Bosonic representation of $SU(N)$: what values can $n_b$ take? In Assa Auerbach's book on page 166, he describes the construction of a bosonic representation of $SU(N)$ where the generators $S^{mn} \rightarrow b^\dagger_m b_n$. I'm a bit confused about the constraint in cases $N\ne 2$:
$$ n_b= \sum_{n}^N b^\dagger_n b_n = NS $$
where $S$ is the spin representation and $n_b$ is the total number of bosons representing that spin. Clearly, in $SU(2)$, $S$ takes on positive integer and half integer values, so that $NS = (0, 1,2 ...)$. 
What about in $N>2$? Wikipedia shows that the adjoint irrep of $SU(3)$ has $16/3$ which is not a half integer. Basically I'm asking what the allowed values of $S$ are for a given $N$, and how to work this out?    
 A: Jordan's (1935) nifty map is a Lie algebra isomorphism of generator matrices (and the vectors they act on) to Fock bilinears (and the Fock states they operate on). It expedites some representation theory, as Schwinger illustrated on SU(2) — after a hiatus of 17 years. 
As you presumably already saw, for SU(2), 
$$
{\vec  J} \equiv    {\mathbf  a}^\dagger \cdot\frac{ {\vec  \sigma } } {2} \cdot    {\mathbf a}^{\,}   ~, \qquad  n  \equiv \sum_{i=1}^2 a^\dagger_i a^{\,}_i   \equiv    {\mathbf  a}^\dagger \cdot    {\mathbf a}^{\,}   ~ ,
$$ 
so the quadratic Casimir operator is 
$$
{\vec  J} \cdot {\vec  J} = \frac{n}{2} \left ( \frac{n}{2}+1\right )~. $$ 
You may thus identify n with 2j, and count aggregate modes, whose net number the Js cannot change. They automatically separate (reduce) the subspaces characterized by a given common j=n/2. That is, they connect only states $a_1^{\dagger ~ k} a_2^{\dagger ~ l}|0\rangle$ among themselves, sharing a common $k+l=n=2j$. You may call this "projection" if you focus on  a j of interest to you. 
On to SU(3), now with three fundamental oscillators, so their dot products range over indices from 1 to 3 (N for general SU(N)). The conventionally normalized generators are the eight 
$$
 F_a \equiv    {\mathbf  a}^\dagger \cdot\frac{ {  \lambda_a } } {2} \cdot    {\mathbf a} ~,
$$
where the $\lambda_a$ s are the Gell—Mann matrices, obeying the Fierz identity where 
$$\lambda^{a ~~i} _j   \lambda^{a~~k} _l   = 2 \delta^i_l  \delta^k _l  -\frac{2}{3}  \delta^i_j  \delta^k _l    ~.
$$
The quadratic Casimir then evaluates to (Einstein summation convention over repeated indices!)
$$
 {\mathbf  a}^\dagger \cdot\frac{ {  \lambda_a } } {2} \cdot    {\mathbf a}~~~ {\mathbf  a}^\dagger \cdot\frac{ {  \lambda_a } } {2} \cdot    {\mathbf a}=-\frac{1}{6}  {\mathbf  a}^\dagger      \cdot    {\mathbf a}~~{\mathbf  a}^\dagger      \cdot    {\mathbf a}+\frac{1}{2}a_i^\dagger a_k a^\dagger_k a_i \\
=\frac{1}{3} n(n+3)~.
$$
You may recognize this as the quadratic Casimir $(p+3p)/3$ of p symmetrized quarks' representation, D(p,0), so n is quark number,   without any antiquarks—no columns in the respective Young tableau. The dimension d(p,0) of these representations is (p+1)(p+2)/2 ; hence p=1 for the fundamental triplet, with Casimir eignevalue 4/3. (Your 16/3 is for Gell—Mann matrix bilinears, which are here normalized by a 2 each, to become SU(3) generators.)
To reassure yourself, inspect states for p=2, so d=3, so C=10/3. Further, p=3, so d=10, so C=6, etc.
NB The adjoint, D(1,1) is not in this symmetrized collection--it has an antiquark as well, hence the telltale Γ Young tableau.
For basics of the direct SU(N) generalization, see Iachello, or Greiner-Mueller ...
