[<< wikibooks] Differentiable Manifolds/Bases of tangent and cotangent spaces and the differentials
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 ==

In the following, we will show that these functionals are a basis of the tangent space.
Theorem 2.2: Let

M

{\displaystyle M}
be a

d

{\displaystyle d}
-dimensional manifold of class

C

n

{\displaystyle {\mathcal {C}}^{n}}
with

n
≥
1

{\displaystyle n\geq 1}
and atlas

{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
, let

(
O
,
ϕ
)
∈
{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle (O,\phi )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
and let

p
∈
O

{\displaystyle p\in O}
. For all

j
∈
{
1
,
…
,
d
}

{\displaystyle j\in \{1,\ldots ,d\}}
:

(

∂

∂

ϕ

j

)

p

∈

T

p

M

{\displaystyle \left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}\in T_{p}M}
i. e. the function

(

∂

∂

ϕ

j

)

p

:

C

n

(
M
)
→

R

{\displaystyle \left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}:{\mathcal {C}}^{n}(M)\to \mathbb {R} }
is contained in the tangent space

T

p

M

{\displaystyle T_{p}M}
.
Proof:
Let

φ
,
ϑ
∈

C

n

(
M
)

{\displaystyle \varphi ,\vartheta \in {\mathcal {C}}^{n}(M)}
.
1. We show linearity.

(

∂

∂

ϕ

j

)

p

(
φ
+
c
ϑ
)

=

(

∂

x

j

(
(
φ
+
c
ϑ
)
∘

ϕ

−
1

)

)

(
ϕ
(
p
)
)

=

(

∂

x

j

(
φ
∘

ϕ

−
1

+
c
ϑ
∘

ϕ

−
1

)

)

(
ϕ
(
p
)
)

=

(

∂

x

j

(
φ
∘

ϕ

−
1

)
+
c

∂

x

j

(
ϑ
∘

ϕ

−
1

)
)

)

(
ϕ
(
p
)
)

=

(

∂

∂

ϕ

j

)

p

(
φ
)
+
c

(

∂

∂

ϕ

j

)

p

(
ϑ
)

{\displaystyle {\begin{aligned}\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\varphi +c\vartheta )&=\left(\partial _{x_{j}}((\varphi +c\vartheta )\circ \phi ^{-1})\right)(\phi (p))\\&=\left(\partial _{x_{j}}(\varphi \circ \phi ^{-1}+c\vartheta \circ \phi ^{-1})\right)(\phi (p))\\&=\left(\partial _{x_{j}}(\varphi \circ \phi ^{-1})+c\partial _{x_{j}}(\vartheta \circ \phi ^{-1}))\right)(\phi (p))\\&=\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\varphi )+c\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\vartheta )\end{aligned}}}
From the second to the third line, we used the linearity of the derivative.
2. We show the product rule.

(

∂

∂

ϕ

j

)

p

(
φ
ϑ
)

=

(

∂

x

j

(
(
φ
ϑ
)
∘

ϕ

−
1

)

)

(
ϕ
(
p
)
)

=

(

∂

x

j

(
(
φ
∘

ϕ

−
1

)
(
ϑ
∘

ϕ

−
1

)
)

)

(
ϕ
(
p
)
)

=
(
φ
∘

ϕ

−
1

)
(
ϕ
(
p
)
)

(

∂

x

j

(
ϑ
∘

ϕ

−
1

)

)

(
ϕ
(
p
)
)
+
(
ϑ
∘

ϕ

−
1

)
(
ϕ
(
p
)
)

(

∂

x

j

(
φ
∘

ϕ

−
1

)

)

(
ϕ
(
p
)
)

=
φ
(
p
)

(

∂

∂

ϕ

j

)

p

(
ϑ
)
+
ϑ
(
p
)

(

∂

∂

ϕ

j

)

p

(
φ
)

{\displaystyle {\begin{aligned}\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\varphi \vartheta )&=\left(\partial _{x_{j}}((\varphi \vartheta )\circ \phi ^{-1})\right)(\phi (p))\\&=\left(\partial _{x_{j}}((\varphi \circ \phi ^{-1})(\vartheta \circ \phi ^{-1}))\right)(\phi (p))\\&=(\varphi \circ \phi ^{-1})(\phi (p))\left(\partial _{x_{j}}(\vartheta \circ \phi ^{-1})\right)(\phi (p))+(\vartheta \circ \phi ^{-1})(\phi (p))\left(\partial _{x_{j}}(\varphi \circ \phi ^{-1})\right)(\phi (p))\\&=\varphi (p)\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\vartheta )+\vartheta (p)\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\varphi )\end{aligned}}}
From the second to the third line, we used the product rule of the derivative.
3. It follows from the definition of

(

∂

∂

ϕ

j

)

p

{\displaystyle \left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}}
, that

(

∂

∂

ϕ

j

)

p

(
φ
)
=
0

{\displaystyle \left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\varphi )=0}
if

φ

{\displaystyle \varphi }
is not defined at

p

{\displaystyle p}
.

◻

{\displaystyle \Box }

Lemma 2.3: Let

M

{\displaystyle M}
be a

d

{\displaystyle d}
-dimensional manifold of class

C

n

{\displaystyle {\mathcal {C}}^{n}}
with atlas

{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
, and let

(
O
,
ϕ
)
∈
{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle (O,\phi )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
. If we write

ϕ
=
(

ϕ

1

,
…
,

ϕ

d

)

{\displaystyle \phi =(\phi _{1},\ldots ,\phi _{d})}
, then we have for each

k
∈
{
1
,
…
,
d
}

{\displaystyle k\in \{1,\ldots ,d\}}
, that

ϕ

k

∈

C

n

(
M
)

{\displaystyle \phi _{k}\in {\mathcal {C}}^{n}(M)}
.
Proof:
Let

(
U
,
θ
)
∈
{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle (U,\theta )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
. Since

{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
is an atlas,

θ

{\displaystyle \theta }
and

ϕ

{\displaystyle \phi }
are compatible. From this follows that the function

ϕ

|

U
∩
O

∘
θ

|

O
∩
U

−
1

{\displaystyle \phi |_{U\cap O}\circ \theta |_{O\cap U}^{-1}}
is of class

C

n

{\displaystyle {\mathcal {C}}^{n}}
. But if we denote by

π

k

{\displaystyle \pi _{k}}
the function

π

k

:

R

d

→

R

,

π

k

(

x

1

,
…
,

x

d

)
=

x

k

{\displaystyle \pi _{k}:\mathbb {R} ^{d}\to \mathbb {R} ,\pi _{k}(x_{1},\ldots ,x_{d})=x_{k}}
, which is also called the projection to the

k

{\displaystyle k}
-th component, then we have:

ϕ

k

|

U
∩
O

∘
θ

|

O
∩
U

−
1

=

π

k

∘
ϕ

|

U
∩
O

∘
θ

|

O
∩
U

−
1

{\displaystyle \phi _{k}|_{U\cap O}\circ \theta |_{O\cap U}^{-1}=\pi _{k}\circ \phi |_{U\cap O}\circ \theta |_{O\cap U}^{-1}}
It is not difficult to show that

π

k

{\displaystyle \pi _{k}}
is contained in

C

∞

(

R

d

,

R

)

{\displaystyle {\mathcal {C}}^{\infty }(\mathbb {R} ^{d},\mathbb {R} )}
, and therefore the function

π

k

∘
ϕ

|

U
∩
O

∘
θ

|

O
∩
U

−
1

{\displaystyle \pi _{k}\circ \phi |_{U\cap O}\circ \theta |_{O\cap U}^{-1}}
is contained in

C

n

(

R

d

,

R

)

{\displaystyle {\mathcal {C}}^{n}(\mathbb {R} ^{d},\mathbb {R} )}
as a composition of

n

{\displaystyle n}
-times continuously differentiable functions (or continuous functions if

n
=
0

{\displaystyle n=0}
).

◻

{\displaystyle \Box }

Lemma 2.4: Let

M

{\displaystyle M}
be a

d

{\displaystyle d}
-dimensional manifold of class

C

n

{\displaystyle {\mathcal {C}}^{n}}
with

n
≥
1

{\displaystyle n\geq 1}
and atlas

{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
, let

(
O
,
ϕ
)
∈
{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle (O,\phi )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
and let

p
∈
O

{\displaystyle p\in O}
. If we write

ϕ
=
(

ϕ

1

,
…
,

ϕ

d

)

{\displaystyle \phi =(\phi _{1},\ldots ,\phi _{d})}
we have:

(

∂

∂

ϕ

j

)

p

(

ϕ

k

)
=

{

1

j
=
k

0

j
≠
k

{\displaystyle \left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\phi _{k})={\begin{cases}1&j=k\\0&j\neq k\end{cases}}}
Note that due to lemma 2.3,

ϕ

k

∈

C

n

(
M
)

{\displaystyle \phi _{k}\in {\mathcal {C}}^{n}(M)}
for all

k
∈
{
1
,
…
,
d
}

{\displaystyle k\in \{1,\ldots ,d\}}
, which is why the above expression makes sense.
Proof:
We have:

(

∂

∂

ϕ

j

)

p

(

ϕ

k

)

=

(

∂

x

j

(

ϕ

k

∘

ϕ

−
1

)

)

(
ϕ
(
p
)
)

=

lim

y
→
0

(

ϕ

k

∘

ϕ

−
1

)
(

x

1

,
…
,

x

j
−
1

,

x

j

+
y
,

x

j
+
1

,
…
,

x

d

)
−
(

ϕ

k

∘

ϕ

−
1

)
(

x

1

,
…
,

x

d

)

y

{\displaystyle {\begin{aligned}\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\phi _{k})&=\left(\partial _{x_{j}}(\phi _{k}\circ \phi ^{-1})\right)(\phi (p))\\&=\lim _{y\to 0}{\frac {(\phi _{k}\circ \phi ^{-1})(x_{1},\ldots ,x_{j-1},x_{j}+y,x_{j+1},\ldots ,x_{d})-(\phi _{k}\circ \phi ^{-1})(x_{1},\ldots ,x_{d})}{y}}\end{aligned}}}
Further,

(

ϕ

k

∘

ϕ

−
1

)
(

x

1

,
…
,

x

d

)
=

x

k

{\displaystyle (\phi _{k}\circ \phi ^{-1})(x_{1},\ldots ,x_{d})=x_{k}}
and

(

ϕ

k

∘

ϕ

−
1

)
(

x

1

,
…
,

x

j
−
1

,

x

j

+
y
,

x

j
+
1

,
…
,

x

d

)
=

{

x

k

+
y

k
=
j

x

k

k
≠
j

{\displaystyle (\phi _{k}\circ \phi ^{-1})(x_{1},\ldots ,x_{j-1},x_{j}+y,x_{j+1},\ldots ,x_{d})={\begin{cases}x_{k}+y&k=j\\x_{k}&k\neq j\end{cases}}}
Inserting this in the above limit gives the lemma.

