How to re-expand the sum of Schur function?

Question

Consider next sum \begin{eqnarray} \label{PF_spindef} Z = \sum_{r=0}^{N N_f} h^{2r} \ Q(r) . \end{eqnarray} and \begin{equation} Q(r) \ = \ \sum_{\sigma \vdash r} s_{\sigma}(1^{N_f}) \ s_{\sigma}(1^{N_f}) \ . \label{QSUN_def} \end{equation} where $s_{\sigma}(1^{N_f})$ is Schur function and $\sigma \vdash r$ run over partition.

Expansion around $h=0$ in the Veneziano limit $N \to \infty$ (where $\frac{N_f}{N }= \kappa $) explicitly gives $$ Z= 1 + h^2 N^2 \kappa^2 + \frac{h^4}{2} N^2 \kappa^2( N^2 \kappa^2 + 1) + \frac{h^6 }{6} N^2 \kappa^2 (N^2 \kappa^2 +1)(N^2 \kappa^2 +2) + ... = e^{ N^2 \kappa^2 \log \frac{1}{1- h^2} } $$ $$ \lim_{N \to \infty} \frac{1}{N^2} Z = \kappa^2 \log \frac{1}{1- h^2} $$

How to re-expand initial $Z$ (in terms of Schur functions) around $h=1$ and takes the same Veneziano limit ?

Answer 1

This is a special case of the generating function for Schur polynomials $$\sum_{\lambda}s_\lambda(x_1,\dots,x_m)s_\lambda(y_1,\dots,y_n)=\prod_{i=1}^m\prod_{j=1}^n\frac 1{1-x_iy_j}.$$ Take $x_1=\dots=x_m=x$ and $y_1=\dots=y_n=1$, gives $$\sum_{\lambda} x^{|\lambda|}s_\lambda(1^m)s_\lambda(1^n)=\frac 1{(1-x)^{mn}}.$$ It seems you take $m=n=N\kappa=N_f$ and $x=h^2$. You have indeed $$\frac 1{N^2}\log \sum_{\lambda}s_\lambda(1^{N_f})^2 h^{2|\lambda|}=\frac 1{N^2}\log \frac 1{(1-h^2)^{N^2\kappa^2}}=\kappa^2\log \frac 1{1-h^2}.$$

Note that $s_\lambda(1^n)$ are the dimensions of the irreducible representations of $\mathrm{GL}(n,\mathbb C)$, so you can probably also give a representation-theoretic proof.