# Cross prod

Homework Helper
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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.

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

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

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

Galileo
Homework Helper
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
Homework Helper
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.

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Gold Member
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.

Galileo