◻

{\displaystyle \Box }

Theorem 2.5: Let

M

{\displaystyle M}
be a

d

{\displaystyle d}
-dimensional manifold of class

C

n

{\displaystyle {\mathcal {C}}^{n}}
with

n
≥
1

{\displaystyle n\geq 1}
and atlas

{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
, let

(
O
,
ϕ
)
∈
{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle (O,\phi )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
and let

p
∈
O

{\displaystyle p\in O}
. The tangent vectors

(

∂

∂

ϕ

j

)

p

∈

T

p

M
,
j
∈
{
1
,
…
,
d
}

{\displaystyle \left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}\in T_{p}M,j\in \{1,\ldots ,d\}}
are linearly independent.
Proof:
We write again

ϕ
=
(

ϕ

1

,
…
,

ϕ

d

)

{\displaystyle \phi =(\phi _{1},\ldots ,\phi _{d})}
.
Let

∑

j
=
1

d

a

j

(

∂

∂

ϕ

j

)

p

=

0

p

{\displaystyle \sum _{j=1}^{d}a_{j}\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}=0_{p}}
. Then we have for all

k
∈
{
1
,
…
,
d
}

{\displaystyle k\in \{1,\ldots ,d\}}
:

0
=

0

p

(

ϕ

k

)
=

∑

j
=
1

d

a

j

(

∂

∂

ϕ

j

)

p

(

ϕ

k

)
=

a

k

{\displaystyle 0=0_{p}(\phi _{k})=\sum _{j=1}^{d}a_{j}\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\phi _{k})=a_{k}}

◻

{\displaystyle \Box }
Lemma 2.6:
Let

M

{\displaystyle M}
be a manifold with atlas

{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
,

p
∈
M

{\displaystyle p\in M}
,

V
⊆
M

{\displaystyle V\subseteq M}
be open, let

V

p

∈

T

p

M

{\displaystyle \mathbf {V} _{p}\in T_{p}M}
and

φ
:
V
→

R

,
φ
(
q
)
=
c

{\displaystyle \varphi :V\to \mathbb {R} ,\varphi (q)=c}
for a

c
∈

R

{\displaystyle c\in \mathbb {R} }
; i. e.

φ

{\displaystyle \varphi }
is a constant function. Then

φ
∈

C

∞

(
M
)

{\displaystyle \varphi \in {\mathcal {C}}^{\infty }(M)}
and

V

p

(
φ
)
=
0

{\displaystyle \mathbf {V} _{p}(\varphi )=0}
.
Proof:
1. We show

φ
∈

C

∞

(
M
)

{\displaystyle \varphi \in {\mathcal {C}}^{\infty }(M)}
.
By assumption,

V
⊆
M

{\displaystyle V\subseteq M}
is open. This means the first part of the definition of a

C

∞

(
M
)

{\displaystyle {\mathcal {C}}^{\infty }(M)}
is fulfilled.
Further, for each

(
U
,
θ
)
∈
{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle (U,\theta )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
and

x
∈
θ
(
V
∩
U
)

{\displaystyle x\in \theta (V\cap U)}
, we have:

φ
∘
θ

|

U
∩
V

(
x
)
=
c

{\displaystyle \varphi \circ \theta |_{U\cap V}(x)=c}
This is contained in

C

∞

(

R

d

,

R

)

{\displaystyle {\mathcal {C}}^{\infty }(\mathbb {R} ^{d},\mathbb {R} )}
.
2. We show that

V

p

(
φ
)
=
0

{\displaystyle \mathbf {V} _{p}(\varphi )=0}
.
We define

ϑ
:
V
→

R

,
ϑ
(
q
)
=
1

{\displaystyle \vartheta :V\to \mathbb {R} ,\vartheta (q)=1}
. Using the two rules linearity and product rule for tangent vectors, we obtain:

V

p

(
φ
)
=

V

p

(
ϑ
φ
)
=
1

V

p

(
φ
)
+
φ
(
p
)

V

p

(
ϑ
)
=

V

p

(
φ
)
+

V

p

(
ϑ
φ
(
p
)
)
=

V

p

(
φ
)
+

V

p

(
φ
)

{\displaystyle \mathbf {V} _{p}(\varphi )=\mathbf {V} _{p}(\vartheta \varphi )=1\mathbf {V} _{p}(\varphi )+\varphi (p)\mathbf {V} _{p}(\vartheta )=\mathbf {V} _{p}(\varphi )+\mathbf {V} _{p}(\vartheta \varphi (p))=\mathbf {V} _{p}(\varphi )+\mathbf {V} _{p}(\varphi )}
Substracting

V

p

(
φ
)

{\displaystyle \mathbf {V} _{p}(\varphi )}
, we obtain

V

p

(
φ
)
=
0

{\displaystyle \mathbf {V} _{p}(\varphi )=0}
.

◻

{\displaystyle \Box }

Proof:
Let

U
⊆
M

{\displaystyle U\subseteq M}
be open, and let

φ
:
U
→

R

{\displaystyle \varphi :U\to \mathbb {R} }
be contained in

C

n

(
M
)

{\displaystyle {\mathcal {C}}^{n}(M)}
.
Case 1:

p
∉
U

{\displaystyle p\notin U}
.
In this case,

V

p

(
φ
)
=
0

{\displaystyle \mathbf {V} _{p}(\varphi )=0}
and

(

∂

∂

ϕ

j

)

p

(
φ
)
=
0

{\displaystyle \left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\varphi )=0}
, since

φ

{\displaystyle \varphi }
is not defined at

p

{\displaystyle p}
and both

V

p

{\displaystyle \mathbf {V} _{p}}
and

(

∂

∂

ϕ

j

)

p

{\displaystyle \left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}}
are tangent vectors. From this follows the formula.
Case 2:

p
∈
U

{\displaystyle p\in U}
.
In this case, we obtain that the set

ϕ
(
U
∩
O
)

{\displaystyle \phi (U\cap O)}
is open in

R

d

{\displaystyle \mathbb {R} ^{d}}
as follows: Since

ϕ
:
O
→
ϕ
(
O
)

{\displaystyle \phi :O\to \phi (O)}
is a homeomorphism by definition of charts, the set

ϕ
(
U
∩
O
)

{\displaystyle \phi (U\cap O)}
is open in

ϕ
(
O
)

{\displaystyle \phi (O)}
. By definition of the subspace topology, we have

ϕ
(
U
∩
O
)
=
V
∩
ϕ
(
O
)

{\displaystyle \phi (U\cap O)=V\cap \phi (O)}
for a

V

{\displaystyle V}
open in

R

d

{\displaystyle \mathbb {R} ^{d}}
. But

V
∩
ϕ
(
O
)

{\displaystyle V\cap \phi (O)}
is open in

R

d

{\displaystyle \mathbb {R} ^{d}}
as the intersection of two open sets; recall that

ϕ
(
O
)

{\displaystyle \phi (O)}
was required to be open in the definition of a chart.
Furthermore, from

p
∈
U

{\displaystyle p\in U}
and

p
∈
O

{\displaystyle p\in O}
it follows that

p
∈
U
∩
O

{\displaystyle p\in U\cap O}
, and therefore

ϕ
(
p
)
∈
ϕ
(
O
∩
U
)

{\displaystyle \phi (p)\in \phi (O\cap U)}
. Since

ϕ
(
O
∩
U
)

{\displaystyle \phi (O\cap U)}
is open, we find an

ϵ
>
0

{\displaystyle \epsilon >0}
such that the open ball

B

ϵ

(
ϕ
(
p
)
)

{\displaystyle B_{\epsilon }(\phi (p))}
is contained in

ϕ
(
O
∩
U
)

{\displaystyle \phi (O\cap U)}
. We define

W
:=

ϕ

−
1

(

B

ϵ

(
ϕ
(
p
)
)
)

{\displaystyle W:=\phi ^{-1}(B_{\epsilon }(\phi (p)))}
. Since

ϕ

{\displaystyle \phi }
is bijective,

W
⊆
U
∩
O

{\displaystyle W\subseteq U\cap O}
, and since

ϕ

{\displaystyle \phi }
is a homeomorphism, in particular continuous,

W

{\displaystyle W}
is open in

O

{\displaystyle O}
with respect to the subspace topology of

O

{\displaystyle O}
. From this also follows

O

{\displaystyle O}
open in

M

{\displaystyle M}
, because if

W

{\displaystyle W}
is open in

O

{\displaystyle O}
, then by definition of the subspace topology it is of the form

V
∩
O

{\displaystyle V\cap O}
for an open set

V
⊆
M

{\displaystyle V\subseteq M}
, and hence it is open as the intersection of two open sets.
We have that

φ

|

W

:
W
→

R

{\displaystyle \varphi |_{W}:W\to \mathbb {R} }
, is contained in

C

∞

(
M
)

{\displaystyle {\mathcal {C}}^{\infty }(M)}
:

W

{\displaystyle W}
is an open subset of

M

{\displaystyle M}
, and if

(
V
,
θ
)
∈
{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle (V,\theta )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
, then

φ

|

W
∩
V

∘
θ

|

W
∩
V

−
1

=
(
φ

|

U
∩
V

∘
θ

|

U
∩
V

−
1

)

|

θ
(
W
∩
V
)

{\displaystyle \varphi |_{W\cap V}\circ \theta |_{W\cap V}^{-1}=(\varphi |_{U\cap V}\circ \theta |_{U\cap V}^{-1})|_{\theta (W\cap V)}}
,(check this by direct calculation!), which is contained in

C

∞

(

R

d

,

R

)

{\displaystyle {\mathcal {C}}^{\infty }(\mathbb {R} ^{d},\mathbb {R} )}
as the restriction of an arbitrarily often continuously differentiable function.
We now define the function

F
:

B

ϵ

(
ϕ
(
p
)
)
→

R

{\displaystyle F:B_{\epsilon }(\phi (p))\to \mathbb {R} }
,

F
(
x
)
=
(
φ
∘

ϕ

−
1

)
(
x
)

