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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:

  1. Which expression of the virtual work is a differental 1-form in SO(3) and why?
  2. 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.

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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, on this site we use MathJaX to format our maths. [Here](http://meta.math.stackexchange.com/q/5020/145141) you can find a basic tutorial. – 2017-02-10
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    Ok, thanks a lot. – 2017-02-10
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    As it stands, neither is a differential form, but rather an output when you apply the desired differential form to a tangent vector $\delta\omega$ as you move along your curve. But you want the first choice, moving along a path in the group. To compute an actual line integral, you'll need either coordinates or an understanding of how to express $\mathbf M$ explicitly as a function of the orthonormal frame $(e_1,e_2,e_3)\in SO(3)$. – 2017-02-10
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    Thank you very much, Ted. I have updated my question. And yes, we can express $\mathbf M$ explicitly as a fucntion of the rotation vector $\mathbf \psi$. You mean we can calculate the line integral on the manifold SO(3) as vector calculus in terms of a set of local coordinates. Does there exist a way of coordinate-free to deal with the line integral? Since if a torque is conservative, then the line integal should be path-independent regardless of the local coordinates (the choice of parameterizations). – 2017-02-11
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    I'm having some trouble parsing, but adding to Ted's comment, the differential one-form is the collection of all $\mathbf{M}$, viewed as a function on $SO(3)$ whose value at each point $p$ is an element of the cotangent space $T_{p}^{*}SO(3)$. It appears that $\mathbf{R}$ is some representation of a path in $SO(3)$; if that's right, the virtual work done "at time $t$" is $$\delta W = \mathbf{M}(\mathbf{R}(t)) \cdot \mathbf{R}'(t)\, dt,$$and the work is the integral over the parameter interval of $\mathbf{R}$. – 2017-02-12
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    Thank you for your explanation, Andrew. $\mathbf R$ is the rotation matrix. In computational mechanics, several parameterizations of finite rotation were often considered, e.g., rotation matrix, rotation vector (axis-angle), quaternions, Euler angles. – 2017-02-12
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    Continued. Your equation is definitely right. However, in most of these parameterizations, the "differential" of the rotation parameter is not a true differental, since there does not exist such a variable from which the "differential" of the parameter can be derived. Consequently, the expression of the virtual work in terms of such rotation parameter cannot be integrated. Here comes my question 1, $\mathbf \delta \omega $ is not a differential, then is $\mathbf \delta W= \mathbf M \cdot \delta \omega$ still a differenrial 1-form? – 2017-02-12
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    I don't have time to write out a more detailed comment/response right now, but regarding your earlier comment, a line integral is *always* well-defined independent of parametrization of the curve; it's independent of *path* when the $1$-form is exact (or the force field conservative). But, yes, there's an invariant way to write $1$-forms on any Lie group. (I do not understand how you're thinking of parametrizing the rotation manifold. Care to elaborate?) – 2017-02-12
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    Thank you for your time, Ted. In the context of computational mechanics, paramterizations of rotation manifold play important roles in both theoretical and computational aspects. For instance, if the rotation vector (axis-angle) $\mathbf \psi=\psi_{i}\mathbf e_{i}$ was used, the rotation matrix can be computed with Rodrigues formula $\mathbf R = I + {sin(\psi) \over \psi} \hat \psi+{(1-cos(\psi)) \over \psi^2} \hat \psi ^{2} $. We can deal with finite rotation with only three components of rotation vector and get some other benifits. – 2017-02-12
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    Continued. The Rodrigues formula is a closed expression of exponential map $exp(\cdot):so(3)\to SO(3)$, $\mathbf R = exp(\hat \psi)$. – 2017-02-12

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