I literally took 2 university courses with all proofs regarding the topic of manifolds and k-curves, etc... But I've never really seen a point in differential forms and integrating over them. I could prove general Stokes theorem for manifolds with boundaries on compact carrier but couldn't see any analogy at all and appreciate what it all means. You have really changed my point of view about whole subject of differential geometry. Thank you so much.
0:32 Integration and Differentiation 1:38 Integration of Differential k-Forms 1:40 Review-Integration of Area 2:59 Review-Integration of Scalar Functions 4:48 Integration of a 2-Form 6:57 Integration of Differential 2-forms-Example 8:55 Integration on Curves 13:08 Integration on Curves-Example 17:15 Stokes’ Theorem 17:31 Boundary 19:50 Boundary of a Boundary 20:55 Review: Fundamental Theorem of Calculus 22:16 Stokes’ Theorem (I) 23:34 Example: Divergence Theorem 27:02 Example: Green’s Theorem 29:49 Stokes’ Theorem (II) 30:17 Fundamental Theorem of Calculus & Stokes’ 31:47 Why is d ◦ d = 0? (I) 34:28 Why is d ◦ d = 0? (II) 35:56 Integration & Stokes’ Theorem - Summary 38:25 Inner Product on Differential k-Forms 38:33 Inner Product-Review 41:06 Euclidean Inner Product-Review 42:18 L2 Inner Product of Functions / 0-forms 46:30 Inner Product on k-Forms 49:57 Inner Product of 1-Forms-Example 52:09 Summary 52:30 Exterior Calculus: Flat vs. Curved 54:28 Exterior Calculus-Summary
Thanks, Prof. Crane. I have enjoyed all the lectures so far. However, I haven't understood some things in this one, and I hope that you (or anyone else) can help me. My questions are related to translating the notation of the covectors dx dy to actual differentials dx,dy, which can be integrated (when and how is this step perfromed) -At 6:23, the motivation behind the integral of a 2-form is that it is the summation of the many 2-forms applied to a pair of vectors along with the whole region \Omega. However, I don't see what the vectors used in the two forms at 8:31 are. Mainly, I'm lost at the motivation behind the interpretation of the 2-form dx^dy as a scalar dxdy inside the integral, given that dx and dy were supposed to be just basis forms (covectors). Maybe I missed it, but when did we start to give the meaning of differentials (little displacements that can be integrated) to the "forms" dx and dy, which were just covectors. My thought is that we make a parallelepiped spanned by the vectors u=(dx)d/dx and v=(dy)d/dy where here dx and dy are small displacements in the directions d/dx and d/dy. This parallelepiped has area dxdy, and we can measure such area by the form dx^dy. Hence, we use dx^dy to measure this parallelepiped obtaining dx^dy(u,v)=dxdy since dx(d/dx)=1, dx(d/dy)=0, etc... However, this reasoning seems artificial to me (since we didnt't take advantage of the notation dx,dy for the basis covectors, instead we used other displacements dxdy different from the basis covectors). It would be better to have defined dx(d/dx)="dx" (where the second dx is the displacement and not the covector) instead of dx/(d/dx)=1 as we did before (move the displacements dx and dy to the definition of the covectors dx and dy instead of the parallelepiped u^v ). -At 14:00 you show the integral of a 1-form \alpha. However, I don't understand what the systematic way of converting the "abstract" integral "\int_{S^1}\alpha" to the integral using the limits and the "differential" ds is. In particular, I don't understand where the ds came from. Note that in this case, we use "dy" not as a differential but as a covector, as opposed to what we did in my previous question (in my previous question I pointed out that in the video dx^dy was interpreted just as dxdy and integrated over). I hope that you see this. These lectures have been very inspiring and have clarified many things I learned many years ago and haven't understood until now. Thank you so much.
I share some of the confusion regarding the integration of differential 2-forms (which should act on the vectors $u_i$ and $v_i$). As for the other point, when integrating the differential 1-form along a curve $\gamma$ (at 9:00), we consider the tangent vectors $t_i$ of gamma at the points $p_i$. Nothing is mentioned about the lengths of the tangent vectors here (it depends on the parametrization of $\gamma$), but perhaps there is the implicit assumption of an arc-length parametrization. But then, for the sum $\sum_i \alpha_{p_i} (t_i)$ to converge to the integral, don't we need a $\Delta s$ (which in the limit becomes ds) here? After all, considering more and more points $p_i$ generally increases the sum (certainly if we assume that the 1-form is more or less aligned with the tangent vectors, which - apart from the start of the curve - seems to be the case in this visual example). Luckily, in the example with the unit circle, the ds makes an appearance.
