More about rotations in 4 dimensions.

It is easy to take a quaternion and recognize the rotation it actually represents. This is about doing the same thing for 4 dimensional rotations. It is easy to multiply two four dimensional rotations together. Can we extract the orthogonal planes of rotation and their angles of rotations from that. We can represent a rotation as a product of two vectors $ab$ via $w\ mapsto ba v ab$ and this will be the rotation of the $ab$ plane which leaves the space orthogonal to that plane pointwise fixed. It will be the rotations from the vector $a$ to the vector $b$ applied twice. This specifies both the direction and that amount of the rotation. Since $ab = g(a,b) + a\wedge b$ then this rotation is $\cos(\alpha/2) + \sin(\alpha/2) \omega$ where $\alpha$ is the amount of rotation and $\omega$ is $a \wedge b$ scaled to have norm one by a positive scalar constant. The product $a\wedge b$ specifies the direction in which the rotation $\alpha$ is applied, since any vectors $c\wedge d$ which equals $a \wedge b$ specifies the direction of the rotation as it is a rotation of angle $\alpha$ rotating starting from $c$ and moving toward $d$. We can extract the plane $a,b$ from the product $a\wedge b$ because it is those vectors $u$ with $u\wedge (a\wedge b) = 0$. This is a system of 4 homogeneous equations which has rank 2 over $\mathbb{R}^4$. We can extract the orthogonal plane $\phi = c \wedge d$ to the $a,b$ plane from $\omega$ alone. Write $\omega = r_{12} e_1e_2 + r_{13} e_1e_3 + r_{14} e_1e_4 + r_{23}e_2e_3 + r_{24}e_2e_4 + r_{34} e_3 e_4$ where $r_{12}r_{34} - r_{13}r_{24} + r_{14}r_{23} = 0$. Then the orthogonal plane is $\psi = r_{34}e_1e_2 - r_{24} e_1e_3 + r_{23} e_1e_4 + r_{14}e_2e_3 - r_{13}e_2e_4 + r_{12}e_3e_4$. Then $\omega \psi = (r_{12} e_1e_2 + r_{13} e_1e_3 + r_{14} e_1e_4 + r_{23}e_2e_3 + r_{24}e_2e_4 + r_{34} e_3 e_4) (r_{34}e_1e_2 - r_{24} e_1e_3 + r_{23} e_1e_4 + r_{14}e_2e_3 - r_{13}e_2e_4 + r_{12}e_3e_4)$. The constant term is $-r_{12}r_{34} + r_{13}r_{24} - r_{14}r_{23} - r_{23}r_{14} + r_{24}r_{13} - r_{34}r_{12} = 0$. The $e_1e_2e_3e_4$ term is $r_{12}^2 + r_{13}^2 + r_{14}^2 + r_{23}^2 + r_{24}^2 + r_{34}^2$. The $e_1e_2$ term of $\omega\psi$ is $r_{13}r_{14} e_1e_3e_2e_3 - r_{14}r_{13}e_1e_4e_2e_4 - r_{23}r_{24}e_2e_3 e_1e_3 + r_{24}r_{23} e_2e_4 e_1e_4$ This simplifies to zero. The other grade 2 terms should simply to zero as well. Then $(a + b \omega) ( c + d \psi) = ac + (bc \omega + ad \psi) + bd \omega\psi$. So if $\omega$ and $\psi$ were norm one then $\omega\psi = e_1e_2e_3e_4$. Thus our product becomes $ac + (bc \omega + ad\psi) + bd e_1e_2e_3e_4$. If $a = \cos(\alpha/2)$ and $b=\sin(\alpha/2)$ and if $c=\cos(\beta/2)$ and $d = \sin(\beta/2)$ then $ac = \cos(\alpha/2)\cos(\beta/2)$ and $bd = \sin(\alpha/2)\sin(\beta/2)$ and consequently $ac - bd = \cos((\beta - \alpha)/2)$ and $ac + bd = \cos((\alpha + \beta)/2)$. Since $ac$ and $bd$ are known (being the constant and the volume coefficients) then we can solve this for $\alpha$ and $\beta$. Now $ad + bc = \sin((\alpha + \beta)/2)$ And $ad - bc = \sin((\beta - \alpha)/2)$. Now $bc\omega + ad \psi = \frac{1}{2}(ad + bc) (\psi + \omega) + \frac{1}{2}(ad - bc) (\psi - \omega)$. This then equals $\frac{1}{2}\sin((\alpha + \beta)/2) (\psi + \omega) + \frac{1}{2}\sin((\beta-\alpha)/2)(\psi - \omega)$ This is useful because $ac$ and $bd$ are not known. But we can find $\sin((\alpha + \beta)/2)$ and $\sin((\beta-\alpha)/2)$ up to sign (of each) using $ac$ and $bd$. This then becomes easy to solve for the actual coefficients $r_{ij}$ of $\omega$ (and $\psi$).