Hey, you probably won't read this anyway since it's been uploaded a year ago, but im currently studying for a theoretical physics exam at the end of semester and these videos help me so much. Thank you so much dude. like seriously, ur a lifesaver...
@@Drachensslay It's not exciting when you're considering just the plane. You can make it more exciting by thinking about the plane with holes in it, or with lines going off to infinity. The interesting thing is that it tells you how many holes your domain has. Basically, the question you want to know the answer to is: How many forms that get killed by differentiating are NOT themselves derivatives? (The question makes sense, because since d^2=0, anything that is a derivative must automatically be killed after taking another derivative. This is where we use this, to even make sense of the question in the first place.) Looking at 0-forms, to answer this question, first of all which 0-forms are killed by differentiating? Only the constants; everything else depends on x or y. And these are certainly not the differential of any lower "-1-form" because there is no such thing. So the de Rham cohomology at level 0 is 1-dimensional. This is true no matter how many holes we put into the domain. Let's jump up two levels. How many 2-forms are killed by differentiating? Well, all of them (assuming we're only thinking about two variables), because the only form is something*dx^dy, and differentiating that means there's nothing left. Then you ask how many of these forms f dx^dy are NOT the derivative of some 1-form? The answer to that question is "none of them", because I can just let g be the antiderivative of f wrt x, and then the form g dy, when differentiated, gives f dx^dy. For example, 3x^2 dx^dy is the differential of x^3 dy. (Here, we're making an assumption, that every function has an antiderivative; but this follows from FTC. The question is mainly "Does the antiderivative exist everywhere that the function itself does?" and the answer is yes.) Finally, what's the 1-D de Rham cohomology? What 1-forms are killed by differentiation? If we're thinking about the plane, then f dx+g dy gives 0 when f_y=g_x. Basically we need the coefficients of dx^dy to cancel. It turns out that given such an f and g, we can pull an F out so that F_x=f and F_y=g; we just integrate f wrt y and integrate g wrt x, and then mix and match. Like if we had 2ycos(xy)dx+(y+2xcos(xy))dy, the differential of this is 0, and so we can find F=y^2/2+2sin(xy) with dF=that. So when the plane is empty of holes, the first cohomology is 0. The issue comes in when you add holes. Like if the domain is the plane minus the origin, then the form (-y dx+x dy)/(x^2+y^2) exists everywhere, and you can check that its differential is 0 - the x and y components cancel out. But you can also check that there is no function that gives this as its differential. The function you might get if you integrate is arctan(y/x), which is the angle that the point makes with the x-axis. To make this continuous, you'd need the function to actually be the angle, but then you come around and you've added 2pi to your angle, so there's no FUNCTION that gives this as its differential. So because there's this hole that we've added, the cohomology is now nontrivial. (It turns out that it actually is still only 1-dimensional. So any form is equal to some multiple of this plus something coming from below.) And so now we can see that given points (x1, y1), (x2, y2), ... (xn, yn) in the plane, we can just take the differentials of arctan((y-yi)/(x-xi)) and get n different forms that are all independent of each other and all not the differential of any function from below. So the de Rham cohomology is n-dimensional. That's what I mean by it counts the number of holes.
@@Drachensslay deRham cohomology in essence measures the failure of certain classical vector calculus theorems on a more general domain. For instance, the theorem that a vector field with zero curl is the gradient of some function is false if your domain has holes in it.
That would be a bit neater, but any sum where the lower bound is higher than the upper bound is zero, so for j=1, the inner sum works out as 0, meaning his notation is correct.
The multi indices notation is a bit overwhelming, is it like elements of partition of , say n terms with p distinct sets of 1,2,... with a condition of sum of p terms add to n? Not necessary to have all 1, 2, .. can be any random distribution. I am thinking something on the lines of coset decomposition of a composite group elements.
How far will these set of videos on differential forms go? Eventually covering the Atiya-Singer Index Theorem? Maybe start into Non-commutative Geometry after the requisite other 10,000 really cool videos cover the "basics". Can't wait! Love it all!
He has not even defined a manifold yet, he has only been doing things for open sets in R^n. And if he wanted to start doing things on a general smooth manifold, he would have to redo everything as the proofs do not directly translate over. And so since Atiyah-Singer does not apply to open subsets of R^n, I think it is safe to say he will not be going anywhere near that
@@willnewman9783 why you got to bust my bubble like that? Figure he will at least get to Generalized Stokes at the very least... If he does not go for the Theorema Egregium and other topological invariant theorems, oh well. Maybe in the next 100,000 vids.
I learned some homology theory as a directed study in my undergraduate degree but didn’t get to cohomology and am unfamiliar with it, and I know there is relation to these differential forms and cohomology but do not know what that relation is. All this to say the property that d(df)=0 is reminiscent of a property of defining the boundary maps in homology theory (the boundary of a boundary is the empty set)
It's explained in the earlier videos in the "differential forms" playlist on this channel. Alternatively you can think of it as just a formal symbol with the formal rule of integration described by the formula he uses.
