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