Dual quaternion inverse 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. 
 A: 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.
A: 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
