While reading Kevin Lynch's "Modern Robotics" [1], I came across a fundamental question concerning the manipulator Jacobian. Upon better reading and further cross-checking with Peter Corke's "Robotics, Vision and Control" [2], I have managed to only confuse myself even more...
Kevin Lynch approaches kinematic representation of open chains with the product of exponentials (PoE) formula. In such representation, the space Jacobian $J_s(\theta) \in \mathbb{R}^{6 \times n}$ relates the joint rate vector $\dot{\theta} \in \mathbb{R}^{n}$ to the end-effector's twist expressed in fixed-frame coordinates $\mathcal{V}_s$ via $$ \mathcal{V}_s = J_s(\theta)\dot{\theta}. $$
A similar concept, the body Jacobian, relates the joint rate vector to the end-effector's twist in the end-effector frame coordinates via $$ \mathcal{V}_b = J_b(\theta)\dot{\theta}. $$
Finally, the relationship between both of the above is given by: $$ J_b(\theta) = [Ad_{T_{bs}}]J_s(\theta); $$ $$ J_s(\theta) = [Ad_{T_{sb}}]J_b(\theta). $$
Given the above, my question is:
Provided $J_b(\theta)$, how can I obtain the geometric Jacobian?
I tried looking into Peter Corke's book and I found the following subsection.
Screenshot taken from Peter Corke's book:
So, apparently, what I am looking for is that final equation. (?)
It seems that $\sideset{^0}{^\mathcal{V}}{J}$ in Corke's book refers to the same as what $J_b(\theta)$ refers to in Lynch's book.
But what does Corke mean exactly with "velocity transformation"?
Specifically, what exactly is the top-right 3x3 block of that 6x6 matrix pre-multiplication?
[1] Kevin M. Lynch and Frank C. Park, "Modern Robotics: Mechanics, Planning, and Control," Cambridge University Press, 2017. (http://modernrobotics.org)
[2] Peter Corke, "Robotics, Vision and Control: Fundamental Algorithms," Completely Revised. Vol. 118. Springer, 2017. (http://petercorke.com/wordpress/rvc/)
