【A Gentle Introduction to Exterior Differentiation】5. Geometric Meaning

【外微分浅谈】5. 几何意义 苏剑林 · 2016-11-06 · Mathematics integral geometry exterior differentiation

The exterior differentiation discussed earlier, along with the integration of differential forms that we'll touch on a bit later, is purely algebraic in nature and carries no intrinsic geometric meaning of its own. However, we can associate certain formulas or definitions with geometric content, which helps us understand them more deeply and use them more flexibly. But this is only a correspondence, and it depends on our interpretation. For example, we say that the exterior differentiation formula

$$\int_{\partial D} Pdx+Qdy = \int_{D} \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\land dy \tag{32} $$

corresponds to Green's formula

$$\int_{\partial D} Pdx+Qdy = \int_{D} \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy \tag{33} $$

.

This is perfectly fine, but they are not equivalent — they merely happen to share the same form. Green's formula describes the relationship between the integral over a closed curve and the surface integral, while the exterior differentiation formula is a purely algebraic operation. In fact, you could just as well map $dx\land dy$ to $-dxdy$ instead of $dxdy$, which would give you a different geometric correspondence.

A deeper question is: why does this particular correspondence hold? That is, why is it that after some adjustments and interpretation, we can arrive at a correspondence with integral formulas? First, it should be made clear that the exterior product differs from ordinary multiplication of numbers in no way other than antisymmetry, so many properties are preserved. Second, we should return to antisymmetry itself: the determinant of a matrix represents the volume of the $n$-dimensional solid spanned by the vectors corresponding to that matrix, and yet the determinant is antisymmetric — which implies that antisymmetric operations have an intrinsic connection to volume, integration, and the like. That said, I myself have not yet arrived at a more refined understanding of this.

Furthermore, when we speak of seeking the geometric meaning of differential forms, we're generally only discussing spaces of dimension no greater than 3 — it's very hard to visualize the geometric picture in higher dimensions, especially for surface integrals in higher dimensions, which are generally handled only by analogy. And whether such an analogy actually holds sometimes requires further scrutiny. So in such cases, it might be simpler to just say: what differential forms describe is geometry, rather than continuing to search for a so-called geometric meaning. In other words, turning things around, we take differential forms and exterior differentiation as axiomatic first principles from which to define geometry.

You could even go so far as to treat exterior differentiation merely as an effective way to remember various differentiation and integration formulas. For instance, if I asked you right now to write down Stokes' formula in three-dimensional space from memory, you'd probably get confused, because you might not remember which term is subtracted from which. But within the framework of exterior differentiation, you can quickly derive it from scratch. Take, for example, equation $(11)$: if you insist on finding a geometric interpretation, it's Kepler's second law — equal areas are swept out in equal times. But even without a geometric interpretation, you can still go ahead and solve the equation. more

The rest of this post will elaborate on geometric interpretation.

Exterior Product: Spanning and Projecting

Let's consider the exterior product of two differential 1-forms, for example

$$\begin{aligned}\alpha_{\mu}dx^{\mu} \land \beta_{\nu}dx^{\nu} =&\alpha_{\mu}\beta_{\nu} dx^{\mu}\land dx^{\nu}\\ =&\sum_{\mu < \nu} (\alpha_{\mu}\beta_{\nu}-\beta_{\mu}\alpha_{\nu}) dx^{\mu}\land dx^{\nu} \\ =&\frac{1}{2}(\alpha_{\mu}\beta_{\nu}-\beta_{\mu}\alpha_{\nu}) dx^{\mu}\land dx^{\nu}\end{aligned} \tag{34} $$

When the summation symbol is omitted, it's understood that $\mu,\nu$ each range independently and unrestrictedly over the summation. Notice that

$$\alpha_{\mu}\beta_{\nu}-\beta_{\mu}\alpha_{\nu}=\det\begin{pmatrix}\alpha_{\mu}&\alpha_{\nu}\\\beta_{\mu}&\beta_{\nu}\end{pmatrix} \tag{35} $$

So, if we regard $dx^{\mu}$ as a basis, then for a chosen pair $\mu,\nu$, $\alpha_{\mu}\beta_{\nu}-\beta_{\mu}\alpha_{\nu}$ corresponds exactly to the signed area of the projection, onto the $dx^{\mu},dx^{\nu}$-plane, of the parallelogram spanned by the vectors $\alpha_{\mu}$ and $\beta_{\nu}$.

