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

Question

I have a question within the framework of spin-coefficient formalism / Newman-Penrose formalism. Suppose I have a space-time metric whose components with respect to some coordinate basis are known. I can then easily construct a null tetrad (to be more precise: the vector components of the tetrad vectors to the coordinate basis).

Since many calculations are easier if the corresponding spinor basis is a normalized spin frame $o_A\iota^A=1$ I would like to know if the spinor basis that corresponds to my null tetrad is such a normalized spin frame.

I am a bit lost at that point. Can I find a spinor basis in spinor space that corresponds to the null tetrad basis in tensor space and will it be a normalized spin frame? Or is there a multitude of possible spinor bases that can be normalized through some procedure? Does it depend on the specific null tetrad (for a given metric there is no unique null tetrad, but a number of independent possibilities)?

Answer 1

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 $$ (l^a,n^a,m^a,\overline{m}^a) = (o^Ao^{A'},\iota^A\iota^{A'},o^A\iota^{A'},\iota^Ao^{A'}). $$ 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 $(S\otimes\overline{S})$ is taken as an alternative representation of $T_pM$ at each point.

Answer 2

I agree with @Erik Jorgenfelt . Besides, I will like to add more about the gauge freedom which preserves the normalization of 2-spinor basis.

For a given spin frame $\{o_A,\iota^A\}$, we can see that linear transformations which preserve the normalization condition is given by : $$o_A\mapsto \lambda o_A \:\:\:\:\:\:\:\:\iota^A\mapsto \lambda^{-1}\iota^A\:\:\:\:\:\:\forall\lambda\in C^{*}$$ Under such transformation, the null tetrads transforms to a new set given by: $$l^a\mapsto \lambda\bar{\lambda}l^a\:\:\:\:\:\:\:\bar{m}^a\mapsto\bar{\lambda}\lambda^{-1}\bar{m}^a$$ $$m^a\mapsto \lambda\bar{\lambda}^{-1}m^a\:\:\:\:n^a\mapsto\lambda^{-1}\bar{\lambda}^{-1}n^a$$ We can set $\lambda^2=Re^{i\theta}$, then the null direction for $\textbf{l},\textbf{n}$ are preserved but not their magnitudes: $$l^a\mapsto R l^a\:\:\:n^a\mapsto R^{-1}n^a$$, but it will induce spatial rotation for complex vectors $\textbf{m},\bar{\textbf{m}}$ : $m^a\mapsto e^{i\theta}m^a$

Here $(R,\theta)$ defines a 2-parameter gauge freedom which preserves the normalization and therefore a certain subgroup of Lorentz transformation.