topological_ques

# topological_ques - MIT OpenCourseWare http/ocw.mit.edu...

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V14. Some Topological Questions We consider once again the criterion for a gradient field. We know that and inquire about the converse. It is natural to try to prove that by using Stokes' theorem: if curl F = 0, then for any closed curve C in space, The difficulty is that we are given C, but not S. So we have to ask: Question. Let D be a region of space in which F is continuously differentiable. Given a closed curve lying in D; is it the boundary of some two-sided surface S lying inside D? We explain the "two-sided" condition. Some surfaces are only one-sided: if you start painting them, you can use only one color, if you don't allow abrupt color changes. An example is S below, formed giving three half-twists to a long strip of paper before joining the ends together. Surface S Boundary C (trefoil knot) S has only one side. This means that it cannot be oriented: there is no continuous choice for the normal vector n over this surface. (If you start with a given n and make it vary continuously, when you return to the same spot after having gone all the way around, you will end up with -n, the oppositely pointing vector.) For such surfaces, it makes no sense to speak of "the flux through S", because there is no consistent way of deciding on the positive direction for flow through the surface. Stokes' theorem does not apply to such surfaces. To see what practical difficulties this causes, even when the domain is all of 3-space, consider the curve C in the above picture. It's called the trefoil knot. We know it is the boundary of the one-sided surface S, but this is no good for equation (3), which requires that we find a two-sided surface which has C for boundary. There are such surfaces; try to find one. It should be smooth and not cross itself. If successful, consider yourself a brown-belt topologist. The preceding gives some ideas about the difficulties involved in finding a two-sided surface whose boundary is a closed curve C when the curve is knotted, i.e., cannot be con- tinuously deformed into a circle without crossing itself at some point during the deformation.
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## This note was uploaded on 05/04/2011 for the course MATH 18.02 taught by Professor Auroux during the Spring '08 term at MIT.

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topological_ques - MIT OpenCourseWare http/ocw.mit.edu...

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