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In physics, standard cosmology is build with simple maximally symmetric 3-manifolds (spacelike time-slices 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.)

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    $\begingroup$ You probably want to read about Thurston geometries. Also about the notion of "locally symmetric space". That $H^3$ is "less popular"? well, hyperbolic 3-dimensional geometry has been a major theme in geometric topology in the last 40 years! $\endgroup$ – YCor May 20 at 16:53
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    $\begingroup$ Sorry, but less popular in standard cosmology, because it is generally believed that space-time in general relativity is not negatively curved. $\endgroup$ – layman May 20 at 16:58
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    $\begingroup$ In any case: finite volume manifolds with curvature $-1$ usually have a very small amount of symmetries, whence the notion of locally symmetric space. $\endgroup$ – YCor May 20 at 17:17
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    $\begingroup$ 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.) $\endgroup$ – AlexArvanitakis May 20 at 19:37
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    $\begingroup$ Equivalently, such locally maximally symmetric spaces admit the maximum number of Killing vector fields on some neighbourhood of every point. $\endgroup$ – AlexArvanitakis May 20 at 19:39
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Kojima has proved that every finite group occurs as the symmetry group of some compact hyperbolic 3-manifold. Hence there is nothing like a hyperbolic 3-manifold of maximal symmetry.

The same result is true in higher dimensions by Belolipetsky-Lubotzky.

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  • $\begingroup$ Sorry, but I need some help to understand your answer. As far as I know there is a hyperbolic 3-manifold 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? $\endgroup$ – layman May 21 at 16:25
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    $\begingroup$ Yes, there is no such manifold. A finite-volume complete hyperbolic manifold has only finitely many symmetries. $\endgroup$ – Bruno Martelli May 21 at 16:44
  • $\begingroup$ @Bruno Martelli thank you for the annotation. $\endgroup$ – layman May 21 at 20:13

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