This answer stated a few facts which I don't understand completely:

The "real part" (the $w$) of the product of two quaternions $pq$ is the same as the "real part of $qp.$

So, what happens when I take a unit quaternion $q$ and a "pure" quaternion $v,$ and calculate $$ p = q^\ast v q.$$

Well, we have $$\parallel p \parallel = 1 \cdot \parallel v > \parallel \cdot 1 = \parallel v \parallel $$

But as to the "real part," we begin with $$ \Re v = 0,$$

then $$ \Re q^\ast (v q) = \Re (v q) q^\ast = \Re v (q q^\ast) = \Re v = 0. $$

My question is not short enough to comment it there, so here it goes as its own question: Why is the real part of $q^*(vq)$ equal to the real part of $(vq)q^*$?
The other steps in the last line seem to be simple enough.

Ultimately, I'm trying to figure out under what preconditions the quaternion multiplication and/or conjugation keep the real part zero. I was unable to find any better resource than the linked answer.

In short:

  1. How to get the equality $\Re {q^*(vq)} = \Re (vq)q^*$ ? ($q^*$ denotes the conjugate of $q$, which is equal to $q^{-1}$ for a unit quaternion)

  2. Which of the preconditions stated in the quote are neccessary for the implication $\Re v = 0 \Rightarrow \Re (qvq^{-1}) = 0$ ?

  • $\begingroup$ Mind that the real part of the quaternion $z$ is $\frac12(z+z^*)$ $\endgroup$ – Andrea Mori Feb 1 '18 at 9:23
  1. Since, for any two quaternions $p$ and $q$, you have $\operatorname{Re}(pq)=\operatorname{Re}(qp)$, then, in particular$$\operatorname{Re}(q^*vq)=\operatorname{Re}(vqq^*).$$
  2. Nothing is needed; it is always true that $\operatorname{Re}(v)=0\implies\operatorname{Re}(q^{-1}vq)=0$ . And I saw no preconditions in the passage that you quote.
  • $\begingroup$ The preconditions in the quote are that q is a unit quaternion and v is pure. the latter is in the implication. Is the first neccessary? Thanks for 1., I get that part now! $\endgroup$ – lucidbrot Feb 1 '18 at 10:08
  • 1
    $\begingroup$ @lucidbrot Note that $v$ is a pure quaternion if and only if $\operatorname{Re}(v)=0$. No, $q$ being a unit quaternion is not necessary. $\endgroup$ – José Carlos Santos Feb 1 '18 at 10:10
  • $\begingroup$ And $Re(pq) = Re(qp)$ holds because the real part of the product is only influenced by the real part of the two factors and by the scalar product of their complex parts as vectors. That's what I didn't get at first, thx. So if $z:=s+\vec{v}$ then $Re(z_1 z_2)= s_1 s_2 + v_1\cdot v_2$, which is commutative $\endgroup$ – lucidbrot Feb 1 '18 at 10:17

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