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Does spinor formalism work just in four dimensions? I am reading the book "Spinors and space-time" written by Penrose. I didn't understand a statement. Penrose says that besides tensor formalism and Cartan formalism, in four dimensions there is another formalism - by accident or providence- for the treatment of manifolds, namely 2-spinor formalism. So does it mean that spinor formalism works just in four dimensions?

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You're in for a real treat -- I really liked those books, and I still use his approach of "use the scalar fields to define everything" to understand differential geometry.

What he's saying is that, indeed, he is going to use 2-spinors to take the "square root" of the $(+~-~-~-)$ metric in the form of an antisymmetric tensor $\epsilon_{AB} = \begin{bmatrix}0&-1\\1&0\end{bmatrix}$ on vectors in $\mathbb C^2$.

In the end you will end up having your 4-vectors be valence-$[2, 0]$ tensors in the spin-space. You can have a bunch of different sign conventions for the metric depending on what subset of tensors you choose for this, but the $(+~-~-~-)$ metric comes about from expecting a sort of "Hermitian" quality to these tensors, where if $$\tau^{A\bar A} = t_{00} ~\omicron^A \omicron^{\bar A} + t_{01} ~\omicron^A \iota^{\bar A} + t_{10} ~\iota^A \omicron^{\bar A} + t_{11}~\iota^A\iota^{\bar A},$$ then we expect $t_{00}$ and $t_{11}$ to be real while $t_{01} = t_{10}^*.$ And for this particular subspace you should get that this inner product, $$\epsilon_{AB}~\epsilon_{\bar A \bar B} ~\tau^{A\bar A}~\omega^{B\bar B}$$works precisely like $g_{ab}~\tau^a ~\omega^b$ with $g_{ab} = \operatorname{diag}(1, -1, -1, -1)$ in the appropriate basis.

In fact the only thing that doesn't get substantially simpler (but neither does it get any harder) is the connection $\nabla_{a}$ which now becomes $\nabla_{A\bar A}$ but remains applicable to spinor expressions. But lots of other expressions, in particular for the curvature tensor and such, become much easier as you can just say, "oh, this world-tensor is (anti-)symmetric in those in those indices, that means that the corresponding spinor breaks into a sum of these these different parts." In particular the vacuum equations for the electromagnetic field simplify to just $\nabla^{A\bar A}\phi_{AB},$ and for the gravitational field it's something similar like $\nabla^{A\bar A}\Psi_{ABCD} = 0$ or so.

So now the quote,

The formalism most commonly used for the mathematical treatment of manifolds and their metrics is, of course, the tensor calculus (or such essentially equivalent alternatives as Cartan's calculus of moving frames). But in the specific case of four dimensions and Lorentzian metric there happens to exist—by accident or providence—another formalism which is in many ways more appropriate, and that is the formalism of 2-spinors.

Now, you are asking, "Does this only work in 4 dimensions?" and the answer is "well, sort of." If you had an 8-dimensional space you might be able to take the square root of some metric tensor, and have it be in terms of vectors in $\mathbb C^4.$ For that matter with a non-Minkowski metric you can use the formalism; you just accept that the matrices are not Hermitian. But I think instead he's probably saying that you do not "get as much out of the formalism" as you do in the 4D case. In other words it's the words "more appropriate" that bear the meaning of the passage: in higher dimensions it does not help to simplify your algebra nearly so much to do all of this spinor decomposition to try to understand the system better; there are probably ways to do it and it might even take a "chunk" of the difficulty away, but there is just so much more difficulty overall that the percentage improvement isn't nearly so impressive.

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  • $\begingroup$ 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 $\mathbb{C}^4$. $\endgroup$ – ACuriousMind Jan 27 '17 at 23:32

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