{\displaystyle F(x)=(\varphi \circ \phi ^{-1})(x)}
, and further for each

x
∈

B

ϵ

(
ϕ
(
p
)
)

{\displaystyle x\in B_{\epsilon }(\phi (p))}
, we define

μ

x

(
ξ
)
:=
F
(
ξ
x
+
(
1
−
ξ
)
ϕ
(
p
)
)

{\displaystyle \mu _{x}(\xi ):=F(\xi x+(1-\xi )\phi (p))}
From the fundamental theorem of calculus, the multi-dimensional chain rule and the linearity of the integral follows for each

x
∈

B

ϵ

(
ϕ
(
p
)
)

{\displaystyle x\in B_{\epsilon }(\phi (p))}
, that

F
(
x
)

=

μ

x

(
1
)

=

μ

x

(
0
)
+

∫

0

1

μ

x

′

(
ξ
)
d
ξ

=
F
(
ϕ
(
p
)
)
+

∑

j
=
1

d

(

x

j

−
ϕ
(
p

)

j

)

∫

0

1

∂

x

j

F
(
ξ
ϕ
(
p
)
+
(
1
−
ξ
)
x
)
d
ξ

{\displaystyle {\begin{aligned}F(x)&=\mu _{x}(1)\\&=\mu _{x}(0)+\int _{0}^{1}\mu _{x}'(\xi )d\xi \\&=F(\phi (p))+\sum _{j=1}^{d}(x_{j}-\phi (p)_{j})\int _{0}^{1}\partial _{x_{j}}F(\xi \phi (p)+(1-\xi )x)d\xi \end{aligned}}}
If one sets

x
=
ϕ
(
q
)

{\displaystyle x=\phi (q)}
for

q
∈
W

{\displaystyle q\in W}
, one obtains, inserting the definition of

F

{\displaystyle F}
:

φ
(
q
)
=
φ
(
p
)
+

∑

j
=
1

d

(
ϕ
(
q

)

j

−
ϕ
(
p

)

j

)

∫

0

1

∂

x

j

(
φ
∘

ϕ

−
1

)
(
ξ
ϕ
(
p
)
+
(
1
−
ξ
)
ϕ
(
q
)
)
d
ξ

{\displaystyle \varphi (q)=\varphi (p)+\sum _{j=1}^{d}(\phi (q)_{j}-\phi (p)_{j})\int _{0}^{1}\partial _{x_{j}}(\varphi \circ \phi ^{-1})(\xi \phi (p)+(1-\xi )\phi (q))d\xi }
Now we define the functions

f

j

:
W
→

R

,

f

j

(
q
)
:=

∫

0

1

∂

x

j

(
φ
∘

ϕ

−
1

)
(
ξ
ϕ
(
p
)
+
(
1
−
ξ
)
ϕ
(
q
)
)
d
ξ

{\displaystyle f_{j}:W\to \mathbb {R} ,f_{j}(q):=\int _{0}^{1}\partial _{x_{j}}(\varphi \circ \phi ^{-1})(\xi \phi (p)+(1-\xi )\phi (q))d\xi }
These are contained in

C

∞

(
M
)

{\displaystyle {\mathcal {C}}^{\infty }(M)}
since they are defined on

W

{\displaystyle W}
which is open, and further, if

(
V
,
θ
)
∈
{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle (V,\theta )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
, then

f

j

|

V
∩
W

∘
θ

|

V
∩
W

−
1

=

∫

0

1

∂

x

j

(
φ
∘

ϕ

−
1

)
(
ξ
ϕ

|

V
∩
W

∘
θ

|

V
∩
W

−
1

+
(
1
−
ξ
)
ϕ

|

V
∩
W

∘
θ

|

V
∩
W

−
1

)
d
ξ

{\displaystyle f_{j}|_{V\cap W}\circ \theta |_{V\cap W}^{-1}=\int _{0}^{1}\partial _{x_{j}}(\varphi \circ \phi ^{-1})(\xi \phi |_{V\cap W}\circ \theta |_{V\cap W}^{-1}+(1-\xi )\phi |_{V\cap W}\circ \theta |_{V\cap W}^{-1})d\xi }
, 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

ϕ
=
(

ϕ

1

,
…
,

ϕ

d

)

{\displaystyle \phi =(\phi _{1},\ldots ,\phi _{d})}
, the functions

ϕ

k

{\displaystyle \phi _{k}}
,

k
∈
{
1
,
…
,
d
}

{\displaystyle k\in \{1,\ldots ,d\}}
are contained in

C

∞

(
M
)

{\displaystyle {\mathcal {C}}^{\infty }(M)}
due to lemma 2.3.
Since

φ

|

W

∈

C

∞

(
M
)

{\displaystyle \varphi |_{W}\in {\mathcal {C}}^{\infty }(M)}
,

V

p

(
φ

|

W

)

{\displaystyle \mathbf {V} _{p}(\varphi |_{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

C

∞

(
M
)

{\displaystyle {\mathcal {C}}^{\infty }(M)}
), and obtain:

V

p

(
φ

|

W

)

=

V

p

(

φ
(
p
)
+

∑

j
=
1

d

(

ϕ

j

−
ϕ
(
p

)

j

)

f

j

)

=

∑

j
=
1

d

(

ϕ

j

(
p
)

V

p

(

f

j

)
+

f

j

(
p
)

V

p

(

ϕ

j

)
−
ϕ
(
p

)

j

V

p

(

f

j

)

)

=

∑

j
=
1

d

f

j

(
p
)

V

p

(

ϕ

j

)

{\displaystyle {\begin{aligned}\mathbf {V} _{p}(\varphi |_{W})&=\mathbf {V} _{p}\left(\varphi (p)+\sum _{j=1}^{d}(\phi _{j}-\phi (p)_{j})f_{j}\right)\\&=\sum _{j=1}^{d}\left(\phi _{j}(p)\mathbf {V} _{p}(f_{j})+f_{j}(p)\mathbf {V} _{p}(\phi _{j})-\phi (p)_{j}\mathbf {V} _{p}(f_{j})\right)\\&=\sum _{j=1}^{d}f_{j}(p)\mathbf {V} _{p}(\phi _{j})\end{aligned}}}
, since due to our notation it's clear that

ϕ

j

(
p
)
=
ϕ
(
p

)

j

{\displaystyle \phi _{j}(p)=\phi (p)_{j}}
.
But

f

j

(
p
)

=

∫

0

1

∂

x

j

(
φ
∘

ϕ

−
1

)
(
ξ
ϕ
(
p
)
+
(
1
−
ξ
)
ϕ
(
p
)
)
d
ξ

=

∫

0

1

∂

x

j

(
φ
∘

ϕ

−
1

)
(
ϕ
(
p
)
)
d
ξ

=

∂

x

j

(
φ
∘

ϕ

−
1

)
(
ϕ
(
p
)
)

=

(

∂

∂

ϕ

j

)

p

(
φ
)

{\displaystyle {\begin{aligned}f_{j}(p)&=\int _{0}^{1}\partial _{x_{j}}(\varphi \circ \phi ^{-1})(\xi \phi (p)+(1-\xi )\phi (p))d\xi \\&=\int _{0}^{1}\partial _{x_{j}}(\varphi \circ \phi ^{-1})(\phi (p))d\xi \\&=\partial _{x_{j}}(\varphi \circ \phi ^{-1})(\phi (p))\\&=\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\varphi )\end{aligned}}}
Thus we have successfully shown

V

p

(
φ

|

W

)
=

∑

j
=
1

d

V

p

(

ϕ

j

)

(

∂

∂

ϕ

j

)

p

(
φ
)

{\displaystyle \mathbf {V} _{p}(\varphi |_{W})=\sum _{j=1}^{d}\mathbf {V} _{p}(\phi _{j})\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\varphi )}
But due to the definition of subtraction on

C

∞

(
M
)

{\displaystyle {\mathcal {C}}^{\infty }(M)}
, due to lemma 2.6, and due to the fact that the constant zero function is a constant function:

V

p

(
φ

|

W

−
φ
)
=

V

p

(
0
)
=
0

{\displaystyle \mathbf {V} _{p}(\varphi |_{W}-\varphi )=\mathbf {V} _{p}(0)=0}
Due to linearity of

V

p

{\displaystyle \mathbf {V} _{p}}
follows

0
=

V

p

(
φ

|

W

)
−

V

p

(
φ
)

{\displaystyle 0=\mathbf {V} _{p}(\varphi |_{W})-\mathbf {V} _{p}(\varphi )}
, i. e.

V

p

(
φ

|

W

)
=

V

p

(
φ
)

{\displaystyle \mathbf {V} _{p}(\varphi |_{W})=\mathbf {V} _{p}(\varphi )}
. Now, inserting in the above equation gives the theorem.

◻

{\displaystyle \Box }

Together with theorem 2.5, this theorem shows that

{

(

∂

∂

ϕ

j

)

p

|

j
∈
{
1
,
…
,
d
}

}

{\displaystyle \left\{\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}{\Bigg |}j\in \{1,\ldots ,d\}\right\}}
is a basis of

T

p

M

{\displaystyle T_{p}M}
, 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

T

p

M

{\displaystyle T_{p}M}
is equal to

d

{\displaystyle d}
.

