# 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 $A=(a_{ij})\in R^{m\times n}$ consisting of the rows $i_1,\ldots,i_k$ and the columns $j_1,\ldots,j_k$ with

$\displaystyle{ [A]{\textstyle\!{\scriptstyle j_1\atop\scriptstyle i_1} \!{\scriptstyle\ldots\atop\scriptstyle\ldots} \!{\scriptstyle j_k\atop\scriptstyle i_k}} := \begin{pmatrix} a_{i_1j_1} & \ldots & a_{i_1j_k}\\ \vdots & \ddots & \vdots\\ a_{i_kj_1} & \ldots & a_{i_kj_k}\\ \end{pmatrix} }$

and its determinant with

$\displaystyle{ A{\textstyle\!{\scriptstyle j_1\atop\scriptstyle i_1} \!{\scriptstyle\ldots\atop\scriptstyle\ldots} \!{\scriptstyle j_k\atop\scriptstyle i_k}} := \det [A]{\textstyle\!{\scriptstyle j_1\atop\scriptstyle i_1} \!{\scriptstyle\ldots\atop\scriptstyle\ldots} \!{\scriptstyle j_k\atop\scriptstyle i_k}}. }$

For example

$\displaystyle{ A = \begin{pmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ \end{pmatrix}, \qquad A_{1,2}^{1,3} = a_{11}a_{23} - a_{21}a_{13}. }$

Suppose

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

and let

$\displaystyle{ f,g\colon U\subseteq R^n\to R, \qquad U \text{ open}, }$

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

$\displaystyle{ f\,H_\alpha^{1\ldots k}, \qquad \alpha=\left(i_1,\ldots,i_k\right) \in\left\{1,\ldots,n\right\}^k, }$

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 $\alpha$ then

$\displaystyle{ \sum_{1\le i_1<\cdots

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

For example for $f\colon R\to R$ and $H\in R^{1\times1}$ 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

$\displaystyle{ \omega = f H^{1\ldots k}_\alpha, \qquad \alpha=\left(i_1,\ldots,i_k\right), }$

we define

$\displaystyle{ d\omega := \sum_{\nu=1}^n {\partial f\over\partial x_\nu} H^{1\ldots k+1}_{\nu,\alpha} }$

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

The $0$-form

$\displaystyle{ \omega = f, \qquad \left|\alpha\right|=k=0 }$

yields

$\displaystyle{ dw = \sum_{\nu=1}^n {\partial f\over\partial x_\nu} H^1_\nu \qquad(1) }$

which corresponds to $\nabla f = \mathop{\rm grad}f$.

In the special case $k=\left|\alpha\right|=1$ we get for

$\displaystyle{ \omega = \sum_{i=1}^n f_i H^1_i }$

the result

$\displaystyle{ d\omega = \sum_{i=1}^n \sum_{j=1}^n {\partial f_i\over\partial x_j} H^{1,2}_{j,i} = \sum_{i

This corresponds to $\mathop{\rm rot} f$.

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

$\displaystyle{ \omega = \sum_{i=1}^n (-1)^{i-1} f_i\,H^{\>1\ldots n-1\>}_{1\ldots\hat\imath\ldots n} }$

delivers

$\displaystyle{ dw = \sum_{i=1}^n \sum_{\nu=1}^n (-1)^{i-1} {\partial f_i\over\partial x_\nu} H^{\>\>1\ldots n}_{\nu,1\ldots\hat\imath\ldots n} = \sum_{i=1}^n {\partial f_i\over\partial x_\nu} H^{1\ldots n}_{1\ldots n} = \left(\sum_{i=1}^n{\partial f_i\over\partial x_i}\right) \det H. }$

This corresponds to $\mathop{\rm div}f$.

Theorem. For $\omega = f H_\alpha^{1\ldots k}$ we have

$\displaystyle{ dd\omega = 0. }$

Proof: With

$\displaystyle{ d\omega = \sum_{\nu=1}^n {\partial f\over\partial x_\nu} H_{\nu,\alpha}^{1\ldots k+1} }$

we get

$\displaystyle{ dd\omega = \sum_{\nu=1}^n \sum_{\mu=1}^n {\partial^2 f\over\partial x_\nu\partial x_\mu} H_{\mu,\nu,\alpha}^{1\ldots k+2} }$

and this is zero, because

$\displaystyle{ H_{\mu,\mu,\alpha}^{1\ldots k+2} = 0, \qquad H_{\mu,\nu,\alpha}^{1\ldots k+2} = -H_{\nu,\mu,\alpha}^{1\ldots k+2}, }$

and

$\displaystyle{ {\partial^2f\over\partial x_\nu\partial x_\mu} = {\partial^2f\over\partial x_\mu\partial x_\nu}. }$

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

$\displaystyle{ \mathop{\rm rot}\mathop{\rm grad} f = 0, \qquad \mathop{\rm div}\mathop{\rm rot} a = 0. }$

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

$\displaystyle{ {n\choose 2} = n \quad (n\in N) \qquad\Leftrightarrow\qquad n = 3. }$

Definition. Suppose

$\displaystyle{ \phi\colon D\to E\subset R^n, \qquad D\subset\!\subset R^k, }$

is differentiable, its derivative denoted by $\phi'$, and

$\displaystyle{ f\colon E\to R. }$

For the differential form $\omega = f H^{1\ldots k}_\alpha$ we define the back-transportation as

