5
$\begingroup$

One way to rigorously define spinor fields on metric manifolds is through the language of associated bundles. Namely, we have a principal bundle $P \overset{\pi}{\longrightarrow} M$ over $\mathrm{Spin}(n)$ (or whatever other signature) that is a $\rho$-equivariant lift of the orthonormal frame bundle $OFM \rightarrow M$ through the map $\phi: P \rightarrow OFM$, where $\rho: \mathrm{Spin}(n) \rightarrow SO(n)$ is the double-cover map. $(P,\phi)$ prescribes the spin structure.

The associated bundle is then constructed with a typical fiber $F$, which is a representation space of $\mathrm{Spin}(n)$, and where the field takes its values. The transformation behavior of the objects is at that point fixed through the left and right actions of the group on these bundles. The fields are then taken as sections of the associated bundle, which can equivalently be expressed as $\mathrm{Spin}(n)$-equivariant functions on the principal bundle.

The part that is unclear to me is how these are related to the usual spinors that we see in physics. In particular:

  1. The general construction doesn't specify the choice of spin structure $(P, \phi)$. Are there any concrete examples for this choice?

  2. To get 'fields' (in the physics sense) defined on $M$ one usually needs a local trivialization of the principal bundle. How would this be provided in practice?

For tangent vectors, for example, the principal bundle is the frame bundle $FM$, and one can induce the trivialization by a chart $(U, x)$ on the base manifold $M$, namely by giving the section $\alpha: U \rightarrow FM,\,\; p \mapsto (\partial / \partial x)_p$, but this doesn't even port over to $OFM$ since $(\partial / \partial x)_p$ are not necessarily orthonormal.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Hi matpisant. Welcome to Phys.SE. Which physics application are you thinking of? $\endgroup$
    – Qmechanic ♦
    Commented Mar 24, 2022 at 16:34
  • 1
    $\begingroup$ @Qmechanic I would welcome any kind of concrete example for a spin structure. But to be more explicit, lets say in the usual QFT on $\mathbb{R}^4$ setting spinors are introduced as functions $\psi: \mathbb{R}^4 \rightarrow \mathbb{C}^4$, whose components transform in a given way (prescribing the group actions in a way), but with no mention of the spin structure that is implied. What would $(P, \phi)$ be in that case and how would the local section $\sigma: U \rightarrow P$ that relates the $\psi$ and the equivariant function $P \rightarrow F$ be chosen? $\endgroup$
    – matpisant
    Commented Mar 24, 2022 at 17:29

1 Answer 1

Reset to default
2
$\begingroup$
  1. The general structure doesn't specify the choice of spin structure.

A spin structure is required to allow for a principal bundle with gauge structure group a spin group. This is already assumed in the general construction you've outlined. Moreover, I expect that this condition is fulfilled over a flat space which is why it isn't usually mentioned. It's only when you begin to do QFT over curved space that this subtlety begins to be important.

  1. To get the fields on $M$, one usually needs a local trivialisation of the principal bundle. How would this be provided in practise?

The bundles in physics are what mathematicians generally call fibre bundles. This means they are exactly bundles that are locally trivialisable. This includes principal bundles. You do not need to specify a local trivialisation to obtain a field. They are what mathematicians call sections and these are defined without asking for a local trivialisation. You do need them, however, to work locally. That is to obtain a bundle chart. Bundle charts are equivalently, local trivialisations.

There isn't a canonical way of choosing a section in a fibre bundle. However, physicists often talk about "choosing a gauge" and this is physics-speak means picking out certain sections that simply the equations of motion. This os amply explored in thr physics literature. See for example, Lorenz or Coulomb gauge.

Improve this answer
$\endgroup$
5
  • $\begingroup$ My question is more about practice than the underlying theory. About point 1, I agree that the condition is probably fulfilled, but my question wasn't about whether some spin structure exists, a fact that is usually explored in the mathematics literature, but rather about concretely writing down one particular structure. If this is completely impossible then could you at least point me to some uniqueness result. $\endgroup$
    – matpisant
    Commented Mar 26, 2022 at 6:39
  • 1
    $\begingroup$ About point 2, my usage of 'field' there was in the physicist's sense, sorry about the misunderstanding. As you elaborate the genuine fields are the sections of the associated bundle (or equivalently $F$-valued, $\mathrm{Spin}(n)$-equivariant functions on $P$) and to get the local version we need a local section $\sigma: U \rightarrow P$. My question, however, is more about whether there is an induced choice for this section (like in the tangent vector case) or do we need to pick it completely independently? $\endgroup$
    – matpisant
    Commented Mar 26, 2022 at 6:46
  • 1
    $\begingroup$ @Matipisant: There isn't generally. However, what is called choosing a gauge physics-speak does pick out certain sections. This is amply explored in the physical literature. For example, Lorenz or Coulomb gauge. This is another different sense in how the term 'gauge' is used in physics. $\endgroup$
    – Mozibur Ullah
    Commented Apr 3, 2022 at 0:06
  • $\begingroup$ Do you want to add this comment to your answer so that I can accept it? $\endgroup$
    – matpisant
    Commented Apr 4, 2022 at 19:04
  • $\begingroup$ @matipisant: Done. $\endgroup$
    – Mozibur Ullah
    Commented Apr 4, 2022 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.