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You can use 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.

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.

You can use 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 $y_1=\dots=y_n=1$, gives $$\sum_{\lambda}s_\lambda(x_1,\dots,x_m)s_\lambda(1^n)=\prod_{i=1}^m\frac 1{(1-x_i)^n}=\sum_{k_1,\dots,k_m=0}^\infty\prod_{i=1}^m\frac{(n)_{k_i}}{k_i!}x_i^{k_i}.$$ Pick out the term of homogeneity $r$ and specialize the $x_j$, $$\sum_{\lambda\vdash r}s_\lambda(1^m)s_\lambda(1^n)=\sum_{k_1+\dots+k_m=r}\prod_{i=1}^m\frac{(n)_{k_i}}{k_i!}=\frac{(nm)_r}{r!}$$ by the multinomial theorem. This gives the generating function $$\sum_{\lambda}s_\lambda(1^m)s_\lambda(1^n)t^{|\lambda|}=\frac 1{(1-t)^{mn}}.$$ It seems you take $m=n=N\kappa=N_f$ and $t=h^2$. 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.

You can use 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.

You can use 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 $y_1=\dots=y_n=1$, gives $$\sum_{\lambda}s_\lambda(x_1,\dots,x_m)s_\lambda(1^n)=\prod_{i=1}^m\frac 1{(1-x_i)^n}=\sum_{k_1,\dots,k_m=0}^\infty\prod_{i=1}^m\frac{(n)_{k_i}}{k_i!}x_i^{k_i}.$$ Pick out the term of homogeneity $r$ and specialize the $x_j$, $$\sum_{\lambda\vdash r}s_\lambda(1^m)s_\lambda(1^n)=\sum_{k_1+\dots+k_m=r}\prod_{i=1}^m\frac{(n)_{k_i}}{k_i!}=\frac{(nm)_r}{r!}$$ by the multinomial theorem. This gives the generating function $$\sum_{\lambda}s_\lambda(1^m)s_\lambda(1^n)t^{|\lambda|}=\frac 1{(1-t)^{mn}}.$$ It seems you take $m=n=N\kappa=N_f$ and $t=h^2$. You have $$\frac 1{N^2}\log \sum_{\lambda}s_\lambda(1^{N_f})^2 h^{2|\lambda|}=\frac 1{N^2}\log \frac 1{(1-h^2)^{N^2\kappa^2}}=\kappa^2\log \frac 1{1-h^2}.$$

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.

You can use 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 $y_1=\dots=y_n=1$, gives $$\sum_{\lambda}s_\lambda(x_1,\dots,x_m)s_\lambda(1^n)=\prod_{i=1}^m\frac 1{(1-x_i)^n}=\sum_{k_1,\dots,k_m=0}^\infty\prod_{i=1}^m\frac{(n)_{k_i}}{k_i!}x_i^{k_i}.$$ Pick out the term of homogeneity $r$ and specialize the $x_j$, $$\sum_{\lambda\vdash r}s_\lambda(1^m)s_\lambda(1^n)=\sum_{k_1+\dots+k_m=r}\prod_{i=1}^m\frac{(n)_{k_i}}{k_i!}=\frac{(nm)_r}{r!}$$ by the multinomial theorem. This gives the generating function $$\sum_{\lambda}s_\lambda(1^m)s_\lambda(1^n)t^{|\lambda|}=\frac 1{(1-t)^{mn}}.$$ It seems you take $m=n=N\kappa=N_f$ and $t=h^2$. 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.