I have exactly the same question! (the first one) The idea of treating (dx ^ dy) as differentials is not clear at all since they were defined as the basis of the 2-forms (in R², in this case). My interpretation when looking at 6:23 is the same as yours: the differentials are actually the vectors 'u' and 'v', not the 2-form itself. The way that I would write the integral would be [integral of e1^e2(du, dv)], where "du,dv" are the vectors being "applied" to the two form and e1^e2 is the 2-form basis (the usual dx^dy). This is my understanding and at this point I'm assuming that the motivation of using (dx ^ dy) as differentials was just skipped by Crane and, in the end, the notation is there because it is something that works algebraically. I would really love to hear from the professor though!
du is the horizontal covector, and dv is the vertical. That means du ∧ dv has a counterclockwise orientation, and these are the little area elements we are adding up during the integration. Since we're applying the Hodge start to a, it gets rotated 90° and is now pointed upwards. So the value we end up calculating is actually v dv ∧ du, because a comes first in the inner product and a = du, so *a = dv. This causes the minus sign to appear because essentially the region we are integrating over has the opposite orientation as the thing we are integrating. And this is why a lot of people use = ∫∫ a ∧ *b for the inner product instead of = ∫∫ *a ∧ b, I think. Because should equal |a|² not -|a|².
At the slide at 52:00 I'm worried that if you follow the same steps with alpha = beta you get -1, while we would like to get a positive value. I wonder if it would be better to define the inner product as 《α、β》= ∬α ∧(＊β)
Beautiful inspiring lecture! Thanks for sharing! Concerning notation. You introduced the L2 inner product of forms as the integral of \star form \wedge form. But if we take in the open rectangle in the plane the form dx, doesn't this give dy\wedge dx as the integrand? That would be in the opposite orientation of the usual one and the inner product would fail to be positive definite, right? Maybe I missed some of the conventions.
Would be great to have a final lecture in the playlist about the history of how the subject was developed. Was it Elie Cartan who majorly did a lot of the legwork in putting this stuff together?
Why at 11:34 the right hand side (discrete sum) is not multiplied by len(curve)/N where N=number of samples. Is it because tangent t_i is not unit but already has length?
Question about the fundamental theorem: how do the dimensions fit there, eg in the one dimensional case, you would get a measure of area, while the difference of the endpoints is a measure of length? I think this generalizes to higher dimensions, the boundary is always one lower. How should we think about this?
First of all, amazing lectures! At 11:49, the professor defines the integral as being the tangent vector applied to the differential form at each point p_i. However, I don't see how this converges when i->∞, since there is no 'differential'! By looking at the sum shown in the slide, it seems that when 'i' goes to infinity (i.e. when the length of each curve segment goes to 0) the sum does not converge since it does not depend on the curve length. Wouldn't it be [sum of alpha_pi(t_i) * s] where s is the length of the curve segment? In the next slide the 'ds' appears in the resolution.
So the "inner product" here given for functions obviously doesn't work as an inner product - integration isn't well behaved enough, there are nonnegative functions other than zero with zero as their integral. They're not continuous though. Question though is: what *is* a sensible inner product?
Sneaky ommission by Keenan here: "does this satisfy the properties of an inner product? I'll let you think about that ;)" You're right that it doesn't! There are two genres of solution to this problem. 1. Consider a smaller class of functions. This is a perfectly good inner product for continuous square-integrable functions! Normally people who aren't analysts don't care for any functions which aren't at least continuous, so this is fine. Differential geometers, for instance, rarely consider functions which aren't infinitely differentiable, and those functions are very much continuous, so there's no loss doing this. 2. The analysts' solution is a bit weirder. Instead of considering the vector space of square-integrable functions, instead consider the vector space that results when we take the vector space of square-integrable functions, and identify two functions when the integral of (the square of) their difference is 0. This is the space analysts call L², and it's a beautiful space for analysts' purposes, since when doing integration you usually don't care about the difference between two functions that the integral thinks are equal, and the resulting vector space is a Hilbert space, which is a very beautiful kind of space that analysts love.
