How can the Universe be infinite and yet have a finite age?

[laymanB]

This conclusion is not correct. If the universe is spatially flat it isn’t neccessarily spatially infinite, its topology can be compact also, e.g. a 3-torus is spatially flat and if large enough we will never be able to confirm it by observation.

I'm confused. I thought that a 3-torus was a closed manifold, meaning that it is spatially finite? And that the only FRW spacetimes with Euclidean space were spatially infinite? What am I missing here?

The FRW model yields the dynamics of the universe, not its topology.

Thanks. Yeah, I'm still learning and trying to get the correct terminology. So what yields the global geometry of the universe? Is it just the RW metric before using the Einstein Equations(EE)? Are topology and geometry synonyms?

 

[PeterDonis]

the only FRW spacetimes with Euclidean space were spatially infinite?

That's how it's often phrased, but it's not strictly correct. The strictly correct statement is that the only FRW spacetimes with flat (Euclidean) spacelike slices of constant FRW coordinate time with trivial topology (i.e., topology ##R^3##) are spatially infinite.

what yields the global geometry of the universe?

Not geometry, topology. The Einstein Field Equation can't tell you the global topology of a spacetime, because the EFE is local; it relates spacetime curvature to stress-energy in a small local region of spacetime. If there are multiple topologically different ways of extending that local region into a global spacetime, the EFE cannot distinguish between them.

 

[PeterDonis]

“trivial” in this case means less assumptions

Not really less assumptions; assuming any global topology, including ##R^3## (for spacelike slices--##R^4## for spacetime), is an assumption. I think the assumption of global ##R^3## spatial topology seems more parsimonious to cosmologists because any local region of spacetime that can be covered by a single coordinate chart without coordinate singularities or ad hoc restrictions on the ranges of the coordinates must have spacetime topology ##R^4##, and any spacelike slice of it must have spatial topology ##R^3##. Therefore those topologies are the obvious ones to assume for the global topology, except in particular cases in which a global ##R^4## topology is impossible (for example, maximally extended Schwarzschild spacetime); but the only FRW spacetime in which that is the case is the one with positive spatial curvature, which can only have global topology ##S^3 \times R##.

 

[timmdeeg]

I'm confused. I thought that a 3-torus was a closed manifold, meaning that it is spatially finite? And that the only FRW spacetimes with Euclidean space were spatially infinite? What am I missing here?

Yes, a 3-torus is spatially finite and flat. “Compact” topology means spatially finite. And no, FRW spacetimes which are spatially flat (euclidean) can be spatially finite or infinite.

So what yields the global geometry of the universe? Is it just the RW metric before using the Einstein Equations(EE)? Are topology and geometry synonyms?

The global geometry or perhaps better the topology of the universe can not be calculated based on the knowledge of its ingredients (the various energy densities) which determine the local spatial curvature, spherical, euclidean or hyperbolic. One needs observational data, as mentioned by @PeterDonis in his previous post.
Topology and geometry are not synonymus. I think topology requires constant local curvature but am not sure if this is the whole story and would leave that to experts around here.

 

[PeterDonis]

I think topology requires constant local curvature

I'm not sure what you mean by this. Topology does not imply any particular curvature. A spacetime with global topology ##R^4##, for example, could be flat Minkowski spacetime, or it could be a curved FRW spacetime with critical density matter and flat infinite spacelike slices, or it could be a different curved FRW spacetime with sub-critical density matter and open (hyperbolic) infinite spacelike slices, or it could be de Sitter spacetime, with just a positive cosmological constant everywhere and nothing else. Or it could be, as our best current model of the actual universe is, a combination of the second and fourth alternatives I gave just now.

 

[timmdeeg]

I'm not sure what you mean by this. Topology does not imply any particular curvature.

Hm, with “constant local curvature” I meant that the curvature is everywhere the same regardless the “particular” curvature.

 

[PeterDonis]

with “constant local curvature” I meant that the curvature is everywhere the same regardless the “particular” curvature.

Topology has nothing to say about that. A spacelike slice could have topology ##R^3##, or ##T^3##, the 3-torus, for that matter, and still not have the same curvature everywhere. There are some particular restrictions that topology can place on possible curvatures: for example, no sphere topology (##S^n## for any ##n##) can be flat (have zero curvature everywhere), and the 2-torus, ##T^2##, also cannot be flat (but the 3-torus can). But those restrictions don't restrict very much--for example, there's nothing preventing a manifold with topology ##S^3## from having curvature that varies from point to point, or even from being flat in some finite region (just not everywhere).

