Differential Forms. Introduction

Transcription

1 ifferential Forms A 305 Kurt Bryan Aesthetic pleasure needs no justification, because a life without such pleasure is one not worth living. ana Gioia, Can Poetry atter? Introduction We ve hit the big three theorems of vector calculus: Green s, Stokes, and the ivergence Theorem (although Green s theorem is really a pretty obvious special case of Stokes theorem). But it s an amazing fact that all of these theorems are really special cases of an even larger theorem that unifies all that we ve studied so far. This new theory also extends what we ve done to spaces of arbitrary (finite) dimension. It rests on the idea of differential forms. ifferential forms are hard to motivate right off the bat. It won t be immediately clear how differential forms are related to what we ve done, but be patient. For now treat differential forms as a new kind of mathematical object; we re going to learn how to manipulate them add, multiply, differentiate, and integrate. It will be quite abstract and formal at first, but you ll soon see how they connect with what we ve already done. anifolds You ve already seen many examples of one and two dimensional manifolds. A one dimensional manifold C is just a curve. Such a curve can exist in any dimension and can be parameterized as X(t) (I m using this instead of R), where t ranges over some interval (a, b) in lr and X(t) (x (t), x 2 (t),..., x n (t)) if the curve exists in n dimensions. You may recall that we required that our curves be smooth, which meant that the curve had to have a proper parameterization X(t) must be differentiable, injective, and X (t) 0 for any t in. Smooth one dimensional manifolds are just smooth curves. Recall that the vector X (t) is tangent to the manifold C. A smooth two dimensional manifold S is just a surface, of which we ve seen many examples (in three dimensions.) Such a surface can be specified as X(u, v), where (u, v) ranges over some appropriate subset of lr 2. The function X(u, v) is of the form X(u, v) (x (u, v), x 2 (u, v),..., x n (u, v)), although we ve mostly considered the case n 3. There were some technicalities in dealing with surfaces and integrating over them the surface had to be orientable, and we did require that the tangent vectors X X and be independent, i.e., neither a multiple of the other. u v If this isn t true then our surface S might end up being only one-dimensional. Also, the surface was not allowed to self-intersect. Based on the above, it s not much of a stretch to consider a k-dimensional manifold. Let be some bounded open region in lr k. We ll use coordinates u (u, u 2,..., u k ) on. Consider a function X(u) (X (u),..., X n (u))

2 defined on, which maps each point in into a point in n dimensional space. X(), the image of in lr n. We will say that is a smooth k-manifold if Take The function X is C. The function X is injective (distinct points in go to distinct points in lr n. The tangent vectors ( X X,..., X ) n u k u k u k are all linearly independent at all points in. This means that the only solution to c X u k + + c n X n u k 0 is c c 2 c n 0. Since this is a semi-intuitive account of differential forms and manifolds, we won t harp too much on this last requirement. Example : Let be the open cube 0 < u, u 2, u 3 < in lr 3. Take the mapping X from lr 3 to lr 5 to be defined by X (u, u 2, u 3 ) u u 2 X 2 (u, u 2, u 3 ) 3u 3 X 3 (u, u 2, u 3 ) u 2 X 4 (u, u 2, u 3 ) u 3 2u 2 X 5 (u, u 2, u 3 ) 3. The function X is obviously C. It s a bit tedious (aple makes it easier) but you can check that no two points in go to the same image point in lr 5, and the tangent vectors are in fact all independent. The image X() is a smooth 3-manifold in lr 5. ifferential Forms Our discussions will take place in lr n. We ll use (x,..., x n ) as coordinates, and write x for the point (x,..., x n ). We will define a 0-form ω to be simply a function on lr n, so f(x,..., x n ). That s all there is to say about 0-forms. A basic or elementary -form in lr n is an expression like dx i, where i n. ore generally, a -form ω is an expression like F (x) dx + F 2 (x) dx 2 + F n (x) dx n 2

