Supersymmetric minimal models - 편집 역사

2020년 11월 16일 (월) 17:49에 Pythagoras0님의 편집

2020-11-16T17:49:24Z

2020년 11월 13일 (금) 07:19에 imported>Pythagoras0님의 편집

2020-11-13T07:19:58Z

2020년 11월 13일 (금) 07:19에 imported>Pythagoras0님의 편집

2020-11-13T07:19:58Z

imported>Pythagoras0: /* examples */

2013-07-15T10:09:29Z

examples

2013년 7월 15일 (월) 10:06에 imported>Pythagoras0님의 편집

2013-07-15T10:06:07Z

imported>Pythagoras0: /* the first type */

2013-07-14T21:35:25Z

the first type

2013년 7월 14일 (일) 21:29에 imported>Pythagoras0님의 편집

2013-07-14T21:29:01Z

imported>Pythagoras0: /* second type */

2013-07-14T21:19:37Z

second type

imported>Pythagoras0: /* second type */

2013-07-14T21:12:35Z

second type

2013년 7월 14일 (일) 21:06에 imported>Pythagoras0님의 편집

2013-07-14T21:06:23Z

← 이전 판	2020년 11월 13일 (금) 07:19 판
1번째 줄:	1번째 줄:
	+	==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
	+	[[분류:migrate]]

← 이전 판		2020년 11월 13일 (금) 07:19 판
1번째 줄:		1번째 줄:
−	~~==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

@@ 81번째 줄: / 81번째 줄: @@
 ==examples==
 * a few modular triples whose matrix part is of the form $A=\mathcal{C}(T_1)\otimes \mathcal{C}(T_n)^{-1}$.
@@ 104번째 줄: / 105번째 줄: @@
 $$\chi _{1,2}^{(3,4)}=\frac{\eta (2\tau )}{\eta (\tau )}=q^{1/24}\sum _{m=-\infty }^{\infty } (-1)^mq^{3 m^2- m }=q^{1/24}\sum_{n=0}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}.$$
-These are in fact, %up to multiples of $q^{\lambda}$, $\lambda\in \mathbb{Q}$
+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}, \\
@@ 132번째 줄: / 132번째 줄: @@
 \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}
-In \cite[p.278-279]{MR1223685}, 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 $$\begin{bmatrix} k\\ r\end{bmatrix}_{q}=\frac{(q)_k}{(q)_r(q)_{k-r}}.$$
@@ 152번째 줄: / 152번째 줄: @@
 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==

@@ 77번째 줄: / 77번째 줄: @@
 $$
 * the matrix $A_k$ for $k$ has size $2(k-1)$
 ==related items==

@@ 17번째 줄: / 17번째 줄: @@
 ==the first type==
 * for $s=1,3\cdots, 2k-1$
 $$
@@ 50번째 줄: / 51번째 줄: @@
     (q)_{m_1}(q)_{m_2} \ldots (q)_{m_{k-1}}} {N_1 \choose m_k}_q ~~.\cr}
 $$
 ==second type==

← 이전 판	2020년 11월 16일 (월) 17:49 판
(차이 없음)

@@ 1번째 줄: / 1번째 줄: @@
 ==introduction==
-* The (normalized) characters of a generic  $N$=1 superconformal
+* two sectors : NS and R
-minimal model $\cal{SM}(p,p')$ are given by
+* 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) =

@@ 53번째 줄: / 53번째 줄: @@
 ==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
+* 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

← 이전 판		2013년 7월 14일 (일) 21:06 판
1번째 줄:		1번째 줄:
		+	==introduction==
		+	* 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==
		+	* 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
		+	$$
		+	M_j = m_j +m_{j+1}+\ldots+m_{2k-2},
		+	\tilde{M}=m_1+m_3+\ldots+m_{2k-3}
		+	$$
		+
		+
		+
	==related items==		==related items==
	* [[Minimal models]]		* [[Minimal models]]