📂Geometry

Operations on Differential Forms: Sum and Wedge Product

Definition¹

Sum $+$

Suppose $\omega = \sum\limits_{I} a_{I} dx_{I}, \varphi = \sum\limits_{I} b_{I} dx_{I}$ is in $k$ notation. Then, the sum of these two is defined as follows.

$$ \omega + \varphi := \sum\limits_{I}\left( a_{I} + b_{I} \right)dx_{I} $$

Wedge Product $\wedge$

Let $\omega = \sum\limits_{I} a_{I}dx_{I}$, $\varphi = \sum\limits_{J} b_{J}dx_{J}$ be in $k$ notation and $s$ notation, respectively. Then, the wedge product of these two is defined as:

$$ \omega \wedge \varphi = \sum \limits_{I,J} a_{I}b_{J} dx_{I} \wedge dx_{J} $$

Explanation

Let $M$ be a $n$-dimensional differential manifold. (If the concept of differential manifold is challenging, think of it as $M=\mathbb{R}^{n}$.) The wedge product $\wedge$ turns two elements $\varphi_{1}, \varphi_{2}$ of $T_{p}^{\ast}M$ into an alternating function $\varphi_{1} \wedge \varphi_{2}$. Moreover, the wedge product itself is an alternating function. That is, for,

$$ \varphi_{i} \wedge \varphi_{j} = - \varphi_{j} \wedge \varphi_{i} \quad \text{and} \quad \varphi_{i} \wedge \varphi_{i} = 0 $$

it is easy to demonstrate according to the definition.

$$ \begin{align*} \left(\varphi_{1} \wedge \varphi_{2}\right)\left(v_{1}, v_{2}\right) =&\ \det \left[\varphi_{i}\left(v_{j}\right)\right] \\ =&\ \begin{vmatrix} \varphi_{1}\left(v_{1}\right) & \varphi_{1}\left(v_{2}\right) \\ \varphi_{2}\left(v_{1}\right) & \varphi_{2}\left(v_{2}\right) \end{vmatrix} \\ =&\ -\begin{vmatrix} \varphi_{2}\left(v_{1}\right) & \varphi_{2}\left(v_{L}\right) \\ \varphi_{1}\left(v_{1}\right) & \varphi_{1}\left(v_{2}\right) \end{vmatrix} \\ =&\ -\left(\varphi_{2} \wedge \varphi_{1}\right)\left(v_{1}, v_{1}\right) \\ \end{align*} $$

$$ \begin{align*} \left(\varphi_{1}, \varphi_{1}\right)\left(v_{1}, v_{2}\right) =&\ \det \left[\varphi_{1}\left(v_{i}\right)\right] \\ =&\ \begin{vmatrix} \varphi_{1}\left(v_{1}\right) & \varphi_{1}\left(v_{2}\right) \\ \varphi_{1}\left(v_{1}\right) & \varphi_{1}\left(v_{2}\right) \end{vmatrix} \\ =&\ 0 \end{align*} $$

By definition, the wedge product of a $k$ notation and a $s$ notation is a $k+s$ notation.

Moreover, while $dx_{i} \wedge dx_{i} = 0$ for all $i$, it is not the case that $\omega \wedge \omega = 0$ always holds. For example, if $\omega = x_{1}dx_{1}\wedge dx_{2} + x2_{2} dx_{3} \wedge dx_{4}$,

$$ \omega \wedge \omega = 2x_{1}x_{2} dx_{1} \wedge dx_{2} \wedge dx_{3} \wedge dx_{4} $$

Example

Let $\omega = x_{1}dx_{1} + x_{2}dx_{2}$ be a first-order form of $\mathbb{R}^{3}$, and $\varphi = x_{1}dx_{1}\wedge dx_{2} + x_{2} dx_{1}\wedge dx_{3} + dx_{2} \wedge dx_{3}$ be a second-order form of $\mathbb{R}^{3}$. Then,

$$ \begin{align*} \omega \wedge \varphi =&\ (x_{1}dx_{1} + x_{2}dx_{2}) \wedge (x_{1}dx_{1}\wedge dx_{2} + x_{2} dx_{1}\wedge dx_{3} + dx_{2} \wedge dx_{3}) \\ =&\ x_{1}^{2} \cancel{dx_{1} \wedge dx_{1}\wedge dx_{2}} + x_{1}x_{2}\cancel{dx_{1} \wedge dx_{1}\wedge dx_{3}} + x_{1}dx_{1} \wedge dx_{2} \wedge dx_{3} \\ & + \cancel{x_{1}x_{2}dx_{2} \wedge dx_{1}\wedge dx_{2}} + x_{2}^{2} dx_{2} \wedge dx_{1}\wedge dx_{3} + x_{2}\cancel{dx_{2} \wedge dx_{2} \wedge dx_{3}} \\ =&\ (x_{1}-x_{2}^{2}) dx_{1} \wedge dx_{2} \wedge dx_{3} \end{align*} $$

