Characterization of cross product (proof that only exists in dimension 3)

Question

In order to show that the cross product exists only in $\mathbb R^3$, our teacher puts this remark, "Exercise left to the reader". I have no idea how to solve it. Here's the problem:

"Let $B:\mathbb R^n \times \mathbb R^n \to \mathbb R^n$ be a surjective bilinear map and antisymmetric so that $B(u,v)=0 \iff \text{u and v are linearly dependent}$. Then $n = 3$."

Answer 1

That's wrong. There is also a 7D binary cross product that satisfies the given properties.

There are exactly two cross products on a 3D inner product space, one for both possible orientations of space. But there are infinitely many 7D cross products - in fact, the space of all cross products on a 7D inner product space is topologically the same as $\mathrm{SO}(7)/G_2\simeq S^7$.

In general, an $n$-dimensional $k$-ary cross product $X$ is defined to be an antisymmetric multilinear function of $k$ vectors in $\mathbb{R}^n$ which also satisfies a volume constraint: namely, $\|X(v_1,\cdots,v_k)\|$ ought to be the volume of the parallelepiped formed from $v_1,\cdots,v_k$ - which may be calculated using the Gramian determinant, $\mathrm{vol}^2=\det(\langle v_i,v_j\rangle)$.

The full classification of $k$-ary $n$-dimensional cross products is as follows:

• Nullary cross product exist in any dimension.
• Unary cross products exist in even dimensions.
• Binary cross products exist in dimensions $0,1,3,7$.
• Ternary cross products exist only in dimensions $4,8$.
• Co-unary cross products exist whenever $k=n-1$.
• Over-ary cross products exist whenever $k>n$.

Some commentary on these:

• The nullary and over-ary cross products are both identically $0$.
• The unary cross product is essentially multiplication-by-$i$ on $\mathbb{C}^n$ as a $2n$-dimensional real inner product space (note that taking the real part of the sesquilinear complex inner product yields a real inner product).
• The binary cross products are given by the imaginary parts $\mathrm{Im}(ab)=a\times b$ in the real normed division algebras $\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}$ respectively. Here, $\mathbb{H}$ is the quaternions and $\mathbb{O}$ is the octonions. The $0$D and $1$D binary cross products are both over-ary.
• The 7D binary and 8D ternary cross product both relate to the octonions (see here).
• The co-unary cross product $X(v_1,\cdots,v_{n-1})$ is the unique vector perpendicular to the span of $\{v_1,\cdots,v_{n-1}\}$, oriented appropriately, of the correct magnitude. The nullary 1D, unary 2D, binary 3D, and ternary 4D cross products are all co-unary.

Answer 2

Write $B_i=\sum_{jk}\varepsilon_{ijk}u_jv_k$. Hereafter, we'll drop the $\sum$. There are constraints on $\varepsilon_{ijk}$, e.g. antisymmetry ($\varepsilon_{ijk}=-\varepsilon_{jik}$) and orthogonality ($\varepsilon_{ijk}=-\varepsilon_{ikj}$). But beyond three dimensions, these leave some degrees of freedom, whereas in $3$ dimensions the symbol $\varepsilon_{ijk}$ is determined up to scaling. The former fact is why the product @runway44 mentioned isn't unique, even up to scaling.