3 where the F i are functions of x. You ve actually encountered -forms before, when we did Green s theorem. Recall that Green s Theorem said ( F2 x F ) dx dy F (x, y) dx + F 2 (x, y) dy. y The integrand on the right is an example of a -form. A differential -form is not a passive object, but in fact can be thought of as a kind of function. The basic -form dx i accepts as input a single vector v and outputs v i, the ith component of v, so dx i (v) v i. A general -form F (x) dx + F n (x) dx n acts on a single input vector v as ω(v) F (x)v + F n (x)v n. You can add two 0-forms in the obvious way (as functions). You can similarly add two -forms, e.g., (x 2 2 dx + e x dx 2 ) + (2 dx x dx 2 ) (x ) dx + (e x x ) dx 2. A basic differential 2-form ω is an expression like dx i dx j where i, j n. The symbol denotes what is called the wedge or exterior product. on t worry too much about what it means (yet); in the end it will mean essentially just dx i dx j, an object suitable to stick under a double integral. Like -forms, 2-forms also act on vectors. A basic two form dx i dx j accepts as input TWO vectors v and v 2. The output is the determinant [ ] dxi (v ω(v, v 2 ) dx i dx j (v, v 2 ) det ) dx i (v 2 ) dx j (v ) dx j (v 2 ) Based on the properties for the determinant that we ve seen, you can immediately conclude that dx i dx j dx j dx i dx i dx i 0 This raises an interesting question: How many different 2-forms are there in n dimensions? If you count all possible combinations for i, j n for dx i dx j you get n 2, but in fact n of these are really zero. That leaves n 2 n possible 2-forms, although in some sense there is only half that many, for the second property above shows that, e.g., dx dx 2 dx 2 dx. There are thus only (n 2 n)/2 half this many independent 2-forms. A (general) differential 2-form is an expression of the form F 2 (x)dx dx 2 + F 3 (x)dx dx 3 + F n,n (x)dx n dx n 3 i<j n F ij (x)dx i dx j.

4 Notice that we can always assume that i < j in the sum above, for if we had j < i we could simply swap the order of dx j dx i and replace it with dx i dx j. You can probably guess how such a 2-form acts on two input vectors: ω(v, v 2 ) i<j n F ij (x)dx i dx j (v, v 2 ) [ dxi (v F ij (x)det ) dx i (v 2 ) dx j (v ) dx j (v 2 ) i,j>i Example 2: Let ω denote the 2-form in three dimensions (x + 2x 3 )dx dx 2 + x 2 dx dx 3. Let s compute ω applied to the vectors v (, 3, 3), v 2 (, 0, 7). You get [ ] [ dx (v ω(v, v 2 ) (x + 2x 3 )det ) dx (v 2 ) dx (v + x dx 2 (v ) dx 2 (v 2 ) 2 det ) dx (v 2 ) dx 3 (v ) dx 3 (v 2 ) [ ] [ ] (x + 2x 3 )det + x det 3 7 3(x + 2x 3 ) + 0x 2 3x + 0x 2 + 6x 3. Now we re ready for the definition of a basic k-form: a basic k-form ω in n dimensions is an expression of the form dx i dx i2 dx ik where i j n for all j. Such a k-form accepts as input k vectors v,..., v k to give output ω(v,..., v k ) det dx i (v ) dx i (v 2 ) dx i (v k ) dx i2 (v ) dx i2 (v 2 ) dx i2 (v k ).... dx ik (v ) dx ik (v 2 ) dx ik (v k ) By using the properties of determinants that we ve already deduced, it s easy to see that dx i dx ij dx ik dx im dx i dx ik dx ij dx im () dx i dx ij dx ij dx im 0 (2) Given the above two facts, it s interesting to contemplate the question how many independent basic k-forms are there in n dimensions? It s not too hard to figure out, so I ll leave it for you. A (general) k-form ω is an expression of the form i,...,i k n Such a k-form accepts k input vectors to produce ω(v,..., v k ) i,...,i k n F i,...,i k (x)dx i dx i2 dx ik. (3) F i,...,i k (x)dx i dx i2 dx ik (v,..., v k ). 4 ] ]

