof:

Let Ξ±p∈TpMβˆ— and 𝐕p∈TpM. Due to theorem 2.7, we have

𝐕p=βˆ‘j=1d𝐕p(Ο•j)(βˆ‚βˆ‚Ο•j)p

Therefore, and due to the linearity of Ξ±p (because TpMβˆ— was the space of linear functions to ℝ):

Ξ±p(𝐕p)=βˆ‘j=1d𝐕p(Ο•j)Ξ±p((βˆ‚βˆ‚Ο•j)p)=βˆ‘j=1dΞ±p((βˆ‚βˆ‚Ο•j)p)(dΟ•j)p(𝐕p)

Since 𝐕p∈TpM was arbitrary, the theorem is proven.β—»

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

{(dΟ•j)p|j∈{1,…,d}}

is a basis for TpMβˆ—, and that the dimension of TpMβˆ— is equal to d, like the dimension of TpM.

Expressing elements of the tangent and cotangent spaces in different bases

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If M is a manifold, p∈M and (O,Ο•),(U,ΞΈ) are two charts in M's atlas such that p∈O and p∈U. Then follows from the last two subsections, that

  • {(βˆ‚βˆ‚Ο•j)p|j∈{1,…,d}} and {(βˆ‚βˆ‚ΞΈj)p|j∈{1,…,d}} are bases for TpM, and
  • {(dΟ•j)p|j∈{1,…,d}} and {(dΞΈj)p|j∈{1,…,d}} are bases for TpMβˆ—.

One could now ask the questions:

If we have an element 𝐕p in TpM given by 𝐕p=βˆ‘j=1daj(βˆ‚βˆ‚Ο•j)p, then how can we represent 𝐕p as linear combination of the basis {(βˆ‚βˆ‚ΞΈj)p|j∈{1,…,d}}?

Or if we have an element Ξ±p in TpMβˆ— given by Ξ±p=βˆ‘j=1daj(dΟ•j)p, then how can we represent Ξ±p as linear combination of the basis {(dΞΈj)p|j∈{1,…,d}}?

The following two theorems answer these questions:

Theorem 2.13:

Let M be a manifold, p∈M and (O,Ο•),(U,ΞΈ) are two charts in M's atlas such that p∈O and p∈U. If 𝐕p∈TpM is given by 𝐕p=βˆ‘j=1daj(βˆ‚βˆ‚Ο•j)p, then we have:

𝐕p=βˆ‘k=1dβˆ‘j=1daj(βˆ‚βˆ‚Ο•j)p(ΞΈk)(βˆ‚βˆ‚ΞΈk)p

Proof:

Due to theorem 2.7, we have for j∈{1,…,d}:

(βˆ‚βˆ‚Ο•j)p=βˆ‘k=1d(βˆ‚βˆ‚Ο•j)p(ΞΈk)(βˆ‚βˆ‚ΞΈk)p

From this follows:

𝐕p=βˆ‘j=1daj(βˆ‚βˆ‚Ο•j)p=βˆ‘j=1dajβˆ‘k=1d(βˆ‚βˆ‚Ο•j)p(ΞΈk)(βˆ‚βˆ‚ΞΈk)p=βˆ‘k=1dβˆ‘j=1daj(βˆ‚βˆ‚Ο•j)p(ΞΈk)(βˆ‚βˆ‚ΞΈk)pβ—»

Theorem 2.14:

Let M be a manifold, p∈M and (O,Ο•),(U,ΞΈ) are two charts in M's atlas such that p∈O and p∈U. If Ξ±p∈TpMβˆ— is given by Ξ±p=βˆ‘j=1daj(dΟ•j)p, then we have:

Ξ±p=βˆ‘k=1dβˆ‘j=1daj(dΟ•j)p((βˆ‚βˆ‚Ο•j)p)(dΞΈk)p

Proof:

Due to theorem 2.12, we have for j∈{1,…,d}:

(dΟ•j)p=βˆ‘k=1d(dΟ•j)p((βˆ‚βˆ‚Ο•k)p)(dΞΈk)p

Thus we obtain:

