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Geometric representation of two-forms.

  1. Nov 2, 2011 #1
    I've been browsing through MTW recently and I found something that puzzles me:

    They claim that if you have two form, call it [itex]\mathbf{T}[/itex], it's value, say [itex]\mathbf{T}(\mathbf{u} , \mathbf{v} ) [/itex] can be represented geometrically as follows: take two vectors [itex]\mathbf{u}[/itex] and [itex]\mathbf{v}[/itex]; the surface containing those two is [itex]\mathbf{u} \bigwedge \mathbf{v}[/itex] (I don't get this, why isn't it just the vector product [itex] \mathbf{u} \times \mathbf{v}[/itex]?) and the value of the two form is just the number of tubes the "egg-crate" structure cuts through this parallelogram. I don't get this.

    They also state that the a basis two-form, say [itex]\mathbf{d}x \bigwedge \mathbf{d}y[/itex] can be represented by just crossing the surfaces of each basis one-form. This is also confusing.
     
    Last edited by a moderator: Nov 2, 2011
  2. jcsd
  3. Nov 3, 2011 #2
    The cross product only works in three dimensions. The cross product of vectors u and v is actuall an axial vector, or pseudovector, rather than a true vector. But in four dimensions, which direction would the cross product point?

    I think for, some of us, it's better not to use the egg crate visual aid but to just to think of [itex]dx \wedge dy[/itex] as an infinitessimal area element--or you may eventually prefer some combination of the two.

    The egg crate cells with 'vortices' in each cell does help visualizing Stoke's theorem.

    Writing out both vectors in two dimension,

    [tex]\mathbf{u} = u_x dx + u_y dy[/tex]
    [tex]\mathbf{v} = v_x dx + v_y dy[/tex]

    [tex]\mathbf{u} \wedge \mathbf{v} = u_x v_y dx \wedge dy - u_y v_x dx \wedge dy[/tex]
    [tex]= (u_x v_y - u_y v_x) dx \wedge dy[/tex]

    This is the parallelogram multiplied by the area infinitessimal [itex]dx \wedge dy[/itex].
     
  4. Nov 3, 2011 #3
    ^OK, sooo how does that relate to the differential forms..? :eek: (Is it that it's just a function of those infinitesimal areas?)
     
  5. Nov 3, 2011 #4

    atyy

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    The wedge product of two one-forms is an area. In 3D and if a metric is present, it is related to the length of a vector given by the cross-product (a special case of "Hodge duality"). I'm not sure I got that right, see http://www.damtp.cam.ac.uk/research/gr/members/gibbons/gwgPartIII_DGeometry2011-1.pdf 5.0.1 Example: The Cross product in R3
     
  6. Nov 3, 2011 #5
    So it's in a way the generalization of the cross product?
     
  7. Nov 3, 2011 #6

    atyy

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    Yes. The wedge product allows one to define "volumes" for integration without a metric in N-dimensions.
     
  8. Nov 3, 2011 #7
    ^All right! ^_^

    How does it allow you to "cross" the surfaces of the one-forms?

    (i.e. dx wedge dy; how does that lead to the honey-comb?)
     
  9. Nov 3, 2011 #8

    atyy

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    I'm not familiar with this analogy, but I do have MTW. What page is it on?
     
  10. Nov 3, 2011 #9
    ^It's in chapter 4, pages 99-101.
     
  11. Nov 4, 2011 #10
    It doesn't. The wedge product is not a cross product. They are similar but not the same. Specifically, in R3 for a cross product w(u,v) and a wedge product z(u,v),

    [tex]w_i = u_j \times v_k = u_j v_k - u_k v_j[/tex]
    [tex]z_{jk} = u_j v_k - u_k v_j[/tex]
    where i,j,k are cyclic permutations of 1,2,3.

    Notice that the tensor entries are similar. There exists a function that relates the wedge product to the cross product.
     
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