Parabolic and hyperbolic Riemann surfaces

Remarks

Equivalently, $X$ is hyperbolic if and only if it has a Green's function. This is the smallest positive harmonic function on $X \setminus \set{a}$ with a logarithmic pole at $a$. A compact Riemann surface is never hyperbolic, because a subharmonic function on it attains its maximum and is therefore constant; such surfaces are sometimes called elliptic. Example: the unit disc $\mathbb{D}$ is hyperbolic, since $u(z) = \abs{z}^2 - 1$ is subharmonic, nonconstant and negative on $\mathbb{D}$. Non-example: the plane $\C$ is not hyperbolic. By the subharmonic form of Liouville's theorem, a subharmonic function on $\C$ that is bounded above is constant. As $\C$ is not compact, it is parabolic. A domain $\Omega \subseteq \C$ is hyperbolic if and only if its boundary is not a polar set. Its Green's function is then the classical Green's function of $\Omega$. This notion differs from the one in the uniformization theorem, which classifies $X$ by its universal cover. For example, $\C \setminus \set{0, 1}$ has the disc as universal cover, but it is parabolic, because a subharmonic function bounded above extends across the two punctures to all of $\C$.