This symbol denotes both derivative and boundary: why?

This symbol is used for two things in maths. One of them is partial derivatives. For a multivariable function f, its partial derivatives are denoted by a curly d. This symbol is also used for boundaries. Curly d u denotes the boundary of u. Is it a coincidence we use the same symbol for both concepts? Well, yes and no. Historically, it really was a fluke we ended up using the same letter, but it's a beautiful coincidence because there's a deep connection between them. We can see this from a one-dimensional integral. Schematically, the integral of a derivative equals the values at the boundary. Here, the boundary is a and b, the endpoints of the interval we're integrating over. If you've studied vector calculus, you've also seen this story before in two theorems, Stokes and divergence. Both theorems state that the integral of some derivatives, which are curl and divergence respectively, equals an integral around the boundary. The reason they all tell the same story is that they are just special cases of a much more general theorem known as the generalized Stokes theorem. To understand it, let's start from the one-dimensional version, approaching it from a slightly different perspective.

First, we write down the definition of a derivative, which is the limit of f of x plus delta x minus f of x over delta x as delta x tends to zero. For small delta x, we can approximate the derivative with this fraction. Then we take the denominator to the other side and draw a little diagram to illustrate the right-hand side. The point on the right is colored red to denote a positive contribution, and the one on the left is colored blue for minus f of x. By summing these over and over, the pairs of pluses and minuses cancel out each other, and at the end, only the endpoints remain. If these endpoints are a and b, then what we have done effectively is to sum this equation over the interval a b. The left-hand side becomes an integral, and the right-hand side is a telescoping sum, giving f of b minus f of a. This telescoping mechanism can be generalized, and let's summarize this in a recipe for future convenience. The first step is to define the derivative, which can be rearranged to basically define the boundary of the small interval x plus delta x. What I mean is that the boundary is actually oriented. The one on the right has a positive orientation, and the one on the left has a negative one. The boundary orientation can be tricky in higher dimensions. Finally, we have the telescoping sum, so that the expression on the bottom only depends on the boundary, and the left-hand side becomes the integral of a derivative f prime of x. We'll use this recipe for the generalized Stokes theorem, but it's written in the language of differential forms. While differential forms have been introduced before, let's reintroduce it to have everyone on the same page.

One way to think about differential forms is that each point in space is associated with a machine. This machine takes in any number of vectors, which is three in this illustration, and outputs a number. It also satisfies some properties. It's multilinear in the sense that if I divide any vector by two, the output is half as much as before. It's also antisymmetric, so that if I swap any two vectors in the input, the output is the negative of what it was before. Overall, a differential form is a collection of these multilinear antisymmetric machines at different points. If all of them take in three vectors as input, we call this collection a three-form. I want to emphasize that even if I send the same set of vectors to each machine, they can give different outputs. Machines at different points are generally different. But what's all this to do with integration? To integrate a differential form omega over a two-dimensional surface s, we need omega to be a two-form. We divide the surface into tiny parallelograms defined by two vectors emanating from a point. Because omega is a two-form, the machine at this point takes in these two vectors and outputs a number. Repeating this for all the parallelograms, we have many different outputs from different machines, and the integral is the sum of all these outputs. There's a weird case I want to highlight. This machine takes in three vectors, but what if it takes in zero vectors? At this point, it's not even a machine, just a number. If we collect all these numbers at different points, we can still call it a zero-form since we take no vectors as inputs, but this collection is really just a function, assigning a value of -3, for example, to a point p. With the introduction of differential forms, we are ready for the generalized Stokes theorem.

First, we define a kind of derivative of a differential form known as the exterior derivative. Let's start with a zero-form f, which is just a function. Its derivative is going to be a one-form, which means it should take in one vector as input with this p reminding us that machines at different points are different. A pretty natural definition of d f acting on v is the derivative of f along the vector v. In symbols, it's the gradient of f dotted with v. As an example, let's take the function f to be just x, and we substitute x into the definition. The gradient vector is just 1 0 for the function x, and so dotting with v gives the x component of v. Similarly, d y and d z acting on the vector v gives the y and z components, respectively. For a general function f, the gradient involves its partial derivatives. On the other hand, each component of v can be written as d x, d y, or d z acting on v. So the action of d f on v is the same as the partial derivatives times d x d y d z acting on v. These so-called differentials now have a rigorous meaning. It's about how they act on vectors. Now we are ready to move on to the second step, defining the boundary. Like what we did before, we treat the derivative as a limit of a fraction and rearrange it to put epsilon on the other side. Since d f is linear, we can put epsilon inside the bracket and illustrate the difference with the little diagram like before. Finally, we do the telescoping sum, but this time the segments have directions given by the vector v. So in general, they carve out a curve. As expected, the pluses and minuses cancel out, and we are left with the endpoints of the curve. The upper expression sums to the integral of d f along the curve gamma, while the lower one telescopes to the {quote} {unquote} integral of f on the boundary of gamma, which is oriented. So far, this is a warm-up because these f's are just zero-forms. For higher forms, let's see what this recipe takes us.

