34
2 Bivector Formalism
which follows from the first of (2.80) and also from
∗ C abcd =
1
2
η ab
rs C rscd =
1
2
η ab
rs C cdrs = C
∗
cdab =
∗ C cdab ,
(2.104)
with the final equality coming from (2.87). If we now substitute (2.93)–(2.95) into
(2.92) we see immediately that (2.103) is satisfied provided
α 1 = β 2 , 2 α 3 = γ 2 , 2 β 3 = γ 1 .
(2.105)
With these satisfied we readily see that (2.90) reduces to (γ 3 + 2 β 2 )g ab = 0, with
g ab given in terms of the null tetrad by (2.14), and so we must also have
γ 3 + 2 β 2 = 0 .
(2.106)
We can verify (2.105) and (2.106) directly from (2.96)–(2.102) using the algebraic
symmetries of C abcd to arrive at
α 1 = −2 C abcd m
a k
b
¯
m
c l
d
= β 2 = −
1
2
γ 3 ,
(2.107)
α 2 = −2 C abcd m
a k
b m
c k
d ,
(2.108)
α 3 = −2 C abcd m
a k
b k
c l
d
=
1
2
γ 2 ,
(2.109)
β 1 = −2 C abcd ¯
m
a l
b
¯
m
c l
d ,
(2.110)
β 3 = −2 C abcd k
a l
b
¯
m
c l
d
=
1
2
γ 1 .
(2.111)
Finally (2.92) can be written as
1
2
C abpq = 0 M ab M pq + 1 (M ab N pq + N ab M pq ) + 2 (−N ab N pq
+ M ab L pq + L ab M pq ) + 3 (L ab N pq + N ab L pq ) + 4 L ab L pq ,
(2.112)
where the five complex-valued coefficients here are given by
0 = −
1
2
α 2 = C abcd m
a k
b m
c k
d ,
(2.113)
1 = −
1
2
α 3 = C abcd m
a k
b k
c l
d ,
(2.114)
2 = −
1
2
α 1 = C abcd m
a k
b
¯
m
c l
d ,
(2.115)