# Understanding Cycles in Hyperbolic Geometry: Definitions and Examples

I'm having trouble understanding the concept of cycles in the hyperbolic geometry, in particular, in the following works.

## Definitions

###### Definition 17. (Cycle) of Umemoto

Choose a vertex or an ideal vertex $$u$$ of $$\bar{D_p}(\Gamma)$$ (a Dirichlet region) and we shall call the set of all vertices of $$\bar{D_p}(\Gamma)$$ congruent to $$u$$ a cycle including $$u$$.

###### Definition 9.3.4 of Beardon

A cycle $$C$$ in $$P$$ (a Dirichlet region) is the intersection of a $$G$$-orbit with $$P$$: this is necessarily a finite set $$\{ z_1, \dots, z_n \}$$ and the length $$|C|$$ of $$C$$ is $$n$$.

## My question

Suppose a following Dirichlet domain (taken from Case 1 of Theorem 24 of Umemoto). What would be a cycle in this figure?

## References

Let $$\triangle$$ be a hyperbolic triangle with vertices $$v_1, v_2, v_3$$, interior angles $$\pi/m_1, \pi/m_2, \pi/m_3$$ at these vertices, and sides $$M_1, M_2, M_3$$ opposite to these vertices. Let $$R_i$$ be the hyperbolic reflection in the geodesic containing $$M_i$$ ($$i = 1, 2, 3$$) and consider the triangle group $$\Gamma$$ generated by $$R_1R_2$$, $$R_2R_3$$, and $$R_3R_1$$ (This group is the same as the group generated by two transformations $$R_iR_j$$ and $$R_jR_k$$, $$i, j, k = \{1, 2, 3\}$$). Then the Dirichlet domain $$D_p(\Gamma)$$ of $$\Gamma$$ centered at $$p$$ is determined as follows:

1. If $$p$$ is on the interior of $$M_i$$, then $$D_p(\Gamma)$$ is a quadrilateral; more precisely $$D_p(\Gamma) = H_{ij} \cap H_{ji} \cap H_{ik} \cap H_{ki}$$ where $$\{i, j, k\} = \{1, 2, 3\}$$, and $$H_{ij}$$ denotes $$H_p(R_iR_j)$$.

The Dirichlet fundamental domain for a Fuchsian group $$\Gamma$$ acting freely on the hyperbolic plane $$\mathbb H^2$$ is a particular type of fundamental domain in hyperbolic geometry. For $$z_0 \in \mathbb H^2$$ we define the Dirichlet domain for $$\Gamma$$ as $$P:=\bigcap_{g \in \Gamma}\{z \in \mathbb H^2 : d(z,z_0)\leq d(gz,z_0)\}$$ and it indeed tessellates $$\mathbb H^2$$, meaning that $$\mathbb H^2 = \bigcup_{g \in \Gamma}g P\quad \text{and} \quad g\, \mathrm{Interior}(P) \cap \mathrm{Interior}(P) = \emptyset \text{ for } g \neq \mathrm{id}.$$

Not necessarily the fundamental domain you are working on is a Dirichlet one. The most common type of fundamental domain comes from Poincaré's theorem for fundamental polygons: consider a convex geodesic polygon $$P$$ with finite sides $$S$$, a bijection $$\overline \bullet :S \mapsto \overline S$$ that maps $$s \to \overline s$$ satisfying $$\overline{\overline{s}} = s$$, and orientation-preserving isometries $$I_s:\mathbb H^2 \to \mathbb H^2$$ for each $$s \in S$$. We also assume that $$I_s(s)=\overline s$$, $$I_s^{-1} = I_{\overline s}$$ and $$I_s P \cap P = \overline s$$. In other words, we have a polygon $$P$$, and for each side $$s$$ we associate an isometry $$I_s$$ such that $$I_s$$ maps $$P$$ to an adjacent polygon $$I_sP$$ sharing $$\overline s$$ with $$P$$.

