Supersymmetric minimal models

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introduction

  • two sectors : NS and R
  • The (normalized) characters of a generic $N$=1 superconformal minimal model $\cal{SM}(p,p')$ are given by

$$ \hat{\chi}_{r,s}^{(p,p')}(q) = \hat{\chi}_{p-r,p'-s}^{(p,p')}(q) = {(-q^{\varepsilon_{r-s}})_\infty \over (q)_\infty} ~\sum_{\ell\in \ZZ} \left( q^{\ell(\ell pp'+rp'-sp)/2} -q^{(\ell p+r)(\ell p'+s)/2} \right)~, $$

where $$ \varepsilon_a= \begin{cases} 1/2, & \text{if $a$ is even$\leftrightarrow$ NS sector}\\ 1, & \text{if $a$ is odd$\leftrightarrow$ ~R~ sector} \\ \end{cases} $$


the first type

  • if \(j\leq k-1\), \(N_j=n_j+\cdots+n_{k-1}\) and \(N_k=0\)
  • for $s=1,3\cdots, 2k-1$

$$ \begin{aligned} \hat{\chi}_{1,s}^{(2,4k)}&(q) ~= \sum_{m_1,\ldots,m_{k-1}=0}^\infty {(-q^{1/2})_{N_1} ~q^{{1\over 2}N_1^2+N_2^2+\ldots+N_{k-1}^2 +N_{(s+1)/2}+N_{(s+3)/2}+\ldots+N_{k-1}} \over (q)_{m_1}(q)_{m_2} \ldots (q)_{m_{k-1}}} \\ &= \sum_{m_1,\ldots,m_{k}=0}^\infty {q^{N_1^2+N_2^2+\ldots+N_{k-1}^2 +N_{(s+1)/2}+N_{(s+3)/2}+\ldots+N_{k-1}-N_1 m_k+{1\over 2}m_k^2} \over (q)_{m_1}(q)_{m_2} \ldots (q)_{m_{k-1}}} {N_1 \choose m_k}_q \end{aligned} $$

  • for $s=2$ and $s=2k$

$$ \eqalign{\hat{\chi}_{1,2}^{(2,4k)}&(q) ~= \sum_{m_1,\ldots,m_{k-1}=0}^\infty {(-q)_{N_1}~q^{{1\over 2}N_1(N_1+1)+N_2(N_2+1)+\ldots+N_{k-1}(N_{k-1}+1)} \over (q)_{m_1}(q)_{m_2} \ldots (q)_{m_{k-1}}} \cr &= \sum_{m_1,\ldots,m_{k}=0}^\infty {q^{N_1(N_1+1)+N_2(N_2+1)+\ldots+N_{k-1}(N_{k-1}+1)-N_1 m_k +{1\over 2}m_k(m_k-1)} \over (q)_{m_1}(q)_{m_2} \ldots (q)_{m_{k-1}}} {N_1 \choose m_k}_q ~~,\cr \hat{\chi}_{1,2k}^{(2,4k)}&(q) ~= \sum_{m_1,\ldots,m_{k-1}=0}^\infty {(-1)_{N_1} ~q^{{1\over 2}N_1(N_1+1)+N_2^2+\ldots+N_{k-1}^2} \over (q)_{m_1}(q)_{m_2} \ldots (q)_{m_{k-1}}} \cr &= \sum_{m_1,\ldots,m_{k}=0}^\infty {q^{N_1^2+N_2^2+\ldots+N_{k-1}^2 -N_1 m_k+{1\over 2}m_k(m_k+1)} \over (q)_{m_1}(q)_{m_2} \ldots (q)_{m_{k-1}}} {N_1 \choose m_k}_q ~~.\cr} $$

second type

  • The second type of fermionic forms for the characters of the same family of models $\cal{SM}(2,4k)$ is presented in the following conjecture:

For $k=2,3,4,\ldots$ and $s=1,2,\ldots,2k$

  • $s$ is odd

$$ \hat{\chi}_{1,s}^{(2,4k)}(q) = \sum_{m_1,\ldots,m_{2k-2}=0}^\infty {q^{{1\over 2}(M_1^2+M_2^2+\ldots+M_{2k-2}^2) +M_s+M_{s+2}+\ldots+M_{2k-3}} \over (q)_{m_1}(q)_{m_2} \ldots (q)_{m_{2k-2}}} $$

  • $s$ is even

$$ \hat{\chi}_{1,s}^{(2,4k)}(q)= \sum_{m_1,\ldots,m_{2k-2}=0}^\infty {q^{{1\over 2}(M_1^2+M_2^2+\ldots+M_{2k-2}^2) +M_s+M_{s+2}+\ldots+M_{2k-2} +{1\over 2}\tilde{M}} \over (q)_{m_1}(q)_{m_2} \ldots (q)_{m_{2k-2}}} $$ where $$ M_j = m_j +m_{j+1}+\ldots+m_{2k-2}, \tilde{M}=m_1+m_3+\ldots+m_{2k-3} $$

  • the matrix $A_k$ for $k$ has size $2(k-1)$


examples

  • a few modular triples whose matrix part is of the form $A=\mathcal{C}(T_1)\otimes \mathcal{C}(T_n)^{-1}$.


