270
IX – Multivariate Differential and Integral Calculus
obviously provided the edge of Ω is oriented properly and a further condition
is imposed; its intuitive meaning is that in the neighbourhood of any of its
border points, the open set Ω is located “ on one side ” of it. Corollary : if X
is an n-dimensional compact manifold, then
X
dω = 0
for any form ω of degree n − 1 on X, since ∂X is empty.
Before moving on to the proof, we make some remarks about what happens in the neighbourhood of a point x 0 ∈ ∂Ω.
Since, by assumption, ∂Ω is an n − 1-dimensional submanifold of X, as
seen at the end of n
◦ 12, there is a cubic chart (U, ϕ) at x 0
70 such that
ϕ(x 0 ) = 0 and ϕ(U ∩ ∂Ω) is the subset of K
n defined by relation ξ
1 = 0. Let
U 0 , U + and U − be the subsets of U defined by the conditions ξ
1 = 0, ξ
1 > 0
and ξ
1 < 0, respectively. So U 0 = U ∩ ∂Ω. Since Ω does not meet its border,
U ∩ Ω ⊂ U + ∪ U − and, for the same reason,
U + ∩ Ω = U + ∩ (Ω ∪ ∂Ω) .
The intersections of Ω with U + and U − are, therefore, both closed and open in
these connected open sets, and so only two cases are possible: (a) U ∩Ω = U +
or U − , (b) U ∩ Ω = U + ∪ U − .
As will be seen, Stokes’ formula supposes that case (a) holds everywhere. However, if one of these two cases holds at x 0 ∈ ∂Ω, then is also
holds at x ∈ ∂Ω sufficiently near x 0 . The set S of x ∈ ∂Ω where case (b)
holds is, therefore, an open subset of ∂Ω. Similarly for the set S
of points
where case (a) holds. This gives a partition of ∂Ω into two open sets, i.e
into two closed ones. Hence, like ∂Ω, S and S
are n − 1-dimensional
compact submanifolds of X. Besides, S ∪ Ω is clearly a connected open
subset of X if Ω is connected. Replacing X by S ∪ Ω, the following question arises : In an n-dimensional connected manifold, can the complement
Ω of an n − 1-dimensional compact submanifold S be connected ? If the
answer was always no, S = ∅ would hold and the “ good ” case (a) would
also hold. If X = R
2 , then S is a smooth curve without multiple points.
Hence, it may be assumed to be the finite union of pairwise disjoint simple closed curves, namely its connected components; the answer to the
question is, therefore, no by Jordan’s theorem, which we alluded to without proof at the end of Chap. IV, § 4 ; the same result holds for a sphere.
But if we remove from the surface a two-dimensional torus, a circle whose
plane contains the rotation axis of the torus, or is orthogonal to it, then
case (b) holds: the complement of such a circle is connected and located
on “ both sides ” of it. To have S = ∅, two circles would need to be
removed from the torus; the open complement then has two connected
70 i.e. such that ϕ(U ) is the cube K
n : |ξ
i | < 1 of R
n .
IX – Multivariate Differential and Integral Calculus
obviously provided the edge of Ω is oriented properly and a further condition
is imposed; its intuitive meaning is that in the neighbourhood of any of its
border points, the open set Ω is located “ on one side ” of it. Corollary : if X
is an n-dimensional compact manifold, then
X
dω = 0
for any form ω of degree n − 1 on X, since ∂X is empty.
Before moving on to the proof, we make some remarks about what happens in the neighbourhood of a point x 0 ∈ ∂Ω.
Since, by assumption, ∂Ω is an n − 1-dimensional submanifold of X, as
seen at the end of n
◦ 12, there is a cubic chart (U, ϕ) at x 0
70 such that
ϕ(x 0 ) = 0 and ϕ(U ∩ ∂Ω) is the subset of K
n defined by relation ξ
1 = 0. Let
U 0 , U + and U − be the subsets of U defined by the conditions ξ
1 = 0, ξ
1 > 0
and ξ
1 < 0, respectively. So U 0 = U ∩ ∂Ω. Since Ω does not meet its border,
U ∩ Ω ⊂ U + ∪ U − and, for the same reason,
U + ∩ Ω = U + ∩ (Ω ∪ ∂Ω) .
The intersections of Ω with U + and U − are, therefore, both closed and open in
these connected open sets, and so only two cases are possible: (a) U ∩Ω = U +
or U − , (b) U ∩ Ω = U + ∪ U − .
As will be seen, Stokes’ formula supposes that case (a) holds everywhere. However, if one of these two cases holds at x 0 ∈ ∂Ω, then is also
holds at x ∈ ∂Ω sufficiently near x 0 . The set S of x ∈ ∂Ω where case (b)
holds is, therefore, an open subset of ∂Ω. Similarly for the set S
of points
where case (a) holds. This gives a partition of ∂Ω into two open sets, i.e
into two closed ones. Hence, like ∂Ω, S and S
are n − 1-dimensional
compact submanifolds of X. Besides, S ∪ Ω is clearly a connected open
subset of X if Ω is connected. Replacing X by S ∪ Ω, the following question arises : In an n-dimensional connected manifold, can the complement
Ω of an n − 1-dimensional compact submanifold S be connected ? If the
answer was always no, S = ∅ would hold and the “ good ” case (a) would
also hold. If X = R
2 , then S is a smooth curve without multiple points.
Hence, it may be assumed to be the finite union of pairwise disjoint simple closed curves, namely its connected components; the answer to the
question is, therefore, no by Jordan’s theorem, which we alluded to without proof at the end of Chap. IV, § 4 ; the same result holds for a sphere.
But if we remove from the surface a two-dimensional torus, a circle whose
plane contains the rotation axis of the torus, or is orthogonal to it, then
case (b) holds: the complement of such a circle is connected and located
on “ both sides ” of it. To have S = ∅, two circles would need to be
removed from the torus; the open complement then has two connected
70 i.e. such that ϕ(U ) is the cube K
n : |ξ
i | < 1 of R
n .