Why do lattice models of fermions need a spin structure?

It is well-known that in order to define a relavistic quantum-field theory containing fermions on a general manifold $M$, the manifold $M$ needs to be equipped with a spin structure. The spin structure is a lift of the frame bundle (which is a principal $SO(n)$-bundle) into a principal $Spin(n)$ bundle, allowing us to define transport of spinor fields.

On the other hand, spin structures also seem to come up in condensed matter physics when studying topological phases of systems involving fermions. These systems are typically defined on a lattice, with no Lorentz-invariance or spin-statistics theorem. There are typically no spinor fields in sight, and certainly no need to define their transformation properties in $SO(n)$. So it is far from obvious why spin-structures would be important. Nevertheless, it is conjectured [1] that topological phases are classified by Spin-TQFTs (Topological quantum field theories over manifolds equipped with a spin structure), suggesting that the lattice models can still only be defined on spaces equipped with a spin structure. In particular models [2], this seems to be borne out (the spin structure enters via the need to choose a "Kasteleyn orientation" on the lattice). But I am looking for a more general explanation.

• It is only a guess, but if you are modeling "real space" that has particles with spin properties, a spin structure would be required. If you are modeling any other "virtual space", it may or may not be required, depending on the properties of their particles. – Guill Sep 27 '17 at 2:13

2 Answers

In the paper https://arxiv.org/abs/1612.01418 , I try to argue, through many examples, that any lattice bosonic model with emergent fermions must have vanishing topological partition function on space-time which is not spin. This supports the notion that fermions need to live on spin manifold. In other words, on space-time which is not spin, there must be loops of emergent fermions.

• That's interesting, but I think that the properties of systems with emergent vs. microscopic fermions could in principle be quite different, no? – Dominic Else Feb 10 '18 at 3:17

In some sense, I was asking the wrong question. A lattice model (bosonic or fermionic) will in general need much more data than just a spin structure to make sense on an arbitrary manifold. At very least, it probably requires a framing on the manifold. Otherwise you can't even construct the lattice, let alone write down a Hamiltonian. However, when describing topological phases, what we believe is that we can write down very special "fixed point" lattice models which capture the low-energy physics. These lattice models will have some kind of "topological liquid" property such that they can be defined on any latttice, and local alterations to the lattice can be effected with local unitaries that only depend on the local geometry. In that way they act as lattice realizations of a topological quantum field theory.

So the spin structure dependence is simply the minimal dependence on the background space which a lattice model of fermions can have, because a topological quantum field theory containing fermonic excitations necessarily requires a spin structure on the manifold. To see this, note that fermionic excitations in a topological quantum field theory necessarily are framed; that is, their configuration is characterized both a point in space and a little vector pointing out from that point. To see why this is necessary, observe that for an unframed excitation, the process of pair-creating two particles, exchanging them and then annihilating them (which is supposed to give an amplitude of $$-1$$ for fermions) is topologically equivalent in space-time to the trivial process, which would be a contradiction for fermions.

These arrows essentially play the role of spinors (because rotating an arrow by $$2\pi$$ -- a Dehn twist -- picks up a factor of topological spin, which is $$-1$$ for fermions). So that is why you need a spin structure.