kth Order Differential Forms
Overview
Just as we defined the second-order differential form, we generalize to define the k-th order forms on the differential manifold $M$.
If the concept of a differential manifold is challenging, it can be simply thought of as $M = \mathbb{R}^{n}$.
Build-up
Let’s say $M$ is a $n$-dimensional differential manifold. $p \in M$ is a point in $M$, and $T_{p}M$ is the tangent space at point $p$ in $M$. $T_{p}^{\ast}M$ is the cotangent space, which is the dual space of the tangent space. Let’s define $\Lambda^{k} (T_{p}^{\ast}M)$ as a set of multilinear alternating functions as follows.
$$ \Lambda^{k} (T_{p}^{\ast}M) := \left\{ \varphi : \underbrace{T_{p}M \times \cdots \times T_{p}M}_{k \text{ times}} \to \mathbb{R}\ | \ \varphi \text{ is k-linear alternating map} \right\} $$
For $\varphi_{1}, \dots, \varphi_{k} \in T_{p}^{\ast}M$, we define the wedge product $\wedge$ as follows, making $(\varphi_{1} \wedge \varphi_{2} \wedge \cdots \wedge \varphi_{k})$ an element of $\Lambda^{k} (T_{p}^{\ast}M)$.
$$ (\varphi_{1} \wedge \varphi_{2} \wedge \cdots \wedge \varphi_{k})(v_{1}, v_{2}, \dots, v_{k}) = \det \left[ \varphi_{i}(v_{j}) \right],\quad i,j=1,\dots,k $$
For convenience, let’s denote it as follows.
$$ (dx_{i_{1}} \wedge dx_{i_{2}} \wedge \cdots \wedge dx_{i_{k}})_{p} \overset{\text{notation}}{=} (dx_{i_{1}})_{p} \wedge (dx_{i_{2}})_{p} \wedge \cdots \wedge (dx_{i_{k}})_{p} \in \Lambda^{k} (T_{p}^{\ast}M) $$
At this point, $i_{1}, i_{2}, \dots, i_{k} = 1, \dots, n$. Thus, $\Lambda^{k} (T_{p}^{\ast}M)$ becomes a vector space.
Theorem
The set below
$$ \mathcal{B} = \left\{ (dx_{i_{1}} \wedge dx_{i_{2}} \wedge \cdots \wedge dx_{i_{k}})_{p} : i_{1} \lt i_{2} \lt \cdots \lt i_{k},\ i_{j}\in \left\{ 1,\dots,n \right\} \right\} $$
is a basis of $\Lambda^{k} (T_{p}^{\ast}M)$.
Proof
According to the definition of a basis, it suffices to show that $\mathcal{B}$ is linearly independent and generates $\Lambda^{k} (T_{p}^{\ast}M)$. For convenience, let’s denote the basis of the tangent space $T_{p}M$ of $M$ as follows.
$$ \left\{ e_{i} \right\} = \left\{ \dfrac{\partial }{\partial x_{i}} \right\} $$
Part 1. Linear Independence
We only need to show that the only solution to the following equation is where all $a_{i_{1}\dots i_{k}}$ are $0$.
$$ \sum \limits_{i_{1} \lt \cdots \lt i_{k} } a_{i_{1}\dots i_{k}}dx_{i_{1}} \wedge dx_{i_{2}} \wedge \cdots \wedge dx_{i_{k}} = 0 $$
Let’s substitute
$$ \left(e_{j_{1}}, \dots, e_{j_{k}} \right),\quad j_{1}\lt \cdots \lt j_{k},\ j_{\ell} \in \left\{ 1,\dots, n \right\} $$
into it.
$$ \begin{align*} 0 =&\ \sum \limits_{i_{1} \lt \cdots \lt i_{k} } a_{i_{1}\dots i_{k}}dx_{i_{1}} \wedge dx_{i_{2}} \wedge \cdots \wedge dx_{i_{k}}\left(e_{j_{1}}, \dots, e_{j_{k}} \right) \\[1em] =&\ \sum \limits_{i_{1} \lt \cdots \lt i_{k} } a_{i_{1}\dots i_{k}} \begin{vmatrix} dx_{i_{1}}(e_{j_{1}}) & dx_{i_{1}}(e_{j_{2}}) & \cdots & dx_{i_{1}}(e_{j_{k}}) \\[1em] dx_{i_{2}}(e_{j_{1}}) & dx_{i_{2}}(e_{j_{2}}) & \cdots & dx_{i_{2}}(e_{j_{k}}) \\[1em] \vdots & \vdots & \ddots & \vdots \\[1em] dx_{i_{k}}(e_{j_{1}}) & dx_{i_{k}}(e_{j_{2}}) & \cdots & dx_{i_{k}}(e_{j_{k}}) \end{vmatrix} \end{align*} $$
Looking at the first column of the determinant, if there’s any element that is not $0$, then $j_{1} \in \left\{ i_{1}, \dots i_{k} \right\}$ must be the case. Applying this condition to subsequent columns yields the following result.
