Supersymmetric Grassmannian Sigma Model
97
β(x + , t)=
(c 0 + F (x + ))
√
2 cos t
2x + cos 2t
√
2 cos t
√
2 sin t
0
+ a 1
−
√
2x 2
+ sin 3 t
0
x +
1
√
2
cos t
+ c 1
√
2x 2
+ cos 3 t
x +
0
− 1
√
2
sin t
.
(22)
The susy invariant solution is obtained when a 1 = c 1 = 0. Again the case Z 2
gives other solutions to our problem than the susy invariant ones.
2. The second case corresponds to cos 2t = 0 or t =
π
4 (the case t =
3π
4 is gauge
equivalent) so that K 2 (x + ,
π
4 ) =
0 x +
x + 0
. Since K 2 (x + ,
π
4 ) is symmetric, we
assume that the matrix β(x + ) is also symmetric, i.e. β 21 (x + ) = β 12 (x + ). These
quantities will remain arbitrary since the condition (10) depends only on β 11 and
β 22 and the susy invariant solutions will be obtained when β 11 = β 22 = 0. The
condition (10) may be written as follows, taking in particular x + = x − = x:
(1 + x
2 )
2
4(x
2
− 1)
(β
11 )
2
+ (β
22 )
2
+ (1 + x
2 )
2
(β
11 )
2
+ (β
22 )
2
−8x(1 + x
2 )(x
2
− 2)
β 11 β
11 + β 22 β
22
+ 4x
2 (1 + x
2 )
2
β 11 β
11 + β 22 β
22
+4(1 − 4x
2
+ x
4 )(β
2
11 + β
2
22 ) − 4x(1 + x
2 )
3
β
11 β
11 + β
22 β
22
= 0 . (23)
Let us first mention the invariance of this equation with respect to the exchange
β 11 ↔ β 22 . After some trials we first get a solution choosing β 22 (x) = xβ 11 (x).
Condition (23) thus becomes very simple (1 + x 2 ) 5 (β
11 (x)) 2 = 0, which implies
that
β 11 (x) = a 0 + d 2 x ,
β 22 (x) = x(a 0 + d 2 x) .
(24)
Using this observation, we assume that β 11 (x) and β 22 (x) are real polynomial
in x. We can easily show that they must be at most of degree 2. If we take
β 11 (x) = a 2 x 2 +a 1 x +a 0 , β 22 (x) = d 2 x 2 +d 1 x +d 0 and identify the coefficients
of different powers of x in (23), we get three independent equations for the
parameters a i and d i ,
a
2
0 − a
2
1 + a
2
2 + d
2
0 − d
2
1 + d
2
2 = 0 ,
a 0 a 2 + d 0 d 2 = 0 ,
a 0 a 1 − a 1 a 2 + d 1 (d 0 − d 2 ) = 0 .
(25)
Let us first assume that a 0 = 0, we then get
β 11 (x) = a 0 + (d 2 − d 0 )x −
d 0 d 2
a 0
x
2 , β 22 (x) = d 0 +
a 0 +
d 0 d 2
a 0
x + d 2 x
2 ,
97
β(x + , t)=
(c 0 + F (x + ))
√
2 cos t
2x + cos 2t
√
2 cos t
√
2 sin t
0
+ a 1
−
√
2x 2
+ sin 3 t
0
x +
1
√
2
cos t
+ c 1
√
2x 2
+ cos 3 t
x +
0
− 1
√
2
sin t
.
(22)
The susy invariant solution is obtained when a 1 = c 1 = 0. Again the case Z 2
gives other solutions to our problem than the susy invariant ones.
2. The second case corresponds to cos 2t = 0 or t =
π
4 (the case t =
3π
4 is gauge
equivalent) so that K 2 (x + ,
π
4 ) =
0 x +
x + 0
. Since K 2 (x + ,
π
4 ) is symmetric, we
assume that the matrix β(x + ) is also symmetric, i.e. β 21 (x + ) = β 12 (x + ). These
quantities will remain arbitrary since the condition (10) depends only on β 11 and
β 22 and the susy invariant solutions will be obtained when β 11 = β 22 = 0. The
condition (10) may be written as follows, taking in particular x + = x − = x:
(1 + x
2 )
2
4(x
2
− 1)
(β
11 )
2
+ (β
22 )
2
+ (1 + x
2 )
2
(β
11 )
2
+ (β
22 )
2
−8x(1 + x
2 )(x
2
− 2)
β 11 β
11 + β 22 β
22
+ 4x
2 (1 + x
2 )
2
β 11 β
11 + β 22 β
22
+4(1 − 4x
2
+ x
4 )(β
2
11 + β
2
22 ) − 4x(1 + x
2 )
3
β
11 β
11 + β
22 β
22
= 0 . (23)
Let us first mention the invariance of this equation with respect to the exchange
β 11 ↔ β 22 . After some trials we first get a solution choosing β 22 (x) = xβ 11 (x).
Condition (23) thus becomes very simple (1 + x 2 ) 5 (β
11 (x)) 2 = 0, which implies
that
β 11 (x) = a 0 + d 2 x ,
β 22 (x) = x(a 0 + d 2 x) .
(24)
Using this observation, we assume that β 11 (x) and β 22 (x) are real polynomial
in x. We can easily show that they must be at most of degree 2. If we take
β 11 (x) = a 2 x 2 +a 1 x +a 0 , β 22 (x) = d 2 x 2 +d 1 x +d 0 and identify the coefficients
of different powers of x in (23), we get three independent equations for the
parameters a i and d i ,
a
2
0 − a
2
1 + a
2
2 + d
2
0 − d
2
1 + d
2
2 = 0 ,
a 0 a 2 + d 0 d 2 = 0 ,
a 0 a 1 − a 1 a 2 + d 1 (d 0 − d 2 ) = 0 .
(25)
Let us first assume that a 0 = 0, we then get
β 11 (x) = a 0 + (d 2 − d 0 )x −
d 0 d 2
a 0
x
2 , β 22 (x) = d 0 +
a 0 +
d 0 d 2
a 0
x + d 2 x
2 ,