Compact-boson spectrum, modular partition function, and T-duality

Put $\rho =\mathrm{\mid }q\mathrm{\mid }<1$ and choose $K$ so $\mathrm{\mid }q{\mathrm{\mid }}^k\le \frac{1}{2}$ for $k\ge K$. For $\mathrm{\mid }w\mathrm{\mid }<1$, the absolutely convergent series $L(w)=-{\sum }_{j\ge 1}{w}^j\mathrm{/}j$ has $L^{\mathrm{\prime }}(w)=-1\mathrm{/}(1-w)$, so $e^{L(w)}\mathrm{/}(1-w)$ has derivative $0$ and value $1$ at $w=0$; hence $e^{L(w)}=1-w$. Also $\mathrm{\mid }L(w)\mathrm{\mid }\le \mathrm{\mid }w\mathrm{\mid }\mathrm{/}(1-\mathrm{\mid }w\mathrm{\mid })$, so ${\sum }_{k\ge K}\mathrm{\mid }L(q^k)\mathrm{\mid }\le 2{\sum }_{k\ge K}{\rho }^k<\mathrm{\infty }$; the tail product is the nonzero exponential $\exp ({\sum }_{k\ge K}L(q^k))$, each earlier factor $1-q^k$ is nonzero, and $e^{\pi i\tau \mathrm{/}12}\mathrm{\ne }0$. The product defining $\eta$ is locally uniformly convergent on $\mathfrak{H}$, since $\mathrm{\mid }q\mathrm{\mid }$ is bounded by a constant less than $1$ on each compact subset, and $\eta (\tau )\mathrm{\ne }0$. In one chiral Fock sector, the oscillator at mode $k\ge 1$ has occupation number $s\in {\mathbb{Z}}_{\ge 0}$ and contributes ${\sum }_{s\ge 0}{q}^{ks}=(1-q^k{)}^{-1}$. Multiplying the independent modes and the vacuum factor $q^{-1\mathrm{/}24}$ from [A2] gives $q^{-1\mathrm{/}24}{\prod }_{k\ge 1}(1-q^k{)}^{-1}=\eta (\tau {)}^{-1}$. The right-moving modes give ${\bar{\eta }}^{-1}$. Each charge sector contributes the zero-mode factor $q^{p_L^2\mathrm{/}2}{\bar{q}}^{p_R^2\mathrm{/}2}$.

The zero-mode theta sum is ${\mathrm{\Theta }}_r(\tau )={\sum }_{x\in {\mathbb{Z}}^2}{e}^{-\pi x^T{C}_{\tau }x}$ by [L1]. The real part of $C_{\tau }$ is the positive diagonal matrix $\mathrm{diag}({\tau }_2\mathrm{/}(2r^2),2r^2{\tau }_2)$, so $x\mapsto e^{-\pi x^T{C}_{\tau }x}$ is a Schwartz function on ${\mathbb{R}}^2$. Applying [F6] at zero is therefore valid under the exact countable-choice assumption [F5].

The product formulas from [F9] give ${\theta }_3={\prod }_{k\ge 1}(1-q^k)(1+q^{k-1\mathrm{/}2}{)}^2$, ${\theta }_4={\prod }_{k\ge 1}(1-q^k)(1-q^{k-1\mathrm{/}2}{)}^2$, and ${\theta }_2=2q^{1\mathrm{/}8}{\prod }_{k\ge 1}(1-q^k)(1+q^k{)}^2$. Absolute convergence permits regrouping; ${\prod }_{k\ge 1}(1+q^k){\prod }_{k\ge 1}(1-q^{2k-1})=1$ because $\prod (1+q^k)=\prod (1-q^{2k})\mathrm{/}\prod (1-q^k)$. Therefore ${\theta }_2{\theta }_3{\theta }_4=2q^{1\mathrm{/}8}{\prod }_{k\ge 1}(1-q^k{)}^3=2\eta (\tau {)}^3$.

