This paper was converted on www.awesomepapers.org from LaTeX by an anonymous user.
Want to know more? Visit the Converter page.

Figure-eight knot is always over there

Jiming Ma School of Mathematical Sciences, Fudan University, Shanghai, China [email protected]  and  Baohua Xie School of Mathematics, Hunan University, Changsha, China [email protected]
(Date: Oct. 4, 2023)
Abstract.

It is well-known that complex hyperbolic triangle groups Δ(3,3,4)\Delta(3,3,4) generated by three complex reflections I1,I2,I3I_{1},I_{2},I_{3} in PU(2,1) has 1-dimensional moduli space. Deforming the representations from the classical \mathbb{R}-Fuchsian one to Δ(3,3,4;)\Delta(3,3,4;\infty), that is, when I3I2I1I2I_{3}I_{2}I_{1}I_{2} is accidental parabolic, the 3-manifolds at infinity change from a Seifert 3-manifold to the figure-eight knot complement.

When I3I2I1I2I_{3}I_{2}I_{1}I_{2} is loxodromic, there is an open set Ω𝐇2=𝕊3\Omega\subset\partial\mathbf{H}^{2}_{\mathbb{C}}=\mathbb{S}^{3} associated to I3I2I1I2I_{3}I_{2}I_{1}I_{2}, which is a subset of the discontinuous region. We show the quotient space Ω/Δ(3,3,4)\Omega/\Delta(3,3,4) is always the figure-eight knot complement in the deformation process. This gives the topological/geometrical explain that the 3-manifold at infinity of Δ(3,3,4;)\Delta(3,3,4;\infty) is the figure-eight knot complement. In particular, this confirms a conjecture of Falbel-Guilloux-Will.

Key words and phrases:
Complex hyperbolic space, CR-spherical uniformization, figure-eight knot complement
2010 Mathematics Subject Classification:
20H10, 57M50, 22E40, 51M10.
Jiming Ma was supported by NSFC (No.12171092), he is also a member of LMNS, Fudan University. Baohua Xie was supported by NSFC (No.11871202).

1. Introduction

Let 𝐇2{\bf H}^{2}_{\mathbb{C}} be the complex hyperbolic plane, the holomorphic isometry group of 𝐇2{\bf H}^{2}_{\mathbb{C}} is PU(2,1). The complex hyperbolic plane 𝐇2{\bf H}^{2}_{\mathbb{C}} can be identified with the unit ball in 2{\mathbb{C}}^{2}, so the ideal boundary 𝐇2\partial{\bf H}^{2}_{\mathbb{C}} of 𝐇2{\bf H}^{2}_{\mathbb{C}} is the 3-sphere 𝕊3\mathbb{S}^{3}.

Thurston’s work on 3-manifolds has shown that geometry has an important role in the study of topology of 3-manifolds. We have three kinds of geometrical structures on 3-manifolds related to the pair (𝐇2,𝐇2)({\bf H}^{2}_{\mathbb{C}},\partial{\bf H}^{2}_{\mathbb{C}}) with increasing group action constraints.

Definition 1.1.

For a smooth 3-manifold MM:

  1. (1)

    A spherical CR-structure on MM is a maximal collection of distinguished charts modeled on the boundary 𝐇2\partial\mathbf{H}^{2}_{\mathbb{C}}, where coordinates changes are restrictions of transformations from PU(2,1). In other words, a spherical CR-structure is a (G,X)(G,X)-structure with G=PU(2,1)G=\mbox{\rm PU(2,1)} and X=𝕊3X=\mathbb{S}^{3};

  2. (2)

    On the other hand, a CR-structure spherical uniformization of MM is a homeomorphism M=Ω/ρ(π1(M))M=\Omega/\rho(\pi_{1}(M)), where Ω\Omega is an open subset of 𝐇2\partial\mathbf{H}^{2}_{\mathbb{C}} on which ρ(π1(M))\rho(\pi_{1}(M)) acts properly discontinuously. See [5, 9];

  3. (3)

    A spherical CR-structure on MM is uniformizable if it is obtained as M=ΩΓ/ΓM=\Omega_{\Gamma}/\Gamma, where ΩΓ𝐇2\Omega_{\Gamma}\subset\partial\mathbf{H}^{2}_{\mathbb{C}} is the discontinuity region of a discrete subgroup Γ\Gamma. The limit set ΛΓ\Lambda_{\Gamma} of Γ\Gamma is 𝐇2ΩΓ\partial\mathbf{H}^{2}_{\mathbb{C}}-\Omega_{\Gamma} by definition.

