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If I have a pure-quaternion $V = 0 + v = v$ where the imaginary part $v = ai + bj + ck$ then why is $v$'s dual-quaternion $v_d$ given as $v_d = 1 + ϵv$?
In notes and articles online, it usually is just stated as such but with no explanation as to why it is defined this way.

mszlazak
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  • Just in case somebody else has not heard of dual quaternions. They are like dual numbers, $\epsilon$ is an element of the center of the ring and satisfies $\epsilon^2=0$. – Jyrki Lahtonen Dec 21 '19 at 22:01
  • Anyway, I don't think there is anything called *the duaternion of a given quaternion, so I don't follow your question. Probably says more about me. The WP article does say that some applications of dual quaternions work better when you constrain the degrees of freedom. May be that is what you have seen? – Jyrki Lahtonen Dec 21 '19 at 22:05
  • Let me state it another way. If you have a $3D$ vector $(a, b, c)$, one can define the associated unit dual quaternion as $v_d = 1 + \epsilon (ai + bj + ck)$. I think I found the reason. A quaternion $v_d = v_0 + \epsilon v_{\epsilon}$ is a unit quaternion if and only if the norm $|| v_0 || = 1$ and the dot product $<v_0, v_{\epsilon}> = 0$. Does that make sense? – mszlazak Dec 23 '19 at 18:50
  • Seems noteworthy that $v\mapsto 1+\epsilon v$ is a one-to-one group homomorphism $(\Bbb R^3,+)\to\Bbb H[\epsilon]^{\times}$. – anon Dec 29 '19 at 02:20

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To see how any $3D$ vector $\pmb{v} = \big(a, b, c\big)$ can be represented as a unit dual quaternion, let's start with the dual quaternion $\pmb{\hat{q}} = \pmb{q_0} + \epsilon\pmb{q_{\epsilon}}$ with quaternions components $\pmb{q_0}$ and $\pmb{q_{\epsilon}}$. The real and imaginary parts of $\pmb{q_0}$ and $\pmb{q_{\epsilon}}$ are given as follows with the imaginary part italicized and an arrow above:

$$\begin{align} \pmb{q_0} &= w_0 + \vec{q}_{0} = w_0 + \big(x_0\pmb{i} + y_0\pmb{j} + z_0\pmb{k}\big)\\ \pmb{q_{\epsilon}} &= w_{\epsilon} + \vec{q}_{\epsilon} = w_{\epsilon} + \big(x_{\epsilon}\pmb{i} + y_{\epsilon}\pmb{j} + z_{\epsilon}\pmb{k} \big) \end{align}$$

Their respective conjugates are:

$$\begin{align} \pmb{q_{0}^{*}} &= w_0 - \vec{q}_{0} = w_0 - \big(x_0\pmb{i} + y_0\pmb{j} + z_0\pmb{k}\big)\\ \pmb{q_{\epsilon}^{*}} &= w_{\epsilon} - \vec{q}_{\epsilon} = w_{\epsilon} - \big(x_{\epsilon}\pmb{i} + y_{\epsilon}\pmb{j} + z_{\epsilon}\pmb{k} \big) \end{align}$$

Unlike regular quaternions, there are three different conjugations of dual quaternions like $\pmb{\hat{q}}$ which depend on whether the dual number is conjugated,

$$\pmb{\bar{\hat{q}}} = \pmb{q_0} - \epsilon\pmb{q_{\epsilon}}$$

the quaternion components are conjugated,

$$\pmb{\hat{q}^{*}} = \pmb{q_{0}^{*}} + \epsilon\pmb{q_{\epsilon}^{*}}$$

or both,

$$\pmb{\bar{\hat{q}}^{*}} = \pmb{q_{0}^{*}} - \epsilon\pmb{q_{\epsilon}^{*}}$$

The conjugate $\pmb{\hat{q}^{*}} = \pmb{q_{0}^{*}} + \epsilon\pmb{q_{\epsilon}^{*}}$ is used to define the norm of $\pmb{\hat{q}}$:

$$\begin{align} \Vert \pmb{\hat{q}} \Vert &= \sqrt{\pmb{\hat{q}\hat{q}^{*}}} \\ &= \sqrt{\big(\pmb{q_0} + \epsilon\pmb{q_{\epsilon}}\big)\big(\pmb{q_{0}^{*}} + \epsilon\pmb{q_{\epsilon}^{*}}\big)} \\ &= \sqrt{\pmb{q_0q_{0}^{*}} + \epsilon\big(\pmb{q_0q_{\epsilon}^{*}} + \pmb{q_{\epsilon}q_0^{*}}\big)} \\ &= \sqrt{\pmb{q_0q_{0}^{*}} + 2\epsilon\big(\pmb{q_0} \cdot \pmb{q_{\epsilon}}\big)} \end{align}$$

where

$$\begin{align} \pmb{q_0q_{0}^{*}} &= \big(w_0 + x_0\pmb{i} + y_0\pmb{j} + z_0\pmb{k}\big)\big( w_0 - x_0\pmb{i} - y_0\pmb{j} - z_0\pmb{k}\big) \\ &= w_0^2 + x_0^2 + y_0^2 + z_0^2 \end{align}$$

and

$$\begin{align} \pmb{q_0q_{\epsilon}^{*}} + \pmb{q_{\epsilon}q_0^{*}} &= 2\big(w_0 w_{\epsilon} + x_0 x_{\epsilon} + y_0 y_{\epsilon} + z_0 z_{\epsilon} \big) \\ &= 2\big(\pmb{q_0} \cdot \pmb{q_{\epsilon}}\big) \end{align}$$

A unit dual quaternion has a norm $\Vert \pmb{\hat{q}} \Vert = 1$. Choosing $\pmb{q_0}$ and $\pmb{q_{\epsilon}}$ to be orthogonal makes the dot-product $\pmb{q_0} \cdot \pmb{q_{\epsilon}} = 0$ which then leaves us with $\sqrt{\pmb{q_0q_{0}^{*}}} = 1$ and therefore $w_0^2 + x_0^2 + y_0^2 + z_0^2 = 1$. Choosing $\vec{q}_{0} = 0$, $\mathrm{q_{\epsilon}} = 0$ and the coefficients of $\vec{q}_{\epsilon}$ to be $\big(a, b, c\big)$ of vector $\pmb{v}$ satisfies the conditions of the $\pmb{q_0} \cdot \pmb{q_{\epsilon}} = 0$ and $w_0^2 + x_0^2 + y_0^2 + z_0^2 = 1$:

$$w_0 w_{\epsilon} + x_0 x_{\epsilon} + y_0 y_{\epsilon} + z_0 z_{\epsilon} = w_0(0) + (0)a + (0)b + (0)c = 0$$
$$w_0^2 + x_0^2 + y_0^2 + z_0^2 = w_0^2 + (0)^2 + (0)^2 + (0)^2 = w_0^2$$
$$w_0^2 = 1$$

Thus $w_0 = 1$ and the vector $\pmb{v} = \big(a, b, c\big)$ is encoded in a dual quaternion as:

$$1 + \epsilon\big( a\pmb{i} + b\pmb{j} + c\pmb{k} \big)$$

mszlazak
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