Perturbative Narain O Dd Lattice Duality

Put $T=\left(\begin{array}{cc}I& 0\\ -B& I\end{array}\right)$ and $D=\mathrm{diag}(G,G^{-1})$. Direct block multiplication gives $\mathcal{M}(E)=T^{T}DT$. The matrix $T$ is a real O$(d,d)$ $B$-shift because $T^T\eta T=\eta$; since it is invertible, this also gives $T\eta T^T=\eta$. Also $D\eta D=\eta$. Hence $\mathcal{M}(E)$ is symmetric and $\mathcal{M}(E)\eta \mathcal{M}(E)=T^{T}D(T\eta T^T)DT=T^T\eta T=\eta$. Since $G$ and $G^{-1}$ are positive definite and $T$ is invertible, $x^T\mathcal{M}(E)x=(Tx{)}^{T}D(Tx)>0$ for every nonzero $x$. By [F3], for $Z=(m,n)$ the lattice norm is $Z^T\eta Z=2m\cdot n$, an even integer. The vectors $(v,v)$ and $(v,-v)$ are $+1$ and $-1$ eigenspaces of $\eta$, each of dimension $d$, so its signature is $(d,d)$; $\mathrm{\mid }\det \eta \mathrm{\mid }=1$, so the integral lattice is unimodular. For $d=1$, $B=0$ and [F2] gives $h_0=\frac{1}{4}(G{m}^2+G^{-1}{n}^2+2mn)=\frac{1}{4}(\sqrt{G}m+n\mathrm{/}\sqrt{G}{)}^2$ and ${\bar{h}}_0=\frac{1}{4}(G{m}^2+G^{-1}{n}^2-2mn)=\frac{1}{4}(\sqrt{G}m-n\mathrm{/}\sqrt{G}{)}^2$. With $m=w$ and $G=R^2\mathrm{/}{\alpha }^{\mathrm{\prime }}$, these are ${\alpha }^{\mathrm{\prime }}{p}_L^2\mathrm{/}4$ and ${\alpha }^{\mathrm{\prime }}{p}_R^2\mathrm{/}4$ for the [F1] momenta.

Write $F=cE+d_0$. From $g^T\eta g=\eta$, its block identities include $a^{T}c+c^{T}a=0$, $b^T{d}_0+d_0^{T}b=0$, and $a^T{d}_0+c^{T}b=I_d$. Expanding $F^T(E^{\mathrm{\prime }}+E^{\mathrm{\prime }T})F=(cE+d_0{)}^T(aE+b)+(aE+b{)}^T(cE+d_0)$ and using these identities reduces the right side to $E+E^T=2G$: the quadratic coefficient is $c^{T}a+a^{T}c=0$, the two mixed coefficients are $c^{T}b+a^T{d}_0=I_d$ and $d_0^{T}a+b^{T}c=I_d$, and the constant coefficient is $d_0^{T}b+b^T{d}_0=0$. Thus $G^{\mathrm{\prime }}=(E^{\mathrm{\prime }}+E^{\mathrm{\prime }T})\mathrm{/}2=F^{-T}G{F}^{-1}>0$, while $B^{\mathrm{\prime }}=(E^{\mathrm{\prime }}-E^{\mathrm{\prime }T})\mathrm{/}2$ is antisymmetric. For the generalized metric define $S(E)=\eta \mathcal{M}(E)=\left(\begin{array}{cc}-G^{-1}B& G^{-1}\\ G-B{G}^{-1}B& B{G}^{-1}\end{array}\right)$. Direct block multiplication using $E=G+B$ gives $S(E)\left(\genfrac{}{}{0px}{}{I}{E}\right)=\left(\genfrac{}{}{0px}{}{I}{E}\right)$, since its upper block is $-G^{-1}B+G^{-1}(G+B)=I$ and its lower block is $G-B{G}^{-1}B+B{G}^{-1}(G+B)=G+B=E$. Also $S(E{)}^2=I$ by $\mathcal{M}(E)\eta \mathcal{M}(E)=\eta$ and ${\eta }^2=I$. Since $\mathcal{M}(E)$ is positive, a $+1$ eigenvector $v$ of $S(E)$ has $v^T\eta v=v^T\mathcal{M}(E)v>0$, while a $-1$ eigenvector has $v^T\eta v=-v^T\mathcal{M}(E)v<0$. The signature $(d,d)$ from step 1.1 therefore makes both eigenspaces dimension $d$; the displayed $d$-dimensional graph of $E$ is exactly the $+1$ eigenspace. The matrix $g$ is invertible because $g^T\eta g=\eta$ with nondegenerate $\eta$, and this identity then also gives $g\eta g^T=\eta$. The O$(d,d)$-congruence ${\mathcal{M}}_{\ast }:=g\mathcal{M}(E)g^T$ is positive and obeys ${\mathcal{M}}_{\ast }\eta {\mathcal{M}}_{\ast }=\eta$. Its associated involution satisfies $\eta {\mathcal{M}}_{\ast }=g^{-T}S(E)g^T$ (using $\eta g=g^{-T}\eta$), hence its $+1$ eigenspace is $g^{-T}\mathrm{graph}(E)$. Orthogonality gives $g^{-T}=\left(\begin{array}{cc}{d}_0& c\\ b& a\end{array}\right)$, so

The factorization and generalized-metric identities of step 1.1 apply to $G^{\mathrm{\prime }},B^{\mathrm{\prime }}$ as well, so $\mathcal{M}(E^{\mathrm{\prime }})$ is a positive symmetric generalized metric; its $+1$ eigenspace is $\mathrm{graph}(E^{\mathrm{\prime }})$ by the same block calculation. The displayed graph identity therefore shows that ${\mathcal{M}}_{\ast }$ and $\mathcal{M}(E^{\mathrm{\prime }})$ have the same $+1$ eigenspace. For any positive symmetric generalized metric $M$, the involution $S=\eta M$ is $\eta$-self-adjoint because $S^T\eta =\eta S=M$. If $Sv=v$, then $Mv=\eta v$, and therefore $v^T\eta v=v^{T}Mv>0$; if $Sv=-v$, then $Mv=-\eta v$ and $v^T\eta v=-v^{T}Mv<0$. The two eigenspaces are $\eta$-orthogonal by self-adjointness. Since $\eta$ has signature $(d,d)$, both eigenspaces have dimension $d$; the negative eigenspace is therefore exactly the $\eta$-orthogonal complement of the positive eigenspace. The positive eigenspace thus determines $S$, and then $M=\eta S$, uniquely. It follows that ${\mathcal{M}}_{\ast }=\mathcal{M}(E^{\mathrm{\prime }})$. Taking determinants in $g^T\eta g=\eta$ gives $(\det g{)}^2=1$, so the integral matrix $g$ is unimodular, $g^{-T}$ is an integral lattice automorphism, and $Z^{\mathrm{\prime }}:=g^{-T}Z$ is integral. Hence

Orthogonality also gives $g^{-1}\eta g^{-T}=\eta$, so $Z^{\mathrm{\prime }T}\eta Z^{\mathrm{\prime }}=Z^T\eta Z$. Together with the generalized-metric quadratic-form identity from step 2.1, the two combinations in [F2] are unchanged separately; thus $h_0^{\mathrm{\prime }}=h_0$ and ${\bar{h}}_0^{\mathrm{\prime }}={\bar{h}}_0$. The calculation concerns only the adopted zero-mode lattice prescription; it does not construct a transformation of oscillator states or imply any of the operator-algebra and full-CFT claims excluded in the statement.