For a discrete group Γ<PU(2,1)\Gamma<\mbox{\rm PU(2,1)}, the open set Ω\Omega in (2) of Definition 1.1 is a subset of the discontinuity region ΩΓ\Omega_{\Gamma} in (3) of Definition 1.1. So for a discrete group Γ\Gamma, there is at most one uniformizable spherical CR-structure associated to it, but there may be infinitely many CR-structure spherical uniformizations associated to it.

For a discrete group Γ<PU(2,1)\Gamma<\mbox{\rm PU(2,1)}, the 3-manifold M=ΩΓ/ΓM=\Omega_{\Gamma}/\Gamma at infinity of the 4-manifold 𝐇2/Γ\mathbf{H}^{2}_{\mathbb{C}}/\Gamma is the analogy of the 2-manifold at infinity of a geometrically finite, infinite volume hyperbolic 3-manifold. In other words, uniformizable spherical CR-structures on 3-manifolds in 𝐇2\mathbf{H}^{2}_{\mathbb{C}}-geometry are the analogies of conformal structures on surfaces in 𝐇3\mathbf{H}^{3}_{\mathbb{R}}-geometry.

In the three kinds of geometrical structures of Definition 1.1, uniformizable spherical CR-structures on 3-manifolds seem to be the most interesting ones. But in contrast to results on other geometric structures carried on 3-manifolds, there are relatively few examples known about them. A possible way to get uniformizable spherical CR-structures is via the deformations of triangle groups in PU(2,1).

Let T(p,q,r)T(p,q,r) be the abstract triangle group with presentation

T(p,q,r)=σ1,σ2,σ3|σ12=σ22=σ32=(σ2σ3)p=(σ3σ1)q=(σ1σ2)r=id,T(p,q,r)=\langle\sigma_{1},\sigma_{2},\sigma_{3}|\sigma^{2}_{1}=\sigma^{2}_{2}=\sigma^{2}_{3}=(\sigma_{2}\sigma_{3})^{p}=(\sigma_{3}\sigma_{1})^{q}=(\sigma_{1}\sigma_{2})^{r}=id\rangle,

where p,q,rp,q,r are positive integers or \infty satisfying

1p+1q+1r<1.\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<1.

We assume that pqrp\leqslant q\leqslant r. If p,qp,q or rr equals \infty, then the corresponding relation does not appear. The ideal triangle group is the case that p=q=r=p=q=r=\infty. A (p,q,r)(p,q,r) complex hyperbolic triangle group is a representation ρ\rho of T(p,q,r)T(p,q,r) into PU(2,1) where the generators fix complex lines. We denote ρ(σi)\rho(\sigma_{i}) by IiI_{i}, and the image group by Δ(p,q,r)=I1,I2,I3\Delta(p,q,r)=\langle I_{1},I_{2},I_{3}\rangle. It is well known [17] that the space of (p,q,r)(p,q,r)-complex reflection triangle groups has real dimension one if 3pqr3\leqslant p\leqslant q\leqslant r.

