§ 3. Integration of Differential Forms
187
where p(x) is a numerical function and det(h 1 , . . . , h n ) is the determinant of
the vectors h i (i.e. of the matrix of their coordinates) with respect to a basis
of E.
Exercise 2. How does the coefficient p change when the basis of E with
respect to which the determinant of n vectors is defined is changed?
In the general case, choosing a basis for the space E, ω can be written as
ω (x; h 1 , . . . , h p ) = a i1...ip (x)h
i1
1 . . . h
ip
p =
=
1
p!
a i1...ip (x) det
i1...ip (h 1 , . . . , h p )
with antisymmetric coefficients
a i1...ip (x) = ω
x; e i1 , . . . , e ip
.
These are the values of ω(x) at the canonical basis vectors, and the upper
indices bearing upon the determinants indicate the p rows that need to be
extracted from the p × n matrix for the coordinates of the h i . The term 1/p!
could be avoided by summing over the systems of strictly increasing indices.
All this, except the variable x which does not play any role in this context,
expresses the standard formulas of multilinear algebra and of the theory of
determinants. It will henceforth be hardly needed since the the way forward
is sufficiently indicated by forms of degree 1,2, and 3.
In the general case, there is also an exterior differentiation operation taking forms of degree p to forms of degree p + 1. For example to understand
what physicists call the “ divergence ” of a vector field, it is necessary to know
how to associate a form dω of degree 3 to a form
ω =
1
2
p ij dx
i
∧ dx
j = p 23 dx
2
∧ dx
3 + p 31 dx
3
∧ dx
1 + p 12 dx
1
∧ dx
2
of degree 2 on R
3 (or on an arbitrary Cartesian space); its value at some
point x ∈ G is an alternating tri linear form, i.e. an antisymmetric function
of three variable vectors h, k, l ∈ E. To define it, we first need to introduce a
covariant derivative
ω
(x; h, k, l) =
d
dt
ω(x + th; k, l) for t = 0
(8.9)
=
1
2
dp ij (x; h)
k
i l
i
− k
j l
j
,
a linear expression in h, k, l and alternating in k, l. Then set
dω(x; h, k, l) = ω
(x; h, k, l) − ω
(x; k, h, l) + ω
(x; l, h, k) =
(8.10)
= ω
(x; h, k, l) + ω
(x; k, l, h) + ω
(x; l, h, k)
=
1
2
D i p jk (x) det
ijk (h, k, l) ,
187
where p(x) is a numerical function and det(h 1 , . . . , h n ) is the determinant of
the vectors h i (i.e. of the matrix of their coordinates) with respect to a basis
of E.
Exercise 2. How does the coefficient p change when the basis of E with
respect to which the determinant of n vectors is defined is changed?
In the general case, choosing a basis for the space E, ω can be written as
ω (x; h 1 , . . . , h p ) = a i1...ip (x)h
i1
1 . . . h
ip
p =
=
1
p!
a i1...ip (x) det
i1...ip (h 1 , . . . , h p )
with antisymmetric coefficients
a i1...ip (x) = ω
x; e i1 , . . . , e ip
.
These are the values of ω(x) at the canonical basis vectors, and the upper
indices bearing upon the determinants indicate the p rows that need to be
extracted from the p × n matrix for the coordinates of the h i . The term 1/p!
could be avoided by summing over the systems of strictly increasing indices.
All this, except the variable x which does not play any role in this context,
expresses the standard formulas of multilinear algebra and of the theory of
determinants. It will henceforth be hardly needed since the the way forward
is sufficiently indicated by forms of degree 1,2, and 3.
In the general case, there is also an exterior differentiation operation taking forms of degree p to forms of degree p + 1. For example to understand
what physicists call the “ divergence ” of a vector field, it is necessary to know
how to associate a form dω of degree 3 to a form
ω =
1
2
p ij dx
i
∧ dx
j = p 23 dx
2
∧ dx
3 + p 31 dx
3
∧ dx
1 + p 12 dx
1
∧ dx
2
of degree 2 on R
3 (or on an arbitrary Cartesian space); its value at some
point x ∈ G is an alternating tri linear form, i.e. an antisymmetric function
of three variable vectors h, k, l ∈ E. To define it, we first need to introduce a
covariant derivative
ω
(x; h, k, l) =
d
dt
ω(x + th; k, l) for t = 0
(8.9)
=
1
2
dp ij (x; h)
k
i l
i
− k
j l
j
,
a linear expression in h, k, l and alternating in k, l. Then set
dω(x; h, k, l) = ω
(x; h, k, l) − ω
(x; k, h, l) + ω
(x; l, h, k) =
(8.10)
= ω
(x; h, k, l) + ω
(x; k, l, h) + ω
(x; l, h, k)
=
1
2
D i p jk (x) det
ijk (h, k, l) ,