Put $r^{\mathrm{\prime }}=(2r{)}^{-1}$, $T(m,n)=(n,m)$ and $\chi (m,n)=(-1{)}^{mn}$. On the algebraic charge-$q_0$ summand define $D_{r}f(x,\bar{x})=\chi (q_0)f(x,-\bar{x})$ in the charge-$T{q}_0$ summand at radius $r^{\mathrm{\prime }}$. Here $f\in P\otimes \bar{P}$ is a polynomial and $-\bar{x}$ replaces every ${\bar{x}}_k$ by $-{\bar{x}}_k$. This preserves the monomial norms, and $D_{r^{\mathrm{\prime }}}{D}_r=I$, so it extends to a unitary map ${\mathcal{H}}_r\to {\mathcal{H}}_{r^{\mathrm{\prime }}}$. Denote target operators by primes. Multiplication and differentiation give $D_r{a}_j{D}_r^{-1}=a_j^{\mathrm{\prime }}$ and $D_r{\bar{a}}_j{D}_r^{-1}=-{\bar{a}}_j^{\mathrm{\prime }}$ for $j\mathrm{\ne }0$; for $j=0$ the same identities follow from $p_L(Tq;r^{\mathrm{\prime }})=p_L(q;r)$ and $p_R(Tq;r^{\mathrm{\prime }})=-p_R(q;r)$. The identity $\chi (q+q_0)\epsilon (q,q_0)=\chi (q)\chi (q_0)\epsilon (Tq,T{q}_0)$ gives $D_r{U}_q{D}_r^{-1}=\chi (q)U_{Tq}^{\mathrm{\prime }}$. Substitution into the oscillator exponentials and zero-mode powers in the lattice definition therefore gives $D_r{V}_q(z,\bar{z})D_r^{-1}=\chi (q)V_{Tq}^{\mathrm{\prime }}(z,\bar{z})$ on their declared radial branches. The same identities intertwine every finite Heisenberg descendant, so the convergent radial operator products in [F3] are preserved.

Since $p_L^2+p_R^2=m^2\mathrm{/}(2r^2)+2n^2{r}^2$, the absolute value of each zero-mode summand is $e^{-\pi {\tau }_2(p_L^2+p_R^2)}$, a product of two convergent Gaussian series for every $r,{\tau }_2>0$. The oscillator products converge because ${\sum }_{k\ge 1}\mathrm{\mid }q{\mathrm{\mid }}^k<\mathrm{\infty }$. Thus the trace is absolutely convergent and equals the displayed expression.

The spin of a state with oscillator levels $N,\bar{N}$ is $h-\bar{h}=\frac{1}{2}(p_L^2-p_R^2)+N-\bar{N}=mn+N-\bar{N}\in \mathbb{Z}$. Under $\tau \mapsto \tau +1$, every summand therefore acquires phase $e^{2\pi i(h-\bar{h})}=1$. Directly from the product definition, $\eta (\tau +1)=e^{\pi i\mathrm{/}12}\eta (\tau )$, whose modulus is unchanged. This proves $T$-invariance.

For $d\in \mathbb{C}$ with $\mathrm{Re}d>0$ and $\xi \in \mathbb{R}$, let $G_d(\xi )={\int }_{\mathbb{R}}{e}^{-\pi d{u}^2-2\pi iu\xi }du$. For $\mathrm{\mid }h\mathrm{\mid }<\mathrm{Re}d\mathrm{/}2$, the difference quotient in $d$ is dominated by $\pi u^2{e}^{-\pi (\mathrm{Re}d)u^2\mathrm{/}2}$, which is integrable by [F7]; dominated convergence [F10] makes $G_d(\xi )$ holomorphic in $d$. The function $d^{-1\mathrm{/}2}{e}^{-\pi {\xi }^2\mathrm{/}d}$ is holomorphic on the right half-plane using the principal logarithm [F11]. They agree for real $d>0$ by [F7], so the identity theorem [F11] gives $G_d(\xi )=d^{-1\mathrm{/}2}{e}^{-\pi {\xi }^2\mathrm{/}d}$ throughout that half-plane.