The isometry group of the real hyperbolic plane 𝐇2{\bf H}^{2}_{\mathbb{R}} is PO(2,1), and it is well known that the ideal triangle group is rigid in PO(2,1). Goldman and Parker [7] initiated the study of the deformations of ideal triangle group into PU(2,1). They gave an interval in the moduli space of complex hyperbolic ideal triangle groups, for points in this interval the corresponding representations are discrete and faithful. They conjectured that a complex hyperbolic ideal triangle group Δ(,,)=I1,I2,I3\Delta(\infty,\infty,\infty)=\langle I_{1},I_{2},I_{3}\rangle is discrete and faithful if and only if I1I2I3I_{1}I_{2}I_{3} is not elliptic. Schwartz proved Goldman-Parker’s conjecture in [15, 18]. Furthermore, Schwartz analyzed the complex hyperbolic ideal triangle group Γ\Gamma when I1I2I3I_{1}I_{2}I_{3} is parabolic, and showed the 3-manifold at infinity of the quotient space 𝐇2/Γ{\bf H}^{2}_{\mathbb{C}}/{\Gamma} is commensurable with the Whitehead link complement in the 3-sphere [16]. In other words, the Whitehead link complement admits uniformizable spherical CR-structure. Seifert 3-manifolds admitting uniformizable spherical CR-structures are rather easy to construct, but the Whitehead link complement is the first example of hyperbolic 3-manifold which admits uniformizable spherical CR-structure.

Richard Schwartz [17] has also conjectured the necessary and sufficient condition for a general complex hyperbolic triangle group

Δ(p,q,r)=I1,I2,I3<PU(2,1)\Delta(p,q,r)=\langle I_{1},I_{2},I_{3}\rangle<\mbox{\rm PU(2,1)}

to be a discrete and faithful representation of T(p,q,r)T(p,q,r). Schwartz’s conjecture has been proved in a few cases [8, 13, 14].

The critical point of the 1-dimensional deformation space of complex hyperbolic triangle groups is a point such that some preferred word WAW_{A} or WBW_{B} is accidental parabolic. For more details, see [17]. People found several more examples of cusped hyperbolic 3-manifolds which admit uniformizable spherical CR-structures at these critical points [1, 3, 4, 8, 10, 11]. Almost all of the examples of uniformizable spherical CR-structures gotten now are via difficult and sophisticated analysis. But we do not know the topological/geometrical reason the 3-manifolds at infinity of the groups associated to critical points should be the ones we got. Falbel-Guilloux-Will [5] proposed a method to predict the 3-manifold when there is an accidental parabolic element.

We now just consider the representations of T(3,3,4)T(3,3,4) into PU(2,1) with complex reflection generators I1,I2,I3I_{1},I_{2},I_{3}. We can parametrize the representations by t[1/3,21]t\in[1/3,\sqrt{2}-1], and the even subgroup of the image group is denoted by Γt\Gamma_{t}, see Section 3 for more details. Moreover,

  • When t=21t=\sqrt{2}-1, the image group lies in PO(2,1). So we have the classical \mathbb{R}-Fuchsian group;

  • For any t(3/8,21]t\in(3/8,\sqrt{2}-1], I3I2I1I2I_{3}I_{2}I_{1}I_{2} is loxodromic;

  • When t=3/8t=3/8, I3I2I1I2I_{3}I_{2}I_{1}I_{2} is parabolic. This is an accidental parabolicity, so t=3/8t=3/8 corresponds to the critical point in the moduli space of Δ(3,3,4)\Delta(3,3,4) in our parameterization;

  • When t[1/3,3/8)t\in[1/3,3/8), I3I2I1I2I_{3}I_{2}I_{1}I_{2} is elliptic. We will not consider representations in this interval.

It is showed by Parker-Wang-Xie [13] for each t[3/8,21]t\in[3/8,\sqrt{2}-1], the corresponding representation is discrete and faithful. Since when t=21t=\sqrt{2}-1, we have a \mathbb{R}-Fuchsian group, so 3-manifold at infinity of the corresponding group is just the unit tangent bundle over the real hyperbolic (3,3,4)(3,3,4)-orbisurface. But when t=3/8t=3/8, there is a new parabolic element I3I2I1I2I_{3}I_{2}I_{1}I_{2}, so the 3-manifold at infinity of the corresponding group must change. It is showed by Deraux-Falbel [4] the 3-manifold at infinity of the even subgroup Γ3/8\Gamma_{3/8} is the figure-eight knot complement. But we do not know the reason that the 3-manifold at infinity of Δ(3,3,4)\Delta(3,3,4) when I3I2I1I2I_{3}I_{2}I_{1}I_{2} is parabolic should be this one. Falbel-Guilloux-Will [5] proposed an explanation of this phenomenon.

