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I know that the hyperbolic plane $\mathbb{H}$ has constant curvature $-1$. I know that if a discrete group $G$ acts properly and freely on $\mathbb{H}$, then $\mathbb{H}/G$ becomes a Riemannian manifold of constant curvature $-1$, too. I am wondering if there exists a Riemannian manifold of constant curvature $-1$ which is not isometric to some quotient $\mathbb{H}/G$? Is this possible?

EDIT: What happens if we have a complete Riemannian manifold $M$ of constant curvature $-1$?

Best regards

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Take an open ball of $\mathbb{H}^n$ endowed with the induced metric, it is not complete, so it cannot be isomorphic to $\mathbb{H}^n/\Gamma$ which is complete.

If you suppose that $M$ is complete connected manifold endowed with a metric of curvature $-1$, its universal cover $\tilde{M}$ is isomorphic to $\mathbb{H}^n$, so $M$ is the quotient of $\mathbb{H}^n$ by a group of isometries that we identifies with its fundamental group.

https://en.wikipedia.org/wiki/Hyperbolic_manifold

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  • $\begingroup$ Thank! This makes a lot of sense. :) What happens if we restrict to complete manifolds? $\endgroup$ – Sammyy Delbrin Apr 8 '17 at 19:00
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Let $M$ be a complete manifold of sectional curvature $-1$. Then its universal cover $\widetilde{M}$ is a simply-connected manifold of sectional curvature $-1$. Thus $\widetilde{M}$ is isometric to the simply-connected model space of sectional curvature $-1$, which is hyperbolic space $\Bbb{H}$.

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