Fubini's theorem says, integrating over the unit square, say, ∫_0^1 ∫_0^1 f dx dy = ∫_0^1 ∫_0^1 f dy dx. But with differential forms, we have dx∧dy = −dy∧dx, so surely ∫_0^1 ∫_0^1 f dx∧dy = −∫_0^1 ∫_0^1 f dy∧dx. Why do we get different answers?
@@Cam-vv5xd Yes, but how do I calculate ∫_0^1 ∫_0^1 1 dy∧dx? I know it should be −1. But calculating it gives ∫_0^1 ∫_0^1 1 dy∧dx = ∫_0^1 y|_0^1 dx = ∫_0^1 (1−0) dx = x|_0^1 = 1−0 = 1
@@columbus8myhw Hm, this is a good question. I have never dealt with oriented integrals like this, but perhaps the limits of integration are also orientation-encoded? The wedge product is, so maybe they are too? Say, if you flip the wedge, then you also flip the limits of integration.
The thing is that while dx∧dy=dxdy, we don't have an equality dy∧dx=dydx, we have dy∧dx=-dydx. It is because dxdy just counts an area of a square, and dx∧dy counts an oriented area of a square, and the equality dx∧dy=dxdy is not natural in any way, it is satisfied just because we fixed some orientation of a square so that this equality is satisfied. If you change the coordinates x->y and y->x, then the orientation will be reversed, and that will affect the wedge product.
Maybe the problem here is assigning limits of integration in the first place. The area element, dx*dy, does require a limit of integration, but the wedge product is a topological idea rather than a geometric one. Perhaps assigning limits it meaningless? That is, we integrate over a region rather than a simple interval in the dA= dx*dy case
I literally took 2 university courses with all proofs regarding the topic of manifolds and k-curves, etc... But I've never really seen a point in differential forms and integrating over them. I could prove general Stokes theorem for manifolds with boundaries on compact carrier but couldn't see any analogy at all and appreciate what it all means. You have really changed my point of view about whole subject of differential geometry. Thank you so much.
Your analogy between an integral and a weighted area is so good.
Since I reached enlightenment @30:10 I will now pursue the mendicant life.
0:32 Integration and Differentiation
1:38 Integration of Differential k-Forms
1:40 Review-Integration of Area
2:59 Review-Integration of Scalar Functions
4:48 Integration of a 2-Form
6:57 Integration of Differential 2-forms-Example
8:55 Integration on Curves
13:08 Integration on Curves-Example
17:15 Stokes’ Theorem
17:31 Boundary
19:50 Boundary of a Boundary
20:55 Review: Fundamental Theorem of Calculus
22:16 Stokes’ Theorem (I)
23:34 Example: Divergence Theorem
27:02 Example: Green’s Theorem
29:49 Stokes’ Theorem (II)
30:17 Fundamental Theorem of Calculus & Stokes’
31:47 Why is d ◦ d = 0? (I)
34:28 Why is d ◦ d = 0? (II)
35:56 Integration & Stokes’ Theorem - Summary
38:25 Inner Product on Differential k-Forms
38:33 Inner Product-Review
41:06 Euclidean Inner Product-Review
42:18 L2 Inner Product of Functions / 0-forms
46:30 Inner Product on k-Forms
49:57 Inner Product of 1-Forms-Example
52:09 Summary
52:30 Exterior Calculus: Flat vs. Curved
54:28 Exterior Calculus-Summary
Thanks, Prof. Crane. I have enjoyed all the lectures so far. However, I haven't understood some things in this one, and I hope that you (or anyone else) can help me. My questions are related to translating the notation of the covectors dx dy to actual differentials dx,dy, which can be integrated (when and how is this step perfromed)
-At 6:23, the motivation behind the integral of a 2-form is that it is the summation of the many 2-forms applied to a pair of vectors along with the whole region \Omega. However, I don't see what the vectors used in the two forms at 8:31 are. Mainly, I'm lost at the motivation behind the interpretation of the 2-form dx^dy as a scalar dxdy inside the integral, given that dx and dy were supposed to be just basis forms (covectors). Maybe I missed it, but when did we start to give the meaning of differentials (little displacements that can be integrated) to the "forms" dx and dy, which were just covectors. My thought is that we make a parallelepiped spanned by the vectors u=(dx)d/dx and v=(dy)d/dy where here dx and dy are small displacements in the directions d/dx and d/dy. This parallelepiped has area dxdy, and we can measure such area by the form dx^dy. Hence, we use dx^dy to measure this parallelepiped obtaining dx^dy(u,v)=dxdy since dx(d/dx)=1, dx(d/dy)=0, etc... However, this reasoning seems artificial to me (since we didnt't take advantage of the notation dx,dy for the basis covectors, instead we used other displacements dxdy different from the basis covectors). It would be better to have defined dx(d/dx)="dx" (where the second dx is the displacement and not the covector) instead of dx/(d/dx)=1 as we did before (move the displacements dx and dy to the definition of the covectors dx and dy instead of the parallelepiped u^v ).