For all t(3/8,21]t\in(3/8,\sqrt{2}-1], I3I2I1I2I_{3}I_{2}I_{1}I_{2} is loxodromic. Let p1p_{1} and p2p_{2} be the attractive and repulsive fixed points of it, they determine a \mathbb{C}-circle. We denote by α1\alpha_{1} a preferred one of the two arcs with end points p1p_{1} and p2p_{2} in the \mathbb{C}-circle (see Section 5 for this arc). Let Λt\Lambda_{t} be the limit set of Γt\Gamma_{t}. Then it is a topological circle. The crown associated to I3I2I1I2I_{3}I_{2}I_{1}I_{2} is the subset of 𝕊3\mathbb{S}^{3} defined as

Crown=CrownΓt,I3I2I1I2=Λt(gΓtgα1).{\rm Crown}={\rm Crown}_{\Gamma_{t},I_{3}I_{2}I_{1}I_{2}}=\Lambda_{t}\cup\Bigl{(}\bigcup_{g\in\Gamma_{t}}g\cdot\alpha_{1}\Bigr{)}.

We denote ΩΓt,I3I2I1I2ΩΓt\Omega_{\Gamma_{t},I_{3}I_{2}I_{1}I_{2}}\subset\Omega_{\Gamma_{t}} as the complement of CrownΓt,I3I2I1I2{\rm Crown}_{\Gamma_{t},I_{3}I_{2}I_{1}I_{2}} in 𝕊3\mathbb{S}^{3}. Recall that ΩΓt=𝕊3Λt\Omega_{\Gamma_{t}}=\mathbb{S}^{3}-\Lambda_{t} is the discontinuous region of Γt\Gamma_{t}’s action on 𝕊3\mathbb{S}^{3}.

It was shown in [2] that ΩΓt,I3I2I1I2/Γt\Omega_{\Gamma_{t},I_{3}I_{2}I_{1}I_{2}}/\Gamma_{t} is homeomorphism to the figure-eight knot complement when t=21t=\sqrt{2}-1. In fact Falbel-Guilloux-Will [5] identified this manifold as drilling out the unit tangent bundle of (3,3,4)(3,3,4)-orbisurface a certain closed orbit associated to I3I2I1I2I_{3}I_{2}I_{1}I_{2}. Moreover, Falbel-Guilloux-Will [5] conjectured that the quotient space of ΩΓt,I3I2I1I2\Omega_{\Gamma_{t},I_{3}I_{2}I_{1}I_{2}} by Γt\Gamma_{t} is always the figure-eight knot complement for any t(3/8,21)t\in(3/8,\sqrt{2}-1). So each of them gives a CR-structure spherical uniformization of figure-eight knot complement. The last one, that is when t=3/8t=3/8, gives the uniformizable spherical CR-structure on the figure-eight knot complement. Which corresponds to pinching on the limit set of Γt\Gamma_{t} to the limit set of Γ3/8\Gamma_{3/8}. So this conjecture explains how to get the 3-manifold at infinity of Γ3/8\Gamma_{3/8} from the 3-manifold at infinity of a \mathbb{R}-Fuchsian group. Falbel-Guilloux-Will [5] confirmed the conjecture when tt is near to 21\sqrt{2}-1.

We certificate Falbel-Guilloux-Will’s conjecture totally in this paper:

Theorem 1.2.

For the parameterazation of complex hyperbolic groups Δ(3,3,4)\Delta(3,3,4) by t(3/8,21]t\in(3/8,\sqrt{2}-1]:

  1. (1)

    The 3-manifold at infinity of the even subgroup Γt\Gamma_{t} is the unit tangent bundle of the (3,3,4)(3,3,4)-orbisurface for all t(3/8,21]t\in(3/8,\sqrt{2}-1];

  2. (2)

    The quotient space of ΩΓt,I3I2I1I2\Omega_{\Gamma_{t},I_{3}I_{2}I_{1}I_{2}} by Γt\Gamma_{t} is always the figure-eight knot complement for all t(3/8,21]t\in(3/8,\sqrt{2}-1].

