13
$\begingroup$

I have the following two impressions about fermions in physics. I'm confused about their accuracy, and their compatibility:

  1. To consider the behavior of a fermion, whose intrinsic spin is described by a representation $V$ of the group $Spin(p,q)$, on a pseudo-Riemannian manifold $M$ of signature $(p,q)$, you first introduce a spin structure on $M$. Then the fermion field is a section of the bundle associated to $V$.

  2. To consider the behavior of a fermion on a pseudo-Riemannian manifold $M$ of signature $(p,q)$, you first turn $M$ into a supermanifold. Then the fermion field is a superfunction on $M$ with some constraints coming from its intrinsic spin.

Question: Is either of (1) or (2) close to accurate? What major points or subtleties have I missed? If both are close to accurate, then how does one "translate" between the formalism of (1) and the formalism of (2)?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
11
$\begingroup$

You've got the right concepts, but they're presented in a way that makes me think some context could be helpful.

In #1, you're really talking about the special case where $V$ is one of the spinor representations. In this case, yes, the fermionic fields are sections of the associated spinor bundle $\mathcal{V}$. Sometimes, people write $\Pi \mathcal{V}$ to emphasize that we take the spinor fields to be anti-commuting when we use them to generate an algebra over the structure sheaf of $M$.

The spin-statistics theorem forces this choice on us in 4+ dimensional relativistic QFT. We get causality violations if we don't choose the spinor fields to be anti-commutative. In lower dimensions, the relation between spin & statistics is more complicated. You can have 2d scalars which anti-commute and spinors which commute.

The connection between 1 & 2 is given by Batchelor's Theorem: The structure sheaf of any supermanifold is (non-canonically) isomorphic to the sections of the exterior algebra of some vector bundle on the underlying manifold.

Locally, the idea is elementary: A function on a supermanifold looks like $$f(x,\theta) = f_0(x) + \sum_i f_1^i(x) \theta_i + \sum_{ij} f_2^{ij}(x) \theta_i\theta_j + ....$$ If the $\theta_i$ form a basis for a spinor representation, then the coefficient functions $f_1$ are the components of a section of the (dual) spinor bundle. The $\theta_i$ anti-commute, so the $f^i_1$ must also anti-commute.

Supermanifolds don't do much for you when you're only thinking about fermions. They're helpful when you want to start packaging the spinors and other fields into representations of a supersymmetry group.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks -- this is exactly the sort of thing I was hoping for, right from the very first sentence! I guess one aspect that puzzles me is that when things are presented as in (2), it's not clear how extract a spin structure from the supermanifold data. And in this respect, the appeal to Batchelor's theorem confuses me because Batchelor relates relates the "super" structure to a plain vector bundle -- with structure group $GL(n)$ -- so where does the spin come from? $\endgroup$
    – Tim Campion
    Commented Sep 12, 2020 at 15:35
  • $\begingroup$ But now I think maybe I see -- when you expand a superfunction in local coordinates, it looks a lot like a section of a bundle of Clifford algebras. So maybe the spin structure is obtained as a subbundle of this associated Clifford bundle (via the multiplicative embedding of $Spin(p,q)$ into $Cl(p,q)$)? Part of what makes it confusing is that ordinary manifolds are special cases of supermanifolds, and not every ordinary manifold admits a spin structure... $\endgroup$
    – Tim Campion
    Commented Sep 12, 2020 at 15:35
  • $\begingroup$ Yes, the structure sheaf is locally isomorphic to $C^\infty \otimes \wedge E$, and the transition functions for E give you the spin structure. But E has rank 0 in the special case of real manifolds, so there's no transition data. $\endgroup$
    – user1504
    Commented Sep 12, 2020 at 15:47
  • 2
    $\begingroup$ @TimCampion For the record, your (2) need not presume that the fermion field is a spinor. The point is to capture the anti-commutativity, not the half-integer spin property. If you want your fermion to also be a spinor, a spin structure has to be provided just as in (1) and you have to parity shift $\Pi\mathcal{V}$ the appropriate spinor bundle, as indicated in this answer. Only when $\mathcal{V}$ is locally trivialized to such spinor fermion fields look like simple superfunctions on $M$. $\endgroup$
    – Igor Khavkine
    Commented Sep 12, 2020 at 18:01
5
$\begingroup$

This is a bit of a repackaging of the same info in the other answer, but maybe it will be more clear.

The short answer is (almost) both: A fermion is a section of the parity shifted spinor bundle on a manifold. As such, you can't have a fermion without a spin structure.

Each aspect of this can be considered separately: there is no classical reason that an anti-commutative field has to be a section of the spinor bundle, and there is no reason that a section of the spinor bundle has to be anti-commutative. However, in physics the Spin-Statistics Theorem says that to have a consistent, Lorentz-invariant theory in >2 spatial dimensions, all anti-commutative fields must be spinors (have half-integral spin).

However, you only need the parity shifted bundle here. The full formalism of supermanifolds is for when you have supersymmetry, which is an odd (ie, anti-commuting) symmetry that relates bosons and fermions.

You can look at this in two ways. The first is as supersymmetric quantum mechanics, where you have maps from, say, the super manifold $\mathbb{R}^{1|1}$ to a Riemannian manifold. Here, the need for a spin-structure arises when you try to quantize theory in order to patch together the Clifford algebras that arise on each local chart.

The second way to look at this is to have your fields be functions on a supermanifold. Here, the supermanifold is modeled on super-Minkowski space, which is acted on by the super-Poincare group. In super-Minkowski space, the odd part is (some number of copies of) the parity shifted the spinor bundle, so the need for the spin-structure is part of the definition.

Dan Freed's notes Classical field theory and supersymmetry on this stuff are very good.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks! So it seems the conclusion I took from the other answer was wrong -- (1) and (2) are not equivalent at all! There are many things I still don't understand, but I think the most pressing one is the phrase "anticommutative field" -- because I don't understand this operation of "multiplying" field configurations. Am I correct in understanding that the phrase "anticommutative field" is meaningless classically, and only has meaning when you are doing quantum physics, where observables are supposed to form an algebra? $\endgroup$
    – Tim Campion
    Commented Sep 14, 2020 at 19:04
  • 1
    $\begingroup$ @TimCampion Yes and no. You can say that fermions are fundamentally quantum mechanical and stop there, but physicists like to do path integrals, which are based on 'classical' actions and involve multiplying (and integrating!) these classical fields. Thus, the formalism of anti-commuting fields/Grassman variables and Berezin integration was used (developed?). Basically, you develop the formalism so that the path integral gives the right answer for fermions, and you can think of the anti commuting fields/Grassman variables as being the (semi-)classical limit of the quantum fields. $\endgroup$
    – Aaron Bergman
    Commented Sep 14, 2020 at 19:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .