Why is the Einstein Static Universe represented as an infinite cylinder when it seems like only half a 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 it would have to go from $-\pi$ to $\pi$ where the end points are identified. So shouldn't it just be half an infinite cylinder (i.e. just a an infinite strip)? If anything it seems to me that the cylinder would be a double covering of $R\times S^3$.

Here a picture to visualise (they use $r′$ instead of $\chi$): https://www.researchgate.net/profile/Alberto_Rozas-Fernandez/publication/267338977/figure/fig1/AS:295687481774090@1447508797819/Fig-2-The-flat-FRW-spacetime-filled-with-a-k-essential-scalar-field-is-conformal-to-the.png

Answer 1

The spatial part of a metric is just a 3-sphere:

$$ d\Omega_3^2 = d\chi^2 + \sin^2\chi d\Omega_2^2 \, ,$$

and in fact the metric on a $n$-dimensional sphere can always be written in a resursive way using the metric on an $(n-1)$-dimensional sphere in the same manner:

$$ d\Omega_n^2 = d\chi^2 + \sin^2\chi d\Omega_{n-1}^2 \, .$$

In these coordinates the $\chi$ variable always lies in the range $\chi \in [0,\pi]$, and you should think of $\chi = 0$ as the north pole and $\chi = \pi$ as the south pole (in these coordinates of course, since there is no preferred coordinate system).

A 2-dimensional cylinder is $\mathbb{R} \times S^1$. The way to see that this space is a cylinder is not by suppressing the $d\Omega_2^2$; $\mathbb{R} \times S^n$ is an $(n+1)$-dimensional cylinder. So rather than taking a circle and adding a line to make a 2-dimensional cylinder, to make a 4-dimensional cylinder one starts with an $S^3$ and adds a line.