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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 $n = 3$."

J.G.
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Bernat Pol
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Dec 25 '20 at 11:28
  • What does exhaustive mean in this case? Surjective (I am guessing this is the meaning)? Or also bijective? And we should have $B:R^n \times R^n \to R^n$. – bonsoon Dec 25 '20 at 11:51

2 Answers2

7

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.
anon
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  • Good answer, just a small nitpick: nullary cross products are not identically zero, they are given by unit vectors (note that the determinant of the corresponding $0\times 0$ Gram matrix is $1$). As a consequence, they exist in any dimension other than $0$. – pregunton Dec 27 '20 at 17:58
3

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.

J.G.
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