[A Brief Introduction to Exterior Differentiation] 4. Differentials Aren't So Small

【外微分浅谈】4. 微分不微 苏剑林 · 2016-11-05 · Mathematics vector symmetry exterior differentiation

Exterior differentiation

The exterior product of vectors is generally only defined in spaces of dimension no greater than 3. To make use of 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 $x$ variables can be written as a linear combination of $dx^{\mu}$. Here each $dx^{\mu}$ effectively plays the role of a basis element, so we may as well regard $dx^{\mu}$ as a set of basis elements, and call any function a differential 0-form, while an expression such as $\omega_{\mu}dx^{\mu}$ is called a differential 1-form.

On top of the basis $dx^{\mu}$, we define the exterior product $\land$, i.e. an antisymmetric operation $dx^{\mu}\land dx^{\nu}$, and we call an expression such as $\omega_{\mu\nu}dx^{\mu}\land dx^{\nu}$ a differential 2-form. Note that this is an exterior product in an $n$-dimensional space, and $dx^{\mu}\land dx^{\nu}$ is in fact a basis element of a new space, one that cannot be represented as a linear combination of $dx^{\mu}$.

Next, we allow $\land$ to be applied repeatedly, i.e. we allow $dx^{\mu}\land dx^{\nu}\land dx^{\lambda}$, and we call an expression such as $\omega_{\mu\nu\lambda}dx^{\mu}\land dx^{\nu}\land dx^{\lambda}$ a differential 3-form. Correspondingly, we can define a general differential $p$-form. As for its geometric meaning, we'll get to that later.

Finally, we define an exterior differentiation operator $d$, which allows us to produce a differential $p+1$-form from a differential $p$-form:

$$\begin{aligned}&d\left(\omega_{\mu_1 \mu_2 \dots \mu_p} dx^{\mu_1}\land dx^{\mu_2} \land \dots\land dx^{\mu_p}\right)\\ =&\frac{\partial \omega_{\mu_1 \mu_2 \dots \mu_p}}{\partial x^{\mu_{1+1}}} dx^{\mu_{p+1}}\land dx^{\mu_1}\land dx^{\mu_2} \land \dots\land dx^{\mu_p}\end{aligned} \tag{25} $$

In fact, this operator $d$ is formally consistent with the ordinary differentiation operator, except that it is allowed to be applied repeatedly. However, it is not hard to prove that for any differential form $\omega$, we always have

$$d^2 \omega = 0 \tag{26} $$

Hence the exterior differentiation operator can be applied to a differential form at most twice in succession. Besides this, the following identity is also not hard to prove: if $\alpha,\beta$ are differential $p,q$-forms respectively, then

$$d(\alpha\land \beta)=d\alpha\land \beta + (-1)^p \alpha\land d\beta \tag{27} $$

The appearance of $(-1)^p$ here is precisely a consequence of antisymmetry. So although it's called exterior "differentiation," its content is anything but "differential" in the trivial sense — there's a lot packed into it.

An immediate application

We know that determinants can be used to determine whether $n$ vectors in $n$-dimensional space are linearly dependent. But what about $k$ vectors?

The exterior product can help us here! Consider $k$ vectors $\alpha_{\mu}^1,\alpha_{\mu}^2,\dots,\alpha_{\mu}^k$. We can construct, in turn, the differential forms $\alpha_{\mu}^1 dx^{\mu},\alpha_{\mu}^2 dx^{\mu},\dots,\alpha_{\mu}^k dx^{\mu}$, 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 — that is, if one of them can be expressed as a linear combination of the remaining $k-1$ vectors — say we assume

$$\alpha_{\mu}^1 = \sum_{i=2}^{k} b_i \alpha_{\mu}^i \tag{29} $$

then

$$\alpha_{\mu}^1 dx^{\mu} = \sum_{i=2}^{k} b_i \alpha_{\mu}^i dx^{\mu} \tag{30} $$

so the exterior product of these $k$ differential forms must necessarily be zero. The converse also holds. That is to say, $k$ 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 simply a consequence of the antisymmetric operation, which shows just how rich the content of antisymmetry really is.