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 $\mathbb{H}^3$.

Fundamental domain

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:

Quotient orbifold

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"?

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