# How to re-expand the sum of Schur function?

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

• $N_f$ is just some number here? The notation is confusing to me... Commented Oct 21, 2021 at 20:34
• Yes $N, N_f$ is some integer. Commented Oct 21, 2021 at 20:36
• Yes, you are right. Commented Oct 21, 2021 at 20:43
• Just a piece of advice: You will get better help here if you try to formulate your questions without unnecessary notation (like calling a number $N_f$ instead of just $n$) and physics lingo (Veneziano limit). Commented Oct 22, 2021 at 5:43

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.
• Thank you for the answer. However we interesting in another case, if we put condition $\lambda_1<N$ for $\lambda=(\lambda_1,..,\lambda_l)$ we arrive to another expression at the $h=1$ (unlike the - $\kappa^2 \log [1-h^2]$ that diverges in that point), and principally we can make expansion around $h=1$ . First terms appears in another topic mathoverflow.net/questions/406707/… . Commented Oct 24, 2021 at 14:19
• So you want to put a bound $\lambda_1\leq N$ on the terms and then look at the asymptotics when $N$ and $N_f$ both tend to $\infty$ with $\kappa=N_f/N$ fixed? Commented Oct 24, 2021 at 14:32