# Is there a way to prove vector triple product from quaternion multiplication?

For pure imaginary quaternions $u, v, w$, is there a way to prove the vector triple product $u\times(v\times w) = v(u\cdot w) - w(u\cdot v)$ from the relation: $$uv = -u\cdot v + u\times v \text{ for u, v \in \mathbb{R}^3 }$$

I tried many times but failed.

• The same identity is not true for the seven-dimensional cross product (which comes from the same exact formula for the octonions), so the identity's proof relies on three dimensions being relatively cramped, something I'm not sure is readily apparent in the purely formal algebraic properties. Also, I normally think of the cross product as more conceptually basic (using geometry) than quaternion multiplication (using abstract algebra), so use the former to prove things about the latter instead of vice-versa.
– anon
Apr 17, 2016 at 9:31

A convenient tool for expressing quaternion statements in terms of dot and cross products is to write the quaternion as a real number and a 3-vector, as $$a = (\alpha, \mathbf{a})$$. Then quaternion multiplication of quaternion $$b = (\beta, \mathbf{b})$$ on the left by $$a$$ is given by $$a b = \begin{bmatrix} \alpha & -\mathbf{a}^T \\ \mathbf{a} & \alpha I + \mathbf{a} \times \end{bmatrix} \begin{pmatrix} \beta \\ \mathbf{b} \end{pmatrix} = \begin{pmatrix} \alpha\beta - \mathbf{a} \cdot \mathbf{b} \\ \beta\mathbf{a} + \alpha \mathbf{b} + \mathbf{a} \times \mathbf{b} \end{pmatrix} .$$
This is much simplified for purely non-real quaternions $$a = (0,\mathbf{a})$$, $$b = (0,\mathbf{b})$$ and $$c = (0,\mathbf{c})$$
$$a b = \begin{bmatrix} 0 & -\mathbf{a}^T \\ \mathbf{a} & \mathbf{a} \times \end{bmatrix} \begin{pmatrix} 0 \\ \mathbf{b} \end{pmatrix} = \begin{pmatrix} -\mathbf{a} \cdot \mathbf{b} \\ \mathbf{a} \times \mathbf{b} \end{pmatrix} .$$ And further, $$a b c = \begin{pmatrix} -(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} \\ -(\mathbf{a} \cdot \mathbf{b}) \mathbf{c} + (\mathbf{a} \times \mathbf{b}) \times \mathbf{c} \end{pmatrix} .$$
To extract the identity from quaternion expressions, for three cyclic permutations of three generic quaternions $$a$$, $$b$$ and $$c$$, write equations expressing associativity, to form a system of equations: $$\left\{ \begin{array}{rcl} ( a b ) c - a ( b c ) &=& 0 \\ ( b c ) d - b ( c d ) &=& 0 \\ ( c a ) b - c ( a b ) &=& 0 . \end{array} \right.$$ In case the three quaternions are purely non-real, write $$a = (0,\mathbf{a})$$, $$b = (0,\mathbf{b})$$ and $$c = (0,\mathbf{c})$$, for which the above system implies $$\left\{ \begin{array}{rcl} -( \mathbf{a} \cdot \mathbf{b} ) \mathbf{c} +\mathbf{a} ( \mathbf{b} \cdot \mathbf{c} ) + ( \mathbf{a} \times \mathbf{b} ) \times \mathbf{c} - \mathbf{a} \times ( \mathbf{b} \times \mathbf{c} ) &=& 0 \\ -( \mathbf{b} \cdot \mathbf{c} ) \mathbf{a} +\mathbf{b} ( \mathbf{c} \cdot \mathbf{a} ) + ( \mathbf{b} \times \mathbf{c} ) \times \mathbf{a} - \mathbf{b} \times ( \mathbf{c} \times \mathbf{a} ) &=& 0 \\ -( \mathbf{c} \cdot \mathbf{a} ) \mathbf{b} +\mathbf{c} ( \mathbf{a} \cdot \mathbf{b} ) + ( \mathbf{c} \times \mathbf{a} ) \times \mathbf{b} - \mathbf{c} \times ( \mathbf{a} \times \mathbf{b} ) &=& 0 . \end{array} \right.$$ The cross product terms can be compared more easily if the anticommutativity of the cross product is applied: $$\left\{ \begin{array}{rcl} -( \mathbf{a} \cdot \mathbf{b} ) \mathbf{c} +\mathbf{a} ( \mathbf{b} \cdot \mathbf{c} ) - \mathbf{c} \times ( \mathbf{a} \times \mathbf{b} ) - \mathbf{a} \times ( \mathbf{b} \times \mathbf{c} ) &=& 0 \\ -( \mathbf{b} \cdot \mathbf{c} ) \mathbf{a} +\mathbf{b} ( \mathbf{c} \cdot \mathbf{a} ) - \mathbf{a} \times ( \mathbf{b} \times \mathbf{c} ) - \mathbf{b} \times ( \mathbf{c} \times \mathbf{a} ) &=& 0 \\ -( \mathbf{c} \cdot \mathbf{a} ) \mathbf{b} +\mathbf{c} ( \mathbf{a} \cdot \mathbf{b} ) - \mathbf{b} \times ( \mathbf{c} \times \mathbf{a} ) - \mathbf{c} \times ( \mathbf{a} \times \mathbf{b} ) &=& 0 . \end{array} \right.$$ The identity results from adding the first two equations of the system, and subtracting the third.