PatrickPowers wrote: "The modes of rotation in 4d, is represented as a point in 6d space, where three of the 6d space forms the L axis (left-clifford) and the other three form the R axis. Any rotation, even at different speeds, can be formed by a simple sum of an L and R axis, where the two circular rotations are L+R and L-R resp."

This I do not understand. It seems it could be quite useful. Care to offer a reference? I'm thinking there must be some buzzword that I can search on.

There's always Wikipedia: https://en.wikipedia.org/wiki/Rotations ... _SO%284%29 . (I don't understand half of it, but there is some useful information there.)

You seem to be familiar with geometric algebra, so I will discuss 4D rotations in terms of bivectors.

Any rotation can be done by the geometric sandwich product with a rotor (a product of an even number of unit vectors), and any rotor is the exponential of a bivector. The set of rotations is a "Lie group", and the set of bivectors is the corresponding "Lie algebra". (But conventionally, these things are described with matrices instead of bivectors.)

The set of all bivectors in 4D is 6-dimensional, being spanned by {e

_{1}e

_{2}, e

_{1}e

_{3}, e

_{1}e

_{4}, e

_{2}e

_{3}, e

_{2}e

_{4}, e

_{3}e

_{4}}. A rotation by angle θ in the x

_{1}x

_{2}-plane corresponds to the bivector θe

_{1}e

_{2}. An isoclinic ("Clifford") rotation would be something like θ(e

_{1}e

_{2}+e

_{3}e

_{4}), and a general double rotation θe

_{1}e

_{2}+ϕe

_{3}e

_{4}.

Instead of writing everything as a sum of these simple bivectors, we could use isoclinic bivectors as a basis:

L

_{1}= e

_{1}e

_{2}+ e

_{3}e

_{4}, L

_{2}= e

_{1}e

_{3}- e

_{2}e

_{4}, L

_{3}= e

_{1}e

_{4}+ e

_{2}e

_{3},

R

_{1}= e

_{1}e

_{2}- e

_{3}e

_{4}, R

_{2}= e

_{1}e

_{3}+ e

_{2}e

_{4}, R

_{3}= e

_{1}e

_{4}- e

_{2}e

_{3}

For example, the simple rotation becomes θe

_{1}e

_{2}= (θ/2)(L

_{1}+R

_{1}).

I haven't studied this in detail, but there is a non-trivial correspondence between operations in the Lie group and the Lie algebra. The Lie group operation is just composition of transformations (like applying several rotations consecutively). The Lie algebra operations are addition and the "commutator" cross product. If two bivectors commute (meaning A×B=0), then the corresponding rotors exp(A) and exp(B) also commute, and exp(A)exp(B)=exp(A+B).

The cross product of two bivectors can be calculated as follows, using the standard basis:

(e

_{1}e

_{2})×(e

_{1}e

_{2}) = 0

(e

_{1}e

_{2})×(e

_{1}e

_{3}) = - e

_{2}e

_{3}

(e

_{1}e

_{2})×(e

_{2}e

_{3}) = e

_{1}e

_{3}

(e

_{1}e

_{2})×(e

_{3}e

_{4}) = 0

etc., and using the isoclinic basis:

L

_{2}×L

_{1}= 2L

_{3}, R

_{1}×R

_{2}= 2R

_{3},

L

_{3}×L

_{2}= 2L

_{1}, R

_{2}×R

_{3}= 2R

_{1},

L

_{1}×L

_{3}= 2L

_{2}, R

_{3}×R

_{1}= 2R

_{2},

L

_{1}×R

_{1}= 0, L

_{1}×R

_{2}= 0, L

_{1}×R

_{3}= 0

Note that the cross product of any two R's is another R, so the right-isoclinic rotations form a group by themselves. (I believe this is the "swirlprism" symmetry group, and the "special unitary group".) And if each R is divided by 2, then these equations are the same as those for bivectors in 3D, or for pure-imaginary quaternions, or the ordinary vector cross product:

i×j = k,

j×k = i,

k×i = j

This isomorphism between bivectors in 4D and 3D is, of course, related to the Hopf fibration between the spheres in 4D and 3D.