Can I integrate the Lie-algebra (body angular velocity) of a quaternion?

Question

This is my first mathematics question here.

So I am trying to model a 3-d rotation rigid body by Euler's equation. Of course quaternion is the place to go. If in each time step I receive the body-fixed angular velocity I can get the time derivative of unit quaternion. $$\boldsymbol{\dot{q}}=\frac{1}{2}\boldsymbol{\omega}_B\boldsymbol{q}$$ However, the integration of $\boldsymbol{\dot{q}}$ $$\boldsymbol{q}_{i+1} = \boldsymbol{q}_i+\boldsymbol{\dot{q}_i}dt$$ does not belong to the multiplication operation defined by unit quaternion group $\mathbb{S}^3$. So by time re-normalization needs to be done, and it gets worse when the angular velocity is large.

So instead I was trying to go to Lie-algebra method. I am in no sense an expert in group theory. So in my mind right now is since the exponential of a pure quatenrion is definitely a unit quaternion $$\boldsymbol{q}=e^{\boldsymbol{\eta}} = \sum_{k=1}^{\infty}\frac{1}{k!}\boldsymbol{\eta}^k=e^{\eta_w}\cdot \begin{bmatrix} cos(|\boldsymbol{\eta_{xyz}}|)\\ \frac{\boldsymbol{\eta_{xyz}}}{|\boldsymbol{\eta_{xyz}}|}sin(|\boldsymbol{\eta_{xyz}}|) \end{bmatrix}$$ The time derivative of unit quatenion can then be expressed as $$\boldsymbol{\dot{q}} = \boldsymbol{e^{\eta}\dot{\eta}} = \boldsymbol{q\dot{\eta}}$$ $\boldsymbol{\eta}$ is by definition the Lie algebra of unit quaternion. It also matches the expression of body-fixed angular velocity $$\boldsymbol{\omega}_B = 2\boldsymbol{\dot{\eta}} = 2\boldsymbol{\overline{q}\dot{q}}$$ So can I simply integrate the body-fixed angular velocity to get $\boldsymbol{\eta}$ and use exponential map back to get unit quaternion? $$\boldsymbol{\eta} = \frac{1}{2}\int_{t0}^{t1}\boldsymbol{\omega}_Bdt$$ I know it looks weird and physically meaningless, it is basically integrating the angular velocity directly. But in the context of lie theory, I can't see the wrong side about it. I also tested on Matlab of a free full inertia tensor object with initial angular velocity. The momentum does not conserve in any of 3 degrees of freedom. Angular momentum on x-axis

So is it because $\boldsymbol{\omega}_B$ is on instantaneous frame so I should integrate world-fixed angular velocity instead. Or there is some deeper knowledge of Lie-group and Lie-algebra that I need to read more about?

I checked a lot of papers, none of which tried the method I described here, the closest on Quaternion Variational Integrators for Spacecraft Dynamics uses the expression as $$^{\epsilon}\boldsymbol{q} = \boldsymbol{q}e^{\epsilon\boldsymbol{\eta}}$$ to describe the varied unit quaternion.

I feel there is surely a mathematical reason.

Thank you for your help.

Answer 1

So I sorta figured it out myself.

The ideal and desired situation is to integrate unit quaternion rate directly

$$\boldsymbol{q} = \int_{t0}^{t1}\boldsymbol{\dot{q}}dt =\int_{t0}^{t1}\boldsymbol{e^\eta\dot{\eta}}dt \neq e^{\int_{t0}^{t1}\boldsymbol{\dot{\eta}}dt}$$

In discrete form $$\boldsymbol{q}_{i+1} = \boldsymbol{q}_i+\boldsymbol{\dot{q}}_i\delta t = e^\boldsymbol{\eta_i}+e^\boldsymbol{\eta_i}\boldsymbol{\dot{\eta}}_i\delta t \neq e^{\boldsymbol{\eta_i+\dot{\eta}_i}\delta t}$$ Please fill me out with more general explanation within the context of Lie theory.

Answer 2

It is correct if ${\pmb\omega}_B$ is constant during the time step. But as that is a zero-order approximation of $\omega_B$, you can only expect a first-order approximation of the trajectory of $q$.

You can get the same order of approximation, or perhaps even one better, by using the first Padé approximation $$ {\bf q}_{i+1}=(1+\tfrac14{\pmb\omega}_B\,dt)(1-\tfrac14{\pmb\omega}_B\,dt)^{-1}{\bf q}_i $$ which is basically the combination of a half-step of implicit Euler with a half-step of explicit Euler, combining to the implicit midpoint method, and in their combination cancelling some of the error that goes in opposite directions between them.