== Some bases of the cotangent space ==

Note that

(
d

ϕ

j

)

p

{\displaystyle (d\phi _{j})_{p}}
is well-defined because of lemma 2.3.
Theorem 2.9: Let

M

{\displaystyle M}
be a

d

{\displaystyle d}
-dimensional manifold of class

C

∞

{\displaystyle {\mathcal {C}}^{\infty }}
and atlas

{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
, let

(
O
,
ϕ
)
∈
{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle (O,\phi )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
and let

p
∈
O

{\displaystyle p\in O}
. For all

j
∈
{
1
,
…
,
d
}

{\displaystyle j\in \{1,\ldots ,d\}}
,

(
d

ϕ

j

)

p

{\displaystyle (d\phi _{j})_{p}}
is contained in

T

p

M

∗

{\displaystyle T_{p}M^{*}}
.
Proof:
By definition,

d

ϕ

k

{\displaystyle d\phi _{k}}
maps from

T

p

M

{\displaystyle T_{p}M}
to

R

{\displaystyle \mathbb {R} }
. Thus, linearity is the only thing left to show. Indeed, for

V

p

,

W

p

∈

T

p

M

{\displaystyle \mathbf {V} _{p},\mathbf {W} _{p}\in T_{p}M}
and

b
∈

R

{\displaystyle b\in \mathbb {R} }
, we have, since addition and scalar multiplication in

T

p

M

{\displaystyle T_{p}M}
are defined pointwise:

(
d

ϕ

j

)

p

(

V

p

+
b

W

p

)

=
(

V

p

+
b

W

p

)
(

ϕ

k

)

=

V

p

(

ϕ

k

)
+
b

W

p

(

ϕ

k

)

=
(
d

ϕ

j

)

p

(

V

p

)
+
b
(
d

ϕ

j

)

p

(

W

p

)

{\displaystyle {\begin{aligned}(d\phi _{j})_{p}(\mathbf {V} _{p}+b\mathbf {W} _{p})&=(\mathbf {V} _{p}+b\mathbf {W} _{p})(\phi _{k})\\&=\mathbf {V} _{p}(\phi _{k})+b\mathbf {W} _{p}(\phi _{k})\\&=(d\phi _{j})_{p}(\mathbf {V} _{p})+b(d\phi _{j})_{p}(\mathbf {W} _{p})\end{aligned}}}

◻

{\displaystyle \Box }
Lemma 2.10: Let

M

{\displaystyle M}
be a

d

{\displaystyle d}
-dimensional manifold of class

C

∞

{\displaystyle {\mathcal {C}}^{\infty }}
and atlas

{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
, let

(
O
,
ϕ
)
∈
{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle (O,\phi )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
and let

p
∈
O

{\displaystyle p\in O}
. For

j
,
k
∈
{
1
,
…
,
d
}

{\displaystyle j,k\in \{1,\ldots ,d\}}
, the following equation holds:

(
d

ϕ

j

)

p

(

(

∂

∂

ϕ

k

)

p

)

=

{

1

k
=
j

0

k
≠
j

{\displaystyle (d\phi _{j})_{p}\left(\left({\frac {\partial }{\partial \phi _{k}}}\right)_{p}\right)={\begin{cases}1&k=j\\0&k\neq j\end{cases}}}
Proof:
We have:

(
d

ϕ

j

)

p

(

(

∂

∂

ϕ

k

)

p

)

=

(

∂

∂

ϕ

k

)

p

(

ϕ

j

)

=
lemma 2.4

{

1

k
=
j

0

k
≠
j

{\displaystyle (d\phi _{j})_{p}\left(\left({\frac {\partial }{\partial \phi _{k}}}\right)_{p}\right)=\left({\frac {\partial }{\partial \phi _{k}}}\right)_{p}(\phi _{j}){\overset {\text{lemma 2.4}}{=}}{\begin{cases}1&k=j\\0&k\neq j\end{cases}}}

◻

{\displaystyle \Box }
Theorem 2.11: Let

M

{\displaystyle M}
be a

d

{\displaystyle d}
-dimensional manifold of class

C

∞

{\displaystyle {\mathcal {C}}^{\infty }}
and atlas

{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
, let

(
O
,
ϕ
)
∈
{
(

O

υ

,

ϕ

υ

)

|

υ
∈
Υ
}

{\displaystyle (O,\phi )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}
and let

p
∈
O

{\displaystyle p\in O}
. The cotangent vectors

(
d

ϕ

j

)

p

,
j
∈
{
1
,
…
,
d
}

{\displaystyle (d\phi _{j})_{p},j\in \{1,\ldots ,d\}}
are linearly independent.
Proof:
Let

0
=

∑

j
=
1

d

a

j

(
d

ϕ

j

)

p

{\displaystyle 0=\sum _{j=1}^{d}a_{j}(d\phi _{j})_{p}}
, where by

0

{\displaystyle 0}
we mean the zero of

T

p

M

∗

{\displaystyle T_{p}M^{*}}
. Then we have for all

k
∈
{
1
,
…
,
d
}

{\displaystyle k\in \{1,\ldots ,d\}}
:

0
=

∑

j
=
1

d

a

j

(
d

ϕ

j

)

p

(

(

∂

∂

ϕ

k

)

p

)

=
lemma 2.10

a

k

{\displaystyle 0=\sum _{j=1}^{d}a_{j}(d\phi _{j})_{p}\left(\left({\frac {\partial }{\partial \phi _{k}}}\right)_{p}\right){\overset {\text{lemma 2.10}}{=}}a_{k}}

◻

{\displaystyle \Box }

Proof:
Let

α

p

∈

T

p

M

∗

{\displaystyle \alpha _{p}\in T_{p}M^{*}}
and

V

p

∈

T

p

M

{\displaystyle \mathbf {V} _{p}\in T_{p}M}
. Due to theorem 2.7, we have

V

p

=

∑

j
=
1

d

V

p

(

ϕ

j

)

(

∂

∂

ϕ

j

)

p

{\displaystyle \mathbf {V} _{p}=\sum _{j=1}^{d}\mathbf {V} _{p}(\phi _{j})\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}}
Therefore, and due to the linearity of

α

p

{\displaystyle \alpha _{p}}
(because

T

p

M

∗

{\displaystyle T_{p}M^{*}}
was the space of linear functions to

R

{\displaystyle \mathbb {R} }
):

α

p

(

V

p

)

=

∑

j
=
1

d

V

p

(

ϕ

j

)

α

p

(

(

∂

∂

ϕ

j

)

p

)

=

∑

j
=
1

d

α

p

(

(

∂

∂

ϕ

j

)

p

)

(
d

ϕ

j

)

p

(

V

p

)

{\displaystyle {\begin{aligned}\alpha _{p}(\mathbf {V} _{p})&=\sum _{j=1}^{d}\mathbf {V} _{p}(\phi _{j})\alpha _{p}\left(\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}\right)\\&=\sum _{j=1}^{d}\alpha _{p}\left(\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}\right)(d\phi _{j})_{p}(\mathbf {V} _{p})\end{aligned}}}
Since

V

p

∈

T

p

M

{\displaystyle \mathbf {V} _{p}\in T_{p}M}
was arbitrary, the theorem is proven.

◻

{\displaystyle \Box }

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

{

(
d

ϕ

j

)

p

|

j
∈
{
1
,
…
,
d
}

}

{\displaystyle \left\{(d\phi _{j})_{p}{\big |}j\in \{1,\ldots ,d\}\right\}}
is a basis for

T

p

M

∗

{\displaystyle T_{p}M^{*}}
, and that the dimension of

T

p

M

∗

{\displaystyle T_{p}M^{*}}
is equal to

d

{\displaystyle d}
, like the dimension of

T

p

M

{\displaystyle T_{p}M}
.

== Expressing elements of the tangent and cotangent spaces in different bases ==
If

M

{\displaystyle M}
is a manifold,

p
∈
M

{\displaystyle p\in M}
and

(
O
,
ϕ
)
,
(
U
,
θ
)

{\displaystyle (O,\phi ),(U,\theta )}
are two charts in

M

{\displaystyle M}
's atlas such that

p
∈
O

{\displaystyle p\in O}
and

p
∈
U

{\displaystyle p\in U}
. Then follows from the last two subsections, that

{

(

∂

∂

ϕ

j

)

p

|

j
∈
{
1
,
…
,
d
}

}

{\displaystyle \left\{\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}{\Bigg |}j\in \{1,\ldots ,d\}\right\}}
and

{

(

∂

∂

θ

j

)

p

|

j
∈
{
1
,
…
,
d
}

}

{\displaystyle \left\{\left({\frac {\partial }{\partial \theta _{j}}}\right)_{p}{\Bigg |}j\in \{1,\ldots ,d\}\right\}}
are bases for

T

p

M

{\displaystyle T_{p}M}
, and

{

(
d

ϕ

j

)

p

|

j
∈
{
1
,
…
,
d
}

}

{\displaystyle \left\{(d\phi _{j})_{p}{\big |}j\in \{1,\ldots ,d\}\right\}}
and

{

(
d

θ

j

)

p

|

j
∈
{
1
,
…
,
d
}

}

{\displaystyle \left\{(d\theta _{j})_{p}{\big |}j\in \{1,\ldots ,d\}\right\}}
are bases for

T

p

M

∗

{\displaystyle T_{p}M^{*}}
.One could now ask the questions:
If we have an element

V

p

{\displaystyle \mathbf {V} _{p}}
in

T

p

M

{\displaystyle T_{p}M}
given by

V

p

=

∑

j
=
1

d

a

j

(

∂

∂

ϕ

j

)

p

{\displaystyle \mathbf {V} _{p}=\sum _{j=1}^{d}a_{j}\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}}
, then how can we represent

V

p

{\displaystyle \mathbf {V} _{p}}
as linear combination of the basis

{

(

∂

∂

θ

j

)

p

|

j
∈
{
1
,
…
,
d
}

}

{\displaystyle \left\{\left({\frac {\partial }{\partial \theta _{j}}}\right)_{p}{\Bigg |}j\in \{1,\ldots ,d\}\right\}}
?
Or if we have an element

α

p

{\displaystyle \alpha _{p}}
in

T

p

M

∗

{\displaystyle T_{p}M^{*}}
given by

α

p

=

∑

j
=
1

d

a

j

(
d

ϕ

j

)

p

{\displaystyle \alpha _{p}=\sum _{j=1}^{d}a_{j}(d\phi _{j})_{p}}
, then how can we represent

α

p

{\displaystyle \alpha _{p}}
as linear combination of the basis

{

(
d

θ

j

)

p

|

j
∈
{
1
,
…
,
d
}

}

{\displaystyle \left\{(d\theta _{j})_{p}{\big |}j\in \{1,\ldots ,d\}\right\}}
?
The following two theorems answer these questions:

Proof:
Due to theorem 2.7, we have for

j
∈
{
1
,
…
,
d
}

{\displaystyle j\in \{1,\ldots ,d\}}
:

(

∂

∂

ϕ

j

)

p

=

∑

k
=
1

d

(

∂

∂

ϕ

j

)

p

(

θ

k

)

(

∂

∂

θ

k

)

p

{\displaystyle \left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}=\sum _{k=1}^{d}\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\theta _{k})\left({\frac {\partial }{\partial \theta _{k}}}\right)_{p}}
From this follows:

V

p

=

∑

j
=
1

d

a

j

(

∂

∂

ϕ

j

)

p

=

∑

j
=
1

d

a

j

∑

k
=
1

d

(

∂

∂

ϕ

j

)

p

(

θ

k

)

