The space of physical states of the spin-$s$ system In the case of a spin-1 particle, the space of its possible (spin) states is a $2$-sphere. Mathematically, it is obtained as follows. We act with the whole $SO(3)$ on a given vector and end up with its orbit under the action of the group. This orbit is not diffeomorphic to $SO(3)$ itself due to existence of the stabilizer subgroup $SO(2)$. Factoring it out gives us the coset space
$$
SO(3)/SO(2) = RP^3/S^1 = S^2 
$$
Now, consider another important case, spin-$1/2$ particle. The orbit of a given spinor under the action of $SU(2)$ is diffeomorphic to $SU(2)$ itself (i.e. to $S^3$) , since this time we don't have a stabilizer subgroup. However, physics tells us that the spinors which differ up to an overall phase correspond to the same physical states, and, so, should be identified. This gives us a reason to quotient the group $SU(2)$ by its $U(1)$ subgroup which is responsible for the phase rotations. Thus, we end up with:
$$
SU(2)/U(1)=S^3/S^1 =S^2 
$$
From the mathematical point of view, in these two cases $S^2$ was obtained in rather different ways. A physicist could, however, just say "oh, c'mon, we're just ignoring the 'rotation of an $s$-spinor around itself' (whatever it means), which in the real case is trivial while gives the phase in the complex case".
To better understand the situation, let me ask few questions about a general rule.


*

*For an arbitrary spin-$s$ particle, what is the space of its physical states? Is it always $S^2$, just because we are dealing with representations of $SO(3)$? To me it's not obvious at all how this happens mathematically.

*(if the answer to the previous question is positive) No matter what $s$ is, the orbit is always of dimension at most $3$. In which cases is the 'extra' $S^1$ factored out as a stabilizer and in which - as a phase?

*How are all these things extended to the relativistic setting? I guess, it should be pretty straightforward (especially, for massive particles), since the spin states of particles are transformed under the action of the little group.
 A: For a spin $s$ particle you have 2s+1 states. An arbitrary state is a unit vector in $C^{2s+1}$, mod the phase so the state space is isomorphic to $CP(2s)$. For spinors it happens that $CP(1)\sim S^2$. For vectors you are only getting $S^2$ because you are effectively considering only a real representation of $SU(2)$. On the full space of states it is $CP(2)$ which is a 4d manifold.
If you have a real unit vector $v_x|x\rangle+v_y|y\rangle+v_z|z\rangle$ you can always find a rotation to take it to $|z\rangle$ by first rotating about $z$ to get rid of $|y\rangle$, then you can do a rotation about $y$ to get rid of $|x\rangle$. This means you can get any real vector by some rotation, up to the stabilizer, as you mentioned.
But if you have say $\frac{1}{\sqrt{2}}(|x\rangle+i|y\rangle)$, rotations about $z$ will just give you a phase factor, and you can't get rid of $|y\rangle$, so the procedure before doesn't work. The action of the group on the spin 1 states is not transitive, you can't reach every state from some other state.
So to sumarize, you always think of what you are modding out as a phase not a stabilizer. You aren't modding out something larger for higher dimensional $s$, it is just that the group action is not transitive. The reason you were able to think about the group manifold and stabilizer in the vector case is because for the real representation it is transitive.
