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Possible Generalization of Cross Product?

  1. Nov 5, 2006 #1
    In R^3 it is easy to compute the cross product, and i know how to compute it in higher dimensions using wedge product and the hodge star, which shows that it only exists in 3n dimensions.

    My question is given two vectors in C^3 (complex), is there a neat way to find one perperdicular to both? Ie - a generalization of the original cross product?
  2. jcsd
  3. Nov 5, 2006 #2
    I don't think that's possible. If you define two vectors that are perpindicular to a third by setting the dot product to zero. Then subtitute and solve the three systems of equations... then you get the cartesian cross product equation. That is the proof that the cross product equation is the general situation of having a vector perpindicular to two others. Therefore, I'll assume the equation is the simplest method.
    Last edited: Nov 5, 2006
  4. Nov 5, 2006 #3


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    And the result turns out to be essentially the same as with real vectors. (I found the formula a different way -- simply starting with the ordinary cross product, and seeing what needs to be changed)
  5. Nov 5, 2006 #4
    So what is the formula guys? I know there must exist one since C^3 is just a six dimensional real space, and i know cross products are defined in 3n space according to the wedge product definition.
  6. Nov 6, 2006 #5
    Don't worry guys i solved it using the wedge product and Hodge star, it turns out that aXb is exactly the same as the normal definition in R^3 except you conjugate both (a) and (b) before you do the cross product.

    where a = (a1,a2,a3) and b = (b1,b2,b3) and the components themselves are elements of C, ie a1 = x+yi
  7. Nov 6, 2006 #6
    Oh, you didn't even know the formula?

    [tex] A X B = [a_{2}b_{3} - a_{3}b_{2}, a_{3}b_{1} - a_{1}b_{3}, a_{1}b_{2} - a_{2}b_{1}] [/tex]

    It may look complicated, but there is a distinct pattern that's very simple to remember. "231".
  8. Nov 6, 2006 #7
    Heh, of course i knew that, but i just wanted to see if there was a complex version for C^3. Btw, its easier to remember by writing in matrix form and taking it as a determinant.
  9. Nov 6, 2006 #8


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    Yay! I figured you could do it.
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