(

∂

∂

θ

k

)

p

=

∑

k
=
1

d

∑

j
=
1

d

a

j

(

∂

∂

ϕ

j

)

p

(

θ

k

)

(

∂

∂

θ

k

)

p

{\displaystyle {\begin{aligned}\mathbf {V} _{p}&=\sum _{j=1}^{d}a_{j}\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}\\&=\sum _{j=1}^{d}a_{j}\sum _{k=1}^{d}\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\theta _{k})\left({\frac {\partial }{\partial \theta _{k}}}\right)_{p}\\&=\sum _{k=1}^{d}\sum _{j=1}^{d}a_{j}\left({\frac {\partial }{\partial \phi _{j}}}\right)_{p}(\theta _{k})\left({\frac {\partial }{\partial \theta _{k}}}\right)_{p}\end{aligned}}}

◻

{\displaystyle \Box }

Proof:
Due to theorem 2.12, we have for

j
∈
{
1
,
…
,
d
}

{\displaystyle j\in \{1,\ldots ,d\}}
:

(
d

ϕ

j

)

p

=

∑

k
=
1

d

(
d

ϕ

j

)

p

(

(

∂

∂

ϕ

k

)

p

)

(
d

θ

k

)

p

{\displaystyle (d\phi _{j})_{p}=\sum _{k=1}^{d}(d\phi _{j})_{p}\left(\left({\frac {\partial }{\partial \phi _{k}}}\right)_{p}\right)(d\theta _{k})_{p}}
Thus we obtain:

α

p

=

∑

j
=
1

d

a

j

(
d

ϕ

j

)

p

=

∑

j
=
1

d

a

j

∑

k
=
1

d

(
d

ϕ

j

)

p

(

(

∂

∂

ϕ

k

)

p

)

(
d

θ

k

)

p

=

∑

k
=
1

d

∑

j
=
1

d

a

j

(
d

ϕ

j

)

p

(

(

∂

∂

ϕ

k

)

p

)

(
d

θ

k

)

p

{\displaystyle {\begin{aligned}\alpha _{p}&=\sum _{j=1}^{d}a_{j}(d\phi _{j})_{p}\\&=\sum _{j=1}^{d}a_{j}\sum _{k=1}^{d}(d\phi _{j})_{p}\left(\left({\frac {\partial }{\partial \phi _{k}}}\right)_{p}\right)(d\theta _{k})_{p}\\&=\sum _{k=1}^{d}\sum _{j=1}^{d}a_{j}(d\phi _{j})_{p}\left(\left({\frac {\partial }{\partial \phi _{k}}}\right)_{p}\right)(d\theta _{k})_{p}\end{aligned}}}

◻

{\displaystyle \Box }

== The pullback and the differentials ==
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

{\displaystyle \mathbb {R} }

functions from an interval

I
⊆

R

{\displaystyle I\subseteq \mathbb {R} }
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:

Lemma 2.16: Let

M

{\displaystyle M}
be a

d

{\displaystyle d}
-dimensional and

N

{\displaystyle N}
be a

b

{\displaystyle b}
-dimensional manifold, let

k
∈

N

0

∪
{
∞
}

{\displaystyle k\in \mathbb {N} _{0}\cup \{\infty \}}
and let

ψ
:
M
→
N

{\displaystyle \psi :M\to N}
be differentiable of class

C

k

{\displaystyle {\mathcal {C}}^{k}}
. Then

ψ

{\displaystyle \psi }
is continuous.
Proof:
We show that for an arbitrary

p
∈
M

{\displaystyle p\in M}
,

ψ

{\displaystyle \psi }
is continuous on an open neighbourhood of

p

{\displaystyle p}
. There is a theorem in topology which states that from this follows continuity.
We choose

(
O
,
ϕ
)

{\displaystyle (O,\phi )}
in the atlas of

M

{\displaystyle M}
such that

p
∈
O

{\displaystyle p\in O}
, and

(
U
,
θ
)

{\displaystyle (U,\theta )}
in the atlas of

N

{\displaystyle N}
such that

ψ
(
p
)
∈
U

{\displaystyle \psi (p)\in U}
. Due to the differentiability of

ψ

{\displaystyle \psi }
, the function

θ
∘
ψ
∘
ϕ

|

ϕ
(
O
∩

ψ

−
1

(
U
)
)

−
1

{\displaystyle \theta \circ \psi \circ \phi |_{\phi (O\cap \psi ^{-1}(U))}^{-1}}
is contained in

C

k

(

R

d

,

R

b

)

{\displaystyle {\mathcal {C}}^{k}(\mathbb {R} ^{d},\mathbb {R} ^{b})}
, and therefore continuous. But

ϕ

{\displaystyle \phi }
and

θ

{\displaystyle \theta }
are charts and therefore homeomorphisms, and thus the function

ψ

|

O
∩

ψ

−
1

(
U
)

:
O
∩

ψ

−
1

(
U
)
→
N
,
ψ
=

θ

−
1

∘
θ
∘
ψ
∘
ϕ

|

O
∩

ψ

−
1

(
U
)

−
1

∘
ϕ

|

O
∩

ψ

−
1

(
U
)

{\displaystyle \psi |_{O\cap \psi ^{-1}(U)}:O\cap \psi ^{-1}(U)\to N,\psi =\theta ^{-1}\circ \theta \circ \psi \circ \phi |_{O\cap \psi ^{-1}(U)}^{-1}\circ \phi |_{O\cap \psi ^{-1}(U)}}
is continuous as the composition of continuous functions.

◻

{\displaystyle \Box }

Lemma 2.17: Let

M
,
N

{\displaystyle M,N}
be two manifolds, let

ψ
:
M
→
N

{\displaystyle \psi :M\to N}
be differentiable of class

C

k

{\displaystyle {\mathcal {C}}^{k}}
, and let

φ
∈

C

k

(
N
)

{\displaystyle \varphi \in {\mathcal {C}}^{k}(N)}
be defined on the open set

U
⊆
N

{\displaystyle U\subseteq N}
. In this case, the function

φ
∘

|

ψ

−
1

(
U
)

{\displaystyle \varphi \circ |_{\psi ^{-1}(U)}}
is contained in

C

k

(
M
)

{\displaystyle {\mathcal {C}}^{k}(M)}
; i. e. the pullback with respect to

ψ

{\displaystyle \psi }
really maps to

C

k

(
M
)

{\displaystyle {\mathcal {C}}^{k}(M)}
.
Proof:
Since

ψ

{\displaystyle \psi }
is continuous due to lemma 2.16,

ψ

−
1

(
U
)

{\displaystyle \psi ^{-1}(U)}
is open in

M

{\displaystyle M}
. Thus

φ
∘
ψ

|

ψ

−
1

(
U
)

{\displaystyle \varphi \circ \psi |_{\psi ^{-1}(U)}}
is defined on an open set.
Let

(
O
,
ϕ
)

{\displaystyle (O,\phi )}
be an arbitrary element of the atlas of

M

{\displaystyle M}
and let

x
∈
ϕ
(
O
)

{\displaystyle x\in \phi (O)}
be arbitrary. We choose

(
V
,
θ
)

{\displaystyle (V,\theta )}
in the atlas of

N

{\displaystyle N}
such that

ψ
(

ϕ

−
1

(
x
)
)
∈
V

{\displaystyle \psi (\phi ^{-1}(x))\in V}
. The function

(
φ
∘
ψ

|

ψ

−
1

(
U
)

∘
ϕ

|

ψ

−
1

(
U
)
∩
O

−
1

)

|

ϕ
(

ψ

−
1

(
U
∩
V
)
∩
O
)

=
φ

|

ψ
(

ψ

−
1

(
U
∩
V
)
∩
O
)

∘
θ

|

ψ
(

ψ

−
1

(
U
∩
V
)
∩
O
)

−
1

∘
θ

|

ψ
(

ψ

−
1

(
U
∩
V
)
∩
O
)

∘
ψ

|

ψ

−
1

(
U
∩
V
)
∩
O

∘
ϕ

|

ϕ
(

ψ

−
1

(
U
∩
V
)
∩
O
)

−
1

{\displaystyle (\varphi \circ \psi |_{\psi ^{-1}(U)}\circ \phi |_{\psi ^{-1}(U)\cap O}^{-1})|_{\phi (\psi ^{-1}(U\cap V)\cap O)}=\varphi |_{\psi (\psi ^{-1}(U\cap V)\cap O)}\circ \theta |_{\psi (\psi ^{-1}(U\cap V)\cap O)}^{-1}\circ \theta |_{\psi (\psi ^{-1}(U\cap V)\cap O)}\circ \psi |_{\psi ^{-1}(U\cap V)\cap O}\circ \phi |_{\phi (\psi ^{-1}(U\cap V)\cap O)}^{-1}}
is

k

{\displaystyle k}
-times continuously differentiable (or continuous if

k
=
0

{\displaystyle k=0}
) at

x

{\displaystyle x}
as the composition of two

k

{\displaystyle k}
times continuously differentiable (or continuous if

k
=
0

{\displaystyle k=0}
) functions. Thus, the function

φ
∘
ψ

|

ψ

−
1

(
U
)

∘
ϕ

|

ψ

−
1

(
U
)
∩
O

−
1

{\displaystyle \varphi \circ \psi |_{\psi ^{-1}(U)}\circ \phi |_{\psi ^{-1}(U)\cap O}^{-1}}
is

k

{\displaystyle k}
-times continuously differentiable (or continuous if

k
=
0

{\displaystyle k=0}
) at every point, and therefore contained in

C

k

(

R

d

,

R

)

{\displaystyle {\mathcal {C}}^{k}(\mathbb {R} ^{d},\mathbb {R} )}
.

◻

{\displaystyle \Box }

Theorem 2.19:
Let

M
,
N

{\displaystyle M,N}
be two manifolds of class

C

n

{\displaystyle {\mathcal {C}}^{n}}
, let

ψ
:
M
→
N

{\displaystyle \psi :M\to N}
be differentiable of class

C

n

{\displaystyle {\mathcal {C}}^{n}}
and let

p
∈
M

{\displaystyle p\in M}
. We have

V

p

∘

ψ

∗

∈

T

p

N

{\displaystyle \mathbf {V} _{p}\circ \psi ^{*}\in T_{p}N}
; i. e. the differential of

ψ

{\displaystyle \psi }
at

p

{\displaystyle p}
really maps to

T

p

N

{\displaystyle T_{p}N}
.
Proof:
Let

O
,
U
⊆
M

{\displaystyle O,U\subseteq M}
be open,

φ
:
O
→

R

,
ϑ
:
U
→

R

∈

C

n

(
M
)

{\displaystyle \varphi :O\to \mathbb {R} ,\vartheta :U\to \mathbb {R} \in {\mathcal {C}}^{n}(M)}
and

c
∈

R

{\displaystyle c\in \mathbb {R} }
be arbitrary. In the proof of the following, we will use that for all open subsets

V
⊆
O

{\displaystyle V\subseteq O}
,

V

p

(
φ

|

V

)
=

V

p

(
φ
)

{\displaystyle \mathbf {V} _{p}(\varphi |_{V})=\mathbf {V} _{p}(\varphi )}
(which follows from the linearity of

V

p

{\displaystyle \mathbf {V} _{p}}
).
1. We prove linearity.

