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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

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

For example

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

Suppose

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

and let

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

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

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

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

$$ \sum_{1\le i_1\lt \cdots\lt i_k\le n} f_{i_1\ldots i_k} H{\textstyle\!{\scriptstyle1\atop\scriptstyle i_1} \!{\scriptstyle\ldots\atop\scriptstyle\ldots} \!{\scriptstyle k\atop\scriptstyle i_k}} $$

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

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

we define

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

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

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

yields

$$ dw = \sum_{\nu=1}^n {\partial f\over\partial x_\nu} H^1_\nu, \tag 1 $$

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

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

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

the result

$$ d\omega = \sum_{i=1}^n \sum_{j=1}^n {\partial f_i\over\partial x_j} H^{1,2}_{j,i} = \sum_{i\lt j} \left({\partial f_i\over\partial x_j} - {\partial f_j\over\partial x_i}\right) H^{1,2}_{j,i}. \tag 2 $$

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

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

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

delivers

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

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

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

$$ dd\omega = 0. $$

Proof: With

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

we get

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

and this is zero, because

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

and

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

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

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

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

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

Definition. Suppose

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

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

$$ f\colon E\to R. $$

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

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

and the integral over $k$-forms as

$$ \int_\phi \omega := \int_D \phi^*\omega. $$

For example the case $k=1$,

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

gives

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

3. The outer product of differential forms

Suppose

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

For the two differential forms

$$ \omega = \sum_{1\le i_1\lt \cdots\lt i_k\le n} f_{i_1\ldots i_k} H{\textstyle\!{\scriptstyle1\atop\scriptstyle i_1} \!{\scriptstyle\ldots\atop\scriptstyle\ldots} \!{\scriptstyle k\atop\scriptstyle i_k}} $$

and

$$ \lambda = \sum_{1\le j_1\lt \cdots\lt j_m\le n} g_{j_1\ldots j_m} H{\textstyle\!{\scriptstyle k+1\atop\scriptstyle j_1} \!{\scriptstyle\ldots\atop\scriptstyle\ldots} \!{\scriptstyle k+m\atop\scriptstyle j_m}} $$

the outer product is defined as

$$ w\land\lambda := \sum _{\scriptstyle1\le i_1\lt \cdots\lt i_k\le n\atop \scriptstyle1\le j_1\lt \cdots\lt j_m\le n} f_{i_1\ldots i_k} g_{j_1\ldots j_m} H{\textstyle\!{\scriptstyle1\atop\scriptstyle i_1} \!{\scriptstyle\ldots\atop\scriptstyle\ldots} \!{\scriptstyle k\atop\scriptstyle i_k} \!{\scriptstyle k+1\atop\scriptstyle j_1} \!{\scriptstyle\ldots\atop\scriptstyle\ldots} \!{\scriptstyle k+m\atop\scriptstyle j_m}} . $$

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

Theorem.

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

Proof: With

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

then

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

due to

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

and

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

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

Theorem. Suppose

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

and

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

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

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

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

Proof:

$$ d\omega = \left|\begin{matrix} \left\langle\nabla f,h_1\right\rangle & \ldots & \left\langle\nabla f,h_k\right\rangle & \left\langle\nabla f,h_{k+1}\right\rangle \\ \left\langle e_{i_1},h_1\right\rangle & \ldots & \left\langle e_{i_1},h_k\right\rangle & \left\langle e_{i_1},h_{k+1}\right\rangle \\ \vdots & \ddots & \vdots & \vdots\cr \left\langle e_{i_k},h_1\right\rangle & \ldots & \left\langle e_{i_k},h_k\right\rangle & \left\langle e_{i_k},h_{k+1}\right\rangle \cr \end{matrix}\right| $$
$$ \qquad = \sum_{\nu=1}^n {\partial f\over\partial x_\nu} \left|\begin{matrix} h_{1,\nu} & h_{1,i_1} & \ldots & h_{1,i_k}\\ \vdots & \vdots & \ddots & \vdots\\ h_{k,\nu} & h_{k,i_1} & \ldots & h_{k,i_k}\\ h_{k+1,\nu}& h_{k+1,i_1} & \ldots & h_{k+1,i_k}\\ \end{matrix}\right| $$

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