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Regarding the topology of the universe, it could be compact like a sphere or open like a Euclidean space, but since the universe started from a single point, doesn't that mean that the shape of the universe must be contractible?

Compact manifolds are not contractible, so how could the universe be compact?

Qmechanic
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    It is not clear whether topology change can happen in the present, and probably even less clear for point-big bangs. But is there anything in the first place making us think that it was point-like? – Anders Sandberg Feb 28 '21 at 12:39
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    The big bang did not happen at a point – jng224 Feb 28 '21 at 13:13
  • Consider a solid cone minus the vertex and minus the boundary. – WillO Feb 28 '21 at 22:22
  • @Jonas Not at a point but everywhere in an unimaginably small yet in some mathematical ways still infinite space. May I call that a point ? – my2cts Mar 16 '21 at 16:31
  • maybe the words "the observable universe 'started from a single point', so it must be contractible, and ~R^4" have some merit. The global topology, outside of the observable universe; who knows - might not even be a sensible thing to ask. You can maybe say some things about its structure (e.g. it probably can't be nonorientable, because you wouldn't be able to define fermionic fields properly), but who knows – QCD_IS_GOOD Mar 16 '21 at 16:40
  • @my2cts Not just "mathematical" ways. If you consider an infinite 3D grid with spacing $\delta$ and then let $\delta \rightarrow 0$, the grid is infinite in extent at every value of $\delta$. It's true that if you pick a finite collection of grid points and then send $\delta\rightarrow 0$ then they will eventually occupy an arbitrarily small volume, but the entire space will be as infinite as it ever was. – J. Murray Mar 16 '21 at 17:15
  • @J.Murray Mathematically fine. If your statement is physical then it follows that space is not expanding, which conflicts with experiment. – my2cts Mar 16 '21 at 18:42
  • @my2cts That's not true. The $\delta$ in my comment is the scale factor $a(t)$, and the grid points are inertial observers with constant comoving coordinates. The "expansion of the universe" simply means that $a'(t)>0$. – J. Murray Mar 16 '21 at 19:36
  • @J.Murray So the universe is scaled up without getting any bigger. OK. – my2cts Mar 16 '21 at 23:05
  • @my2cts I'm not sure what it would mean for an already-spatially-infinite universe to get bigger. The (inertial) comoving observers are all getting further apart at a rate which is proportional to their present proper distance ($v = \frac{\dot a}{a} d\equiv H d$), which is what we observe. – J. Murray Mar 16 '21 at 23:12

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The main point is: don't make the mistake of extending classical GR all the way to a singular point! In sufficiently extreme conditions the physics of the universe is not known, but since quantum physics is involved, it is likely that no classical manifold is adequate to describe the early universe.

Andrew Steane
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In big bang cosmologies with an initial singularity, spacetime only exists at $t>0$. Spacetime manifolds don't actually have singular points, they just have open edges beyond which they can't be extended. I'm not sure that the spatial topology at $t=0$ is a well defined concept, but the horizon problem suggests that it shouldn't be viewed as a single point in general. Cosmologies with positive spatial curvature can have a horizon problem, so even a finite total spatial volume that goes to zero as $t\to 0$ isn't enough to guarantee that the initial singularity can be reasonably treated as a point.

In cosmologies (probably based on quantum gravity) that really do start at a nonsingular point, there is no obvious obstacle to having noncontractible spatial slices at later times. Take the Euclidean plane with cosmological time being distance from the origin. Space is a point at $t=0$ and a circle at all $t>0$. This example has the wrong metric signature, but there is no obvious reason why a real quantum-gravitational cosmology couldn't have broadly similar features.

benrg
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I am unfamiliar with the usage "compact manifold", and the clearest definition I could find is "A compact manifold is a manifold which is compact as a topological space," in https://topospaces.subwiki.org/wiki/Compact_manifold . Since this was not helpful, I then found that the terminology is quite ambiguous in https://topospaces.subwiki.org/wiki/Compact_space . The clearest definition example there is: "For a metric space to be compact with the induced topology is equivalent to a condition on it called being totally bounded."

OK then, with this definition it seems that a compact universe is a finite space with a boundary. However, a cosmological model of a finite universe has no boundary. So, the vocabulary used in the question you are asking seems to not be applicable to cosmological models.

Buzz
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  • A compact space need not have a boundary. The surface of a sphere or a torus is compact, for example. – J. Murray Mar 16 '21 at 17:12
  • I don't think the universe could be shrunk to a point or else the cosmic microwave background would be uniform. –  Mar 16 '21 at 19:20
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    If you're not familiar with compact manifolds then perhaps this shouldn't be an answer to a question on compact manifolds? – Eletie Mar 16 '21 at 19:36
  • @Eletie Thank you for your suggestion. I now see that I should not posted an answer. Perhaps a comment would have been OK. – Buzz Mar 16 '21 at 19:49
  • @Buzz No worries at all – Eletie Mar 16 '21 at 19:53
  • @J. Murray I do not grasp the meaning of your two comments. The first comment says, "A compact space need not have a boundary." The second says, "it can be contained in a closed and bounded region." Does this mean that in this context that a bounded region need not have a boundary? – Buzz Mar 16 '21 at 19:56
  • @Buzz A bounded subset of $\mathbb R^n$ need not have a boundary, no. – J. Murray Mar 16 '21 at 22:15
  • @Buzz I have deleted my previous comment because it is incorrectly phrased - it should have said "Broadly speaking, a manifold is compact if, upon embedding it into $\mathbb R^n$ for some $n$, its image under the embedding is a closed and bounded region of the embedding space." – J. Murray Mar 16 '21 at 22:28