Simple triplet endpoints and the actual lattice quotient

First prove stability of $S$. For $X\in \{F,H,E\}$, each simple source factor $L(c,a_n)$ and each quotient factor $L(c,b_m)$ define, by [F2], a formal intertwiner after restricting to that source factor and projecting to that quotient factor. When $n\ge 1$, [F1] allows target weights only $a_{n-1},a_n,a_{n+1}$. These do not equal any $b_m$: writing their Kac momenta as $-1-2pk$ and $1-2pm$, equality of squares implies either $k=-m$ or $p(m-k)=1$, impossible for $k\ge 0,m\ge 1,p\ge 2$. Thus all quotient-factor projections vanish. For $n=0$, the only extra numerical channel is $h_{p-1,-1}=b_1$. Let $w=e^{\beta \alpha }$; [A4] gives $Qw=0$. Every coefficient of $Y(F,z)w$ has charge $\beta -1$, whereas the $b_1$ quotient factor has highest charges $\beta$ and $\beta +1$, and Virasoro descendants preserve charge. Hence its $b_1$ projection vanishes. Since $Q$ is a derivation, $Y(H,z)=[Q,Y(F,z)]$ and $Y(E,z)=[Q,Y(H,z)]$. Projection onto the $b_1$ isotypic quotient factor commutes with $Q$ by [A4], so these projections also vanish on $w$. The primary commutator from [F2] propagates each zero projection from $w$ to its Virasoro descendants: commuting past $L(-r)$ introduces only a scalar multiple of the projected field and its derivative, both still zero. Therefore every mode of $F,H,E$ preserves $S$.

Now prove stability of $T$. For $n\ge 2$, the channels from $b_n$ are $b_{n-1},b_n,b_{n+1}$, none equal to any quotient weight $a_k$: equality of the Kac momenta $1-2pj$ and $-1-2pk$ forces either $j=-k$ or $p(j-k)=1$, impossible for $j\ge 1,k\ge 0,p\ge 2$. At $n=1$ the only extra channel is $b_0=a_0$. Write $t=e^{-\alpha \mathrm{/}2}$; the top of this factor is $⟨t,Qt⟩$ and $Q^{2}t=0$ by [A4]. The $a_0$ quotient top has charge $+\frac{1}{2}\alpha$. The fields $Y(F,z)t$ and $Y(F,z)Qt$ have charges $-\frac{3}{2}\alpha$ and $-\frac{1}{2}\alpha$, so both projections to this factor vanish. Let $P$ denote that projection; [A4] gives $PQ=QP$. From $Y(H)=[Q,Y(F)]$ and $Y(E)=[Q,Y(H)]$ it follows, using $Q^{2}t=0$, that $PY(H,z)t=QPY(F,z)t-PY(F,z)Qt=0,PY(H,z)Qt=QPY(F,z)Qt=0,$ $PY(E,z)t=Q^{2}PY(F,z)t-2QPY(F,z)Qt=0,PY(E,z)Qt=Q^{2}PY(F,z)Qt=0.$ The primary commutator propagates these vanishings to the Virasoro descendants in the $b_1$ source factor. Thus every mode of $F,H,E$ preserves $T$.

For $x=2pt$, define the polynomial $f_p(x)={\sum }_{i=1}^{2p-1}(-1{)}^i\left(\genfrac{}{}{0px}{}{2p}{i}\right)\left(\genfrac{}{}{0px}{}{x}{2p-1-i}\right)\left(\genfrac{}{}{0px}{}{x}{i-1}\right)\mathrm{.}$ The coefficient of the bare target vacuum $e^{(t+1)\alpha }$ in $Y(E,z)e^{t\alpha }$ is a single homogeneous power of $z$ times $f_p(2pt)$. Indeed, insert $E=Q^{2}F={\mathrm{Res}}_u{\mathrm{Res}}_{v}Y(e^{\alpha },u)Y(e^{\alpha },v)e^{-\alpha }$ in the actual normal-order formula. After translating the two screening positions relative to the field position and factoring its homogeneous power, the scalar residue is ${\mathrm{Res}}_u{\mathrm{Res}}_v(uv{)}^{-2p}(u-v{)}^{2p}(1+u{)}^x(1+v{)}^x\mathrm{.}$ The factors are respectively the two screen-to-$F$ contractions, the screen-to-screen contraction, and the contractions with the input charge $t$; taking the oscillator-vacuum coefficient removes all creation terms. Expanding $(u-v{)}^{2p}$ and extracting $u^{-1}{v}^{-1}$ gives exactly $f_p(x)$; the generalized binomials are polynomials in $x$.

