proof of general Stokes theorem

proof of general Stokes theorem

We divide the proof in several steps.

Step One.

Suppose

M

=

(

0

,

1

]

×

(

0

,

1

)

n

-

1

and

ω

⁢

(

x

1

,

…

,

x

n

)

=

f

⁢

(

x

1

,

…

,

x

n

)

⁢

d

⁢

x

1

∧

⋯

∧

d

⁢

x

j

^

∧

⋯

∧

d

⁢

x

n

(i.e. the term

d

⁢

x

j

is missing).
Hence we have

d

⁢

ω

⁢

(

x

1

,

…

,

x

n

)

=

(

∂

⁡

f

∂

⁡

x

1

⁢

d

⁢

x

1

+

⋯

+

∂

⁡

f

∂

⁡

x

n

⁢

d

⁢

x

n

)

∧

d

⁢

x

1

∧

⋯

∧

d

⁢

x

j

^

∧

⋯

∧

d

⁢

x

n

=

(

-

1

)

j

-

1

⁢

∂

⁡

f

∂

⁡

x

j

⁢

d

⁢

x

1

∧

⋯

∧

d

⁢

x

n

and from the definition of

integral

on a

manifold

we get

∫

M

𝑑

ω

=

∫

0

1

⋯

⁢

∫

0

1

(

-

1

)

j

-

1

⁢

∂

⁡

f

∂

⁡

x

j

⁢

𝑑

x

1

⁢

⋯

⁢

𝑑

x

n

.

From the

fundamental theorem of Calculus

we get

∫

M

𝑑

ω

=

(

-

1

)

j

-

1

⁢

∫

0

1

⋯

⁢

∫

0

1

^

⁢

⋯

⁢

∫

0

1

f

⁢

(

x

1

,

…

,

1

,

…

,

x

n

)

-

f

⁢

(

x

1

,

…

,

0

,

…

,

x

n

)

⁢

d

⁢

x

1

⁢

⋯

⁢

d

⁢

x

j

^

⁢

⋯

⁢

d

⁢

x

n

.

Since

ω

and hence

f

have compact support in

M

we obtain

∫

M

𝑑

ω

=

{

∫

0

1

⋯

⁢

∫

0

1

f

⁢

(

1

,

x

2

,

…

,

x

n

)

⁢

𝑑

x

2

⁢

⋯

⁢

𝑑

x

n

if

j

=

1

0

if

j

>

1

.

On the other hand we notice that

∫

∂

⁡

M

ω

is to be understood as

∫

∂

⁡

M

i

*

⁢

ω

where

i

:

∂

⁡

M

→

M

is the

inclusion map

.
Hence it is

trivial

to verify that when

j

≠

1

then

i

*

⁢

ω

=

0

while if

j

=

1

it holds

i

*

⁢

ω

⁢

(

x

)

=

f

⁢

(

1

,

x

2

,

…

,

x

n

)

⁢

d

⁢

x

2

∧

…

∧

d

⁢

x

n

and hence, as wanted

∫

∂

⁡

M

i

*

⁢

ω

=

∫

0

1

⋯

⁢

∫

0

1

f

⁢

(

1

,

x

2

,

…

,

x

n

)

⁢

𝑑

x

2

⁢

⋯

⁢

𝑑

x

n

.

Step Two.

Suppose now that

M

=

(

0

,

1

]

×

(

0

,

1

)

n

-

1

and let

ω

be any

differential form

.
We can always write

ω

⁢

(

x

)

=

∑

j

f

j

⁢

(

x

)

⁢

d

⁢

x

1

∧

⋯

∧

d

⁢

x

j

^

∧

⋯

∧

d

⁢

x

n

and by the

additivity

of the integral we can reduce ourself to the previous case.

Step Three.

When

M

=

(

0

,

1

)

n

we could follow the proof as in the first case
and end up with

∫

M

𝑑

ω

=

0

while, in fact,

∂

⁡

M

=

∅

.

Step Four.

Consider now the general case.

First of all we consider an oriented atlas

(

U

i

,

ϕ

i

)

such that either

U

i

is the cube

(

0

,

1

]

×

(

0

,

1

)

n

-

1

or

U

i

=

(

0

,

1

)

n

.
This is always possible. In fact given any

open set

U

in

[

0

,

+

∞

)

×

ℝ

n

-

1

and a point

x

∈

U

up to translations and rescaling it is possible
to find a “cubic”

neighbourhood

of

x

contained

in

U

.

Then consider a

partition of unity

α

i

for this atlas.

From the properties of the integral on manifolds we have

∫

M

𝑑

ω

=

∑

i

∫

U

i

α

i

⁢

ϕ

*

⁢

𝑑

ω

=

∑

i

∫

U

i

α

i

⁢

d

⁢

(

ϕ

*

⁢

ω

)

=

∑

i

∫

U

i

d

⁢

(

α

i

⋅

ϕ

*

⁢

ω

)

-

∑

i

∫

U

i

(

d

⁢

α

i

)

∧

(

ϕ

*

⁢

ω

)

.

The second integral in the last equality is zero since

∑

i

d

⁢

α

i

=

d

⁢

∑

i

α

i

=

0

, while applying the previous steps to the first integral we have

∫

M

𝑑

ω

=

∑

i

∫

∂

⁡

U

i

α

i

⋅

ϕ

*

⁢

ω

.

On the other hand, being

(

∂

⁡

U

i

,

ϕ

|

∂

U

i

)

an oriented atlas for

∂

⁡

M

and being

α

i

|

∂

U

i

a partition of unity, we have

∫

∂

⁡

M

ω

=

∑

i

∫

∂

⁡

U

i

α

i

⁢

ϕ

*

⁢

ω

and the

theorem

is proved.

Title

proof of general Stokes theorem

Canonical name

ProofOfGeneralStokesTheorem

Date of creation

2013-03-22 13:41:43

Last modified on

2013-03-22 13:41:43

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

9

Author

paolini (1187)

Entry type

Proof

Classification

msc 58C35

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