对偶四元数是一种结合位置和姿态表示的工具，乘法遵循四元数的乘法规则，但涉及到额外的对偶部分。文章详细展开并展示了两个对偶四元数相乘的具体计算过程，包括标量部分和向量部分的运算。通过对偶四元数，可以简洁地处理三维空间中的变换组合。
前言
对偶四元数也称为螺旋算子，能够有效描述位置和姿态的耦合关系。
一、对偶四元数是什么？
对偶四元数是同时描述位置和姿态的一种工具，本文对其乘法进行讲解。
二、对偶四元数乘法
给定两个对偶四元数，其乘法按照四元数乘法进行处理。
a = [ a i ] ∈ R 4 \boldsymbol{a}=[a_i]\in\mathbb{R}^4 a=[ai]∈R4， b = [ b i ] ∈ R 4 \boldsymbol{b}=[b_i]\in\mathbb{R}^4 b=[bi]∈R4， c = [ c i ] ∈ R 4 \boldsymbol{c}=[c_i]\in\mathbb{R}^4 c=[ci]∈R4， d = [ d i ] ∈ R 4 \boldsymbol{d}=[d_i]\in\mathbb{R}^4 d=[di]∈R4为四个四元数，其中 a 1 a_1 a1， b 1 b_1 b1， c 1 c_1 c1， d 1 d_1 d1为相应的四个标量， ε \varepsilon ε 为对偶四元数的单位。
[ a 1 + ε b 1 a 2 + ε b 2 a 3 + ε b 3 a 4 + ε b 4 ] ∘ [ c 1 + ε d 1 c 2 + ε d 2 c 3 + ε d 3 c 4 + ε d 4 ] \left[\begin{array}{c} a_1+\varepsilon b_1\\ a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\varepsilon b_4 \end{array}\right] \circ \left[\begin{array}{c} c_1+\varepsilon d_1\\ c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] a1+εb1a2+εb2a3+εb3a4+εb4 ∘ c1+εd1c2+εd2c3+εd3c4+εd4
参考四元数的乘法，可以对上式进行展开。
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\left[\begin{array}{c} a_1+\varepsilon b_1\\ a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\varepsilon b_4 \end{array}\right] \circ \left[\begin{array}{c} c_1+\varepsilon d_1\\ c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] =\left[\begin{array}{c} (a_1+\varepsilon b_1)(c_1+\varepsilon d_1)-(a_2+\varepsilon b_2)(c_2+\varepsilon d_2)-(a_3+\varepsilon b_3)(c_3+\varepsilon d_3)-(a_4+\varepsilon b_4)(c_4+\varepsilon d_4)\\ (c_1+\varepsilon d_1)\left[\begin{array}{c} a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\varepsilon b_4 \end{array}\right] +(a_1+\varepsilon b_1)\left[\begin{array}{c} c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] + \left[\begin{array}{c} a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\varepsilon b_4 \end{array}\right]^{\times} \left[\begin{array}{c} c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] \end{array}\right]
a1+εb1a2+εb2a3+εb3a4+εb4
∘
c1+εd1c2+εd2c3+εd3c4+εd4
=
(a1+εb1)(c1+εd1)−(a2+εb2)(c2+εd2)−(a3+εb3)(c3+εd3)−(a4+εb4)(c4+εd4)(c1+εd1)
a2+εb2a3+εb3a4+εb4
+(a1+εb1)
c2+εd2c3+εd3c4+εd4
+
a2+εb2a3+εb3a4+εb4
×
c2+εd2c3+εd3c4+εd4
同时，有
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\left[\begin{array}{c} a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\varepsilon b_4 \end{array}\right]^\times=\left[\begin{array}{ccc} 0 & -(a_4+\varepsilon b_4) & a_3+\varepsilon b_3 \\ a_4+\varepsilon b_4 & 0 & -(a_2+\varepsilon b_2)\\ -(a_3+\varepsilon b_3) & a_2+\varepsilon b_2 & 0 \end{array}\right]
a2+εb2a3+εb3a4+εb4
×=
0a4+εb4−(a3+εb3)−(a4+εb4)0a2+εb2a3+εb3−(a2+εb2)0
进一步展开，
