# Positive local spatial curvature of the universe implies that the universe is compact (i.e. finite)?

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?

## 1 Answer

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.