The screening string spaces have dimensions $m+1$ and bases $z_j=Q^j{z}_0$ for $0\le j\le m$. The state $H=QF$ is $S_{2p-1}({\alpha }_{-1},{\alpha }_{-2},\dots )e^0$, where ${\sum }_{k\ge 0}{S}_k{u}^k=\exp ({\sum }_{r\ge 1}{\alpha }_{-r}{u}^r\mathrm{/}r)$. In its field on a Fock vacuum $e^{t\alpha }$, the oscillator-free part of that generating series replaces ${\alpha }_{-r}$ by the zero-mode contribution $(-1{)}^{r-1}(2pt)z^{-r}$, giving $(1+u\mathrm{/}z{)}^{2pt}$. Therefore $H_0{e}^{t\alpha }=\left(\genfrac{}{}{0px}{}{2pt}{2p-1}\right)e^{t\alpha },H_0:=H_{(2p-2)}\mathrm{.}$ The screening commutators give $[Q,F_0]=H_0$, $[Q,H_0]=E_0$, $[Q,E_0]=0$, with $X_0=X_{(2p-2)}$. Project each weight-preserving generator mode to a fixed Virasoro isotypic summand and restrict to its highest multiplicity space; [A4] makes these projections commute with $Q$. Charge shifts are $-1,0,+1$, so write $F_0{z}_j=f_j{z}_{j-1}$, $H_0{z}_j=h_j{z}_j$, and $E_0{z}_j=e_j{z}_{j+1}$, with absent end vectors zero and $f_0=0$.

Virasoro already preserves $S,T$. For either subspace, the set of ambient states whose modes preserve it contains $\mathbf{1},\omega ,F,H,E$ and is closed under every state product by the module iterate identity; each iterate sum is finite on a fixed vector by truncation. It is therefore a vertex subalgebra acting on that subspace. In particular $S,T$ are $W$-submodules, and $B\mathrm{/}S$ is an actual quotient $W$-module.

The degree of $f_p$ is at most $2p-2$. For integers $x=0,1,\dots ,p-2$, the polynomial $(u-v{)}^{2p}(1+u{)}^x(1+v{)}^x$ has total degree at most $2p+2x\le 4p-4$, below the degree $4p-2$ required to extract $u^{2p-1}{v}^{2p-1}$, so $f_p(x)=0$. In the residue above make the invertible formal substitutions $u=-r\mathrm{/}(1+r)$ and $v=-s\mathrm{/}(1+s)$. Residue invariance and the Jacobians transform its integrand to $(rs{)}^{-2p}(r-s{)}^{2p}(1+r{)}^{-x-2}(1+s{)}^{-x-2},$ so $f_p(x)=f_p(-x-2)$ for these integer inputs. Thus $-2,-3,\dots ,-p$ are also roots. At $x=p-1$ only the summand $i=p$ is nonzero, with value $(-1{)}^p\left(\genfrac{}{}{0px}{}{2p}{p}\right)$. The $2p-2$ distinct roots and this value give $f_p(x)=\frac{(-1{)}^{p}p\left(\genfrac{}{}{0px}{}{2p}{p}\right)}{(2p-1)!}{\prod }_{r=0}^{p-2}(x-r){\prod }_{s=2}^p(x+s)\mathrm{.}$ It follows that in $S$ the $E$-mode taking $e^{(\beta -n)\alpha }$ to $e^{(\beta -n+1)\alpha }$ has nonzero coefficient $f_p(p-2-2pn)$ for every $n\ge 1$; in $T$ the corresponding coefficient is $f_p(p-2pn)$, nonzero for $n\ge 2$ and zero at the boundary $n=1$; and in $B\mathrm{/}S$ the $E$-mode taking $[e^{(\beta +n)\alpha }]$ to $[e^{(\beta +n+1)\alpha }]$ has nonzero coefficient $f_p(p-2+2pn)$ for all $n\ge 1$. These are nonzero because their arguments lie outside the root set just found.

