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Why $3$ and $7$? I know from some reading that Hurwitz's Theorem explains this, but can someone help me build some intuition behind this or perhaps provide a simpler explanation? It still seems mysterious to me.

William Chang
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  • How is it defined in $7$ dimensions? – Berci Mar 09 '14 at 23:48
  • It is defined in $\mathbb{R^7}$. See here: http://en.wikipedia.org/wiki/Seven-dimensional_cross_product. – William Chang Mar 09 '14 at 23:49
  • The tag "number-theory" is misleading. –  Mar 10 '14 at 00:01
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    I believe this article talks about it http://www.jstor.org/stable/2323537 – Ian Coley Mar 10 '14 at 00:02
  • @SanathDevalapurkar I deleted it. The reason I originally had it is because I predicted an intuitive explanation would invoke the notion of quaternions. – William Chang Mar 10 '14 at 00:02
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    The dim 3 cross product is to the quaternions as the dim 7 cross product is to the octonions. In some sense, the uniqueness of these cross products is equivalent to some uniqueness (but don't quote me on that) of these two division algebras – Ian Coley Mar 10 '14 at 00:03
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    I had never seen that vector product before - I suspect that in order to construct a cross-product with sensible properties you need to use the properties of a division algebra which would explain why only $R^3$ and $R^7$.

    But I would like to remark that as quoted in the wikipedia article you linked, there are 480 different cross products in $\mathbb{R}^7$. This suggests to me that it is not a very natural operation - at least if interpreted as a cross product.

    The natural extension of the cross product to higher dimensions is what is called wedge product.

     – GFR Mar 10 '14 at 00:04
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    Since the only normed division algebras are the quaternions and the octonions, the cross product is formed from the product of the normed division algebra by restricting it to the $0, 1, 3, 7$ imaginary dimensions of the algebra. This gives nonzero products in only three and seven dimensions. This gives nonzero products in only three and seven dimensions and not in dimension $0$ or $1$ because in zero dimensions there is only the zero vector, so the cross product is identically zero. In one dimension all vectors are parallel, so in this case also the product is identically zero. –  Mar 10 '14 at 00:04
  • @IanColey That's what I thought.. and cross product is anticommutative like quaternion multiplication. But the connection doesn't quite convince me that one follows from the other or explain why Hurwitz's Theorem is true. – William Chang Mar 10 '14 at 00:06
  • GFR and SanathDevalapurkar, I like both of those explanations! Thank you. – William Chang Mar 10 '14 at 00:13
  • @IlmariKaronen I saw that as I was posting. I feel that my question is related but different. – William Chang Mar 10 '14 at 01:01

2 Answers2

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Since the only normed division algebras are the real numbers, the complex numbers, the quaternions and the octonions, the cross product is formed from the product of the normed division algebra by restricting it to the $0, 1, 3, 7$ imaginary dimensions of the algebra. This gives nonzero products in only three and seven dimensions.

Now why, you may ask, does this give nonzero products in only three and seven dimensions? Why not in dimension $0$ or $1$? That is because in zero dimensions there is only the zero vector, so the cross product is identically zero. In one dimension all vectors are parallel, so in this case also the product is identically zero.

silvascientist
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I like Sanath Devalapurkar's explanation!

But also, after some more research, I see that if you identify $\mathbb{R^7}$ with the strictly imaginary octonions, you can explicitly define the cross product in terms of octonion multiplication with the following: $$x \times y = \operatorname{Im}(xy)=\frac{1}{2}(xy-yx).$$

We can conversely construct a Euclidean space with the cross product isomorphic to the octonions. If $V$ is a seven-dimensional Euclidean space with a given cross product, we can have bilinear multiplication on $\mathbb{R} \oplus V$ like this: $$(a,x)(b,y)=(ab-x \cdot y, ay+bx+x\times y)$$

forming an isomorphism $\psi:\mathbb{R} \oplus V \to \mathbb{O}$.

We can do the same thing in three dimensions, and in any $n-1 > 2$ dimensions such that an division algebra over $\mathbb{R}$ exists for $n$ dimensions - so cross product is defined only $3$ and $7$ dimensions.

cqfd
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William Chang
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