15th March 2015, 7 min read

On Differential Forms

Original post is here eklausmeier.goip.de/blog/2015/03-15-on-differential-forms-2.

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.
2. Differentiation of $k$ -forms.
3. The outer product of differential forms.

1. Basic definitions

We denote the submatrix of $A = (a_{i j}) \in R^{m \times n}$ consisting of the rows $i_{1}, \dots, i_{k}$ and the columns $j_{1}, \dots, j_{k}$ with

[A] \binom{j_{1}}{i_{1}} \binom{\dots}{\dots} \binom{j_{k}}{i_{k}} := (\begin{matrix} a_{i_{1} j_{1}} & \dots & a_{i_{1} j_{k}} \\ ⋮ & ⋱ & ⋮ \\ a_{i_{k} j_{1}} & \dots & a_{i_{k} j_{k}} \end{matrix})

and its determinant with

A \binom{j_{1}}{i_{1}} \binom{\dots}{\dots} \binom{j_{k}}{i_{k}} := det [A] \binom{j_{1}}{i_{1}} \binom{\dots}{\dots} \binom{j_{k}}{i_{k}} .

For example

A = (\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{matrix}), A_{1, 2}^{1, 3} = a_{11} a_{23} - a_{21} a_{13} .

Suppose

H \in R^{n \times (n + 1)}

and let

f, g : U \subseteq R^{n} \to R, U open,

be two functions which are two-times continuously differentiable. Then we call for a fixed $k$ the expression

f H_{α}^{1 \dots k}, α = (i_{1}, \dots, i_{k}) \in {1, \dots, n}^{k},

a basic $k$ -form or basic differential form of order $k$ . It's a real function of $n + k^{2}$ variables. For $k > n$ the expression is defined to be zero. If $f$ also depends on $α$ then

\sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} H \binom{1}{i_{1}} \binom{\dots}{\dots} \binom{k}{i_{k}}

is called a $k$ -form. It's a real function of $n + k n$ variables which is $k$ -linear in the $k$ column-vectors of $H$ .

For example for $f : R \to R$ and $H \in R^{1 \times 1}$ we have $f (x), H$ . This is a linear function in $H$ and a possibly non-linear function in $x$ .

2. Differentiation of $k$ -forms

For the differential form

ω = f H_{α}^{1 \dots k}, α = (i_{1}, \dots, i_{k}),

we define

d ω := \sum_{ν = 1}^{n} \frac{\partial f}{\partial x_{ν}} H_{ν, α}^{1 \dots k + 1}

as the outer differentiation of $ω$ . This is a $(k + 1)$ -form. It's a function of $n + (k + 1) n$ variables.

The $0$ -form

ω = f, | α | = k = 0

yields

\begin{matrix} (1) & d w = \sum_{ν = 1}^{n} \frac{\partial f}{\partial x_{ν}} H_{ν}^{1}, \end{matrix}

which corresponds to $\nabla f = grad f$ .

In the special case $k = | α | = 1$ we get for

ω = \sum_{i = 1}^{n} f_{i} H_{i}^{1}

the result

\begin{matrix} (2) & d ω = \sum_{i = 1}^{n} \sum_{j = 1}^{n} \frac{\partial f_{i}}{\partial x_{j}} H_{j, i}^{1, 2} = \sum_{i < j} (\frac{\partial f_{i}}{\partial x_{j}} - \frac{\partial f_{j}}{\partial x_{i}}) H_{j, i}^{1, 2} . \end{matrix}

This corresponds to $rot f$ .

