Is it possible to use the imaginary components of quaternions to facilitate calculation of vector cross products?

Question

It has come to my attention that the cross products of the vectors $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are almost identical to the products of the imaginary components of quaternions $i$, $j$, and $k$. If one were to ignore the real portion of quaternions, could one find cross products by representing the vectors as quaternions?

Another question: would it be possible to use quaternions to find cross products of Cartesian vectors in $(x, y, z, t)$? If so, could this be used as a representation of physical entities such as space-time?

Answer 1

Yes: Given writing $3$-space vectors $a=(a_1,a_2,a_3)$ and $b=(b_1,b_2,b_3)$ as pure quaternions $a'=a_1i+a_2j+a_3k$ and $b'=b_1i+b_2j+b_3k$, then the pure quaternion part of $a'b'$ is the pure quaternion representing $a\times b$.

Using quaternions doesn't really save any complexity, however, since the quaternion multiplication computes these components plus an extra real part.

---

Indeed, the quaternions $i,j,k$ span a subspace of $\Bbb H$ (the pure quaternions) which can be naturally interpreted as $3$-space vectors. There is even a way to perform rotations and reflections on this space using quaternion mulitplication. That is probably the most valuable application of quaternions to $3$-space.

There are also ways to relate the quaternions to Minkowski $4$-space, but apparently it is not ideally suited for the task. Those who apply Clifford algebra in physics have some sort of nicer scheme worked out, apparently.

Answer 2

No, this is not possible. There is a superficial simmilarity but there is also a big difference: while ${\bf i} \times {\bf i} = 0$, $i \cdot i = -1$. In other words, imaginary quaternions are not closed under multiplication and thus don't form an algebraic structure.

Nevertheless, there is a relation to vector products, albeit a little bit deeper. The unit quaternions $a + ib + jc + kd$ such that $a^2 + b^2 + c^2 + d^2 = 0$ with multiplication are isomorphic to a Lie group knows as $SU(2)$. This group has as a Lie algebra the algebra of traceless skew-hermitian matrices ${\frak su}(2)$ with operation that of commutator which as a matter of fact is isomorphic to $\mathbb R^3$ with vector product as its operation.

Regarding the other question of $(x,y,z,t)$, the answer is again no. The vector product structure is only available in three-dimensions and that's because fundamentally it comes from a commutator operation on skew-symmetric matrices (this gives isomorphism with another Lie algebra, this time ${\frak so}(3)$) and the dimension of this algebra coincides with the dimension of the space only for $d=3$.