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?

  • $\begingroup$ Hi Aidan, welcome to the site. I took the liberty of editing your question's formatting and choosing more appropriate tags. If you'd like to edit it further then you can click the edit link near the bottom left of the question body. $\endgroup$ Commented Oct 1, 2013 at 15:33

2 Answers 2


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.


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 $d=3$.

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    $\begingroup$ But if you "ignore the real portion of quaternions", as the OP suggests, then it does work: If $q_1$ and $q_2$ are 2 purely imaginary quaternions, then $Im(q_1 q_2)$ is the same as the cross product $q_1 \times q_2$. $\endgroup$
    – Ted
    Commented Oct 1, 2013 at 16:12
  • $\begingroup$ @Ted: sure, that works. But in that case you are not really working with quaternions anymore. There isn't any difference from using just $\bf i, j, k$ beyond not writing bold... $\endgroup$
    – Marek
    Commented Oct 1, 2013 at 16:22
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    $\begingroup$ That's a pretty vague dismissal :) I definitely know a few authors would look at "not really working with quaternions anymore" and counter with something like "What do you mean? That shows the cross product is built into the quaternions!" In Clifford algebra, the wedge product of two vectors is a reasonable analogue of this function of the cross-product, but that is admittedly not a product on $3$-space alone. $\endgroup$
    – rschwieb
    Commented Oct 8, 2013 at 16:17

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