Cross prod

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  • #1
quasar987
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Simple question: is the cross product defined in R^n ? In my linear algebra textbook, they talk about the dot product in lenght but don't even mention the cross product.
 

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No, in general it is not (and cannot be) defined for [itex]\mathbb{R}^n[/itex]. An analogous product does exist in [itex]\mathbb{R}^7[/itex], though, constructed using the multiplication table for octonions.
 
  • #3
jcsd
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No the cross product is only defined in R^3.

It's defintion doesn't lend itself well to generalizations other spaces of different dimenion (especially if you want a binary operation). Of course thta's not to say that generalizations are impossible.
[tex]a\times b = \left|\begin{array}{ccc}\hat{x}&\hat{y}&\hat{z}\\a_x&a_y&a_z\\b_x&b_y&b_z\end{array}\right|[/tex]

Which matrix would you take the determinant of when n is not equal to 3?
 
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quasar987
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What is the easiest route to showing the equivalence of the algebraic and geometric definitions of the cross product?
 
  • #5
Galileo
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That's a fairly difficult problem in general. Usually they start with the geometric definition, then show that the cross-product is distributive: A X (B+C)=(A X B)+(A X C)
Then you can derive the algebraic definition by writing the vectors out in components and use distributivity. Proving distributivity is not very easy, but certainly doable.
 
  • #6
dextercioby
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The name "cross product" refers to the simple Euclidean 3D case.But since this "cross product" is nothing but a Hodge dual of a wedge product between two 1-forms,going to p-forms on arbitrary manifolds gives you the desired generalization...



Daniel.
 
  • #7
quasar987
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Galileo said:
That's a fairly difficult problem in general. Usually they start with the geometric definition, then show that the cross-product is distributive: A X (B+C)=(A X B)+(A X C)
Then you can derive the algebraic definition by writing the vectors out in components and use distributivity. Proving distributivity is not very easy, but certainly doable.
That's exactly what I was trying. And I think I'm on the right track to proving distributivity. :cool:
 
  • #8
Galileo
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quasar987 said:
That's exactly what I was trying. And I think I'm on the right track to proving distributivity. :cool:
Good luck :biggrin:
 

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