﻿ 态场对应与共形超代数

# 态场对应与共形超代数State-Field Correspondences and Conformal Super-Algebras

Abstract: In the present paper, the concept of a conformal super-algebra is introduced and its relationship with a state-field correspondence is given. The locality is defined for any two Laurent-series with well-defined coefficients and some examples are discussed. In particular, the Loop state-field cor-respondence is constructed and some properties are obtained.

1. 引言

2. 态场对应的预备知识

$V\to glf\left(V\right)$ , $a↦Y\left(a,z\right)=\underset{n\in Ζ}{\sum }{a}_{\left(n\right)}{z}^{-n-1}$ ,

$T\left({a}_{\left(n\right)}b\right)={\left(Ta\right)}_{\left(n\right)}b+{a}_{\left(n\right)}\left(Tb\right)$ , ${\left(Ta\right)}_{\left(n\right)}=-n{a}_{\left(n-1\right)}$ , $\forall a,b\in V$ .

${Y}^{op}:V\to glf\left(V\right),a↦{Y}^{op}\left(a,z\right)={\left(-1\right)}^{p\left(a\right)p\left(b\right)}{\text{e}}^{zT}Y\left(b,-z\right)a$

3. 共形超代数

$\left[\partial {a}_{\lambda }b\right]=-\lambda \left[{a}_{\lambda }b\right]$ , $\left[{a}_{\lambda }\partial b\right]=\left(\partial +\lambda \right)\left[{a}_{\lambda }b\right]$ ,

$\left[{a}_{\lambda }b\right]={\sum }_{j=0}^{\infty }\frac{{\lambda }^{j}}{j!}{a}_{\left(j\right)}b$ 是关于变量 $\lambda$ 的多项式， $\left(j\right)$ 也可以看成向量超空间R上的一个双线性运算，且

${a}_{\left(j\right)}b\ge 0$$j\gg 0$

${\left[.\lambda .\right]}^{op}:R×R\to R\left[\lambda \right],\left(a,b\right)↦{\left[{a}_{\lambda }b\right]}^{op}$ ,

$\left[{u}_{\lambda }v\right]=Re{s}_{z}{\text{e}}^{\lambda z}Y\left(u,z\right)v$ ,

${\left[{u}_{\lambda }v\right]}^{op}=Re{s}_{z}{\text{e}}^{\lambda z}{Y}^{op}\left(u,z\right)v$ ,

$\begin{array}{c}\left[\partial {u}_{\lambda }v\right]=Re{s}_{z}{\text{e}}^{\lambda z}Y\left(\partial u,z\right)v\\ =-\lambda Re{s}_{z}{\text{e}}^{\lambda z}Y\left(u,z\right)v\\ =-\lambda \left[{u}_{\lambda }v\right]\end{array}$ ,

$\begin{array}{c}\left[{u}_{\lambda }\partial v\right]=Re{s}_{z}{\text{e}}^{\lambda z}Y\left(u,z\right)\partial v\\ =\underset{n=0}{\overset{\infty }{\sum }}\frac{{\lambda }^{n}}{n!}{u}_{\left(n\right)}\partial v\\ =\underset{n=0}{\overset{\infty }{\sum }}\frac{{\lambda }^{n}}{n!}\left(\partial \left({u}_{\left(n\right)}v\right)+n{u}_{\left(n-1\right)}v\right)\\ =\left(\partial +\lambda \right)\left[{u}_{\lambda }v\right]\end{array}$ .

$\stackrel{˜}{R}=R\otimes F\left[t,{t}^{-1}\right]=R\left[t,{t}^{-1}\right]$ ,  ,

$\left[a{t}^{m},b{t}^{n}\right]=\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ j\end{array}\right)\left({a}_{\left(j\right)}b\right){t}^{m+n-j}$ .

