Representations of U(n) using bosons and fermions I would be glad if someone can help me understand the argument in the first paragraph of page 4 of this paper. 
Especially I don't understand their first sentence,
"Using N bosons (fermions) distributed over m states, one can construct completely symmetric (antisymmetric) irreducible representations of the group U(m) associated with Young tableaux with N boxes in a row (column)"
(I am quite familiar with the Quantum Statistics concepts being alluded to but not so much with the Young-Tableux technology being used) 
All I can see is that $U(n)$ can act on the space of $c_i$ and keep the operators defined in their equation 2.9 and 2.10 unchanged. 
Also on page 4 is there a typo in Equation 2.13? 
It doesn't seem to follow from the line previous to it and the line previous to it makes no sense to me. I guess there should have been a "=" between the $\lambda^N$ and $exp$ in the line just before 2.13. 
Even if I make the above "correction" I don't see how 2.13 follows from it. 
 A: I see this paper was written by a physicist.  Here's what he's trying to say.  Let $T$ be a linear operator $V \to V$ with eigenvalues $\lambda_1, ... \lambda_n$.  $T$ defines linear operators $\Lambda^N T : \Lambda^N V \to \Lambda^N V$ and $S^N T : S^N V \to S^N V$ for all $N$.  Let us suppose that $T$ is diagonalizable with eigenvectors $v_1, ... v_n$.  Then a basis of eigenvectors for $\Lambda^N T$ is given by the set of all exterior products of $N$ distinct eigenvectors, and a basis of eigenvectors for $S^N T$ is given by the set of all symmetric products of $N$ eigenvectors.  
It follows that $\text{tr } \Lambda^N T$ is the $N^{th}$ elementary symmetric polynomial in the eigenvalues (this determines the character of the corresponding representation of $U(n)$) and $\text{tr } S^N T$ is the $N^{th}$ complete homogeneous symmetric polynomial in the eigenvalues (same).  If $T$ is the time evolution operator $U(t)$ for a quantum system, then the eigenvalues of $T$ are complex exponentials of the eigenvalues of the Hamiltonian.
I can't help you with the rest; I don't understand the notation.  As far as I can tell, he's just using the identity described in this blog post:
$$\frac{1}{\det(I - Tz)} = \frac{1}{\sum_{n \ge 0} \text{tr } \Lambda^n T z^n} = \sum_{n \ge 0} \text{tr } S^n T z^n = \exp \left( \sum \frac{\text{tr } T^n}{n} z^n \right).$$
This identity expresses, in a way I don't really understand, boson-fermion duality.
