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Differential geometry:2-form computation on a pair of tangent vectors

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data

    I am given a one form Psi = zdx -xydy and two vectors v(1,1,-2) and w(-2,1,1) both tangent vectors of R3 at point P(2,-1,0).

    I am asked to find dPsi(v,w).

    2. Relevant equations

    Lie bracket?

    3. The attempt at a solution

    I know how to computer Psi(v) at p but this is a 1-form.
    I am not sure how to compute the 2 form though.

    I was able to find dPsi:

    dPsi = d(z)^dx +d(-xy)^dy (^== wedge product)

    = -ydxdy -dxdz

    Now I have looked it up in the Internet. The only thing I found was dPsi of two vector fields. But here I just have 2 vectors.
    I am a bit lost.

    Any hints?

    Thanks
     
  2. jcsd
  3. Jun 16, 2009 #2
    I have actually also found this formula on wikipedia:

    (A^B)(v,w)p = A(vp)B(wp) -A(wp)B(vp)

    So I tried to rewrite my two form as a wedge of two 1-forms:

    A = dx
    B= -ydy -dz

    Finally i applied the formula and I got
    A(vp)=1
    B(wp)=1
    A(wp)=-2
    B(vp)=4

    And

    dPsi(v,w) = (A^B)(v,w) = 9

    I am not sure if my work is correct...
    I would appreciate a good enlightment..
     
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