By 苏剑林 | November 06, 2016
Regarding the previously described exterior differentiation, as well as the integration of differential forms that will be briefly mentioned later, these are purely algebraic definitions and do not inherently possess any geometric meaning. However, we can associate certain formulas or definitions with geometric content to help us understand them more deeply and apply them more flexibly. However, this is merely a correspondence, and it depends on our interpretation. For example, we say that the exterior derivative formula
$$\int_{\partial D} Pdx+Qdy = \int_{D} \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx \wedge dy \tag{32}$$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 fine, but they are not equivalent; they are merely identical in form. Green's formula describes the relationship between the line integral along a closed curve and the double integral over the area, while the exterior derivative formula is a purely algebraic operation. You could just as easily define the correspondence such that $dx \wedge dy$ corresponds to $-dxdy$ instead of $dxdy$, which would yield a different geometric correspondence.
A deeper question is: why does this correspondence exist at all? That is to say, why is it that after some adjustments and interpretations, we can obtain a correspondence with integral formulas? First, it must be clear that the exterior product and the ordinary product of numbers, aside from antisymmetry, have no difference, so many properties are preserved. Second, one must return to the concept of antisymmetry itself: the determinant of a matrix represents the volume of the $n$-dimensional solid spanned by the vectors in the matrix. Since the determinant is antisymmetric, this implies that antisymmetric operations have an inherent connection with volume and integration. Of course, the author has yet to achieve a more detailed realization of this.
Furthermore, when we seek the geometric meaning of differential forms, we are usually discussing spaces with no more than 3 dimensions. Geometric images of higher-dimensional spaces are difficult to visualize, especially higher-dimensional surface integrals, which are generally handled by analogy—though whether the analogy holds sometimes requires further scrutiny. Therefore, in such cases, it might be better to simply state that what differential forms describe is geometry, rather than looking for a so-called geometric meaning. In other words, conversely, one can take differential forms and exterior differentiation as axiomatic first principles to define geometry.
You can even treat exterior differentiation merely as an effective way to memorize various differential and integral formulas. For instance, if I were to ask everyone to write down the 3D Stokes' formula from memory right now, people would likely get confused because they might not remember which term subtracts which. But within the framework of exterior calculus, it can be derived very quickly. It is much like formula $(11)$; even if one does not seek a geometric explanation (which would be Kepler's Second Law: equal areas swept in equal time), one can still solve the equations.
The following discussion will revolve around geometric interpretations.
Consider the exterior product of two differential 1-forms, for example:
$$\begin{aligned}\alpha_{\mu}dx^{\mu} \wedge \beta_{\nu}dx^{\nu} =&\alpha_{\mu}\beta_{\nu} dx^{\mu}\wedge dx^{\nu}\\ =&\sum_{\mu < \nu} (\alpha_{\mu}\beta_{\nu}-\beta_{\mu}\alpha_{\nu}) dx^{\mu}\wedge dx^{\nu} \\ =&\frac{1}{2}(\alpha_{\mu}\beta_{\nu}-\beta_{\mu}\alpha_{\nu}) dx^{\mu}\wedge dx^{\nu}\end{aligned} \tag{34}$$When the summation symbol is omitted, it indicates that $\mu, \nu$ are summed over their range without constraint. Notice that
$$\alpha_{\mu}\beta_{\nu}-\beta_{\mu}\alpha_{\nu}=\det\begin{pmatrix}\alpha_{\mu}&\alpha_{\nu}\\\beta_{\mu}&\beta_{\nu}\end{pmatrix} \tag{35}$$If we treat $dx^{\mu}$ as a basis, then for a selected pair $\mu, \nu$, $\alpha_{\mu}\beta_{\nu}-\beta_{\mu}\alpha_{\nu}$ corresponds exactly to the oriented area of the projection of the parallelogram spanned by vector $\alpha_\mu$ and vector $\beta_\nu$ onto the $dx^{\mu}, dx^{\nu}$ plane.
For general differential $p$-forms and $q$-forms, their exterior product can be constructed similarly, though higher dimensions are harder to visualize. For instance, the exterior product of a 1-form and a 2-form can be imagined as an ordinary vector and an "area vector" (effectively a tensor) spanning a cube. Each term in the result of the exterior product is the volume of the projection of that cube onto the corresponding three-dimensional subspace, and so on. In particular, in an $n$-dimensional space, the exterior product of $n$ differential 1-forms will be:
$$\alpha_{\mu_1}^{1} dx^{\mu_1}\wedge \dots \wedge \alpha_{\mu_n}^{n} dx^{\mu_n}=\det(\alpha_{\mu}^{\nu}) dx^1 \wedge \dots \wedge dx^n \tag{36}$$That is, it exactly produces the determinant of a matrix, which is remarkable and a direct manifestation of antisymmetry. Antisymmetry also exists in determinants; swapping two rows or columns changes the sign of the determinant. Let $f$ be any function; we have $df=\frac{\partial f}{\partial x^{\mu}} dx^{\mu}$, so
$$df^1 \wedge \dots \wedge df^n = \det\left(\frac{\partial f^{\mu}}{\partial x^{\nu}}\right) dx^1 \wedge \dots \wedge dx^n \tag{37}$$From a transformation perspective, $\det\left(\frac{\partial f^{\mu}}{\partial x^{\nu}}\right)$ is the Jacobian determinant of the integral transformation. This tempts us to ignore the $\wedge$ symbol and treat $dx^1 \wedge \dots \wedge dx^n$ directly as the integral element $dx^1 \dots dx^n$. In fact, this is exactly what is done! We define $dx^{\mu_1}\wedge \dots \wedge dx^{\mu_k}=\pm dx^{\mu_1}\dots dx^{\mu_k}$, where the sign depends on the specific geometric interpretation we wish to apply. This allows us to represent integral theory using exterior calculus.
