I've read in some works related to state estimation in robotics (e.g. I've found "A micro Lie theory
for state estimation in robotics" to be greatly useful), that the structure of the tangent space of a Lie Group, e.g. $SO(3)$ is the same everywhere.
But I'm not sure if I have fully understood this.
Focusing on $SO(3)$, its Lie algebra (tangent plane at the identity) is the vector space of all 3×3 skew symmetric matrices, and everywhere else the tangent plane can be computed as:
$$\dot{R} = [\omega]_x R$$
Where $R\in SO(3)$, $\dot{R}$ is its time derivative, and $[\omega]_x$ is a skew-symmetric matrix.
Thereby the elements lying on the tangent plane at everywhere else than the identity don't have to be skew-symmetric. So I'm not sure what the author is refering to when noting that structure of the tangent planes is the same at any point or if it has something to do with what I expressed before,
Any help is highly appreciated.
Best Answer
The idea is that there is a natural way to identify the tangent space at any point $M$ with that of the identity - To do this pick a basis of the tangent space and then consider the image under multiplication by $M$, typically on the left. This means that the tangent bundle is trivial and you can pick nowhere vanishing everywhere linearly independent vector fields and use these to identify different tangent spaces. But you're right that if you imagine a matrix group being embedded in Euclidean space then these tangent spaces won't look the same in that embedding.
(Note that if $\gamma$ is any curve in your Lie group then $M\gamma$ is another curve with derivative $M\gamma'$so multiplying by $M$ should, and indeed does, give isomorphisms of the tangent spaces).