§ 4. Differential Manifolds
271
components, and case (a) holds for each of them. Algebraic topology (duality theory) has long resolved and generalized this problem: the relation
between the topology of a submanifold and that of its complement.
In what follows, case (a) will be assumed to always hold. ∂Ω is then said
to be the boundary of Ω, a more restrictive expression than “ border ”, and
that Ω is an open set with boundary in X. It may then be assumed that ξ
1 > 0
on U ∩ Ω, if need be by replacing the function ξ
1 by −ξ
1 and by carrying
out an odd permutation on the other coordinates, operations that leave the
orientation of the chart considered and its cubic character unchanged. Then
U ∩ Ω = U + .
Having said that, the compact set Ω ∪ ∂Ω can be covered by a finite
number of cubic charts (U, ϕ) compatible with the orientation of X and such
that U ∩Ω = U or U + . Let G be the union of these charts; it is open in X. By
lemma 4 above, in G, ω can be decomposed into a sum of differential forms
whose supports are compact sets contained in the charts considered. By the
additive formula (15), it, therefore, suffices to prove Stokes’ formula for these
forms. In other words, the support of ω can be assumed to be compact and
contained in one of these cubic charts (U, ϕ).
Let us first consider the case U ∩ Ω = U . As the support of ω does not
meet ∂Ω, the left hand side of Stokes’ formula is zero. To show that the same
is true for the right hand side, consider the open cube ϕ(U ) = {|ξ
i
| < 1} of
R
n . Thus ω is replaced by a form
p i (ξ)dξ
1
∧ . . . ∧
dξ i ∧ . . . ∧ dξ
n ,
(18.2)
where the accent indicates that dξ
i must be omitted, and dω by
(−1)
i−1 D i p i (ξ)dξ
1
∧ . . . ∧ dξ
n .
(18.3)
To find the Lebesgue-Fubini integral of the i
th term of (3), we can first integrate with respect to ξ
i , which gives the variation over ] − 1, +1[, up to sign,
of the function
t −→ p i
ξ
1 , . . . , ξ
i−1 , t, ξ
i+1 , . . . , ξ
n
;
(18.4)
as the support of ω is compact in the open set ϕ(U ), this function vanishes
if t is sufficiently near 1 or −1. The result follows.
The U ∩ Ω = U + remains to be proved. Once again, there is a form (2)
on ϕ(U ), but now we integrate its exterior derivative (3) over the open set
ξ
1 > 0, so that integration with respect to the variable ξ
i must be extended
to the interval ] − 1, +1[ if i = 1 and to the interval ]0, 1[ if i = 1. If i = 1, the
result is zero as in case (a). If, however, i = 1, the result is the variation of (4)
over ]0, 1[ ; the support of dω being compact in ϕ(U ), function (4) vanishes
for t in the neighbourhood of 1 as in case (a), but not in the neighbourhood
of ξ
1 = 0 ; so, applying the FT,
271
components, and case (a) holds for each of them. Algebraic topology (duality theory) has long resolved and generalized this problem: the relation
between the topology of a submanifold and that of its complement.
In what follows, case (a) will be assumed to always hold. ∂Ω is then said
to be the boundary of Ω, a more restrictive expression than “ border ”, and
that Ω is an open set with boundary in X. It may then be assumed that ξ
1 > 0
on U ∩ Ω, if need be by replacing the function ξ
1 by −ξ
1 and by carrying
out an odd permutation on the other coordinates, operations that leave the
orientation of the chart considered and its cubic character unchanged. Then
U ∩ Ω = U + .
Having said that, the compact set Ω ∪ ∂Ω can be covered by a finite
number of cubic charts (U, ϕ) compatible with the orientation of X and such
that U ∩Ω = U or U + . Let G be the union of these charts; it is open in X. By
lemma 4 above, in G, ω can be decomposed into a sum of differential forms
whose supports are compact sets contained in the charts considered. By the
additive formula (15), it, therefore, suffices to prove Stokes’ formula for these
forms. In other words, the support of ω can be assumed to be compact and
contained in one of these cubic charts (U, ϕ).
Let us first consider the case U ∩ Ω = U . As the support of ω does not
meet ∂Ω, the left hand side of Stokes’ formula is zero. To show that the same
is true for the right hand side, consider the open cube ϕ(U ) = {|ξ
i
| < 1} of
R
n . Thus ω is replaced by a form
p i (ξ)dξ
1
∧ . . . ∧
dξ i ∧ . . . ∧ dξ
n ,
(18.2)
where the accent indicates that dξ
i must be omitted, and dω by
(−1)
i−1 D i p i (ξ)dξ
1
∧ . . . ∧ dξ
n .
(18.3)
To find the Lebesgue-Fubini integral of the i
th term of (3), we can first integrate with respect to ξ
i , which gives the variation over ] − 1, +1[, up to sign,
of the function
t −→ p i
ξ
1 , . . . , ξ
i−1 , t, ξ
i+1 , . . . , ξ
n
;
(18.4)
as the support of ω is compact in the open set ϕ(U ), this function vanishes
if t is sufficiently near 1 or −1. The result follows.
The U ∩ Ω = U + remains to be proved. Once again, there is a form (2)
on ϕ(U ), but now we integrate its exterior derivative (3) over the open set
ξ
1 > 0, so that integration with respect to the variable ξ
i must be extended
to the interval ] − 1, +1[ if i = 1 and to the interval ]0, 1[ if i = 1. If i = 1, the
result is zero as in case (a). If, however, i = 1, the result is the variation of (4)
over ]0, 1[ ; the support of dω being compact in ϕ(U ), function (4) vanishes
for t in the neighbourhood of 1 as in case (a), but not in the neighbourhood
of ξ
1 = 0 ; so, applying the FT,