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In $\mathbb R^4$ a reflect in a hyperplane through the origin $O$ is, $\forall q \in \mathbb R^4$ $$q \mapsto -u\overline{q}u$$ , where $u$ is a unit quaternion.

In $\mathbb R^3$ a rotation fixing the origin $O$ is $$p \mapsto x^{-1}px$$ , where $p$ is a pure quaternion, and $x$ is a unit quaternion.

Since a rotation is equivalent to 2 reflects chained up, given unit quaternion $u$ for a rotation $p \mapsto u^{-1}pu$, I'm trying to find unit quaternions $x$, $y$ so that the rotation can be decomposed to two reflects, first via $x$, then via $y$, i.e.

$$p \mapsto -x\overline{p}x \mapsto -y \overline{-x\overline{p}x} y = yx^{-1}px^{-1}y$$

Let this equals to $u^{-1}pu$ we get

$$x^{-1}y=u, \qquad yx^{-1}=u^{-1}$$

Please enlighten me how to find such a pair of $x$ and $y$ ?

athos
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    Don't know if it helps, but taking inverse of second equation and equating leads to $x^2=y^2.$ – coffeemath Oct 04 '22 at 23:16
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    @coffeemath since $x, y$ stand for reflections, $x^2=y^2=1$ are expected. – Just a user Oct 05 '22 at 00:50
  • You should make it clear that in $\mathbb R^3$ when you say'rotation' you mean 'rotation about an axis throgh the origin' and when you say reflection you mean 'reflection about a plane passing through the origin.' – P. Lawrence Oct 05 '22 at 01:30
  • @Justauser to be fair , I updated the question, the origin version didn’t make that obvious – athos Oct 05 '22 at 08:47
  • @P.Lawrence thanks, I’ve updated the question accordingly – athos Oct 05 '22 at 08:47

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In 2D the rotation by an angle of $\theta$ is achieved by reflecting across two lines (intersecting at the center of rotation) separated by an angle of $\frac{1}{2}\theta$. (The order in which you do the reflections determines the orientation of the rotation.) This generalizes to 3D with two planes intersecting at the axis of rotation with a dihedral angle of $\frac{1}{2}\theta$, and it generalizes to any number of dimensions with hyperplanes (i.e. codimension $1$ subspaces) intersecting at the orthogonal complement of a 2D plane of rotation.

Every unit quaternion, aka versor, has a polar form $e^{\theta\mathbf{u}}$ with $\mathbf{u}\in S^2\subset\mathbb{R}^3$ and $0\le\theta\le\pi$ is convex, which is unique except for real versors $\pm1$ (corresponding to $\theta=0,\pi$ respectively and with $\mathbf{u}\in S^2$ arbitrary), and by Euler equals $\cos(\theta)+\sin(\theta)\mathbf{u}$. (Note the square roots of $-1$ in the quaternions are precisely the unit vectors.) Conjugating a 3D vector by $e^{\theta\mathbf{u}}$ has the effect of rotating it around the $\mathbf{u}$-axis by $2\theta$. To see this, describe everything wrt an orthonormal basis $\{1,\mathbf{u},\mathbf{v},\mathbf{w}\}$.

When $p$ is a versor we can show $x\mapsto -p\overline{x}p$ is a reflection across the 3D hyperplane perpendicular to $p$. (We can get this formula by seeing $-\overline{x}$ works for $p=1$ and then conjugating this map by either left- or right-multiplication by $p$.) So to rotate 3D space around an axis $\mathbf{u}$ by an angle $\theta$, we can pick two vectors $\mathbf{v}$ and $\mathbf{w}$ which are orthogonal to $\mathbf{u}$ but at an (correctly oriented) angle of $\frac{1}{2}\theta$ to each other, then reflect across $\mathbf{u}^\perp$ followed by reflecting across $\mathbf{v}^\perp$.

anon
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    @athos Yes, I meant $p=1$. If $F$ is any reflection ("flip") about a hyperplane $H$ and $R$ is any rotation, the conjugate $RFR^{-1}$ is a reflection across the hyperplane $RH$ (where $RH$ is the hyperplane got by applying $R$ to $H$). This can be seen by simply applying $RFR^{-1}$ directly to an element assumed to be from $RH$ or its complement. (This applies here with $R(x)=px$ or $R(x)=xp$ turning $1$ into $p$, hence $1^\perp$ into $p^\perp$.) – anon Oct 06 '22 at 01:44
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    In fact, the same idea holds for conjugating reflections by other reflections, or plane rotations by reflections, or by rotations, or... Really, this idea applies to almost any transformation in almost any symmetry group. It's present in change-of-basis, in permutation cycle structures, in homotopy, etc. – anon Oct 06 '22 at 01:49