Ξ±p=βˆ‘j=1daj(dΟ•j)p=βˆ‘j=1dajβˆ‘k=1d(dΟ•j)p((βˆ‚βˆ‚Ο•k)p)(dΞΈk)p=βˆ‘k=1dβˆ‘j=1daj(dΟ•j)p((βˆ‚βˆ‚Ο•k)p)(dΞΈk)pβ—»

The pullback and the differentials

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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 ℝ
  • functions from an interval IβŠ†β„ 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 M,N be two manifolds of class π’žn and ψ:Mβ†’N be differentiable of class π’žk. We define the pullback with respect to ψ of Ο†, where Ο†βˆˆπ’žk(N) as

Οˆβˆ—:π’žk(N)β†’π’žk(M),Οˆβˆ—(Ο†):=Ο†βˆ˜Οˆ|Οˆβˆ’1(O),

where OβŠ†N is the open set on which Ο† is defined.

Lemma 2.16: Let M be a d-dimensional and N be a b-dimensional manifold, let kβˆˆβ„•0βˆͺ{∞} and let ψ:Mβ†’N be differentiable of class π’žk. Then ψ is continuous.

Proof:

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

We choose (O,Ο•) in the atlas of M such that p∈O, and (U,ΞΈ) in the atlas of N such that ψ(p)∈U. Due to the differentiability of ψ, the function

ΞΈβˆ˜Οˆβˆ˜Ο•|Ο•(Oβˆ©Οˆβˆ’1(U))βˆ’1

is contained in π’žk(ℝd,ℝb), and therefore continuous. But Ο• and ΞΈ are charts and therefore homeomorphisms, and thus the function

ψ|Oβˆ©Οˆβˆ’1(U):Oβˆ©Οˆβˆ’1(U)β†’N,ψ=ΞΈβˆ’1βˆ˜ΞΈβˆ˜Οˆβˆ˜Ο•|Oβˆ©Οˆβˆ’1(U)βˆ’1βˆ˜Ο•|Oβˆ©Οˆβˆ’1(U)

is continuous as the composition of continuous functions.β—»

Lemma 2.17: Let M,N be two manifolds, let ψ:Mβ†’N be differentiable of class π’žk, and let Ο†βˆˆπ’žk(N) be defined on the open set UβŠ†N. In this case, the function Ο†βˆ˜|Οˆβˆ’1(U) is contained in π’žk(M); i. e. the pullback with respect to ψ really maps to π’žk(M).

Proof:

Since ψ is continuous due to lemma 2.16, Οˆβˆ’1(U) is open in M. Thus Ο†βˆ˜Οˆ|Οˆβˆ’1(U) is defined on an open set.

Let (O,Ο•) be an arbitrary element of the atlas of M and let xβˆˆΟ•(O) be arbitrary. We choose (V,ΞΈ) in the atlas of N such that ψ(Ο•βˆ’1(x))∈V. The function

(Ο†βˆ˜Οˆ|Οˆβˆ’1(U)βˆ˜Ο•|Οˆβˆ’1(U)∩Oβˆ’1)|Ο•(Οˆβˆ’1(U∩V)∩O)=Ο†|ψ(Οˆβˆ’1(U∩V)∩O)∘θ|ψ(Οˆβˆ’1(U∩V)∩O)βˆ’1∘θ|ψ(Οˆβˆ’1(U∩V)∩O)∘ψ|Οˆβˆ’1(U∩V)∩Oβˆ˜Ο•|Ο•(Οˆβˆ’1(U∩V)∩O)βˆ’1

is k-times continuously differentiable (or continuous if k=0) at x as the composition of two k times continuously differentiable (or continuous if k=0) functions. Thus, the function

Ο†βˆ˜Οˆ|Οˆβˆ’1(U)βˆ˜Ο•|Οˆβˆ’1(U)∩Oβˆ’1

is k-times continuously differentiable (or continuous if k=0) at every point, and therefore contained in π’žk(ℝd,ℝ).β—»

Definition 2.18:

Let M,N be two manifolds of class π’žn, let ψ:Mβ†’N be differentiable of class π’žn and let p∈M. The differential of ψ at p shall be defined as the function

dψp:TpMβ†’Tψ(p)N,dψp(𝐕p):=𝐕pβˆ˜Οˆβˆ—

Theorem 2.19:

Let M,N be two manifolds of class π’žn, let ψ:Mβ†’N be differentiable of class π’žn and let p∈M. We have 𝐕pβˆ˜Οˆβˆ—βˆˆTpN; i. e. the differential of ψ at p really maps to TpN.

