§ 2. Differential Forms of Degree 1
171
Integrating by parts the second term, it becomes
ω(tx; h)t
1
0
−
1
0
d
dt
[ω(tx; h)] .tdt = ω(x; h) −
1
0
. . . .
But
d
dt
ω(tx; h) =
d
ds
ω(tx + sx; h)
s=0
= ω
(tx; x, h) ,
so that finally,
dF (x; h) = ω(x; h) +
1
0
[ω
(tx; h, x) − ω
(tx; x, h)] dt .
(5.13)
This involves what is called the exterior derivative
dω(x; h, k) = ω
(x; h, k) − ω
(x; k, h)
(5.14)
of the form ω ; for given x, it is an alternating or antisymmetric bilinear form
of the vectors h, k. Without going further into this topic which we will return
to later, we find that the formula
dF (x; h) = ω(x; h) −
1
0
dω(tx; x, h)dt
(5.13’)
holds without any assumptions on ω. On the other hand, setting
ω(x; h) = p i (x)h
i ,
it is easy to first calculate
ω
(x; h, k) = dp j (x; h)k
j = D i p j (x)h
i k
j ,
ω
(x; k, h) = dp i (x; k)h
i = D j p i (x)h
i k
j ,
and then
dω(x; h, k) = (D i p j − D j p i ) h
i k
j .
(5.15)
This formula shows that
ω is closed ⇐⇒ dω = 0 .
(5.16)
Relation (13’) then reduces to
dF (x; h) = ω(x; h) ,
ending the proof.
Relation
ω
(x; h, k) = ω
(x; k, h)
is still well-defined in Banach spaces where calculations in terms of coordinates is no longer possible; it then serves as a definition for closed forms.
Exercise. Suppose ω = dF ; calculate ω
(x; h, k) directly, without coordinates, and show that (16) reduces to the formula d
2 /dsdt = d
2 /dtds.
171
Integrating by parts the second term, it becomes
ω(tx; h)t
1
0
−
1
0
d
dt
[ω(tx; h)] .tdt = ω(x; h) −
1
0
. . . .
But
d
dt
ω(tx; h) =
d
ds
ω(tx + sx; h)
s=0
= ω
(tx; x, h) ,
so that finally,
dF (x; h) = ω(x; h) +
1
0
[ω
(tx; h, x) − ω
(tx; x, h)] dt .
(5.13)
This involves what is called the exterior derivative
dω(x; h, k) = ω
(x; h, k) − ω
(x; k, h)
(5.14)
of the form ω ; for given x, it is an alternating or antisymmetric bilinear form
of the vectors h, k. Without going further into this topic which we will return
to later, we find that the formula
dF (x; h) = ω(x; h) −
1
0
dω(tx; x, h)dt
(5.13’)
holds without any assumptions on ω. On the other hand, setting
ω(x; h) = p i (x)h
i ,
it is easy to first calculate
ω
(x; h, k) = dp j (x; h)k
j = D i p j (x)h
i k
j ,
ω
(x; k, h) = dp i (x; k)h
i = D j p i (x)h
i k
j ,
and then
dω(x; h, k) = (D i p j − D j p i ) h
i k
j .
(5.15)
This formula shows that
ω is closed ⇐⇒ dω = 0 .
(5.16)
Relation (13’) then reduces to
dF (x; h) = ω(x; h) ,
ending the proof.
Relation
ω
(x; h, k) = ω
(x; k, h)
is still well-defined in Banach spaces where calculations in terms of coordinates is no longer possible; it then serves as a definition for closed forms.
Exercise. Suppose ω = dF ; calculate ω
(x; h, k) directly, without coordinates, and show that (16) reduces to the formula d
2 /dsdt = d
2 /dtds.