# The topology of a “closed” universe - is it really closed?

The spatial part of the positive curvature FRW metric has the form $$ds^2=\frac{dr^2}{1-(r/R)^2}+r^2d\Omega^2$$ or $$ds^2=R^2(d\chi^2+\sin{\chi}^2d\Omega^2)$$

This is described as "closed", as it has the metric of a three-sphere, but I want to know what this actually means. Is $\chi$ not limited to the interval $[0,\pi/2)$, since beyond that we are simply reproducing the same $r$? In which case, is the universe not really one "hemi-threesphere" and not closed in the same way as a full sphere is?

(Edit: I note that Carroll states that the "only possible global structure is the complete three-sphere", but he doesn't go into why.)

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"Is $\chi$ not limited to the interval $[0,\pi/2[$, since beyond that we are simply reproducing the same $r$?" If we are reproducing the same thing, we are in the same equivalency class, which is pretty much exactly what makes a topology closed. – leftaroundabout Oct 13 '11 at 11:47

The first metric form uses coordinates that are valid only in a patch of the whole space. This is a fairly common situation. Consider the example of a 2-sphere projected onto a tangent plane from the center. This is the gnomonic projection which maps all great circles into straight lines. The $(x,y)$ coordinate system on the tangent plane define a coordinate system on a patch of the sphere but not the whole sphere. The sphere is still all there though. Projecting from infinity onto the tangent plane maps the the sphere onto the unit disk, and $(r,\theta)$ on this disk are the equivalent to the first metric form. But each point on the disk represents two points on the sphere, so you restrict the mapping to a patch which is a hemisphere.
OK, thanks, although I still wonder about the Carroll statement "only possible global structure is the complete three-sphere" - i.e. how do we know we should complete the sphere if we start off we the $r$ coordinates? – James Oct 14 '11 at 15:11