My calculus professor would always correct me when I said "vector field" with "vector space", even though I'm fairly certain the vector spaces we were discussing at the time were, in fact, fields. Does anyone know why he was so adamant about making this correction? Is there something I could have mistaken? We did not cover fields in calculus 3 until the very end, but that only involved a definition and some examples.
Vector fields, Vector spaces and fields are 3 different objects. A field is a set with addition, multiplication, subtraction, division. A vector space is an abelian group with an action of a field on it. A vector field is a function from one vector space to another (over the same field). You can think of it as associating a vector with every point in a vector space.
@@lakshaymd a vector field doesn't have to have a vector field as its domain. Many spaces will do, one natural and useful generalization is the domain being a topologycal manifold.
@@zoltankurti I see, I've seen vector fields over differential manifolds in my differential equations course, but just forgot about that completely while writing the answer. Is the concept still useful when you don't have a differentiable structure?
@@lakshaymd no, it's not useful, in fact now that I think about it a differentiable structure is needed to define vector fields. So not topological manifolds, but differentiable manifolds.
a vector field is a map, a vector space is an algebraic structure, they're not the same, but definitely are related. for example a vector field on R^n is a map f : R^n -> V (where V is a vector space) it maps points on the space you're working with to vectors in whichever vector space you want
@@sergioh5515 Are you familiar with what dx and dy do in this context? This isn't calculus and you need to know that in order to understand the wedge product.
Hey, you probably won't read this anyway since it's been uploaded a year ago, but im currently studying for a theoretical physics exam at the end of semester and these videos help me so much. Thank you so much dude. like seriously, ur a lifesaver...
Nice
19:38
I love your pseudo
Hello Professor,
Thanks for the brilliant explanation,
I enjoyed it a lot.
very motivating!
Will you be going into how d^2 = 0 gives rise to a cochain complex? And deRham cohomology?
It would be amazing if he did.
@Jon Snow sorry, I'm unfamiliar with this area of math :( anyone mind explaining why deRahm cohomology is exciting?
@@Drachensslay It's not exciting when you're considering just the plane. You can make it more exciting by thinking about the plane with holes in it, or with lines going off to infinity. The interesting thing is that it tells you how many holes your domain has.
Basically, the question you want to know the answer to is: How many forms that get killed by differentiating are NOT themselves derivatives? (The question makes sense, because since d^2=0, anything that is a derivative must automatically be killed after taking another derivative. This is where we use this, to even make sense of the question in the first place.) Looking at 0-forms, to answer this question, first of all which 0-forms are killed by differentiating? Only the constants; everything else depends on x or y. And these are certainly not the differential of any lower "-1-form" because there is no such thing. So the de Rham cohomology at level 0 is 1-dimensional. This is true no matter how many holes we put into the domain.
Let's jump up two levels. How many 2-forms are killed by differentiating? Well, all of them (assuming we're only thinking about two variables), because the only form is something*dx^dy, and differentiating that means there's nothing left. Then you ask how many of these forms f dx^dy are NOT the derivative of some 1-form? The answer to that question is "none of them", because I can just let g be the antiderivative of f wrt x, and then the form g dy, when differentiated, gives f dx^dy.
For example, 3x^2 dx^dy is the differential of x^3 dy. (Here, we're making an assumption, that every function has an antiderivative; but this follows from FTC. The question is mainly "Does the antiderivative exist everywhere that the function itself does?" and the answer is yes.)
Finally, what's the 1-D de Rham cohomology? What 1-forms are killed by differentiation? If we're thinking about the plane, then f dx+g dy gives 0 when f_y=g_x. Basically we need the coefficients of dx^dy to cancel. It turns out that given such an f and g, we can pull an F out so that F_x=f and F_y=g; we just integrate f wrt y and integrate g wrt x, and then mix and match. Like if we had 2ycos(xy)dx+(y+2xcos(xy))dy, the differential of this is 0, and so we can find F=y^2/2+2sin(xy) with dF=that. So when the plane is empty of holes, the first cohomology is 0.
The issue comes in when you add holes. Like if the domain is the plane minus the origin, then the form (-y dx+x dy)/(x^2+y^2) exists everywhere, and you can check that its differential is 0 - the x and y components cancel out. But you can also check that there is no function that gives this as its differential. The function you might get if you integrate is arctan(y/x), which is the angle that the point makes with the x-axis. To make this continuous, you'd need the function to actually be the angle, but then you come around and you've added 2pi to your angle, so there's no FUNCTION that gives this as its differential. So because there's this hole that we've added, the cohomology is now nontrivial. (It turns out that it actually is still only 1-dimensional. So any form is equal to some multiple of this plus something coming from below.)