Properties

Let $\omega$ be a $k$ notation, $\varphi$ be a $s$ notation, and $\theta$ be a $r$ notation. Then,

$$ \begin{align} (\omega \wedge \varphi) \wedge \theta =&\ \omega \wedge (\varphi \wedge \theta) \\ (\omega \wedge \varphi) =&\ (-1)^{ks} (\varphi \wedge \omega) \\ \omega \wedge (\varphi + \theta) =&\ \omega \wedge \varphi + \omega \wedge \theta,\quad \text{if } r=s \end{align} $$

Proof

Proof(1)

Let $\omega = \sum\limits_{I} a_{I} dx_{I}$, $\varphi = \sum\limits_{J} b_{J} dx_{J}$, $\vartheta = \sum\limits_{L} c_{L} dx_{L}$.

$$ \begin{align*} (\omega \wedge \varphi) \wedge \vartheta =&\ \left( \sum\limits_{I}a_{I}dx_{I} \wedge \sum\limits_{J}b_{J}dx_{J} \right) \wedge \sum\limits_{L} c_{L}dx_{L} \\ =&\ \left( \sum\limits_{I,J}a_{I}b_{J} dx_{I} \wedge dx_{J} \right) \wedge \sum\limits_{L} c_{L}dx_{L} & \text{by definition of } \wedge\\ =&\ \sum\limits_{I,J,L} (a_{I}b_{J})c_{L} (dx_{I} \wedge dx_{J}) \wedge dx_{L} & \text{by definition of } \wedge\\ =&\ \sum\limits_{I,J,L} (a_{I}b_{J})c_{L} dx_{I} \wedge (dx_{J} \wedge dx_{L}) & \text{by property of } \wedge\\ \end{align*} $$

Since the functions $a, b, c$ are real-valued, the associative law with respect to multiplication holds. Therefore,

$$ \begin{align*} (\omega \wedge \varphi) \wedge \vartheta =&\ \sum\limits_{I,J,L} a_{I}(b_{J}c_{L}) dx_{I} \wedge (dx_{J} \wedge dx_{L}) \\ =&\ \sum\limits_{I} a_{I} dx_{I} \wedge \sum\limits_{J,L} b_{J}c_{L} dx_{J} \wedge dx_{L} & \text{by property of } \wedge\\ =&\ \omega \wedge ( \varphi \wedge \vartheta) & \text{by property of } \wedge\\ \end{align*} $$

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Proof(2)

Let $\omega = \sum\limits_{I} a_{I} dx_{I}$, $\varphi = \sum\limits_{J} b_{J} dx_{J}$.

$$ \begin{align*} \omega \wedge \varphi =&\ \sum\limits_{I,J} a_{I}b_{J} dx_{I} \wedge dx_{J} \\ =&\ \sum\limits_{I,J} a_{I}b_{J} dx_{i_{1}} \wedge \dots \wedge dx_{i_{k}} \wedge dx_{j_{1}} \wedge \dots \wedge dx_{j_{s}} \\ =&\ \sum\limits_{I,J} a_{I}b_{J} (-1)^{k} dx_{j_{1}} \wedge dx_{i_{1}} \wedge \dots \wedge dx_{i_{k}} \wedge dx_{j_{2}} \wedge \dots \wedge dx_{j_{s}} \\ =&\ \sum\limits_{I,J} a_{I}b_{J} (-1)^{k}(-1)^{k} dx_{j_{1}} \wedge dx_{j_{2}} \wedge dx_{i_{1}} \wedge \dots \wedge dx_{i_{k}} \wedge dx_{j_{3}} \wedge \dots \wedge dx_{j_{s}} \\ &\ \vdots \\ =&\ \sum\limits_{I,J} a_{I}b_{J} (-1)^{ks} dx_{j_{1}} \wedge dx_{j_{2}} \wedge \cdots \wedge dx_{j_{s}} \wedge dx_{i_{1}} \wedge \dots \wedge dx_{i_{k}}\\ =&\ (-1)^{ks} \sum\limits_{I,J} a_{I}b_{J} dx_{I} \wedge dx_{J} \\ =&\ (-1)^{ks} \varphi \wedge \omega \end{align*} $$

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Proof(3)

Let $\omega = \sum\limits_{I} a_{I} dx_{I}$, $\varphi = \sum\limits_{J} b_{J} dx_{J}$, $\vartheta = \sum\limits_{J} c_{J} dx_{J}$.

$$ \begin{align*} \omega =&\ \omega \wedge (\varphi + \vartheta) \\ =&\ \omega \wedge \left( \sum\limits_{J} (b_{J} + c_{J}) dx_{J} \right) & \text{by definition of } + \\ =&\ \left( \sum\limits_{I} a_{I} dx_{I} \right)\wedge \left( \sum\limits_{J} (b_{J} + c_{J}) dx_{J} \right) \\ =&\ \sum\limits_{I,J} a_{I}(b_{J} + c_{J})dx_{I}\wedge dx_{J} & \text{by definition of } \wedge\\ \end{align*} $$

Since the functions $a, b, c$ are real-valued, the distributive law holds. Therefore,

$$ \begin{align*} \omega =&\ \sum\limits_{I,J} a_{I}b_{J}dx_{I}\wedge dx_{J} + \sum\limits_{I,J} a_{I}c_{J}dx_{I}\wedge dx_{J} \\ =&\ \omega \wedge \varphi + \omega \wedge \vartheta & \text{by definition of } \wedge\\ \end{align*} $$

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