Contents

3.6 Spinors 41spinors, but two-index spinors v AḂ, because the sign ambiguity arisingfrom the map of SL(2, C) toL ↑ + must be removed and it must be possibleto relate the length of a vector in the Minkowski metric (quadratic inthe components v a ) to a determinant (also quadratic in the entries of a(2 × 2) matrix). Such a map will be given by any set of spin tensors σ aA Ḃsatisfyingσ aA Ḃ σaCḊ = −δ C A δḊḂ⇔ σ aA Ḃ σbAḂ = −δ b a. (3.69)Then v a corresponds to v AḂ byv a = −σ a AḂvAḂ⇔ v AḂ = σ aAḂv a . (3.70)(Note that the formulae here exhibit some sign changes as compared withPenrose (1960), due to a change of convention about the signature of thespace-time metric.) The spin tensors will be Hermitian,σ aAḂ = σ aB ˙ A ≡ σ aAḂ.(3.71)Given a null vector, its spinor counterpart must be an outer productζ AηḂ since the matrix v A Ḃmust have determinant zero and thus be ofrank 1. Given a null tetrad (m, m, l, k) and a pair of basis spinors o A , ι Asuch that o A ι A =1(adyad), one can choose the map σAḂ a so that inthe orthonormal tetrad associated with (m, m, l, k) by (3.12), and in thespin basis consisting of o A , ι A themselves (so o A =(1, 0) and ι A =(0, 1)),one hasσAḂ 1 = √ 1 ( ) 0 1,σAḂ 2 1 0 2 = √ 1 ( ) 0 i, 2 −i 0σAḂ 3 = √ 1 ( ) 1 0,σAḂ 2 0 −1 4 = √ 1 ( ) (3.72)1 0.2 0 1Thenm a ↔ o A ῑḂ, m a ↔ ι A ōḂ, l a ↔ ι A ῑḂ, k a ↔ o A ōḂ. (3.73)One can check from (3.69), (3.70) that this is consistent with the normalizationof (m, m, l, k). Conversely (3.73) could be used to define a dyadso that (3.72) arises.The null rotations (3.14), (3.15) correspond to the transformationso ′A = o A + Eι A , ι ′A = ι A + Bo A . (3.74)In Table 3.1 we give some examples of spinor equivalents of tensors, constructedaccording to the relation (3.70). The spinor form of the decomposition(3.45) of the curvature tensor is obtained from the spinor