Why is the Einstein Static Universe an infinite cylinder?

Question

The Einstein static universe metric is $$ds^2=-dt^2 + d\chi^2 + \sin(\chi)^2d\Omega^2$$ where $-\infty<t<\infty$ , $0<\chi<\pi$ and $d\Omega^2$ is the metric on a $S^2$. It describes the topology of $R\times S^3$. Suppressing the $S^2$ coordinates, why is it always represented as an infinite cylinder if $\chi$ only goes from $0$ to $\pi$? to be a cylinder is would have to have some periodicity $\phi \sim \phi + 2\pi$. Shouldn't it just be half an infinite cylinder (i.e. just a rectangle)?

Answer 1

The "vertical" axis of the infinite cylinder is designated by $t$, which goes from $-\infty$ to $+\infty$. The variable $\chi$, in contrast, is one component of spherical coordinates on $\mathbb S^3$. If we designate standard spherical coordinates on $\mathbb S^2$ by $(\phi,\theta)$ and those on $\mathbb S^3$ by $(\chi,\phi,\theta)$, then the standard round metric on $\mathbb S^2$ is $$ d\Omega^2 = d\phi^2 + \sin(\phi)^2 d\theta^2, $$ and the standard round metric on $\mathbb S^3$ is $$ d\chi^2 + \sin(\chi)^2 d\phi^2 + \sin(\chi)^2\sin(\phi)^2 d\theta^2 = d\chi^2 + \sin(\chi)^2 d\Omega^2. $$ The variable $\chi$ only goes from $0$ to $\pi$ because it represents the angle downward from the "north pole" of $\mathbb S^3$.