Proof:

Let O,UβŠ†M be open, Ο†:O→ℝ,Ο‘:Uβ†’β„βˆˆπ’žn(M) and cβˆˆβ„ be arbitrary. In the proof of the following, we will use that for all open subsets VβŠ†O, 𝐕p(Ο†|V)=𝐕p(Ο†) (which follows from the linearity of 𝐕p).

1. We prove linearity.

(𝐕pβˆ˜Οˆβˆ—)(Ο†+cΟ‘)=𝐕p(Οˆβˆ—(Ο†+cΟ‘))=𝐕p((Ο†+cΟ‘)∘ψ|Οˆβˆ’1(O∩U))=𝐕p(Ο†|O∩U∘ψ|Οˆβˆ’1(O∩U)+cΟ‘|O∩U∘ψ|Οˆβˆ’1(O∩U))=𝐕p(Ο†|O∩U∘ψ|Οˆβˆ’1(O∩U))+c𝐕p(Ο‘|O∩U∘ψ|Οˆβˆ’1(O∩U))=𝐕p(Οˆβˆ—(Ο†))+c𝐕p(Οˆβˆ—(Ο‘))=(𝐕pβˆ˜Οˆβˆ—)(Ο†)+c(𝐕pβˆ˜Οˆβˆ—)(Ο‘)

2. We prove the product rule.

(𝐕pβˆ˜Οˆβˆ—)(φϑ)=𝐕p(Οˆβˆ—(φϑ))=𝐕p((Ο†|O∩U∘ψ|Οˆβˆ’1(O∩U))(Ο‘|O∩U∘ψ|Οˆβˆ’1(O∩U)))=(Ο†|O∩U∘ψ|Οˆβˆ’1(O∩U))(p)𝐕p(Ο‘|O∩U∘ψ|Οˆβˆ’1(O∩U))+(Ο‘|O∩U∘ψ|Οˆβˆ’1(O∩U))(p)𝐕p(Ο†|O∩U∘ψ|Οˆβˆ’1(O∩U))=Ο†(ψ(p))𝐕p(Οˆβˆ—Ο‘)+Ο‘(ψ(p))𝐕p(Οˆβˆ—Ο†)=Ο†(ψ(p))(𝐕pβˆ˜Οˆβˆ—)(Ο‘)+Ο‘(ψ(p))(𝐕pβˆ˜Οˆβˆ—)(Ο†)β—»

Definition 2.20:

Let M be a manifold of class π’žn, let p∈M and let Ο†βˆˆπ’žn(M). The differential of Ο†, denoted by dΟ†p, is defined as the function

dΟ†p:TpM→ℝ,dΟ†p(𝐕p):=𝐕p(Ο†)

Definition 2.21:

Let M be a manifold of class π’žn, nβ‰₯1, let IβŠ†β„ be an interval, let y∈I and let Ξ³:Iβ†’M be a differentiable curve of class π’žn. The differential of Ξ³ at y shall be defined as the function

Ξ³'y:π’žn(M)→ℝ,Ξ³'y(Ο†):={(Ο†βˆ˜Ξ³)′(y)if Ο† is defined at Ξ³(y)0else

Theorem 2.22: Let M be a manifold of class π’žn, nβ‰₯1, let IβŠ†β„ be an interval, let y∈I and let Ξ³:Iβ†’M be a differentiable curve of class π’žn. Then Ο†βˆ˜Ξ³ is contained in π’žn(ℝ,ℝ) for every Ο†βˆˆπ’žn(M) and Ξ³'y is a tangent vector of M at Ξ³(y).

Proof:

1. We show βˆ€Ο†βˆˆπ’žn(M):βˆ˜Ξ³βˆˆπ’žn(ℝ,ℝ)

Let x∈I be arbitrary, and let U be the set where Ο† is defined (U is open by the definition of π’žn(M) functions. We choose (O,Ο•) in the atlas of M such that Ξ³(x)∈O. Then the function

(Ο†βˆ˜Ξ³)|Ξ³βˆ’1(O∩U)∩I=Ο†βˆ˜Ο•βˆ’1βˆ˜Ο•βˆ˜Ξ³|Ξ³βˆ’1(O∩U)∩I

is contained in π’žn(ℝ,ℝ) as the composition of two n times continuously differentiable (or continuous if n=0) functions.

