How to prove that a hyperbolic geodesic triangle induces a tesselation

71 Views Asked by At

Let $m_1,m_2,m_3$ be integers greather than $2$ such that $\frac{1}{m_1}+\frac{1}{m_2}+\frac{1}{m_3}<1$. Consider a hyperbolic triangle $\tau$ with internal angles given by $\frac{\pi}{m_1},\frac{\pi}{m_2}$ and $\frac{\pi}{m_3}$ on each respectively vertex. I would like to prove that if $G$ is the subgroup of $Aut(\mathbb{H})$ generated by the three reflexions along the sides of the triangle, say $r_1,r_2$ and $r_3$, then $\{g(\tau)\}_{g\in G}$ is a tesselation of the Hyperbolic Plane $\mathbb{H}$.

My approach was to define the quotient $$X:=\frac{\bigcup_{g\in G}\overline{\tau}\times\{g\}}{\sim},$$ where $(x,h)\sim (y,hr_i)$ if $i=1,2,3$ and $y=r_{i}^{-1}x$, in order to prove that the function $f:X\rightarrow\mathbb{H}$ given by $f([x,g])=gx$ is an $G$-equivariant homeomorphism. Intuitively, the quotient makes the desired tesselation identifying the correct sides.

There's a way to give a metric to $X$ in a way that its completness is equivalent to $f$ being a surjective covering map, but i could not be able to fulfill the details. Any hint or an easier aproach is appreciated!.

1

There are 1 best solutions below

0
On BEST ANSWER

This is a straightforward application of the Poincare polygon theorem, which is a substantial theorem. For references, take a look at the answer to this mathoverflow question, which includes generalizations to higher dimensions and other geometries.