oriented surface of genus g and m punctures

Question

Let $S_{g,m}$ be an oriented surface of genus g and m punctures, what's the condition to ensure $S_{g,m}$ is hyperbolic? If $g\ge 2$, I know it is hyperbolic, how about g=0 and g=1? Thanks in advance.

Answer 1

For $g=0$ and any $m \ge 1$, the surface $S_{0,m}$ has a complete hyperbolic metric:

• The surface $S_{0,1}$ is the open disc. It is homeomorphic to the whole hyperbolic plane, and that is the only possibility. In particular $S_{0,1}$ does not have a finite area complete hyperbolic structure.
• The surface $S_{0,2}$ is the open annulus. It is homeomorphic to the hyperbolic plane modulo an infinite cyclic group generated by a parabolic isometry. Also, $S_{0,2}$ is homeomorphic to the hyperbolic plane modulo the infinite cyclic group generated by a hyperbolic isometry. Again these are the only possibilities, and $S_{0,2}$ does not have a finite area complete hyperbolic structure.

According to some ways of defining a "hyperbolic surface", one might require the metric to be complete and of finite area and one would therefore not say that $S_{0,1}$ and $S_{0,2}$ are hyperbolic.

• For $m \ge 3$ there is a finite area complete hyperbolic metric on $S_{0,m}$, and so $S_{0,m}$ is unambiguously hyperbolic.

Finally, for $g=1$ and any $m \ge 1$ the surface $S_{1,m}$ has a complete hyperbolic metric of finite area, and so these are unambiguously hyperbolic.