So in the deformation process, the figure-eight knot is always over there! This explains the 3-manifold at infinity of the even subgroup Γ3/8\Gamma_{3/8} (with accidental parabolic element) is the figure-eight knot [4].

We prove Theorem 1.2 in the following steps:

  • For Δ(3,3,4)=I1,I2,I3\Delta(3,3,4)=\langle I_{1},I_{2},I_{3}\rangle depends on t(3/8,21]t\in(3/8,\sqrt{2}-1], I1I2I_{1}I_{2} has order 4, and I1I2I_{1}I_{2} has fixed point p0𝐇2p_{0}\in\mathbf{H}^{2}_{\mathbb{C}};

  • Consider the Dirichlet domain DtD_{t} of Γt<Δ(3,3,4)\Gamma_{t}<\Delta(3,3,4) with center p0p_{0}, DtD_{t} has eight facets [13];

  • The ideal boundary Dt=Dt𝐇2\partial_{\infty}D_{t}=D_{t}\cap\partial\mathbf{H}^{2}_{\mathbb{C}} is a solid torus. Moreover, the boundary of Dt\partial_{\infty}D_{t} consists of eight annuli, the side-pairing pattern on them is independent of t(3/8,21]t\in(3/8,\sqrt{2}-1]. So the 3-manifold at infinity of the group Γt\Gamma_{t} is independent of tt. This proves (1) of Theorem 1.2;

  • We then consider the complement of the crown in Dt\partial_{\infty}D_{t}, that is,

    DtCrownΓt,I3I2I1I2.\partial_{\infty}D_{t}-{\rm Crown}_{\Gamma_{t},I_{3}I_{2}I_{1}I_{2}}.

    Which is a fundamental domain of Γt\Gamma_{t}’s action on ΩΓt,I3I2I1I2\Omega_{\Gamma_{t},I_{3}I_{2}I_{1}I_{2}}. In fact DtCrownΓt,I3I2I1I2\partial_{\infty}D_{t}\cap{\rm Crown}_{\Gamma_{t},I_{3}I_{2}I_{1}I_{2}} are exactly eight arcs. We will show the topology and the side-pairing pattern on DtCrownΓt,I3I2I1I2\partial_{\infty}D_{t}-{\rm Crown}_{\Gamma_{t},I_{3}I_{2}I_{1}I_{2}} are independent of tt. This in turn proves (2) of Theorem 1.2.

Acknowledgement: Part of the work was carried out when Jiming Ma was visiting Hunan University in the summer of 2022, the hospitality is gratefully appreciated. The second author thanks John Parker for a useful discussion about the parametrization of the deformation space of the triangle group Δ(3,3,4)\Delta(3,3,4) several years ago.

2. Background

We will briefly introduce some background of complex hyperbolic geometry in this section. One can refer to Goldman’s book [6] for more details.

2.1. Complex projective space and complex hyperbolic plane

The projective space 2\mathbb{CP}^{2} is the quotient of the complex space 3\mathbb{C}^{3} minus the origin, by the non-zero complex numbers. We denote by \mathbb{P} the projectivisation map :3\{0}2\mathbb{P}:\mathbb{C}^{3}\backslash\{0\}\to\mathbb{CP}^{2}. We will constantly use points in the projective space 2\mathbb{CP}^{2} and lifts to 3\mathbb{C}^{3}(or in 2,1\mathbb{C}^{2,1}, see below) throughout this paper. In this situation, points in 3\mathbb{C}^{3} will be denoted by 𝐳{\bf z}, and zz will denote the image in 2\mathbb{CP}^{2} under projectivisation.

