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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$?

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    $\begingroup$ unizar.es/matematicas/algebra/elduque/Talks/crossproducts.pdf is an interesting reference to this, which shows that cross products which are a function of more than 2 vectors (e.g. 3-fold) exist for other dimensions. Also from Wikipedia: "The nonexistence of nontrivial vector-valued cross products of two vectors in other dimensions is related to the result from Hurwitz's theorem that the only normed division algebras are the ones with dimension 1, 2, 4, and 8." This is related to the real line(1), complex plane (2), quaternions(4), and octonions(8). $\endgroup$
    – Ross B.
    Jun 19, 2013 at 13:51
  • $\begingroup$ @RossB.: Okay, so the magical number $7$ shows up because it's of the form $2^n-1$? $\endgroup$
    – Nikolaj-K
    Jun 19, 2013 at 14:03
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    $\begingroup$ With my limited knowledge I'd hate to generalize. I'd only go so far to say the dimensions where cross products exists $(0,1,3,7)$ are $(1,2,4,8)-1$. $\endgroup$
    – Ross B.
    Jun 19, 2013 at 14:29
  • $\begingroup$ @RossB. Unfortunately, the link is dead... :( $\endgroup$
    – byk7
    Apr 11 at 21:07
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    $\begingroup$ @byk7 personal.unizar.es/elduque/Talks/crossproducts.pdf new link. $\endgroup$
    – Ross B.
    Apr 12 at 22:03

4 Answers 4

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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.

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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...

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    $\begingroup$ +1. This is the same intuition I have (which also accounts for the fact that there is a cross product of $n-1$ vectors in $\mathbb{R}^n$), but I wonder why it breaks in $\mathbb{R}^7$? (One can use the structure of the purely imaginary octonians to define the cross product on a pair of vectors in $\mathbb{R}^7$). $\endgroup$ Jun 19, 2013 at 14:24
  • $\begingroup$ @JasonDeVito I've written an answer that gives a little more explanation of why $3$ and $7$ are the only exceptional values---in short, it's because bilinear cross products can only arise from composition algebras in one dimension higher, but composition algebras only exist in dimensions $1$, $2$, $4$, and $8$. The notion "perpendicular space" in this answer leads to the so-called $(n - 1)$-fold cross products on vectors spaces of arbitrary positive $n$-dimensional vector spaces. $\endgroup$ Jul 5, 2015 at 11:24
  • $\begingroup$ Hi @MartinSleziak please can you help with math.stackexchange.com/q/4396103/585488 $\endgroup$
    – LiNKeR
    Mar 5 at 15:41
  • $\begingroup$ Is there a way to find the magnitude of the cross product in any dimension, even if you can't reduce your answer down to one vector with that magnitude? $\endgroup$ Apr 27 at 1:24
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For what it's worth: I asked Mark Green this question when I was a student in his Differential Geometry course at Ucla, and his reply was that that's when you get into the grassmanian and such.

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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.

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  • $\begingroup$ Is there a version that generalizes the magnitude of the result? For example, the magnitude of the cross product in 3D, and the equation a.x * b.y - a.y * b.x for 2D generalizes this for 2D and 3D, but I don't know how to generalize this in 4D, 5D, etc. $\endgroup$ Apr 27 at 1:26

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