The actual endpoint equations are as follows. From $[Q,E_0]=0$ we get $e_j=e_{j+1}$ only for $0\le j\le m-2$, so $e_0=\cdots =e_{m-1}=:e$; at $j=m-1$ the commutator is zero on both sides and gives no extra boundary equation. From $[Q,H_0]=E_0$ we get $h_j-h_{j+1}=e$ for $0\le j\le m-1$. From $[Q,F_0]=H_0$ we get $f_j-f_{j+1}=h_j$ for $0\le j<m$, with $f_0=0$, and at the top $j=m$ it gives $h_m=f_m$. Write $c_0=h_0$. Summing the recurrences yields $h_j=c_0-je$, $f_m=-m{c}_0+em(m-1)\mathrm{/}2$, and the top equation $h_m=f_m$ gives $e=2c_0\mathrm{/}m$. Hence $F_0{z}_j=-\frac{c_0}{m}j(m-j+1)z_{j-1},H_0{z}_j=c_0(1-\frac{2j}{m})z_j,E_0{z}_j=\frac{2c_0}{m}{z}_{j+1}\mathrm{.}$ If $c_0\mathrm{\ne }0$, the $H_0$ eigenvalues are distinct and each adjacent $E_0,F_0$ coefficient is nonzero. Polynomial spectral projections in $H_0$ isolate any nonzero basis component of an invariant subspace, and the adjacent shifts then generate every $z_j$; thus this zero-mode representation is irreducible. For $S$ at $n\ge 1$, $m=2n$ and $c_0=\left(\genfrac{}{}{0px}{}{p-2-2pn}{2p-1}\right)\mathrm{\ne }0$. For $T$ at $n\ge 1$, $m=2n-1$ and $c_0=\left(\genfrac{}{}{0px}{}{p-2pn}{2p-1}\right)\mathrm{\ne }0$. For $B\mathrm{/}S$ at $n\ge 1$, $m=2n-1$ and terminal $z_m=[e^{(\beta +n)\alpha }]$ has $H_0$ eigenvalue $\left(\genfrac{}{}{0px}{}{p-2+2pn}{2p-1}\right)\mathrm{\ne }0$; the formula gives $h_m=-c_0$, so $c_0$ is the negative of this value and is nonzero. The $n=0$ top of $S$ is one-dimensional.

Let $R$ be a nonzero $W$-submodule of $S,T$, or $B\mathrm{/}S$. These modules are algebraic direct sums of simple Virasoro modules with diagonal $L(0)$ and a lower-bounded, finite-dimensional-at-each-grade decomposition by [A3]–[A4]. Since every vector has finitely many weight components, polynomial interpolation in $L(0)$ shows that $R$ is graded. Choose its least occupied weight $h$. Every vector in $R_h$ is Virasoro-highest because positive Virasoro modes lower weight and $R$ has no lower occupied weight. For primary $X\in \{F,H,E\}$, the commutator of $L(n)$ with $X_0$ is a scalar multiple of a mode of $X$ of degree $-n$; its value on a vector of $R_h$ lies in $R_{h-n}=0$. Thus each $X_0$ preserves $R_h$. In the simple Virasoro decomposition, the highest vectors at that weight are exactly one index's displayed multiplicity space; no positive-grade singular vectors occur in a simple factor. Therefore $R_h$ is a nonzero invariant subspace for the corresponding projected matrices in step 2.3, and equals the whole multiplicity space. If this is index $n\ge 1$ in $S$, $R$ contains $e^{(\beta -n)\alpha }$ and the nonzero lowering mode from step 2.2 gives $e^{(\beta -n+1)\alpha }$ at the lower weight $a_{n-1}$, contradiction. If it is index $n\ge 2$ in $T$, the corresponding lowering mode gives the lower $b_{n-1}$ vector, also contradiction. If it is index $n\ge 2$ in $B\mathrm{/}S$, $R$ contains the terminal class $[e^{(\beta +n)\alpha }]$ and the leading coefficient of $Y(F,z)e^{(\beta +n)\alpha }$ is the nonzero class $[e^{(\beta +n-1)\alpha }]$ at weight $b_{n-1}$, again a contradiction. Hence every nonzero submodule meets the lowest multiplicity space.

Let $A=\left(\genfrac{}{}{0px}{}{3p-2}{2p-1}\right)>0$. In the $b_1$ bottom of $T$, the basis $(t,Qt)$ has $H_0$ eigenvalues $(-A,A)$, $F_0(Qt)=At$, and $E_0(t)=-2AQt$, with the other adjacent actions zero; this is step 2.3 with $m=1$ and $\left(\genfrac{}{}{0px}{}{-p}{2p-1}\right)=-A$. In the $b_1$ bottom of $B\mathrm{/}S$, choose $u$ with $Qu=v=[e^{(\beta +1)\alpha }]$, possible by the length-two $Q$ chain in [A4]. Since $2p(\beta +1)=3p-2$, $H_{0}v=Av$; the same recurrences give the identical matrices on $(u,v)$. Both bottoms have the same scalar $L(0)$ weight $b_1$. The strong-generator zero-mode reduction [F4] therefore makes the degree-zero action of every state of $W$ identical on these two bottom spaces.

