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Homework Help: Differential Geometery: 3-forms on a surface

  1. Jun 6, 2008 #1
    3-forms on a surface. They are equal to 0. i know this. What i dont know is how to prove it.

    This is not a homework question, but my final is tomorrow, and this is a fair question he might ask.

    All i can think of is the following

    Let dx and dy be one forms

    dx^dy would be two form

    if we take the differential of that it would be a 3 form, so we get the following

    d(dx^dy) = d(dx)^dy + dx^d(dy) by leibniz rule, which is 0 +0.

    I know this proof is not good enough cause it is only a special case, and doesnt even take into account that we are on a surface.

    So could somebody please help me outline a proof?
  2. jcsd
  3. Jun 6, 2008 #2


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    A 3-form is a totally antisymmetric linear form on vectors. E.g. f(x,y,z)=-f(y,x,z) and etc for all transpositions of the vectors x,y,z. If x,y,z are from a two dimensional space then at most two of them are linearly independent. Why does this mean f must be 0?
  4. Jun 6, 2008 #3
    Well if at most 2 of them are linearly independent, then this implies that in 2d space

    z= av + bu

    Thus f(u,v, av+bu) = f(u,v,av) + f(u,v,bu) ...

    actually thats probably not right, and im stuck.
  5. Jun 6, 2008 #4


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    What would f(u,v,v) and f(u,v,u) be if f is totally antisymmetric? Remember f is linear, you can pull the constants out.
  6. Jun 6, 2008 #5
    If it is anti-symetric then

    f(u,v,v)= - f(u,v,v) (i switched the v's there)

    Thus f must be 0

    That good enough?
  7. Jun 6, 2008 #6


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    Well, f(u,v,v) is always zero if it's antisymmetric (symmetric has 2 m's in it). f doesn't need to be zero on any combination of vectors if the underlying space has more than two dimensions. Look at it this way. f is determined by the values of f(ei,ej,ek) where ei,ej,ek are basis vectors. If the vector space is two dimensional, it only has two basis vectors. Do you see why that means a 3-form on a surface vanishes?
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