Definition and Existence of Angular Velocity $\newcommand{\R}{\mathbf R}$ Let $f:\R\to SO(3)$ be a smooth map whose derivative never vanishes. So $f(t)$ can be thought of as the snapshot at time $t$ of a rigid pivoted at the origin and rotating smoothly about the pivot.
In my high school physics books, the angular velocity at time $t$ of a rigid body pivoted at a point was defined as a vector $\vec\omega(t)\in \R^3$ such that if $\vec p(t)$ is the position vector of a particle $P$ on the rigid body at time $t$, then the velocity of $P$ at $t$ is $\vec\omega(t)\times \vec p(t)$.
Under this definition, I want to know if we can prove that the angular velocity of a smoothly rotating rigid body always exists.
If $f:\mathbf R\to SO(3)$ is a smoothly rotating rigid body with $f(0)=Id$, and $v\in \R^3$ is a point, then define $P_v(t)=f(t)v$. (Here $f(t)$ is a linear map which is begin fed $v$).
Basically $P_v(t)$ is the position of the paritcle of the rigid body at time $t$ which was at $v$ at time $0$. So the velocity of this particle at time $t$ is $\dot P_v(t)$.

Can somebody see as to why at any given time $t$ we have a vector $\omega(t)\in \R^3$ such that $\omega(t)\times P_v(t)=\dot P_v(t)$ for all $v\in \R^3$?

Further, does $\omega(t)$ vary smoothly with $t$?
Thank you.
 A: By Euler's rotation theorem, any value of $f$ can be represented by a single rotation about a fixed axis,
$$
f=e^A,
$$
where $A$ is an antisymmetric matrix. In your application, $A$ is a continuous function of time with $A(0)=0$. Although the matrix $A$ described by Euler's theorem may not be unique, it is unique in this application because $A$ is continuous and connected to the zero matrix. The position of a point on the body is
$$
P_v = fv,
$$
and differentiation gives
$$
\frac{dP_v}{dt} = e^A\frac{dA}{dt}v.
$$
To prove what you want to prove, it's sufficient to prove it at $t=0$, where this simplifies to
$$
\left.\frac{dP_v}{dt}\right|_0 = \left.\frac{dA}{dt}\right|_0v.
$$
Let's drop the $|_0$ symbols for convenience. The matrix $dA/dt$ is antisymmetric, and its components can be identified with the components of the angular velocity vector:
$$
  A =\begin{pmatrix}
  0 & \omega_z & -\omega_y \\
  -\omega_z & 0 & \omega_x \\
  \omega_y & -\omega_x & 0
  \end{pmatrix}.
$$
This value of the vector $\omega$ was calculated in a way that didn't depend on the choice of $v$, so that completes the proof.
Stepping back and looking at the big picture here, there are really two things that make all this work. (1) In 3 dimensions, we have an isomorphism between antisymmetric matrices and vectors. This isomorphism only works in 3 dimensions. (2) Rotations are not commutative, but infinitesimal rotations are, so that there is no $\Delta \theta$ vector, but there is a $d\theta$ vector.

Further, does $\omega(t)$ vary smoothly with $t$?

You say that $f$ is a smooth map, so yes, because all we did in order to find $\omega$ was differentiate and pick components out of a matrix. Actually all you really need is that $f$ be twice differentiable. Otherwise, for example, I can simply make up an example of rotation in the x-y plane by an angle $\theta(t)$, where $\theta''(t)$ doesn't exist at a certain time.
