📂Geometry

Differential Forms of Type k

Definition¹

Let $\omega = \sum\limits_{I} a_{I} dx_{I}$ be a $k$-form on the $n$-dimensional differential manifold $M$. The exterior differential $d\omega$ of $\omega$ is defined as follows:

$$d\omega := \sum\limits_{I} da_{I} \wedge dx_{I}$$

Here, $\wedge$ is the wedge product.

Description

Since $da_{I}$ is a $1$-form and $dx_{I}$ is a $k$-form, $d\omega$ is a $(k+1)$-form.

Example

Let $\omega$ be a $1$-form given in $\mathbb{R}^{3}$ as follows:

$$\omega = xyz dx + yz dy + (x + z)dz$$

Then, $d\omega$ is as follows:

$$\begin{align*} d\omega &= d(xyz) \wedge dx + d(yz) \wedge dy + d(x+z)\wedge dz \\ &= (yzdx + xzdy + xydz) \wedge dx + (zdy + ydz) \wedge dy + (dx + dz) \wedge dz \\ &= xzdy \wedge dx + xydz \wedge dx + ydz \wedge dy + dx \wedge dz \\ &= -xzdx \wedge dy - xydx \wedge dz - ydy \wedge dz + dx \wedge dz \\ &= -xzdx \wedge dy + (1 - xy) dx \wedge dz - ydy \wedge dz \end{align*}$$

Properties

(a) When $\omega_{1}, \omega_{2}$ is a $k$-form,

$$d(\omega_{1} + \omega_{2}) = d\omega_{1} + d\omega_{2}$$

(b) When $\omega$ is a $k$-form and $\varphi$ is a $s$-form,

$$d(\omega \wedge \varphi) = d\omega \wedge \varphi + (-1)^{k}\omega \wedge d\varphi$$

(c) $d(d\omega) = d^{2}\omega = 0$

(d) When $N, M$ is a $n, m$-dimensional differential manifold respectively, $\omega$ is a $k$-form on $M$, and $f : N \to M$ is a differentiable function,

$$d(f^{\ast} \omega) = f^{\ast}(d\omega)$$

Here, $f^{\ast}\omega$ is the pullback of $\omega$.

Proof

(a)

Let $\omega_{1} = \sum\limits_{I}a_{I}dx_{I}$ and $\omega_{2} = \sum\limits_{I}b_{I}dx_{I}$. Then, by the properties of the sum and wedge product of differential forms,

$$\begin{align*} d(\omega_{1} + \omega_{2}) &= d\left( \sum\limits_{I}(a_{I} + b_{I})dx_{I} \right) \\ &= \sum\limits_{I} d (a_{I} + b_{I}) \wedge dx_{I} \\ &= \sum\limits_{I} (d a_{I} + d b_{I}) \wedge dx_{I} \\ &= \sum\limits_{I} (d a_{I} \wedge dx_{I} + d b_{I} \wedge dx_{I}) \\ &= \sum\limits_{I} d a_{I} \wedge dx_{I} + \sum\limits_{I} d b_{I} \wedge dx_{I} \\ &= d\omega_{1} + d\omega_{2} \end{align*}$$

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(b)

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

$$\begin{align*} d(\omega \wedge \varphi) &= d\left( \sum\limits_{I}a_{I}dx_{I} \wedge \sum\limits_{J}b_{J}dx_{J} \right) \\ &= d\left( \sum\limits_{I,J}a_{I}b_{J} dx_{I} \wedge dx_{J} \right) \\ &= \sum\limits_{I,J} d\left( a_{I}b_{J} \right) \wedge dx_{I} \wedge dx_{J} \\ &= \sum\limits_{I,J} \left( b_{J}da_{I} + a_{I}db_{J} \right) \wedge dx_{I} \wedge dx_{J} \\ &= \sum\limits_{I,J} b_{J}da_{I} \wedge dx_{I} \wedge dx_{J} + \sum\limits_{I,J} a_{I}db_{J}\wedge dx_{I} \wedge dx_{J} \\ &= \left( \sum\limits_{I} da_{I} \wedge dx_{I} \right) \wedge \sum\limits_{J}b_{J} dx_{J} + (-1)^{k}\sum\limits_{I,J} a_{I}dx_{I}\wedge db_{J} \wedge dx_{J} \\ &= d\omega \wedge \varphi + (-1)^{k} \left( \sum\limits_{I}a_{I}dx_{I} \right)\wedge \left( \sum\limits_{J} db_{J}\wedge dx_{J} \right) \\ &= d\omega \wedge \varphi + (-1)^{k} \omega \wedge d\varphi \\ \end{align*}$$

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(c)

• Case 1. $k=0$

Let $\omega$ be a $0$-form.

$$\omega = f : M \to \mathbb{R}$$

Then,

$$\begin{align*} d(d\omega) = d(df) &= d\left( \sum\limits_{i=1}^{n} \dfrac{\partial f}{\partial x_{i}}dx_{i} \right) \\ &= \sum_{i} d\left( \dfrac{\partial f}{\partial x_{i}} \right) \wedge dx_{i} \\ &= \sum_{i} \left( \sum_{j} \dfrac{\partial }{\partial x_{j}} \dfrac{\partial f}{\partial x_{i}} dx_{j}\right) \wedge dx_{i} \\ &= \sum_{i, j} \left( \dfrac{\partial^{2} f}{\partial x_{j} \partial x_{i}} dx_{j}\right) \wedge dx_{i} \\ &= \begin{cases} 0 & i = j \\ 0 & i \ne j \end{cases} \\ &= 0 \end{align*}$$

In the case of $i=j$, since $dx_{i} \wedge dx_{i} = 0$, it follows that $0$. In the case of $i \ne j$,

$$\begin{align*} \left( \dfrac{\partial^{2} f}{\partial x_{j} \partial x_{i}} dx_{j}\right) \wedge dx_{i} + \left( \dfrac{\partial^{2} f}{\partial x_{i} \partial x_{j}} dx_{i}\right) \wedge dx_{j} \end{align*}$$