20220621 Three Conjugates of Dual Quaternions

Dual Quaternions. Yan-Bin Jia

If we view a dual quaternion $\sigma =p+ϵq$ as just an 8-tuple, its conjugate could be a vector generated from $\sigma$ by flipping the signs of some of its elements. The dual quaternion indeed has three conjugates, the first of which is from the conjugate of a dual number:
$${\sigma }^{\bullet }=p-ϵq.$$Or, viewed as an 8-tuple, it is
$${\sigma }^{\bullet }=\left(p_0,p_1,p_2,p_3,-q_0,-q_1,-q_2,-q_3\right)$$In this sense, the product of $\sigma$ with ${\sigma }^{\bullet }$ ought to be treated as an inner product to yield a scalar. This is not the case, however, if under the multiplication rule (14), for we have
$$\begin{array}{rl}\sigma \otimes {\sigma }^{\bullet }& =(p+ϵq)(p-ϵq)\\ & =pp+ϵ(qp-pq)\end{array}$$Generally, the quaternion product $pp$ is not a scalar, and $qp\ne pq$ following the non-commutativity of quaternion multiplication. The first conjugate is seldom used other than for deriving the third conjugate to be introduced shortly.
The second conjugate of $\sigma$ follows from the classical quaternion conjugation:
$${\sigma }^{\ast }=p^{\ast }+ϵq^{\ast }$$
where $p^{\ast }$ and $q^{\ast }$ are the conjugates of the quaternions $p$ and $q$, respectively. The conjugate ${\sigma }^{\ast }$ can also be viewed as an 8-tuple:
$${\sigma }^{\ast }=\left(p_0,-p_1,-p_2,-p_3,q_0,-q_1,-q_2,-q_3\right).$$
Let us calculate the product of $\sigma$ with ${\sigma }^{\ast }$ below:
$$\begin{array}{rl}\sigma \otimes {\sigma }^{\ast }& =(p+ϵq)\left(p^{\ast }+ϵq^{\ast }\right)\\ & =p{p}^{\ast }+ϵ\left(p{q}^{\ast }+q{p}^{\ast }\right)\\ & =\left(p_0^2+p_1^2+p_2^2+p_3^2\right)+2ϵ\left(p_0{q}_0+p_1{q}_1+p_2{q}_2+p_3{q}_3\right)\\ & =\Vert p{\Vert }^2+2ϵ\left(p_0{q}_0+\mathbf{p}\cdot \mathbf{q}\right)\end{array}$$

It is a dual scalar but not a scalar unless $p_0{q}_0+\mathbf{p}\cdot \mathbf{q}=0$, that is, unless $p$ and $q$ are orthogonal 4-tuples.
The third conjugate of $\sigma$ combines the first two conjugation operators:
$${\sigma }^{\diamond }=p^{\ast }-ϵq^{\ast }$$
As an 8-tuple, it takes the form
$${\sigma }^{\circ }=\left(p_0,-p_1,-p_2,-p_3,-q_0,q_1,q_2,q_3\right).$$
Similarly, we calculate the product of $\sigma$ with the above conjugate:
$$\begin{array}{rl}\sigma \otimes {\sigma }^{\diamond }& =(p+ϵq)\left(p^{\ast }-ϵq^{\ast }\right)\\ & =p{p}^{\ast }+ϵ\left(q{p}^{\ast }-p{q}^{\ast }\right).\end{array}$$ The product is a dual quaternion whose real part is a scalar and dual part is a vector.

For each of the three introduced conjugates, the conjugate of the conjugate of $\sigma$ is $\sigma$ itself. The conjugate of the product of dual quaternions equals the product of the individual conjugates of these dual quaternions in the reverse order. This is established below for all three conjugates:
$$\begin{array}{rl}{\left({\sigma }_1\otimes {\sigma }_2\right)}^{\bullet }& ={\left(p_1{p}_2+ϵ\left(p_1{q}_2+q_1{p}_2\right)\right)}^{\bullet }\\ & =p_1{p}_2-ϵ\left(p_1{q}_2+q_1{p}_2\right)\\ & =\left(p_1-ϵq_1\right)\left(p_2-ϵq_2\right)\\ & ={\sigma }_1^{\bullet }\otimes {\sigma }_2^{\bullet };\end{array}$$

$$\begin{array}{rl}{\left({\sigma }_1\otimes {\sigma }_2\right)}^{\ast }& ={\left(p_1{p}_2+ϵ\left(p_1{q}_2+q_1{p}_2\right)\right)}^{\ast }\\ & ={\left(p_1{p}_2\right)}^{\ast }+ϵ{\left(p_1{q}_2+q_1{p}_2\right)}^{\ast }\\ & ={\left(p_1{p}_2\right)}^{\ast }+ϵ\left({\left(p_1{q}_2\right)}^{\ast }+{\left(q_1{p}_2\right)}^{\ast }\right)\\ & =p_2^{\ast }{p}_1^{\ast }+ϵ\left(q_2^{\ast }{p}_1^{\ast }+p_2^{\ast }{q}_1^{\ast }\right)\\ & =\left(p_2^{\ast }+ϵq_2^{\ast }\right)\left(p_1^{\ast }+ϵq_1^{\ast }\right)\\ & ={\sigma }_2^{\ast }\otimes {\sigma }_1^{\ast }\end{array}$$

$$\begin{array}{rl}{\left({\sigma }_1\otimes {\sigma }_2\right)}^{\diamond }& ={\left(p_1{p}_2+ϵ\left(p_1{q}_2+q_1{p}_2\right)\right)}^{\diamond }\\ & ={\left(p_1{p}_2\right)}^{\ast }-ϵ{\left(p_1{q}_2+q_1{p}_2\right)}^{\ast }\\ & ={\left(p_1{p}_2\right)}^{\ast }-ϵ\left({\left(p_1{q}_2\right)}^{\ast }+{\left(q_1{p}_2\right)}^{\ast }\right)\\ & =p_2^{\ast }{p}_1^{\ast }-ϵ\left(q_2^{\ast }{p}_1^{\ast }+p_2^{\ast }{q}_1^{\ast }\right)\\ & =\left(p_2^{\ast }-ϵq_2^{\ast }\right)\left(p_1^{\ast }-ϵq_1^{\ast }\right)\\ & ={\sigma }_2^{\circ }\otimes {\sigma }_1^{\circ }\end{array}$$.