$N=2$ superconformal mode algebra in NS and Ramond sectors

The field [F2] supplies the coefficient scalars for the bilinear brackets. Triples containing only $L$ modes satisfy Jacobi by [F1]. Centrality handles any triple containing $C$.

For the other even triples, $LLJ$ reduces to $p(n+p)-p(m+p)=-p(m-n)$; both sides multiply $J_{m+n+p}$. For $LJJ$, the only possible central coefficient is $-n(m+n)\mathrm{/}3-pn\mathrm{/}3=-n(m+n+p)\mathrm{/}3,$ which vanishes when its Kronecker delta is nonzero. $JJJ$ has only central inner brackets and hence vanishes. These exhaust the even triples not covered by 1.1.

For $LL{G}^q$ with $q=\pm 1$, the coefficient identity is $a(n,r)a(m,n+r)-a(m,r)a(n,m+r)=(m-n)a(m+n,r)\mathrm{.}$ Expansion verifies it. For $LJ{G}^q$, the difference of the two iterated brackets has coefficient $q(a(m,n+r)-a(m,r))=-qn$, equal to the coefficient from $[-n{J}_{m+n},G_r^q]$. For $JJ{G}^q$, the two iterated terms both have coefficient $q^2$ and cancel, while $[J_m,J_n]$ is central, so its bracket with $G_r^q$ vanishes.

With one even mode and two equal-charge odd modes, every term vanishes. For mixed charges, $LGG$ has matching $L$ coefficient $2(a(m,r)+a(m,s))=2(m-r-s)$. Its $J$ coefficient identity is $a(m,r)(m+r-s)+a(m,s)(r-m-s)=-(r-s)(r+s),$ which expands to $s^2-r^2$. When $m+r+s=0$, the central terms agree because, on substituting $s=-m-r$, $a(m,r)b(m+r)+a(m,s)b(r)=\frac{(m\mathrm{/}2-r)((m+r{)}^2-1\mathrm{/}4)+(3m\mathrm{/}2+r)(r^2-1\mathrm{/}4)}{3}=\frac{m^3-m}{6}\mathrm{.}$ This is the central term in $[L_m,2L_{r+s}]$. For $JGG$, the $J$ coefficient on both sides is $2m$; the central coefficients agree since $b(m+r)-b(r)=m(2r+m)\mathrm{/}3=(r-s)m\mathrm{/}3$ on the support $r+s+m=0$. Thus the two-odd cases satisfy Jacobi.

For three odd modes with charges $+,+,-$, the two nonzero cyclic terms have coefficients $2(r-s)$ and $2(s-r)$ on $G_{r+s+t}^{+}$; with charges $+,-,-$, they have coefficients $2(s-t)$ and $2(t-s)$ on $G_{r+s+t}^{-}$. The third term in each case is zero because equal-charge anticommutators vanish. All-same-charge triples vanish term by term. Hence the odd triples satisfy Jacobi, completing the graded Jacobi check.

The adjoint reverses the order of brackets. For $LJ$, $[L_m,J_n{]}^{\ast }=-n{J}_{-m-n}=[J_{-n},L_{-m}]$. For $LG$ it follows from $a(m,r)=-a(-m,-r)$ and $[G_{-r}^{\mp },L_{-m}]=-a(-m,-r)G_{-m-r}^{\mp }$; for $JG$ the charge changes sign together with $G^{+}\leftrightarrow G^{-}$; for $GG$ the two indices are exchanged and $b(-r)=b(r)$. For $LL$, direct substitution gives $[L_{-n},L_{-m}]=(m-n)L_{-m-n}+\frac{(-n{)}^3-(-n)}{12}{\delta }_{m+n,0}C,$ whose central coefficient equals $(m^3-m)\mathrm{/}12$ on $m+n=0$, so $[L_m,L_n{]}^{\ast }=[L_n^{\ast },L_m^{\ast }]$ including the central term. For $JJ$, $[J_{-n},J_{-m}]=(-n){\delta }_{m+n,0}C\mathrm{/}3=m{\delta }_{m+n,0}C\mathrm{/}3$, matching $[J_m,J_n{]}^{\ast }$. Thus the displayed operation is an involutive anti-linear anti-automorphism. In a nonzero $\ast$-representation with $\rho (C)=cI$, the adjoint identity for $C=C^{\ast }$ gives $\bar{c}⟨u,u⟩=c⟨u,u⟩$ for nonzero $u\in \mathcal{D}$, hence $c\in \mathbb{R}$.

The spectral-flow map preserves $[L,L]$ because its additional central term is ${\eta }^{2}m\mathrm{/}3{\delta }_{m+n,0}C$, equal to the shift from $(m-n){\phi }_{\eta }(L_{m+n})$. It preserves $[L,J]$ since on $m+n=0$ both central shifts are $\eta m\mathrm{/}3$, and leaves $[J,J]$ unchanged. For $q=\pm 1$, $[L_m+\eta J_m,G_{r+q\eta }^q]$ has coefficient $a(m,r+q\eta )+q\eta =a(m,r)$, and the $JG$ bracket preserves the charge. For the mixed anticommutator, its noncentral coefficients match because the transformed $J$ coefficient is $(r+\eta )-(s-\eta )=r-s+2\eta$, equal to the shift from $2{\phi }_{\eta }(L)+(r-s){\phi }_{\eta }(J)$. On $r+s=0$, the image central coefficient is $b(r+\eta )=b(r)+\frac{2r\eta +{\eta }^2}{3},$ which equals the central shift from $2{\phi }_{\eta }(L)+(r-s){\phi }_{\eta }(J)$ because $s=-r$. Equal-charge brackets and centrality remain zero and central, respectively. Therefore ${\phi }_{\eta }$ preserves every bracket; the lattice statement follows from adding an integer or half-integer to $r$. It also preserves the *-structure: for example, ${\phi }_{\eta }(G_r^{+}{)}^{\ast }=G_{-r-\eta }^{-}={\phi }_{\eta }(G_{-r}^{-})$, and the $L,J,C$ formulas are similarly compatible.

The barred copy has the same internal Jacobi identities and spectral-flow checks as the unbarred copy. Every cross-copy bracket is zero by definition, so all mixed Jacobi expressions vanish term by term; independent spectral-flow maps on the two copies therefore preserve the full $N=(2,2)$ product algebra.