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I wanted to give a beginning student of vector calculus a deeper understanding of dot and cross products and fell into some questions for which I have no ready answers. In some sense one feels that dot and cross products exhaust the possibilities of important products of vectors, and I expect there is a theorem here somewhere.

Suppose we have a "product" P of two vectors which is a vector. So we say that it is a bilinear map of $V \times V \to V$. Of course so far it is just an arbitrary map of pairs of basis vectors to vectors which extends by linearity to all vectors. Now assume also it is associative. We can define $P_+$ and $P_-$ to be the even and odd parts of $P$, so $P_+$ is commutative and $P_-$ is anticommutative. Can we relate $P_+$ to dot and $P_-$ to cross in $3$ dimensions or is more needed? Characterize all such functions $P_+$ and $P_-$.

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  • $\begingroup$ What do you mean by "odd and even" parts of $P$? $\endgroup$ – Aweygan Nov 11 '16 at 20:57
  • $\begingroup$ P+(u,v) = [P(u,v) + P(v,u)]/2 and P-(u,v) = [P(u,v) - P(v,u)]/2 $\endgroup$ – berkeleychocolate Nov 11 '16 at 21:19
  • $\begingroup$ The dot product isn't even of type $V\times V\to V$. $\endgroup$ – Bobbie D Nov 11 '16 at 22:29
  • $\begingroup$ Don't be so nitpicking. I didn't say P+ is the dot product - just that it is closely related to it, for example, some constant vector times the dot product or something like that. $\endgroup$ – berkeleychocolate Nov 12 '16 at 1:25
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Maybe this will help. Let's take the outer product of $(a,b,c)$ and $(x,y,z)$:

$$\pmatrix{a \\ b \\ c}\pmatrix{x & y & z} = \pmatrix{ax & ay & az \\ bx & by & bz \\ cx & cy & cz}$$

Now we decompose this matrix into it's antisymmetric, traceless symmetric, and diagonal parts:

$$\pmatrix{a \\ b \\ c}\pmatrix{x & y & z} = \frac12\pmatrix{0 & ay-bx & az-cx \\ bx-ay & 0 & bz-cy \\ cx-az & cy-bz & 0} + \frac12\pmatrix{0 & ay+bx & az+cx \\ bx+ay & 0 & bz+cy \\ cx+az & cy+bz & 0} + \pmatrix{ax & 0 & 0 \\ 0 & by & 0 \\ 0 & 0 & cz}$$

The first matrix in this decomposition has the same components as the cross product and the last matrix has trace equal to the dot product. The traceless symmetric part has components of the symmetric product, but is less useful because, as far as I know, there's no clear geometric meaning of that product (it also doesn't have the nice invariance properties of the cross and dot products).

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