Topologically speaking, our universe is either open (topologically $E^3$) or closed (topologically $S^3$). Then with time we'd have another factor of $E^1$ and a metric connection would determine the curvature. If curvature was non-positive, then it's easy to see that the universe would be open. On the other hand, the 3-sphere has positive curvature. My question is: if the universe globally has positive curvature, does that also mean that the universe is closed? Could it be possible for the universe to be both open and have positive curvature? If this were the case, the universe would not be able to be embedded in Euclidean 4-space, just like how the Klein bottle cannot be embedded into Euclidean 3-space even though its surface is only two dimensional. Conceptually with the 2-sphere, I'm picturing a parameterization where the sphere is drawn from the top down and at the bottom point, instead of connecting it, we continue to go around and overlap the same points on the sphere, but instead of overlapping, these points are disjoint from the previous sphere like how an immersion of the Klein bottle has intersections but the actual Klein bottle itself has no intersections. We could continue overlapping this sphere forever creating an open universe. My question again: is this mathematically possible? And is this a physical possibility if space-time were positive in global curvature.

  • $\begingroup$ the 3-sphere has negative curvature. No, it doesn’t. $\endgroup$
    – G. Smith
    Jun 13, 2020 at 20:44
  • $\begingroup$ You confuse negative with positive curvature. Do you think a 2-sphere with a hole has the same structure as a Möbius strip? Obviously, when you stand on the outside (one side) of the sphere and go around the edge of the hole you arrive at the inside (opposite) of the sphere? The orientation stays the same, contrary to the Ms. The trip has to be continuous. A point on the inside of the sphere is not disjoint from the same point on the outside. A sphere with a hole is open but doesn't correspond to an open universe. $\endgroup$ Jun 13, 2020 at 22:03
  • $\begingroup$ @descheleschilder oops that's embarrassing yeah I meant positive... I was thinking positive I don't know why I wrote negative. Fixing it now $\endgroup$
    – Cam White
    Jun 13, 2020 at 22:56
  • $\begingroup$ @G.Smith my bad I meant positive like I said in the title $\endgroup$
    – Cam White
    Jun 13, 2020 at 22:58
  • $\begingroup$ @CamWhite I think I know what you mean now. When you arrive at the hole (let's keep it 2-d so we can envision it) in the sphere, you make a continuous turn and start making a new sphere on the inside (the distance between the two approaches zero, so the points of the two are disjoint), after which you make a new continuous turn on the opposite side of the hole (creating a new hole), making a new sphere on the inside of the second one, and so on to the center. Is this what you envision? It makes me think about the opposite: the trumpet Universe. $\endgroup$ Jun 14, 2020 at 13:54

1 Answer 1


It is believed in cosmology that the universe is homogeneous and isotropic. (I'm not qualified to comment on the plausibility of this belief, being a pure math topologist, but let's proceed from that current standard assumption.) Informally, that is to say that it looks the same from every point, and that from every point it looks the same in every direction. This implies constant sectional curvature, an even stronger condition than constant curvature, as it means that cross sections have the same curvature no matter how/where you slice them.

Under that setup, positive curvature would indeed imply the universe is closed.

A Klein bottle embedded into 3-space does not have constant curvature (let alone constant sectional curvature). Your self-intersecting sphere parametrization would also fail to have constant curvature. Perhaps an easy way to see this is to take it down to $1$-D and imagine how you would need to curve an immersed self-intersecting circle.

On the other hand, a positively curved universe does not necessarily imply that it's $S^3$. It could be multiconnected, as in it could be a fundamental domain that tessellates $S^3$ under the action of an isometry group. Similarly, a multiconnected flat universe need not be open. For example, it could be a cubical fundamental domain that tessellates $\mathbb{R}^3$, and is topologically a flat $3$-torus (or some other face-gluing on the cube). The same goes for hyperbolic space. Such multiconnected universe models are consistent with the assumptions of homogeneity and isotropy, as well as other aspects of general relativity.

  • $\begingroup$ How about an infinite spring? It is homogenous, isotrope, it has a constant positive curvature - and yet it is open. Why would it be impossible with a 3D internal space? Internal measurements would show a constant positive curvature with a constant radius - except if you go $2\pi r$ into a direction, you do not come back. (Although it is currently not possible due to the accelerating expansion) $\endgroup$
    – peterh
    Jan 16, 2023 at 11:58

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