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Jan 27, 2017 at 23:32 comment added ACuriousMind The trick cannot work for dimension 8: The Dirac spinors in $d$ dimensions have dimension $2^{\lfloor d/2\rfloor}$, and this must match the dimensions of the vectors in order to have the two Weyl spinor be half the dimension of the ordinary vectors, hence to be able to take the sqaure root. But for $d=8$, you have already Dirac spinors of $2^4 = 16$, so the Weyl spinors are equal in dimension to the vectors - you don't get a $\mathbb{C}^4$.
Jan 27, 2017 at 17:48 history answered CR Drost CC BY-SA 3.0