From: foltinek@math.utexas.edu (Kevin Foltinek) Subject: Re: Vector Calc Help!! Date: 14 Dec 1999 12:40:56 -0600 Newsgroups: sci.math Keywords: algebra of differential forms In article "Joe Smith" writes: > I looked up the meaning of "^" and found it to be the wedge product. > I read the properties of the wedge product, and I now see how they get > d(alpha) =0, but I still don't get the concept of closed and exact > differential forms. Not sure what you mean by "get the concept of". Presumably you can look up the definitions yourself, but here they are anyway: alpha closed: d(alpha)=0. alpha exact: there is a beta such that alpha=d(beta). Maybe you want to know why we care about these things. d is a generalization of the gradient, curl, and divergence. (You can write those three in terms of d if you're careful and you also know about something called the Hodge star operator. grad=d, curl=*d, div=*d*.) Remember the identities curl(grad)=0 and div(curl)=0? Well, these are both special cases of d(d)=0. That is, if alpha is exact, then it is closed. Do you remember that you have to be careful when you say "curl(F)=0, therefore F=grad(f)"? Specifically, you have to make sure about the domain of F; to keep things simple, let's just say that if F is defined on an open disk in 2 dimensions, and curl(F)=0 everywhere in that disk, then F=grad(f) for some function f defined on that disk. Puncture the disk, and the conclusion might be false. The generalization of this is: if d(alpha)=0, i.e., if alpha is closed, is it exact? The answer, again, depends on the domain of alpha; if it is an open n-dimensional ball, then the answer is yes, but in general the answer may be no. The study of this question is related to algebraic topology (de Rham cohomology); very loosely, what kind of holes are in the domain of alpha, and what its general "shape" is (sphere, donut, cylinder, etc.). Another use of closed and exact: the fact that d^2=0 is another way of expressing the fact that (mixed) partial derivatives commute. There is a way of saying some things about partial differential equations by using the exterior derivative d, and studying closed and/or exact forms plays a role in this. I hope this helped somehow. Kevin. ============================================================================== From: jacheann@aol.com (JacheAnn) Subject: Re: Vector Calc Help!! Date: 15 Dec 1999 06:09:31 GMT Newsgroups: sci.math Let's pretend that @ is a "theta" for now. Then d@ =(-y/[x^2+y^2]) dx + (x/[x^2+y^2]) dy is a closed form on the plane minus the origin. But it's not exact on that domain, essentially because the function it would be the differential of has to differ from @ by an additive constant on the plane minus the polar axis (y=0,x>0). Oh, and @ isn't even close to being a continuous function on that domain. For higher degree forms, the same phenomenon happens. NOT, however, if the domain is simply connected (has no holes or points missing). The closed but non-exact forms then may be integrated to study the (co)homology of the domain, a type of topological classification. Help this hopes.(sic) Len ============================================================================== From: foltinek@math.utexas.edu (Kevin Foltinek) Subject: Re: Vector Calc Help!! Date: 15 Dec 1999 12:23:08 -0600 Newsgroups: sci.math In article <19991215010931.11857.00000979@ng-cl1.aol.com> jacheann@aol.com (JacheAnn) writes: > Let's pretend that @ is a "theta" for now. > Then d@ =(-y/[x^2+y^2]) dx + (x/[x^2+y^2]) dy > is a closed form on the plane minus the origin. > > But it's not exact on that domain, A technical point of some importance: by calling something "d\theta" (d@), you are saying that it *is* exact; it is the exterior derivative of \theta. The expression on the right-hand-side of your equation, let's call it \phi = (-y/(x^2+y^2))dx + (x/(x^2+y^2))dy , is, as you say, not exact on the punctured plane. \theta is defined on the plane minus a ray; d\theta has the same domain; \phi is the continuous extension of d\theta to the punctured plane. > For higher degree forms, the same phenomenon happens. NOT, however, > if the domain is simply connected (has no holes or points missing). This latter sentence is not true; there are many examples of simply connected domains with non-trivial cohomology. (Simply connected does not imply "no holes or points missing".) It is true if you replace "simply connected" by "star-shaped". Kevin.