$\mathcal{C}(T_1)\otimes \mathcal{C}(T_1)^{-1}$

The matrix $\mathcal{C}\left(T_1\right)\otimes \mathcal{C}\left(T_1\right)^{-1}=(1)$ of rank 1 is related to the identity of Euler, \begin{equation} \prod_{n=0}^{\infty}(1+zq^n)=\sum_{n=0}^{\infty}\frac{q^{n(n-1)/2}}{(q)_n} z^n. \end{equation} When $z=1$, it becomes \begin{equation}\label{2:weber1} \prod_{n=0}^{\infty} (1+q^{n})=\sum_{n=0}^{\infty}\frac{q^{n(n-1)/2}}{(q)_n}. \end{equation} If we specialize $z=q^{1/2}$, we get \begin{equation} \prod_{n=1}^{\infty} (1+q^{n-\frac{1}{2}})=\sum_{n=0}^{\infty}\frac{q^{n^2/2}}{(q)_n}. \end{equation} Another specialization $z=q$ gives \begin{equation} \label{2:weber3} \prod_{n=1}^{\infty} (1+q^{n})=\sum_{n=0}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}. \end{equation}

Note that (\ref{2:weber1}) and (\ref{2:weber3}) are just scalar multiples of each other. In terms of minimal model characters, we have $$\chi _{1,2}^{(3,4)}=\frac{\eta (2\tau )}{\eta (\tau )}=q^{1/24}\sum _{m=-\infty }^{\infty } (-1)^mq^{3 m^2- m }=q^{1/24}\sum_{n=0}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}.$$

These are in fact, with a minor modification, known as Weber's modular functions \begin{align} \mathfrak{f}(\tau)&=\frac{e^{-\frac{\pi i}{24}}\eta(\frac{\tau+1}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1+q^{n-\frac{1}{2}})=q^{-1/48}\sum_{n=0}^{\infty}\frac{q^{n^2/2}}{(q)_n}, \\ \mathfrak{f}_1(\tau)&=\frac{\eta(\frac{\tau}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1-q^{n-\frac{1}{2}}), \\ \mathfrak{f}_2(\tau)&=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n}) =\sqrt{2}q^{1/24} \sum_{n=0}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}. \label{3:Weber3} \end{align}

From these considerations, we can obtain three modular triples \begin{align} \left((1),(0), -1/48\right) \\ \left((1), (1/2) ,1/24\right) \\ \left((1), (-1/2),1/24\right). \end{align}

The equation $x=1-x$ has the unique solution $x=1/2$. From (\ref{3:Weber3}), one can derive the identity $$L(\frac{1}{2})=\frac{1}{2}L(1)=\frac{\pi^2}{12}.$$ This is consistent with the identity $$L(\frac{1}{2})=\frac{h(T_1) r(T_1) r(T_1)}{h(T_1)+h(T_1)}L(1)=\frac{3}{6}L(1)=\frac{1}{2}L(1).$$


$\mathcal{C}(T_1)\otimes \mathcal{C}(T_2)^{-1}$

For the matrix $\mathcal{C}(T_1)\otimes \mathcal{C}(T_2)^{-1}$, let us take a look at the $q$-series identity $$f(q,z)=\sum_{k\geq 0}\frac{q^{k(k+1)/2}(-zq;q)_{k}}{(q)_{k}}=(-zq^2;q^2)_{\infty}(-q;q)_{\infty}=\prod_{m=1}^{\infty} (1+zq^{2m})(1+q^{m}).$$ This is called the Lebesgue's identity\index{Lebesgue's identity} and the left-hand side of it can be rewritten as \begin{equation}\label{3:Lebeq} \sum_{k\geq 0}\frac{q^{k(k+1)/2}(-zq;q)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{z^j q^{\frac{i^2}{2}+i j+j^2+\frac{i}{2}+j}}{(q)_{i}(q)_{j}}. \end{equation}

This identity was studied from the viewpoint of partition identities and continued fractions. To prove (\ref{3:Lebeq}), one can use the $q$-binomial identity $$(-zq;q)_{k}= \sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r$$ where $$\begin{bmatrix} k\\ r\end{bmatrix}_{q}=\frac{(q)_k}{(q)_r(q)_{k-r}}.$$

Thus from this, one can produce modular triples with its matrix part given by $\begin{pmatrix}1&1\\1&2\end{pmatrix}=\mathcal{C}(T_1)\otimes \mathcal{C}(T_2)^{-1}$ by specializing $z$ appropriately. For example, $z=1$ gives $$f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i+2j}{2}}}{(q)_{i}(q)_{j}}=(-q^2;q^2)_{\infty}(-q;q)_{\infty}=\frac{(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}.$$ So $$q^{1/8}\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i+2j}{2}}}{(q)_{i}(q)_{j}}=\frac{\eta(4\tau)}{\eta(\tau)}$$ gives a modular triple $(\mathcal{C}(T_1)\otimes \mathcal{C}(T_2)^{-1},(1/2,1),1/8)$. By solving the system of equations $$ \left\{ \begin{array}{lll} x_1&=&(1-x_1) (1-x_2) \\ x_2&=&(1-x_1) (1-x_2)^2 \end{array} \right., $$ we can get the corresponding dilogarithm identity $$L(\sqrt{2}-1)+L \left(\frac{1}{2}(2-\sqrt{2})\right)=\frac{3}{4}L(1).$$ Note also that $$\frac{h(T_1) r(T_1) r(T_2)}{h(T_1)+h(T_2)}=\frac{6}{8}=\frac{3}{4}.$$

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