By 苏剑林 | November 05, 2016
Exterior Differentiation
The exterior product of vectors is generally only defined for spaces of no more than 3 dimensions. To use antisymmetric operations in higher-dimensional spaces, we need the differential forms and exterior differentiation described below.
We know that the differential of any function of $x$ can be written as a linear combination of $dx^{\mu}$. Here, each $dx^{\mu}$ actually plays the role of a basis. Therefore, we might as well regard $dx^{\mu}$ as a set of basis vectors, and we call any arbitrary function a differential 0-form, while expressions like $\omega_{\mu}dx^{\mu}$ are called differential 1-forms.
On the basis of $dx^{\mu}$, we define the wedge product $\land$, which provides an antisymmetric operation $dx^{\mu}\land dx^{\nu}$, and we call expressions such as $\omega_{\mu\nu}dx^{\mu}\land dx^{\nu}$ differential 2-forms. Note that this is the wedge product in an $n$-dimensional space; $dx^{\mu}\land dx^{\nu}$ is actually a basis for a new space and cannot be represented as a linear combination of $dx^{\mu}$.
Next, we allow $\land$ to be applied repeatedly, allowing for $dx^{\mu}\land dx^{\nu}\land dx^{\lambda}$, and we call expressions such as $\omega_{\mu\nu\lambda}dx^{\mu}\land dx^{\nu}\land dx^{\lambda}$ differential 3-forms. Correspondingly, one can define a general differential $p$-form. As for its geometric meaning, we will discuss that later.
Finally, we define an exterior derivative operator $d$, which allows us to generate a differential $(p+1)$-form from a differential $p$-form:
\begin{aligned}
&d\left(\omega_{\mu_1 \mu_2 \dots \mu_p} dx^{\mu_1}\land dx^{\mu_2} \land \dots\land dx^{\mu_p}\right)\\
=&\frac{\partial \omega_{\mu_1 \mu_2 \dots \mu_p}}{\partial x^{\mu_{p+1}}} dx^{\mu_{p+1}}\land dx^{\mu_1}\land dx^{\mu_2} \land \dots\land dx^{\mu_p}
\end{aligned} \tag{25}
In fact, this operator $d$ is formally consistent with the ordinary differential operator, except that it allows for repeated application. However, it is not difficult to prove that for any differential form $\omega$, we always have:
\[d^2 \omega = 0 \tag{26}\]
Therefore, the exterior derivative operator acts on a differential form twice at most. Furthermore, the following identity is also not difficult to prove: if $\alpha, \beta$ are differential $p$-forms and $q$-forms respectively, then:
\[d(\alpha\land \beta)=d\alpha\land \beta + (-1)^p \alpha\land d\beta \tag{27}\]
The appearance of $(-1)^p$ is precisely due to antisymmetry. Although it is called exterior "micro"-differentiation (外“微”分), its implications are not "micro" (tiny) at all.
Immediate Application
We know that determinants can be used to determine whether $n$ vectors in an $n$-dimensional space are linearly dependent. But what about $k$ vectors?
The exterior product can help us! Consider $k$ vectors $\alpha_{\mu}^1,\alpha_{\mu}^2,\dots,\alpha_{\mu}^k$. We can construct the differential forms $\alpha_{\mu}^1 dx^{\mu},\alpha_{\mu}^2 dx^{\mu},\dots,\alpha_{\mu}^k dx^{\mu}$ in sequence, and then consider the exterior product:
\[(\alpha_{\mu}^1 dx^{\mu}) \land (\alpha_{\mu}^2 dx^{\mu}) \land \dots \land (\alpha_{\mu}^k dx^{\mu}) \tag{28}\]
If these $k$ vectors are linearly dependent, meaning one of them can be expressed as a linear combination of the remaining $k-1$ vectors, for example, assuming:
\[\alpha_{\mu}^1 = \sum_{i=2}^{k} b_i \alpha_{\mu}^i \tag{29}\]
then
\[\alpha_{\mu}^1 dx^{\mu} = \sum_{i=2}^{k} b_i \alpha_{\mu}^i dx^{\mu} \tag{30}\]
In this way, the exterior product of these $k$ differential forms must be zero, and the converse is also true. That is, $k$ vectors are linearly dependent if and only if:
\[(\alpha_{\mu}^1 dx^{\mu}) \land (\alpha_{\mu}^2 dx^{\mu}) \land \dots \land (\alpha_{\mu}^k dx^{\mu})=0 \tag{31}\]
Of course, strictly speaking, this is a result of antisymmetric operations, showing that the connotations of antisymmetric properties are quite rich.