User:Rschwieb/GA Discussion/Comparison of methods: rotations

Description

The subsequent sections first describe how a rotation is represented, then describe the procedures to solve basic tasks.

Basic rotation tasks:

Problem A: Given two vectors, find a rotation with plane of rotation containing the vectors that turns the first vector in the direction of the second.

Problem B: Given two sets of vectors, one of which is known to be obtained from the other through rotation, find a rotation that carries each vector in the first set to the corresponding vector in the second set (again keeping vectors orthogonal to all those involved unchanged).

Problem C: in 3-D, given a plane and a specified angle of rotation, find a rotor representing the rotation.

Problem D: (n-D analogues of C)

Rotation matrix procedure

• Problem A: The matrix b a^T rotates a unit vector a to a unit vector b More complicated than this: matrix needs to be full rank. Without thinking very hard I suspect you may have to construct full before-and-after orthonomal basis sets A and B, then form B A^T similar to Problem B, but there may be some short-cuts for the invariant subspace.
• Problem B: see e.g Triad method for 3d case; in higher dimensions, one could use Gram-Schmidt to enforce orthogonalisation.

GA procedure

In a geometric algebra, a rotation is represented directly by an element of the algebra, called a rotor. A rotor R acts upon any element X of the algebra to produce its rotated equivalent X′:

$\,X'=RXR^{-1}$.

In 3 dimensions, this is exactly the quaternion method. (Found at Euvision Technologies: “Fontijne’s Formula” (apparently derived from GA): if 3D vector p rotates to p’, and q to q’, the quaternion of the rotation is [(q’+q)·(p’−p), (q’−q)×(p’−p)]. — Quondum^☏✎ 20:34, 6 February 2012 (UTC))

It is customary (but not necessary) to normalize R, so that RR^~ = 1, i.e. R⁻¹ = R^~. R is always an even multivector. Here the notation A^~ (alternately A^†) denotes the reversion of A.

Solution of Problems A and B in n-D (in 3-D this is no different)

Given two vectors a and b with a² = b² = ±1, the rotation that rotates a into b and keeps their orthogonal complement unchanged is:

$R=\pm {\frac {1+ba}{|a+b|}}$

The solution has been normalized. The distinction of sign is topologically significant. The case of (a + b)² = 0 corresponds to no solution.

This rotation is associated with a plane of rotation given by b∧a. If b = −a, this is not sufficient to define a plane; there are many planes that could achieve such a rotation, but such rotations will also cause some vectors orthogonal to a and b to be transformed.

Example solution of Problem B

Given two sets of k vectors {e_i} and {f_i} related by a rotation, find a rotor R that rotates the one into the other (so that f_i = Re_iR⁻¹), leaving all orthogonal vectors unchanged. Doran and Lazenby give the normalized solution for where the vectors form a 3-D basis in Euclidean space:

$\,R=\pm {\frac {1+f_{j}e^{j}}{|1+f_{i}e^{i}|}}$,

where {e^k} is the dual basis to {e_k}. The case of f_ke^k = −1 requires special care.

A solution for n-D and n independent vectors related by a rotation (not verified):

$\,R=\pm {\frac {1+f_{k}e^{k}}{\sqrt {2+2f_{k}\cdot e^{k}}}}$

Example 3

Given a plane and an angle of rotation in n-D, give the bivector describing the rotation and how R is related to it.

Given two vectors a and b spanning the plane and with the magnitude giving angle of rotation of θ in the direction from a to b, the bivector describing the plane and angle of rotation (picture a unit vector sweeping out a pizza slice) is

$\,B={\frac {\theta (a\wedge b)}{|a^{2}b^{2}-(a\cdot b)^{2}|}}$

and the rotor is

$\,R=\exp(-B/2)$.

This can be easily generalized to independent angles with respect orthogonal planes of rotation in 4-D and above: just add the bivectors.

Example 4

Smooth rotation: expression of the SLERP in GA.

Quaternion procedure

Example "baby" solution in quaternions

Problem: Demonstrate how to rotate the vector [0,1,1] 90 degrees about the y-axis to lie over [1,1,0] using a quaternion.

I am inclined to think the problem through in 3D GA, and apply the following mapping (treating quaternions (underlined) as being the even subalgebra of the GA), with the correspondence i = −e₂e₃, j = −e₃e₁, k = −e₁e₂; also I = e₁e₂e₃:

MAPPING : QUATERNION ↔ G_3,0

identity : quaternion scalar ↔ scalar [a = a]

dual : quaternion vector ↔ vector [a = −aI, a = aI]

identity : quaternion pseudovector ↔ bivector [a = a]

identity : quaternion rotor ↔ even grade versor (i.e. proper rotor) [R = R]

dual : quaternion reflection ↔ odd grade versor (i.e. improper rotor)

dual : quaternion pseudoscalar ↔ pseudoscalar

In GA, the rotor for the rotation (rotating ${\hat {z}}=e_{3}$ into ${\hat {x}}=e_{1}$ the short way around the y-axis; for the long way negate the rotor) is

${\underline {R}}=R={1+e_{1}e_{3} \over \left|e_{1}+e_{2}\right|}={\frac {1}{\sqrt {2}}}+{j \over {\sqrt {2}}}$

$a=e_{2}+e_{3}\qquad b=e_{1}+e_{2}$

${\underline {a}}=-aI=-(e_{2}+e_{3})e_{1}e_{2}e_{3}=j+k\qquad {\underline {b}}=-bI=-(e_{1}+e_{2})e_{1}e_{2}e_{3}=i+j$

So working purely with quaternions,

${\underline {a}}=j+k\qquad {\underline {b}}=i+j$

${\underline {R}}={\tfrac {1}{\sqrt {2}}}+{\tfrac {j}{\sqrt {2}}}\qquad {\underline {R}}^{\dagger }={\tfrac {1}{\sqrt {2}}}-{\tfrac {j}{\sqrt {2}}}$

Checking, we find

${\underline {R}}{\underline {a}}{\underline {R}}^{\dagger }=({\tfrac {1}{\sqrt {2}}}+{\tfrac {j}{\sqrt {2}}})(j+k)({\tfrac {1}{\sqrt {2}}}-{\tfrac {j}{\sqrt {2}}})=i+j={\underline {b}}$

as expected.

Beware the normal convention with quaternions is not the GA-centric approach I have used, though I have deliberately chosen the dual mapping for vectors to have matching signs, so it may be directly equivalent. Note that I have used the GA vectors (not underlined) rather than the corresponding quaternion vectors to calculate the rotor, though finding the expression for ${\underline {R}}$ from the quaternions ${\hat {\underline {x}}}=-{\hat {x}}I=i$ etc. will be simple enough. Feel free to trim this down to a concise version. Also, no guarantees that this is correct. — Quondum^☏ 21:29, 28 April 2012 (UTC)