Thus, Ο†βˆ˜Ξ³ is n times continuously differentiable (or continuous if n=0) at every point, and hence n times continuously differentiable (or continuous if n=0).

2. We show that γ'y∈Tγ(y)M in three steps:

Let Ο†,Ο‘βˆˆπ’žn(M) and cβˆˆβ„.

2.1 We show linearity.

We have:

Ξ³'y(Ο†+cΟ‘)=((Ο†+cΟ‘)∘γ)′(y)=(Ο†βˆ˜Ξ³+cΟ‘βˆ˜Ξ³)′(y)=(Ο†βˆ˜Ξ³)′(y)+c(Ο‘βˆ˜Ξ³)′(y)=Ξ³'y(Ο†)+cΞ³'y(Ο‘)

2.2 We prove the product rule.

Ξ³'y(φϑ)=((φϑ)∘γ)′(y)=((Ο†βˆ˜Ξ³)(Ο‘βˆ˜Ξ³))′(y)=(Ο†βˆ˜Ξ³)(y)(Ο‘βˆ˜Ξ³)′(y)+(Ο‘βˆ˜Ξ³)(y)(Ο†βˆ˜Ξ³)′(y)=Ο†(Ξ³(y))Ξ³'y(Ο‘)+Ο‘(Ξ³(y))Ξ³'y(Ο†)

2.3 It follows from the definition of Ξ³'y that Ξ³'y(Ο†) is equal to zero if Ο† is not defined at Ξ³(y).β—»

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

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

Theorem 2.23:

Let M be a manifold, kβˆˆβ„•0βˆͺ{∞}, Ο†,Ο‘βˆˆπ’žk(M) and cβˆˆβ„. Then Ο†+cΟ‘βˆˆπ’žk(M) and

d(φ+cϑ)=dφ+cdϑ

Proof:

1. We show that Ο†+cΟ‘βˆˆπ’žk(M).

Let U be the (open as intersection of two open sets) set on which Ο†+cΟ‘ is defined, and let (O,Ο•) be contained in the atlas of M. The function

(Ο†+cΟ‘)|O∩Uβˆ˜Ο•|O∩Uβˆ’1=Ο†|O∩Uβˆ˜Ο•|O∩Uβˆ’1+cΟ‘|O∩Uβˆ˜Ο•|O∩Uβˆ’1

is contained in π’žn(ℝd,ℝ) as the linear combination of two π’žn(ℝd,ℝ) functions.

2. We show that d(φ+cϑ)=dφ+cdϑ.

For all p∈M and 𝐕p∈TpM, we have:

d(Ο†+cΟ‘)p(𝐕p)=𝐕p(Ο†+cΟ‘)=𝐕p(Ο†)+c𝐕p(Ο‘)=dΟ†p(𝐕p)+cdΟ‘p(𝐕p)β—»

Remark 2.24: This also shows that for all Ο†βˆˆπ’žn(M), dΟ†p∈TpMβˆ—.

Theorem 2.25:

Let M be a manifold, kβˆˆβ„•0βˆͺ{∞} and Ο†,Ο‘βˆˆπ’žk(M). Then Ο†Ο‘βˆˆπ’žk(M) and

d(φϑ)=Ο†dΟ‘+Ο‘dΟ†

Proof:

1. We show that Ο†Ο‘βˆˆπ’žk(M).

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

(φϑ)|O∩Uβˆ˜Ο•|O∩Uβˆ’1=Ο†|O∩Uβˆ˜Ο•|O∩Uβˆ’1Ο‘|O∩Uβˆ˜Ο•|O∩Uβˆ’1

is contained in π’žn(ℝd,ℝ) as the product of two π’žn(ℝd,ℝ) functions.

