Exterior Differentiation
The exterior product of vectors is generally only defined for spaces of no more than 3 dimensions. To utilize antisymmetric operations in higher-dimensional spaces, we require 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 treat dx^{\mu} as a set of basis vectors, calling any function a differential 0-form, and expressions such as \omega_{\mu}dx^{\mu} a differential 1-form.
On this basis dx^{\mu}, we define the exterior product \wedge, which involves antisymmetric operations dx^{\mu} \wedge dx^{\nu}. Expressions like \omega_{\mu\nu}dx^{\mu} \wedge dx^{\nu} are called differential 2-forms. Note that this is an exterior product in n-dimensional space; dx^{\mu} \wedge dx^{\nu} is actually a basis for a new space and cannot be represented as a linear combination of dx^{\mu}.
Next, we allow \wedge to be applied repeatedly, such as dx^{\mu} \wedge dx^{\nu} \wedge dx^{\lambda}, and call expressions like \omega_{\mu\nu\lambda}dx^{\mu} \wedge dx^{\nu} \wedge dx^{\lambda} differential 3-forms. Correspondingly, we 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}\wedge dx^{\mu_2} \wedge \dots\wedge dx^{\mu_p}\right)\\ =&\frac{\partial \omega_{\mu_1 \mu_2 \dots \mu_p}}{\partial x^{\mu_{p+1}}} dx^{\mu_{p+1}}\wedge dx^{\mu_1}\wedge dx^{\mu_2} \wedge \dots\wedge 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 have: d^2 \omega = 0 \tag{26} Therefore, the exterior derivative operator acts on a differential form at most twice (in the sense that the second application yields zero). Furthermore, the following identity is also easy to prove: if \alpha and \beta are differential p and q forms respectively, then: d(\alpha\wedge \beta)=d\alpha\wedge \beta + (-1)^p \alpha\wedge d\beta \tag{27} The appearance of (-1)^p is precisely due to antisymmetry. Although it is called exterior “differential” (literally “tiny division” in Chinese), its implications are not “tiny” at all.
Immediate Applications
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 sequentially construct differential forms \alpha_{\mu}^1 dx^{\mu}, \alpha_{\mu}^2 dx^{\mu}, \dots, \alpha_{\mu}^k dx^{\mu}, and then consider their exterior product: (\alpha_{\mu}^1 dx^{\mu}) \wedge (\alpha_{\mu}^2 dx^{\mu}) \wedge \dots \wedge (\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 case, 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}) \wedge (\alpha_{\mu}^2 dx^{\mu}) \wedge \dots \wedge (\alpha_{\mu}^k dx^{\mu})=0 \tag{31}
Of course, strictly speaking, this is a result of antisymmetric operations, showing that the implications of antisymmetric properties are quite rich.
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