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$\DeclareMathOperator{\SO}{\operatorname{SO}}$I am trying to understand the properties of the $\SO(3)$ Lie Group but when expressed via Euler angles instead of rotation matrix or quaternions. I am building an Invariant Extended Kalman Filter (IEKF), which exploits the invariance property of $\SO(3)$ dynamics $\mathbf{\dot{R}} = \mathbf{R}[\omega]_\times$, where $[\cdot]_\times$ it the skew-symmetric operator.

The original filter is based on the left-invariance of the dynamics equation under constant rotations $\mathbf{R}_0$, which translates into \begin{equation} \mathbf{\bar{R}} \doteq \mathbf{R}_0\mathbf{R} \Rightarrow \dot{\mathbf{\bar{R}}} = \mathbf{R}_0\mathbf{R}[\omega]_\times=\mathbf{\bar{R}}[\omega]_\times \end{equation} I have also applied the same property with the quaternion representation of $\SO(3)$.

What I am trying to understand is if it is possible to apply a (non-linear) left-invariance property to the Roll-Pitch-Yaw representation of the dynamic equation, that is \begin{cases} \dot{\phi} = \omega_x + \text{sin}(\phi)\text{tan}(\theta)\omega_y+\text{cos}(\phi)\text{tan}(\theta)\omega_z \\ \dot{\theta} = \text{cos}(\phi)\omega_y-\text{sin}(\phi)\omega_z \\ \dot{\psi} = \frac{\text{sin}(\phi)}{\text{cos}(\theta)}\omega_y+\frac{\text{cos}(\phi)}{\text{cos}(\theta)}\omega_z \\ \end{cases} In particular, I first defined some fixed rotations about xyz axes as $(\phi_0,\theta_0,\psi_0)$ in analogy to the matrix case.

My issue is that I am not able to find a transformation which defines $(\bar{\phi},\bar{\theta},\bar{\psi}) = f(\phi,\theta,\psi,\phi_0,\theta_0,\psi_0)$ (and eventually $(\bar{\omega}_x,\bar{\omega}_y,\bar{\omega}_z)$) and satisfies the previous dynamics equation, as done previously with the rotation matrix.

Thank you all for your support

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    $\begingroup$ There are many question about Euler angles on our sister site, math.stackexchange.com, see e.g. math.stackexchange.com/search?q=Roll-Pitch-Yaw, but I couldn't quickly find one that answers your question. So I do hope you get a good answer here, but it might be worthwhile to take a look over there also. $\endgroup$
    – Vincent
    Commented Oct 27, 2020 at 10:14

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