The compact-boson lattice is even, unimodular, and mutually local

For $q=(m,n)$ and $q^{\mathrm{\prime }}=(m^{\mathrm{\prime }},n^{\mathrm{\prime }})$, expansion of the two squares gives $B({\lambda }_r(q),{\lambda }_r(q^{\mathrm{\prime }}))=p_L(q;r)p_L(q^{\mathrm{\prime }};r)-p_R(q;r)p_R(q^{\mathrm{\prime }};r)=m{n}^{\mathrm{\prime }}+n{m}^{\mathrm{\prime }}$. In the basis $e_1={\lambda }_r(1,0)$, $e_2={\lambda }_r(0,1)$, the Gram matrix is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Thus every pairing is integral and $B({\lambda }_r(q),{\lambda }_r(q))=2mn\in 2\mathbb{Z}$. If $x=a{e}_1+b{e}_2$ pairs integrally with both basis vectors, then $B(x,e_1)=b\in \mathbb{Z}$ and $B(x,e_2)=a\in \mathbb{Z}$; hence $x\in {\mathrm{\Gamma }}_r$. This proves self-duality. For the oscillator calculation, the unbarred modes satisfy the relation in [F2]. Realize the independent right-moving copy on $\bar{P}=\mathbb{C}[{\bar{x}}_1,{\bar{x}}_2,\dots ]$ by ${\bar{a}}_{-k}g={\bar{x}}_{k}g$, ${\bar{a}}_{k}g=k\mathrm{\partial }g\mathrm{/}\mathrm{\partial }{\bar{x}}_k$ for $k\ge 1$, and ${\bar{a}}_{0}g=p_{R}g$. The same multiplication-differentiation calculation gives $[{\bar{a}}_m,{\bar{a}}_n]=m{\delta }_{m+n,0}I$; on $P\otimes \bar{P}$, the unbarred and barred modes act on separate factors, so $[a_m,{\bar{a}}_n]=0$.

Bilinearity of the exponent gives $\epsilon (q,q^{\mathrm{\prime }})\epsilon (q+q^{\mathrm{\prime }},q^{\mathrm{\prime }\mathrm{\prime }})=(-1{)}^{m{n}^{\mathrm{\prime }}+(m+m^{\mathrm{\prime }})n^{\mathrm{\prime }\mathrm{\prime }}}=(-1{)}^{m^{\mathrm{\prime }}{n}^{\mathrm{\prime }\mathrm{\prime }}+m(n^{\mathrm{\prime }}+n^{\mathrm{\prime }\mathrm{\prime }})}=\epsilon (q^{\mathrm{\prime }},q^{\mathrm{\prime }\mathrm{\prime }})\epsilon (q,q^{\mathrm{\prime }}+q^{\mathrm{\prime }\mathrm{\prime }})$. Therefore the charge-shift products are associative.

For $u=w\mathrm{/}z$ with $\mathrm{\mid }u\mathrm{\mid }<1$, the unbarred and barred oscillator commutators from [F1] and [F2] give $E_s^{+}(z)E_t^{-}(w)=E_t^{-}(w)E_s^{+}(z)\exp (-st{\sum }_{k\ge 1}{u}^k\mathrm{/}k)=E_t^{-}(w)E_s^{+}(z)(1-u{)}^{st}$. The scalar series is normally convergent on every compact subset of $\mathrm{\mid }u\mathrm{\mid }<1$, since ${\sum }_{k\ge 1}\mathrm{\mid }u{\mathrm{\mid }}^k\mathrm{/}k$ converges uniformly when $\mathrm{\mid }u\mathrm{\mid }\le c<1$. For fixed polynomial input, each annihilation exponential has finite differential action; pairing against a fixed polynomial output selects only finitely many creation coefficients. Hence every finite-energy matrix coefficient of an ordered product is a finite combination of derivatives of these normally convergent contraction factors and polynomial powers. It is absolutely and locally uniformly convergent on compact nested radial domains. Moving the zero-mode factors through the charge shifts supplies $z^{st}$ and its barred analogue, so the leading factor is $(z-w{)}^{st}(\bar{z}-\bar{w}{)}^{s^{\mathrm{\prime }}{t}^{\mathrm{\prime }}}$ on the branches in [F1]. The remaining normal-ordered exponential is the field of charge $q+q^{\mathrm{\prime }}$ and its descendants.

Exchanging two insertions by a half-turn multiplies the left/right branch factors by $e^{\pi iB(q,q^{\mathrm{\prime }})}$. The cocycle ratio is $\epsilon (q,q^{\mathrm{\prime }})\mathrm{/}\epsilon (q^{\mathrm{\prime }},q)=(-1{)}^{m{n}^{\mathrm{\prime }}-m^{\mathrm{\prime }}n}=(-1{)}^{m{n}^{\mathrm{\prime }}+n{m}^{\mathrm{\prime }}}$, and $B(q,q^{\mathrm{\prime }})=m{n}^{\mathrm{\prime }}+n{m}^{\mathrm{\prime }}\in \mathbb{Z}$, so the cocycle and branch factors multiply to $1$. Thus the finite-energy radial matrix coefficients agree after analytic continuation exchanging the insertions away from the diagonal. The field spin is $\frac{1}{2}(p_L^2-p_R^2)=mn\in \mathbb{Z}$. The two-cocycle identity in step 2.1 and the normally convergent contraction factors in step 3.1 make finite iterated radial OPEs independent of bracketing on nested radial domains.

With $r^{\mathrm{\prime }}=(2r{)}^{-1}$ and $T(m,n)=(n,m)$, $p_L(Tq;r^{\mathrm{\prime }})=n\mathrm{/}(2r^{\mathrm{\prime }})+m{r}^{\mathrm{\prime }}=nr+m\mathrm{/}(2r)=p_L(q;r)$ and $p_R(Tq;r^{\mathrm{\prime }})=nr-m\mathrm{/}(2r)=-p_R(q;r)$. The phase $\chi (q)=(-1{)}^{mn}$ satisfies $\chi (q+q^{\mathrm{\prime }})\mathrm{/}(\chi (q)\chi (q^{\mathrm{\prime }}))=(-1{)}^{m{n}^{\mathrm{\prime }}+m^{\mathrm{\prime }}n}=\epsilon (Tq,T{q}^{\mathrm{\prime }})\mathrm{/}\epsilon (q,q^{\mathrm{\prime }})$. Consequently $V_q\mapsto \chi (q)V_{Tq}$ together with sign reversal of every right-moving oscillator preserves the OPE, the vacuum, and both conformal weights. It is bijective because $T$ is an involution of ${\mathbb{Z}}^2$.