I am new to the topic of Lie groups and Lie algebra, and have been studying them a lot in the last couple days. I am particularly interest in the $SE(3)$ Lie groups and $se(3)$ lie algebra as I am looking for ways to describe rigid body motions in an optimization problem.
My specific questions now is: Given a function $f(\pmb{x}^o): \mathbb{R}^3 \to \mathbb{R} $ and each points local spatial gradient $\nabla f \bigg|_{\pmb{x}^o}$, (the reference frame points are given in is denoted with supscript $^o$), I am looking for the derivative of this function after the $se(3)$ twist coordinates $\tau_c^o \in \mathbb{R}^6$ so that $\hat{\tau_c^o} \in se(3)$, describing the lie group element $\mathcal{M}_c^o \in SE(3)$ transforming points between two frames ($\mathcal{M}_c^o \pmb{x}^c = \pmb{x}^o$) in the tangent space of the identity. So far I have: $$\frac{\partial f(\pmb{x}^o)}{\partial \tau_c^o} = \frac{\partial f(\pmb{x}^o)}{\partial \pmb{x}^o}\frac{\partial \pmb{x}^o}{\partial \tau_c^o}= \nabla f \bigg|_{\pmb{x}^o} \frac{\partial \pmb{x}^o}{\partial \tau_c^o}$$ Now with the exponential map $\exp(\hat{\tau_c^o})$ that maps the twist coordinates to $SE(3)$, the point $\pmb{x}^o$ can be expressed as $\pmb{x}^o = \mathcal{M}_c^o \pmb{x}^c = \exp(\hat{\tau_c^o} ) \pmb{x}^c$ : $$ \frac{\partial \pmb{x}^o}{\partial \tau_c^o} = \frac{\partial( \exp(\hat{\tau_c^o} ) \pmb{x}^c)}{\partial \tau_c^o} = \frac{\partial \exp(\hat{\tau_c^o})}{\partial \tau_c^o} \bigg|_{\tau_c^o}\pmb{x}^c = \frac{\partial \exp(\hat{\delta \tau})}{\partial (\delta \tau)} \bigg|_{\delta \tau = 0} \exp(\hat{\tau_c^o}) \pmb{x}^c = \frac{\partial \exp(\hat{\delta \tau})}{\partial (\delta \tau)} \bigg|_{\delta \tau = 0} \pmb{x}^o $$ From here [1] I found that this equals a left hand side multiplication with the infinitessimal generators of $SE(3)$: $$\frac{\partial \pmb{x}^o}{\partial \tau_c^o} = [G_1\pmb{x}^o, G_2\pmb{x}^o, G_3\pmb{x}^o, G_4\pmb{x}^o, G_5\pmb{x}^o, G_6\pmb{x}^o] = (I \mid (-\pmb{x}^o)_\times )$$ where $(.)_\times$ denotes the screw matrix and $I$ the identity matrix. Putting it all together this would give the following result: $$ \frac{\partial f(\pmb{x}^o)}{\partial \tau_c^o} = (\nabla f)^T \bigg|_{\pmb{x}^o} (I \mid (-\pmb{x}^o)_\times ) $$ Is this the correct result? does the left hand side multiplication with the infinitessimal generators really work like this? How does this result describe the correct derivative in the tangent space, when the spatial gradient given in the specific coordinate frame is just (partly) multiplied with the identity?
Basically I am looking for, what is the correct way to express a spatial gradient given in a specific coordinate reference frame, as a gradient in the $se(3)$ tangent space.
I welcome any kind of feedback. Thank you.