I noticed the other day that you can compute the conjugate and extract the "high" and "low" complex parts of a quaternion using linear functions. This made we wonder what surfaces you can hit by taking images of quaternion polynomials.
For my own background, I'm studying quaternions as part of a math hobby group. I also know as a vague historical fact that quaternions were more heavily used before the introduction of modern linear algebra in the last hundred years or so.
First, let me define an affine quaternion function as functions of the form $q \mapsto a + bqc$ where $a, b, c$ are quaternions.
I will define a quaternion polynomial as the ring of functions generated by the affine quaternion functions, where multiplication is pointwise multiplication rather than composition.
There's a linear formula for the quaternion conjugate:
$$ -2q^* = q + iqi + jqj + kqk $$
As proof, consider:
$$ q = a + bi + cj + dk \\ iqi = -a - bi + cj + dk \\ jqj = -a + bi - cj + dk \\ kqk = -a + bi + cj - dk $$
Additionally, there's a linear formula for extracting the $a+bi$ portion of a quaternion:
$$ 2(a+bi) = 2L(q) = q - iqi $$
And the $cj+dk$ portion
$$ 2(cj+dk) = 2H(q) = q + iqi $$
The real part of a quaternion would then be $\Re(q) = \frac{1}{2}(q + q^*)$
Since all these functions are polynomials (and in fact linear), it seems clear that we can make:
- a line with $q \mapsto u\Re(q)v$
- a plane with $q \mapsto uL(q)v$
- a 3d-hyperplane with $q \mapsto u(\Re(q) + H(q))v$
I find it very hard to visualize, though, what something like $q \mapsto qaq$ where $a$ is some quaternion is actually doing in $\mathbb{R}^4$.
I can work out for a handful of examples what I get if I take a quaternion function and precompose it with one of the subspace operations defined above.
$q \mapsto q^2$ gives you back the plane parameterized in an interesting way, since $z^2$ is a surjective complex function.
$q \mapsto qjq$ gives you $a+bi \mapsto (a^2+b^2)j$, which is a ray, which is kind of cool.