$\displaystyle{ \phi^*\omega := (f\circ\phi) \, (\phi')^{1\ldots k}_{\alpha} }$

and the integral over $k$-forms as

$\displaystyle{ \int_\phi \omega := \int_D \phi^*\omega. }$

For example the case $k=1$,

$\displaystyle{ \omega = \sum_{i=1}^n f_i H^1_i }$

gives

$\displaystyle{ \phi^*\omega = \sum_{i=1}^n (f_i\circ\phi) \, (\phi')_i^1 . }$

# 3. The outer product of differential forms.

Suppose

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

For the two differential forms

$\displaystyle{ \omega = \sum_{1\le i_1<\cdots

and

$\displaystyle{ \lambda = \sum_{1\le j_1<\cdots

the outer product is defined as

$\displaystyle{ w\land\lambda := \sum _{\scriptstyle1\le i_1<\cdots

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

Theorem.

$\displaystyle{ d(\omega\land\lambda) = d\omega\land\lambda + (-1)^k\omega\land d\lambda }$

Proof: With

$\displaystyle{ \omega = \sum_\alpha f_\alpha H_\alpha^{1\ldots k}, \qquad \lambda = \sum_\beta g_\beta H_\beta^{1\ldots m} }$

then

$\displaystyle{ d(\omega\land\lambda) = \sum_{\alpha,\beta} \sum_{\nu=1}^n \left( {\partial f_\alpha\over\partial x_\nu} g_\beta + f_\beta {\partial g_\beta\over\partial x_\nu} \right) H_{\nu,\alpha,\beta}^{1\ldots k+m+1} }$

$\displaystyle{ .\qquad\qquad\qquad\qquad\qquad = \sum_{\alpha,\beta} \sum_{\nu=1}^n {\partial f_\alpha\over\partial x_\nu} g_\beta H_{\nu,\alpha,\beta}^{1\ldots k+m+1} + \sum_{\alpha,\beta} \sum_{\nu=1}^n f_\alpha {\partial g_\beta\over\partial x_\nu} H_{\nu,\alpha,\beta}^{1\ldots k+m+1} }$

$\displaystyle{ = d\omega\land\lambda + (-1)^k\omega\land d\lambda, }$

due to

$\displaystyle{ H_{\nu,\alpha,\beta}^{1\ldots k+m+1} = (-1)^k H_{\nu,\beta,\alpha}^{1\ldots k+m+1} }$

and

$\displaystyle{ d\lambda = \sum_\beta \sum_{\nu=1}^n {\partial g_\beta\over\partial x_\nu} H_{\nu,\beta}^{1\ldots m+1} . }$

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

Theorem. Suppose

$\displaystyle{ \omega = f H_\alpha^{1\ldots k}, \qquad0\le\left|\alpha\right|\le k, }$

and

$\displaystyle{ H = \left(h_1,\ldots,h_n,h_{n+1}\right) \in R^{n\times(n+1)} }$

with $\alpha=\left(i_1,\ldots,i_k\right)$ we have

$\displaystyle{ d\omega = \det\left( \mathop{\rm col}\left( \nabla f, [{\rm Id}_n]_\alpha^{1\ldots n} \right) [H]_{1\ldots n}^{1\ldots k+1} \right) = \sum_{\nu=1}^n {\partial f\over\partial x_\nu} H_{\nu,\alpha}^{1\ldots k+1}, }$

where $\rm col$ just stacks matrices one above another and ${\rm Id}_n$ is the identity matrix in $R^n$.

Proof:

$\displaystyle{ d\omega = \left|\begin{matrix} \left\langle\nabla f,h_1\right\rangle & \ldots & \left\langle\nabla f,h_k\right\rangle & \left\langle\nabla f,h_{k+1}\right\rangle \\ \left\langle e_{i_1},h_1\right\rangle & \ldots & \left\langle e_{i_1},h_k\right\rangle & \left\langle e_{i_1},h_{k+1}\right\rangle \\ \vdots & \ddots & \vdots & \vdots\cr \left\langle e_{i_k},h_1\right\rangle & \ldots & \left\langle e_{i_k},h_k\right\rangle & \left\langle e_{i_k},h_{k+1}\right\rangle \cr \end{matrix}\right| }$

$\displaystyle{ \qquad = \sum_{\nu=1}^n {\partial f\over\partial x_\nu} \left|\begin{matrix} h_{1,\nu} & h_{1,i_1} & \ldots & h_{1,i_k}\\ \vdots & \vdots & \ddots & \vdots\\ h_{k,\nu} & h_{k,i_1} & \ldots & h_{k,i_k}\\ h_{k+1,\nu}& h_{k+1,i_1} & \ldots & h_{k+1,i_k}\\ \end{matrix}\right| }$

since

${ \left\langle\nabla f,h_1\right\rangle = \sum_{\nu=1}^n {\partial f\over\partial x_\nu} h_{1,\nu}, }$

${ \vdots\qquad\qquad\vdots }$

${ \left\langle\nabla f,h_{k+1}\right\rangle = \sum_{\nu=1}^n {\partial f\over\partial x_\nu} h_{k+1,\nu}. }$

REFERENCES.

1. Walter Rudin, Principles of Mathematical Analysis, Second Edition, McGraw-Hill, New York, 1964

2. Otto Forster, Analysis 3: Integralrechnung im $R^n$ mit Anwendungen, Third Edition, Friedrich Vieweg & Sohn, Braunschweig/Wiesbaden, 1984