Geometric Meaning of Complex Null Vector in Newman-Penrose Formalism

[RiccardoVen]

Reading Chandrasekhar's The mathematical theory of black holes, I reached the point in which the Newman-Penrose GR formalism is explained. Actually I'm able to grasp and understand the usage of null tetrads in GR, but The null tetrads used in this formalism, are very special, since are made by a couple or real null vectors, and another couple of complex null vector.
I'm struggling a bit in understanding the "geometric" meaning of a complex null vector, in the context of differential geometry. Actually I could take a couple of real vectors and getting their dot product using the metric. If the we take a Lorenz metric, for instance, we can use that metric and check if two 4-vectors are orthogonal having the dot product = 0.
But how a complex vector can be seen as a 4-vector? My problem is I can "see" a 4-vector as a "real vector", but how a complex vector is "embedded" into a 4D spacetime?
I have the feeling, since this refer to Penrose as well, this is something related to the spinorial description of 4D spacetime. But in the original Penrose formulation, the spinors were not there yet.
For example, we may a real null vector n and a real complex vector m, and Penrose requires their product to be 0. How this can expanded, in tensor notation, without using spinors?

How my question makes sense to you.

Thanks

 

[Mentz114]

Reading Chandrasekhar's The mathematical theory of black holes, I reached the point in which the Newman-Penrose GR formalism is explained. Actually I'm able to grasp and understand the usage of null tetrads in GR, but The null tetrads used in this formalism, are very special, since are made by a couple or real null vectors, and another couple of complex null vector.
I'm struggling a bit in understanding the "geometric" meaning of a complex null vector, in the context of differential geometry. Actually I could take a couple of real vectors and getting their dot product using the metric. If the we take a Lorenz metric, for instance, we can use that metric and check if two 4-vectors are orthogonal having the dot product = 0.
But how a complex vector can be seen as a 4-vector? My problem is I can "see" a 4-vector as a "real vector", but how a complex vector is "embedded" into a 4D spacetime?
I have the feeling, since this refer to Penrose as well, this is something related to the spinorial description of 4D spacetime. But in the original Penrose formulation, the spinors were not there yet.
For example, we may a real null vector n and a real complex vector m, and Penrose requires their product to be 0. How this can expanded, in tensor notation, without using spinors?

How my question makes sense to you.

Thanks

I can only comment on the geometrical implications of introducing a complex vector from a rather basic understanding. It is like introducing another dimension which is orthogonal to the other spatial dimensions and operations like conjugation or multiplying by $i$ make a π/2 rotation.

Can the complex vector be related to some kind of dual form or even replaced by it ? I'm not sure if dual tensors have a clear geometrical interpretation so this is a shot in the dark.

 

[RiccardoVen]

I can only comment on the geometrical implications of introducing a complex vector from a rather basic understanding. It is like introducing another dimension which is orthogonal to the other spatial dimensions and operations like conjugation or multiplying by $i$ make a π/2 rotation.

Can the complex vector be related to some kind of dual form or even replaced by it ? I'm not sure if dual tensors have a clear geometrical interpretation so this is a shot in the dark.

Thanks for your opinion. Actually is by far more complex than this. The PI/2 trick, gotten from applying the "i" is actually working on 2D, i.e. when you relate a 2D vector to a complex number. Here we are in 4D, and it's totally different.
Actually I found partially an answer to this accessing my Penrose's book about spinors (the 1st volume). Here you can read in many places "a complex null vector corresponds to a couple of real null vector". Always remember "null" here has a very specific and technical definition, i.e. "a vector which norm is 0", where the norm is here computed taking the scalar prod of a vector with itself (using the correct metric, of course).
So, since this pertaining to the Newman-Penrose formalism it makes sense you have to rely to Penrose's invention about representing the 4D spacetime by spinors. In this sense, a spinor is the "square root" of a vector, so this could answer to my question (even if a spinor is not really a complex null vector, indeed).
But my original question was about to keep a tensorial approach and understanding which a complex null vector is in that context.