2. We show that d(φϑ)=Ο†dΟ‘+Ο‘dΟ†.

For all p∈M and 𝐕p∈TpM, we have:

d(φϑ)p(𝐕p)=𝐕p(φϑ)=Ο†(p)𝐕p(Ο‘)+Ο‘(p)𝐕p(Ο†)=Ο†(p)dΟ‘p+Ο‘(p)dΟ†pβ—»

Theorem 2.26:

Let M be a d-dimensional manifold of class π’žn and let Ο†,Ο‘βˆˆπ’žn(M) such that Ο‘ is zero at no point. Then Ο†Ο‘βˆˆπ’žn(M) and

d(φϑ)=Ο‘dΟ†βˆ’Ο†dϑϑ2

Proof:

1. We show that Ο†Ο‘βˆˆπ’žn(M):

Let U be the (open as the intersection of two open set) set on which φϑ is defined, and let (O,Ο•) be in the atlas of M such that O∩Uβ‰ βˆ…οΈ€. The function

φϑ|O∩Uβˆ˜Ο•|O∩Uβˆ’1=Ο†|O∩Uβˆ˜Ο•|O∩Uβˆ’1Ο‘|O∩Uβˆ˜Ο•|O∩Uβˆ’1

is contained in π’žn(ℝd,ℝ) as the quotient of two π’žn(ℝd,ℝ) from which the function in the denominator vanishes nowhere.

2. We show that d(φϑ)=Ο‘dΟ†βˆ’Ο†dϑϑ2:

Choosing Ο† as the constant one function, we obtain from 1. that the function 1Ο‘ is in π’žn(M). Hence follows from the product rule:

0=d(Ο‘1Ο‘)=Ο‘d(1Ο‘)+1Ο‘dΟ‘

which, through equivalent transformations, can be transformed to

d(1Ο‘)=βˆ’dϑϑ2

From this and from the product rule we obtain:

d(Ο†1Ο‘)=1Ο‘dΟ†βˆ’Ο†dϑϑ2=Ο‘dΟ†βˆ’Ο†dϑϑ2β—»

Theorem 2.27:

Let M,N be manifolds of class π’žn, let ψ:Mβ†’N be differentiable of class π’žn and let Ο†:N→ℝ be differentiable of class π’žn. Then Ο†βˆ˜Οˆ=Οˆβˆ—Ο† is differentiable of class π’žn and for all p∈M we have the equation:

d(Οˆβˆ—Ο†)=dΟ†Οˆ(p)∘dψp

Proof:

1. We already know that Ο†βˆ˜Οˆ=Οˆβˆ—Ο† is differentiable of class π’žn; this is what lemma 2.17 says.

2. We prove that d(Οˆβˆ—Ο†)p=dΟ†Οˆ(p)∘dψp.

Let 𝐕p∈TpM. Then we have:

(dΟ†Οˆ(p)∘dψp)(𝐕p)=dΟ†Οˆ(p)(dψp(𝐕p))=dΟ†Οˆ(p)(𝐕p∘ψ)=(𝐕pβˆ˜Οˆβˆ—)(Ο†)=𝐕p(Οˆβˆ—(Ο†))=𝐕p(Ο†βˆ˜Οˆ)=d(Ο†βˆ˜Οˆ)p(𝐕p)β—»

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 M,N,K be three manifolds, and let ψ:Mβ†’N and Ο‡:Nβ†’K be two functions differentiable of class π’žk. Then

(Ο‡βˆ˜Οˆ)βˆ—=Οˆβˆ—βˆ˜Ο‡βˆ—

Proof: Let Ο†βˆˆπ’žk(K). Then we have:

(Ο‡βˆ˜Οˆ)βˆ—(Ο†)=Ο†βˆ˜(Ο‡βˆ˜Οˆ)=(Ο†βˆ˜Ο‡)∘ψ=Ο‡βˆ—(Ο†)∘ψ=Οˆβˆ—(Ο‡βˆ—(Ο†))β—»

Theorem 2.29:

Let M,N,K be manifolds of class π’žn, and let ψ:Mβ†’N and Ο‡:Nβ†’K be two functions which are differentiable of class π’žn. Then Ο‡βˆ˜Οˆ is differentiable of class π’žn and

βˆ€p∈M:d(Ο‡βˆ˜Οˆ)p=dΟ‡Οˆ(p)∘dψp

Proof:

1. We prove that Ο‡βˆ˜Οˆ is differentiable of class π’žk.

Let (O,Ο•) be contained in the atlas of M and let (U,ΞΈ) be contained in the atlas of K such that Oβˆ©Οˆβˆ’1(Ο‡βˆ’1(U))β‰ βˆ…οΈ€, and let xβˆˆΟ•βˆ’1(O)βˆ©Οˆβˆ’1(Ο‡βˆ’1(U)) be arbitrary. We choose (V,Ξ·) in the atlas of N such that ψ(Ο•(x))∈V.

