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Differential form

  1. Aug 6, 2007 #1
    1. The problem statement, all variables and given/known data
    Let w be the form w= xdydz in R^3. Let S^2 be the unit sphere in R^3.
    If we restrict w on S^2, is w exact?


    2. Relevant equations



    3. The attempt at a solution
    My guess is w is not exact on S^2.
    Suppose w is exact on S^2. Then w=da for some 1-form a=fdx+gdy+hdz.
    Then by definition of exterior derivative, we get
    w=(-df/dy+dg/dx)(dx^dy)+(-df/dz+dh/dx)(dx^dz)+(-dg/dz+dh/dy)(dy^dz)
    So we get the conditions:
    df/dy=dg/dx, df/dz=dh/dx, x=-dg/dz+dh/dy.
    I think I should use a fact that I am working on a unit sphere. Could anybody help me?
     
  2. jcsd
  3. Aug 6, 2007 #2
    I did not tried but it would be useful to write everything in spherical coordinates and to take radius as 1.
     
  4. Aug 6, 2007 #3
    streographic projection

    Using stereographic projection,(say (s,t)) I attained follwoing condition if I assume xdydz is exact on S^2,

    for some smooth function g and f,
    -df/dt+dg/ds = (-24(s^2)(t^2)-8(s^4)-8s)/(1+s^2+t^2)^4.
    Now in order to show xdydz is not exact, it suffices to show such f and g does not exist. How can I show it?
     
  5. Aug 7, 2007 #4
    Different way: I tried to use the thm saying if a differential form is exact then
    its closed.
    i.e.if you can show dw is not equal to zero(meaning not closed)
    you can conclude w is not exact.If i didn't make a mistake dw=dx dy dz
    But still don't now how it is related to S^2
     
  6. Aug 7, 2007 #5

    Dick

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    dw IS zero when restricted to S^2, it's a three form on a two manifold. But you can't apply the Poincare lemma to show it's exact since S^2 isn't contractible.
     
  7. Aug 7, 2007 #6

    Dick

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    Try it in spherical coordinates, as matness suggested. It's pretty straight forward to find a solution to w=da, I think.
     
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