# Branched 2-fold covering over edge of a 3-orbifold

I am reading the paper "On some generalized triangle groups and three-dimensional orbifolds" by Vinberg, Mennike and Khelling (Tran. Moscow Math. Soc. 1995 (56)).

Let $$k,l,m>0$$, at most one of them equal to $$2$$ and $$(k,l,m)\neq(2,3,3)$$. Let $$\tilde\Gamma=\langle X,Y,T\ |\ X^k=Y^l=(X^Y Y^X)^m=T^2=1, X^T=X^{-1}, Y^T=Y^{-1}\rangle$$.

The paper claims the following complex is a fundamental domain for the action of $$\Gamma$$ on hyperbolic $$3$$-space $$\mathbb{H}^3$$.

Vertices are denoted $$O_X$$, $$O_Y$$, $$O_X'$$, $$O_Y'$$, $$P$$, $$Q$$ and $$O$$ and stabilisers of edges are denoted by the generating element of the subgroup.

It goes on to describe how the action of $$\tilde\Gamma$$ identifies faces edges and vertices.

• $$XO_Y=O_Y'$$,
• $$YO_X=O_X'$$,
• $$X^Y[OP]=[OQ]$$ so $$P\sim Q$$,
• $$X^Y[OO_Y'P]=[OO_Y'Q]$$
• $$Y^X[OO'_XQ]=[OO_X'P]$$
• $$(XYT)[O_YO_X'P]=[O_Y'O_XQ]$$

It also states that $$T$$ is a rotation through the angle $$\pi$$ about the line passing through $$O_X$$ and $$O_Y$$.

From this one can deduce vertex stabilisers in $$\tilde\Gamma$$ are isomorphic to

• $$Stab(O)\cong Tr(k,l,m)=\langle a, b\ |\ a^l=b^k=(ab^{-1})^m=1 \rangle$$,
• $$Stab(O_X)\cong D_{2k}$$,
• $$Stab(O_Y)\cong D_{2l}$$,
• $$Stab(P)\cong D_{2m}$$ (I think).

Now $$\Gamma = \langle x,y\ |\ x^k=y^l=(x^yy^x)^m=1\rangle$$ is isomorphic to an index $$2$$ subgroup of $$\Gamma$$. How exactly can I picture the fundamental domain for this group acting on $$\mathbb{H}^3$$.

Note that when passing to the quotient space $$\mathbb{H}^3/\tilde\Gamma$$, the paper states that the orbifold corresponding to $$\Gamma$$ is a branched covering over the unlabelled edges" of:

To summarise my three questions are:

1. What is a cell structure for the fundamental domain of $$\Gamma$$ and how can it be determined?
2. Are the vertex stabilisers correct?
3. What exactly is meant by a branched covering over the unlabelled edges"?