In $\mathbb{R}^3$, we can take both dot products and cross products. These are related by the formula $$|x\times y|^2+|x\bullet y|^2 = |x|^2|y|^2,$$ which tells you the the length of one in terms of the length of the other. This formula follows from: $$|x \times y| = |x||y|\sin \theta_{x,y}, \qquad |x \bullet y| = |x||y||\cos \theta_{x,y}|$$

Suppose we want to generalize to $\mathbb{R}^n$ for $n > 3$. We can still take dot products of pairs of vectors, and it's still true that $$|x \bullet y| = |x||y||\cos \theta_{x,y}|,$$ almost by definition. But since the higher-dimensional cross product requires more than $2$ inputs, there aren't any obvious formulae involving it that might be true. I was thinking that perhaps we can use wedge products instead. And, I've been gazing at the wikipedia article on Clifford algebra trying to find something relevant, but to no avail.

(I don't really understand Clifford algebras...)


Question. Is there a reasonable generalization of the formula $|x\times y|^2+|x\bullet y|^2 = |x|^2|y|^2$ to higher dimensions?


Yes, it is $$|x|^2|y|^2=(x\cdot y)^2+|x\wedge y|^2.$$ But here $x\wedge y$ is an element of the exterior square $\bigwedge^2\Bbb R^n$. If $e_1,\ldots,e_n$ is the standard basis of $\Bbb R^n$ then $e_1\wedge e_2,e_1\wedge e_3,\ldots,e_1\wedge e_n, e_2\wedge e_3, \ldots,e_{n-1}\wedge e_n$ is a orthonormal basis of $\bigwedge^2\Bbb R^n$.

We have the relations $u\wedge u=0$ and $u\wedge v=-v\wedge u$.

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  • $\begingroup$ I see. And I guess the formula $|x \wedge y| = |x||y| \sin \theta_{x,y}$ is still true at this level of generality? $\endgroup$ – goblin Aug 10 '17 at 6:49

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