Exact lengths of boundary long-screening strings

Commuting each annihilation substitution through the later creation exponential contributes $\exp (-2p{\sum }_{m\ge 1}(z_s\mathrm{/}{z}_r{)}^m\mathrm{/}m)=(1-z_s\mathrm{/}{z}_r{)}^{2p}$. The subsequent charge shift contributes $z_r^{2p}$ for that pair. Consequently repeated substitution in [F1] gives the exact finite-residue identity $Q^j{e}^{t\alpha }={\mathrm{Res}}_{z_1}\cdots {\mathrm{Res}}_{z_j}{\mathrm{\Delta }}_j(z{)}^{2p}{\prod }_r{z}_r^{2pt}\exp \left({\sum }_{m\ge 1}\frac{x_m}{m}{\sum }_r{z}_r^m\right)e^{(t+j)\alpha }\mathrm{.}$ Only the symmetric homogeneous degree $D=j(-2pt-1)-pj(j-1)$ in the exponential can contribute. Its oscillator coefficients span the symmetric polynomials of degree $D$ in the $j$ variables. Indeed, the coefficient of ${\prod }_m{x}_m^{q_m}$, where ${\sum }_{m}m{q}_m=D$, is ${\prod }_m{p}_m(z{)}^{q_m}\mathrm{/}(m^{q_m}{q}_m!)$. Products of power sums span every orbit sum in [L1]: in $p_{{\lambda }_1}\cdots p_{{\lambda }_{\mathrm{\ell }}}$ the terms using $\mathrm{\ell }$ distinct variables give a positive integer multiple of the orbit sum for $\lambda$, while every term with repeated variables has fewer than $\mathrm{\ell }$ distinct variables and is an orbit sum for a partition obtained by merging parts. Induction on $\mathrm{\ell }$ therefore solves for every orbit sum as a linear combination of products of power sums. The degree-$D$ oscillator space is finite-dimensional, so a linear functional on it can select any desired symmetric polynomial from those coefficient functions.

In the first family take $j=2n>0$, $t=\beta -n$; then $D=jd$. Use the symmetric polynomial $({\prod }_r{z}_r{)}^d$ in step 1.1. In the second family take $j=2n-1$, $t=\delta -n$; then $D=j(i+1)$, and use $({\prod }_r{z}_r{)}^{i+1}$. Both resulting residues equal the constant term of ${\mathrm{\Delta }}_j(z{)}^{2p}{\prod }_r{z}_r^{-p(j-1)}$. If $V(z)={\mathrm{\Delta }}_j(z{)}^p={\sum }_{\mu }{c}_{\mu }{z}^{\mu }$, reciprocal substitution gives ${\mathrm{\Delta }}_j(z{)}^{2p}{\prod }_r{z}_r^{-p(j-1)}=(-1{)}^{pj(j-1)\mathrm{/}2}V(z)V(z^{-1})\mathrm{.}$ The constant term is $(-1{)}^{pj(j-1)\mathrm{/}2}{\sum }_{\mu }{c}_{\mu }^2\mathrm{\ne }0$, since $V$ is a nonzero polynomial with real coefficients. This proves both asserted nonvanishing claims, including $n=0$ in the first family because $Q^0$ is the identity.

Give a homogeneous input polynomial oscillator degree $N$. In a term of the creation exponential, choosing powers with total oscillator degree $c$ adds total $z$-degree $c$; choosing substitution terms that lower oscillator degree by $a$ adds total $z$-degree $-a$. The coefficient selected by $Q_t={\mathrm{Res}}_z{\mathcal{E}}_{1,t}(z)$ satisfies $2pt+c-a=-1$, hence $a-c=2pt+1$ and the output degree is $N-a+c=N-2pt-1$. Since $h(t+1)-h(t)=2pt+1$, $Q_t$ preserves the explicit weight $N+h(t)$. Put $a_n=((2pn+d{)}^2-(p-1{)}^2)\mathrm{/}(4p)$ and $b_n=((2pn-d{)}^2-(p-1{)}^2)\mathrm{/}(4p)$. The first family's initial weight is $a_n$; its charge after $2n+1$ steps would be $\beta +n+1$, whose minimum weight is $b_{n+1}$. But $b_{n+1}-a_n=(i+1)(2n+1)>0$, so this output is zero. The second family's weight is $b_n$; after $2n$ steps its charge would be $\delta +n$, whose minimum weight is $a_n$, and $a_n-b_n=2nd>0$. This output is also zero. If any earlier power vanished, all its later composites would vanish, contradicting step 2.1. This proves all claimed exact lengths.