Re: N Dimensional Rotor Spherical Interpolation

[Fred Lunnon]

  It isn't immediately clear what the question intends by the term "rotor" ---
    https://en.wikipedia.org/wiki/Rotor_(mathematics)
lists three overlapping definitions.  Reading between the lines, it seems that
the target universe comprises proper (continuous rigid) transformations fixing
the origin in Euclidean space with  n  dimensions.  But bearing in mind that
similar elementary practical algorithms remain currently (as far as I know)
almost entirely absent from published sources, it might be worthwhile to float
a more general discussion here.

  So let me assume initially that "rotor" means the same as "even versor",
ie. product of an even number of vectors in some given complex Cifford algebra;
and (initially at any rate) exclude parabolic, isoclinic, or half-turn factors,
along with any other nasties that I may have overlooked ...

  Example: quaternions in 3-space.  Clifford signature equals (+1,+1,+1) ,
or  (-1,-1,-1)  if you insist (sigh!); quaternion axes  i, j, k  map to  
g f, e g, f e  in terms of Clifford generators  e, f, g .  

  Remark: it is advisable to abandon the almost universal convention mapping
vectors to points, which though sustainable in 3-space (at the cost of some
confusion), causes increasing inconvenience elsewhere.  Here instead generators  
u, v, w, ...  represent coordinate hyperplanes, and the product of orthogonal
blades represents subspace intersection.

  In  n-space an elementary rotation  X , through angle  s , fixing axial coline
represented by normalised bivector  L , is represented by
      X  =  cos(s/2) + sin(s/2) L ;
and analogous representations apply to translation and boost.

  [ If the sign of  L  is ambiguous, it is always possible instead to select  
s = min(s, Pi-s) .  However, beware that in dimension  n > 3  it is in general
impossible to transform some coline through the origin into another via a single
elementary rotation! ]

  Interpolation by versors between rotors  X, Y  is in principle achievable
via product
      Z(t)  =  X^(1-t) Y^t  in range  0 <= t <= 1 ;
where  ^  denotes multivector exponentiation, implemented via exponential & 

logarithmic series.

  [ This is an active area of research: see eg.
    https://math.stackexchange.com/questions/1030737/exponential-function-of-quaternion-derivation
    https://link.springer.com/chapter/10.1007/978-3-0348-0622-0_5
    https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0116943
]

  I should be happy to investigate or explain more detail, if there is continuing
interest.

WFL 

On Sat, Nov 18, 2023 at 12:03 AM Joe Subbiani <joe.su...@gmail.com> wrote:

I'm using Rotors in a project and am trying to spherically interpolate (slerp) between 2 rotors.

All the resources I can find either don't cover how to check if a Rotor between two Rotors is a "long" or "short" path, or they are limited to using a dot product between 4 dimensional vectors, essentially treating Rotors as quaternions.

Any help with this would be appreciated!

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[Manfred]

The original question refers to "spherically interpolate (slerp) between 2 rotors".  This does not make sense to me: slerp is defined as an interpolation between two vectors (not rotors), so let's assume that this is what was meant.

General rotors (i.e. even versors) are more general than needed in this instance.  The slerp interpolation between two vectors needs only rotors that are the product of two vectors that are in the plane of the vectors that are being interpolated (irrespective of the number of dimensions).  This excludes nasties such as isoclinic rotations (or, more generally, any multi-plane rotations).  Thus, the problem in any number of dimensions reduces to a simple expression for the rotor that is identical to the expression for two dimensions (a cosine scalar term and a sine bivector term).

Manfred

[Fred Lunnon]

  It did occur to me too to wonder why anyone should want to do it 

--- but having become interested in how to how to solve the problem 

as stated, I complelely forgot to follow up that last clue.  

  The unfortunate enquirer must have found my response mystifying. 

Oops! 

WFL 

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[alexander arsenovic]

you should join the discord server.  seel link on this page 

https://bivector.net/

 lot of people on there can answer your ?