Maximum distance in a closed static universe I'm confused about a detail in cosmology.  Consider a static closed universe, of the following metric (consider $a$ as a simple constant with units of length) :
\begin{equation}\tag{1}
ds^2 = dt^2 - a^2 \big( d\chi^2 + \sin^2 {\chi} \; (d\vartheta^2 + \sin^2 {\vartheta} \; d\varphi^2) \big).
\end{equation}
Here, the radial coordinate takes values on a bounded domain : $0 \le \chi \le \pi$.  The proper radial lenght is defined by this line element ($d\vartheta = d\varphi = 0$) :
\begin{equation}\tag{2}
d\ell^2 = a^2 \, d\chi^2.
\end{equation}
Integrating gives trivially $\ell = a \, \chi$.
The proper volume of the whole space is easily found to be $\mathcal{V} = 2 \pi^2 a^3$, and the area of a sphere of coordinate radius $\chi$ is given by $\mathcal{A}(\chi) = 4 \pi a^2 \sin^2 {\chi}$.  Thus $\mathcal{A}(0) = \mathcal{A}(\pi) = 0$ and $\mathcal{A}_{\text{max}} = \mathcal{A}(\frac{\pi}{2}) = 4 \pi a^2$.
In a closed universe, it is important to not confuse length and distance.
The question is this :

What is the maximal proper distance from a given stationary observer in this space : $\mathcal{D}_{\text{max}} = \pi \, a$, or $\mathcal{D}_{\text{max}} = \frac{\pi}{2} \; a$ ?

I'm confused because of the area behavior, and I believed that the maximal distance is $\mathcal{D}_{\text{max}} = \pi \, a$ and not $\mathcal{D}_{\text{max}} = \frac{\pi}{2} \; a$, despite the fact that $\mathcal{A}(\pi) = 0$.  I'm not sure anymore that it's making sense !  I may have confused distance with length and I need a confirmation.
If you built a linear structure in that space, its maximal length should be $2 \pi \, a$, and the distance between both extremities should be 0, right ?  Or is the lenght actually $\pi \, a$ ??
 A: The spatial component of the spacetime you describe is a 3-sphere. The largest distance is obtained when $\Delta t=0$ so we can ignore the time direction of the spacetime.
A 3-sphere is natural extension of a 2-sphere. Where a 2-sphere consists of a circle with radius $\sin(\theta)$ for every value of $\theta$, a 3-sphere consists of 2-sphere with radius $\sin(\chi)$ for every value of $\chi$. The largest distance on a 2-sphere is the distance from the north to the south-pole. This corresponds to varying $\theta$ by $\pi$. When the radius is $a$, this gives a distance of $\pi a$. 
A similar reasoning lets you travel from one side of the 3-sphere to the other by varying $\chi$ by $\pi$. Here the distance is again $\pi a$.
For the second part of your question. If by maximal length you mean the distance traveled along a space-like path, then there is no maximal length. A path can curl around the 3-sphere and get an infinite length. Even when you demand the path to be straight, i.e., a geodesic, there is no maximal length as the path could travel around the 3-sphere with a small time-like component, thereby forever circling the 3-sphere and obtaining infinite length.
The length of a straight path around the 3-sphere without a time-like component in this coordinate system is $2\pi a$, just as for the 2-sphere.