Under these conditions, we can talk about the cycle of vertices. For each vertex $$v_1$$ we have a cycle constructed in the following way. Take $$s_1$$ to an adjacent side to $$v_1$$. We have that $$v_2:=I_{s_1}v_1$$ has $$\overline s_1$$ as an adjacent side. We define $$s_2$$ as the other adjacent side. Then we do the same for $$v_2$$ and $$s_2$$, we have $$v_3:=I_{s_2} v_2$$ and $$I_{s_2}s_2$$. We define $$s_3$$ to be the other side adjacent to $$v_3$$, and so on. Following this procedure we eventually obtain $$v_{n+1}=v_1$$ again and $$s_{n+1}=s_1$$. Thus, we obtain the cycle $$(v_1,s_1)\to (v_2,s_2) \to (v_3,s_3) \to \cdots \to (v_n,s_n) \to (v_1,s_1).$$

The idea is to impose that the sum of inner angles on the vertex $$v_1, \ldots,v_n$$ is $$2\pi/m$$ for some positive integer $$m$$. If for all vertices cycle this condition is met, then the group generated by the isometries $$I_s$$ tessellates $$\mathbb H^2$$ with fundamental domain $$P$$.

What all this means is the following. Assume for simplicity that $$m=1$$. Then, around $$v_1$$ we have the polygons $$P,\quad I_{s_1}^{-1}P,\quad I_{s_1}^{-1}I_{s_2}^{-1}P,\quad \ldots, \quad I_{s_1}^{-1}I_{s_2}^{-1}\cdots I_{s_{n-1}}^{-1}P.$$ tessellating around $$v_1$$. The idea is that tessellation around vertices implies the tessellation of the whole hyperbolic plane. Additionally, by studying the tesselation around each vertex we obtain $$I_{s_n} \cdots I_{s_2}I_{s_1} =\mathrm{id}$$ (and $$I_{s_n} \cdots I_{s_2}I_{s_1}$$ is a $$\pm 2\pi/m$$ rotation centered in $$v_1$$ for $$m> 1$$).

Now, about your particular problem. We have the polygon you described and we need isometries that generate the tesselation. Define $$v_4 = R_1 v_1$$ and consider the isometries $$I_2 = R_1 R_3$$ and $$I_3 = R_2 R_1$$. The isometry $$I_2$$ is the $$-2\pi/m_2$$ rotation centered at $$v_2$$ and $$I_3$$ the $$-2\pi/m_3$$ rotation centered at $$v_3$$. Note that $$I_2(v_2v_1) = v_2v_4$$ and $$I_3(v_3v_4) = v_3v_1$$. The cycle of vertices:

• for vertex $$v_2$$: $$(v_2,v_2v_1)\to(v_2,v_2v_1);$$
• for vertex $$v_3$$: $$(v_3,v_3v_4)\to(v_3,v_3v_4);$$
• for vertex $$v_1$$: $$(v_1,v_1v_2)\to(v_4,v_4v_3)\to (v_1,v_1v_2);$$
• for vertex $$v_4$$: $$(v_4,v_4v_3)\to(v_1,v_1v_2)\to (v_4,v_4v_3).$$

The cycle for $$v_2$$ has only $$v_2$$ as vertex and the sum of angles is $$2\pi/m_2$$ and $$I_2^{m_2}= \mathrm{id}$$. The same goes for $$v_3$$. For $$v_1$$, the cycle has two points, $$v_1,v_4$$, and the sum of angles is $$2\pi/m_1$$. Additionally, $$(I_3I_2)^{m_1} = \mathrm{id}$$ (the isometry $$I_1:=(I_3I_2)^{-1}=R_3R_2$$ is a rotation centered at $$v_1$$ and angle $$-2\pi/m_1$$). Thus, we obtain that your quadrilateral tesselates under the action of the group $$\Gamma := \langle I_1,I_2,I_3: I_j^{m_j} = \mathrm{id} \quad \text{and}\quad I_3I_2I_1 = \mathrm{id} \rangle.$$