四元数において，可逆元とその逆元を左右から作用させる

\(\mathbf{q}\mathbf{p}\mathbf{q}^{-1}\)

$$ \begin{eqnarray} \mathbf{q}\mathbf{p}\mathbf{q}^{-1}&=& \mathbf{q}\mathbf{p}\frac{\overline{\mathbf{q}}}{|\mathbf{q}|^2}=\frac{1}{|\mathbf{q}|^2}\mathbf{q}\mathbf{p}\overline{\mathbf{q}} \\&=&\frac{1}{w^2+|\mathbf{V}|^2}\left(w,\;\mathbf{V}\right)\left(w_p,\;\mathbf{V}_p\right)\left(w,\;-\mathbf{V}\right) \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/quaternion.html}{\overline{\mathbf{q}}=w-x\mathbf{i}-y\mathbf{j}-z\mathbf{k}=\left(w,\;-\mathbf{V}\right),\;\left(実部,\mathbf{i}\mathbf{j}\mathbf{k} 部\right),\;\mathbf{V}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}} \\&=&\frac{1}{w^2+|\mathbf{V}|^2}\left( ww_p-\mathbf{V}\cdot\mathbf{V}_p ,\; w\mathbf{V}_p+w_p\mathbf{V}+\mathbf{V}\times\mathbf{V}_p \right)\left(w,\;-\mathbf{V}\right) \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/quaternion.html}{\mathbf{q}_1\mathbf{q}_2=(w_1,\mathbf{V}_1)(w_2,\mathbf{V}_2)=\left(w_1w_2-\mathbf{V}_1\cdot\mathbf{V}_2,\;w_1\mathbf{V}_2+w_2\mathbf{V}_1+\mathbf{V}_1\times\mathbf{V}_2\right)} \\&=&\frac{1}{w^2+|\mathbf{V}|^2}\left( (ww_p-\mathbf{V}\cdot\mathbf{V}_p)w -(w\mathbf{V}_p+w_p\mathbf{V} +\mathbf{V}\times\mathbf{V}_p)\cdot(-\mathbf{V}) ,\; (ww_p-\mathbf{V}\cdot\mathbf{V}_p)(-\mathbf{V}) +w(w\mathbf{V}_p+w_p\mathbf{V}+\mathbf{V}\times\mathbf{V}_p) +(w\mathbf{V}_p+w_p\mathbf{V}+\mathbf{V}\times\mathbf{V}_p)\times(-\mathbf{V}) \right) \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/quaternion.html}{\mathbf{q}_1\mathbf{q}_2=(w_1,\mathbf{V}_1)(w_2,\mathbf{V}_2)=\left(w_1w_2-\mathbf{V}_1\cdot\mathbf{V}_2,\;w_1\mathbf{V}_2+w_2\mathbf{V}_1+\mathbf{V}_1\times\mathbf{V}_2\right)} \\&=&\frac{1}{w^2+|\mathbf{V}|^2}\left( w^2w_p -w\mathbf{V}\cdot\mathbf{V}_p +w\mathbf{V}_p\cdot\mathbf{V} +w_p\mathbf{V}\cdot\mathbf{V} +(\mathbf{V}\times\mathbf{V}_p)\cdot\mathbf{V}) ,\; -ww_p\mathbf{V}+(\mathbf{V}\cdot\mathbf{V}_p)\mathbf{V} +w^2\mathbf{V}_p+ww_p\mathbf{V}+w\mathbf{V}\times\mathbf{V}_p +\mathbf{V}\times(w\mathbf{V}_p+w_p\mathbf{V}+\mathbf{V}\times\mathbf{V}_p) \right) \\&&\;\cdots\;\mathbf{A}\times\mathbf{B}=-\mathbf{B}\times\mathbf{A} \\&=&\frac{1}{w^2+|\mathbf{V}|^2}\left( w^2w_p +w_p|\mathbf{V}|^2 +(\mathbf{V}\times\mathbf{V})\cdot\mathbf{V}_p ,\; +(\mathbf{V}\cdot\mathbf{V}_p)\mathbf{V} +w^2\mathbf{V}_p +w\mathbf{V}\times\mathbf{V}_p +w\mathbf{V}\times\mathbf{V}_p +w_p\mathbf{V}\times\mathbf{V} +\mathbf{V}\times(\mathbf{V}\times\mathbf{V}_p) \right) \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/06/blog-post_42.html}{(\mathbf{A}\times\mathbf{B})\cdot\mathbf{C}=(\mathbf{B}\times\mathbf{C})\cdot\mathbf{A}=(\mathbf{C}\times\mathbf{A})\cdot\mathbf{B}} ,\;\mathbf{A}\times\mathbf{A}=0 ,\;\mathbf{A}\cdot\mathbf{A}=|\mathbf{A}|^2 \\&=&\frac{1}{w^2+|\mathbf{V}|^2}\left( w_p(w^2+|\mathbf{V}|^2) ,\; (\mathbf{V}\cdot\mathbf{V}_p)\mathbf{V} +w^2\mathbf{V}_p +2w\mathbf{V}\times\mathbf{V}_p +(\mathbf{V}\cdot\mathbf{V}_p)\mathbf{V} -(\mathbf{V}\cdot\mathbf{V})\mathbf{V}_p \right) \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/06/blog-post_35.html}{\mathbf{A}\times(\mathbf{B}\times\mathbf{C})=(\mathbf{A}\cdot\mathbf{C})\mathbf{B}-(\mathbf{A}\cdot\mathbf{B})\mathbf{C}} \\&=&\frac{1}{w^2+|\mathbf{V}|^2}\left( w_p(w^2+|\mathbf{V}|^2) ,\; 2(\mathbf{V}\cdot\mathbf{V}_p) \mathbf{V} +w^2 \mathbf{V}_p +2w \mathbf{V}\times\mathbf{V}_p -|\mathbf{V}|^2\mathbf{V}_p \right) \\&&\;\cdots\;\mathbf{A}\cdot\mathbf{A}=|\mathbf{A}|^2 \\&=&\frac{1}{w^2+|\mathbf{V}|^2}\left( w_p(w^2+|\mathbf{V}|^2) ,\; 2(\mathbf{V}\cdot\mathbf{V}_p) \mathbf{V} +(w^2-|\mathbf{V}|^2) \mathbf{V}_p +2w \mathbf{V}\times\mathbf{V}_p \right) \end{eqnarray} $$

特殊なp, qを考える

ここで，\(\mathbf{q}=\left(\cos{\left(\frac{\theta}{2}\right)}, \sin{\left(\frac{\theta}{2}\right)\mathbf{V}}\right)(ただし|\mathbf{V}|=1), \mathbf{p}=(0,\mathbf{V}_p)\)とする． $$ \begin{eqnarray} \mathbf{q}\mathbf{p}\mathbf{q}^{-1} &=&\frac{1}{\cos^2{\left(\frac{\theta}{2}\right)}+\left|\sin{\left(\frac{\theta}{2}\right)}\mathbf{V}\right|^2}\left( 0\left(\cos^2{\left(\frac{\theta}{2}\right)}+\left|\sin{\left(\frac{\theta}{2}\right)}\mathbf{V}\right|^2\right) ,\; 2\left(\sin{\left(\frac{\theta}{2}\right)}\mathbf{V}\cdot\mathbf{V}_p\right)\sin{\left(\frac{\theta}{2}\right)} \mathbf{V} +\left(\cos^2{\left(\frac{\theta}{2}\right)}-\left|\sin{\left(\frac{\theta}{2}\right)}\mathbf{V}\right|^2\right) \mathbf{V}_p +2\cos{\left(\frac{\theta}{2}\right)} \sin{\left(\frac{\theta}{2}\right)}\mathbf{V}\times\mathbf{V}_p \right) \\&=&\frac{1}{\cos^2{\left(\frac{\theta}{2}\right)}+\sin{\left(\frac{\theta}{2}\right)}\left|\mathbf{V}\right|^2}\left(0,\; 2\sin^2{\left(\frac{\theta}{2}\right)} \left(\mathbf{V}\cdot\mathbf{V}_p\right)\mathbf{V} +\left(\cos^2{\left(\frac{\theta}{2}\right)}-\sin^2{\left(\frac{\theta}{2}\right)}\left|\mathbf{V}\right|^2\right) \mathbf{V}_p +2\cos{\left(\frac{\theta}{2}\right)}\sin{\left(\frac{\theta}{2}\right)} \mathbf{V}\times\mathbf{V}_p \right) \\&=&\frac{1}{\cos^2{\left(\frac{\theta}{2}\right)}+\sin{\left(\frac{\theta}{2}\right)}\cdot 1}\left(0,\; 2\sin^2{\left(\frac{\theta}{2}\right)} \left(\mathbf{V}\cdot\mathbf{V}_p\right)\mathbf{V} +\left(\cos^2{\left(\frac{\theta}{2}\right)}-\sin^2{\left(\frac{\theta}{2}\right)}\cdot 1\right) \mathbf{V}_p +2\cos{\left(\frac{\theta}{2}\right)}\sin{\left(\frac{\theta}{2}\right)} \mathbf{V}\times\mathbf{V}_p \right) \\&=&\frac{1}{1}\left(0,\; 2\sin^2{\left(\frac{\theta}{2}\right)} \left(\mathbf{V}\cdot\mathbf{V}_p\right)\mathbf{V} +\left(\cos^2{\left(\frac{\theta}{2}\right)}-\sin^2{\left(\frac{\theta}{2}\right)}\right) \mathbf{V}_p +2\cos{\left(\frac{\theta}{2}\right)}\sin{\left(\frac{\theta}{2}\right)} \mathbf{V}\times\mathbf{V}_p \right) \\&=&\left(0,\; \left(1-\cos{\left(\theta\right)}\right)\left(\mathbf{V}\cdot\mathbf{V}_p\right) \mathbf{V} +\cos{\left(\theta\right)} \mathbf{V}_p +\sin{\left(\theta\right)} \mathbf{V}\times\mathbf{V}_p \right) \\&&\;\cdots\;2\sin^2{\left(\frac{\theta}{2}\right)} =\sin^2{\left(\frac{\theta}{2}\right)}+\sin^2{\left(\frac{\theta}{2}\right)} =\left(1-\cos^2{\left(\frac{\theta}{2}\right)}\right)+\sin^2{\left(\frac{\theta}{2}\right)} =1-\left(\cos^2{\left(\frac{\theta}{2}\right)}-\sin^2{\left(\frac{\theta}{2}\right)}\right) =1-\cos{\left(\theta\right)} \\&&\;\cdots\;\cos^2{\left(\frac{\theta}{2}\right)}-\sin^2{\left(\frac{\theta}{2}\right)} =\cos{\left(\frac{\theta}{2}+\frac{\theta}{2}\right)} =\cos{\left(\theta\right)} \\&&\;\cdots\;2\cos{\left(\frac{\theta}{2}\right)}\sin{\left(\frac{\theta}{2}\right)} =\sin{\left(\frac{\theta}{2}+\frac{\theta}{2}\right)} =\sin{\left(\theta\right)} \\&=&\left(0,\; \left(\mathbf{V}\cdot\mathbf{V}_p\right)\mathbf{V} -\cos{\left(\theta\right)} \left(\mathbf{V}\cdot\mathbf{V}_p\right)\mathbf{V} +\cos{\left(\theta\right)} \mathbf{V}_p +\sin{\left(\theta\right)} \mathbf{V}\times\mathbf{V}_p \right) \\&=&\left(0,\; \left(\mathbf{V}\cdot\mathbf{V}_p\right) \mathbf{V} +\cos{\left(\theta\right)} \left(\mathbf{V}_p-\left(\mathbf{V}\cdot\mathbf{V}_p\right)\mathbf{V}\right) +\sin{\left(\theta\right)} \mathbf{V}\times\mathbf{V}_p \right) \end{eqnarray} $$ 位置ベクトル\(\mathbf{V}_p\)を任意の回転軸\(\mathbf{V}\)周りで\(\theta\)だけ回転させる計算に対応する．