# Conjugation of Quaternions as Rotations in $\mathbb{R}^3$

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?

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I don't know the theory. And I don't know what "conjugacy map" means. But I do regularly use quaternions to represent 3d rotations. If the rotating quaternion is not normalized, it may scale the vector you're rotating. The vector itself, however, can be any length without any issues. But if it's a unit-length vector, then it happens to look just like a regular unit-length quaternion during the rotation, which probably makes it easier, in a text, to illustrate how the quaternion/conjugate sandwich multiplication produces a pure rotation of the augmented 3d vector. – JCooper Nov 3 '11 at 1:35
@JCooper: the map $q\to w^{-1}qw$ is a conjugation; the norm of $w$ doesn't matter in that unit quaternions are sent to unit quaternions regardless. – anon Nov 3 '11 at 2:05
If I had to guess, I'd say that whatever you're reading simply defined $q\to w^{-1}qw$ with $\|w\|=1$ in order to buckle down on redudancy. Since conugation with $w$ is independent of $\|w\|$ (so long as its nonzero), we can parametrize the family of conjugations with unit quaternions nicely and know that no two parameters represent the same conjugation. – anon Nov 3 '11 at 2:10

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 $\|w\|=1$.