Each of these bottom spaces generates its whole module. The lowest weights increase strictly: $a_{n+1}-a_n=p(2n+1)+1>0$ and $b_{n+1}-b_n=p(2n+1)-1>0$ in their indicated ranges. For $S$ and $T$, the leading coefficient of $Y(F,z)e^{t\alpha }$ is $e^{(t-1)\alpha }$ with coefficient one, producing the next bare highest vector as the index increases. For $B\mathrm{/}S$, the nonzero $E$ coefficient in step 2.2 sends each terminal bare class to the next terminal class. Inductively suppose all lower-index Virasoro factors have been generated. The produced vector is a highest vector at the next index. Applying $F_0,H_0,E_0$ to it gives vectors whose components lie in the already-generated lower-index descendants and the new highest multiplicity space; no higher-index factor can occur at that weight. Subtract the lower-index components, which are already in the generated submodule. The irreducible zero-mode matrices of step 2.3 then generate the complete new multiplicity space, and negative Virasoro modes generate all its simple factors. This induction proves generation by the bottom in each case. By step 3.1, every nonzero submodule contains the whole bottom; it then equals the full module. Thus $S,T,B\mathrm{/}S$ are irreducible as $W$-modules, and $S=W{e}^{\beta \alpha }=N_2$, $T=W{e}^{-\alpha \mathrm{/}2}=\mathrm{\Pi }(1)$.

Write $J=B\mathrm{/}S$, and identify the actual two-dimensional bottoms by the map $\varphi :T_{b_1}\to J_{b_1}$ matching $(t,Qt)$ to $(u,v)$ in step 3.2. This map intertwines the four generator zero modes by their explicit matrices, hence the full bottom action by [F4]. In the actual direct-sum module $T\oplus J$, put $\mathrm{\Delta }=\{(x,\varphi x):x\in T_{b_1}\}$ and choose a nonzero $\xi \in \mathrm{\Delta }$. Give both summands the inherited nonnegative integer grade $L(0)-b_1$, and let $R=W\xi$. The matched generator matrices preserve $\mathrm{\Delta }$ and make it irreducible, so their finite zero-mode words on $\xi$ span $\mathrm{\Delta }$. To verify that [F4] applies to $R$, note that $R$ is a $W$-submodule and $\omega \in W$, so $L(0)={\omega }_{(1)}$ preserves $R$. Each vector $r\in T\oplus J$ has finite support in the ambient grading. If $r={\sum }_{k\in K}{r}_k$ with finite $K\subseteq {\mathbb{Z}}_{\ge 0}$ and $r_k$ of grade $k$, then for each $k\in K$ the polynomial $P_k(t)={\prod }_{\begin{array}{c}\mathrm{\ell }\in K\\ \mathrm{\ell }\mathrm{\ne }k\end{array}}\frac{t-(b_1+\mathrm{\ell })}{k-\mathrm{\ell }}\mathrm{.}$ satisfies $P_k(L(0))r=r_k$. Since $R$ is $L(0)$-invariant, $r_k\in R$. Thus $R={⨁}_{k\ge 0}R(k),R(k):=R\cap (T\oplus J)(k)\mathrm{.}$ For homogeneous $a\in W$ of weight $d$, the module commutator with $L(0)$ gives $a_{(m)}R(k)\subseteq R(k+d-m-1)$, with negative grades zero. Consequently $R$ is an admissibly graded $W$-module, cyclic on $\xi \in R(0)$, and [F4] applies. It shows that $R(0)$ is precisely the span of generator zero-mode words on $\xi$. Those words span $\mathrm{\Delta }$, so $R(0)=\mathrm{\Delta }$. Both coordinate projections $R\to T$ and $R\to J$ are module maps with nonzero image, and are surjective because their actual targets are simple by step 4.1. The kernel of the first is a submodule of $0\oplus J$; if it were nonzero, simplicity of $J$ would make it $0\oplus J$, whose nonzero bottom vectors contradict $R_{b_1}=\mathrm{\Delta }$ and $\mathrm{\Delta }\cap (0\oplus J_{b_1})=0$. Thus the first projection is an isomorphism. In particular $R$ is simple, so the nonzero second projection is an isomorphism as well. Composing the second projection with the inverse of the first gives $T\cong J=B\mathrm{/}S$ as actual $W$-modules. Finally $a_0=(2-p)\mathrm{/}4$ and $b_1-a_0=p-1$; [A4] gives the one-dimensional $N_2$ bottom and the two-dimensional $b_1$ bottoms at the claimed common bottom degree.