 

[laymanB]

or it could be a curved FRW spacetime with critical density matter and flat infinite spacelike slices

If we take a spacelike slice of the universe now in this constant FRW coordinate time and at the time of the CMB, does this let us extrapolate anything about the global topology?

Or is it that we cannot fully determine whether the topology of these spacelike slices are trivial or not?

 

[PeterDonis]

If we take a spacelike slice of the universe now in this constant FRW coordinate time and at the time of the CMB, does this let us extrapolate anything about the global topology?

We can't "take a spacelike slice" because we can't observe an entire spacelike slice; we can only observe the portion of it that is in our past light cone. That is the fundamental restriction that prevents us from proving that the global topology is one thing or another. We can only collect data and place limits, such as, if the global topology of a spacelike slice is a 3-torus, or a 3-sphere, instead of ##R^3##, its "size" (roughly the maximum possible distance between distinct points in the slice) must be much larger than the size of our observable universe (which is what we can see in our past light cone, and which looks like flat Euclidean space with no sign of non-trivial topology).

 

[laymanB]

We can't "take a spacelike slice" because we can't observe an entire spacelike slice; we can only observe the portion of it that is in our past light cone. That is the fundamental restriction that prevents us from proving that the global topology is one thing or another. We can only collect data and place limits, such as, if the global topology of a spacelike slice is a 3-torus, or a 3-sphere, instead of ##R^3##, its "size" (roughly the maximum possible distance between distinct points in the slice) must be much larger than the size of our observable universe (which is what we can see in our past light cone, and which looks like flat Euclidean space with no sign of non-trivial topology).

I see. Thanks.

 

[kimbyd]

So essentially we are left to conclude that the universe is much larger than the volume that we can observe, but the observational evidence gives greater credence to the hypothesis that the global geometry is flat and infinite in extent, assuming isotropy and homogeneity for the universe as a whole?

The universe is much larger than the volume we observe (most likely a few hundred times larger at the minimum). There is no observational preference whatsoever given to "infinite" vs. "very big".

Do you foresee any future experiments or modifications to theory that would give greater confidence to this hypothesis?

Our only real chance is learning more about the event that kicked off our early universe. If learning more about that event doesn't tell us, then there's probably no way to know.

 

[laymanB]

If we can calculate the density parameter ##\Omega## was even closer to 1 in the very early universe than it is now, would this not rule out a finite, closed global topology? In other words, if the global topology was an spherical or hyperbolic topology now, shouldn't we be able to see that by calculating ##\Omega## for the early universe? Is this possible?

 

[PeterDonis]

If we can calculate the density parameter ##\Omega## was even closer to 1 in the very early universe than it is now, would this not rule out a finite, closed global topology?

No. The density parameter being 1, even if it is exactly 1, does not specify the global topology. It might rule out certain topologies (such as ##S^3##), but does not narrow it down to only one.

 

[kimbyd]

If we can calculate the density parameter ##\Omega## was even closer to 1 in the very early universe than it is now, would this not rule out a finite, closed global topology? In other words, if the global topology was an spherical or hyperbolic topology now, shouldn't we be able to see that by calculating ##\Omega## for the early universe? Is this possible?

The topology isn't really measurable.

It could be measurable if the universe wrapped back on itself near the horizon, but current observations seem to rule this out.

In the end, the topology is just the overall connectedness of the universe. If you have a flexible surface, then anything you can do to that surface that doesn't cause a cut or tear (stretching, contracting, bulging outward, etc.) will not do anything to change the topology. If that surface is the surface of a sphere, then no matter what you do to it it will still be a spherical topology. If it's the surface of a torus, it will always have a toroidal topology.

This means that unless the universe wraps back on itself pretty close to (or within) the cosmological horizon, any local geometry will be consistent with any global topology. Since we don't see the impact of the topology within the observable universe, we can't say anything about it.

Our one hope of learning something about the topology would be learning more about the very early universe. But that may also tell us nothing.

 

[laymanB]

Thanks everybody.

 

[timmdeeg]

Topology has nothing to say about that. A spacelike slice could have topology ##R^3##, or ##T^3##, the 3-torus, for that matter, and still not have the same curvature everywhere. There are some particular restrictions that topology can place on possible curvatures: for example, no sphere topology (##S^n## for any ##n##) can be flat (have zero curvature everywhere), and the 2-torus, ##T^2##, also cannot be flat (but the 3-torus can). But those restrictions don't restrict very much--for example, there's nothing preventing a manifold with topology ##S^3## from having curvature that varies from point to point, or even from being flat in some finite region (just not everywhere).

