2
$\begingroup$

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 ?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ $N_f$ is just some number here? The notation is confusing to me... $\endgroup$ Commented Oct 21, 2021 at 20:34
  • $\begingroup$ Yes $N, N_f$ is some integer. $\endgroup$ Commented Oct 21, 2021 at 20:36
  • $\begingroup$ Yes, you are right. $\endgroup$ Commented Oct 21, 2021 at 20:43
  • 4
    $\begingroup$ 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). $\endgroup$ Commented Oct 22, 2021 at 5:43

1 Answer 1

Reset to default
6
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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/… . $\endgroup$ Commented Oct 24, 2021 at 14:19
  • $\begingroup$ 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? $\endgroup$ Commented Oct 24, 2021 at 14:32
  • $\begingroup$ Yes! This is the case. $\endgroup$ Commented Oct 24, 2021 at 14:46

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.