222
V. K. Dobrev
Imposing (11a) on (18) we obtain:
Λ(H 2 ) =
3
2 − Λ(H 1 ) , κ 1 = (3 − 4h(1))κ 4 ,
κ 2 = −2κ 4 , κ 3 = (2h(1) −
3
2 )κ 4 .
(19)
Thus, the singular vector is:
v
δ 1 +δ 2
s
= κ 4
(
3
2 − 2h(1))(2c
+
− a
+
1 a
+
2 ) + ((a
+
2 )
2
− 2b
+
2 )d
+
v 0 .
(20)
Weight δ 1 − δ 2
Next we try a singular vector of weight Λ ∼ δ 1 − δ 2 . The only possible singular
vector is:
v
δ 1 −δ 2
s
= λd
+ v 0 .
(21)
Imposing (11a) on (21) we obtain that v
δ 1 −δ 2
s
is a singular vector iff:
Λ(H 2 ) = Λ(H 1 ).
(22)
Weight δ 1
Next we try a singular vector of weight Λ ∼ δ 1 . The possible singular vector is:
v
δ 1
s =
λ 1 a
+
1 + λ 2 a
+
2 d
+
v 0 .
(23)
Imposing (11a) on (23) we obtain:
λ 1 = λ 2 = 0.
(24)
Thus, there is no singular vector of weight δ 1 .
Weight δ 2
Finally, we try a singular vector of weight Λ ∼ δ 2 . The only possible singular
vector is:
v
δ 2
s = μa
+
2 v 0 .
(25)
Imposing (11a) on (25) we obtain:
μ = 0.
(26)
Thus, there is no singular vector of weight δ 2 .