$\begin{array}{l}\left[\left(\partial a\right){t}^{m}+ma{t}^{m-1},b{t}^{n}\right]\\ =\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ j\end{array}\right)\left({\left(\partial a\right)}_{\left(j\right)}b\right){t}^{m+n-j}+m\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m-1\\ j\end{array}\right)\left({a}_{\left(j\right)}b\right){t}^{m+n-j-1}\\ =-\underset{j=1}{\overset{\infty }{\sum }}j\left(\begin{array}{c}m\\ j\end{array}\right)\left({a}_{\left(j-1\right)}b\right){t}^{m+n-j}+m\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m-1\\ j\end{array}\right)\left({a}_{\left(j\right)}b\right){t}^{m+n-j-1}\\ =-\underset{j=0}{\overset{\infty }{\sum }}\left(j+1\right)\left(\begin{array}{c}m\\ j+1\end{array}\right)\left({a}_{\left(j\right)}b\right){t}^{m+n-j-1}+\underset{j=0}{\overset{\infty }{\sum }}m\left(\begin{array}{c}m-1\\ j\end{array}\right)\left({a}_{\left(j\right)}b\right){t}^{m+n-j-1}\\ =0\end{array}$

$\begin{array}{l}\left[a{t}^{m},\left(\partial b\right){t}^{n}+nb{t}^{n-1}\right]\\ =\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ j\end{array}\right)\left({a}_{\left(j\right)}\partial b\right){t}^{m+n-j}+n\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ j\end{array}\right)\left({a}_{\left(j\right)}b\right){t}^{m+n-j-1}\\ =\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ j\end{array}\right)\partial \left({a}_{\left(j\right)}b\right){t}^{m+n-j}+\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ j\end{array}\right)j\left({a}_{\left(j-1\right)}b\right){t}^{m+n-j}+n\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ j\end{array}\right)\left({a}_{\left(j\right)}b\right){t}^{m+n-j-1}\\ =\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ j\end{array}\right)\left(\partial \left({a}_{\left(j\right)}b\right){t}^{m+n-j}+\left(m+n-j\right)\left({a}_{\left(j\right)}b\right){t}^{m+n-j-1}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ j\end{array}\right)\left(m+n-j\right)\left({a}_{\left(j\right)}b\right){t}^{m+n-j-1}+\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ j\end{array}\right)j\left({a}_{\left(j-1\right)}b\right){t}^{m+n-j}+n\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ j\end{array}\right)\left({a}_{\left(j\right)}b\right){t}^{m+n-j-1}\end{array}$ .

$\left[{a}_{\left(m\right)},{b}_{\left(n\right)}\right]=\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ j\end{array}\right){\left({a}_{\left(j\right)}b\right)}_{\left(m+n-j\right)}$ .

$\left[{b}_{\lambda }a\right]=-{\left(-1\right)}^{p\left(a\right)p\left(b\right)}{\left[{a}_{\left(-\lambda -\partial \right)}b\right]}^{op}$ ,

$\begin{array}{c}\underset{j=0}{\overset{\infty }{\sum }}\frac{{\lambda }^{j}}{j!}{b}_{\left(j\right)}a=-{\left(-1\right)}^{p\left(a\right)p\left(b\right)}\underset{m=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{m}}{m!}{\left(\lambda +\partial \right)}^{m}{a}_{\left(m\right)}b\\ =-{\left(-1\right)}^{p\left(a\right)p\left(b\right)}\underset{m=0}{\overset{\infty }{\sum }}\underset{k=0}{\overset{m}{\sum }}\frac{{\left(-1\right)}^{m}}{m!}\left(\begin{array}{c}m\\ k\end{array}\right){\lambda }^{m-k}{\partial }^{k}\left({a}_{\left(m\right)}b\right)\\ =-{\left(-1\right)}^{p\left(a\right)p\left(b\right)}\underset{j=0}{\overset{\infty }{\sum }}\underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{j+k}}{k!j!}{\lambda }^{j}{\partial }^{k}\left({a}_{\left(j+k\right)}b\right)\end{array}$ ,

${b}_{\left(j\right)}a=-{\left(-1\right)}^{p\left(a\right)p\left(b\right)}\underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{j+k}}{k!}{\partial }^{k}\left({a}_{\left(j+k\right)}b\right)$ .