Let 2,1\mathbb{C}^{2,1} denote a copy of 3\mathbb{C}^{3} equipped with a Hermitian form ,\langle\cdot,\cdot\rangle of signature (2,1)(2,1) on 3\mathbb{C}^{3}, and define

V\displaystyle V_{-} ={Z3:Z,Z<0},\displaystyle=\{Z\in\mathbb{C}^{3}:\langle Z,Z\rangle<0\},
V+\displaystyle V_{+} ={Z3:Z,Z>0},\displaystyle=\{Z\in\mathbb{C}^{3}:\langle Z,Z\rangle>0\},
V0\displaystyle V_{0} ={Z3:Z,Z=0}.\displaystyle=\{Z\in\mathbb{C}^{3}:\langle Z,Z\rangle=0\}.

The complex hyperbolic plane 𝐇2\mathbf{H}^{2}_{\mathbb{C}} is the projectivsation of the cone VV_{-} in 2,1\mathbb{C}^{2,1}, equipped with a Hermitian metric induced by the Hermitian form ,\langle\cdot,\cdot\rangle. The projection to 2\mathbb{CP}^{2} of the quadratic V0V_{0} can be thought of as the boundary at infinity of 2,1\mathbb{C}^{2,1}, and we will denote it as 𝐇2\partial\mathbf{H}^{2}_{\mathbb{C}}. The space 𝐇2\mathbf{H}^{2}_{\mathbb{C}} is homeomorphic to a ball B4B^{4}, and 𝐇2\partial\mathbf{H}^{2}_{\mathbb{C}} is homeomorphic to 3-sphere 𝕊3\mathbb{S}^{3}.

The complex hyperbolic distance on 𝐇2\mathbf{H}^{2}_{\mathbb{C}} is given by

cosh(d(p,q)2)=|𝐩,𝐪|2|𝐩,𝐩||𝐪,𝐪|.\cosh\left(\frac{d(p,q)}{2}\right)=\frac{\left|\langle{\bf p},{\bf q}\rangle\right|^{2}}{\left|\langle{\bf p},{\bf p}\rangle\right|\left|\langle{\bf q},{\bf q}\rangle\right|}.

The subgroup of SL(3,\mathbb{C}) of maps that preserve the Hermitian form ,\langle\cdot,\cdot\rangle is by definition SU(2,1) and its projectivisation PU(2,1) the group of holomorphic isometries of 𝐇2\mathbf{H}^{2}_{\mathbb{C}}. We will often work with SU(2,1), which is a 3-fold cover of PU(2,1).

2.2. Two models

There are two special choices of the Hermitian forms

J1=[100010001]andJ2=[001010100].J_{1}=\left[\begin{array}[]{ccc}1&0&0\\ 0&1&0\\ 0&0&-1\\ \end{array}\right]\quad{\rm and}\quad J_{2}=\left[\begin{array}[]{ccc}0&0&1\\ 0&1&0\\ 1&0&0\\ \end{array}\right].

Note that they are conjugate by the Cayley transformation

Cay=12[101020101].Cay=\frac{1}{\sqrt{2}}\left[\begin{array}[]{ccc}1&0&1\\ 0&\sqrt{2}&0\\ 1&0&-1\\ \end{array}\right].

By using the Hermitian form given by J1J_{1}, we obtain the ball model of 𝐇2\mathbf{H}^{2}_{\mathbb{C}}. With this model, 𝐇2\mathbf{H}^{2}_{\mathbb{C}} can be seen as the unit ball in 2\mathbb{C}^{2}, where 2\mathbb{C}^{2} itself is seen as the affine chart z3=1z_{3}=1 of 2\mathbb{CP}^{2}. Any point in 𝐇2\mathbf{H}^{2}_{\mathbb{C}} can be lifted to 3\mathbb{C}^{3} in a unique way as a vector [z1,z2,1]T[z_{1},z_{2},1]^{T}, where ziz_{i}\in\mathbb{C} and |z1|2+|z2|2<1\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}<1. The boundary 𝐇2\partial\mathbf{H}^{2}_{\mathbb{C}} is just the 3-sphere 𝕊3\mathbb{S}^{3} defined by |z1|2+|z2|2=1\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}=1.

