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To try to understand the derivations of quaternion multiplication rules, I was conducting rotation experiments to actually verify the meaning of these rules.

In practice, according to this source (Quaternions: why does ijk = -1 and ij=k and -ji=k), it is stated that:

A right angle rotation about the X-axis followed by an equal amount of rotation about the Y-axis corresponds to an overall effective rotation of 90 degrees about the Z-axis. So, ij = k. Similarly, you can physically verify the quaternion multiplication laws.

So, I took a point A = (1, 1, 1) and tried applying the composition of ij and separately k. Here's what I did:

  1. I rotated A around the X-axis by 90 degrees, resulting in $A' = (1, -1, 1)$.
  2. I rotated A' around the Y-axis by 90 degrees, resulting in $A'' = (1, -1, -1)$.
  3. Then, I separately rotated A around the Z-axis by 90 degrees, resulting in $A'_1 = (-1, 1, 1)$.

According to the relation ij = k, I should have obtained $A'' = A'_1$, but this didn't happen in my experiment. Perhaps there's something I'm missing, and I haven't fully grasped the theory behind quaternions. What am I doing wrong?

Huntwer
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    The convention for group actions like this is normally that they are like functions. So $ij(P)=i(j(P))$ probably means do $j$ and then do $i$. – Eric Jun 10 '23 at 17:22
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    Also, his exact analogy is wrong. The exact relationship between quaternions and rotations is kinda tricky. $i$ doesn’t correspond to a 90 degree rotation. It corresponds to a 180 degree rotation and $-1$ happens to correspond to the identity rotation where you do a rotation of point $p$ by quaternion $q$ by calculating $qpq^{-1}$. This is all really nonobvious and still confuses me sometimes. – Eric Jun 10 '23 at 17:34
  • @Eric So when he talks about 90-degree rotations around the axes in the link, either he was mistaken or he wasn't referring to rotations relative to i, j, and k. However, as confirmed by the evidence, it is incorrect to claim that rotations around x and then y are equivalent to a rotation around z, if we are talking about a 90° rotation. It's worth noting that considering ij(P) = i(j(P)) is more helpful in understanding the composition of rotations when visualized in this way. – Huntwer Jun 10 '23 at 19:43

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To summarize @Eric's comments:

  • order matters: $\mathbf{i\,j}$ means rotate first around $y$-axis then around $x$-axis (group acts from the left)

  • angle matters: each of $\mathbf{i\,,j\,,k}$ represents a rotation by $180^\circ$ degrees as we can learn from here.

Check: \begin{align} \begin{pmatrix}-1&0&0\\0&1&0\\0&0&-1\end{pmatrix}\begin{pmatrix}1\\1\\1\end{pmatrix}&=\begin{pmatrix}-1\\1\\-1\end{pmatrix}\quad\text{($180^\circ$ around $y$-axis)}\\[2mm] \begin{pmatrix}1&0&0\\0&-1&0\\0&0&-1\end{pmatrix}\begin{pmatrix}-1\\1\\-1\end{pmatrix}&= \begin{pmatrix}-1\\-1\\1\end{pmatrix}\quad\text{($180^\circ$ around $x$-axis)}\\[2mm] \begin{pmatrix}-1&0&0\\0&-1&0\\0&0&1\end{pmatrix}\begin{pmatrix}1\\1\\1\end{pmatrix}&=\begin{pmatrix}-1\\-1\\1\end{pmatrix}\quad\text{($180^\circ$ around $z$-axis)}\\[2mm] \end{align} Or more generally: \begin{align} \underbrace{ \begin{pmatrix}1&0&0\\0&-1&0\\0&0&-1\end{pmatrix}}_{\textstyle\mathbf{i}}\underbrace{\begin{pmatrix}-1&0&0\\0&1&0\\0&0&-1\end{pmatrix}}_{\textstyle\mathbf{j}}=\underbrace{\begin{pmatrix}-1&0&0\\0&-1&0\\0&0&1\end{pmatrix}}_{\textstyle\mathbf{k}}\,. \end{align}

Kurt G.
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  • Perfect! Now it's clear. However, if $i$ represents a 180-degree rotation around the $i$-axis, then $i^2 = i \cdot i$ implies two 180-degree rotations around the $i$-axis, resulting in 360 degrees or $0°$, a null rotation. In that case, it should be $i^2 = 1$ as a null rotation is an identity like 1 ($1 \cdot i = i \cdot 1$). So, why does $i^2 = -1$? I understand the analogy with complex numbers, but doesn't $i^2 = -1$ change the rules? – Huntwer Jun 10 '23 at 20:13
  • @Huntwer No. In the Wikipedia article to which I linked you see that a unit quaternion $q$ rotates a pure quaternion $p$ by $p'=qpq^{-1},.$ Therefore the sign of $q$ does not matter. In other words: $SU(2)$ is the double cover of $SO(3),.$ – Kurt G. Jun 10 '23 at 20:43
  • Yes. If you want a visual explanation, it’s because -1 represents a 360 degree rotation which in a weird way twists space which 720 doesn’t. See Dirac’s belt trick for an explanation: https://m.youtube.com/watch?v=EgsUDby0X1M – Eric Jun 10 '23 at 20:49
  • @KurtG. Are you referring to this?: "A physical rotation about $\vec {u}$ by $\theta$ and a physical rotation about $-\vec {u}$ by $\displaystyle 2\pi -\theta$ both achieve the same final orientation by disjoint paths through intermediate orientations. By inserting those vectors and angles into the formula for $q$ above, one finds that if $q$ represents the first rotation, $-q$ represents the second rotation. This is a geometric proof that conjugation by $q$ and by $−q$ must produce the same rotational transformation matrix." – Huntwer Jun 10 '23 at 21:20
  • @Huntwer . More or less yes. Everyone describes this in their own favourite way. Digest it patiently using your favourite source. One last hint: if you understood what $p'=qpq^{-1}$ means it is trivial that $q$ and $-q$ produce the same $p',.$ Quaternion rotations are neither tricky nor weird. Otherwise, not so many game programmers would use them. – Kurt G. Jun 11 '23 at 04:56
  • @KurtG. Applying and understanding them may not be complicated. However, if you want to know the underlying theory and the mathematical derivations, they can seem a bit difficult to grasp. In practice, at first glance, it appears that the rules are taken for granted, as if someone has made numerous attempts and found the right one. But in mathematics, nothing is assumed, or at least everything that is taken as true is proven unless it is an axiom derived from logical deductions. Thank you for the clarification. – Huntwer Jun 11 '23 at 23:53
  • @Huntwer . Mathematics and Physics goes like this: a genius discovers something that makes a difficult task simpler. All that remains to do for the rest of the people is to learn the rules and take them for granted. You may want to read the chapter "The Discovery of Quaternions" in A. Hanson, Visualizing Quaternions. – Kurt G. Jun 12 '23 at 09:02
  • @KurtG. You are right about what you said, but rather than saying "the rest of the people," we should talk about a certain category of people. Curious ones like me want to fully understand something because I believe mathematicians like Hamilton and Euler didn't know things by taking them for granted. Thank you for the source :) – Huntwer Jun 12 '23 at 11:44