"Supersymmetric minimal models"의 두 판 사이의 차이

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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 <math>j\leq k-1</math>, <math>N_j=n_j+\cdots+n_{k-1}</math> and <math>N_k=0</math>
-* 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 $$\frac{h(T_1) r(T_1) r(T_2)}{h(T_1)+h(T_2)}=\frac{6}{8}=\frac{3}{4}.$$
-==related items==
-* [[Minimal models]]
-* [[Ramanujan-Göllnitz-Gordon continued fraction]]
-==computational resource==
-* https://docs.google.com/file/d/0B8XXo8Tve1cxZl9xd1p3aUFVSUU/edit
-==articles==
-* Melzer, Ezer. 1994. “Supersymmetric Analogs of the Gordon-Andrews Identities, and Related TBA Systems”. ArXiv e-print hep-th/9412154. http://arxiv.org/abs/hep-th/9412154.
-** [[파일:Supersymmetric Analogs of the Gordon-Andrews Identities, and Related TBA Systems.tex]]
-* A. Meurman and A. Rocha-Caridi, Highest weight representations of the Neveu-Schwarz and Ramond algebras Commun. Math. Phys. 107:263 (1986) http://link.springer.com/article/10.1007/BF01209395

"Supersymmetric minimal models"의 두 판 사이의 차이

2020년 11월 13일 (금) 00:19 판

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