Are spinors intrinsically nonlocal?

Question

I would prefer a purely classical answer since I don't think quantum mechanics (quantum field theory etc.) are necessary to answer this question and such answers will likely complicate matters. If you disagree with this sentiment then please explain why you disagree.

One motivation for asking is the Dirac belt trick. My reading of the Dirac belt trick is if you have two ends of an orientable belt and you rotate one end by 360 degrees you don't get back where you started. Rotating one end is a local operation. But the belt is physically extended and your rotation of one end was not a rotation of the complete system, so it's really not so surprising. The belt is a "non-local" object with respect to rotation of only one of the ends. So the point is: "is the case for spinors somehow similar?". i.e. the strange behavior of spinors is related to the fact that they actually encode something non-local.

Another motivation for asking is that to define spinors at all it seems to be very necessary to ensure you are working in the right sort of global space. I.e. it has the right structure, transformation properties etc. This is in contrast to regular tensors which can be defined on smooth manifolds. That is, on a manifold I can define vector and tensor spaces starting by defining vectors as derivations at a point. I think this definition is entirely local (the only possible non-local thing is that derivations are defined as derivatives in the coordinate space which involves limits of slightly non-local displacements). For a tensor field I can always straightforwardly answer the question "what is the tensor at point $p$"? Can the same be said for spinor fields? Do spinor fields even make sense in the sense the tensor fields do?

Another point which may not be directly related to the stated question, but is related to why spinors are so complicated is this. To define vectors in the tangent spaces of a smooth manifold I only need to know some calculus. But to define spinors it feels like I need to know all of differential geometry and a bunch of group theory and representation theory. And a bunch of stuff has to a-priori "transform in the right ways" to get spinors. But none of that was needed for vectors. Why do we need so much structure and representation theory to understand spinors when nothing of the sort is needed for vectors?

To extend the last point a little bit: I can also define vectors without any reference to a manifold. I can just lay out the axioms of a vector space and speak generally about a vector detached from any kind of physical space. It doesn't seem like there's an analogous "set of axioms about a spinor space" and then we can pick out spinors from that space detached from any physical space. It somehow feels like the ambient physical space/manifold is critical to the definition of and existence of manifolds. This is another difference with vectors that makes me feel like spinors are about something more global than a vector is.

This question is related to Are fermions intrinsically non-local? however this question focuses on spinors specifically (not fermions) and asks different specific questions.

Answer 1

No, spinors are not non-local in any of the usual senses of the word.

The correct formalization of spinor fields is exactly analogous to that of ordinary vector and tensor fields - they are sections of a bundle, the spinor bundle, and one of the equivalent ways of representing such a section is as a function (or set of local functions for non-trivial bundles) that assigns to every point on the manifold a spinor, exactly like a vector field is a section of the tangent bundle that can be written as a function (or set of local functions for non-trivial bundles) that assigns to every point on the manifold a tangent vector.

A "spinor" for a spacetime with signature $p,q$ is just an element of the Dirac representation of $\mathrm{SO}(p,q)$, which is the unique irreducible representation (or one of the two irreps, depending on the signature) of the Clifford algebra ("$\gamma$-matrices") associated with signature $p,q$.

So the spinor bundle is just a vector bundle with the Dirac representation ("spinor space") as its fiber. While there is a global obstruction to the existence of a spin structure, spinors are therefore not somehow a new kind of mathematical object - they are mathematically just vector fields transforming in a particular representation of the generalized Lorentz group. There is nothing non-local about the individual spinors - that there are some sort of global obstructions to the existence of particular fields is not unique to spin structures, e.g. spacetimes for which you can choose a consistent time direction also have to have a particular global structure, but no one thinks this makes a time-like vector field "non-local".

Answer 2

The other answer is correct that spinors are just like tensor fields, except that they don't really come from a representation of the Lorentz group. Rather, they come from a representation of the spin group, which is a 2-to-1 cover of the Lorentz group. In particular, it doesn't make sense to ask how a spinor transforms under (say) a rotation. Instead, you need to pick an element of the spin group which covers your rotation, and ask how the spinor transforms under that. You can't necessarily pick that element in a coherent, continuous way everywhere for every Lorentz transformation. Roughly speaking, this would be what we call a choice of spin structure: depending on the topology of space-time, this may or may not exist, and if it exists, it may not be unique. For this to be an issue, there has to be some interesting cohomology in degree 2 (the second Stiefel-Whitney class is the obstruction to finding a spin structure). So the existence of a spin structure is highly nonlocal.

EDIT: The problem of finding spin structures is somewhat like the problem of defining an orientation (handedness). In that setting, you can always do it locally, but sometimes you go around a loop and find that your choice switched. The problem for spin structures is that you might try to define it over a whole sphere, starting at the North pole, only to discover that you can't extend your choice over the South pole. END EDIT

The reason for the existence of the spin group is that the Lorentz group (or its maximal compact the rotation group) is not simply-connected: the loop of rotations which turns 360 degrees around a given axis cannot be deformed continuously to the "constant loop" which does nothing. Upstairs, in the spin group, this 360-degree loop is not a loop at all: each rotation along the loop is covered by two elements of the spin group, and when you go all the way around the loop, these switch over. They switch back if you rotate by a further 360 degrees.

The belt trick is just illustrating this: think of the belt as having an s-coordinate (long direction) and a t-coordinate (short direction). As you go along the s-direction, the t-direction twists, and by the time you get to the end you've twisted by 360 degrees. The twisted belt as a whole represents a path of rotations. As long as you keep the endpoints fixed relative to one another you can't undo the twist. You could undo it if you had twisted by 720 degrees instead.

One nice thing you can do is to think about "projective spinors"; for simplicity let's work with SO(3) or SO(3,1). Instead of your spinor being a complex 2-vector, you think of it as a nonzero complex 2-vector up-to-scale (where scale means a complex rescaling). The space of projective spinors is the Riemann sphere, and rotations/Lorentz transformations do act in a well-defined way on that (as rotations/Moebius transformations respectively). This works because the two different spin group elements giving the same rotation just differ by a sign, which then gets ignored when you take the projectivisation.