138
4 Four-Dimensional Superfield Supersymmetry
Moving the term with W
γ K α (t) to the left-hand side of the expression, multiplying
this equation by the matrix inverse to (
e
t N −1
N
), and calculating the trace (with imposing
the restriction N
α
α = 0), we arrive at
W
α K α (t) = tr(
N
e t N − 1
) ˜
K (t).
(4.353)
Proceeding in a similar way for the identity conjugated to (4.350), we find
¯
W ˙
α
¯
K
˙
α
(t) = tr(
¯
N
e t ¯
N − 1
) ˜
K (t).
(4.354)
We note that this term was absent in [74] where only the contribution dependent on
W α , but not on ¯
W ˙
α was considered.
Finally, one can identically repeat the calculation performed in [74], to obtain the
K ab (t). One starts with the identity
0 =
d
4 k
(2π) 4
∂
∂k b
(X a e
t ˜
) = iδ ab ˜
K (t) +
d
4 k
(2π) 4 X a
∂
∂k b
e
t ˜
,
(4.355)
and, similarly to the calculations above, one finds
∂
∂k b
e
t ˜
=
∞
n=0
t
n
(n + 1)!
ad
(n)
(X · X )(X b ) = 2it B bc (t)X c ,
(4.356)
where
B bc =
e
−t ( ¯
M−M)
− 1
−t ( ¯
M − M)
bc
.
(4.357)
Therefore, restoring the K ac with use of its definitions following from (4.345), one
finds that the identity (4.355) leads to
0 = iδ ab ˜
K (t) + 2it B bc (t)K ac (t)
(4.358)
(the term involving K a (t) will be irrelevant just as in [74]), so, one has
K ab (t) = −
1
2t
(B
−1
) ba (t) ˜
K (t) =
1
2
¯
M − M
e −t ( ¯
M−M) − 1
ba
˜
K (t).
(4.359)
For the sake of simplicity, we suggested within these calculations that N
α
α = 0,
as well as N
˙
α
˙
α = 0. These identities can be imposed since these expressions do not
contribute to the degrees of freedom of the stress tensor F ab ; actually, for the Abelian
background superfield they are just equivalent to the Bianchi identities {D
α
, W α } =