Differential Forms | The exterior derivative.

here we're gonna begin looking at the notion of the derivative of a differential M form on our end so let's recall a few things so let's say we've got Omega which is this sum over multi indices capital I of F sub I D X sub I like I said I is a multi index that means it's an M tuple and all of those things are between 1 and n so in other words we have 1 is less than or equal to I 1 which is strictly less than I 2 all the way up to I M which is less than or equal to n and the F sub i's are differentiable functions and this is known as a differential M form on our end and let's recall that it's got this two-stage evaluation so the first thing that you do is you plug in a point from RN and that turns this differential M form to just an M form and you generally call that Omega P and that is a multi linear map and I should say it's an alternating multi linear map from the tangent space at Point P of R in down to the real numbers and furthermore these d x i's are described by this action so we have D X I is equal to DX I one wedge all the way up to D X I am and they have the following action on em different vectors from our n so DX I one wedge all the way up to DX I M evaluated at these M vectors from RN gives you the determinant of a certain M by M matrix so the first row of this matrix is given by the first vector with the components I 1 up to I am and then the second row of this matrix is given by the second vector with those components and so on and so forth so in the end you have an M by M matrix and you can take a determinant of a square matrix ok so our goal for this video is to define some sort of derivative which takes us from the space of differential M forms on RN to the space of differential M one forms on our end and we're gonna start looking at what it would take to go from zero forms to one forms and we really want to think about these as differential zero forms to differential one forms so in other words given a zero form which is just a function so let's go ahead and write that so a zero form in other words a function on RN so we'll just call that F x1 xn we really want to answer the question what is DF good so what should the notion of the derivative of this 0 form be so first of all we need it to be a one form so let's go ahead and write that down DF is a one form and that's because our goal is to create a one form out of the zero form so we're given a zero form which is this function and we want to somehow construct a one form out of that in a nice geometric way okay so good and so number two is how can we evaluate this one form so let's just go ahead and write that so how to evaluate so in other words we'll say some point from RN and then let's say we have some vector in T PRN because remember if we've got a one form it's got a two stage evaluation you first print in a point from RN and then you put in a single vector from t PRN because over here in the general case you've got a two stage evaluation on an M form first a single point and then M vectors so now let's like look at this for a little bit and think of some nice geometric thing that we could do given a point in RN and a vector in RN and this function related to the derivative and if you sit around and think about it for a little bit you realize you've probably already heard of this and it's the directional derivative so let's go ahead and write that so what I mean by that is if we take D F and then we evaluate it at P so that's our first stage evaluation and then we further evaluate it at V so that's the second stage of evaluation what we should get is the derivative in the direction of V of F evaluated at x equals P and we've got actually a nicer calculation formula for the directional derivative and that's using the gradient so the directional derivative in the direction of V of F at the point P can also be described by the gradient of F evaluated at P so I'll just put that and then dotted with V okay fantastic but now what we can do from there is expand that so notice that's exactly equal to partial F with respect to x1 evaluated at P but I'll go ahead and leave that out for now and then times v1 plus up to the partial derivative of F with respect to X n times V n because that just follows from the definition of the dot product and there we've got the gradient is given by the column vector of these partial derivatives and then V is given by the vector v1 up to the N okay fantastic now the next thing that I want to notice is our elementary one form DX I acts on vectors and just pulls out their components because an elementary one form is essentially giving you the determinant of a one by one matrix formed by just pulling a single component out of a vector so now I can rewrite in that in that language so this is the same thing as the partial of F with respect to X 1 DX 1 evaluated at V so let's talk our way through that so DX 1 forms a 1 by 1 matrix from the first entry of V and then takes its determinate but the determinant of 1 by 1 matrix is just the number so that gives you V 1 and then plus all the way up to the partial with 2 X n of D X n evaluated at V good and so now if we look at the extreme left and right hand side of the equation we have some sort of feel for how we should define this DF operator so notice this DF operator can logically be described as this sum of elementary one forms built by the partial of F with respect to X 1 DX 1 plus all the way up to the partial of F with respect to X and DX n great so let's start with that at the top of the board and then we'll move on to a more arbitrary so far we motivated the derivative of a zero form in other words a function and we said that DF where f was that 0 form or that function is equal to the partial of F with respect to X 1 DX where that's the elementary 1 form plus all the way up to the partial of F with respect to X in DX in and again that is another elementary one form now we want to say well what about a differential in forms and we'll do this term by term so I'll start just by defining what happens to F I D X I so we're essentially just taking a natural extension of what we saw right here so let's go ahead and write this