During the study of the motion of a rigid body, in Arnold's book, two coordinates systems are introduced: one is fixed $k=\{O',\hat e_1',\hat e_2',\hat e_3'\}$ and one is inside the rigid body $K=\{O,\hat e_1,\hat e_2,\hat e_3\}$. We suppose that the rigid body is rotating around a line passing through $O$. We also have a transformation $B:K\to k$, which is the matrix of change of coordinates. One can prove that $B\in\operatorname{SO}(3)$.
I read the chapter concerning the rigid body but I didn't find an explicit definition of angular velocity of the rigid body. I think that the angular speed of a rigid body can be defined as the unique differentiable vector $\vec\omega(t)$ such that $\forall \vec u$ belonging to a coordinate system rotating with the rigid body, $\frac{d}{dt}\vec u=\vec \omega(t)\times \vec u$, $\forall t\in I \subseteq \mathbb R$.
Second thing I'd like to check is the following. Given a point $P$ belonging to the rigid body, variables in moving coordinate system are named $\vec Q=P-O,\vec V,\vec \Omega,\vec M$ and variables in the fixed coordinate system are named $\vec q=P-O',\vec v,\vec \omega,\vec m$ and it holds the relation $\forall \vec X\in K$, $B\vec X=\vec x\in k$.
Observing that the angular momentum of the point $P$ relative to the pole $O$ is given by $$M_O=(P-O)\times m \vec V=\vec Q\times m\cdot d\vec Q/dt=m \vec Q\times(\vec \Omega\times\vec Q)$$
the book introduces an operator $A:K\to K$ such that $A\vec\Omega=\vec M$, called inertia operator. I think this is not the standard definition of inertia operator. Infact, reading different notes, I think that one can can define the inertia operator in the following way. Given a real affine space $\mathbb A^n=(V,V(\mathbb R),-)$ the inertia operator relative to a pole $O\in V$ (for a discrete rigid body) is defined as
$$A_O:V\to V\text{ such that}$$
$$A_O \vec u:=\sum_i m_i(P_i-O)\times[\vec u\times (P_i-O)].$$
So the operator defined in Arnold's textbook it's like a "punctual inertia operator" $(A_O)_i$ and instead of defining the operator on $\mathbb R^3$, it takes the restriction to the moving frame $K$. Is that right?
Thank you for your attention.