Angular velocity in body-frame vs inertial frame This question is based on the nomenclature and definitions in Structure and Interpretation of Classical Mechanics.
In Section 2.2, we start with a rotation $\mathscr{M}(q(t))$ where $q$ is the path describing the motion of the body. The rotation is later represented by the rotation matrix $\mathbf{M}$.
To paraphrase:

Let $\vec{\xi}_\alpha(t)$ be the vector to some constituent particle with the body in the orientation specified by q(t) given by $q(t)$, and $\vec{\xi'}_\alpha$ be the vector to the same constituent with the body in the reference orientation, then:


$$
\vec{\xi}_\alpha(t) = \mathscr{M}(q(t)) \vec{\xi'}_\alpha\tag{2.11}
$$

Through a series of steps, it shows how the angular velocity vector consists of the components that can be extracted from the skew-symmetric matrix $D\mathbf{M} \mathbf{M}^\intercal$. Or
$$
\mathbf{\omega} = \mathscr{A}^{-1}( D\mathbf{M} \mathbf{M}^\intercal )
$$
$\mathscr{A}^{-1}$ is defined as the function that can extract the components of the angular velocity vector from an anti-symmetric matrix. In Eq. 2.20, it defines the derivative of the position of any constituent particle in the rigid body as:
$$
\dot{\vec{\xi}}_\alpha = \vec{\omega} \times \vec{\xi}_\alpha\tag{2.20}
$$
While it is not explicitly mentioned anywhere, I assume that the angular velocity components mentioned above are in terms of the inertial reference frame. So after this, it goes on to state the following:

The components $\omega'$ of the angular velocity vector on the body axes are $\omega'= \mathbf{M}^\intercal \omega$, so


$$
\mathbf{\omega}' =\mathbf{M}^\intercal \mathscr{A}^{-1}( D\mathbf{M} \mathbf{M}^\intercal )\tag{2.21}
$$

So here is my question:
In Eq. 2.11, $\mathbf{M}_\alpha$ seems to transform $\vec{\xi}$ from its reference frame components $\vec{\xi}'$ to its body-fixed frame components at time $t$. But in the last equation, angular velocity is transformed from reference to body-frame components as $\omega'= \mathbf{M}^\intercal \omega$. Why is it $\mathbf{M}^\intercal$ for angular velocity and not $\mathbf{M}$?
Is this somehow inferred from the prior steps?
 A: The angular velocity skew matrix  is:
$$ \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}}
\\ \omega_{{z}}&0&-\omega_{{x}}\\ 
-\omega_{{y}}&\omega_{{x}}&0\end {array} \right]_I=
\mathbf{\dot{M}}\,\mathbf{M}^T\tag 1$$
where $~\mathbf M~$ is the transformation matrix between body system (B-system) and inertial system (I-system)
from equation (1) you obtain the components of the angular velocity vector in inertial system$~(\vec\omega)_I~$ , the components of the angular velocity vector in B-system
are
$$(\vec\omega)_B=\mathbf M^T\,(\vec\omega)_I$$

the components of the angular velocity vector in B-system  can also obtained  with this equation
\begin{align*} \mathbf M^T\,&\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}}
\\ \omega_{{z}}&0&-\omega_{{x}}\\ 
-\omega_{{y}}&\omega_{{x}}&0\end {array} \right]_I\,\mathbf M=
\mathbf M^T\,\mathbf{\dot{M}}\,\mathbf{M}^T\,\mathbf M\quad \Rightarrow\\\\
 &\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}}
\\ \omega_{{z}}&0&-\omega_{{x}}\\ 
-\omega_{{y}}&\omega_{{x}}&0\end {array} \right]_B=
\mathbf M^T\,\mathbf{\dot{M}}\end{align*}