$$ j_{1}, \dots, j_{k} \in \left\{ i_{1}, \dots i_{k} \right\} $$
However, since there are conditions that both indices $i$ and $j$ are $i_{1}\lt \cdots \lt i_{k}$, $j_{1}\lt \cdots \lt j_{k}$, $i_{\ell} = j_{\ell}$ holds.
$$ 0 = a_{j_{1}\dots j_{k}} $$
By the same reasoning, it’s found that all coefficients $a$ must be $0$.
Part 2. Spanning
If $f \in \Lambda^{k} (T_{p}^{\ast}M)$, we need to show that $f$ can be expressed as a linear combination of $\mathcal{B}$, as follows.
$$ f = \sum \limits_{i_{1} \lt \cdots \lt i_{k} } a_{i_{1} \dots i_{k}} dx_{i_{1}} \wedge dx_{i_{2}} \wedge \cdots \wedge dx_{i_{k}} $$
Let’s define $g$ as follows.
$$ g = \sum \limits_{i_{1} \lt \cdots \lt i_{k} } f(e_{i_{1}},\dots,e_{i_{k}}) dx_{i_{1}} \wedge dx_{i_{2}} \wedge \cdots \wedge dx_{i_{k}} $$
Then, it becomes clear that $g$ is $f$. After substituting $(e_{i_{1}},\dots,e_{i_{k}})$,
$$ \begin{align*} g(e_{i_{1}},\dots,e_{i_{k}}) =&\ \sum \limits_{i_{1} \lt \cdots \lt i_{k} } f(e_{i_{1}},\dots,e_{i_{k}}) dx_{i_{1}} \wedge dx_{i_{2}} \wedge \cdots \wedge dx_{i_{k}}(e_{i_{1}},\dots,e_{i_{k}}) \\ =&\ f(e_{i_{1}},\dots,e_{i_{k}}) \end{align*} $$
Therefore, if we let,
$$ f = g = \sum \limits_{i_{1} \lt \cdots \lt i_{k} } a_{i_{1} \dots i_{k}} dx_{i_{1}} \wedge dx_{i_{2}} \wedge \cdots \wedge dx_{i_{k}} $$
■
Definition
The function $\omega : M \to \Lambda^{k} (T_{p}^{\ast}M)$ that maps the point $p \in M$ as follows is defined as the k-th order form in $M$.
$$ \omega (p) = \sum \limits_{i_{1} \lt \cdots \lt i_{k} } a_{i_{1}\dots i_{k}}(p)(dx_{i_{1}} \wedge dx_{i_{2}} \wedge \cdots \wedge dx_{i_{k}})_{p},\quad i_{j} \in \left\{ 1, \dots, n \right\} $$
$$ \omega = \sum \limits_{i_{1} \lt \cdots \lt i_{k} } a_{i_{1}\dots i_{k}}dx_{i_{1}} \wedge dx_{i_{2}} \wedge \cdots \wedge dx_{i_{k}} $$
At this point, $a_{i_{1}\dots i_{k}} : M \to \mathbb{R}$ holds. If each $a_{i_{1}\dots i_{k}}$ is differentiable, then $\omega$ is called a k-th order differential form. Also, for convenience, let’s denote it as $I = (i_{1},\dots,i_{k})$ and represent it as follows.
$$ \omega = \sum \limits_{I} a_{I}dx_{I} $$
■
Explanation
By definition, in a $n$-dimensional manifold, there can exist forms up to the k-th order. Moreover, a k-th order form in a $n$-dimensional manifold has $\binom{n}{k}$ terms. Therefore, $\Lambda^{k} (T_{p}^{\ast}M)$ is a $\binom{n}{k}$-dimensional vector space. Notably, a $0$-form on the differential manifold $M$ is defined by the function $f : M \to \mathbb{R}$ defined on $M$.
For example, up to 3rd order forms exist in $\mathbb{R}^{3}$.
- 0th order form: Functions on $\mathbb{R}^{3}$
- 1st order form: $a_{1}dx_{1} + a_{2}dx_{2} + a_{3}dx_{3}$
- 2nd order form: $a_{12}dx_{1}\wedge dx_{2} + a_{13}dx_{1}\wedge dx_{3} + a_{23} dx_{2} \wedge dx_{3}$
- 3rd order form: $a_{123}dx_{1} \wedge dx_{2} \wedge dx_{3}$