First step, define the derivative of a one-form. We know this is going to be a two-form taking in two vectors, but how exactly is it defined? A good guess is some derivative along one of the vectors. Since omega is a one-form, we let it eat a vector and output a number. This number depends on where you are. In other words, omega of v2 is a function of p. Then, take the derivative of this function along v1. However, differential forms should be antisymmetric in its arguments, and this isn't. So, we make it antisymmetric by swapping V1 and V2 and taking the difference. This is all quite abstract, so let's visualize it in the second step where we define boundaries. Let's focus on one of the terms for now. We let omega act on the vector V2 at point P. To take the derivative along V1, we consider the setup translated along a small step of epsilon V1. Then, the derivative is approximately the difference between omega acting on the two vectors divided by epsilon. The colored vectors here actually mean omega acting on the vectors, but for simplicity of notation, we get rid of the omega. Now, we adjust this diagram a little. We scale both V2 vectors down by a factor of epsilon. And we will obtain a difference divided by epsilon squared instead. Because these differences are also scaled down by a factor of epsilon, and we compensate by dividing another factor of epsilon. This is just the first term in the expression for D omega, but the other one is very similar. It just swaps V1 and V2. We can recycle this diagram, just that we need to swap the roles of V1 and V2. Similar to the previous expression, this derivative can also be approximated by the difference divided by epsilon squared. Just that these are the V1 vectors at different locations versus the previous V2 vectors. We also have the minus sign because we are subtracting off this term in D omega. What we need to do now is to combine the two contributions. There are four terms altogether, and we separate them into two kinds. The first kind has a positive sign in front. The bottom one has two negative signs, so an overall positive sign. We'll come back to them in a moment, but for now, focus on the others which have a negative sign in front. If we flip the vector around, the sign also flips from minus to plus. In the diagram, the vector flips like this. For the other term with the negative sign in front, we can do the same flipping vector and flipping sign again in this expression and in the diagram. After all the flipping, all four terms have a positive sign, and on the diagram, the vectors form a nice cycle around the parallelogram. We also unify the colors since they all have a positive sign now. All in all, D omega is applying omega to vectors in this order divided by epsilon squared. Finally, move the epsilon squared to the other side and use linearity of D omega to distribute the two copies of epsilon to the vectors. That concludes the second step of defining the boundary and its orientation. The final step is now a lot easier. Summing up the contributions, we get the integral of D omega. On the other side, we get some sort of telescoping sum because the interior edges are all traversed twice in opposite directions and cancel out. So, we are left with the exterior edges. At the end, it sums up to the integral of omega along the oriented boundary. This is the generalized Stokes' theorem for one forms. Let's go to two forms.

First step, define its derivative. We know it's going to be a three form which should take in three vectors at every point P. From our experience earlier, it should be some derivative along one of the vectors, and then we somehow make it antisymmetric. Now, this expression is already antisymmetric in the pair V2 and V3 since omega was already a differential form and is antisymmetric by definition. So, we just need to make it antisymmetric in the other pairs as well, which are V1 V2 and V1 V3. If you really want to see those terms written out, here they are. Now, we move on to the second step of defining the boundary. Previously, we ended up with this boundary orientation of the parallelogram from epsilon V1 and epsilon V2. So, we expect this step gives a parallelopiped from epsilon V1, epsilon V2, and epsilon V3. But the orientations of the faces can be messy. You can start from this expression for D omega and work like before, but there is a shortcut. The idea is that the three vectors you put into D omega determine the orientation of the 3D parallelopiped. In this illustration, it's a so-called right-handed set that follows the right-hand rule. Now, there are two possible orientations of this face, either V2 V3 or the other way around. To see which one we choose, we add any outward pointing vector in front. We added negative V1 here because negative V1 on this face points out of the parallelopiped. We choose whichever gives the same orientation as V1 V2 V3. Since there is an extra minus sign in front of V1, this is a left-handed set opposite to the parallelopiped's orientation. So, the other one is the correct choice. And you can check that it is a right-handed set, and we choose V3 V2 as the orientation of this face. — [snorts] — Therefore, this face should contribute omega of epsilon V3, epsilon V2 instead of the other way around. Repeat this for all the faces, and we get six contributions of these omega along the boundary. To reiterate, you can obtain all these from the expression for D omega from the start. But if you are fine with orientation, this serves as a quick mnemonic. The final step, as always, is a telescoping sum where we sum the contributions from neighboring parallelopipeds. On the right, we have omega of epsilon V3, epsilon V2 because, as we discussed, negative V1 V3 V2 is a right-handed set where negative V1 points out of this parallelopiped. For the other one, we have omega of epsilon V2, epsilon V3 instead. Because this time, it's V1 that points outside. So, we need to choose the other orientation to make this a right-handed set. Summing these contributions from each parallelopiped, the interior faces contribute in opposite orientations and cancel out each other. As expected, only the exterior faces remain, and this is the generalized Stokes' theorem for two forms. We've established zero, one, and two forms, and you can continue this pattern if you want. With these three cases, we recover the fundamental theorem of calculus for zero forms, the usual Stokes' theorem for one forms, and the divergence theorem for two forms. But it doesn't end here. We started this video wondering about the relationship between derivative and boundary. The generalized Stokes' theorem is about integral of derivative equals boundary, but there is a sense in which derivative is literally boundary. Let's explore this sense in the next video, where we'll also see why multiplication is literally intersection. By the way, I've made another video on the second channel about a better way of visualizing the power rule and other derivatives. Check it out. As always, thanks to the patrons, and don't forget to like, comment, and subscribe. See you next time.