Let hat ( $\hat{}$ ) mean exclusion from the index list. The case $k = n - 1$ for

ω = \sum_{i = 1}^{n} (- 1)^{i - 1} f_{i} H_{1 \dots \hat{ı} \dots n}^{1 \dots n - 1}

delivers

d w = \sum_{i = 1}^{n} \sum_{ν = 1}^{n} (- 1)^{i - 1} \frac{\partial f_{i}}{\partial x_{ν}} H_{ν, 1 \dots \hat{ı} \dots n}^{1 \dots n} = \sum_{i = 1}^{n} \frac{\partial f_{i}}{\partial x_{ν}} H_{1 \dots n}^{1 \dots n} = (\sum_{i = 1}^{n} \frac{\partial f_{i}}{\partial x_{i}}) det H .

This corresponds to $div f$ .

Theorem. For $ω = f H_{α}^{1 \dots k}$ we have

d d ω = 0.

Proof: With

d ω = \sum_{ν = 1}^{n} \frac{\partial f}{\partial x_{ν}} H_{ν, α}^{1 \dots k + 1}

we get

d d ω = \sum_{ν = 1}^{n} \sum_{μ = 1}^{n} \frac{\partial^{2} f}{\partial x_{ν} \partial x_{μ}} H_{μ, ν, α}^{1 \dots k + 2}

and this is zero, because

H_{μ, μ, α}^{1 \dots k + 2} = 0, H_{μ, ν, α}^{1 \dots k + 2} = - H_{ν, μ, α}^{1 \dots k + 2},

and

\frac{\partial^{2} f}{\partial x_{ν} \partial x_{μ}} = \frac{\partial^{2} f}{\partial x_{μ} \partial x_{ν}} .

Application of this theorem to an $0$ -form with an $f : U \subseteq R^{n} \to R$ and a $1$ -form with an $a : U \to R^{n}$ reading (1) and then (2) yields

rot grad f = 0, div rot a = 0.

The second equation is only true for $n = 3$ because

(\binom{n}{2}) = n (n \in N) \Leftrightarrow n = 3.

Definition. Suppose

ϕ : D \to E \subset R^{n}, D \subset \subset R^{k},

is differentiable, its derivative denoted by $ϕ^{'}$ , and

f : E \to R .

For the differential form $ω = f H_{α}^{1 \dots k}$ we define the back-transportation as

ϕ^{*} ω := (f \circ ϕ) (ϕ^{'})_{α}^{1 \dots k}

and the integral over $k$ -forms as

\int_{ϕ} ω := \int_{D} ϕ^{*} ω .

For example the case $k = 1$ ,

ω = \sum_{i = 1}^{n} f_{i} H_{i}^{1}

gives

ϕ^{*} ω = \sum_{i = 1}^{n} (f_{i} \circ ϕ) (ϕ^{'})_{i}^{1} .

3. The outer product of differential forms

Suppose

H \in R^{n \times (n + 1)}, k + m \leq n .

For the two differential forms

ω = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} H \binom{1}{i_{1}} \binom{\dots}{\dots} \binom{k}{i_{k}}

and

λ = \sum_{1 \leq j_{1} < \dots < j_{m} \leq n} g_{j_{1} \dots j_{m}} H \binom{k + 1}{j_{1}} \binom{\dots}{\dots} \binom{k + m}{j_{m}}

the outer product is defined as

w \land λ := \sum_{\binom{1 \leq i_{1} < \dots < i_{k} \leq n}{1 \leq j_{1} < \dots < j_{m} \leq n}} f_{i_{1} \dots i_{k}} g_{j_{1} \dots j_{m}} H \binom{1}{i_{1}} \binom{\dots}{\dots} \binom{k}{i_{k}} \binom{k + 1}{j_{1}} \binom{\dots}{\dots} \binom{k + m}{j_{m}} .

This is a differential form of order $k + m$ . It's a function in $n + (k + m) n$ variables.