The second model that one will consider is the Siegel model if one uses the form J2J_{2}. It will be more convenient to analyze Heisenberg geometry and draw pictures. In this model, the projection of VV0V_{-}\cup V_{0} to 2\mathbb{CP}^{2} is contained in the affine chart z3=1z_{3}=1, except for the projection of [1,0,0]T[1,0,0]^{T}, which is at infinity. Thus any point in the closure of 𝐇2\mathbf{H}^{2}_{\mathbb{C}} admits a unique lift to 3\mathbb{C}^{3}, which is given by

ψ(z,t,u)=[|z|2u+it2z1]and[100],\psi(z,t,u)=\left[\begin{array}[]{c}\frac{-|z|^{2}-u+it}{2}\\ z\\ 1\\ \end{array}\right]\quad{\rm and}\quad\left[\begin{array}[]{c}1\\ 0\\ 0\\ \end{array}\right],

where z,tz\in\mathbb{C},t\in\mathbb{R} and u0u\geqslant 0. There coordinates are often called horospherical coordinates. When necessary, we will call the vector given above the standard lift of a point in 𝐇2\mathbf{H}^{2}_{\mathbb{C}}. We will denote by [z,t][z,t] the point in 𝐇2\partial\mathbf{H}^{2}_{\mathbb{C}} which is the projection of ψ(z,t,0)\psi(z,t,0). Then one can identify 𝐇2\partial\mathbf{H}^{2}_{\mathbb{C}} with ×{}}\mathbb{C}\times\mathbb{R}\cup\{\infty\}\}. Removing the point at infinity, we obtain the Heisenberg group, defined as ×\mathbb{C}\times\mathbb{R} with multiplication

(w,s)(z,t)=(w+z,s+t+2Im(wz¯)).\left(w,s\right)\ast\left(z,t\right)=\left(w+z,s+t+2\mbox{\rm Im}\,(w\bar{z})\right).

2.3. Two totally geodesic submanifolds and their boundarys

There are two kinds of totally geodesic submanifolds of real dimension 2 in 𝐇2\mathbf{H}^{2}_{\mathbb{C}}: complex lines in 𝐇2\mathbf{H}^{2}_{\mathbb{C}} are complex geodesics(represented by 𝐇𝟏\mathbf{H^{1}_{\mathbb{C}}}) and Langrangian planes in 𝐇2\mathbf{H}^{2}_{\mathbb{C}} are totally real geodesic 2-planes(represented by 𝐇𝟐\mathbf{H^{2}_{\mathbb{R}}}). Each of these totally geodesic submanifolds is a model of the hyperbolic plane. A polar vector of a complex line is the unique vector(up to scaling) in V+V_{+} perpendicular to this complex line.

A discrete subgroup of PU(2,1) preserving a complex line is called \mathbb{C}-Fuchsian and is isomorphic to a subgroup of P(U(1)×U(1,1))PU(2,1)P(U(1)\times U(1,1))\subset\mbox{\rm PU(2,1)}. A discrete subgroup of PU(2,1) preserving a Langrangian plane is called \mathbb{R}-Fuchsian and is isomorphic to a subgroup of SO(2,1)SU(2,1)SO(2,1)\in SU(2,1).

Consider the complex hyperbolic space 𝐇2\mathbf{H}^{2}_{\mathbb{C}} and its boundary 𝐇2\partial\mathbf{H}^{2}_{\mathbb{C}}. We define the \mathbb{C}-circle in 𝐇2\partial\mathbf{H}^{2}_{\mathbb{C}} to be the boundary of a complex geodesic in 𝐇2\mathbf{H}^{2}_{\mathbb{C}}. Analogously, we define the \mathbb{R}-circle in 𝐇2\partial\mathbf{H}^{2}_{\mathbb{C}} to be the boundary of a Langrangian plane in 𝐇2\mathbf{H}^{2}_{\mathbb{C}}.

Definition 2.1.

