Newman-Penrose formalism: Can a normalized spin-frame be found for every null tetrad?

The Newman-Penrose formalism replaces tensors with spinors by introducing vector space isomorphisms $varsigma_p: T_pM to (S otimes overline{S})$, where $S$ is our spinor space and $overline{S}$ is its complex conjugate; corresponding maps for $T_p^*M$ are of course introduced via the canonical isometries. These maps are required to preserve the inner product at each point, so that omitting the $p$ subscript and adopting index notation we write: $$tag{1} g_{ab} = varsigma_a{}^{AA'}varsigma_b{}^{BB'}epsilon_{AB}epsilon_{A'B'}, $$ where $epsilon_{AB}$ ($epsilon_{A'B'}$) is the inner product on $S$ ($overline{S}$). This makes $varsigma$ an isometry at each point.

Thus for each choice of $varsigma$ a null frame corresponds to a unique combination of spinors (at each point). Usually we choose $varsigma$ and a basis $(o^A,iota^A)$ on $S$ such that $$ (l^a,n^a,m^a,overline{m}^a) = (o^Ao^{A'},iota^Aiota^{A'},o^Aiota^{A'},iota^Ao^{A'}). $$ Because of $(1)$, any such spinor basis will be normalized, since the null frame is. Of course, other spinor bases can also be chosen and each one can be made to correspond to the same null frame by redefining $varsigma$.

The crucial point to notice is that with this approach $S$ is not intrinsic to the geometry, but rather $(Sotimesoverline{S})$ is taken as an alternative representation of $T_pM$ at each point.