(

V

p

∘

ψ

∗

)
(
φ
+
c
ϑ
)

=

V

p

(

ψ

∗

(
φ
+
c
ϑ
)
)

=

V

p

(
(
φ
+
c
ϑ
)
∘
ψ

|

ψ

−
1

(
O
∩
U
)

)

=

V

p

(
φ

|

O
∩
U

∘
ψ

|

ψ

−
1

(
O
∩
U
)

+
c
ϑ

|

O
∩
U

∘
ψ

|

ψ

−
1

(
O
∩
U
)

)

=

V

p

(
φ

|

O
∩
U

∘
ψ

|

ψ

−
1

(
O
∩
U
)

)
+
c

V

p

(
ϑ

|

O
∩
U

∘
ψ

|

ψ

−
1

(
O
∩
U
)

)

=

V

p

(

ψ

∗

(
φ
)
)
+
c

V

p

(

ψ

∗

(
ϑ
)
)

=
(

V

p

∘

ψ

∗

)
(
φ
)
+
c
(

V

p

∘

ψ

∗

)
(
ϑ
)

{\displaystyle {\begin{aligned}(\mathbf {V} _{p}\circ \psi ^{*})(\varphi +c\vartheta )&=\mathbf {V} _{p}(\psi ^{*}(\varphi +c\vartheta ))\\&=\mathbf {V} _{p}((\varphi +c\vartheta )\circ \psi |_{\psi ^{-1}(O\cap U)})\\&=\mathbf {V} _{p}(\varphi |_{O\cap U}\circ \psi |_{\psi ^{-1}(O\cap U)}+c\vartheta |_{O\cap U}\circ \psi |_{\psi ^{-1}(O\cap U)})\\&=\mathbf {V} _{p}(\varphi |_{O\cap U}\circ \psi |_{\psi ^{-1}(O\cap U)})+c\mathbf {V} _{p}(\vartheta |_{O\cap U}\circ \psi |_{\psi ^{-1}(O\cap U)})\\&=\mathbf {V} _{p}(\psi ^{*}(\varphi ))+c\mathbf {V} _{p}(\psi ^{*}(\vartheta ))\\&=(\mathbf {V} _{p}\circ \psi ^{*})(\varphi )+c(\mathbf {V} _{p}\circ \psi ^{*})(\vartheta )\end{aligned}}}
2. We prove the product rule.

(

V

p

∘

ψ

∗

)
(
φ
ϑ
)

=

V

p

(

ψ

∗

(
φ
ϑ
)
)

=

V

p

(
(
φ

|

O
∩
U

∘
ψ

|

ψ

−
1

(
O
∩
U
)

)
(
ϑ

|

O
∩
U

∘
ψ

|

ψ

−
1

(
O
∩
U
)

)
)

=
(
φ

|

O
∩
U

∘
ψ

|

ψ

−
1

(
O
∩
U
)

)
(
p
)

V

p

(
ϑ

|

O
∩
U

∘
ψ

|

ψ

−
1

(
O
∩
U
)

)
+
(
ϑ

|

O
∩
U

∘
ψ

|

ψ

−
1

(
O
∩
U
)

)
(
p
)

V

p

(
φ

|

O
∩
U

∘
ψ

|

ψ

−
1

(
O
∩
U
)

)

=
φ
(
ψ
(
p
)
)

V

p

(

ψ

∗

ϑ
)
+
ϑ
(
ψ
(
p
)
)

V

p

(

ψ

∗

φ
)

=
φ
(
ψ
(
p
)
)
(

V

p

∘

ψ

∗

)
(
ϑ
)
+
ϑ
(
ψ
(
p
)
)
(

V

p

∘

ψ

∗

)
(
φ
)

{\displaystyle {\begin{aligned}(\mathbf {V} _{p}\circ \psi ^{*})(\varphi \vartheta )&=\mathbf {V} _{p}(\psi ^{*}(\varphi \vartheta ))\\&=\mathbf {V} _{p}((\varphi |_{O\cap U}\circ \psi |_{\psi ^{-1}(O\cap U)})(\vartheta |_{O\cap U}\circ \psi |_{\psi ^{-1}(O\cap U)}))\\&=(\varphi |_{O\cap U}\circ \psi |_{\psi ^{-1}(O\cap U)})(p)\mathbf {V} _{p}(\vartheta |_{O\cap U}\circ \psi |_{\psi ^{-1}(O\cap U)})+(\vartheta |_{O\cap U}\circ \psi |_{\psi ^{-1}(O\cap U)})(p)\mathbf {V} _{p}(\varphi |_{O\cap U}\circ \psi |_{\psi ^{-1}(O\cap U)})\\&=\varphi (\psi (p))\mathbf {V} _{p}(\psi ^{*}\vartheta )+\vartheta (\psi (p))\mathbf {V} _{p}(\psi ^{*}\varphi )\\&=\varphi (\psi (p))(\mathbf {V} _{p}\circ \psi ^{*})(\vartheta )+\vartheta (\psi (p))(\mathbf {V} _{p}\circ \psi ^{*})(\varphi )\end{aligned}}}

◻

{\displaystyle \Box }

Theorem 2.22: Let

M

{\displaystyle M}
be a manifold of class

C

n

{\displaystyle {\mathcal {C}}^{n}}
,

n
≥
1

{\displaystyle n\geq 1}
, let

I
⊆

R

{\displaystyle I\subseteq \mathbb {R} }
be an interval, let

y
∈
I

{\displaystyle y\in I}
and let

γ
:
I
→
M

{\displaystyle \gamma :I\to M}
be a differentiable curve of class

C

n

{\displaystyle {\mathcal {C}}^{n}}
. Then

φ
∘
γ

{\displaystyle \varphi \circ \gamma }
is contained in

C

n

(

R

,

R

)

{\displaystyle {\mathcal {C}}^{n}(\mathbb {R} ,\mathbb {R} )}
for every

φ
∈

C

n

(
M
)

{\displaystyle \varphi \in {\mathcal {C}}^{n}(M)}
and

γ

y

′

{\displaystyle \gamma '_{y}}
is a tangent vector of

M

{\displaystyle M}
at

γ
(
y
)

{\displaystyle \gamma (y)}
.
Proof:
1. We show

∀
φ
∈

C

n

(
M
)
:
∘
γ
∈

C

n

(

R

,

R

)

{\displaystyle \forall \varphi \in {\mathcal {C}}^{n}(M):\circ \gamma \in {\mathcal {C}}^{n}(\mathbb {R} ,\mathbb {R} )}

Let

x
∈
I

{\displaystyle x\in I}
be arbitrary, and let

U

{\displaystyle U}
be the set where

φ

{\displaystyle \varphi }
is defined (

U

{\displaystyle U}
is open by the definition of

C

n

(
M
)

{\displaystyle {\mathcal {C}}^{n}(M)}
functions. We choose

(
O
,
ϕ
)

{\displaystyle (O,\phi )}
in the atlas of

M

{\displaystyle M}
such that

γ
(
x
)
∈
O

{\displaystyle \gamma (x)\in O}
. Then the function

(
φ
∘
γ
)

|

γ

−
1

(
O
∩
U
)
∩
I

=
φ
∘

ϕ

−
1

∘
ϕ
∘
γ

|

γ

−
1

(
O
∩
U
)
∩
I

{\displaystyle (\varphi \circ \gamma )|_{\gamma ^{-1}(O\cap U)\cap I}=\varphi \circ \phi ^{-1}\circ \phi \circ \gamma |_{\gamma ^{-1}(O\cap U)\cap I}}
is contained in

C

n

(

R

,

R

)

{\displaystyle {\mathcal {C}}^{n}(\mathbb {R} ,\mathbb {R} )}
as the composition of two

n

{\displaystyle n}
times continuously differentiable (or continuous if

n
=
0

{\displaystyle n=0}
) functions.
Thus,

φ
∘
γ

{\displaystyle \varphi \circ \gamma }
is

n

{\displaystyle n}
times continuously differentiable (or continuous if

n
=
0

{\displaystyle n=0}
) at every point, and hence

n

{\displaystyle n}
times continuously differentiable (or continuous if

n
=
0

{\displaystyle n=0}
).
2. We show that

γ

y

′

∈

T

γ
(
y
)

M

{\displaystyle \gamma '_{y}\in T_{\gamma (y)}M}
in three steps:
Let

φ
,
ϑ
∈

C

n

(
M
)

{\displaystyle \varphi ,\vartheta \in {\mathcal {C}}^{n}(M)}
and

c
∈

R

{\displaystyle c\in \mathbb {R} }
.
2.1 We show linearity.
We have:

γ

y

′

(
φ
+
c
ϑ
)

=
(
(
φ
+
c
ϑ
)
∘
γ

)
′

(
y
)

=
(
φ
∘
γ
+
c
ϑ
∘
γ

)
′

(
y
)

=
(
φ
∘
γ

)
′

(
y
)
+
c
(
ϑ
∘
γ

)
′

(
y
)

=

γ

y

′

(
φ
)
+
c

γ

y

′

(
ϑ
)

{\displaystyle {\begin{aligned}\gamma '_{y}(\varphi +c\vartheta )&=((\varphi +c\vartheta )\circ \gamma )'(y)\\&=(\varphi \circ \gamma +c\vartheta \circ \gamma )'(y)\\&=(\varphi \circ \gamma )'(y)+c(\vartheta \circ \gamma )'(y)\\&=\gamma '_{y}(\varphi )+c\gamma '_{y}(\vartheta )\end{aligned}}}
2.2 We prove the product rule.