Even more worthy of profound understanding is formula $(25)$, which describes what happens when moving from a $p$-form to a $p+1$ form, or rather, what geometric content it corresponds to.
Starting again from a differential 1-form, consider $\omega_{\nu} dx^{\nu}$. Under the action of the operator $d$, we have:
$$\begin{aligned}d(\omega_{\nu} dx^{\nu}) =& \frac{\partial \omega_{\nu}}{\partial x^{\mu}} dx^{\mu} \wedge dx^{\nu}\\ =&\sum_{\mu < \nu} \left(\frac{\partial \omega_{\nu}}{\partial x^{\mu}}-\frac{\partial \omega_{\mu}}{\partial x^{\nu}}\right) dx^{\mu} \wedge dx^{\nu}\\ =&\frac{1}{2} \left(\frac{\partial \omega_{\nu}}{\partial x^{\mu}}-\frac{\partial \omega_{\mu}}{\partial x^{\nu}}\right) dx^{\mu} \wedge dx^{\nu}\end{aligned} \tag{38}$$What is the geometric correspondence for something of this shape? We can view $\omega_{\nu} dx^{\nu}$ as the increment of a quantity $\Omega$ from $x$ to $x+dx$, i.e.,
$$\Omega (x+dx) = \Omega (x) + \omega_{\nu}(x) dx^{\nu} \tag{39}$$Then, what if we go from $x+dx$ to $x+dx+\delta x$? Naturally, we have:
$$\begin{aligned}\Omega_1 (x+dx+\delta x) =& \Omega (x+dx) + \omega_{\nu}(x+dx) \delta x^{\nu}\\ =&\Omega (x) + \omega_{\nu}(x) dx^{\nu} + \omega_{\nu}(x) \delta x^{\nu} + \frac{\partial \omega_{\nu}}{\partial x^{\mu}} dx^{\mu} \delta x^{\nu} \end{aligned} \tag{40}$$This follows the path $x\to x+dx\to x+dx+\delta x$. Swapping $dx$ and $\delta x$, i.e., following the path $x\to x+\delta x\to x+\delta x+ dx$, yields:
$$\Omega_2 (x+dx+\delta x) =\Omega (x) + \omega_{\nu}(x) \delta x^{\nu} + \omega_{\nu}(x) d x^{\nu} + \frac{\partial \omega_{\nu}}{\partial x^{\mu}} \delta x^{\mu} d x^{\nu} \tag{41}$$The difference between the two:
\begin{equation}\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}\end{equation}is the variation produced after wandering around the closed path $x\to x+dx\to x+dx+\delta x\to x+\delta x\to x$.
If we let
$$dx^{\mu} \wedge dx^{\nu} = \det\begin{pmatrix}dx^{\mu} & \delta x^{\mu}\\ dx^{\nu} & \delta x^{\nu}\end{pmatrix} \tag{43}$$then formula $(42)$ is exactly $d(\omega_{\nu} dx^{\nu})$. And $\det\begin{pmatrix}dx^{\mu} & \delta x^{\mu}\\ dx^{\nu} & \delta x^{\nu}\end{pmatrix}$ is the area of the projection of the parallelogram spanned by the two vectors $dx$ and $\delta x$ onto the $x^{\mu}, x^{\nu}$ plane, which is also antisymmetric. From this perspective, $dx^{\mu} \wedge dx^{\nu}$ can be interpreted as an oriented area element, and the meaning of $d(\omega_{\nu} dx^{\nu})$ is the change in a quantity after traversing a small circle!
Through the "going in circles" approach, we explained the meaning of moving from a differential 1-form to a 2-form. Unfortunately, moving from a general $p$-form to a $p+1$ form is not as easy to visualize; furthermore, it is inherently difficult for us to imagine the geometric imagery of integrals in spaces exceeding 3 dimensions. Therefore, we use a "backward" (putting the cart before the horse) path here. 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 is equal to the integral of $d\omega$ within the region. This is the "Stokes' formula" in differential forms, or one could say the fundamental theorem of calculus in exterior calculus.
The beauty of this formula lies in its unification and generalization of the Newton-Leibniz formula, Green's theorem, Gauss's divergence theorem, and Stokes' theorem. One might be confused by "What is the integration of a differential form?" In fact, there is nothing particularly special about the integration of differential forms, because expressions like $dx^{\mu}\wedge dx^{\nu}$, aside from being antisymmetric, are no different from ordinary differential elements $dx^{\mu}dx^{\nu}$, and the definition of an integral (such as using a simple Riemann integral definition) is independent of symmetry or antisymmetry.
In this way, we can imagine that moving from a general $p$-form to a $p+1$ form, or from $\omega$ to $d\omega$, actually does something similar to "going in circles," just like the transition from the 1-form to the 2-form. That is, $\omega$ is interpreted as the motion/change on the boundary, while $d\omega$ is the variation produced after traversing a small region. Thus, it is natural that in a closed region $D$, we have $\int_{\partial D} \omega = \int_{D} d\omega$.
Of course, as mentioned before, this is a "backward" approach. This integral theorem is actually a "result" rather than a "cause," and it requires a lengthy proof. We have cited it here without proof simply to provide everyone with the quickest and clearest possible intuition for the meaning of $d\omega$.