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\left[\begin{array}{c} a_1+\varepsilon b_1\\ a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\varepsilon b_4 \end{array}\right] \circ \left[\begin{array}{c} c_1+\varepsilon d_1\\ c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] =\left[\begin{array}{c} a_1c_1-a_2c_2-a_3c_3-a_4c_4+\varepsilon(b_1c_1+a_1d_1-b_2c_2-a_2d_2-b_3c_3-a_3d_3-b_4c_4-a_4d_4) \\ \left[\begin{array}{c} c_1a_2+\varepsilon (c_1b_2+d_1a_2)\\ c_1a_3+\varepsilon (c_1b_3+d_1a_3)\\ c_1a_4+\varepsilon (c_1b_4+d_1a_4) \end{array}\right] + \left[\begin{array}{c} a_1c_2+\varepsilon (a_1d_2+b_1c_2)\\ a_1c_3+\varepsilon (a_1d_3+b_1c_3)\\ a_1c_4+\varepsilon (a_1d_4+b_1c_4) \end{array}\right] + \left[\begin{array}{ccc} 0 & -(a_4+\varepsilon b_4) & a_3+\varepsilon b_3 \\ a_4+\varepsilon b_4 & 0 & -(a_2+\varepsilon b_2)\\ -(a_3+\varepsilon b_3) & a_2+\varepsilon b_2 & 0 \end{array}\right] \left[\begin{array}{c} c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] \end{array}\right]
a1+εb1a2+εb2a3+εb3a4+εb4
∘
c1+εd1c2+εd2c3+εd3c4+εd4
=
a1c1−a2c2−a3c3−a4c4+ε(b1c1+a1d1−b2c2−a2d2−b3c3−a3d3−b4c4−a4d4)
c1a2+ε(c1b2+d1a2)c1a3+ε(c1b3+d1a3)c1a4+ε(c1b4+d1a4)
+
a1c2+ε(a1d2+b1c2)a1c3+ε(a1d3+b1c3)a1c4+ε(a1d4+b1c4)
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0a4+εb4−(a3+εb3)−(a4+εb4)0a2+εb2a3+εb3−(a2+εb2)0
c2+εd2c3+εd3c4+εd4
再进行整理，可以得到
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\left[\begin{array}{c} a_1+\varepsilon b_1\\ a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\varepsilon b_4 \end{array}\right] \circ \left[\begin{array}{c} c_1+\varepsilon d_1\\ c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] =\left[\begin{array}{c} a_1c_1-a_2c_2-a_3c_3-a_4c_4+\varepsilon(b_1c_1+a_1d_1-b_2c_2-a_2d_2-b_3c_3-a_3d_3-b_4c_4-a_4d_4) \\ \left[\begin{array}{c} c_1a_2+a_1c_2+\varepsilon (c_1b_2+d_1a_2+a_1d_2+b_1c_2)\\ c_1a_3+a_1c_3+\varepsilon (c_1b_3+d_1a_3+a_1d_3+b_1c_3)\\ c_1a_4+a_1c_4+\varepsilon (c_1b_4+d_1a_4+a_1d_4+b_1c_4) \end{array}\right] + \left[\begin{array}{c} -(a_4+\varepsilon b_4)(c_3+\varepsilon d_3)+(a_3+\varepsilon b_3)(c_4+\varepsilon d_4)\\ (a_4+\varepsilon b_4)(c_2+\varepsilon d_2)-(a_2+\varepsilon b_2)(c_4+\varepsilon d_4)\\ -(a_3+\varepsilon b_3)(c_2+\varepsilon d_2)+(a_2+\varepsilon b_2)(c_3+\varepsilon d_3) \end{array}\right] \end{array}\right]
a1+εb1a2+εb2a3+εb3a4+εb4
∘
c1+εd1c2+εd2c3+εd3c4+εd4
=
a1c1−a2c2−a3c3−a4c4+ε(b1c1+a1d1−b2c2−a2d2−b3c3−a3d3−b4c4−a4d4)
c1a2+a1c2+ε(c1b2+d1a2+a1d2+b1c2)c1a3+a1c3+ε(c1b3+d1a3+a1d3+b1c3)c1a4+a1c4+ε(c1b4+d1a4+a1d4+b1c4)
+
−(a4+εb4)(c3+εd3)+(a3+εb3)(c4+εd4)(a4+εb4)(c2+εd2)−(a2+εb2)(c4+εd4)−(a3+εb3)(c2+εd2)+(a2+εb2)(c3+εd3)
总结
对偶四元数的乘法与四元数的乘法形式一致。
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