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Spivak (Calculus on Manifolds) proof of stolkes theorem

  1. Aug 22, 2010 #1
    that's pretty much the proof of Stolkes Theorem given in Spivak
    but I'm having a lot of difficulty understanding the details

    specifically...when the piecewise function is defined for j>1 the integral is 0
    and for j=1 the integral is nontrivial...why is it defined like that?

    Also, Im having difficulty understanding what the inclusion map does (spivak defines it

    as I(j,alpha) which is a continuous function or a chain of some sort) but the pull back

    I*(j,alpha)fdx^1....dx^n is taken and integrated over in that piecewise function

    could someone shed some light on that?

  2. jcsd
  3. Aug 23, 2010 #2


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    There is no definition involved here.. It is just a fact. Recall that at this stage, they are working on the special case M=(0,1]x(0,1)n-1. For simplicity, let me assume that that n=2 here, so that M=(0,1]x(0,1). The hypothesis is that f has compact support on M. So for any given y in (0,1), f(0,y)=0. And for any x in (0,1], f(x,1)=0. But f(1,y) need not be 0.

    Hopefully, you can demonstrate these little assertions if they are not clear to you, and you can generalize my reasoning to higher n, and conclude to the formula for when j=1, and when j>1.
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