Possible Generalization of Cross Product?

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Discussion Overview

The discussion revolves around the possibility of generalizing the cross product to complex vectors in C^3, exploring theoretical frameworks and mathematical formulations related to this concept.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant notes that the cross product in R^3 can be computed using the wedge product and Hodge star, suggesting a generalization might exist in C^3.
  • Another participant argues that defining perpendicular vectors through the dot product leads back to the Cartesian cross product, implying that this method is the simplest approach.
  • A different participant claims that the result for complex vectors is essentially the same as for real vectors, having derived a formula from the ordinary cross product.
  • One participant expresses curiosity about the specific formula for the cross product in C^3, asserting that C^3 can be viewed as a six-dimensional real space.
  • Another participant claims to have solved the problem using the wedge product and Hodge star, stating that the cross product in C^3 involves conjugating the vectors before applying the standard cross product formula.
  • A participant provides the formula for the cross product in C^3, highlighting a pattern in its components.
  • One participant mentions an alternative way to remember the formula by expressing it in matrix form and taking the determinant.

Areas of Agreement / Disagreement

The discussion contains multiple competing views regarding the generalization of the cross product to complex vectors, with no consensus reached on a singular approach or formula.

Contextual Notes

Participants reference the relationship between complex and real vector spaces, and the implications of using the wedge product and Hodge star, but do not resolve the mathematical complexities involved.

SeReNiTy
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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?
 
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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.
 
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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)
 
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.
 
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
 
Oh, you didn't even know the formula?

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}]

It may look complicated, but there is a distinct pattern that's very simple to remember. "231".
 
Sane said:
Oh, you didn't even know the formula?

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}]

It may look complicated, but there is a distinct pattern that's very simple to remember. "231".

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.
 
Don't worry guys i solved it using the wedge product and Hodge star
Yay! I figured you could do it.
 

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