I'm taking a course this semester, and in it we proved that any complete hyperbolic surface is universally covered by $\mathbb{H}^2$. The text, found at http://www.math.brown.edu/~res/Papers/surfacebook.pdf, that we were given for the course, after proving the theorem, discusses only how this result opens the door to many beautiful tilings of the hyperbolic plane.

I asked my professor what the mathematical significance of the result is (I'm still new with the material so I haven't really gotten a hold of the big picture just yet). I can't remember exactly what he said, but it was something along the lines of

"Because of the result, if we wish to study complete hyperbolic surfaces, we can instead study the quotients $\mathbb{H}^2/\Gamma$, where $\Gamma$ is a torsion-free, discrete group of isometries of $\mathbb{H}^2$".

I looked up the Wikipedia article on "Hyperbolic space" http://en.wikipedia.org/wiki/Hyperbolic_space, and in it, it is said (under "Hyperbolic manifolds"),

Every complete, connected, simply-connected manifold of constant negative curvature $−1$ is isometric to the real hyperbolic space $\mathbb{H}^n$. As a result, the universal cover of any closed manifold $M$ of constant negative curvature $−1$, which is to say, a hyperbolic manifold, is $\mathbb{H}^n$. Thus, every such $M$ can be written as $\mathbb{H}^n/\Gamma$ where $\Gamma$ is a torsion-free, discrete group of isometries on $\mathbb{H}^n$.

(I checked Wikipedia to make sure I wasn't remembering incorrectly what my professor told me.)

My question is, how does the fact that the universal cover of any complete hyperbolic surface is $\mathbb{H}^2$ imply that any suchcomplete hyperbolic surface can be written $\mathbb{H}^2/\Gamma$?

  • $\begingroup$ It reminds me something like that the fundamental group $\pi_1(M)$ of a manifold $M$ acts on its universal cover $H$, such that $H/\pi_1(M) \cong M$, and probably the loops in $\pi_1(M)$ may be represented also by corresponding isometries. $\endgroup$
    – Berci
    Oct 27, 2012 at 13:34

1 Answer 1


If $M$ is a closed, complete hyperbolic manifold, then the hyperbolic structure lifts to its universal cover, $\widetilde{M}$. This means that $\Gamma = \pi_1(M)$ acts on $\widetilde{M}$ by isometries, and $M = \widetilde{M}/\Gamma$. Since $\widetilde{M}$ is isometric to $\mathbb{H}^n$, we have $M = \mathbb{H}^n/\Gamma$. If $\Gamma$ had torsion, then the action would not be fixed-point free and the quotient would have a cone point and not be a manifold, so we can conclude that $\Gamma$ has no torsion.

(Here's a short discussion of bigger picture because it's interesting! Study of hyperbolic surfaces is especially relevant for dimensions $2$ and $3$, where "most" manifolds are hyperbolic. For orientable closed surfaces, the Euler characteristic and Gauss-Bonnet imply that only $S^2$ and $T^2$ are not hyperbolic. Geometrization (Thurston-Perelman are the two biggest names here) states that if a closed $3$-manifold $M$ has: $\pi_1M$ infinite (i.e., $M$ is not covered by $S^3$), $\pi_1(M)$ not a free product (i.e., no incompressible spheres - $M$ is prime), and $\pi_1(M)$ contains no $\mathbb{Z}^2$ (i.e., no incompressible tori), then $M$ is hyperbolic. In three dimensions, once a space is hyperbolic, its geometric invariants become topological invariants by Mostow rigidity: If two hyperbolic manifolds are homotopy-equivalent, then that homotopy-equivalence was induced by an isometry.)

  • 2
    $\begingroup$ If you would like, I can add more information about why the hyperbolic structure lifts to the universal cover and why the deck transformations are isometries. $\endgroup$
    – Neal
    Oct 27, 2012 at 13:52
  • 2
    $\begingroup$ Sorry for the necropost; I would love it if you did add info., and/or refs. $\endgroup$
    – RikOsuave
    Dec 21, 2014 at 8:43

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