$\mathbf{Background:}$
The following is paraphrased from ``Representations of the rotation and Lorentz groups and their applications,'' by Gel'fand.
Consider a finite-dimensional representation $T: SO_3 \to GL_n$ of the rotation group. Label elements of $SO_3$ as $(\theta_1,\theta_2,\theta_3)$. We have (for real $s,t$): \begin{align} T_{jk}[s \cdot (\theta_1,\theta_2,\theta_3)] \cdot T_{kl}[t \cdot (\theta_1,\theta_2,\theta_3)] = T_{jl}[(s+t)\cdot (\theta_1,\theta_2,\theta_3)] \end{align} Applying $\frac{d}{ds} \rvert_{s=0}$ to both sides (assuming $T$ is such that we can do this), we get: \begin{align} \theta_a \cdot A^a_{jk} \cdot T_{kl}[t \cdot (\theta_1,\theta_2,\theta_3)] = \frac{d}{dt} T_{jl} [t\cdot (\theta_1,\theta_2,\theta_3)] \end{align} where: \begin{align} A^a = \frac{\partial}{\partial \theta_a} \rvert_{\theta_1 = \theta_2 = \theta_3 = 0}\ T[(\theta_1,\theta_2,\theta_3)] \end{align} The solution of these differential equations is: \begin{align} T_{jk}[t\cdot (\theta_1,\theta_2,\theta_3)] = (\exp [ t \cdot \theta_a \cdot A^a ])_{jk} \end{align} Setting $t=1$, we see that the representation $T$ is completely determined if we are given the matrices $A^1,A^2,A^3$.
$\mathbf{Question:}$
Consider an infinite dimensional rep $T : SO_3 \to C^\infty(\mathbb{R}^3)$. Can the above argument be made rigorous in this case, with $A^1, A^2, A^3$ as differential operators?
One problem is that the equation \begin{align} \theta_a \cdot A^a \cdot T[t \cdot (\theta_1,\theta_2,\theta_3)] = \frac{d}{dt} T [t\cdot (\theta_1,\theta_2,\theta_3)] \end{align} is no longer a system of (finitely many) ordinary differential equations. Also, I'm not sure if $A^a = \frac{\partial}{\partial \theta_a} \rvert_{\theta_1 = \theta_2 = \theta_3 = 0}\ T[(\theta_1,\theta_2,\theta_3)]$ is well-defined.
Writing rotation vectors ($\theta_1,\theta_2,\theta_3$) in spherical coordinates instead of cartesian, Gel'fand gets, for instance, $A^z \sim \frac{\partial}{\partial \phi}$, where $\phi$ is the azimuthal angle. That's what I'm trying to justify.
I don't know where gel'fands answer comes from; I believe that the initial equation should be placed by composition, wich generalizes matrix multiplication: $$T(s \cdot (\theta_1,\theta_2,\theta_3))\circ T(t \cdot (\theta_1,\theta_2,\theta_3))=T((s+t) \cdot (\theta_1,\theta_2,\theta_3))$$
This time taking the derivative gives
\begin{align} \theta_a \cdot A^a \circ T[t \cdot (\theta_1,\theta_2,\theta_3)] = \frac{d}{dt} T [t\cdot (\theta_1,\theta_2,\theta_3)] \end{align}
Where $A^a$ is given by the same equation. Does this sound right? If so, $A^a$ is a function and not an operator.