We have ψ(Ο•(x))∈Vβˆ©Ο‡βˆ’1(U); indeed, ψ(Ο•(x))∈V due to the choice of (V,Ο•) and ψ(Ο•(x))βˆˆΟ‡βˆ’1(U) because Ο•(x)βˆˆΟˆβˆ’1(Ο‡βˆ’1(U)). Further, we choose W:=Oβˆ©Οˆβˆ’1(Vβˆ©Ο‡βˆ’1(U)). Then the function

ΞΈβˆ’1∘(Ο‡βˆ˜Οˆ)βˆ˜Ο•|Wβˆ’1=ΞΈβˆ’1βˆ˜Ο‡βˆ˜Ξ·βˆ’1∘η∘ψ|Wβˆ˜Ο•|Wβˆ’1

is contained in π’žn(ℝd,ℝd) as the composition of two π’žn(ℝd,ℝd) functions.

Thus, ΞΈβˆ’1∘(Ο‡βˆ˜Οˆ)βˆ˜Ο•|Oβˆ©Οˆβˆ’1(Ο‡βˆ’1(U))βˆ’1 is n times continuously differentiable (or continuous if n=0) at every point, and thus n times continuously differentiable (or continuous if n=0).

2. We prove that βˆ€p∈M:d(Ο‡βˆ˜Οˆ)p=dΟ‡Οˆ(p)∘dψp.

For all p∈M and 𝐕p∈TpM, we have:

(dΟ‡Οˆ(p)∘dψp)(𝐕p)=dΟ‡Οˆ(p)(dψp(𝐕p))=dΟ‡Οˆ(p)(𝐕pβˆ˜Οˆβˆ—)=𝐕pβˆ˜Οˆβˆ—βˆ˜Ο‡βˆ—=theorem 2.26𝐕p∘(Ο‡βˆ˜Οˆ)βˆ—=d(Ο‡βˆ˜Οˆ)p(𝐕p)β—»

Theorem 2.30:

Let M be a manifold of class π’žn, nβ‰₯1 (!) IβŠ†β„ an open interval, Ξ³:Iβ†’M a curve in M which is differentiable of class π’žn and Ο†βˆˆπ’žn(M). Then Ο†βˆ˜Ξ³βˆˆπ’žn(ℝ,ℝ) and

(Ο†βˆ˜Ξ³)′(y)=dφγ(y)(Ξ³'y)

Proof:

1. Among another thing, theorem 2.22 states that Ο†βˆ˜Ξ³ is contained in π’žn(ℝ,ℝ).

2. We show that (Ο†βˆ˜Ξ³)′(y)=dφγ(y)(Ξ³'y):

dφγ(y)(Ξ³'y)=Ξ³'y(Ο†)=(Ο†βˆ˜Ξ³)′(y)β—»

Intuition behind the tangent space

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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 f:ℝ→ℝ.

We start by proving the following lemma from linear algebra:

Lemma 2.33:

Let 𝐔,𝐕,𝐖 be vector spaces and let L:𝐔→𝐕, T:𝐕→𝐖 and S:𝐖→𝐔 be linear functions such that

L∘S∘T=Id𝐕, T∘L∘S=Id𝐖 and S∘T∘L=Id𝐔.

Then L, T and S are all vector space isomorphisms.

Proof:

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

From L∘S∘T=Id𝐕 and T∘L∘S=Id𝐖 follows that L∘S is the inverse function of T.β—»

Definition 2.34:

Let M be a manifold of class π’žn, nβ‰₯1, and let p∈M. Then the space of velocities of M at p is defined to be the vector space consisting of the equivalence classes of the equivalence relation of the set of curves {Ξ³:Iβ†’M|I=(a,b),a,bβˆˆβ„}

γ∼pρ:⇔γ′(p)=ρ′(p).

Sources

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