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I cannot wrap my head around something that seems pretty basic.

In "A micro Lie theory for state estimation in robotics", the authors derive the structure of the Lie algebra of $SO(3)$ by taking the time derivative of the group constraint. Then, assuming a constant angular velocity $\omega$, they conclude that the solution to the ordinary differential equation $\dot{R} = R[\omega]_\times$is the exponential map:

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where $[\omega]_\times$ is a skew-symmetric matrix.

However, if I derive $R(t) = \exp([\omega]_\times t)$ using the derivative of the exponential map, I don't get the same expression:

$$ \begin{align} \frac{d}{dt} R(t) &= \frac{d}{dt} \exp([\omega]_\times t) = \exp([\omega]_\times t) \; J_r([\omega]_\times t) \; [\omega]_\times \mspace{40mu} J_r(\tau) \triangleq \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^k} (ad_\tau)^k \\ &= R(t) \; J_r([\omega]_\times t) \; [\omega]_\times \\ &\neq R(t)\; [\omega]_\times \mspace{40mu} \text{when $t>0$} \end{align} $$

Where did I go wrong?

Eskilade
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  • $R(t)$ is a path in the Lie group, not the Lie algebra; the derivative formula you used is for paths in the Lie algebra. – kobe Jun 25 '20 at 16:58
  • Thanks @kobe, though I suspect I'm not getting the nuance correctly. Given $R(t) = \exp(\tau(t))$ such that $R:\mathbb{R} \to SO(3)$ and $\tau:\mathbb{R} \to \mathfrak{so}(3)$, I would expected $\frac{d}{dt} \tau(t)$ to be the derivative of the path Lie algebra, and $\frac{d}{dt} R(t)$ to be the derivative of the path in the group. It is also possible that I am not correctly grasping the domain and image of the derivative in the Wikipedia article, i.e. that $\frac{d}{dt} ; \exp(X(t)): T\mathfrak{g} \to TG$. Could you elaborate, please? – Eskilade Jun 28 '20 at 12:25

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