Jacobi Algebra
221
where ν k are numerical coefficients which may be fixed when we impose (11a)
on (12). (Note that (11b) is fulfilled by every term of (12).)
After we impose (11a) on (12) we find the solution:
Λ(H 1 ) =
3
4 , ν 3 = − 2ν 6 ,
ν 1 = −ν 6 (Λ(H 2 ) − Λ(H 1 ))(2Λ(H 2 ) − 2Λ(H 1 ) − 1),
ν 2 = 2ν 6 (2Λ(H 2 ) − 2Λ(H 1 ) − 1),
ν 4 = ν 6 (Λ(H 2 ) − Λ(H 1 ))(Λ(H 2 ) − Λ(H 1 ) −
1
2 ),
ν 5 = −ν 6 (2Λ(H 2 ) − 2Λ(H 1 ) − 1).
(13)
Thus, the singular vector is:
v
2δ 1
s
= ν 6
(Λ(H 2 ) −
3
4 )(Λ(H 2 ) −
5
4 )((a
+
1 )
2
− 2b
+
1 ) +
+ 2(Λ(H 2 ) −
5
4 )(2c
+
− a
+
1 a
+
2 )d
+
+
+ ((a
+
2 )
2
− 2b
+
2 )(d
+ )
2
v 0 , Λ(H 1 ) =
3
4 .
(14)
Weight 2δ 2
As next example we try to find a singular vector of weight Λ ∼ 2δ 2 . The possible
singular vector is:
v
2δ 2
s
=
μ 1 b
+
2 + μ 2 (a
+
2 )
2
v 0 .
(15)
Imposing (11a) on (15) we obtain:
Λ(H 2 ) =
1
4 , μ 1 = − 2μ 2 ,
(16)
Thus, the singular vector is:
v
2δ 2
s
= μ 2 ((a
+
2 )
2
− 2b
+
2 )v 0 , Λ(H 2 ) =
1
4 .
(17)
Weight δ 1 + δ 2
Next we try a singular vector of weight Λ ∼ δ 1 + δ 2 . The possible singular
vector is:
v
δ 1 +δ 2
s
=
κ 1 c
+
+ κ 2 b
+
2 d
+
+ κ 3 a
+
1 a
+
2 + κ 4 (a
+
2 )
2 d
+
v 0 .
(18)