Theorem.

d (ω \land λ) = d ω \land λ + (- 1)^{k} ω \land d λ

Proof: With

ω = \sum_{α} f_{α} H_{α}^{1 \dots k}, λ = \sum_{β} g_{β} H_{β}^{1 \dots m}

then

\begin{aligned} d (ω \land λ) & = \sum_{α, β} \sum_{ν = 1}^{n} (\frac{\partial f_{α}}{\partial x_{ν}} g_{β} + f_{β} \frac{\partial g_{β}}{\partial x_{ν}}) H_{ν, α, β}^{1 \dots k + m + 1} \\ = \sum_{α, β} \sum_{ν = 1}^{n} \frac{\partial f_{α}}{\partial x_{ν}} g_{β} H_{ν, α, β}^{1 \dots k + m + 1} + \sum_{α, β} \sum_{ν = 1}^{n} f_{α} \frac{\partial g_{β}}{\partial x_{ν}} H_{ν, α, β}^{1 \dots k + m + 1} \\ = d ω \land λ + (- 1)^{k} ω \land d λ, \end{aligned}

due to

H_{ν, α, β}^{1 \dots k + m + 1} = (- 1)^{k} H_{ν, β, α}^{1 \dots k + m + 1}

and

d λ = \sum_{β} \sum_{ν = 1}^{n} \frac{\partial g_{β}}{\partial x_{ν}} H_{ν, β}^{1 \dots m + 1} .

An alternative definition for the differentiation of $k$ -forms could be given.

Theorem. Suppose

ω = f H_{α}^{1 \dots k}, 0 \leq | α | \leq k,

and

H = (h_{1}, \dots, h_{n}, h_{n + 1}) \in R^{n \times (n + 1)}

with $α = (i_{1}, \dots, i_{k})$ we have

d ω = det (col (\nabla f, [{I d}_{n}]_{α}^{1 \dots n}) [H]_{1 \dots n}^{1 \dots k + 1}) = \sum_{ν = 1}^{n} \frac{\partial f}{\partial x_{ν}} H_{ν, α}^{1 \dots k + 1},

where $c o l$ just stacks matrices one above another and ${I d}_{n}$ is the identity matrix in $R^{n}$ .

Proof:

d ω = | \begin{matrix} ⟨ \nabla f, h_{1} ⟩ & \dots & ⟨ \nabla f, h_{k} ⟩ & ⟨ \nabla f, h_{k + 1} ⟩ \\ ⟨ e_{i_{1}}, h_{1} ⟩ & \dots & ⟨ e_{i_{1}}, h_{k} ⟩ & ⟨ e_{i_{1}}, h_{k + 1} ⟩ \\ ⋮ & ⋱ & ⋮ & ⋮ \\ ⟨ e_{i_{k}}, h_{1} ⟩ & \dots & ⟨ e_{i_{k}}, h_{k} ⟩ & ⟨ e_{i_{k}}, h_{k + 1} ⟩ \end{matrix} |

= \sum_{ν = 1}^{n} \frac{\partial f}{\partial x_{ν}} | \begin{matrix} h_{1, ν} & h_{1, i_{1}} & \dots & h_{1, i_{k}} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ h_{k, ν} & h_{k, i_{1}} & \dots & h_{k, i_{k}} \\ h_{k + 1, ν} & h_{k + 1, i_{1}} & \dots & h_{k + 1, i_{k}} \end{matrix} |

since

⟨ \nabla f, h_{1} ⟩ = \sum_{ν = 1}^{n} \frac{\partial f}{\partial x_{ν}} h_{1, ν},

⋮ ⋮

⟨ \nabla f, h_{k + 1} ⟩ = \sum_{ν = 1}^{n} \frac{\partial f}{\partial x_{ν}} h_{k + 1, ν} .

REFERENCES.

Walter Rudin, Principles of Mathematical Analysis, Second Edition, McGraw-Hill, New York, 1964
Otto Forster, Analysis 3: Integralrechnung im $R^{n}$ mit Anwendungen, Third Edition, Friedrich Vieweg & Sohn, Braunschweig/Wiesbaden, 1984

On Differential Forms

1. Basic definitions

2. Differentiation of k-forms

3. The outer product of differential forms

2. Differentiation of $k$ -forms