Spin coherent state for general spin $S$ In Wen's book on Many-body QFT, he claimed that the coherent state for a spin-$S$ particle can be written as a tensor product of $2S$ spin-1/2 coherent states: 
$$|\hat{n}\rangle=|z\rangle\otimes|z\rangle...\otimes|z\rangle$$
where each $|z\rangle$ is a spin-1/2 spinor. 
If we look at the degrees of freedom of both sides of the above equation, the spin-$S$ coherent state has $2S+1$ free parameters, whereas the $2S$ spin-1/2 have $4S$ free parameters in total. How do we understand this?
 A: By definition, a spin coherent state $\vert \theta,\phi\rangle_S$ is just a rotation of the $\vert S,S\rangle$ state.  
What makes the state "coherent" is that all the individual spin states point in the same  $(\theta,\phi)$ direction (are "coherently aligned"), thus 
\begin{align}
\vert \theta,\phi\rangle_S
&:= \vert \theta,\phi\rangle_1\otimes\ldots\otimes 
\vert \theta,\phi\rangle_{2S} \tag{1} \\
&=R_z(\phi)R_y(\theta)\vert S,S\rangle \\
&
\; = [R_z(\phi)R_y(\theta)\vert +\rangle_1]\otimes \ldots \otimes
[R_z(\phi)R_y(\theta)\vert +\rangle _{2S}]
\end{align}
$\vert \theta,\phi\rangle_S$ clearly depends on only 2 parameters, as it should, and indeed there is no freedom in the orientation of any of the particle spins (except the first) are they must all be aligned in the  same $(\theta,\phi)$ direction as the first.  Thus, the right hand side of (1) depends on only two angles also.
The state in Eq.(1) is clearly symmetric under permutation of the particle indices, so taking $\theta=\phi=0$ gives
$$
\vert S,S\rangle = \vert +\rangle_1\otimes \vert +\rangle_2\otimes\ldots \otimes \vert +\rangle_{2S} \tag{2}
$$
This state is an eigenstate of $L_z$ 
$$
L_z= L_z^{(1)}+L_z^{(2)}+\ldots +L_z^{(2S)}
$$
with eigenvalue $M=S$, and 
is killed by the raising operator 
$$
L_+= L_+^{(1)}+L_+^{(2)}+\ldots +L_+^{(2S)}
$$
where $L_+^{(k)}$ acts on the state of particle $k$ alone.
A: A coherent spin state, when expressed in terms of composite spin 1/2 particles, is the tensor product of all spin 1/2 particles where each spinor is in the same state. In your expression, each $|z\rangle$ is identical, so there are in fact only two degrees of freedom on the right hand side (simply the degrees of freedom of a single spin half particle).
On the left hand side, the coherent spin state is not just any state with total spin S, but a highly constrained combination (as you see from the righthand side decomposition into spin half particles). There are again only two degrees of freedom.
The usual picture is that the coherent spin state lives on a collective Bloch sphere of radius S. The position on the collective Bloch sphere, parametrized by angles $\theta, \phi$, similarly describe the Bloch sphere position of each spin half particle in the decomposition.
Mathematically: a single spin half can be parametrized by angles $\theta, \phi$ as: $$|\psi(\theta, \phi)\rangle = \cos (\theta/2) |\downarrow\rangle + e^{i\phi} \sin(\theta/2) |\uparrow\rangle$$
Then the collective spin state is
$$
|\Psi(\theta, \phi)\rangle  = |\psi(\theta,\phi)\rangle^{\otimes N}
$$
A: You need to look only at the symmetric combinations (under particle index permutation) of spin 1/2. So for instance, with $N=2$, the familiar triplet state has the same multiplicity as spin 1 vector boson. For $N=3$, you find:
$$ |+\frac 3 2\rangle = |\uparrow\uparrow\uparrow \rangle $$
$$ |+\frac 1 2\rangle = (|\downarrow\uparrow\uparrow \rangle + |\uparrow\downarrow\uparrow \rangle +|\uparrow\uparrow\downarrow \rangle)/\sqrt 3$$
$$ |-\frac 1 2\rangle = (|\downarrow\downarrow\uparrow \rangle + |\uparrow\downarrow\downarrow \rangle +|\downarrow\uparrow\downarrow \rangle)/\sqrt 3$$
$$ |-\frac 3 2\rangle = |\downarrow\downarrow\downarrow \rangle $$
and so on. There are many more combos of 3 spin 1/2 particles, but only those four are symmetric under interchange.
As you increase $N=2S$, you'll aways find $N+1$ states that are totally symmetric and they are in a 1-to-1 correspondence with the $2S+1$ states of a spin $S$ particle.