For general differential $p$-forms and differential $q$-forms, their exterior product can be constructed similarly, only it's harder to visualize in higher dimensions. For instance, the exterior product of a differential 1-form and a differential 2-form can be imagined as an ordinary vector and an "area vector" (actually a tensor) spanning a parallelepiped; each term of the exterior product result is the volume of the projection of that parallelepiped onto the corresponding three-dimensional subspace, and so on. In particular, in $n$-dimensional space, if we take the exterior product of $n$ differential 1-forms, the result is

$$\alpha_{\mu_1}^{1} dx^{\mu_1}\land \dots \land \alpha_{\mu_n}^{n} dx^{\mu_n}=\det(\alpha_{\mu}^{\nu}) dx^1 \land \dots \land dx^n \tag{36} $$

which is precisely the determinant of a matrix — quite remarkable, and a direct manifestation of antisymmetry. Antisymmetry also exists in determinants: swapping two rows or two columns of a determinant flips its sign. Let $f$ be an arbitrary function, and suppose $df=\frac{\partial f}{\partial x^{\mu}} dx^{\mu}$; then we have

$$df^1 \land \dots \land df^n = \det\left(\frac{\partial f^{\mu}}{\partial x^{\nu}}\right) dx^1 \land \dots \land dx^n \tag{37} $$

From the point of view of change of variables, $\det\left(\frac{\partial f^{\mu}}{\partial x^{\nu}}\right)$ is exactly the Jacobian determinant of the integral transformation. This tempts us to drop $\land$ altogether and simply regard $dx^1 \land \dots \land dx^n$ directly as the integration element $dx^1\dots dx^n$. And indeed, that's exactly what we do! We define $dx^{\mu_1}\land \dots \land dx^{\mu_k}=\pm dx^{\mu_1}\dots dx^{\mu_k}$, where whether it's positive or negative depends on what specific geometric content we want to interpret it as. In this way we can express the theory of integration using exterior differentiation.

The Differential Operator: Going Around a Loop

Something worth understanding more deeply is equation $(25)$ — that is, exactly what happens when going from a $p$-form to a $p+1$-form, or in other words, what geometric content this corresponds to.

Let's again start from a differential 1-form, and consider $\omega_{\nu} dx^{\nu}$. Under the operator $d$, we have

$$\begin{aligned}d(\omega_{\nu} dx^{\nu}) =& \frac{\partial \omega_{\nu}}{\partial x^{\mu}} dx^{\mu} \land dx^{\nu}\\ =&\sum_{\mu < \nu} \left(\frac{\partial \omega_{\nu}}{\partial x^{\mu}}-\frac{\partial \omega_{\mu}}{\partial x^{\nu}}\right) dx^{\mu} \land dx^{\nu}\\ =&\frac{1}{2} \left(\frac{\partial \omega_{\nu}}{\partial x^{\mu}}-\frac{\partial \omega_{\mu}}{\partial x^{\nu}}\right) dx^{\mu} \land dx^{\nu}\end{aligned} \tag{38} $$

What geometric correspondence does something of this shape have? We can think of $\omega_{\nu} dx^{\nu}$ as the increment of the quantity $\Omega$ going from $x$ to $x+dx$, that is,

$$\Omega (x+dx) = \Omega (x) + \omega_{\nu}(x) dx^{\nu} \tag{39} $$

Then, what if we go further from $x+dx$ to $x+dx+\delta x$? Naturally,

$$\begin{aligned}\Omega_1 (x+dx+\delta x) =& \Omega (x+dx) + \omega_{\nu}(x+dx) \delta x^{\nu}\\ =&\Omega (x) + \omega_{\nu}(x) dx^{\nu} + \omega_{\nu}(x) \delta x^{\nu} + \frac{\partial \omega_{\nu}}{\partial x^{\mu}} dx^{\mu} \delta x^{\nu} \end{aligned} \tag{40} $$

This corresponds to walking along the path $x\to x+dx\to x+dx+\delta x$. If we swap $dx$ and $\delta x$ — that is, walk the path $x\to x+\delta x\to x+\delta x+ dx$ instead — we get

$$\Omega_2 (x+dx+\delta x) =\Omega (x) + \omega_{\nu}(x) \delta x^{\nu} + \omega_{\nu}(x) d x^{\nu} + \frac{\partial \omega_{\nu}}{\partial x^{\mu}} \delta x^{\mu} d x^{\nu} \tag{41} $$

