Free Majorana and Dirac torus spin-structure characters

Put $\rho =\mathrm{\mid }q\mathrm{\mid }<1$. The product defining $\eta$ is nonzero: 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 ${\sum }_{k\ge K}\mathrm{\mid }L(q^k)\mathrm{\mid }\le 2{\sum }_{k\ge K}{\rho }^k<\mathrm{\infty }$. Thus the tail product is the nonzero exponential $\exp ({\sum }_{k\ge K}L(q^k))$, every preceding factor $1-q^k$ is nonzero, and $e^{\pi i\tau \mathrm{/}12}\mathrm{\ne }0$. The product also converges locally uniformly on $\mathfrak{H}$, since $\mathrm{\mid }q\mathrm{\mid }$ is bounded above by a constant less than $1$ on each compact subset. Therefore $\eta (\tau )\mathrm{\ne }0$ and the ratios in the statement are defined. For one real chiral Majorana, the double contraction in $T(z)T(w)$ from $T=-\frac{1}{2}:\psi \mathrm{\partial }\psi :$ is $\frac{1}{4}(z-w{)}^{-4}$, so the Virasoro central term $c\mathrm{/}(2(z-w{)}^4)$ has $c=1\mathrm{/}2$. The cylinder trace factor is therefore $q^{-c\mathrm{/}24}=q^{-1\mathrm{/}48}$. In a chiral NS sector each positive half-integer mode is either empty or occupied, so its trace contributes $1+q^{k+1\mathrm{/}2}$; insertion of fermion parity changes this to $1-q^{k+1\mathrm{/}2}$. For the two independent real components in one chiral Dirac field, the two traces are $q^{-1\mathrm{/}24}{\prod }_{k\ge 0}(1+q^{k+1\mathrm{/}2}{)}^2$ and $q^{-1\mathrm{/}24}{\prod }_{k\ge 0}(1-q^{k+1\mathrm{/}2}{)}^2$.

In the R sector the two real zero modes form the two-dimensional complex Clifford module specified in [F1]. Each real Ramond ground has weight $1\mathrm{/}16$ by the declared spin-field normalization, so the two-component chiral ground has weight $1\mathrm{/}8$; subtracting $c\mathrm{/}24=1\mathrm{/}24$ gives the factor $q^{1\mathrm{/}12}$. Setting $z=\tau \mathrm{/}2$ in [F2] and using ${\theta }_2=e^{\pi i\tau \mathrm{/}4}\mathrm{\Theta }(\tau \mathrm{/}2\mid \tau )$ gives ${\theta }_2=2q^{1\mathrm{/}8}{\prod }_{n\ge 1}(1-q^n)(1+q^n{)}^2$. Including the two ground states and the positive modes therefore gives $2q^{1\mathrm{/}12}{\prod }_{n\ge 1}(1+q^n{)}^2={\theta }_2\mathrm{/}\eta$. The parity-inserted trace is zero because the two zero-mode parity eigenspaces have equal dimension and the positive-mode factors are paired. Taking the right-moving trace gives $Z_{R,NS}=a_2^2$ and $Z_{R,R}=0$.

Setting $z=0$ and $z=1\mathrm{/}2$ in [F2] gives ${\theta }_3={\prod }_{n\ge 1}(1-q^n)(1+q^{n-1\mathrm{/}2}{)}^2$ and ${\theta }_4={\prod }_{n\ge 1}(1-q^n)(1-q^{n-1\mathrm{/}2}{)}^2$. Together with $\eta =q^{1\mathrm{/}24}{\prod }_{n\ge 1}(1-q^n)$ these identify the two NS products in step 1.1 with ${\theta }_3\mathrm{/}\eta$ and ${\theta }_4\mathrm{/}\eta$, respectively. Taking the independent right-moving trace gives $Z_{NS,NS}=a_3^2$ and $Z_{NS,R}=a_4^2$.

The four common-spin traces are averaged with coefficient $1\mathrm{/}2$ in the defined GSO projection, hence $Z_D=\frac{1}{2}(a_3^2+a_4^2+a_2^2)$. For one real Majorana, the NS, NS-with-parity, R, and R-with-parity nonchiral traces are $a_3,a_4,a_2,0$; averaging its four spin structures gives $Z_M=\frac{1}{2}(a_3+a_4+a_2)$. Independence of two separately spin-summed Majorana systems makes their joint trace the product $Z_M^2$.

