Reflected oscillator pairing and transpose of the long screening

On each finite oscillator degree the monomial pairing is diagonal with nonzero diagonal entries, so it is nondegenerate. Multiplication by $x_m$ transposes to $-2pm{\mathrm{\partial }}_{x_m}$: on monomials the only nonzero comparison is between exponents $r_m+1$ and $r_m+1$, when $(-2pm{)}^{r_m+1}(r_m+1)!=(-2pm)(r_m+1)(-2pm{)}^{r_m}{r}_m!$. Differentiation gives the reverse identity. Hence ${\alpha }_m^{\ast }=-{\alpha }_{-m}$ for $m\mathrm{\ne }0$; the reflected charge gives ${\alpha }_0^{\ast }=2(p-1)-{\alpha }_0$. Different oscillator degrees are orthogonal, and $h(t)=p{t}^2-(p-1)t$ satisfies $h(t)=h(t^{\ast })$.

Transpose the normally ordered quadratic expression for $L_t(n)$. For $n\mathrm{\ne }0$, a term with two nonzero indices becomes the reversed product of the modes with both indices negated. Their new indices sum to $-n\mathrm{\ne }0$, so their commutator is zero by [F1]; reordering them therefore gives exactly the normally ordered quadratic term of index $-n$. The terms with one zero index contribute the correction $-(p-1){\alpha }_{-n}\mathrm{/}p$ from the reflection of ${\alpha }_0$. The transposed improvement term contributes $(p-1)(n+1){\alpha }_{-n}\mathrm{/}(2p)$; the sum of these two corrections is $(p-1)(n-1){\alpha }_{-n}\mathrm{/}(2p)$, exactly the improvement term of $L_{t^{\ast }}(-n)$. For $n=0$, the explicit diagonal form is $L_t(0)={\sum }_{m\ge 1}m{x}_m{\mathrm{\partial }}_{x_m}+h(t)$, which is its own transpose into the reflected space by step 1.1. This proves the claimed adjoint identity for every $n$.

Put $s=(t+1{)}^{\ast }=(p-1)\mathrm{/}p-t-1$. In the explicit field ${\mathcal{E}}_{1,t}(z)$, the creation exponential transposes to the negative annihilation exponential and the annihilation exponential transposes to the positive creation exponential, by step 1.1. Transpose reverses their order. The resulting expression has precisely the creation-then-annihilation order of ${\mathcal{E}}_{1,s}(z^{-1})$. The charge-shift operators have scalar coefficient one. Their zero-mode powers satisfy $2pt=-2-2ps$. Therefore the matrix coefficient identity is $⟨{\mathcal{E}}_{1,t}(z)u,v⟩=⟨u,z^{-2}{\mathcal{E}}_{1,s}(z^{-1})v⟩\mathrm{.}$ Each matrix coefficient against fixed polynomials involves only finitely many creation degrees, so transposition and coefficient extraction are algebraic operations on a Laurent polynomial. The coefficient of $z^{-1}$ on the right is exactly the coefficient of the inverse variable to power $-1$ in ${\mathcal{E}}_{1,s}$; taking that coefficient proves the screening transpose identity. For a homogeneous input of oscillator degree $N$, the residue condition in ${\mathcal{E}}_{1,t}$ is $2pt+c-a=-1$, where creation adds degree $c$ and substitution removes degree $a$. Thus $Q_t$ has shift $c-a=-2pt-1$. Its pairing adjoint reverses that shift, taking degree $N-2pt-1$ to degree $N$, so its shift is $2pt+1$. Since $s=-t-1\mathrm{/}p$, the shift of $Q_s$ is $-2ps-1=2pt+1$, as required.