0
$\begingroup$

In a lot of published texts, when I read about the conjugacy class of pure quaternions being rotations, they assume that the norm of the non-pure quaternion is 1. Is there a reason that this is necessary? Doesn't the conjugacy map preserve length of the pure quaternion it's rotating no matter what since the norm of a quaternion is the reciprocal of the norm of its inverse? Or is it just that this assumption about the norm gives some nicer looking resultant pure quaternion?

  • 0
    @JCooper is actually right — as is anon. It's just an unfortunate difference of terminology. See my answer for details.2017-11-20

2 Answers 2

1

The problem is just that there's a slightly different meaning to "conjugation" when using unit quaternions. You are correct in that mathematicians mean something like $R v R^{-1}$ when they say "conjugation". But frequently when discussing quaternions (or more general geometric algebras), people mean just $R v \bar{R}$, where $\bar{R}$ is the quaternion conjugate (or reverse in more general geometric algebras). The reason they do this is that for a quaternion \begin{equation} R^{-1} = \frac{\bar{R}} {\lVert R \rVert^2}. \end{equation} But of course, if $\lVert R \rVert = 1$, there's no difference. And the quaternion conjugate is slightly easier to compute (just flip the sign of the "vector" part) than the inverse (which also requires computing and dividing by the squared norm). So quaternion users frequently conflate the two operations. Personally, I prefer to call $R v R^{-1}$ conjugation and $R v \bar{R}$ "sandwiching".

You are also correct that true conjugation will necessarily preserve the norm of the object being conjugated. A related important fact is that "sandwiching" a vector by a general quaternion necessarily preserves the vectorial nature. That is, if $v$ is a pure vector quaternion (the scalar part is zero), then $R v \bar{R}$ is also a pure vector.

For what it's worth, in practice I actually find better numerical results at negligible cost if I use $R v R^{-1}$. As long as I can ensure that $R \neq 0$, this works out much better all around. But it is occasionally easier to do analysis with sandwiching.

1

Leaving anon's comment as a CW answer to get this off the list of unanswered questions.

If $r$ is a non-zero real number, it makes no difference whether you conjugate with $w$ a or $wr$. This is because all the quaternions commute with $r$ so $ wqw^{-1}=wqrr^{-1}w^{-1}=wrqr^{-1}w^{-1}=(wr)q(wr)^{-1}. $ Therefore we are at liberty to conjugate with $w/\|w\|$ instead of $w$, and may assume that $\|w\|=1$.