Cross Product in R^n: Defined or Undefined?

[quasar987]

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

 

[Data]

No, in general it is not (and cannot be) defined for $\mathbb{R}^n$. An analogous product does exist in $\mathbb{R}^7$, though, constructed using the multiplication table for octonions.

 

[jcsd]

No the cross product is only defined in R^3.

It's definition 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.
$$a\times b = \left|\begin{array}{ccc}\hat{x}&\hat{y}&\hat{z}\\a_x&a_y&a_z\\b_x&b_y&b_z\end{array}\right|$$

Which matrix would you take the determinant of when n is not equal to 3?

 

[quasar987]

What is the easiest route to showing the equivalence of the algebraic and geometric definitions of the cross product?

 

[Galileo]

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.

 

[dextercioby]

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.

 

[quasar987]

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.

 

[Galileo]

That's exactly what I was trying. And I think I'm on the right track to proving distributivity.

Good luck