The difference between the two

$$\label{bihecha}\begin{aligned}\left(\frac{\partial \omega_{\nu}}{\partial x^{\mu}}-\frac{\partial \omega_{\mu}}{\partial x^{\nu}}\right)dx^{\mu} \delta x^{\nu}=&\frac{1}{2}\left(\frac{\partial \omega_{\nu}}{\partial x^{\mu}}-\frac{\partial \omega_{\mu}}{\partial x^{\nu}}\right)(dx^{\mu} \delta x^{\nu}-dx^{\nu} \delta x^{\mu})\\ =&\frac{1}{2}\left(\frac{\partial \omega_{\nu}}{\partial x^{\mu}}-\frac{\partial \omega_{\mu}}{\partial x^{\nu}}\right) \det\begin{pmatrix}dx^{\mu} & \delta x^{\mu}\\ dx^{\nu} & \delta x^{\nu}\end{pmatrix}\end{aligned} \tag{42} $$

is exactly the change produced after strolling once around the closed path $x\to x+dx\to x+dx+\delta x\to x+\delta x\to x$.

If we let

$$dx^{\mu} \land dx^{\nu} = \det\begin{pmatrix}dx^{\mu} & \delta x^{\mu}\\ dx^{\nu} & \delta x^{\nu}\end{pmatrix} \tag{43} $$

then equation $(42)$ is exactly $d(\omega_{\nu} dx^{\nu})$. And $\det\begin{pmatrix}dx^{\mu} & \delta x^{\mu}\\ dx^{\nu} & \delta x^{\nu}\end{pmatrix}$ is precisely the area of the projection, onto the $x^{\mu},x^{\nu}$-plane, of the parallelogram spanned by the two vectors $dx$ and $\delta x$ — which is itself antisymmetric. From this point of view, we can interpret $dx^{\mu} \land dx^{\nu}$ as a signed area element, and the meaning of $d(\omega_{\nu} dx^{\nu})$ is exactly the change in a quantity after going around a small loop and returning to the starting point!

The Fundamental Theorem of Calculus

Through this "going around a loop" approach, we've explained the meaning of going from a differential 1-form to a 2-form. Unfortunately, going from a general $p$-form to a $p+1$-form isn't so easy to visualize, and in fact, the geometric picture of integration in spaces of more than 3 dimensions is also very hard for us to imagine. So here we take a path that "puts the cart before the horse," so to speak. If $\omega$ is a differential $p$-form and $D$ is a given region, then

$$\int_{\partial D} \omega = \int_{D} d\omega \tag{44} $$

That is to say, the integral of $\omega$ over the boundary equals the integral of $d\omega$ over the region — this is the "Stokes' formula" of differential forms, and it can also be considered the fundamental theorem of calculus within exterior differentiation.

What makes this formula so celebrated is that it unifies the Newton–Leibniz formula, Green's formula, Gauss's theorem, and Stokes' formula, generalizing all of them at once. You might be puzzled about "what is the integral of a differential form?" In fact, there's nothing especially unusual about the integral of a differential form, because an expression such as $dx^{\mu}\land dx^{\nu}$ is, aside from its antisymmetry, no different from an ordinary differential element $dx^{\mu}dx^{\nu}$, and the definition of integration (say, taking the simple Riemann integral definition) has nothing to do with whether something is symmetric or antisymmetric.

In this way, we can imagine that going from a general $p$-form to a $p+1$-form, or in other words from $\omega$ to $d\omega$, is in fact doing something analogous to "going around a loop," just like going from 1-forms to 2-forms: we understand $\omega$ as the change occurring through motion along the boundary, while $d\omega$ is the change produced after traversing a small region and returning to the start. Then it follows naturally that, over a closed region $D$, we have $\int_{\partial D} \omega = \int_{D} d\omega$.

Of course, as mentioned earlier, this is a path that "puts the cart before the horse." This integral theorem is actually the "result," not the "cause" — it requires a lengthy proof. Here we've cited it without proof, using it instead, in reverse, to explain the meaning of $d\omega$. This is only meant to give everyone a way of thinking that is as quick and as clear as possible.