Here the attempt is made to tie this formalism to the usual things studied in calculus. It is
not sufficient to simply re define something like flux when there already exists an
understanding of its meaning. This is why what follows will seem a lot less elegant than what
has just been presented.
First let m = n = 1. What does Stoke’s theorem say? In this case, dω is a 1 form and so ω
should be a 0 form which is just a function.
′
ω = a(x), dω = a(x)dx
Then if r :
[0,1]
→ ℝ is twice continuously differentiable.
∫ ∫ ∫ 1 dr
dω ≡ a′(x)dx ≡ a′(r(t))dt dt = a(r(1)) − a(r(0))
r r 0
which is the same thing. It is just the fundamental theorem of calculus essentially. What if
m = 3,n = 1? This is similar. You need to have ω a zero form. Thus ω = a
(x,y,z)
.
Then
dω = ax(x,y,z)dx+ ay(x,y,z)dy+ az(x,y,z)dz
Letting r :
[0,1]
→ ℝ^{3},
∫ ∫ 1( dx dy dz)
dω = ax(r(t))dt-+ay (r (t))dt + az(r(t))dt dt
r ∫01
= ∇a (r (t))⋅r′(t)dt = a (r (1))− a(r(0))
0
∫
ω = (− 1)2 a(r(1))− (− 1)2a (r(0))
∂r
which is the same thing. This is the case of a conservative vector field, the potential function
being a.
Next let n = m = 2 so dω is a 2 form and ω is a 1 form. Say
ω = P (x,y)dx+ Q (x,y)dy
dω = Py (x,y)dy ∧dx + Qxdx ∧dy
Recall that terms like dx ∧ dx are zero because they result in a determinant which equals 0.
Then let r be smooth and map
This is computing a line integral by summing the contributions around the edges of
[0,1]
^{2}. It
is
∫
C Pdx +Qdy
where the orientation on this curve C comes from the counter clockwise orientation on the
boundary of
[0,1]
^{2} as shown in the picture.
PICT
Thus if x_{s}y_{t}− x_{t}y_{s}> 0, this is just Green’s theorem from calculus. Thus this gives a
proof of this important theorem provided the region U is the C^{2} image of the unit
square. This falls very short of the best results for Green’s theorem which involve a
rectifiable simple closed curve with U its interior and illustrates the mathematical
weakness of this approach. However, it can include things like circular disks. Imagine
squashing a square onto a disk for example. The limitation can also be partly remedied
by considering chains of k cells although even this is not completely satisfactory.
Next let n = 2 and m = 3. Thus dω is a 2 form so ω is a 1 form. Say
, see Definition 10.4.1 the area increment
on the surface r
( )
[0,1]2
is
∘ --------------------------------------
|y z − y z|2 + |x z − x z |2 +|x y − x y|2dsdt
s( t ts ts) s t s t ts
= det Dr (t,s)∗ Dr (s,t)1∕2dsdt
also the vector
(Ry − Qz)i+ (Pz − Rx )j+(Qx − Py)k
is the curl of the vector field F ≡ Pi + Qj + Rj. Thus in more familiar calculus notation, the
above integral is of the form
∫
2 ∇ ×F ⋅ndS
S≡r([0,1])
where n is a vector which is in the direction of
(yszt − ytzs)
i +
(xtzs − xszt)
j +
(xsyt − xtys)
k.
It happens that this vector is perpendicular to the surface S at every point where r_{s}× r_{t}≠0,
this vector equalling
(yszt − ytzs)i+ (xtzs − xszt)j+ (xsyt − xtys)k
which is seen to occur in the above formula. Now, as (hopefully) discussed in beginning
calculus, this vector is perpendicular to the surface. Recall the reason for this. You have
r_{s}⋅ r_{s}× r_{t} = r_{t}⋅ r_{s}× r_{t} = 0 and this is sufficient to claim, based on geometric reasoning that
it is perpendicular to the surface. Thus ∫_{r}dω is one side of the usual Stoke’s theorem
from calculus. Next consider what happens with ∫_{∂r}ω. This is just like it was with
Green’s theorem but with more terms. To save on space, P
(x(1,t),y (1,t),z(1,t))
is
denoted as P
(1,t)
with similar considerations for Q and R. Then this results in
where 0 is in the j^{th} slot. Now the k^{th}
component of ∂r
(u1,⋅⋅⋅,0,⋅⋅⋅,un)
∕∂u_{i} times
(− 1)
^{k+j} multiplied by the above and added
over k gives the expansion of a determinant along the j^{th} column which has two equal
columns. Therefore, this equals 0. Thus the vector whose k^{th} component is given in 14.5 is
really normal to r
(Sj0)
and similarly when r
(u1,⋅⋅⋅,0,⋅⋅⋅,un )
is replaced with
r
(u1,⋅⋅⋅,1,⋅⋅⋅,un)
, it is normal to r
(Sj1)
. This is why ∫_{∂r}ω represents a flux integral across
the boundary of r
∫ ∫ ∂ (x ⋅⋅⋅x )
dω = n ∇ ⋅a(r(u))---1----n du
r ∫[0,1] ∂ (u(1⋅⋅⋅un ) )
= ∇ ⋅a(x)sgn ∂-(x1-⋅⋅⋅xn) dV
r([0,1]n) ∂ (u1 ⋅⋅⋅un)
where a has the k^{th} component equal to a_{k}. If det
(Dr (u))
≥ 0, then this yields the
divergence theorem from calculus. In other words, in this situation, the statement of Stoke’s
theorem ∫_{∂r}ω = ∫_{r}dω is a disguised version of the divergence theorem at least for the
situation discussed here.