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Fun fact: Spinor calculus already contains tensor calculus. Every tensor has a spinor analogue. A spin-vector multiplied by its complex conjugate produces a null vector.
by null vector you mean SU(N) invariant? Then i agree. What's more is that you can construct all tensor reps of SU(N) by suitable direct products of the fundamental representation!
I'm coming from GR here. There is a isomorphism between the space of real [1,1]-index spinors and the tangent space of 4d space-time. (By (1,1) I mean 1 index from a spin-vector space and one from it's complex conjugate). 1/~
yea! we're saying the same thing! What you're calling a [1,1] spinor is actually just the (1/2, 1/2) representation of the Poincare group!
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Then after "walking in a circle" (i.e. 2π radians), you will actually be upside-down.
