Cross product in $\mathbb R^n$

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$?

Solution 1:

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 somtimes 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 $$A \times B := -\text{tf}(AB),$$ where $\text{tf } C$ denotes the tracefree part $$C - \tfrac{1}{2} (\text{tr } C) \Bbb I_2$$ 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 $$\langle A, B \rangle := \text{tr}(AB);$$ explicitly, this is $$\pmatrix{a & b \\ c & -a} \cdot \pmatrix{a' & b' \\ c' & -a'} = -a a' - \tfrac{1}{2}(b c' + c b').$$

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 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.

Solution 2:

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...