Manifolds 55 | Proof of Stokes's Theorem (General Case)

インデックス作成: 2026-05-19

[00:00:00]Hello and welcome back to Manifolds the video series where we talk about generalized surfaces and the integration we can do on them and in that regard you already know the famous theorem of Stokes and in today's part 55 we will finally prove it. This means that we will write down the proof for the general case where we can also use the results from the last videos. Therefore, this video gets a little bit technical because we will also use something that we call petition of unity. And what this actually means, I will tell you soon.

[00:00:33]But first, I want to thank all the nice people who support the channel on Steady here on YouTube or via other means. And please don't forget that Steady and Patreon members can easily download the additional material for all the videos with the link in the description. So, please click there to find PDF versions, quizzes, and books. And then I would say without further ado, let's quickly write down Stokes's theorem again. So what we have is a manifold m with boundary dell m and the differential form omega. And now the claim of Stokes's theorem is simply the equality of these two integrals.

[00:01:11]However, the essential ingredient here is that both integrals live on a compact set because our omega should have a compact support. This means our manifold M does not have to be compact at all. It can definitely stretch to infinity. And this is not important for our proof at all because we will restrict ourselves to the support of omega anyway. So this is already a key insight. We just have to look at a compact set. And now to keep it simple, let's call this set A.

[00:01:43]So it's the support of omega and a compact set on the manifold. And moreover, please recall that we have already proven Stokes's theorem in the case that this set A lies completely in one chart. Therefore, now we have to tackle the case where we don't have that. So one chart is not enough. So let's simply say that A is covered by our atlas of M. So we have a lot of charts consisting of open sets UI and maps HI. And indeed it does not matter how many charts we have here because we know we can cover the whole manifold.

[00:02:21]Hence for a given point P in our set A we definitely find a chart where P lies in UI. And to make this clear let's say we have the chart UIP and HIPP. So there could be different possibilities but now we fix one. And as always for a manifold with boundary we have two cases. either we have a boundary point or not. This means we either map into Rn or into the half space with our map H. However, the difference is not so huge. So it does not matter in our sketch here. The important thing is that we just end up in our flat Rn. And in this one we find our image point which we could call P prime as always. And now around this point P prime we can define an ordinary C infinity function. So we just have an ordinary function with n variables that should map into R. Or more precisely I want to have positive values between zero and one. And moreover this function should also have compact support in Rn or in the half space. And in fact this is quite simple to construct because we have the exponential function. So for example, if we have a function with two variables, we can sketch the graph here.

[00:03:43]So it's zero here and then comes a bump at the origin and then it goes back to zero. And now we can just imagine that everything is symmetric around this axis here. So this means what we have as the support is a two-dimensional ball here in the domain. And now if you have never seen that I can quickly show you how you would define such a bump function here.

[00:04:08]Indeed we can just use the exponential function because this one gives us the C infinity property. And then in the exponent we just write -1 / 1 minus the norm of the point we put in. So this one is the standard ukidian norm of x. And if we square it there's no square root involved as well. And we can simply use this definition for every x inside the unit ball. And for all the other cases, we just define the function to be zero.

[00:04:40]And this immediately gives us the compact support. And the C infinity property on the boundary. You can also easily check. So this is the basic definition of the function. And the only thing that remains to do is now to scale it and move it around. So what I want is that the bump at the highest point is just equal to one. and that p prime is mapped to this one. So let's simply say the map that we get out here is called kip and as already mentioned we can scale as much as we want. So we scale this function such that the support we have here lies completely in our uy prime and this also implies that we have a compact set here in Rn or in the half space and we can translate it back to the manifold m and with that our picture is almost complete but let's not forget that s of p prime should be equal to one and moreover in the case of the half space this bump here would also be cut in half but otherwise everything works exactly the same for the half space as well. And now as already mentioned we can just pull back this compact set by using the inverse of our chart which just means that we also get a compact set here and most importantly by the whole construction here it is a neighborhood of our point P and in addition we can also form the composition of the two maps. So it's a new map and defined on the manifold and I called it lambda p or more precisely we should say it's defined on the open set uip simply because this is the domain of definition for hip and moreover the composition now says that we map into 01 and now as already mentioned the definition of the map should just be the composition. So we first have our hip of x and then we apply our kip and then clearly this also implies that our p is mapped to one and this is already one of the properties of lambda p we should write down. So lambda P of the same point P is equal to 1. And the second important property is of course that our support of this map is a compact set in the open set UIP.

[00:07:01]And maybe now at this point you ask yourself why we even do all of that.

