Dual quaternion inverse

Question

Is it true that for every dual quaternion $Q$ I can find it's inverse such that $QQ^{-1} = 1?$ Using the usual definition $Q^{-1}=\frac{Q^{*}}{||Q||^2}$ doesn't work for me, since the dual part doesn't generally cancel out, and also $QQ^{-1}Q {\neq} {Q} $

At least, that's what my implementation in code says.

Answer 1

No, the dual quaternions contain zero divisors, one of which is a nonzero element $\epsilon$ with square zero. Such elements cannot have inverses.

It turns out that the units are exactly the $a+\epsilon b$ where $a$ is nonzero. To see this, first notice that $1+\epsilon b$ has the obvious inverse $1-\epsilon b$. Suppose now $a\neq0$. Then $1+\epsilon ba^{-1}$ is a unit, and so is its product with $a$, which is $a+\epsilon b$ . Working through this, it's easy to see why $a^{-1}(1-\epsilon ba^{-1})$ is the inverse of $a+\epsilon b$ .

Trying to establish an inverse with the formula you gave would be circular since you need to establish the inverse of $\|Q\|^2$ before you can divide by it, and that is, generally, just some other dual quaternion.

If the inverse of Q exists, then $QQ^{-1}Q=Q$. The product of the first two things is $1$, and the product of $1$ with Q is Q. If you believe in associativity and multiplicative identity in the dual quaternions, and you know the definition of inverse, it has to hold.

Answer 2

There are useful applications of dual quaternions to geometry and mechanics that use assignments and interpretations that constrain the forms and tame the unruliness.

Nomenclature:

• dual quantity d = a + ε b, ε ≠ 0, ε² = 0.
  The monstrously confusing industry standard is to call a and b the 'real' and 'dual' parts, but I'll call them the 'primary' and 'secondary' parts.
• dual sum d₁ + d₂ = (a₁ + a₂) + ε (b₁ + b₂)
• dual product d₁ d₂ = (a₁ a₂) + ε (a₁ b₂ + b₁ a₂)
• dual conjugate d* = (a + ε b)* = (a - ε b)
• dual norm = |d| = √(d d*) = √(a² + ε (a b - b a))
  If (a b) is commutative then |d| = √a²
• dual inverse 1/d = (1/a + ε ((1/a) b (1/a)) if a has an inverse
• quaternion U = {u, u}, u a one-dimension form, u a three-vector.
• quaternion conjugate U^* = {u, -u}
• quaternion norm |U| = √(U U^*)
• unit quaternion |U| = 1
• unit vector quaternion {0, u}, |u|=1.
• dual quaternion Q = U + ε V
• dual quaternion inverse Q^-1 = (U₀ + ε V₀)
  (U + ε V) (U₀ + ε V₀) = 1
  1.) U U₀ = 1, U₀ = U*
  2.) V U₀ + U V₀ = 0 = V U* + U V₀,
  V₀ = - U* V U*
  Q^-1 = (U* - ε U* V U*)

An example is the system described by Michael McCarthy here and loosely interpreted below.

In the usual way, a unit vector quaternion U expresses a direction, R its spherical transform about the origin, a rotation expressed as a quaternion product:

U₂ = R₁₂ U₁ R₁₂^*
R₁₂ = {cos(θ/2), sin(θ/2) n}, where θ = the rotation angle and n = the axis vector direction.

The norm of R = 1 and R^-1 = R^*

Exploiting the constrained forms and the condition that the rotation axis n is normal to both u₁ and u₂, the rotation quaternion R₁₂ can be computed directly, given U₁ and U₂ as

R₁₂ = √(U₂ U₁^*)

A dual quaternion line has both orientation and position in space. It's defined as L = U + ε V, where U as above is a dimensionless unit vector quaternion with u expressing the line's orientation and where V is a vector quaternion with units of length, the vector v expressing the moment of the line about the origin.

By the nature of the geometrical definition, u.v = 0., |L| = 1 and L^-1 = L^*.

As an example of the form, a line in direction u passing through point p, the dual quaternion expression is L = {0, u} + ε {0, p × u}

Any rigid spatial transform can be resolved to a screw transform - a combined rotation about and translation along an axis. Dual quaternion lines can undergo rigid transformations, combining rotation and translation, by a screw quaternion of the form Q = {cos(θ/2), sin(θ/2) n), where θ is a dual angle, the primary part being the angle of rotation, the secondary part being the translation distance measured along the axis, and n a dual vector, the primary part being the direction of the rotation axis and the secondary part the moment of the axis line about the origin.

L₂ = Q₁₂ L₁ Q₁₂^*

Again, because of the constrained forms and recognizing that the screw axis {0, n} intersects and is normal to both L₁ and L₂, this equation can be solved directly for Q₁₂

Q₁₂ = √(L₂ L₁*)

In computation, it seems to be an advantage to perform these functions on their matrix forms, expressing both the geometric objects and their transformations as octonions.

--
Fred Klingener