Differential forms and line integral in rotation group SO(3) My major is mechanical engineering. Recently, I am working on some subject involving three-dimensional finite rotations. More specially, the necessary and sufficient conditions for an applied torque/moment be conservative in the finite rotation range. I have tried to read some math books, but I got more confused. The following is the description of the background.
In mechanics, an externally torque generally exhibits unusual property of configuration-dependent, which means the torque varies from its initial value $\mathbf M_0$ to its current value $\mathbf M$ when moving along a curve lying on SO(3) staring form the identity $\mathbf I$ to the current position $\mathbf R$. In other words, the current counterpart $\mathbf M$ can be viewed as a $\mathbf explicit function$ of the rotation $\mathbf R \in SO(3) $.
Let $\mathbf \delta \omega$ be the spatial spin (an element which belongs to the tangent space of SO(3) at the base point $\mathbf R$, i.e., $\mathbf \delta \omega \in T_{R}SO(3)$). Then the virtual work done by the torque over the spin is given by
$$\delta W = \mathbf M \cdot \delta \omega$$
where $\delta W $ is a real number, and "$\cdot$" means dot product.
In mathematics, $\mathbf M$ is an element of cotangent space of SO(3) at the base point $\mathbf R$, i.e., $\mathbf M \in T^{*}_{R}SO(3)$. 
On the other hand, if the rotation vector (axis-angle representation) $\mathbf \psi = \psi_{i} \mathbf e_{i}$ was used to parameterize the rotation manifold, $\mathbf R = exp(\hat \psi)$, then we can express the torque as $\mathbf M=\mathbf Q \mathbf M_0$ explicitly, where $\mathbf Q=\mathbf Q(\psi)$ is the transformation matrix relating the initial and current values of the torque.
We also can represent the virtual rotation by $\mathbf \delta \psi$, the variation of rotation vector $\mathbf \psi$, $\mathbf \delta \psi \in T_{I}SO(3)$. The relation between $\mathbf \delta \omega$ and $\mathbf \delta \psi$ is given by $ \delta \omega = \mathbf L \delta \psi$, where $\mathbf L= \mathbf L(\psi)$ is the tangential operator, $\mathbf L:T_ISO(3)\to T_RSO(3)$. Thus, the virtual work can be rewritten as
$$ \mathbf \delta W = \mathbf L^T \mathbf M \cdot \delta \psi$$
My questions are:


*

*Which expression of the virtual work is a differental 1-form in SO(3) and why? 

*How to calculate the line integral of the virtual work over a curve lying on SO(3) in terms of a differential 1-form?


