This is sort of a continuation of this and this previous discussions.
In the first of my links one sees the surjective isometry between real or complex $(1,3)$ signature Minkowski space and the real or complex (respectively) span of Pauli matrices to be given as, $V_\mu \mapsto V_{\alpha \dot{\alpha}} = V^\mu \sigma _{\mu \alpha \dot{\alpha}}$ using the standard Pauli matrices.
- In reference to the second of the above links one would need that another set of matrices (say A) to "reverse" the above map as $V^\mu = A^{\mu \dot{\alpha} \alpha} V_{\alpha \dot{\alpha}}$. What would be these matrices $A^\mu$?
(..in the expression similar to the above in Lubos's answer he has $V_\mu = \sigma _ \mu ^{\alpha \dot{\alpha}} V_{\alpha \dot{\alpha}}$..which seems to have the $\alpha$ and $\dot{\alpha}$ indices in unexpected (wrong?) places and doesn't seem to the inverse of $V_{\alpha \dot{\alpha}} = V^\mu \sigma _{\mu \alpha \dot{\alpha}}$..)
In the second of my links the answer by Lubos says that the relationship between $V_\mu$ and $V_{\alpha \dot{\alpha}}$ is a reflection of the fact that $SL(2,C)$ is a double cover (hence locally isomorphic) of $SO^+(1,3)$.
This is confusing to me - since as also pointed out in the same answer a $4$ vector can be thought of as a tensor product of the $2$ and $\bar{2}$ representations of $SL(2,C)$ (..the left and the right handed Weyl fermions..)
Hence isn't the relationship between $V_\mu$ and $V_{\alpha \dot{\alpha}}$ and the above interpretation coming from the fact that the complexified $(1,3)$ signature M inkowski space supports a $(\frac{1}{2},\frac{1}{2})$ representation of $SL(2,C)\times SL(2,C)$?
- In the light of the above isn't it necessary that one things of complex space-time vectors $V_\mu$ to do the interpretation of them as lying in tensor products of $2$ and $\bar{2}$ representations of $SL(2,C)$?
Or is it there in some implicit form since Lubos doesn't seem to need the null twistor condition as required in the answer by Roy Simpson to map to real Minkowski space.
The above mapping between $V_\mu$ and $V_{\alpha \dot{\alpha}}$ seems to have in it built the most negative sign convention for the metric. ($det(V_{\alpha \dot{\alpha}}) = (V^0)^2 - \sum _{i =1} ^{i=3} (V^i)^2$) How does one change the mapping if one wants to work in the most positive sign convention?
Once one has mapped a space-time vector $V^\mu$ to $V_{\alpha \dot{\alpha}}$ when is it true that one can now find a left and right chiral Weyl spinors $\lambda_\alpha$ and $\bar{\lambda}_\dot{\alpha}$ such that $V_{\alpha \dot{\alpha}} = \lambda_\alpha \bar{\lambda}_\dot{\alpha}$ ?
Is it only when $V$ is a null vector as is the case for modern application of scattering of fast gluons where one neglects their mass? (..but there I am confused to see why not both $\lambda$ and $\bar{\lambda}$ is used for each of the gluons but only one depending on identifications about incoming/outgoing (anti)article nature..)
I am using the convention of having both $\alpha$ and $\dot{\alpha}$ being down as I see in many recent string theory papers. I guess sometimes one wants to write the index of the conjugate representations ($\dot{\alpha}$) upstairs and $\alpha$ downstairs.