= Solution
A torque-free <rigid body> has constant <angular momentum> $\mathbf L$ in inertial space. Its <angular velocity> need not be parallel to $\mathbf L$, because $\mathbf L=\mathbf I\boldsymbol\omega$. In body principal axes, <Euler equations for a torque-free rigid body> are
$$
I_1\dot\omega_1=(I_2-I_3)\omega_2\omega_3,\quad I_2\dot\omega_2=(I_3-I_1)\omega_3\omega_1,\quad I_3\dot\omega_3=(I_1-I_2)\omega_1\omega_2.
$$
For a <symmetric top> $I_1=I_2=I_\perp$, $I_3=I_\parallel$, the component $\omega_3$ is constant and
$$
\boxed{\omega_1+i\omega_2=C e^{i\Omega_b t},\qquad\Omega_b=\frac{I_\parallel-I_\perp}{I_\perp}\omega_3.}
$$
This is the signed rotation frequency of the transverse angular-velocity components in the body frame. In space the symmetry axis instead obeys
$$
\dot{\mathbf e}_3=\frac{\mathbf L}{I_\perp}\times\mathbf e_3,\qquad\boxed{\Omega_{\rm space}=L/I_\perp.}
$$
Its angle to $\mathbf L$ is constant, with $\cos\theta=I_\parallel\omega_3/L$. The <body-frame and space-frame frequencies of free symmetric-top precession> describe different frames. A spherical body has $\boldsymbol\omega\parallel\mathbf L$ and no corresponding wobble. For a triaxial body, rotations about the largest and smallest principal-inertia axes are stable, whereas rotation about the intermediate axis is unstable.<Classical Dynamics by David Tong>[Sections 3.4.2–3.4.3: free symmetric top and stability of the asymmetric top]{ref}
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