Consider the hamiltonian \begin{align} H& = H_0 + a\, \vec{s} \cdot \vec{l} \\& = \frac{1}{2m}p^2+ V(r) + a\, \vec{s} \cdot \vec{l}, \end{align} where $V(r)$ denotes an arbitrary central potential.
What is the symmetry of this hamiltonian? $\rm SU(2)$ or $\rm SO(3)$? Without the spin-orbit coupling part (or even the spin-degree of freedom), the symmetry of $H_0 $ should be $\rm SO(3)$, right?
I suspect that it is $\rm SU(2)$, but cannot prove it.