down so if we take DF DX I so let's read this so this D is the operator this exterior derivative and this DXi is our elementary M form on RN so this is going to be equal to the partial of F with respect to X 1 DX 1 wedge DX I good and then plus all the way up to the partial derivative of F with respect to X in DX in wedge DX I good so that's kind of a natural extension of what we did up there and then after that we can define D Omega just linearly so in other words we'll define D Omega as this sum over I and then the some little J goes from 1 to n sir that we can collapse this into some more compact notation and then we have the partial of F sub I with respect to XJ and then the elementary one form DX J wedge this elementary M form DX I and so notice this most definitely gives us a differential M plus one form if Omega itself was a differential M form so we have our goal in other words so let's go ahead and just write this down this is a differential M plus one form again given Omega is a differential M form okay so let's go ahead and clean this up and then we'll do kind of a general but small example so let's look at a general but fairly small example so let's say we're working with differential forms over r3 so we have choices of zero one two or three forms and zero forms are just functions so in other words let's say we've just got a function I'll call it F and then we can take the derivative of this function and this is this special like exterior derivative so DF so in this case this will be f sub X so I'll use that notation for the partial of F with respect to X just to make it a little bit shorter and then DX plus F sub y dy plus F sub Z DZ so we're DX dy and DZ are just the elementary one forms and so now let's say we've got a one form we'll call it alpha and so that means that's going to be a sum of F DX plus G dy plus h DZ where F G and H are differentiable functions of three variables XY and Z and then these are the elementary one forms okay so let's see what we can get for the derivative of this so d alpha in this case will be f sub X DX wedge DX plus F sub y dy wedge DX plus F sub Z DZ wedge DX so that's what we get from running through all the f' parts and then from running through all the g parts we'll have G sub X DX wedge dy plus G sub y dy wedge dy plus G sub Z DZ wedge D Y good so that's what we get from running through all of the derivatives of G and then finally for H we'll have h sub X DX wedge D Z plus H sub y dy wedge DZ and then finally plus H sub Z DZ wedge DZ now we use the fact that this wedge product is anti commutative so that means we have things like DX wedge DX is going to be equal to 0 and then dy wedge DX will be negative DX wedge dy so let's go ahead and write that down so DX wedge dy because there's some sort of canonical order for these things and then also this guy is going to be negative DX wedge D Z and then this dy wedge dy will be 0 and then this guy will be negative dy wedge DZ and then finally this guy DZ wedge DZ will be equal to 0 and so now we can go ahead and combine like terms so notice that like this term right here and this term right here are both attached to DX wedge D Y so we can write that as G sub X minus F sub y DX wedge dy good and now that we can keep going so then for the D Y wedge DZ part notice we'll have h sub y minus G sub Z and like I said that's going to be dy wedge DZ and then finally for the last part we'll have H sub X minus F sub Z and then DX wedge DZ good okay so this is the derivative of the one form and notice it has turned it into a two form okay great so now let's move on to this two form we should take the derivative of this and get a three form so let's see what we get there so d beta so notice here we're gonna have F sub X DX wedge DX wedge D Y so that's what we get from this first derivative so we've got to do the partial with respect to X but notice we can immediately zero that out because we've got two DX's wedge together but we know again that this is an anti commutative operator this wedge product so that cancels out immediately in fact everything is going to cancel except for the Z component because the next one will have to D Y is in it so I'll just go ahead and write only that one so that's going to be F sub Z DZ wedge DX wedge dy and now the only one that we keep from here is the derivative with respect to Y and so that's going to give us plus G sub y and then we have dy wedge DX wedge DZ so we've got to make sure and do it in the correct order will reorder it later though good and then finally we have plus the derivative of H with respect to X D H wedge dy wedge DZ that's the only thing that survives over there so now let's see how we can reorder this so notice that will move this dy past this DX and we'll pick up a minus sign so this is going to turn into minus DX wedge dy wedge DZ but now in this case we actually pick up two minus signs which cancel each other out so this DZ passes past the DX and we pick up a minus sign but then it passes past the dy and we pick up another minus sign so in the end there's no minus sign and we can just put this in the correct order like alphabetical order DX wedge dy wedge DZ and so now let's see what we have so we're gonna have the partial of F with respect to Z minus the partial of G respect to Y plus the partial of H with respect to X and then DX wedge dy wedge DZ and so we've done it we've taken a differential to form and taken its exterior derivative to produce a differential three form now let's say we've got a three form and this one's actually going to be very quick because no matter what partial derivative here we add something that is already part of this wedge product so the derivative with respect to X will give us a DX wedge this whole thing that's going to be zero similarly for the partial with respect to Y and the partial with respect to Z so we have in this case that D gamma is equal to zero the derivative of three form on our three is going to be equal to zero okay so I think that's a good place to stop

Channel: Michael Penn

Views: 35,002

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn

Id: 4--S4N8YDmE

Length: 16min 27sec (987 seconds)

Published: Mon Jul 20 2020