維拉宿代數可以被認為是以下Witt 代數 中心拓展
[
l
m
,
l
n
]
=
(
m
−
n
)
l
m
+
n
{\displaystyle [l_{m},l_{n}]=(m-n)l_{m+n}}
[
l
¯
m
,
l
¯
n
]
=
(
m
−
n
)
l
¯
m
+
n
{\displaystyle [{\bar {l}}_{m},{\bar {l}}_{n}]=(m-n){\bar {l}}_{m+n}}
[
l
m
,
l
¯
n
]
=
0
{\displaystyle [l_{m},{\bar {l}}_{n}]=0}
對於一李代數
g
{\displaystyle \ {\bf {g}}}
C
{\displaystyle \ {\bf {C}}}
g
~
{\displaystyle \ {\tilde {g}}}
交換子 :
[
x
~
,
y
~
]
g
~
=
[
x
,
y
]
g
+
c
p
(
x
,
y
)
,
{\displaystyle [{\tilde {x}},{\tilde {y}}]_{\tilde {g}}=[x,y]_{g}+cp(x,y),}
[
x
~
,
c
]
g
~
=
0
,
{\displaystyle [{\tilde {x}},c]_{\tilde {g}}=0,}
[
c
,
c
]
g
~
=
0
,
{\displaystyle [c,c]_{\tilde {g}}=0,}
其中
x
~
,
y
~
∈
g
~
,
x
,
y
∈
g
,
c
∈
C
,
p
:
g
~
×
g
~
→
C
{\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}}}
[
L
m
,
L
n
]
=
(
m
−
n
)
L
m
+
n
+
c
p
(
m
,
n
)
{\displaystyle [L_{m},L_{n}]=(m-n)L_{m+n}+cp(m,n)}
p
(
m
,
n
)
{\displaystyle p(m,n)}
交換子必須是反對易的, 所以
p
(
m
,
n
)
=
−
p
(
n
,
m
)
{\displaystyle p(m,n)=-p(n,m)}
可以觀察到, 如果定義以下生成元
L
^
n
=
L
n
+
c
p
(
n
,
0
)
n
,
n
≠
0
{\displaystyle {\hat {L}}_{n}=L_{n}+{\frac {cp(n,0)}{n}},n\neq 0}
L
^
0
=
L
0
+
c
p
(
1
,
−
1
)
2
,
{\displaystyle {\hat {L}}_{0}=L_{0}+{\frac {cp(1,-1)}{2}},}
它們滿足
[
L
^
n
,
L
^
0
]
=
n
L
n
+
c
p
(
n
,
0
)
=
n
L
^
n
,
{\displaystyle [{\hat {L}}_{n},{\hat {L}}_{0}]=nL_{n}+cp(n,0)=n{\hat {L}}_{n},}
[
L
^
1
,
L
^
−
1
]
=
2
L
0
+
c
p
(
1
,
−
1
)
=
2
L
^
0
.
{\displaystyle [{\hat {L}}_{1},{\hat {L}}_{-1}]=2L_{0}+cp(1,-1)=2{\hat {L}}_{0}.}
比較函數
p
(
m
,
n
)
{\displaystyle p(m,n)}
p
(
1
,
−
1
)
{\displaystyle p(1,-1)}
p
(
n
,
0
)
{\displaystyle p(n,0)}
0
=
[
[
L
m
,
L
n
]
,
L
0
]
+
[
[
L
n
,
L
0
]
,
L
m
]
+
[
[
L
0
,
L
m
]
,
L
n
]
{\displaystyle \ 0=[[L_{m},L_{n}],L_{0}]+[[L_{n},L_{0}],L_{m}]+[[L_{0},L_{m}],L_{n}]}
=
(
m
−
n
)
c
p
(
m
+
n
,
0
)
+
n
c
p
(
n
,
m
)
−
m
c
p
(
m
,
n
)
{\displaystyle \ =(m-n)cp(m+n,0)+ncp(n,m)-mcp(m,n)}
=
(
m
+
n
)
p
(
n
,
m
)
{\displaystyle \ =(m+n)p(n,m)}
所以
p
(
n
,
m
)
=
0
{\displaystyle p(n,m)=0}
n
≠
−
m
{\displaystyle n\neq -m}
p
(
n
,
−
n
)
{\displaystyle p(n,-n)}
|
n
|
>=
2
{\displaystyle |n|>=2}
0
=
[
[
L
−
n
+
1
,
L
n
]
,
L
−
1
]
+
[
[
L
n
,
L
−
1
]
,
L
−
n
+
1
]
+
[
[
L
−
1
,
L
−
n
+
1
]
,
L
n
]
{\displaystyle \ 0=[[L_{-n+1},L_{n}],L_{-1}]+[[L_{n},L_{-1}],L_{-n+1}]+[[L_{-1},L_{-n+1}],L_{n}]}
=
(
−
2
n
+
1
)
c
p
(
1
,
−
1
)
+
(
n
+
1
)
c
p
(
n
−
1
,
−
n
+
1
)
+
(
n
−
1
)
c
p
(
−
n
,
n
)
{\displaystyle \ =(-2n+1)cp(1,-1)+(n+1)cp(n-1,-n+1)+(n-1)cp(-n,n)}
可知
p
(
m
,
n
)
{\displaystyle p(m,n)}
p
(
n
,
−
n
)
=
n
+
1
n
−
2
p
(
n
−
1
,
−
n
+
1
)
{\displaystyle \ p(n,-n)={\frac {n+1}{n-2}}p(n-1,-n+1)}
=
n
+
1
n
−
2
n
n
−
3
p
(
n
−
2
,
−
n
+
2
)
{\displaystyle \ ={\frac {n+1}{n-2}}{\frac {n}{n-3}}p(n-2,-n+2)}
=
n
+
1
n
−
2
n
n
−
3
.
