# Why is the fundamental group of a closed hyperbolic $n$-manifold is a (uniform) lattice in $SO(n,1)$?

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?

• 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 $SO(n,1)$. – Charlie Frohman Mar 12 at 2:38
• For dimension $3$ download Thurston’s notes from MSRI. – Charlie Frohman Mar 12 at 2:39