γ

y

′

(
φ
ϑ
)

=
(
(
φ
ϑ
)
∘
γ

)
′

(
y
)

=
(
(
φ
∘
γ
)
(
ϑ
∘
γ
)

)
′

(
y
)

=
(
φ
∘
γ
)
(
y
)
(
ϑ
∘
γ

)
′

(
y
)
+
(
ϑ
∘
γ
)
(
y
)
(
φ
∘
γ

)
′

(
y
)

=
φ
(
γ
(
y
)
)

γ

y

′

(
ϑ
)
+
ϑ
(
γ
(
y
)
)

γ

y

′

(
φ
)

{\displaystyle {\begin{aligned}\gamma '_{y}(\varphi \vartheta )&=((\varphi \vartheta )\circ \gamma )'(y)\\&=((\varphi \circ \gamma )(\vartheta \circ \gamma ))'(y)\\&=(\varphi \circ \gamma )(y)(\vartheta \circ \gamma )'(y)+(\vartheta \circ \gamma )(y)(\varphi \circ \gamma )'(y)\\&=\varphi (\gamma (y))\gamma '_{y}(\vartheta )+\vartheta (\gamma (y))\gamma '_{y}(\varphi )\end{aligned}}}
2.3 It follows from the definition of

γ

y

′

{\displaystyle \gamma '_{y}}
that

γ

y

′

(
φ
)

{\displaystyle \gamma '_{y}(\varphi )}
is equal to zero if

φ

{\displaystyle \varphi }
is not defined at

γ
(
y
)

{\displaystyle \gamma (y)}
.

◻

{\displaystyle \Box }

== Linearity of the differential for Ck(M), product, quotient and chain rules ==
In this subsection, we will first prove linearity and product rule for functions from a manifold to

R

{\displaystyle \mathbb {R} }
.

Proof:
1. We show that

φ
+
c
ϑ
∈

C

k

(
M
)

{\displaystyle \varphi +c\vartheta \in {\mathcal {C}}^{k}(M)}
.
Let

U

{\displaystyle U}
be the (open as intersection of two open sets) set on which

φ
+
c
ϑ

{\displaystyle \varphi +c\vartheta }
is defined, and let

(
O
,
ϕ
)

{\displaystyle (O,\phi )}
be contained in the atlas of

M

{\displaystyle M}
. The function

(
φ
+
c
ϑ
)

|

O
∩
U

∘
ϕ

|

O
∩
U

−
1

=
φ

|

O
∩
U

∘
ϕ

|

O
∩
U

−
1

+
c
ϑ

|

O
∩
U

∘
ϕ

|

O
∩
U

−
1

{\displaystyle (\varphi +c\vartheta )|_{O\cap U}\circ \phi |_{O\cap U}^{-1}=\varphi |_{O\cap U}\circ \phi |_{O\cap U}^{-1}+c\vartheta |_{O\cap U}\circ \phi |_{O\cap U}^{-1}}
is contained in

C

n

(

R

d

,

R

)

{\displaystyle {\mathcal {C}}^{n}(\mathbb {R} ^{d},\mathbb {R} )}
as the linear combination of two

C

n

(

R

d

,

R

)

{\displaystyle {\mathcal {C}}^{n}(\mathbb {R} ^{d},\mathbb {R} )}
functions.
2. We show that

d
(
φ
+
c
ϑ
)
=
d
φ
+
c
d
ϑ

{\displaystyle d(\varphi +c\vartheta )=d\varphi +cd\vartheta }
.
For all

p
∈
M

{\displaystyle p\in M}
and

V

p

∈

T

p

M

{\displaystyle \mathbf {V} _{p}\in T_{p}M}
, we have:

d
(
φ
+
c
ϑ

)

p

(

V

p

)
=

V

p

(
φ
+
c
ϑ
)
=

V

p

(
φ
)
+
c

V

p

(
ϑ
)
=
d

φ

p

(

V

p

)
+
c
d

ϑ

p

(

V

p

)

{\displaystyle d(\varphi +c\vartheta )_{p}(\mathbf {V} _{p})=\mathbf {V} _{p}(\varphi +c\vartheta )=\mathbf {V} _{p}(\varphi )+c\mathbf {V} _{p}(\vartheta )=d\varphi _{p}(\mathbf {V} _{p})+cd\vartheta _{p}(\mathbf {V} _{p})}

◻

{\displaystyle \Box }
Remark 2.24: This also shows that for all

φ
∈

C

n

(
M
)

{\displaystyle \varphi \in {\mathcal {C}}^{n}(M)}
,

d

φ

p

∈

T

p

M

∗

{\displaystyle d\varphi _{p}\in T_{p}M^{*}}
.

Proof:
1. We show that

φ
ϑ
∈

C

k

(
M
)

{\displaystyle \varphi \vartheta \in {\mathcal {C}}^{k}(M)}
.
Let

U

{\displaystyle U}
be the (open as intersection of two open sets) set on which

φ
ϑ

{\displaystyle \varphi \vartheta }
is defined, and let

(
O
,
ϕ
)

{\displaystyle (O,\phi )}
be contained in the atlas of

M

{\displaystyle M}
. The function

(
φ
ϑ
)

|

O
∩
U

∘
ϕ

|

O
∩
U

−
1

=
φ

|

O
∩
U

∘
ϕ

|

O
∩
U

−
1

ϑ

|

O
∩
U

∘
ϕ

|

O
∩
U

−
1

{\displaystyle (\varphi \vartheta )|_{O\cap U}\circ \phi |_{O\cap U}^{-1}=\varphi |_{O\cap U}\circ \phi |_{O\cap U}^{-1}\vartheta |_{O\cap U}\circ \phi |_{O\cap U}^{-1}}
is contained in

C

n

(

R

d

,

R

)

{\displaystyle {\mathcal {C}}^{n}(\mathbb {R} ^{d},\mathbb {R} )}
as the product of two

C

n

(

R

d

,

R

)

{\displaystyle {\mathcal {C}}^{n}(\mathbb {R} ^{d},\mathbb {R} )}
functions.
2. We show that

d
(
φ
ϑ
)
=
φ
d
ϑ
+
ϑ
d
φ

{\displaystyle d(\varphi \vartheta )=\varphi d\vartheta +\vartheta d\varphi }
.
For all

p
∈
M

{\displaystyle p\in M}
and

V

p

∈

T

p

M

{\displaystyle \mathbf {V} _{p}\in T_{p}M}
, we have:

d
(
φ
ϑ

)

p

(

V

p

)
=

V

p

(
φ
ϑ
)
=
φ
(
p
)

V

p

(
ϑ
)
+
ϑ
(
p
)

V

p

(
φ
)
=
φ
(
p
)
d

ϑ

p

+
ϑ
(
p
)
d

φ

p

{\displaystyle d(\varphi \vartheta )_{p}(\mathbf {V} _{p})=\mathbf {V} _{p}(\varphi \vartheta )=\varphi (p)\mathbf {V} _{p}(\vartheta )+\vartheta (p)\mathbf {V} _{p}(\varphi )=\varphi (p)d\vartheta _{p}+\vartheta (p)d\varphi _{p}}

◻

{\displaystyle \Box }

Proof:
1. We show that

φ
ϑ

∈

C

n

(
M
)

{\displaystyle {\frac {\varphi }{\vartheta }}\in {\mathcal {C}}^{n}(M)}
:
Let

U

{\displaystyle U}
be the (open as the intersection of two open set) set on which

φ
ϑ

{\displaystyle {\frac {\varphi }{\vartheta }}}
is defined, and let

(
O
,
ϕ
)

{\displaystyle (O,\phi )}
be in the atlas of

M

{\displaystyle M}
such that

O
∩
U
≠
∅

{\displaystyle O\cap U\neq \emptyset }
. The function

φ
ϑ

|

O
∩
U

∘
ϕ

|

O
∩
U

−
1

=

φ

|

O
∩
U

∘
ϕ

|

O
∩
U

−
1

ϑ

|

O
∩
U

∘
ϕ

|

O
∩
U

−
1

{\displaystyle {\frac {\varphi }{\vartheta }}{\big |}_{O\cap U}\circ \phi |_{O\cap U}^{-1}={\frac {\varphi |_{O\cap U}\circ \phi |_{O\cap U}^{-1}}{\vartheta |_{O\cap U}\circ \phi |_{O\cap U}^{-1}}}}
is contained in

C

n

(

R

d

,

R

)

{\displaystyle {\mathcal {C}}^{n}(\mathbb {R} ^{d},\mathbb {R} )}
as the quotient of two

C

n

(

R

d

,

R

)

{\displaystyle {\mathcal {C}}^{n}(\mathbb {R} ^{d},\mathbb {R} )}
from which the function in the denominator vanishes nowhere.
2. We show that

d

(

φ
ϑ

)

=

ϑ
d
φ
−
φ
d
ϑ

ϑ

2

{\displaystyle d\left({\frac {\varphi }{\vartheta }}\right)={\frac {\vartheta d\varphi -\varphi d\vartheta }{\vartheta ^{2}}}}
:
Choosing

φ

{\displaystyle \varphi }
as the constant one function, we obtain from 1. that the function

1
ϑ

{\displaystyle {\frac {1}{\vartheta }}}
is in

C

n

(
M
)

{\displaystyle {\mathcal {C}}^{n}(M)}
. Hence follows from the product rule:

0
=
d

(

ϑ

1
ϑ

)

=
ϑ
d

(

1
ϑ

)

+

1
ϑ

d
ϑ

{\displaystyle 0=d\left(\vartheta {\frac {1}{\vartheta }}\right)=\vartheta d\left({\frac {1}{\vartheta }}\right)+{\frac {1}{\vartheta }}d\vartheta }
which, through equivalent transformations, can be transformed to

d

(

1
ϑ

)

=
−

d
ϑ

ϑ

2

{\displaystyle d\left({\frac {1}{\vartheta }}\right)=-{\frac {d\vartheta }{\vartheta ^{2}}}}
From this and from the product rule we obtain:

d

(

φ

1
ϑ

)

=

1
ϑ

d
φ
−

φ
d
ϑ

ϑ

2

=

ϑ
d
φ
−
φ
d
ϑ

ϑ

2

{\displaystyle d\left(\varphi {\frac {1}{\vartheta }}\right)={\frac {1}{\vartheta }}d\varphi -{\frac {\varphi d\vartheta }{\vartheta ^{2}}}={\frac {\vartheta d\varphi -\varphi d\vartheta }{\vartheta ^{2}}}}

◻

{\displaystyle \Box }

Proof:

