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I've just been introduced to the Kronecker delta, $\delta_{ij}$, along with the alternating tensor, $\varepsilon_{ijk}$ (in vector calculus).

Motivation for the question:

I've been introduced to some properties of $\varepsilon_{ijk}$, e.g. antisymmetry (i.e. if you swap two indices, then the $\varepsilon_{...}$ is negated). Also, cyclic permutations of indices are allowed ($\varepsilon_{ijk}=\varepsilon_{kij}=\varepsilon_{jki}$).

These properties seemed familiar: namely, they satisfy two of the conditions for unit quaternion multiplication. When multiplying unit quaternions, $i, j, k$, we may cyclically permute them ($ijk=kij=jki=-1$), and swapping the order in which we multiply unit quaternions negates the result (e.g. $ij=-ji, jk=-kj, ...$).

Question:

Since $\varepsilon_{ijk}$ is used to represent the cross product of two vectors, is there some inherent relationship between the cross product and quaternion multiplication (and if not/so, why?), or is this just a coincidence?

Any links to some already-established result (if it exists) would also be very helpful!

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    $\begingroup$ If you represent a quaternion using the scalar plus a vector convection $a = a_0 + \vec{a}$, you will find $$a b = (a_0 + \vec{a})(b_0 + \vec{b}) = ( a_0b_0 - \vec{a}\cdot\vec{b} ) + a_0 \vec{b} + b_0 \vec{a} + \vec{a} \times \vec{b}\\ \implies\frac12 (ab - ba) = \vec{a} \times \vec{b}$$ $\endgroup$
    – achille hui
    Oct 21, 2014 at 16:32
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    $\begingroup$ The details are given in Arkamis' post below, but the connection you give is correct. Identifying the space of pure quaternions with $\mathbb{R}^3$ and taking $e_1, e_2, e_3$ as a basis of the space, multiplication in the quaternions is defined by $e_i e_j = \epsilon^{ijk} e_k - \delta_{ij}$, and the cross product is defined by $e_i \times e_j = \epsilon^{ijk} e_k$. $\endgroup$
    – anomaly
    Oct 21, 2014 at 16:47

2 Answers 2

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Yes. If we let a vector be written as the coefficients of a quaternion with zero "real" component, i.e. $[a,\ b,\ c] = ai+bj+ck$, then the cross product is simply the quaternion product with the real part omitted. See also: http://en.wikipedia.org/wiki/Cross_product#Quaternions

You may have heard along the way that vector cross products only exist in 3 and 7 dimensions. Why 3 and 7? Because we can "mimic" the cross product in $\mathbb{R}^3$ with quaternions, and likewise we can use octonions to mimic the cross product in $\mathbb{R}^7$.

Why not any other dimension? As it turns out, octonions represent the highest-dimension normed division algebra. So we can go no higher!

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    $\begingroup$ We can, however, generalize the cross product to other dimensions if we relax some of the restrictions. $\endgroup$
    – Emily
    Oct 21, 2014 at 16:38
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See the product formula at the end of this section about quaternions. Note that (following the same notation) in particular for $v,w\in \mathbb{R}^3$ $$(0,v)(0,w) = (-v\cdot w, v\times w).$$

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