[Math] Intuition behind $dx \wedge dy=-dy \wedge dx$

differential-forms

I was re-reading this old book of mine; and I noticed that in defining the rules of differential forms, it "makes sense" that we have the rule $dx \wedge dx=0$ because if $dx$ is infinitesimal, then to first order approximations we can ignore powers of $dx$. Similarly, the definition for the exterior derivative $d$, of a differential form $\omega=Adx+Bdy+Cdz$, $d\omega=\frac{dA}{dx}dx + \frac{dB}{dy}dy + \frac{dC}{dz}dz $ "makes sense" because it feels like we are just multiplying the top and bottom by the differentials $dx,dy,$ and $dz$.

But it is practically a miracle that by introducing the simple anti-symmetrical commutation relations for differential forms, and applying very elementary operations, we can arrive at all the results of vector calculus such as gradient and cross product, among a large amount of other well known results.

In this particular book, the authors motivate the anti-symmetry condition by properties of determinants and Jacobian's for change of variables in integration. But I was wondering if there are other ways to think about why differential forms should commute anti-symmetrically which might provide some more intuition on just why this "miracle" works.

Thanks!

Best Answer

One way of looking at the antisymmetric relation is a consequence of $dx∧dx=0$ (which feels intuitive to you). Applied to $(dx+dy)∧(dx+dy)=0$, we get $(dx∧dx)+(dx∧dy)+(dy∧dx)+(dy∧dy)=0$. So, $(dx∧dy)+(dy∧dx)=0$, so $(dx∧dy)=-(dy∧dx)$

Related Solutions

Differential Forms – Geometrical Intuition Behind Them

A 2-form is a function that eats a parallelogram (technically it eats 2 vectors, which you should think of as spanning a parallelogram) and spits out a number proportional to its area. A 3-form eats a parallelepiped (the 3-dimensional analog of a parallelogram) and spits out a number proportional to its volume. A 4-form eats a 4-dimensional parallelotope and spits out a number proportional to its hypervolume. A 1-form eats a line segment (which you can think of as a 1-dimensional parallelogram) and spits out a number proportional to its length. A 0-form eats a single point (which you can think of as a 0-dimensional parallelogram) and spits out a number, though there's nothing for it to be proportional to since a point has no extension in space. I think you get the picture. In general an n-form eats n vectors, which you should think of as spanning an n-dimensional parallelotope, and spits out a number proportional to its hypervolume.

Usually books that teach differential forms obscure this. They will define an n-form as a "real-valued multilinear, skew-symmetric function of n vectors". But it means the same thing. Multilinearity and skew-symmetry = output is proportional to length/area/volume/hypervolume. The determinant, which is used to compute the volume of a parallelepiped (and its higher and lower dimensional analogs), has the same two properties.

So why do we require forms to have this property? Well it's just because it's needed for integration. Imagine a curve you want to integrate over. The first step is to approximate it with line segments. Then you apply some function to each line segment in order to get a number. You need that number to shrink as the size of the line segment shrinks otherwise the sum won't converge. Think about it, if the output of the function was independent of the length of the input, then as more segments were added to the approximation the sum would just shoot up to infinity. Now think of a surface you want to integrate over. You can approximate it with parallelograms, imagine the scales of an armadillo. Then for each parallelogram you apply some function that spits out a number. We need the numbers to shrink as the scales do so the sum actually converges. If you want to integrate over some 3-dimensional volume, approximate it with parallelepipeds and again evaluate a function for each parallelepiped. The output of this function needs to shrink with its input for the sum to converge. These functions that we integrate over curves/surfaces/volumes/hypervolumes are forms.

Now let me explain why you write forms as linear combinations of elementary forms. It has to do with the generalized Pythagorean theorem, which I'll just call the GPT. In the same way that the length of a line segment is equal to the sum of the squared lengths of its projections onto the various coordinate axes, the area of an arbitrary parallelogram is equal to the sum of the squared areas of its projections onto the various coordinate planes. And the volume of a parallelepiped is equal to the sum of the squared volumes of its projections onto the various 3-dimensional subspaces. And so on. So the Pythagorean theorem applies to more than just line segments.

So let's look at the example of a 1-form that eats line segments embedded in 3-dimensional space. In general it's gonna look like $adx + bdy + cdz$ (if you forgot, $dx$, $dy$, and $dz$ are just functions that eat a line segment and spit out its projections on the x axis, y axis, and z axis respectively). All that's happening is you're taking the dot product of a vector $(a,b,c)$ with another vector $(dx,dy,dz)$ which equals the projection of $(a,b,c)$ onto $(dx,dy,dz)$ times the length of $(dx,dy,dz)$ (the length of $(dx,dy,dz)$ is $\sqrt{dx^2 + dy^2 + dz^2}$ ie the length of the line segment by the GPT). In other words $adx + bdy + cdz$ is literally just another way of writing: (projection of $(a,b,c)$ onto $(dx,dy,dz)$) times (length of the line segment). Since the length of the line segment is a factor in this product, the function is obviously proportional to the length of the line segment. Any 1-form can be written like this.