Write $s(\tau )=\exp (\frac{1}{2}\mathrm{Log}(\tau \mathrm{/}i))$, so $\mathrm{\mid }s(\tau )\mathrm{\mid }=\mathrm{\mid }\tau {\mathrm{\mid }}^{1\mathrm{/}2}$. From the theta definitions, ${\theta }_3(\tau )=\mathrm{\Theta }(0\mid \tau )$, ${\theta }_4(\tau )=\mathrm{\Theta }(1\mathrm{/}2\mid \tau )$, and ${\theta }_2(\tau )=e^{\pi i\tau \mathrm{/}4}\mathrm{\Theta }(\tau \mathrm{/}2\mid \tau )$. The inversion formula in [F3] at $z=0$ gives ${\theta }_3(-1\mathrm{/}\tau )=s(\tau ){\theta }_3(\tau )$; at $z=1\mathrm{/}2$ it gives ${\theta }_4(-1\mathrm{/}\tau )=s(\tau ){\theta }_2(\tau )$. Apply the shifted-theta identity in [F3] with $\tau$ replaced by $-1\mathrm{/}\tau$: its left side is ${\theta }_4(\tau )$ by the series definition, and its right side is $s(-1\mathrm{/}\tau ){\theta }_2(-1\mathrm{/}\tau )$. The holomorphic nonzero function $s(\tau )s(-1\mathrm{/}\tau )$ has square $1$ and equals $1$ at $\tau =i$, so connectedness of $\mathfrak{H}$ gives $s(-1\mathrm{/}\tau )=s(\tau {)}^{-1}$ and therefore ${\theta }_2(-1\mathrm{/}\tau )=s(\tau ){\theta }_4(\tau )$. To obtain the eta factor, apply the triple product [F2] at $z=0$, $z=1\mathrm{/}2$, and $z=\tau \mathrm{/}2$ (with $Q=e^{\pi i\tau }$ and $q=Q^2$). Absolute convergence permits regrouping and gives ${\theta }_3={\prod }_{k\ge 1}(1-q^k)(1+q^{k-1\mathrm{/}2}{)}^2$, ${\theta }_4={\prod }_{k\ge 1}(1-q^k)(1-q^{k-1\mathrm{/}2}{)}^2$, and ${\theta }_2=2q^{1\mathrm{/}8}{\prod }_{k\ge 1}(1-q^k)(1+q^k{)}^2$. Since ${\prod }_{k\ge 1}(1+q^k){\prod }_{k\ge 1}(1-q^{2k-1})=1$, multiplication yields ${\theta }_2{\theta }_3{\theta }_4=2q^{1\mathrm{/}8}{\prod }_{k\ge 1}(1-q^k{)}^3=2\eta (\tau {)}^3$. The three theta inversion formulas then imply $\eta (-1\mathrm{/}\tau {)}^3=s(\tau {)}^3\eta (\tau {)}^3$. The continuous nonzero ratio $\eta (-1\mathrm{/}\tau )\mathrm{/}(s(\tau )\eta (\tau ))$ has cube $1$, so it is constant on the connected upper half-plane; its value at $\tau =i$ is $1$. Thus $\mathrm{\mid }\eta (-1\mathrm{/}\tau )\mathrm{\mid }=\mathrm{\mid }\tau {\mathrm{\mid }}^{1\mathrm{/}2}\mathrm{\mid }\eta (\tau )\mathrm{\mid }$, and the three $S$ laws exchange $a_2,a_4$ and fix $a_3$. Directly from the series, ${\theta }_3(\tau +1)={\theta }_4(\tau )$, ${\theta }_4(\tau +1)={\theta }_3(\tau )$, and ${\theta }_2(\tau +1)=e^{\pi i\mathrm{/}4}{\theta }_2(\tau )$; from the product, $\eta (\tau +1)=e^{\pi i\mathrm{/}12}\eta (\tau )$. Hence $T$ exchanges $a_3,a_4$ and fixes $a_2$. Both displayed spin sums are symmetric in $a_2,a_3,a_4$, so they are invariant.