维拉宿代数

维拉宿代數是一李代數，生成元是

• ${\displaystyle {L_{n}:n\in \mathbb {Z} }}$,
• c ，
• 符合：${\displaystyle [L_{m},L_{n}]=(m-n)L_{m+n}+\delta _{m+n}{\frac {(m^{3}-m)}{12}}c}$

维拉宿代數可以被认为是以下Witt 代数 的 中心拓展:

${\displaystyle [l_{m},l_{n}]=(m-n)l_{m+n}}$,

${\displaystyle [{\bar {l}}_{m},{\bar {l}}_{n}]=(m-n){\bar {l}}_{m+n}}$,

${\displaystyle [l_{m},{\bar {l}}_{n}]=0}$.

对于一李代数${\displaystyle \ {\bf {g}}}$, 其在复数域${\displaystyle \ {\bf {C}}}$的 central extension${\displaystyle \ {\tilde {g}}}$ 满足下列交换子:

${\displaystyle [{\tilde {x}},{\tilde {y}}]_{\tilde {g}}=[x,y]_{g}+cp(x,y),}$

${\displaystyle [{\tilde {x}},c]_{\tilde {g}}=0,}$

${\displaystyle [c,c]_{\tilde {g}}=0,}$

其中${\displaystyle \ {\tilde {x}},{\tilde {y}}\in {\tilde {g}},x,y\in g,c\in {\bf {C}},p:{\tilde {g}}\times {\tilde {g}}\rightarrow {\bf {C}}}$. 由此定义, 维拉宿代數的生成元满足以下交换子

${\displaystyle [L_{m},L_{n}]=(m-n)L_{m+n}+cp(m,n)}$.

${\displaystyle p(m,n)}$可以由以下条件决定:

• 交换子必须是反对易的, 所以${\displaystyle p(m,n)=-p(n,m)}$
• 可以观察到, 如果定义以下生成元

${\displaystyle {\hat {L}}_{n}=L_{n}+{\frac {cp(n,0)}{n}},n\neq 0}$

${\displaystyle {\hat {L}}_{0}=L_{0}+{\frac {cp(1,-1)}{2}},}$

它们满足

${\displaystyle [{\hat {L}}_{n},{\hat {L}}_{0}]=nL_{n}+cp(n,0)=n{\hat {L}}_{n},}$

${\displaystyle [{\hat {L}}_{1},{\hat {L}}_{-1}]=2L_{0}+cp(1,-1)=2{\hat {L}}_{0}.}$

比较函数${\displaystyle p(m,n)}$的定义可以得知,${\displaystyle p(1,-1)}$与${\displaystyle p(n,0)}$总是可以被设为0.

• 交换子满足雅可比恒等式,即

${\displaystyle \ 0=[[L_{m},L_{n}],L_{0}]+[[L_{n},L_{0}],L_{m}]+[[L_{0},L_{m}],L_{n}]}$

	${\displaystyle \ =(m-n)cp(m+n,0)+ncp(n,m)-mcp(m,n)}$
	${\displaystyle \ =(m+n)p(n,m)}$

所以${\displaystyle p(n,m)=0}$如果${\displaystyle n\neq -m}$, 即唯一的非零 central extension为${\displaystyle p(n,-n)}$且${\displaystyle |n|>=2}$.

• 最后计算以下雅克比恒等式

${\displaystyle \ 0=[[L_{-n+1},L_{n}],L_{-1}]+[[L_{n},L_{-1}],L_{-n+1}]+[[L_{-1},L_{-n+1}],L_{n}]}$

${\displaystyle \ =(-2n+1)cp(1,-1)+(n+1)cp(n-1,-n+1)+(n-1)cp(-n,n)}$

可知${\displaystyle p(m,n)}$满足以下递推公式

${\displaystyle \ p(n,-n)={\frac {n+1}{n-2}}p(n-1,-n+1)}$

${\displaystyle \ ={\frac {n+1}{n-2}}{\frac {n}{n-3}}p(n-2,-n+2)}$=...

${\displaystyle \ ={\frac {n+1}{n-2}}{\frac {n}{n-3}}...{\frac {4}{1}}p(2,-2)}$

${\displaystyle \ ={n+1 \choose 3}{\frac {1}{2}}}$

${\displaystyle \ ={\frac {1}{12}}(n+1)n(n-1),}$

其中归一化条件为${\displaystyle p(2,-2)={\frac {1}{2}}}$.综上所述, Witt algebra在复数域唯一非零的central extension, 即维拉宿代数的生成元满足以下交换子

${\displaystyle [L_{m},L_{n}]=(m-n)L_{m+n}+c{\frac {1}{12}}(n+1)n(n-1)\delta _{m+n,0}}$.

• V.G. Kac: "Infinite dimensional Lie algebras", Cambridge University Press
• V.G. Kac / A.K. Raina : "Bombay Lectures on highest weight representations" , World Scientific, Singapore
• Di Francesco / Mathieu / Senechal : "Conformal field theory", Springer Verlag
• Wakimoto: "Infinite-dimensional Lie algebras" （日語書《無限次元環》的譯本）, American Mathematical Society
• Ralph Blumenhagen/ Erik Plauschinn : "Introduction to conformal field theory: with applications to string theory", Springer Lecture notes in physics 779, Page 15