Another example: A 2-form that eats parallelograms embedded in 3-dimensional space is gonna have the form $a(dx \wedge dy) + b(dx \wedge dz) + c(dy \wedge dz)$ (if you forgot, $dx \wedge dy$, $dx \wedge dz$, and $dy \wedge dz$ are just functions that eat parallelograms and spit out the areas of their projections on the xy, xz, and yz planes respectively). So this is just another way of writing the dot product of $(a,b,c)$ and $(dx \wedge dy, dx \wedge dz, dy \wedge dz)$ which is just the projection of $(a,b,c)$ onto $(dx \wedge dy, dx \wedge dz, dy \wedge dz)$ times the length of $(dx \wedge dy, dx \wedge dz, dy \wedge dz)$ (which is $\sqrt{(dx \wedge dy)^2 + (dx \wedge dz)^2 + (dy \wedge dz)^2}$ ie the area of the parallelogram by the GPT). In other words the linear combination is just equal to: (projection of $(a,b,c)$ onto $(dx \wedge dy, dx \wedge dz, dy \wedge dz)$) times (area of the parallelogram). Which is clearly a function proportional to the area of the parallelogram.

Another example: A 2-form that eats parallelograms in the plane. It has the general form $a(dx \wedge dy)$. You only need one term because $dx \wedge dy$ already gives you the area of the parallelogram. In the same way $dx$ gives you the length of your line segment if you're only in 1 dimension. It's only when you're in a dimension higher than the dimension of the line segment/parallelogram/parallelepiped/parallelotope that you're gonna have to invoke the GPT ie have a linear combination of multiple elementary forms.

So hopefully you see that differential forms are actually very simple objects. They're merely generalized integrands. Other things in exterior calculus like the exterior derivative, the generalized stokes theorem, etc are similarly very simple when explained properly.

edit: a slightly cleaned up version of this post with some pictures can be found here: https://simplermath.wordpress.com/2020/02/13/understanding-differential-forms/

Differential Forms – Why is the Exterior Product an Antiderivation Intuitively?

I hope you're fine with accepting that we should impose some sort of product rule in order to define $d$ (clearly product rule holds for functions, so we would like something similar to hold for higher order forms as well).

So, suppose the exterior derivative is such that for each integer $k,l\geq 0$, there exist $a_{kl},b_{kl}\in\Bbb{R}$ such that for all $k$-forms $\omega$ and $l$-forms $\eta$, \begin{align} d(\omega\wedge \eta)=a_{kl}\,d\omega\wedge\eta+ b_{kl}\,\omega\wedge d\eta.\tag{$*$} \end{align} Of course if $k=l=0$ then $a_{00}=b_{00}=1$. On the other hand, let us invoke the alternating nature of wedge products: \begin{align} d(\omega\wedge \eta)&=d\left[(-1)^{kl}\eta\wedge \omega\right]\\ &=(-1)^{kl}[a_{lk}\,d\eta\wedge \omega + b_{lk}\eta\wedge d\omega]\\ &=(-1)^{lk}[a_{lk}(-1)^{(l+1)k}\omega\wedge d\eta + b_{lk}(-1)^{l(k+1)}d\omega\wedge \eta]\\ &= (-1)^lb_{lk}\,d\omega\wedge \eta+ (-1)^{k}a_{lk}\,\omega\wedge d\eta\tag{$**$} \end{align} So, by comparing the coefficients in $(*)$ and $(**)$, we see that for all $k,l\geq 0$, we must have \begin{align} a_{kl}&= (-1)^lb_{lk}\quad \text{and}\quad b_{kl}=(-1)^ka_{lk} \end{align} By simple index manipulation you can see that these two conditions are equivalent. Therefore, the conditions on the coefficients are as follows:

$a_{00}=1$.
For all integers $k,l\geq 0$, $b_{kl}=(-1)^ka_{lk}$.

From these, we get $b_{00}=1$, so it is consistent with the product rule of functions. Notice that thus far, we have only derived the natural consistency condition arising purely because of how the wedge product behaves; and that's where the factor of $(-1)^k$ creeps in. The usual exterior derivative is obtained by setting $a_{kl}=1$ for all integers $k,l\geq 0$, so that $b_{kl}=(-1)^k$.