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I quote from the Wikipedia page about the shape of the universe:

If the spatial geometry [of the universe] is spherical, i.e., possess positive curvature, the topology is compact.

I'm trying to understand whether the quoted statement is true, in light of this simplified example:

Suppose that you have some coordinate system for the universe, $t,x,y,z$, and the spatial curvature is given by a function $C(t,x,y,z)$. Let's assume that for all $x$, the two events $(t,x,y,z)$ and $(t,x+L,y,z)$ are identical. In other words, the dimension $x$ is "compact" with "period" $L$ (In yet other words, the topology of this space has the circle $S^1$ as a factor, I think).

Now, I can define another universe, with coordinates $t,x,y,z$, such that $x$ is not compact anymore, that is, $(t,x,y,z)$ is different from any event $(t,x+L',y,z)$ for all $x$ and $L'\neq 0$. For this universe, let its curvature be given by the same function as above $C(t,x,y,z)$. In other words, this new universe has the exact same local curvature structure as the original universe, but it is not compact anymore.

Now, I understand that in this example I did not mention curvature (and in particular, $S^1$ is flat). But, can't the same "uncompactification" be done also to positively curved spaces, showing that a positive curvature does not necesarily imply a compact universe?


As an additional thought, if I imagine space to be a $2$-sphere, with spherical coordinates $(\varphi,\theta)$ ($\varphi$ is azimuth, $\theta$ is inclination, $(\varphi,\theta)=(\varphi+2\pi,\theta)$ ), then I could do an "uncompactification" by making $(\varphi,\theta)$ and $(\varphi +2\pi,\theta)$ be distinct points in space. I see, though, that this breaks down at the pole $\theta=0$, because all points with $\theta=0$ are identical. But is this a problem?

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I see, though, that this breaks down at the pole θ=0, because all points with θ=0 are identical. But is this a problem?

Yes, it is a problem, because if you omit the poles from a sphere, then it has the topology of a cylinder, not a sphere.

There is a theorem in differential geometry called the Myers theorem. This is a theorem in Riemannian geometry (not semi-Riemannian geometry), but we can apply it to the metric of a spacelike slice of a cosmological spacetime, which is Riemannian. The Myers theorem basically tells us that if the Ricci curvature is bounded below by a positive bound, then the space is compact (and the theorem also puts a bound on its diameter).

Note that it doesn't suffice to have a curvature that's everywhere positive. You also need a global lower bound. But we normally make our cosmological models homogeneous, so the Ricci curvature is in fact the same everywhere.

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