Understaind configuration space of rigid body, necessity of integral frame

I'd like to undertand more deeply the concept of configuration space of a rigid body. The question arises from the formula's regarding know physics quantities for rigid bodies. Let's took in exam a random one, for example the angular moment $$M_Q$$, with respect to $$Q \in \mathbb{E}^3$$, where $$\mathbb{E}^3$$ is the underlying affine space :

$$M_Q = \int_C \rho(x')(\chi(q;x')-x_Q) \times v(q,\dot{q},x')dx' = \int_C \rho(x')(x_{O'}+Rx'-x_Q)\times(v_{O'}+\omega \times R x')dx'$$

Here $$C$$ denotes the rigid body,$$\Sigma'$$, $$\rho$$ the density (integrable on the coordinate of $$C$$ in $$\Sigma'$$), $$q = (x_{O'},\alpha)$$, $$R = R(\alpha)$$ where $$\alpha = (\varphi,\theta,\psi)$$ are the Euler's angles.

What I asked myself is why do we need to integrate on a solidal frame (I don't the terminology here could be integral with the rigid body, what I mean is that the velocity of the points of the rigid body in $$\Sigma'$$ are $$0$$, so costant coordinates). Running through my notes I think the reason is due to the following theorem :

Theorem : Consider a rigid body with three points $$P_1,P_2,P_3$$ not aligned. The map $$\phi : \mathbb{R}^3 \times SO(3) \longrightarrow \mathbb{R}^9$$ such that $$\phi(x_{O'},R) = (x_{O'}+Rx_1',x_{O'}+Rx_2',x_{O'}+Rx_3')$$ where $$x_j'$$ are the coordinates of $$P_j$$ in the solidal frame, is a diffeomorphism on the image $$\phi(\mathbb{R}^3 \times SO(3))$$.

Doubts I'd like to clear :

$$\bullet \hspace{0.1cm}$$ I do understand the proof of the theorem, which depends on taking a solidal frame $$\Sigma'$$, what I don't understand is why is necessary. In general can't be possible to determine the coordinates of the rigid body in a fixed frame $$\Sigma$$, without passing through a solidal frame $$\Sigma'$$? If so, does a manageable counterexample exists ?

(Specifically I don't see how the Euler angles play a special role when integral frames are involved, I think the same Euler angles could describe passing from two orthonormal basis to another, without the second be "fixed" to the body, am I wrong ?)

$$\bullet \hspace{0.1cm}$$ In general, if I don't take a solidal frame, the space of configuration could not be a manifold ? So it shouldn't make sense integrating on that ? Is this correct ? Otherwise, why bother to define the integrals passing through an integral frame ?

I'm new to Physics Stack Exchange so I apologize if I made some mistake about the policy of the questions, any help would be appreciated.

Edit : The definition of "solidal" frame I'm using is : given a frame $$\Sigma' = O' e_1',e_2',e_3'$$, this is called solidale to the rigid body $$C$$ is all the points $$P_j$$ of the body have $$0$$ velocity respect to $$\Sigma'$$, i.e the coordinates of the points $$P_j$$ are costant in $$\Sigma'$$.

• Welcome to Physics! I'm unfamiliar with the terms "solidal frame" and "integral frame"; can you define them? In particular, if you learned these terms in another language, it might help to provide them in that language, as English translations of technical terms are not always straightforward. May 7 at 15:56
• @MichaelSeifert Added the definitions with an edit, could you suggest me the correct term in english ? May 7 at 16:04
• That's helpful; I think what you're calling the "solidal frame" is more commonly called the "body frame" in English. See page 2 of these notes. Can you provide a similar definition for "integral frame"? May 7 at 17:17
• @MichaelSeifert I used integral frame as an equivalent word of solidal because I thought the name could be evocative May 7 at 17:18
• @MichaelSeifert Citing your notes "In order to describe this, ,we specify positions in a fixed space frame". This exactly what I'am asking in my question that I don't understand May 7 at 17:20

2. The space of configuration couldn't be a manifold in general. But in the case of a rigid body, it is. It's your mentioned $$\mathbb R^3\times SO(3)$$.