is the the rotation matrix corresponding to the quaternion different from the rotation matrix corresponding to the quaternion error (between desired and actual value) ?
Let $\mathrm{Q}$ be the quaternion and the dynamic attitude system represented as $$ \dot{\mathrm{Q}}(t)=\frac{1}{2}\mathrm{Q}(t)\otimes\overline{\Omega}(t)\ ,\hspace{1cm}(1) $$
$$ J\dot{\Omega}(t)=-\Omega(t)\times J\Omega(t)+u(t)+d(t)\ , $$ where $J\in R^{3\times 3}$ denotes the inertia matrix of the body and satisfies $J=J^{T}>0, \overline{\Omega}=(0,\ \Omega)$ , and $\Omega\in R^{3}$ is the angular velocity vector of the body in the body-fixed frame, $u(t)\in R^{3}$ is the control torque vector, and $d(t)\in R^{3}$ is the external
disturbance vector. The attitude quaternion $\mathrm{Q}(t) \in R^{4}$ is defined by $\mathrm{Q}(t)=(q_{0}(t),\ q_{1}(t),\ q_{2}(t),\ q_{3}(t))^{T}=(q_{0}(t),\ q_{v}(t))^{T}$ and the Euclidean norm $\Vert \mathrm{Q}(t)\Vert_{2}=1, \forall t\geq 0$. Jf $\mathrm{Q}_{d}$ is the desired quaternion written in dynamic form as $$ \displaystyle \dot{\mathrm{Q}}_{d}(t)=\frac{1}{2}\mathrm{Q}_{d}(t)\otimes\overline{\Omega}_{d}(t)\hspace{1cm}(2) $$ with $\overline{\Omega}_{d}=(0,\ \Omega_{d}), \Omega_{d}\in R^{3}$ is the desired angular velocity The quaternion error in multiplicative form is $$ \mathrm{Q}_{e}(t)=\mathrm{Q}_{d}^{-1}(t)\otimes \mathrm{Q}(t) \hspace{1cm}(3) $$ or $$ \mathrm{Q}_{d}(t)\otimes \mathrm{Q}_{e}(t)=\mathrm{Q}(t) . \hspace{1cm}(4) $$ Then the derivative of the above equation gives $$ \dot{\mathrm{Q}}_{d}(t)\otimes \mathrm{Q}_{e}(t)+\mathrm{Q}_{d}(t)\otimes\dot{\mathrm{Q}}_{e}(t)=\dot{\mathrm{Q}}(t) , \hspace{1cm} (5) $$ which leads to
$\displaystyle \dot{\mathrm{Q}}_{d}(t)\otimes \mathrm{Q}_{e}(t)+\mathrm{Q}_{d}(t)\otimes\dot{\mathrm{Q}}_{e}(t)=\frac{1}{2}\mathrm{Q}(t)\otimes\overline{\Omega}(t)$ ,
$\dot{\mathrm{Q}}_{e}(t)$ $$ =\displaystyle \frac{1}{2}(\mathrm{Q}_{d}^{-1}(t)\otimes \mathrm{Q}(t)\otimes\overline{\Omega}(t))-\mathrm{Q}_{d}^{-1}(t)\otimes\dot{\mathrm{Q}}_{d}(t)\otimes \mathrm{Q}_{e}(t)\ , \hspace{1cm} (6) $$
$$ \displaystyle \dot{\mathrm{Q}}_{e}(t)=\frac{1}{2}(\mathrm{Q}_{e}(t)\otimes\overline{\Omega}(t)-\overline{\Omega}_{d}(t)\otimes \mathrm{Q}_{e}(t)) ; $$ then, $$ \displaystyle \dot{\mathrm{Q}}_{e}(t)=\frac{1}{2}\mathrm{Q}_{e}(t)\otimes(\overline{\Omega}(t)-\mathrm{Q}_{e}^{-1}(t)\otimes\overline{\Omega}_{d}(t)\otimes \mathrm{Q}_{e}(t)) . \hspace{1cm}(7) $$ Let $$ \mathrm{Q}_{e}^{-1}(t)\otimes\overline{\Omega}_{d}(t)\otimes \mathrm{Q}_{e}(t)=\overline{\Omega}_{d}^{*}(t) \hspace{1cm}(8) $$ with $$ \Omega_{d}^{*}(t)=R^{T}(\mathrm{Q}_{e}(t))\Omega_{d}(t) . \hspace{1cm} (9) $$ Using Rodriguez formula one can define the rotation matrix in quaternion representation [15, 16]: $$ R^{T}(\mathrm{Q}_{e}(t))=I+2S(\mathrm{Q}_{e}(t))+2S^{2}(\mathrm{Q}_{e}(t)) . \hspace{1cm}(10) $$ $S$ is skew-symmetric which satisfies the condition $-S=S^{T}.$ Let an auxiliary angular velocity be defined as $$ \Omega_{\mathrm{a}\mathrm{u}\mathrm{x}}(t)=\Omega(t)-\Omega_{d}^{*}(t)\ ,\hspace{1cm}(11) $$
$$ \dot{\Omega}_{\mathrm{a}\mathrm{u}\mathrm{x}}(t)=\dot{\Omega}(t)-\dot{\Omega}_{d}^{*}(t)\ , $$ so the system in quaternion error can be represented as $$ \displaystyle \dot{\mathrm{Q}}_{e}(t)=\frac{1}{2}\mathrm{Q}_{e}(t)\otimes\overline{\Omega}_{\mathrm{a}\mathrm{u}\mathrm{x}}(t) ; \hspace{1cm}(12) $$
in equation (10) why the rotation matrix of quaternion error is different from the rotation matrix of the quaternion ?
let $I_{3}$ is the $3\times 3$ identity matrix, and the matrix $$ q\times=\left(\begin{array}{lll} 0 & -q_{3} & q_{2}\\ q_{3} & 0 & -q_{\mathrm{l}}\\ -q_{2} & q_{\mathrm{l}} & 0 \end{array}\right) $$ carries out the cross product. the rotation matrix corresponding to $q$ is then $$ R=(q_{0}^{2}-\Vert q\Vert^{2})I_{3}+2qq^{T}+2q_{0}q\times $$
equations (1) to (12) are from this research paper https://www.hindawi.com/journals/mpe/2016/8573235/
another question this symbol $\otimes$ denotes matrix multiplication form or the Kronecker product ?