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Why is the fundamental group of a closed hyperbolic $n$-manifold is a (uniform) lattice in $SO(n,1)$? Why is $SO(n,1)$ the (orientation-preserving) isometry group of real hyperbolic $n$-space?

Is there any reference for a proof?

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  • $\begingroup$ That is the group of congruence transformations of hyperbolic $n$-space. Since the geometric structure is complete the deck transformations are a discrete subgroup of $SO(n,1)$. $\endgroup$ – Charlie Frohman Mar 12 at 2:38
  • $\begingroup$ For dimension $3$ download Thurston’s notes from MSRI. $\endgroup$ – Charlie Frohman Mar 12 at 2:39

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