Cross product in $\mathbb R^n$

Question

I read that the cross product can't be generalized to $\mathbb R^n$. Then I found that in $n=7$ there is a Cross product: https://en.wikipedia.org/wiki/Seven-dimensional_cross_product

Why is it not possible to define a cross product for other dimensions $ \ge 4$?

Answer 1

The issue is the problem of choice.

Given two linearly independent vectors in $\mathbb R^3$, the dimension of the space perpendicular on both is $1$. This means that up to scalar multiplication, you know the perpendicular direction.

The only issue is is choosing the one of the opposing directions and magnitude, and there is a simple way of doing this, the known way, which in some sense comes out in a natural way from the Cramer's rule. Moreover, this choice works nicely in the case of linearly dependent vectors.

In higher dimensions the problem becomes much more complicated since the perpendicular space on two vectors has higher dimension. Then, if one tries to define the cross product, one has to chose one of infinitely many directions in a consistent way.

Also, $\mathbb R^3$ can be identified in a "natural" way with a subspace of $\mathbb R^4$ in many ways, for example $\mathbb R^3 \times \{0\}$ or $\{0\} \times \mathbb R^3$. But no matter how you define the cross product in $\mathbb R^4$, it won't be consistent with one of these identifications...

Answer 2

What property do you want to generalize? If you insist on a binary operator on vectors that produces a unique vector perpendicular to the two arguments, and you dream of it operating in spaces of dimensions other than 3, you are going to suffer disappointment.

An obvious generalization to the vector product, which in an $n+1$ dimensional vector space is an $n$-ary operator, producing a vector perpendicular to its arguments, is calculated with the familiar symbolic determinant procedure.

For instance, in $R^4$, with axes labeled $w$, $x$, $y$, $z$, this trinary form
$$ \begin{vmatrix} w & x & y & z \\ a_w & a_x & a_y & a_z \\ b_w & b_x & b_y & b_z \\ c_w & c_x & c_y & c_z \end{vmatrix} $$ produces the 4-vector perpendicular to the three 4-vectors $a$, $b$, and $c$.

This generalization also works for $n = 2$: $$ \begin{vmatrix} x & y \\ a_x & a_y \end{vmatrix} = (a_y, -a_x), $$ which, regarded as a complex number is $$ -i (a_x + a_y i ), $$ that is, a rotation of $a$ by $\pi/2$.

(Now I see this is the thing Mr. Willse was talking about in the Remark in his post. Maybe my explicit examples will be helpful. I think some of the other posters discussed other directions of generalization, which maintain the property of being a binary operator on a vector space, but abandon the property that the result of the operator is a single vector of the same space.)

About generalizations: consider that, if all algebras were the same, all or most mathematicians would be bored out of their minds and out of a job. I am pleased to report they're not. This amounts to the reality that not all properties generalize as one might hope. Some properties, for instance, the linear algebra properties of addition and scalar multiplication, generalize directly because linear algebras were constructed that way.