φ
∘
ψ
=

ψ

∗

φ

{\displaystyle \varphi \circ \psi =\psi ^{*}\varphi }
is differentiable of class

C

n

{\displaystyle {\mathcal {C}}^{n}}
; this is what lemma 2.17 says.
2. We prove that

d
(

ψ

∗

φ

)

p

=
d

φ

ψ
(
p
)

∘
d

ψ

p

{\displaystyle d(\psi ^{*}\varphi )_{p}=d\varphi _{\psi (p)}\circ d\psi _{p}}
.
Let

V

p

∈

T

p

M

{\displaystyle \mathbf {V} _{p}\in T_{p}M}
. Then we have:

(
d

φ

ψ
(
p
)

∘
d

ψ

p

)
(

V

p

)

=
d

φ

ψ
(
p
)

(
d

ψ

p

(

V

p

)
)

=
d

φ

ψ
(
p
)

(

V

p

∘
ψ
)

=
(

V

p

∘

ψ

∗

)
(
φ
)

=

V

p

(

ψ

∗

(
φ
)
)

=

V

p

(
φ
∘
ψ
)

=
d
(
φ
∘
ψ

)

p

(

V

p

)

{\displaystyle {\begin{aligned}(d\varphi _{\psi (p)}\circ d\psi _{p})(\mathbf {V} _{p})&=d\varphi _{\psi (p)}(d\psi _{p}(\mathbf {V} _{p}))\\&=d\varphi _{\psi (p)}(\mathbf {V} _{p}\circ \psi )\\&=(\mathbf {V} _{p}\circ \psi ^{*})(\varphi )\\&=\mathbf {V} _{p}(\psi ^{*}(\varphi ))\\&=\mathbf {V} _{p}(\varphi \circ \psi )\\&=d(\varphi \circ \psi )_{p}(\mathbf {V} _{p})\end{aligned}}}

◻

{\displaystyle \Box }
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

{\displaystyle M,N,K}
be three manifolds, and let

ψ
:
M
→
N

{\displaystyle \psi :M\to N}
and

χ
:
N
→
K

{\displaystyle \chi :N\to K}
be two functions differentiable of class

C

k

{\displaystyle {\mathcal {C}}^{k}}
. Then

(
χ
∘
ψ

)

∗

=

ψ

∗

∘

χ

∗

{\displaystyle (\chi \circ \psi )^{*}=\psi ^{*}\circ \chi ^{*}}
Proof:
Let

φ
∈

C

k

(
K
)

{\displaystyle \varphi \in {\mathcal {C}}^{k}(K)}
. Then we have:

(
χ
∘
ψ

)

∗

(
φ
)

=
φ
∘
(
χ
∘
ψ
)

=
(
φ
∘
χ
)
∘
ψ

=

χ

∗

(
φ
)
∘
ψ

=

ψ

∗

(

χ

∗

(
φ
)
)

{\displaystyle {\begin{aligned}(\chi \circ \psi )^{*}(\varphi )&=\varphi \circ (\chi \circ \psi )\\&=(\varphi \circ \chi )\circ \psi \\&=\chi ^{*}(\varphi )\circ \psi \\&=\psi ^{*}(\chi ^{*}(\varphi ))\end{aligned}}}

◻

{\displaystyle \Box }

Proof:
1. We prove that

χ
∘
ψ

{\displaystyle \chi \circ \psi }
is differentiable of class

C

k

{\displaystyle {\mathcal {C}}^{k}}
.
Let

(
O
,
ϕ
)

{\displaystyle (O,\phi )}
be contained in the atlas of

M

{\displaystyle M}
and let

(
U
,
θ
)

{\displaystyle (U,\theta )}
be contained in the atlas of

K

{\displaystyle K}
such that

O
∩

ψ

−
1

(

χ

−
1

(
U
)
)
≠
∅

{\displaystyle O\cap \psi ^{-1}(\chi ^{-1}(U))\neq \emptyset }
, and let

x
∈

ϕ

−
1

(
O
)
∩

ψ

−
1

(

χ

−
1

(
U
)
)

{\displaystyle x\in \phi ^{-1}(O)\cap \psi ^{-1}(\chi ^{-1}(U))}
be arbitrary. We choose

(
V
,
η
)

{\displaystyle (V,\eta )}
in the atlas of

N

{\displaystyle N}
such that

ψ
(
ϕ
(
x
)
)
∈
V

{\displaystyle \psi (\phi (x))\in V}
.
We have

ψ
(
ϕ
(
x
)
)
∈
V
∩

χ

−
1

(
U
)

{\displaystyle \psi (\phi (x))\in V\cap \chi ^{-1}(U)}
; indeed,

ψ
(
ϕ
(
x
)
)
∈
V

{\displaystyle \psi (\phi (x))\in V}
due to the choice of

(
V
,
ϕ
)

{\displaystyle (V,\phi )}
and

ψ
(
ϕ
(
x
)
)
∈

χ

−
1

(
U
)

{\displaystyle \psi (\phi (x))\in \chi ^{-1}(U)}
because

ϕ
(
x
)
∈

ψ

−
1

(

χ

−
1

(
U
)
)

{\displaystyle \phi (x)\in \psi ^{-1}(\chi ^{-1}(U))}
. Further, we choose

W
:=
O
∩

ψ

−
1

(
V
∩

χ

−
1

(
U
)
)

{\displaystyle W:=O\cap \psi ^{-1}(V\cap \chi ^{-1}(U))}
. Then the function

θ

−
1

∘
(
χ
∘
ψ
)
∘
ϕ

|

W

−
1

=

θ

−
1

∘
χ
∘

η

−
1

∘
η
∘
ψ

|

W

∘
ϕ

|

W

−
1

{\displaystyle \theta ^{-1}\circ (\chi \circ \psi )\circ \phi |_{W}^{-1}=\theta ^{-1}\circ \chi \circ \eta ^{-1}\circ \eta \circ \psi |_{W}\circ \phi |_{W}^{-1}}
is contained in

C

n

(

R

d

,

R

d

)

{\displaystyle {\mathcal {C}}^{n}(\mathbb {R} ^{d},\mathbb {R} ^{d})}
as the composition of two

C

n

(

R

d

,

R

d

)

{\displaystyle {\mathcal {C}}^{n}(\mathbb {R} ^{d},\mathbb {R} ^{d})}
functions.
Thus,

θ

−
1

∘
(
χ
∘
ψ
)
∘
ϕ

|

O
∩

ψ

−
1

(

χ

−
1

(
U
)
)

−
1

{\displaystyle \theta ^{-1}\circ (\chi \circ \psi )\circ \phi |_{O\cap \psi ^{-1}(\chi ^{-1}(U))}^{-1}}
is

n

{\displaystyle n}
times continuously differentiable (or continuous if

n
=
0

{\displaystyle n=0}
) at every point, and thus

n

{\displaystyle n}
times continuously differentiable (or continuous if

n
=
0

{\displaystyle n=0}
).
2. We prove that

∀
p
∈
M
:
d
(
χ
∘
ψ

)

p

=
d

χ

ψ
(
p
)

∘
d

ψ

p

{\displaystyle \forall p\in M:d(\chi \circ \psi )_{p}=d\chi _{\psi (p)}\circ d\psi _{p}}
.
For all

p
∈
M

{\displaystyle p\in M}
and

V

p

∈

T

p

M

{\displaystyle \mathbf {V} _{p}\in T_{p}M}
, we have:

(
d

χ

ψ
(
p
)

∘
d

ψ

p

)
(

V

p

)

=
d

χ

ψ
(
p
)

(
d

ψ

p

(

V

p

)
)

=
d

χ

ψ
(
p
)

(

V

p

∘

ψ

∗

)

=

V

p

∘

ψ

∗

∘

χ

∗

=
theorem 2.26

V

p

∘
(
χ
∘
ψ

)

∗

=
d
(
χ
∘
ψ

)

p

(

V

p

)

{\displaystyle {\begin{aligned}(d\chi _{\psi (p)}\circ d\psi _{p})(\mathbf {V} _{p})&=d\chi _{\psi (p)}(d\psi _{p}(\mathbf {V} _{p}))\\&=d\chi _{\psi (p)}(\mathbf {V} _{p}\circ \psi ^{*})\\&=\mathbf {V} _{p}\circ \psi ^{*}\circ \chi ^{*}\\&{\overset {\text{theorem 2.26}}{=}}\mathbf {V} _{p}\circ (\chi \circ \psi )^{*}\\&=d(\chi \circ \psi )_{p}(\mathbf {V} _{p})\end{aligned}}}

◻

{\displaystyle \Box }

Proof:
1. Among another thing, theorem 2.22 states that

φ
∘
γ

{\displaystyle \varphi \circ \gamma }
is contained in

C

n

(

R

,

R

)

{\displaystyle {\mathcal {C}}^{n}(\mathbb {R} ,\mathbb {R} )}
.
2. We show that

(
φ
∘
γ

)
′

(
y
)
=
d

φ

γ
(
y
)

(

γ

y

′

)

{\displaystyle (\varphi \circ \gamma )'(y)=d\varphi _{\gamma (y)}(\gamma '_{y})}
:

d

φ

γ
(
y
)

(

γ

y

′

)

=

γ

y

′

(
φ
)

=
(
φ
∘
γ

)
′

(
y
)

{\displaystyle {\begin{aligned}d\varphi _{\gamma (y)}(\gamma '_{y})&=\gamma '_{y}(\varphi )\\&=(\varphi \circ \gamma )'(y)\end{aligned}}}

◻

{\displaystyle \Box }

== Intuition behind the tangent space ==
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

{\displaystyle f:\mathbb {R} \to \mathbb {R} }
.
We start by proving the following lemma from linear algebra:

Proof:
We only prove that

T

{\displaystyle T}
is a vector space isomorphism; that

S

{\displaystyle S}
and

L

{\displaystyle L}
are also vector space isomorphisms will follow in exactly the same way.
From

L
∘
S
∘
T
=

Id

V

{\displaystyle L\circ S\circ T={\text{Id}}_{\mathbf {V} }}
and

T
∘
L
∘
S
=

Id

W

{\displaystyle T\circ L\circ S={\text{Id}}_{\mathbf {W} }}
follows that

L
∘
S

{\displaystyle L\circ S}
is the inverse function of

T

{\displaystyle T}
.

◻

{\displaystyle \Box }

== Sources ==
Torres del Castillo, Gerardo (2012). Differentiable Manifolds. Boston: Birkhäuser. ISBN 978-0-8176-8271-2.