Thank you very much for this detailed and enlightening answer.

 

[kimbyd]

There are some particular restrictions that topology can place on possible curvatures: for example, no sphere topology (##S^n## for any ##n##) can be flat (have zero curvature everywhere), and the 2-torus, ##T^2##, also cannot be flat (but the 3-torus can).

I don't think you can extend this to the observable universe. These topologies cannot be globally flat. But any topology can be locally flat. As we can only see a finite distance away, I'm pretty sure it's possible that any topology combined with a universe significantly larger than the observable universe can be flat within the observable universe.

 

[timmdeeg]

A sphere might seem to be flat locally. But doesn’t that depend on the precision of our measurement? Isn’t flat in this context true only for an infinitesimally small area?

 

[PeterDonis]

I don't think you can extend this to the observable universe.

Agreed; the observable universe could be perfectly flat without excluding a global topology like ##S^3##. That's why I specified that ##S^3## cannot be flat everywhere; it can be flat in a finite region, just not everywhere.

 

[PeterDonis]

A sphere

You are confusing two meanings of the word "sphere". The relevant one for this thread is "topology ##S^n##". But the one you are implicitly using is "topology ##S^n## plus constant curvature". The two are not the same, and only the former meaning is intended in this thread.

 

[kimbyd]

A sphere might seem to be flat locally. But doesn’t that depend on the precision of our measurement? Isn’t flat in this context true only for an infinitesimally small area?

To expand a bit upon what PeterDonis said, a blobby ball has a spherical topology. So does a sea cucumber. Or a sea urchin. Or an octopus. It is very possible for a local region with spherical topology to be perfectly flat, at any measurement accuracy.

The main argument against such a thing is that it requires fine tuning: if there's a spherical topology, there will be many regions on the surface that will not be flat. So you would need fine-tuning to land in a place on the surface that is.

 

[rootone]

,,, if there's a spherical topology, there will be many regions on the surface that will not be flat. So you would need fine-tuning to land in a place on the surface that is.

Apollo 11.

 

[timmdeeg]

You are confusing two meanings of the word "sphere". The relevant one for this thread is "topology ##S^n##". But the one you are implicitly using is "topology ##S^n## plus constant curvature". The two are not the same, and only the former meaning is intended in this thread.

Thanks. Yes indeed, I was thinking of the latter. Is that, "topology ##S^n## plus constant curvature", restricted to a FRW universe or generally to a universe which obeys the cosmological principle?

 

[timmdeeg]

To expand a bit upon what PeterDonis said, a blobby ball has a spherical topology. So does a sea cucumber. Or a sea urchin. Or an octopus. It is very possible for a local region with spherical topology to be perfectly flat, at any measurement accuracy.

The main argument against such a thing is that it requires fine tuning: if there's a spherical topology, there will be many regions on the surface that will not be flat. So you would need fine-tuning to land in a place on the surface that is.

That’s helpful, thanks!

 

[Ibix]

a blobby ball has a spherical topology. So does a sea cucumber. Or a sea urchin. Or an octopus.

Point of pedantry - an octopus at least has a digestive tract, so is topologically a torus (or possibly something more complex if it has something like gills as well - I'm no marine biologist) since it effectively has a hole right through it. A stuffed toy octopus, though, lacking such biological messiness, would be topologically a sphere.

 

[PeterDonis]

Is that, "topology ##S^n## plus constant curvature", restricted to a FRW universe or generally to a universe which obeys the cosmological principle?

It's restricted to a spacetime that has spatial topology ##S^n## and constant spatial curvature. Such spacetimes are a subset of FRW spacetimes or spacetimes that obey the cosmological principle (since there are other topologies besides ##S^n## that allow constant spatial curvature).

 

[kimbyd]

Point of pedantry - an octopus at least has a digestive tract, so is topologically a torus (or possibly something more complex if it has something like gills as well - I'm no marine biologist) since it effectively has a hole right through it. A stuffed toy octopus, though, lacking such biological messiness, would be topologically a sphere.

Ah, yes, you're right. I didn't realize it had a full digestive tract. I misremembered.

 

[Fervent Freyja]

If the universe does indeed bounce in between cycles of beginning and ending, then it can be considered infinite, with a finite "time" in between each cycle. Conceptually, this makes sense, more so than one universe having a single beginning and end, and nothing else existing after that one cycle. With so many possible combinations with particles at each beginning, each new cycle can be different than prior.