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I'm trying to rotate points in 4D. in fact i have been able to rotate using 6 rotation matrices. I've managed to perform 3d rotations using rotation quaternions. i understand that i can use plane angle rotations using a pair of quaternions and the vector(used as a quaternion).

ZlVZr

i understand that that Zl and Zr i,j,k represent the 6 rotational planes but how do i construct Zl and Zr ? and can i simply add the rotations together like quaternion 3d rotations?

thankyou for your help.

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  • $\begingroup$ What do you mean by "the 6 rotational planes"? If you use $\{1,i,j,k\}$ as a basis for the quaternions (which is standard), then there are six "basis" plane rotations in the six planes spanned by $\{1,i\},$ $\{1,j\},$ $\{1,k\},$ $\{i,j\},$ $\{j,k\},$ and $\{k,i\}$. What is "ZlVZr" supposed to mean? What are "Zl" and "Zr"? Reading your question and figuring out what you're asking should be easier than answering the question itself, not harder $\endgroup$ – anon Aug 25 '17 at 16:03
  • $\begingroup$ Zl is the left rotation, Zr is the right rotation and v is the 4 vector used as a quaternion so Zl * v * Zr rotates gives a quaternion representing the new vector. my question is how do i construct the i j k values for the left and right rotations. example if i wanted to rotate 45 degrees around the xy plane and 20 degrees around the zw. what how would i construct the right and left rotation quaternions to give me that rotation. $\endgroup$ – aceofjohnonlone Aug 26 '17 at 16:52
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Suppose $R$ is a rotation of $\mathbb{H}\cong\mathbb{R}^4$ in the oriented planes with orthonormal bases $\{a,b\}$ and $\{c,d\}$ by angles $\theta$ and $\phi$ respectively. (Assume the ordered basis $\{a,b,c,d\}$ induces the same orientation of space as does $\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$.) There exist unit quaternions $u,v\in S^3$ such that $R(x)=uxv$.

First, solve the system of congruences

$$ \begin{cases} \alpha+\beta\equiv \theta \\ \alpha-\beta\equiv\phi \end{cases} \pmod{2\pi}$$

Up to integer multiples of $2\pi$ in each coordinate, the solution is $(\frac{1}{2}(\theta+\phi),\frac{1}{2}(\theta-\phi))$.

Then $R=ST$ where $S$ is the left isoclinic rotation by $\theta$ and $T$ is the right isoclinic rotation by $\phi$ in the $\{a,b\}$- and $\{c,d\}$-planes. That is, they may be represented by matrices

$$ \begin{array}{l} S=\begin{bmatrix}\cos(\theta) & -\sin(\theta) & 0 & 0 \\ \sin(\theta) & \phantom{-}\cos(\theta) & 0 & 0 \\ 0 & 0 & \cos(\phi) & -\sin(\phi) \\ 0 & 0 & \sin(\phi) & \phantom{-}\cos(\phi) \end{bmatrix}, \\[7pt] T=\begin{bmatrix}\cos(\theta) & -\sin(\theta) & 0 & 0 \\ \sin(\theta) & \phantom{-}\cos(\theta) & 0 & 0 \\ 0 & 0 & \cos(-\phi) & -\sin(-\phi) \\ 0 & 0 & \sin(-\phi) & \phantom{-}\cos(-\phi) \end{bmatrix} \end{array}$$

with respect to the orthonormal basis $\{a,b,c,d\}$ (not with respect to the usual basis $\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$).

Note $ST=TS$. Define $u$ and $v$ such that $ua=b$ and $cv=d$ (these are easy to solve: $u=b\overline{a}$ and $v=\overline{c}d$). Then $S(x)=\exp(\alpha u)x$ and $T(x)=x\exp(\beta v)$ so we conclude that

$$ R(x)=\exp\left(\frac{\theta+\phi}{2}b\overline{a}\right)\,x\,\exp\left(\frac{\theta-\phi}{2}\overline{c}d\right). $$

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  • $\begingroup$ This is what I've been looking for. The math hear is slightly above me (Which i don't mind because i will learn a great deal from it) but I'm positive i can. but that's the hole point. This is the clearest and most useful answer I've come across. Thank you very much. $\endgroup$ – aceofjohnonlone Aug 26 '17 at 22:13

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