Note how the orientation of the curve comes from the unit interval and the choice of
parameterization.
You can think of the differential form a
(x,y,z)
dx + b
(x,y,z)
dy + c
(x,y,z)
dz as a
symbol which represents something which makes vector valued functions r defined on
[0,1]
into numbers according to the above procedure. You might also have written the line integral
in the form
∫
ra(x,y,z)dx+ b(x,y,z)dy+ c(x,y,z)dz
Thus it is a functional which makes vector valued functions defined on
[0,1]
into numbers. It
is desired to extend this simple idea to functions of k > 1 variables. First here is some
notation.
Notation 14.1.1Let x :
[0,1]
^{k}→ ℝ^{n}with x ∈ C^{1}
( )
[0,1]k,ℝn
. This last symbol meansthat x is the restriction to
[0,1]
^{k}of a C^{1}function defined on all of ℝ^{k}. Now let Idenote an ordered list of k indices taken from
{1,2,⋅⋅⋅,n}
. Thus I =
(i ,⋅⋅⋅,i )
1 k
.Then
( x x ⋅⋅⋅ x )
( ) | xi1,t1 xi1,t2 ⋅⋅⋅ xi1,tk |
det dxI- ≡ det|| i2.,t1 i2.,t2 i2,.tk || ≡ ∂(xi1,⋅⋅⋅xik)-
dt ( .. .. .. ) ∂ (t1,⋅⋅⋅,tk)
xik,t1 xik,t2 ⋅⋅⋅ xik,tk
It is the same as
det(DxI )
where x_{I}has values in ℝ^{n}and is obtained by permuting the rows of x to be in the orderdetermined by I and leaving out the other rows. More generally, suppose I is an ordered list ofl indices and that J is an ordered list of l indices. Then
Now with this definition, here is the generalization of the differential forms defined in
calculus.
Definition 14.1.2A differential form is
∑
ω ≡ aI (x)dxI
I
where
dxI ≡ dxi ∧ dxi ∧⋅⋅⋅∧ dxi
1 2 k
for I =
(i1,⋅⋅⋅,ik)
. It is a function mapping functions in C^{1}
( k n)
[0,1] ,ℝ
to ℝ , denotedas
∫
ω
(⋅)
if, for r :
[0,1]
^{k}→ ℝ^{n}, define
∫ ∫ ∑ ( dxI (t))
rω ≡ [0,1]k aI (r(t))det dt dmk
∫ I ( )
= ∑ aI (r(t))det drI (t) dmk
[0,1]k I dt
The sum is taken over all ordered lists of indices from
{1,⋅⋅⋅,n}
. Note that if there are anyrepeats in an ordered list I, then
( drI (t))
det --dt-- = 0
and so it suffices to consider the sum only over lists of indices in which there are no repeats.Thus the sum can be considered to consist of no more than P
(n,k)
terms where this denotesthe permutations of n things taken k at a time.
It will always be assumed that a_{I} is as smooth as desired to make everything work. As to
r ∈ C^{1}
([0,1]k,ℝn )
, it is not required to be one to one or have nonzero determinant. This
means r
( )
[0,1]k
can be various sets which have points and edges.
Note that if n < k, then the functional ∑_{I}a_{I}
(x)
dx_{I} must equal the zero function because
there must be two equal rows in Dx_{I} and so its determinant must equal 0. However, more can
be said.
Lemma 14.1.3Each differential form ω can be written in a unique way as
Proof: Suppose you have a differential form ω = ∑_{I}a_{I}
(x )
dx_{I}. Then for r ∈ C^{1}
( )
[0,1]k ,ℝn
and
{I}
=
{J}
meaning that
{i ,⋅⋅⋅,i} = {j,⋅⋅⋅,j}
1 k 1 k
so that J is a permutation of I, it follows that
∫ ( ) ∫ ( )
aJ (r(t))det drJ (t) dmk = sgn(σ) aJ (r (t))det drI-(t)- dmk
r dt r dt
where sgn
(σ)
is the sign of the permutation which takes I to J. This follows from the
alternating property of determinants. This verifies the second claim. Thus, considering the
two terms
aI (x)dxI,aJ (x)dxJ
in the sum defining the differential form, the effect on r would be the same as that of the
single term
(aI (x)+ aJ (x) sgn (σ)) dxI
combining these in this way leads to a sum of the form
ω = ∑ a (x)dx ∧ dx ∧ ⋅⋅⋅∧ dx
i1<i2<⋅⋅⋅<ik i1,i2,⋅⋅⋅,ik i1 i2 ik
where the a_{i1,i2,}
⋅⋅⋅
,i_{k} are obtained from the original a_{I} by combining them as just
explained.
Now it will be shown that the above sum is uniquely determined. In fact, the coefficients
a_{i1,i2,}
⋅⋅⋅
,i_{k} are determined uniquely. This can be shown by considering a particular function in
C^{1}
( k )
[0,1] ,ℝn
which is chosen auspiciously to reveal a_{i1,i2,⋅⋅⋅
,ik}