Isotopy class of spacetime

Question

We know that spacetime is an orientable manifold:

Can spacetime be non-orientable?

But supposing that spacetime is an orientable closed 2D surface, one might envision a variety of non-equivalent solutions in the following sense:

Given a 2D strip, by one rotation(twist), one can create a Moebius strip (it's non-orientable so discarded), but by another rotation (360 degrees) one finds an orientable 2D surface. Suppose one can repeat this for arbitrary many times(integer multiples of 360 degrees), then one has a countable set of possible orientable spacetimes

Is there any way to determine which spacetime relates to ours(2D), given the fact that Einstein's Field equations are pretty much open-minded regarding the topology of spacetime?

Can I find any physical observable in QFTs on such spacetime that is related to the number of turns in general?

If not, can one hypothetically say that the real spacetime is a superposition of all these possibilities?

Is it possible to extend the idea of twist to 3D hypersurfaces?

Answer 1

QFTs are definitely sensitive to the topology of spacetime. As a matter of principle, you can of course measure this topology. You can even do it classically: get on a rocket and keep going; if the universe turns out to be a torus, you'll eventually come back home (and fly arbitrarily close to Henri Poincaré, infinitely many times). In practice, of course, there is no real way of doing this: in real life you can only perform experiments over finite regions of spacetime, which by definition are only sensitive to local properties (e.g. curvature).

Same goes for QFTs. As a matter of principle, they are sensitive to topology. In practice, it is very hard to use this to measure the topology. You would need to perform a quantum experiment over a region of spacetime of cosmological scales. Unlikely to get proper funding.

How exactly QFTs notice the topology is quite a subtle subject. The cleanest and most explicit answers come from TQFTs, where the QFT is only sensitive to topology. For example, Chern-Simons theories in 3d or Donaldson in 4d. For the former you could e.g. perform a quantum-hall-effect on a surface of genus $g$. The vacuum degeneracy is $k^g$, where $k$ is the number of vacua on the torus. So by measuring the ground state degeneracy you can "measure" $g$, the number of holes of your surface. I'm not sure how to do something similar in 4d, if you know how to engineer supersymmetric gauge theories in the lab let me know.

But anyway, physical predictions, quantum or otherwise, really do depend on the topology of spacetime. You can use this, in principle, to measure the topology, but it's hard to imagine how to do this in practice. Topology is, after all, that what cannot be measured locally.

If you want to have a dynamical spacetime, where topology can fluctuate, then you are doing quantum gravity, for which we have little to say at the moment. But the expectation is that, indeed, you would sum over all topologies.

Answer 2

Is there any way to determine which spacetime relates to ours(2D)

No: all these spaces are (globally) homeomorphic. They are not isotopic, but that is a property of embedded manifolds, not of all of spacetime, and general relativity depends only on the intrinsic geometry of spacetime.

Can I find any physical observable in QFTs on such spacetime that is related to the number of turns in general?

For the same reason as the previous, not as such. The tangent bundle and all derived tensor bundles and principal bundles will be the same.

Is it possible to extend the idea of twist to 3D hypersurfaces?

A very direct generalization would be too start out with a solid cube, and identify two opposite faces via one of their four orientation preserving isometries. This should give you three different spaces. Embedding this in $\mathbb R^3$ (though not isometrically) you can twist it as many times as you want, but it will be homeomorphic to one of the three. Likewise for other solid prisms or the solid cylinder (the latter having infinitely many non-equivalent twists).

Answer 3

The orientability of space (rather than spacetime) was first raised by Kant in his 1768 paper, Concerning the Ultimate Foundations in the Differentiation of Regions in Space.

We generally take space to be orientable because experiment shows that it is and Kant alluded to this in his paper, but his paper was really about highlighting this property of space.

In a non-orientable space we can't define the Levi-Civita tensor that is used to integrate volumes in the usual presentation of GR. So GR itself requires an orientable spacetime.

Mobius strips, intrinsically speaking, are the same as a twisted line bundles over a circle. In fact, intrisically speaking, there are only two such line bundles, classifed by no twists or just a single half-twist. Extrinsically speaking, and by this I mean the line bundle is embedded in a vector space, we can have any number of half-twists.

We can generalise the intrinsic picture to twisted vector bundle over spacetime. Instantons, which are certain solutions to the Yang Mills equations are described by these and are used to describe vacuum condensates, which are a superposition of vacua, say in QCD.