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The space-time metric decomposes into the antisymmetric epsilon spinor and it's complex conjugate. (2 st-indices -> 4 spinor indices). Therefore the contraction of the product of two spin-vectors with itself will give 0 (because the cont. with the epsilons are symmetric). 2/~
i think maybe you're mixing "components" with indices. One spacetime index μ <--> two weyl spinor indices α\dot{α}. So that way they both have 4 degrees of freedom.
I'm not sure if I understand that correctly. I think, I am talking about indices. 1 spinor index has two complex components. 1 space-time index corresponds to 2 spinor indices -> 4 complex components -> 8 real components 2.5/~
but the tangent space only corresponds to real [1,1] spinors, so we are back to our 4 real components
Ah yes! I think I know what you are referring to. I meant exactly that. I consider a Dirac/Weyl spinor as consisting of two 2-component spinors.
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Then after "walking in a circle" (i.e. 2π radians), you will actually be upside-down.