-At 14:00 you show the integral of a 1-form \alpha. However, I don't understand what the systematic way of converting the "abstract" integral "\int_{S^1}\alpha" to the integral using the limits and the "differential" ds is. In particular, I don't understand where the ds came from. Note that in this case, we use "dy" not as a differential but as a covector, as opposed to what we did in my previous question (in my previous question I pointed out that in the video dx^dy was interpreted just as dxdy and integrated over).
I hope that you see this. These lectures have been very inspiring and have clarified many things I learned many years ago and haven't understood until now. Thank you so much.
I share some of the confusion regarding the integration of differential 2-forms (which should act on the vectors $u_i$ and $v_i$).
As for the other point, when integrating the differential 1-form along a curve $\gamma$ (at 9:00), we consider the tangent vectors $t_i$ of gamma at the points $p_i$. Nothing is mentioned about the lengths of the tangent vectors here (it depends on the parametrization of $\gamma$), but perhaps there is the implicit assumption of an arc-length parametrization. But then, for the sum $\sum_i \alpha_{p_i} (t_i)$ to converge to the integral, don't we need a $\Delta s$ (which in the limit becomes ds) here? After all, considering more and more points $p_i$ generally increases the sum (certainly if we assume that the 1-form is more or less aligned with the tangent vectors, which - apart from the start of the curve - seems to be the case in this visual example). Luckily, in the example with the unit circle, the ds makes an appearance.
I have exactly the same question! (the first one) The idea of treating (dx ^ dy) as differentials is not clear at all since they were defined as the basis of the 2-forms (in R², in this case). My interpretation when looking at 6:23 is the same as yours: the differentials are actually the vectors 'u' and 'v', not the 2-form itself. The way that I would write the integral would be [integral of e1^e2(du, dv)], where "du,dv" are the vectors being "applied" to the two form and e1^e2 is the 2-form basis (the usual dx^dy). This is my understanding and at this point I'm assuming that the motivation of using (dx ^ dy) as differentials was just skipped by Crane and, in the end, the notation is there because it is something that works algebraically. I would really love to hear from the professor though!
At 52:00 the negative value makes me uncomfortable, I feel like something went wrong :\
du is the horizontal covector, and dv is the vertical. That means du ∧ dv has a counterclockwise orientation, and these are the little area elements we are adding up during the integration.
Since we're applying the Hodge start to a, it gets rotated 90° and is now pointed upwards. So the value we end up calculating is actually v dv ∧ du, because a comes first in the inner product and a = du, so *a = dv. This causes the minus sign to appear because essentially the region we are integrating over has the opposite orientation as the thing we are integrating. And this is why a lot of people use = ∫∫ a ∧ *b for the inner product instead of = ∫∫ *a ∧ b, I think. Because should equal |a|² not -|a|².
At the slide at 52:00 I'm worried that if you follow the same steps with alpha = beta you get -1, while we would like to get a positive value. I wonder if it would be better to define the inner product as 《α、β》= ∬α ∧(＊β)
I also checked that it should be as Mr or Ms T said.
Beautiful inspiring lecture! Thanks for sharing! Concerning notation. You introduced the L2 inner product of forms as the integral of \star form \wedge form. But if we take in the open rectangle in the plane the form dx, doesn't this give dy\wedge dx as the integrand? That would be in the opposite orientation of the usual one and the inner product would fail to be positive definite, right? Maybe I missed some of the conventions.
Would be great to have a final lecture in the playlist about the history of how the subject was developed. Was it Elie Cartan who majorly did a lot of the legwork in putting this stuff together?