5 With this notation, it s easy to see why a function is called a 0-form it doesn t accept any input vectors. Integrating ifferential Forms Integrating differential forms is easy. The general rule is that one integrates a k-form over a k-manifold. Let s first look at how to integrate a -form over a -manifold. Accordingly, let X(t) with a t b be a smooth -manifold in lr n and let ω be a -form defined in a neighborhood of. We use ω to denote the integral of ω over and define this integral by b a ω(x (t)) dt. (4) Example 3: Let be the -manifold in lr 3 defined by X(t) (3t, t 2, 5 t) for 0 t 2. You can see that is just a curve in three dimensions. Let ω be the -form 2x 2 dx x x 3 dx 2 + dx 3. Using equation (4) and recalling how a -form acts on input vectors we obtain 2 0 ω((3, 2t, )) dt ((2x 2 )(3) x x 3 (2t) ) dt (6t 3 24t 2 ) dt 42. Example 4: Suppose that F(x) (F, F 2,..., F n ) is a vector field in lr n (so each of the F i are functions of x). efine the differential form F dx + F 2 dx F n dx n. If we compute the integral of ω over some -manifold defined by R(t) (x (t),..., x n (t)) we get b a b a (F dx + F 2 dx F n dx n )(R (t)) dt (F x t + + F n x n t ) dt F dr. In other words, the integral of the -form ω over is just the familiar line integral of the vector field F over the curve. One can integrate a 2-form ω over a 2-manifold. Suppose that is parameterized by X(u), where u (u, u 2 ) ranges over, a region in lr 2. The integral is defined by ω( X, X ) du du 2. Example 5: Let be a 2-manifold in lr 4 parameterized by X(u) (u, u u 2, 3 u + u u 2, 3u 2 ) 5

6 where u 2 + u 2 2 <. Take x 2 dx dx 3 x 4 dx 3 dx 4. You can easily compute that X (,, u 2, 0) X (0,, u, 3) and that ω( X, X [ 0 ) x 2 det u 2 u x 2 u + 3x 4 (u 2 ) As a result u 2 u u 2 9u u 2. ] x 4 det [ u2 u 0 3 u 2 (u 2 u 2 u u 2 9u u 2 ) du 2 du 2π. Example 6: Consider a smooth 2-manifold in lr 3 parameterized as X(u) for u in lr 2. Let F dx 2 dx 3 + F 2 dx dx 3 + F 3 dx dx 2 be a 2-form defined in a neighborhood of. Then F (dx 2 dx 3 )( X, X ) + F 2 (dx dx 3 )( X, X ) + F 3 (dx dx 2 )( X, X ) ( [ X2 ] [ X 2 X ] [ X X ]) X u F det u + F 2 det u + F 3 det du du 2 F ( X 2 X 2 ) + F 2 ( X X ) +F 3 ( X X 2 X X 2 ) du du 2 u F da. ] X 2 X 2 In other words, the integral of ω over is just the flux of the vector field F (F, F 2, F 3 ) over. If the step between the last two equations looks mysterious, all I ve done is use the fact that da ( X/ ) ( X/ ) and dotted this with F; it s the usual procedure for computing flux over a parameterized surface. You can probably see how we should integrate a general k-form over a k-manifold. If the manifold is parameterized by X(u) for u lr k, and if we have a general k-form ω given by an equation like (3) then i,...,i k n F i,...,i k (X(u))(dx i dx ik )( X,..., X u k ) du du k. 6

7 Example 7: Let be a 3-manifold in lr 4 parameterized as for u 2 + u u 2 3 <. Take It s easy to compute that Then ω( X, X, X ) x 3 det u 3 Finally, we find that X(u, u 2, u 3 ) (u 2 u 3, u 2, 3u 2 + u 3, u u 2 ) x 3 dx dx 2 dx 4. X (0, 2u, 0, u 2 ) X (u 3, 0, 3, u ) X u 3 (u 2, 0,, 0). u 3 u 2 2u 0 0 u 2 u 0 2u 2 u 2 6u 2 u u 2 u 2 u 3. (2u 2 u 2 6u 2 u u 2 u 2 u 3 ) du du 2 du π. The last triple integral is easier if you convert it to spherical. Example 8: Let be an open region in lr n and suppose that is parameterized as X(u) where u, where is another region in lr n. In other words, is an n-manifold in n-dimensional space. Of course, we could very easily parameterize by taking and using X(u) u, the identity map, but this isn t necessary, and by not doing so you ll see some important issues in integrating differential forms over manifolds. Let f(x) dx dx 2 dx n (this is the most general n-form in lr n why?) Then f(x(u))(dx dx n )( X,..., X ) u n X X n f(x(u))det... du... du n X X u n n u n ± f(x,..., x n ) dx... dx n where the last equality follows from the change of variables formula in n dimensions (notice that the absolute value signs on the determinant are missing). The integral of ω over is PLUS OR INUS the integral of f over. It could be either, depending on the parameterization. 7

