Complete boundary Fock submodules, quotients and screening chains

Write $B_k={\mathcal{F}}_{\beta +k}$ and $D_k={\mathcal{F}}_{\delta +k}$. For a fixed $B_k$, the allowed terms in $S_B$ have $n\ge \mathrm{\mid }k\mathrm{\mid }$ and are generated by $Q^{n+k}{e}^{(\beta -n)\alpha }$. For a fixed $D_k$, the allowed terms in $S_D$ have $n\ge \max (-k,k+1)$ and are generated by $Q^{n+k}{e}^{(\delta -n)\alpha }$. For fixed $n$, the B terms have charges $k=-n,-n+1,\dots ,n$, giving $2n+1$ factors; the D terms have $k=-n,-n+1,\dots ,n-1$, giving $2n$ factors. Their screening powers are nonzero by [F3]. By [F1], each map $Q^j$ commutes with the Virasoro action, and by [F2] its source cyclic module is simple; since $Q^j$ sends the vacuum to a nonzero vector by [F3], its kernel is zero. Its image is exactly the cyclic module generated by $Q^j{e}^{(\beta -n)\alpha }$ or $Q^j{e}^{(\delta -n)\alpha }$, so each displayed copy is simple. In a fixed charge, distinct $n$ have distinct highest weights $a_n$ (respectively $b_n$), so those factors are nonisomorphic; copies at different charges lie in different direct summands of the charge decomposition. A sum of pairwise nonisomorphic simple submodules in a fixed charge is direct: for a finite sum this follows by projecting a simple intersection to its direct summands; every vector belongs to a finite sum, so the infinite sum is direct. Counting the allowed charges gives multiplicities $2n+1$ and $2n$.

The pairing [F4] identifies each $B_k$ with the restricted graded dual of $D_{-k}$, because the reflected charge of $\beta +k$ is $\delta -k$. Every submodule of a restricted graded dual is graded: each vector has finite grade support, and polynomial interpolation in $L_0$ projects onto each grade. A restricted graded dual of a simple highest-weight module with finite-dimensional graded pieces is simple: a nonzero proper dual submodule has a nonzero proper gradewise annihilator in the original module, by finite-dimensional nondegenerate duality, contradicting simplicity. The functional on the original lowest grade is the dual top: the adjoint identity makes every positive mode annihilate it, and its $L_0$ weight is unchanged. The weight sets $\{a_n:n\ge 0\}$ and $\{b_m:m\ge 1\}$ are disjoint: equality would imply $p(m-n)=d$ or $m+n=0$, impossible for $1\le d<p$. A nonzero restricted pairing would define a Virasoro homomorphism from one simple factor to the restricted dual of the other. Such a homomorphism between simple modules is zero unless they are isomorphic, which is impossible for distinct highest weights. Thus every simple summand of $S_B\cap B_k$ is orthogonal to every simple summand of $S_D\cap D_{-k}$.

A monomial ${\prod }_m{x}_m^{k_m}{e}^{t\alpha }$ has oscillator degree ${\sum }_{m}m{k}_m$, and these monomials are in bijection with partitions of that degree. Thus [F5] gives $\mathrm{ch}{\mathcal{F}}_t=q^{h(t)}R(q)$, where $R(q)={\prod }_{m\ge 1}(1-q^m{)}^{-1}$. For $k\ge 1$, both paired socles have indices $n\ge k$, and [F2] gives their combined character ${\prod }_{m\ge 1}(1-q^m{)}^{-1}{\sum }_{n\ge k}(q^{a_n}-q^{b_{n+1}}+q^{b_n}-q^{a_n})=q^{b_k}{\prod }_{m\ge 1}(1-q^m{)}^{-1}=\mathrm{ch}{B}_k\mathrm{.}$ For $k\le 0$, the two ranges are $n\ge -k$ and $n\ge 1-k$, respectively, so their character sum is $R(q){\sum }_{n\ge -k}(q^{a_n}-q^{b_{n+1}})+R(q){\sum }_{n\ge 1-k}(q^{b_n}-q^{a_n})=q^{a_{-k}}R(q)=\mathrm{ch}{B}_k$. These are coefficientwise finite sums because $a_n,b_n$ grow quadratically. The orthogonality in step 2.1 gives $S_B\cap B_k\subseteq {\mathrm{ann}}_{B_k}(S_D\cap D_{-k})$. The character equality and finite-dimensional nondegenerate pairing in every grade give equality; interchanging $B,D$ gives the other annihilator equality. Thus $B_k\mathrm{/}(S_B\cap B_k)\cong (S_D\cap D_{-k}{)}^{\vee },D_{-k}\mathrm{/}(S_D\cap D_{-k})\cong (S_B\cap B_k{)}^{\vee }\mathrm{.}$ The restricted duals preserve the direct-sum decompositions because only finitely many highest weights occur in each fixed grade. This proves both quotient decompositions with the stated multiplicities.

Under transposition of $Q$ in [F4], each dual of a screening string in $S_D\cap D_{-k}$ becomes the reverse chain of highest vectors in $B\mathrm{/}{S}_B$. For fixed $n\ge 1$, these occur at charges $\beta +n-j$ for $0\le j\le 2n-1$, so there are $2n$ entries; the terminal bare exponential $e^{(\beta +n)\alpha }$ has weight $b_n$ and cannot lie in $S_B$, whose summands in that charge begin at weights $a_m$ for $m\ge n$. Direct subtraction gives $a_{m+1}-a_m=2pm+d+p>0$ and $a_n-b_n=2nd>0$, so all these weights exceed $b_n$. Hence its quotient class is nonzero and is the terminal vector. Interchanging the two paired spaces gives the $2n+1$-entry chain in $D\mathrm{/}{S}_D$, including $n=0$.

For each quotient highest vector, choose the dual functional on the corresponding simple summand of the opposite socle to take a nonzero value on the specified screening string vector, and zero on the other summands in that grade; lift it homogeneously through the annihilator quotient. This gives lifts $w_B(j,n)$ and $w_D(j,n)$ of the charges and weights in the Statement. Transposing $Q^j$ shows that $Q^j{w}_B(j,n)$ pairs nontrivially with $e^{(\delta -n)\alpha }$. It has charge $\beta +n$; degreewise nondegeneracy and its nonzero pairing with the grade-zero vector $e^{(\delta -n)\alpha }$ force it to have oscillator grade zero. Its weight is therefore the vacuum weight $b_n$ at that charge, so it lies in the one-dimensional line $\mathbb{C}{e}^{(\beta +n)\alpha }$. Rescale to obtain the stated equality. The same argument gives $Q^j{w}_D(j,n)=e^{(\delta +n)\alpha }$. Their classes generate each simple quotient summand; together with the socle they generate the whole Fock direct sums.