Why at 11:34 the right hand side (discrete sum) is not multiplied by len(curve)/N where N=number of samples. Is it because tangent t_i is not unit but already has length?
So in 3D standard surface integrals of vector fields equal the integrals of 2-forms (Hodge-star of the corresponding 1-form) on surfaces. Am I right?
Question about the fundamental theorem: how do the dimensions fit there, eg in the one dimensional case, you would get a measure of area, while the difference of the endpoints is a measure of length? I think this generalizes to higher dimensions, the boundary is always one lower. How should we think about this?
Your example in 43:00, is that inspired by KL Divergence? Maybe that's why we can say that it doesn't obey inner product?
First of all, amazing lectures! At 11:49, the professor defines the integral as being the tangent vector applied to the differential form at each point p_i. However, I don't see how this converges when i->∞, since there is no 'differential'! By looking at the sum shown in the slide, it seems that when 'i' goes to infinity (i.e. when the length of each curve segment goes to 0) the sum does not converge since it does not depend on the curve length. Wouldn't it be [sum of alpha_pi(t_i) * s] where s is the length of the curve segment? In the next slide the 'ds' appears in the resolution.
I think at 35:50 there is a typo in the product rule. There should be a + just before (-1)^k. Great lecture series nevertheless!
i love you
So the "inner product" here given for functions obviously doesn't work as an inner product - integration isn't well behaved enough, there are nonnegative functions other than zero with zero as their integral. They're not continuous though. Question though is: what *is* a sensible inner product?
Sneaky ommission by Keenan here: "does this satisfy the properties of an inner product? I'll let you think about that ;)"
You're right that it doesn't! There are two genres of solution to this problem.
1. Consider a smaller class of functions. This is a perfectly good inner product for continuous square-integrable functions! Normally people who aren't analysts don't care for any functions which aren't at least continuous, so this is fine. Differential geometers, for instance, rarely consider functions which aren't infinitely differentiable, and those functions are very much continuous, so there's no loss doing this.
2. The analysts' solution is a bit weirder. Instead of considering the vector space of square-integrable functions, instead consider the vector space that results when we take the vector space of square-integrable functions, and identify two functions when the integral of (the square of) their difference is 0. This is the space analysts call L², and it's a beautiful space for analysts' purposes, since when doing integration you usually don't care about the difference between two functions that the integral thinks are equal, and the resulting vector space is a Hilbert space, which is a very beautiful kind of space that analysts love.
@@Boarbarktree Wow, the one and only BBT himself!
43:37
23:29
Fubini's theorem says, integrating over the unit square, say,
∫_0^1 ∫_0^1 f dx dy = ∫_0^1 ∫_0^1 f dy dx.
But with differential forms, we have dx∧dy = −dy∧dx, so surely
∫_0^1 ∫_0^1 f dx∧dy = −∫_0^1 ∫_0^1 f dy∧dx.
Why do we get different answers?
the wedge encodes orientation, so dx∧dy does equal −dy∧dx.
@@Cam-vv5xd Yes, but how do I calculate ∫_0^1 ∫_0^1 1 dy∧dx? I know it should be −1. But calculating it gives
∫_0^1 ∫_0^1 1 dy∧dx
= ∫_0^1 y|_0^1 dx
= ∫_0^1 (1−0) dx
= x|_0^1
= 1−0 = 1
@@columbus8myhw Hm, this is a good question. I have never dealt with oriented integrals like this, but perhaps the limits of integration are also orientation-encoded? The wedge product is, so maybe they are too? Say, if you flip the wedge, then you also flip the limits of integration.
The thing is that while dx∧dy=dxdy, we don't have an equality dy∧dx=dydx, we have dy∧dx=-dydx. It is because dxdy just counts an area of a square, and dx∧dy counts an oriented area of a square, and the equality dx∧dy=dxdy is not natural in any way, it is satisfied just because we fixed some orientation of a square so that this equality is satisfied. If you change the coordinates x->y and y->x, then the orientation will be reversed, and that will affect the wedge product.
Maybe the problem here is assigning limits of integration in the first place. The area element, dx*dy, does require a limit of integration, but the wedge product is a topological idea rather than a geometric one. Perhaps assigning limits it meaningless? That is, we integrate over a region rather than a simple interval in the dA= dx*dy case