8 The last example raises an important issue that we won t entirely resolve. You might hope the quantity ω should depend only on and ω, but this is true only up to a minus sign. The problem is that any manifold which is orientable has two possible orientations, and this is where the sign ambiguity arises. You ve already encountered this in line and surface integrals: if C is a curve in two or three dimensions and F is a vector field, the quantity C F dr is not well-defined. It depends on which direction you parameterize or traverse the curve C; if you consider this integral physically, as work done, it clearly depends on whether you re moving from point A to point B or the reverse. The sign ambiguity also arose in doing flux integrals over a surface. Which direction you choose for the normal affects the sign of the answer. It is a general fact (that I ll let you check for yourself) that the quantity ω does not depend on how is parameterized, except for the sign of the answer. Proving this really comes down to the change of variable formula for integral of n variables. The different signs are a manifestation of how we chose to orient (implicitly) when we parameterized it. The Wedge Product The wedge (or exterior product allows us to multiply differential forms. Suppose that f is a 0-form (a function) and ω is a k-form as given in equation (3). The wedge product of f with ω is just f ff i,...,i k (x)dx i dx i2 dx ik. i,...,i k Suppose that η is an m-form, η G j,...,j m (x)dx j dx j2 dx jm. j,...,j m The wedge product of η with ω is the k + m form given by adjoining the two forms, as η F i,...,i k (x)g j,...,j m (x)dx i dx ik dx j dx jm. As complicated as this looks, it s pretty easy in any specific example, and the rules () and (2) make the result a lot smaller than you might think. Example 9: Let x 2 dx dx 3 + dx dx 4 and η (x + )dx 2 dx 4. Then ω η x 2 (x + ) dx dx 3 dx 2 dx 4 + (x + ) dx dx 4 dx 2 dx 4. However, by equation (2) the second term above (which contains two copies of dx 4 ) is zero. Also, by property () we can flip the dx 2 and dx 3 in the first term, which will introduce a minus sign, so ω η x 2 (x + ) dx dx 2 dx 3 dx 4. It s not hard to prove the following properties for the wedge product: If f is a function, ω and ω 2 are k-forms, η an m-form, and τ any form then (ω + ω 2 ) η ω η + ω 2 η 8

9 (ω η) τ ω (η τ) (fω ) η f(ω η) ω (fη) ω η ( ) km η ω Only the last property is at all nonobvious, but a little experimentation will show you how to prove it. In summary, this last property says that ω η η ω UNLESS both forms are of odd degree, in which case the sign flips. The Exterior erivative Given that differential forms can be integrated, it stands to reason that they can also be differentiated. The exterior derivative d is an operator that turns k-forms into k + -forms, according to very simple rules: The exterior derivative of a 0-form f(x) in lr n (recall, f is just a function) is The exterior derivative of the k-form is given by df f x dx + + f x n dx n. (5) F i,...,i k dx i dx ik i,...,i k d df i,...,i k dx i dx ik (6) i,...,i k where df i,...,i k is computed according to equation (5). Example 0: Let (x + x 2 3)dx dx 2. Then d d(x + x 2 3) dx dx 2 dx dx dx 2 + 2x 3 dx 3 dx dx 2 2x 3 dx 3 dx dx 2 2x 3 dx dx 3 dx 2 2x 3 dx dx 2 dx 3 Notice the term with two dx s is zero, and in the last two steps I just put dω in standard form. Generalized Stokes Theorem The generalized version of Stokes Theorem embodies almost everything we ve done in this course; the divergence theorem and the original Stokes Theorem are special cases of this more general theorem. Before stating this theorem we need a few preliminary remarks. 9