$\begin{array}{c}\left[{b}_{\left(n\right)},{a}_{\left(m\right)}\right]=\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}n\\ j\end{array}\right){\left({b}_{\left(j\right)}a\right)}_{\left(m+n-j\right)}\\ =-{\left(-1\right)}^{p\left(a\right)p\left(b\right)}\underset{j=0}{\overset{\infty }{\sum }}\underset{k=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\left(-1\right)}^{j+k}}{k!}{\left({\partial }^{k}\left({a}_{\left(j+k\right)}b\right)\right)}_{\left(m+n-j\right)}\\ =-{\left(-1\right)}^{p\left(a\right)p\left(b\right)}\underset{j=0}{\overset{\infty }{\sum }}\underset{k=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}n\\ j\end{array}\right){\left(-1\right)}^{j}\left(\begin{array}{c}m+n-j\\ k\end{array}\right){\left({a}_{\left(j+k\right)}b\right)}_{\left(m+n-j-k\right)}\\ =-{\left(-1\right)}^{p\left(a\right)p\left(b\right)}\underset{r=0}{\overset{\infty }{\sum }}\underset{k=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}n\\ r-k\end{array}\right){\left(-1\right)}^{r+k}\left(\begin{array}{c}m+n-r+k\\ k\end{array}\right){\left({a}_{\left(r\right)}b\right)}_{\left(m+n-r\right)}\end{array}$ .

$\begin{array}{l}\underset{k=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}n\\ r-k\end{array}\right){\left(-1\right)}^{r+k}\left(\begin{array}{c}m+n-r+k\\ k\end{array}\right)\\ ={\left(-1\right)}^{r}\underset{k=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}n\\ r-k\end{array}\right){\left(-1\right)}^{k}\left(\begin{array}{c}m+n-r+k\\ k\end{array}\right)\\ ={\left(-1\right)}^{r}\underset{k=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}n\\ r-k\end{array}\right)\left(\begin{array}{c}-m-n+r-1\\ k\end{array}\right)\\ ={\left(-1\right)}^{r}\left(\begin{array}{c}-m+r-1\\ r\end{array}\right)=\left(\begin{array}{c}m\\ r\end{array}\right)\end{array}$ .

$-{\left(-1\right)}^{p\left(a\right)p\left(b\right)}\underset{r=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ r\end{array}\right){\left({a}_{\left(r\right)}b\right)}_{\left(m+n-r\right)}=-{\left(-1\right)}^{p\left(a\right)p\left(b\right)}{\left[{a}_{\left(m\right)},{b}_{\left(n\right)}\right]}^{op}$ .

${\left(z-w\right)}^{N}a\left(z\right)b\left(w\right)=0$ .

$\left[Y\left(a,z\right),Y\left(b,z\right)\right]=\underset{j=0}{\overset{\infty }{\sum }}Y\left({a}_{\left(j\right)}b,w\right)\frac{{\partial }_{w}^{j}}{j!}\delta \left(z,w\right)$ .

${\left(z-w\right)}^{N}\left[Y\left(a,z\right),Y\left(b,w\right)\right]=0$ .

${a}_{\left(m\right)}\otimes {b}_{\left(n\right)}-{\left(-1\right)}^{p\left(a\right)p\left(b\right)}{b}_{\left(n\right)}\otimes {a}_{\left(m\right)}-\left[{a}_{\left(m\right)},{b}_{\left(n\right)}\right]$ ,

$B=T\left(A\right)/I$ 是相应的商代数，它有生成元集： ${a}_{\left(n\right)}+I$，这里a是向量超空间R的齐次元素，n是任意整数。把等价类 ${a}_{\left(n\right)}+I$ 简记为 ${a}_{\left(n\right)}$，从而得到下列括积的等式：

$\left[{a}_{\left(m\right)},{b}_{\left(n\right)}\right]=\underset{j=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}m\\ j\end{array}\right){\left({a}_{\left(j\right)}b\right)}_{\left(m+n-j\right)}$ .

$g=span\left\{{a}_{\left(n\right)};a\in {R}_{\stackrel{¯}{0}}\cup {R}_{\stackrel{¯}{1}},n\in Ζ\right\}$ ,

${g}_{-}=span\left\{{a}_{\left(n\right)};a\in {R}_{\stackrel{¯}{0}}\cup {R}_{\stackrel{¯}{1}},n\ge 0\right\}\subset g$ .

${a}_{\left(-{m}_{1}\right)}^{1}{a}_{\left(-{m}_{2}\right)}^{2}\cdots {a}_{\left(-{m}_{r}\right)}^{r}\cdot 1$ , ${m}_{1}\ge {m}_{2}\ge \cdots \ge {m}_{r}\ge 1$ .

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