Content uploaded by Elmar Klausmeier
Author content
All content in this area was uploaded by Elmar Klausmeier on Jan 02, 2019
Content may be subject to copyright.
On Differential Forms
Abstract. This article will give a very simple definition of k-forms or differential forms. It just requires
basic knowledge about matrices and determinants. Furthermore a very simple proof will be given for the
proposition that the double outer differentiation of k-forms vanishes.
MSC 2010: 58A10
1. Basic definitions.
We denote the submatrix of A= (aij )∈Rm×nconsisting of the rows i1, . . . , ikand the columns j1,...,jk
with
[A]j1
i1
...
...
jk
ik:=
ai1j1. . . ai1jk
.
.
.....
.
.
aikj1. . . aikjk
and its determinant with
Aj1
i1
...
...
jk
ik:= det[A]j1
i1
...
...
jk
ik.
For example
A=a11 a12 a13
a21 a22 a23 , A1,3
1,2=a11a23 −a21 a13.
Suppose
H∈Rn×(n+1)
and let
f, g:U⊆Rn→R, U open
be two functions which are two-times continuously differentiable. Then we call for a fixed kthe expression
f H1...k
α, α = (i1,...,ik)∈ {1,...,n}k,
abasic k-form or basic differential form of order k. It’s a real function of n+k2variables. For k > n the
expression is defined to be zero. If falso depends on αthen
X
1≤i1<···<ik≤n
fi1...ikH1
i1
...
...
k
ik
is called a k-form. It’s a real function of n+kn variables which is k-linear in the kcolumn-vectors of H.
For example for f:R→Rand H∈R1×1we have f(x)H. This is a linear function in Hand a possibly
non-linear function in x.
– 1 –
2. Differentiation of k-forms.
For the differential form
ω=fH1...k
α, α = (i1,...,ik),
we define
dω :=
n
X
ν=1
∂f
∂xν
H1...k+1
ν,α
as the outer differentiation of ω. This is a (k+ 1)-form. It’s a function of n+ (k+ 1)nvariables.
The 0-form
ω=f, |α|=k= 0
yields
dw =
n
X
ν=1
∂f
∂xν
H1
ν(1)
which corresponds to ∇f= grad f.
In the special case k=|α|= 1 we get for
ω=
n
X
i=1
fiH1
i
the result
dω =
n
X
i=1
n
X
j=1
∂fi
∂xj
H1,2
j,i =X
i<j ∂fi
∂xj
−∂fj
∂xiH1,2
j,i .(2)
This corresponds to rot f.
Let hat (ˆ) mean exclusion from the index list. The case k=n−1 for
ω=
n
X
i=1
(−1)i−1fiH1...n−1
1...ˆı...n
delivers
dw =
n
X
i=1
n
X
ν=1
(−1)i−1∂fi
∂xν
H1...n
ν,1...ˆı...n =
n
X
i=1
∂fi
∂xν
H1...n
1...n = n
X
i=1
∂fi
∂xi!det H.
This corresponds to div f.
Theorem. For ω=fH1...k
αwe have
ddω = 0.
Proof: With
dω =
n
X
ν=1
∂f
∂xν
H1...k+1
ν,α
we get
ddω =
n
X
ν=1
n
X
µ=1
∂2f
∂xν∂xµ
H1...k+2
µ,ν,α
and this is zero, because
H1...k+2
µ,µ,α = 0, H1...k+2
µ,ν,α =−H1...k+2
ν,µ,α ,
and ∂2f
∂xν∂xµ
=∂2f
∂xµ∂xν
.
– 2 –
Application of this theorem to an 0-form with an f:U⊆Rn→Rand a 1-form with an a:U→Rn
reading (1) and then (2) yields
rot grad f= 0,div rot a= 0.
The second equation is only true for n= 3 because
n
2=n(n∈N)⇔n= 3.
Definition. Suppose
φ:D→E⊂Rn, D ⊂⊂ Rk,
is differentiable, its derivative denoted by φ′, and
f:E→R.
For the differential form ω=fH1...k
αwe define the back-transportation as
φ∗ω:= (f◦φ) (φ′)1...k
α
and the integral over k-forms as
Zφ
ω:= ZD
φ∗ω.
For example the case k= 1,
ω=
n
X
i=1
fiH1
i
gives
φ∗ω=
n
X
i=1
(fi◦φ) (φ′)1
i.
– 3 –
3. The outer product of differential forms.
Suppose
H∈Rn×(n+1), k +m≤n.
For the two differential forms
ω=X
1≤i1<···<ik≤n
fi1...ikH1
i1
...
...
k
ik
and
λ=X
1≤j1<···<jm≤n
gj1...jmHk+1
j1
...
...
k+m
jm
the outer product is defined as
w∧λ:= X
1≤i1<···<ik≤n
1≤j1<···<jm≤n
fi1...ikgj1...jmH1
i1
...
...
k
ik
k+1
j1
...
...
k+m
jm.
This is a differential form of order k+m. It’s a function in n+ (k+m)nvariables.
Theorem.
d(ω∧λ) = dω ∧λ+ (−1)kω∧dλ
Proof: With
ω=X
α
fαH1...k
α, λ =X
β
gβH1...m
β
then
d(ω∧λ) = X
α,β
n
X
ν=1 ∂fα
∂xν
gβ+fβ
∂gβ
∂xνH1...k+m+1
ν,α,β
=X
α,β
n
X
ν=1
∂fα
∂xν
gβH1...k+m+1
ν,α,β +X
α,β
n
X
ν=1
fα
∂gβ
∂xν
H1...k+m+1
ν,α,β
=dω ∧λ+ (−1)kω∧dλ,
due to
H1...k+m+1
ν,α,β = (−1)kH1...k+m+1
ν,β,α
and
dλ =X
β
n
X
ν=1
∂gβ
∂xν
H1...m+1
ν,β .
An alternative definition for the differentiation of k-forms could be given.
Theorem. Suppose
ω=fH1...k
α,0≤ |α| ≤ k,
and
H= (h1,...,hn, hn+1)∈Rn×(n+1)
with α= (i1,...,ik)we have
dω = det col ∇f, [Idn]1...n
α[H]1...k+1
1...n =
n
X
ν=1
∂f
∂xν
H1...k+1
ν,α ,
– 4 –
where col just stacks matrices one above another and Idnis the identity matrix in Rn.
Proof:
dω =
h∇f, h1i... h∇f , hki h∇f, hk+1i
hei1, h1i... hei1, hki hei1, hk+1i
.
.
.....
.
..
.
.
heik, h1i... heik, hki heik, hk+1i
=
n
X
ν=1
∂f
∂xν
h1,ν h1,i1. . . h1,ik
.
.
..
.
.....
.
.
hk,ν hk,i1. . . hk,ik
hk+1,ν hk+1,i1. . . hk+1,ik
since
h∇f, h1i=
n
X
ν=1
∂f
∂xν
h1,ν ,
.
.
..
.
.
h∇f, hk+1 i=
n
X
ν=1
∂f
∂xν
hk+1,ν .
REFERENCES.
1. Walter Rudin, Principles of Mathematical Analysis, Second Edition, McGraw-Hill, New York, 1964
2. Otto Forster, Analysis 3: Integralrechnung im Rnmit Anwendungen, Third Edition, Friedrich Vieweg
& Sohn, Braunschweig/Wiesbaden, 1984
Author’s address:
Elmar Klausmeier
Goethestrasse 4
D-63128 Dietzenbach
Germany
http://eklausmeier.wordpress.com
– 5 –