10 A smooth k-manifold in lr n may or may not be orientable. If it is orientable then it has two possible orientations, as we saw in examples above. It s not hard to believe (though we won t stop to prove) that the boundary of a k-manifold is itself a manifold of dimension k. The boundary may or may not be orientable. We will assume that the manifolds that follow, as well as their boundaries, are orientable. Recall that in the divergence theorem we related a volume integral to a surface integral, and for things to work out properly we needed the surface (in particular, da) to be oriented properly. Similarly in Stokes theorem, after choosing a direction for the unit normal n on a surface S, we then had to orient S with the right hand rule. Similar considerations hold in higher dimensions. If we have a manifold and we have chosen an orientation for, then this induces a consistent orientation on, somewhat akin to the right hand rule. How this can be done we won t go into for now. The down side is that our integrals may occasionally be off by a minus sign. Stokes Theorem: (Generalized version) Let be a smooth oriented k-manifold with consistently oriented smooth boundary. Let ω be a k form defined in a neighborhood of. Then d ω. Example : Let be as in Example 5. Given how was parameterized, you will probably believe that can be described as the image of the boundary of the unit disk, u 2 + u 2 2, under the mapping from lr 2 to lr 4 that parameterized. Given that the boundary of the disk can be parameterized by u cos(t), u 2 sin(t), it seems reasonable that the one dimensional curve can be parameterized as Y(t) (cos(t), cos(t) sin(t), 3 cos(t) + cos(t) sin(t), 3 sin(t)) (notice I m using Y for.) Now let x 2 3 dx. It s easy to compute that d 2x 3 dx dx 3. Let s first compute dω: Now for d ω we have u 2 u 2 u 2 u 2 u 2 2π 0 2π 0 π 2. u 2 dω( X, X ) 2x 3 det ω(y (t)) dt [ 0 u 2 u ] du 2 du 2(3 u + u u 2 )u du 2 du π 2. (3 cos(t) + cos(t) sin(t)) 2 ( sin(t)) dt 0

11 Example 2: Let F F i + F 2 j + F 3 k be a vector field in three dimensions and let F dx + F 2 dx 2 + F 3 dx 3, a -form. Suppose that is an orientable 2-manifold (a surface) in three dimensions with one dimensional boundary. As we saw in Example 4, F dr. Now let s compute dω and integrate it over. If you write out dω you find that many terms cancel (they contain a wedge product with identical dx i s) and you obtain d + ( F2 F ) dx dx 2 + x x 2 ( F3 F ) 2 dx 2 dx 3 x 2 x 3 ( F3 F ) dx dx 3 x x 3 Now define a vector field G G i + G 2 j + G 3 k by taking ( F3 G F ) 2 x 2 x 3 ( F3 G 2 F ) x x 3 ( F2 G 3 F ). x x 2 In short, G F. Then we can write d G dx 2 dx 3 + G 2 dx dx 3 + G 3 dx dx 2. If you look at Example 6 (and replace F there by G here), you find that d G da. But given that G F, we conclude from the generalized Stokes Theorem that F dr ( F) da which is the usual version of Stokes Theorem in three dimensions! Example 2: Let be a bounded region in three dimensions with orientable two dimensional boundary. Again, let F F i + F 2 j + F 3 k be a smooth vector field defined on and define the 2-form F dx 2 dx 3 + F 2 dx dx 3 + F 3 dx dx 2 Again referring back to Example 6, we find that F da,

12 the flux of F over. Now compute dω. ost of the terms are zero and we end up with d ( F + F 2 + F ) 3 dx dx 2 dx 3 ( F)dx dx 2 dx 3. x x 2 x 3 Now from Example 8 we know that (if is oriented correctly) d F dv. From the general Stokes Theorem we conclude that which is the divergence theorem! F da F dv. 2