.
.
4
1
p
(
2
,
−
2
)
{\displaystyle \ ={\frac {n+1}{n-2}}{\frac {n}{n-3}}...{\frac {4}{1}}p(2,-2)}
=
(
n
+
1
3
)
1
2
{\displaystyle \ ={n+1 \choose 3}{\frac {1}{2}}}
=
1
12
(
n
+
1
)
n
(
n
−
1
)
,
{\displaystyle \ ={\frac {1}{12}}(n+1)n(n-1),}
其中歸一化條件為
p
(
2
,
−
2
)
=
1
2
{\displaystyle p(2,-2)={\frac {1}{2}}}
[
L
m
,
L
n
]
=
(
m
−
n
)
L
m
+
n
+
c
1
12
(
n
+
1
)
n
(
n
−
1
)
δ
m
+
n
,
0
{\displaystyle [L_{m},L_{n}]=(m-n)L_{m+n}+c{\frac {1}{12}}(n+1)n(n-1)\delta _{m+n,0}}
![{\displaystyle [L_{m},L_{n}]=(m-n)L_{m+n}+\delta _{m+n}{\frac {(m^{3}-m)}{12}}c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0c415f6a2ef15042cf67d166d9614181c6bb952)
![{\displaystyle [l_{m},l_{n}]=(m-n)l_{m+n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d18f9fe8d45442cad2ae220d5d1239268ef3f338)
![{\displaystyle [{\bar {l}}_{m},{\bar {l}}_{n}]=(m-n){\bar {l}}_{m+n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bcb01ab0c4f0f7d6990f00d2a2095f38404c7cb0)
![{\displaystyle [l_{m},{\bar {l}}_{n}]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc7fffb1dddb429f21412d24760e4b664bad1b37)
![{\displaystyle [{\tilde {x}},{\tilde {y}}]_{\tilde {g}}=[x,y]_{g}+cp(x,y),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09cb08877adfeb7d0065008367f13d3ff446b830)
![{\displaystyle [{\tilde {x}},c]_{\tilde {g}}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0226376ead4bbd631eeb4b262e1bd22f8a375dec)
![{\displaystyle [c,c]_{\tilde {g}}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b15b772a9e0302c7ad60323bc29ebd7b66160873)
![{\displaystyle [L_{m},L_{n}]=(m-n)L_{m+n}+cp(m,n)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a23655eb84b5cb6103524849a39811f06b4347f7)
![{\displaystyle [{\hat {L}}_{n},{\hat {L}}_{0}]=nL_{n}+cp(n,0)=n{\hat {L}}_{n},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b10ac217e70e38c9fdb5ea5986c09696a1b544ea)
![{\displaystyle [{\hat {L}}_{1},{\hat {L}}_{-1}]=2L_{0}+cp(1,-1)=2{\hat {L}}_{0}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/161413898619103153aed63c0405853f5954f451)
![{\displaystyle \ 0=[[L_{m},L_{n}],L_{0}]+[[L_{n},L_{0}],L_{m}]+[[L_{0},L_{m}],L_{n}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2654d812d0ba50df6f921fc0b0d383d3ebeedb67)
![{\displaystyle \ 0=[[L_{-n+1},L_{n}],L_{-1}]+[[L_{n},L_{-1}],L_{-n+1}]+[[L_{-1},L_{-n+1}],L_{n}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f5b2771a593374029138567427cbaa7d6470b3d)
![{\displaystyle [L_{m},L_{n}]=(m-n)L_{m+n}+c{\frac {1}{12}}(n+1)n(n-1)\delta _{m+n,0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1d698d7f5c22bb99064f39f84123edf9e95d6a3)