The theta transformation [F8] gives ${\theta }_3(-1\mathrm{/}\tau )=s(\tau ){\theta }_3(\tau )$ and ${\theta }_4(-1\mathrm{/}\tau )=s(\tau ){\theta }_2(\tau )$, where $s(\tau )=\sqrt{-i\tau }$ is the branch positive at $\tau =i$. Applying the second relation at $-1\mathrm{/}\tau$ gives ${\theta }_2(-1\mathrm{/}\tau )=s(\tau ){\theta }_4(\tau )$ because $s(\tau )s(-1\mathrm{/}\tau )=1$. Thus $\eta (-1\mathrm{/}\tau {)}^3=s(\tau {)}^3\eta (\tau {)}^3$. The ratio $\eta (-1\mathrm{/}\tau )\mathrm{/}(s(\tau )\eta (\tau ))$ is holomorphic, nonzero, and takes values in the three cube roots of $1$, so it is constant on the connected upper half-plane; at $\tau =i$ it equals $1$. Hence $\mathrm{\mid }\eta (-1\mathrm{/}\tau ){\mathrm{\mid }}^2=\mathrm{\mid }\tau \mathrm{\mid }\mathrm{\mid }\eta (\tau ){\mathrm{\mid }}^2$. The transformations are used with the countable-choice hypothesis [F5].

Change variables $y=A_{r}x$ in the Fourier integral of the function in step 1.2. The Jacobian has absolute value $\mathrm{\mid }\det A_r\mathrm{\mid }=1$. The two factors are the one-dimensional transforms in step 2.3 with $d_1=-i{\tau }^{\mathrm{\prime }}$, $d_2=i{\bar{\tau }}^{\mathrm{\prime }}$ and ${\tau }^{\mathrm{\prime }}=-1\mathrm{/}\tau$. Their real parts are ${\tau }_2\mathrm{/}\mathrm{\mid }\tau {\mathrm{\mid }}^2>0$, and $d_1{d}_2=\mathrm{\mid }\tau {\mathrm{\mid }}^{-2}$ with conjugate principal roots. Hence the Poisson prefactor is $\mathrm{\mid }\tau \mathrm{\mid }$. Matrix multiplication gives $\det C_{{\tau }^{\mathrm{\prime }}}=\mathrm{\mid }\tau {\mathrm{\mid }}^{-2}$ and $C_{{\tau }^{\mathrm{\prime }}}^{-1}=P{C}_{\tau }P$, where $P=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Since $P{\mathbb{Z}}^2={\mathbb{Z}}^2$, [F6] yields ${\mathrm{\Theta }}_r(-1\mathrm{/}\tau )=\mathrm{\mid }\tau \mathrm{\mid }{\mathrm{\Theta }}_r(\tau )$.

Combining steps 3.1 and 2.4, the zero-mode theta factor gains $\mathrm{\mid }\tau \mathrm{\mid }$ under $S$, while $\mathrm{\mid }\eta {\mathrm{\mid }}^{-2}$ gains $\mathrm{\mid }\tau {\mathrm{\mid }}^{-1}$. The full $Z_r$ is therefore invariant under $\tau \mapsto -1\mathrm{/}\tau$.

The charge squares and oscillator levels in each chiral weight are unchanged by $D_r$. Consequently $D_r{H}_r=H_{r^{\mathrm{\prime }}}{D}_r$, so $D_r$ also preserves the common-domain seminorms of [F1]. It fixes the vacuum and bijects the trace summands, including the shifts in [A2]. Finally $r=(2r{)}^{-1}$ has the unique positive solution $r=1\mathrm{/}\sqrt{2}$.