How to get the form of elements of the Lie algebra? I'm leaning Lie theory in robotics. My question comes from a paper: A micro Lie theory for state estimation in robotics by Joan Sola et al. at page 4:

For multiplicative groups this yields the new constraint $\mathcal{X}^{-1} \dot{\mathcal{X}} + \dot{\mathcal{X}^{-1}} \mathcal{X} = 0$, which applies to the elements tangent at $\mathcal{X}$ (the term $\dot{\mathcal{X}^{-1}}$ is the derivative of the inverse). The elements of the Lie algebra are therefore of the form
$$
  \mathbf{v}^\wedge
  = \mathcal{X}^{-1} \dot{\mathcal{X}}
  = -\dot{\mathcal{X}^{-1}} \mathcal{X} \,.
  \tag{$9$}
$$
(Original screenshot)

To my understanding, the symbol $\mathbf{v}^{\wedge}$ in equation (9) should more precisely be $^\mathcal{X}\mathbf{v}^{\wedge}\in T_{\mathcal{X}}\mathcal{M}$, which is an element in the local tangent space of $\mathcal{X}$. But I didn’t get how the form of $\mathbf{v}^{\wedge}$ was derived from the new constraint.
I have 2 questions:

*

*Could anyone tell me in detail what is going on in the underlined sentence?


*Why isn’t $\mathbf{v}^{\wedge}$ equal to $\dot{\mathcal{X}}=\partial\mathcal{X}/\partial t$, which I thought was just the tangent element at $\mathcal{X}$ in  $T_{\mathcal{X}}\mathcal{M}$?
Thank you.
 A: I'm not totally familiar with the nomenclature used here, but I think your statement

To my understanding, the symbol $\mathbf{v}^{\wedge}$ in eq(9) should more precisely be $^\mathcal{X}\mathbf{v}^{\wedge}\in T_{\mathcal{X}}\mathcal{M}$, which is an element in the local tangent space of $\mathcal{X}$.

is right.
I don't fully understand what you mean with your first question, but I think that your confusion in the second questions comes from not taking into account the corresponding reference frames. In "II-D. The exponential map", you can find equation (12):
equation (12)
If I have interpreted it correctly, we can rewrite this equation to explicitly state the reference frame each variable is referred to:
$^\mathcal{E}\dot{\mathcal{X}} =\ ^\mathcal{E}\mathcal{X}\ ^\mathcal{X}\mathbf{v}^{\wedge}$
So  they both  refer to the same velocity but in different reference frames:

*

*$\dot{\mathcal{X}}$ is the time derivative of $\mathcal{X}$ in the global reference frame of the manifold $\mathcal{M}$, whereas

*$\mathbf{v}^{\wedge}$, as you very well said, is defined at the local tangent space at $\mathcal{X}$: $T_{\mathcal{X}}\mathcal{M}$.

Cheers.
