In their classic paper, Vector cross products, Brown and Gray define a (bilinear) cross product on a finite-dimensional vector space $\def\bfv{{\bf v}} \def\bfw{{\bf w}} \Bbb V$ equipped with a nondegenerate, symmetric bilinear form $\langle \,\cdot\, , \,\cdot\,\rangle$ to be a bilinear map
$$\times: \Bbb V \times \Bbb V \to \Bbb V$$ satisfying the identities
\begin{align}
\langle \bfv \times \bfw , \bfv \rangle &= 0 \\
\langle \bfv \times \bfw , \bfw \rangle &= 0 \\
\quad \langle \bfv \times \bfw, \bfv \times \bfw \rangle &= (\bfv \cdot \bfv)(\bfw \cdot \bfw) - (\bfv \cdot \bfw)^2.
\end{align}
Composition algebras It turns out that bilinear cross products are intimately related to composition algebras: Given a field $\Bbb F$, a composition algebra is an $\Bbb F$-algebra $\Bbb A$ equipped with a quadratic form $Q$ satisfying the following:
- (nondegeneracy) the associated bilinear form
$$\langle x, y \rangle := \tfrac{1}{2}(Q(x + y) - Q(x) - Q(y))$$ is nondegenerate
- (identity) there is an element $e \in \Bbb A$ such that $ea = a = ae$ for all $a \in \Bbb A$
- (multiplicity) for all $x, y \in \Bbb A$ we have $Q(xy) = Q(x) Q(y)$ (This is a modest weakening of the notion of [not necessarily associative] division algebra.)
It turns out that any division algebra has dimension $1$, $2$, $4$, or $8$.
The composition algebra-cross product correspondence Now, given any composition algebra $\Bbb A$ (perhaps assuming $\text{char } F \neq 2$), we can build a cross product on a codimension-$1$ subspace of $\Bbb A$: Let $\bar{\cdot} : \Bbb A \to \Bbb A$ be the linear map that restricts to the identity on $\langle e \rangle$ and the negative of the identity on $\Bbb I := \langle e \rangle^{\perp}$; $\Bbb I$ is sometimes called the imaginary part of $\Bbb A$, and $\bar{\cdot}$ map is called conjugation (and is just the usual complex conjugation map in the case $\Bbb F = \Bbb R$, $\Bbb A = \Bbb C$), and let $\Im : \Bbb A \to \Bbb I$ denote the orthogonal projection. Then, it's not hard to show that the map
$$\times: \Bbb I \times \Bbb I \to \Bbb I$$
defined by
$$x \times y := \Im(xy)$$
is a cross product on $\Bbb I$ (endowed with the restriction of $\langle \,\cdot\, , \,\cdot\, \rangle$).
On the other hand, it's a nice exercise to show that we can reverse this construction, that is, starting with a bilinear cross product $\times$ on a space $(\Bbb I, \langle \,\cdot\, , \,\cdot\, \rangle)$, we can build a natural composition algebra on $\Bbb F \oplus \Bbb I$ that yields $\times$ when applying the above construction.
In particular, the above dimensional constraints on composition algebras mean that cross products can only exist on vector spaces of dimension $0$, $1$, $3$, and $7$. By the first identity at the beginning of this answer, any bilinear cross product on a $0$- or $1$-dimensional vector space is trivial, so nontrivial cross products only exist in dimension $3$ and $7$. (This observation is essentially the content of the proof of Theorem 4.2(ii) in the mentioned paper.)
It is perhaps not well known that, in fact, over $\Bbb R$ there are two nonisomorphic cross products in each of those dimensions.
Example: The split quaternions In dimension $3$, the less familiar cross product can be constructed as follows: We can identify $\Bbb R^3$ with the space
$$\Bbb M := \left\{\pmatrix{a & b \\ c & -a} : a, b, c \in \Bbb R\right\}$$
of tracefree $2 \times 2$ real matrices, and define the map
$$\times : \Bbb M \times \Bbb M \to \Bbb M$$
by
$$A \times B := \operatorname{tf}(AB) = \frac{1}{2}(AB - BA),$$
where $\text{tf } C$ denotes the tracefree part
$$C - \tfrac{1}{2} (\operatorname{tr} C) \Bbb I_2$$
of $C$. Explicitly, the map is
$$\pmatrix{a & b \\ c & -a} \times \pmatrix{a' & b' \\ c' & -a'}
= \pmatrix{\frac{1}{2}(bc' - b'c) & a b' - b a' \\ c a' - a c' & -\frac{1}{2}(b c' - c b')}.$$ Computing directly shows that this is a cross product on $\Bbb M$ endowed with the (indefinite) symmetric bilinear form
$$\langle A, B \rangle := -\operatorname{tr}(AB) = -\frac{1}{2}\operatorname{tr}(A B + B A);$$
explicitly, this is
$$\pmatrix{a & b \\ c & -a} \cdot \pmatrix{a' & b' \\ c' & -a'} = -2 a a' - b c' - c b'.$$
This cross product corresponds to a composition algebra called the split quaternions, $\widetilde{\Bbb H}$: We can identify it with the algebra $M(2, \Bbb R)$ of real $2 \times 2$ matrices; in this case, the quadratic form is just (up to sign) the determinant.
Remark Brown and Gray actually define a cross product more generally, namely, they allow any number of arguments, and require a natural generalization of the identities at the beginning of the answer: More precisely, the define a cross product on a finite-dimensional vector space $\Bbb V$ equipped with a nondegenerate, symmetric bilinear form $\langle \,\cdot\, , \,\cdot\, \rangle$ to be an $r$-fold multilinear map
$$\times: \Bbb V^r \to \Bbb V$$
satisfying the identities
\begin{align}
\langle \times(\bfv_1, \ldots, \bfv_r), \bfv_a \rangle &= 0, \qquad\qquad\qquad a \in \{1, \ldots, r\} \\
\langle \times(\bfv_1, \ldots, \bfv_r), \times(\bfv_1, \ldots, \bfv_r) \rangle &= \det [\langle \bfv_a, \bfv_b \rangle],
\end{align}
where $[\langle \bfv_a, \bfv_b \rangle]$ denotes the matrix with $(a, b)$ entry $\langle \bfv_a, \bfv_b \rangle$.
This generalization allows for many more possibilities: The $r$, $n$ for which there may be a cross product (at least for $\text{char } \Bbb F \neq 2$) are:
- $n$ even, $r = 1$ (these are the complex structures)
- $n$ arbitrary, $r = n - 1$ (these are the special cases of the Hodge Star operators)
- $n = 3 \text{ or } 7$, $r = 2$ (the cases that come from composition algebras)
- $n = 4 \text{ or } 8$, $r = 3$.
Brown, Robert B.; Gray, Alfred, "Vector cross products". Commentarii Mathematici Helvetici 42 1 (1967): 222–236. doi:10.1007/BF02564418.