And so now we can see that given points (x1, y1), (x2, y2), ... (xn, yn) in the plane, we can just take the differentials of arctan((y-yi)/(x-xi)) and get n different forms that are all independent of each other and all not the differential of any function from below. So the de Rham cohomology is n-dimensional. That's what I mean by it counts the number of holes.
@@noahtaul Wow. That seriously cleared up so much for me. Thank you times a million for the very down to earth explanation!
@@Drachensslay deRham cohomology in essence measures the failure of certain classical vector calculus theorems on a more general domain. For instance, the theorem that a vector field with zero curl is the gradient of some function is false if your domain has holes in it.
Edited because of Bart's comment: At 15:23, You can let j = 1 even though 1
Loving this series by the way. Keep up the good work Michael!
That would be a bit neater, but any sum where the lower bound is higher than the upper bound is zero, so for j=1, the inner sum works out as 0, meaning his notation is correct.
@@barutjeh Yeah, you're right. Taking a second look at that, my argument for how to change the left sum works equally well for the right sum. Edited.
This fucking blew my mind. AWESOME.
brilliant just brilliant !
you're my savior
The multi indices notation is a bit overwhelming, is it like elements of partition of , say n terms with p distinct sets of 1,2,... with a condition of sum of p terms add to n? Not necessary to have all 1, 2, .. can be any random distribution. I am thinking something on the lines of coset decomposition of a composite group elements.
How far will these set of videos on differential forms go? Eventually covering the Atiya-Singer Index Theorem? Maybe start into Non-commutative Geometry after the requisite other 10,000 really cool videos cover the "basics". Can't wait! Love it all!
He has not even defined a manifold yet, he has only been doing things for open sets in R^n. And if he wanted to start doing things on a general smooth manifold, he would have to redo everything as the proofs do not directly translate over.
And so since Atiyah-Singer does not apply to open subsets of R^n, I think it is safe to say he will not be going anywhere near that
@@willnewman9783 why you got to bust my bubble like that? Figure he will at least get to Generalized Stokes at the very least... If he does not go for the Theorema Egregium and other topological invariant theorems, oh well. Maybe in the next 100,000 vids.
I learned some homology theory as a directed study in my undergraduate degree but didn’t get to cohomology and am unfamiliar with it, and I know there is relation to these differential forms and cohomology but do not know what that relation is. All this to say the property that d(df)=0 is reminiscent of a property of defining the boundary maps in homology theory (the boundary of a boundary is the empty set)
Cohomologia de de rham, a Cohomologia é bem melhor que a homologia em alguns casos
@@isabelshurmanfeitoza9897 language?
I Normally watch this videos just for fun, don't know much about high level math, just Engineering type things.
I just understood when he said "Hi"
What's the video that explains the wedge operator?
It's explained in the earlier videos in the "differential forms" playlist on this channel. Alternatively you can think of it as just a formal symbol with the formal rule of integration described by the formula he uses.
Love your videos!
My calculus professor would always correct me when I said "vector field" with "vector space", even though I'm fairly certain the vector spaces we were discussing at the time were, in fact, fields.
Does anyone know why he was so adamant about making this correction? Is there something I could have mistaken? We did not cover fields in calculus 3 until the very end, but that only involved a definition and some examples.
Vector fields, Vector spaces and fields are 3 different objects. A field is a set with addition, multiplication, subtraction, division. A vector space is an abelian group with an action of a field on it. A vector field is a function from one vector space to another (over the same field). You can think of it as associating a vector with every point in a vector space.
@@lakshaymd a vector field doesn't have to have a vector field as its domain. Many spaces will do, one natural and useful generalization is the domain being a topologycal manifold.
@@zoltankurti I see, I've seen vector fields over differential manifolds in my differential equations course, but just forgot about that completely while writing the answer. Is the concept still useful when you don't have a differentiable structure?
@@lakshaymd no, it's not useful, in fact now that I think about it a differentiable structure is needed to define vector fields. So not topological manifolds, but differentiable manifolds.
a vector field is a map, a vector space is an algebraic structure, they're not the same, but definitely are related. for example a vector field on R^n is a map f : R^n -> V (where V is a vector space) it maps points on the space you're working with to vectors in whichever vector space you want
why it is called exterior?
Vem da algebra exterior, a algebra dos tensores alternados
I don't understand : why to deal with second derivatives if they all are zero ?
This is about *exterior* derivatives, not normal derivatives
@VeryEvilPettingZoo great comment
@VeryEvilPettingZoo Thank you. Now I have understood that df was a form and not a function.
squares to zero?
What does "wedge" mean? I love your videos but I don't entirely understand these videos with my pathetic math education :)
He defines it in early videos in this series
@@willnewman9783 sorry I don't understand it then
@@sergioh5515 Are you familiar with what dx and dy do in this context? This isn't calculus and you need to know that in order to understand the wedge product.
@@barutjeh I don't understand 😔 but I still watch
@@sergioh5515 I get it. The start of this series was pretty abstract. That doesn't help with retention.
Hype