In physics, standard cosmology is build with simple maximally symmetric 3manifolds (spacelike timeslices of constant curvature, e.g. $S^3$ or less popular the hyperbolic space $H^3$). Since $S^3$ has a finite volume it seems natural to ask whether there also exists a maximally symmetric hyperbolic counterpart which also has a finite volume? (An answer in layman's terms would be fine, if possible.)

1$\begingroup$ You probably want to read about Thurston geometries. Also about the notion of "locally symmetric space". That $H^3$ is "less popular"? well, hyperbolic 3dimensional geometry has been a major theme in geometric topology in the last 40 years! $\endgroup$– YCorMay 20, 2020 at 16:53

1$\begingroup$ Sorry, but less popular in standard cosmology, because it is generally believed that spacetime in general relativity is not negatively curved. $\endgroup$– laymanMay 20, 2020 at 16:58

1$\begingroup$ In any case: finite volume manifolds with curvature $1$ usually have a very small amount of symmetries, whence the notion of locally symmetric space. $\endgroup$– YCorMay 20, 2020 at 17:17

1$\begingroup$ In GR we sometimes say "locally maximally symmetric" to mean: for every point on the manifold there exists an open neighbourhood of said point which is isometrically diffeomorphic to an open set on a "maximally symmetric" space. So for the purposes of this question said maximally symmetric space would be $H^3$. There should be lots of examples* which are not $H^3$ itself, but whether any have finite volume, I do not know. (*In the Lorentzian case the BTZ black hole geometry is some quotient of maximally symmetric $AdS_3$ but I do not remember the details now.) $\endgroup$– AlexArvanitakisMay 20, 2020 at 19:37

1$\begingroup$ Equivalently, such locally maximally symmetric spaces admit the maximum number of Killing vector fields on some neighbourhood of every point. $\endgroup$– AlexArvanitakisMay 20, 2020 at 19:39
1 Answer
Kojima has proved that every finite group occurs as the symmetry group of some compact hyperbolic 3manifold. Hence there is nothing like a hyperbolic 3manifold of maximal symmetry.
The same result is true in higher dimensions by BelolipetskyLubotzky.

$\begingroup$ Sorry, but I need some help to understand your answer. As far as I know there is a hyperbolic 3manifold which satisfies homogenity and isotropy (space of constant sectional curvature 1), but it has an infinite volume. Do I understand right that there is no such manifold with finite volume? $\endgroup$– laymanMay 21, 2020 at 16:25

1$\begingroup$ Yes, there is no such manifold. A finitevolume complete hyperbolic manifold has only finitely many symmetries. $\endgroup$ May 21, 2020 at 16:44

$\begingroup$ @Bruno Martelli thank you for the annotation. $\endgroup$– laymanMay 21, 2020 at 20:13