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Forms and determinants

  1. Jul 30, 2014 #1
    Hi, 2-forms are defined as

    [itex]du^{j} \wedge du^{k}(v,w) = v^{j}w^{k}-v^{k}w^{j} = \begin{vmatrix} du^{j}(v) & du^{j}(w) \\ du^{k}(v) & du^{k}(w) \end{vmatrix}[/itex]

    But what if I have two concret 1-forms in [itex]R^{3}[/itex] like [itex](2dx-3dy+dz)\wedge (dx+2dy-dz)[/itex] and then I calculte [itex](2dx-3dy+dz)\wedge (dx+2dy-dz)=-7dy \wedge dx +3dz \wedge dx - dy \wedge dz= 7 dx \wedge dy + 3 dz \wedge dx + dy \wedge dz[/itex] (now a 2-form)
    I know this is the same as the cross product between [itex](2,-3,1)^{T}[/itex] and [itex](1,2,-1)^{T}[/itex]
    What has this to do with the determinant? If I calculate this for 1-forms in [itex]R^{2}[/itex] like [itex](2dx+4dx)\wedge (3dx+9dy) = -18 dx\wedge dy +12 dx \wedge dy = 6 dx \wedge dy[/itex] which equals the determinant of the vectors [itex](2,4)[/itex] and [itex](3,9)[/itex]. But this is not like 1-forms in [itex]R^{3}[/itex]. And is the geometric interpretation right that [itex]7 dx \wedge dy + 3 dz \wedge dx + dy \wedge dz[/itex] means that 7 is the part of the area projected to the xy axis, 3 the part projecte to the zx axis and 1 the part projected onto the yz axis? Or is it a different coordinate system like dxdy, dzdx and dydz ?
    Now actually these are all functions of vectors like [itex]v[/itex] or [itex]w[/itex] as in the definition of 2-forms. What happens if they come into this picture?
  2. jcsd
  3. Aug 14, 2014 #2
    I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
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