[00:07:06]Therefore, please recall what we want to use is Stokes's theorem just in one chart. And this means we want to split up our differential form omega without losing the C infinity property. And now as you can see we have these nice bump functions but we have them defined on the manifold. And in fact by this property here we can cover the whole manifold with them. Or to be more efficient the only thing we need to cover is the set A. But we can also be efficient and just cover the set A because that's what we need. And now you see we can go through all the points P by just looking at the pre-im images of one. So in other words, this one here is just a fancy way to write the point P.

[00:07:53]But this means the whole union covers the set A. And here please don't forget A is a compact set because it's the support of omega. However, now to actually use the compactness, we should cover the set with open sets. And we can just do that by increasing the pre-im images here. So we take the interval 0 to 1 where 0 is not included. Of course here we could also go to infinity but because our co- doain was chosen as the unit interval I should stop at one here.

[00:08:24]However also in this case this set we get here is an open set in our topological space. This is important because by definition our lambda p is a continuous map. So the pre-im images of open sets are still open. Hence here we have a union of open sets. So an open cover for our compact set. Therefore a finite subcover is already sufficient to cover the whole compact set. So at this point here we finally use the compactness of our support of omega. And in our description here this just implies that we only need a finite amount of points from a. So let's simply say we take capital n many points and then we can just write the cover as the union of j that goes from one to capital n. And now this open cover here gives us an open set and I want to call it b.

[00:09:20]Okay. So now the picture is quite simple. Here's our set A. And now we have n open sets that form the set B and it covers the whole set A. And now finally the partition of unity comes in because we will define it for the set B.

[00:09:36]It just means that we will define n functions where the combination of all the functions will give us the constant one on B. And this is actually quite easy because we already have all the lambda functions from before. And perhaps here let's simply call the functions we define FJ. So every FJ is defined on B and it also maps into the unit interval. And now for the definition we say that X is mapped to the corresponding lambda PJ. So please don't forget now we have exactly end points we care about and therefore we can also definitely just form the finite sum of all the values of the lambda functions. More concretely here we can go with an index k from one to capital n. And here you can see the point x has to lie at least in one of the open sets of the union. So we know that one of these values is non zero. And since there are no negative values anyway, we know that the sum is always positive.

[00:10:43]Therefore, we don't have a problem at all to form the inverse of this number.

[00:10:48]And with that we have the whole definition of our FJ function and we immediately see it's still a C infinity function and it still has compact support. In fact the support still lies in one chart as we want it but moreover it's also a partition of unity because if we form the whole sum we get out the constant one. This means we just take f_sub_1 of x plus f_sub_2 of x and so on. And then we immediately see that all the lambda functions add up and cancel with the inverse here. And this cancelling means that we get out exactly one no matter which point x from b we put in. So visually speaking the bumps of all these functions just add up perfectly such that we get out one. But this is exactly the trick we need to do to finally split up our differential form omega. So please note we cannot just split up omega in an arbitrary way because we are not allowed to lose the differentiability.

[00:11:51]This is because we don't just want to integrate omega we also want to calculate the cau derivative of it. So this is the key point why we need this special partition of unity here. But by using it we can actually decompose omega into smaller differential forms where we can still form the cau derivative without any problems. So first of all for any point q on the manifold omega at q is well defined. In fact omega q is actually just zero if q is not in our compact set a. So the only interesting case is when Q comes from A and there we can put Q into FJ and sum all these values up and there we already know this is just the value one which means we don't change anything at all and that's already the whole idea. We get out n differential forms. So in order to keep it simple, we can just say this is our new differential form omega J at the point Q and each support there lies completely in one chart. This means we can finally use our special version of Stokes's theorem for just one chart. The only thing we need to do now is to apply it n times. Or more concretely here, let's start with the integral of d omega and let's substitute omega by the whole new sum. So we have the cau derivative of the finite sum of the new differential forms omega j. And since the ctor derivative and the integral are linear, we can just pull out the sum all together. And this immediately tells us that now we can use Stokes's theorem because the support of omega J lies just in one chart. And if you don't remember that, please check out the last video where we have proven Stokes's theorem just in one chart. In any case, what we get out is the integral over the boundary of M of omega J. Hence, the only thing missing now is to push in the sum again to get out our omega again.

[00:14:00]And reading that from left to right is exactly Stokes's theorem as we have formulated at the beginning. And now we finally have proven it under the general assumption. And I cannot stretch it enough. The most important ingredient in this proof was that we have a compact support for omega. Otherwise, the whole thing simply does not work. And we cannot expect that Stokes's theorem holds. So please remember that the compact support of omega is essential for Stokes's theorem. Indeed, sometimes the whole manifold m is already compact and then obviously the support of omega is compact as well. So you already see it might be really helpful to look at more examples for Stokes's theorem. So I would say let's do that in the next videos and I wish you a nice day.

[00:14:49]Bye-bye.

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