Thank you very much! 
EDIT 1: In the above description, the spin $\mathbf \delta \omega$ is not a differential, since there does not exist a variable from which the spin can be derived. It comes from the variation of the orthogonality condition of rotation matrix, $\mathbf \delta(\mathbf R \mathbf R^T=\mathbf I)=0$,  $\mathbf \delta \mathbf R=\widehat (\delta\omega) \mathbf R$.
However, the variation $\mathbf \delta \psi$ of rotation vector is a differential.
 A: $\vec{r}: [0,2\pi] \times [0,\pi] \times [0,2\pi] \to \mathbb{R}^4$ given by $$\vec{r}(\psi,\theta,\phi)=\cos(\frac{\phi + \psi}{2})\cos(\frac{\theta}{2})c\hat{t}+\cos(\frac{\phi - \psi}{2})\sin(\frac{\theta}{2})\hat{i}+\sin(\frac{\phi - \psi}{2})\sin(\frac{\theta}{2})\hat{j}+\sin(\frac{\phi + \psi}{2})\cos(\frac{\theta}{2})\hat{k}$$ is a parametrization of $SO(3)$ by the angles yaw ($\psi$), pitch ($\theta$), and roll ($\phi$) https://pasteboard.co/JeFt2eK.png. (Actually, it is a parametrization of half of $S^3$, but, as long as we stay inside the coordinate patch, it will suffice. Note $S^3$ is a double cover, and the universal cover, of $SO(3)$; this is a single "sheet" in the covering space.) (Note $\hat{t}$ is the unit vector in the ''time direction'' and $c$ is the speed of light, converting time dimensions to space dimensions. You do not actually include $c$ in your calculations, $c\hat{t}$ is just heuristic for indicating a fourth spatial dimension.) (I forget which of the standard parametrizations -- rotation matrix, rotation vector (axis-angle), quaternions, Euler angles, etc. -- this is, but you probably know it off the top of your head. I suppose it is the quaternion parametrization.)
We need $\mathbf{R}$ to be oriented to do the integral (a change of orientation only changes the sign of the integral's value); call the oriented path $[\mathbf{R}]$. Parametrize $\mathbf{R}$ in $SO(3)$ by the path $\gamma: [a,b] \to SO(3)$, consistent with the orientation of $[\mathbf{R}]$. Pull $[\mathbf{R}]$ to $[\mathbf{R}^*]$ in $[0,2\pi] \times [0,\pi] \times [0,2\pi]$ and $\gamma$ back to $\gamma^*: [a,b] \to [0,2\pi] \times [0,\pi] \times [0,2\pi]$, with $\gamma(t) = \vec{r}[\gamma^*(t)]$ the desired parametrization of $\mathbf{R}$ in $SO(3)$, consistent with the orientation of $[\mathbf{R}]$. Now, suppose we extend $\mathbf{M}$ to a neighborhood of $S^3$ in $\mathbb{R}^4$ and write $\mathbf{M}(ct, x, y, z) = $ $\mathbf{M}_1(ct, x, y, z)d\{ct\} + \mathbf{M}_2(ct, x, y, z)dx + \mathbf{M}_3(ct, x, y, z)dy + \mathbf{M}_4(ct, x, y, z)dz$, then pull the restriction of this to $S^3$ back via $\vec{r}$ to be $\mathbf{M}^*(\psi,\theta,\phi) \in T^*([0,2\pi] \times [0,\pi] \times [0,2\pi])$ via formulas like $\displaystyle \mathbf{M}_1[\vec{r}(\phi,\theta,\psi)] \ \left(\frac{\partial \{ct\}}{\partial \psi}d\psi + \frac{\partial \{ct\}}{\partial \theta}d\theta+ \frac{\partial \{ct\}}{\partial \phi}d\phi\right)$; alternatively, we could write $\mathbf{M}^*(\psi,\theta,\phi) = \mathbf{M}^*_1(\psi,\theta,\phi)d\psi + \mathbf{M}^*_2(\psi,\theta,\phi)d\theta + \mathbf{M}^*_3(\psi,\theta,\phi)d\phi$ directly. Next, take $\gamma^{*'}(t)$ and $\mathbf{M}^*[\gamma^*(t)]$.
Then $$ \int_{[\mathbf{R}]} \mathbf{M} = \int_a^b \langle \mathbf{M}^*[\gamma^*(t)]\ |\ \gamma^{*'}(t)\rangle\ dt$$, where $\langle\ |\ \rangle$ denotes the action of a differential 1-form on $[0,2\pi] \times [0,\pi] \times [0,2\pi]$ on a tangent vector to $[0,2\pi] \times [0,\pi] \times [0,2\pi]$, is the work done by the torque along the path https://en.wikipedia.org/wiki/Torque#Relationship_between_torque,_power,_and_energy; one could turn the RHS over to a talented Calc II student, in principle.
Here is a Mathematica notebook that does an example; you should only have to change the formulas in the top lines, where $\gamma^*(t)$ and $\mathbf{M}$ are defined, then drag down to "Evaluation $\to$ Evaluate Notebook" to do any reasonable computation you desire. (There could be a teensy problem if your path "leaks out of" or "wraps around" the coordinate patch; you would then just have to break your integral into "sub-integrals" over portions of the oriented path contained entirely in the coordinate patch to do the full integral.)
A: (I think I've edited the first answer enough; we'll try another answer for the potential function solution.)
Here is a Mathematica notebook that does the same example using a potential function to compute the integral. Many special thanks to @Michael_E2 on mathematica.stackexchange.com for posting this answer to "help" with the technical details of computing a potential function as a function in Mathematica. It is still caveat emptor for the user to make sure the exterior derivative of $\mathbf{M}$ is 0; recall that part of what makes this work is that $SO(3)$ is a "de Rham cohomology sphere" (see the comments above the first answer).
Here is a Mathematica notebook that uses the DifferentialForms.m package https://library.wolfram.com/infocenter/MathSource/482 to compute the integral directly and using an antiderivative (there's a kludge on the HomotopyOperator I'm not sure why I need).
