6
$\begingroup$

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?

$\endgroup$
11
  • 1
    $\begingroup$ Aren't fermions famously invariant under 720 degree rotations but not 360 degree rotations? That may make orientable spacetimes with even or odd number of double-twists distinguishable. $\endgroup$ Apr 19, 2021 at 20:06
  • $\begingroup$ Is there any topological difference between 0 twists and 2 twists? $\endgroup$
    – geshel
    Apr 19, 2021 at 22:02
  • 2
    $\begingroup$ @BastamTajik Two spaces which are homeomorphic are topologically the same. $\endgroup$
    – J. Murray
    Apr 20, 2021 at 1:55
  • 2
    $\begingroup$ @BastamTajik - If you have a space with a 360 degree twist, if you send a spin up electron around it it will come back with spin up, but the opposite phase of when it started. So you could do an interference experiment with a pair of correlated electrons to tell whether the twist was 360(2n+1) or 360(2n). $\endgroup$ Apr 20, 2021 at 8:37
  • 3
    $\begingroup$ Okay, but note that a strip with 2 twists is homeomorphic to a strip with 0 twists. $\endgroup$
    – J. Murray
    Apr 20, 2021 at 19:02

3 Answers 3

6
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Thanks. But one thing, I understood well in the other question(attached) is that the local observation of CP violation constraints the spacetime to be orientable which means local observations can tell things about the global topology of spacetime! Moreover, what if one tries to use Rovelli's argument that "spacetime=gravitational field". Can't one look at the twist of the gravitational field tensor? And apply the constraint of orientability so that the torsion of the gravitational field becomes quantized so that it will be either an odd or an even number? $\endgroup$ Apr 20, 2021 at 18:53
  • $\begingroup$ Very naively it means that there might be two possibilities for spacetime twist, and can be interpreted as the existence of two types of particles, fermions, and bosons. Only because of two possibilities for spacetime. In other words, fermions and bosons become two eigenstates of spacetime. $\endgroup$ Apr 20, 2021 at 19:00
  • $\begingroup$ @BastamTajik Local observations can indeed constraint the topology. For example, a local observation of a fermion means that the manifold must admit a spin structure, so the first two Stiefel-Whitney classes must vanish. This doesn't change my claim: you cannot determine the topology via local observations. You can only rule out possibilities. $\endgroup$ Apr 20, 2021 at 19:06
  • $\begingroup$ But isn't all science falsification rather than proof? I guess it's what we do in general. We rule out things by observation, and "then establish a model which is merely a hypothesis and there's no proof for it(in our case a model that assumes a certain topolgy for spacetime, or in QCD quarks, etc...) that is compatible with its previous observations and then predicts a number and makes itself falsifiable". I think there's no way to understand the true essence of anything like if quarks really exist or not, strings do or not, and so on...These are merely theoretical assumptions. $\endgroup$ Apr 20, 2021 at 19:19
  • $\begingroup$ This means your statement of "we can never understand the true topology of spacetime with local observation" is not specific to the topology. And mathematically at least by now, we have no theorem that determines the topology. Homeomorphism mathematically can't be determined due to an infinite number of homotopy groups, which gives an infinite number of observables at best that should be experimented with! which means it's fundamentally impossible even from the mathematical point of view to prove a homeomorphism "in general" unless there's some global information as you mentionde. $\endgroup$ Apr 20, 2021 at 19:26
2
+100
$\begingroup$

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).

$\endgroup$
8
  • $\begingroup$ Perfect answer. So if I've understood correctly, you mean that the number of twists is not a topologically unique number( since a 360-degree twist and 720-degree twist are basically homeomorphic). Can you let me know if isotopy classes can be changed with respect to the embedding space (like R^3 for the case of a 2d strip) or they don't depend on such space fundamentally and the embedding space can be forgotten after determining the isotopy classes? $\endgroup$ Apr 21, 2021 at 10:13
  • 1
    $\begingroup$ @BastamTajik I'm afraid the isotopy class of an embedding in a given manifolds depends on a very fundamental way on the ambient manifold. That said, if for a given ambient and submanifold you have a classification of embeddings, I guess you might in some cases just take the isotopy class, represented in any way, as an additional extrinsic property of your spacetime without explicitly referring to the ambient manifold. For example for a torus in $\mathbb R^3$ (or rather $S^3$) I think the isotopy classes correspond to knots. $\endgroup$
    – doetoe
    Apr 21, 2021 at 20:55
  • 1
    $\begingroup$ This means that you could consider any knot invariant as a generalization of a twist (in the sense that it says something, though not everything, about the embedding) in the case of a torus in the 3-sphere. $\endgroup$
    – doetoe
    Apr 21, 2021 at 20:59
  • 1
    $\begingroup$ @BastamTajik is your question whether they become isotopic when embedded in 4-space? If so, I am not sure, though that sounds reasonable. To see that a homeomorphism exists is not hard though, especially when you think of a homeomorphism as a bijection that also induces a bijection on open subsets. As for the orientability, all of these are. If you also consider identifications between opposite faces by orientation reversing isometries of the square, you construct the non-orientable ones. These can not be embedded in 3-space. $\endgroup$
    – doetoe
    Apr 21, 2021 at 21:52
  • 1
    $\begingroup$ As for a reference, what comes to mind is that for the case of embeddings in 3-space you might consider the literature on knots and links. This link will give you some more references and context for the problem of classification of embeddings of manifolds up to isotopy. $\endgroup$
    – doetoe
    Apr 22, 2021 at 10:29
-4
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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