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<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

\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<j} \left({\partial f_i\over\partial x_j}                  - {\partial f_j\over\partial x_i}\right) H^{1,2}_{j,i}. \qquad(2)  }

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<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

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

\displaystyle{          w\land\lambda :=                  \sum                          _{\scriptstyle1\le i_1<\cdots<i_k\le n\atop                          \scriptstyle1\le j_1<\cdots<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.

\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

Advertisements

Leave a Reply

Please log in using one of these methods to post your comment:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s