For a given complex geodesic CC, a complex reflection with minor CC is the isometry ιC\iota_{C} in PU(2,1) given by

ιC=z+2z,cc,cc,\iota_{C}=-z+2\frac{\langle z,c\rangle}{\langle c,c\rangle}c,

where cc is a polar vector of CC.

Definition 2.2.

The contact plane at M=(a,b,c)M=(a,b,c) is the plane P(M):=Zc+2aY2bXP(M):=Z-c+2aY-2bX.

The \mathbb{C}-circle of center M=(a,b,c)M=(a,b,c) and radius RR is the intersection of the contact plane at MM and the cylinder (Xa)2+(Yb)2=R2(X-a)^{2}+(Y-b)^{2}=R^{2}.

Proposition 2.3.

In the Heisenberg group, \mathbb{C}-circles are either vertical lines or ellipses whose projections on the zz-plane are circles.

For a given pair of distinct points in 𝐇2\partial\mathbf{H}^{2}_{\mathbb{C}}, there is a unique \mathbb{C}-circle passing through them. A finite \mathbb{C}-circle is determined by a center and a radius. For example, the finite \mathbb{C}-circle with center (z0,t0)(z_{0},t_{0}) and radius R>0R>0 has a polar vector

[(R2|z0|2+it0)/2z01],\left[\begin{array}[]{c}(R^{2}-\left|z_{0}\right|^{2}+it_{0})/2\\ z_{0}\\ 1\\ \end{array}\right],

and in it any point (z,t)(z,t) satisfies the equations

{|zz0|=R,t=t0+2Im(z¯z0).\begin{cases}\left|z-z_{0}\right|=R,\\ t=t_{0}+2\mbox{\rm Im}\,(\bar{z}z_{0}).\\ \end{cases}
Figure 1. The affine disk bounded by the finite \mathbb{C}-circle in the contact plane.
Definition 2.4.

We define the \mathbb{C}-disk to be the affine disk bounded by the finite \mathbb{C}-circle in the contact plane, see Figure 1.

The condition for self-intersection between the complex lines defined by polar vectors v1v_{1} and v2v_{2} is

(2.1) L(v1,v2)=|v1,v2|2v1,v1v2,v2<0.L(v_{1},v_{2})=\left|\langle v_{1},v_{2}\rangle\right|^{2}-\langle v_{1},v_{1}\rangle\langle v_{2},v_{2}\rangle<0.

This condition L(v1,v2)>0L(v_{1},v_{2})>0 was also known as a non-linking condition for two \mathbb{C}-circles with polar vectors v1v_{1} and v2v_{2}, see [12].

2.4. Bisectors and Dirichlet domain

There are no totally geodesic real hypersurface 𝐇2\mathbf{H}^{2}_{\mathbb{C}}, and so we must choose hypersurfaces for sides of our polyhedron. We choose to work with bisector. A bisector in 𝐇2\mathbf{H}^{2}_{\mathbb{C}} is the locus of points equidistant (with respect to the Bergman metric) from a given pair of points in 𝐇2\mathbf{H}^{2}_{\mathbb{C}}. Suppose that these points are uu and vv. Choose lifts 𝐮\mathbf{u}, 𝐯{\mathbf{v}} of uu and vv so that 𝐮,𝐮=𝐯,𝐯\langle\mathbf{u},\mathbf{u}\rangle=\langle\mathbf{v},\mathbf{v}\rangle. Then the bisector equidistant from uu and vv is

=(u,v)={p𝐇2:|𝐩,𝐮|=|𝐩,𝐯|}.\mathcal{B}=\mathcal{B}(u,v)=\{p\in\mathbf{H}^{2}_{\mathbb{C}}:|\langle\mathbf{p},\mathbf{u}\rangle|=|\langle\mathbf{p},\mathbf{v}\rangle|\}.

Suppose that we are given three points u,v1u,v_{1} and v2v_{2} in 𝐇2\mathbf{H}^{2}_{\mathbb{C}}. If the three corresponding vectors u,v1u,v_{1} and v2v_{2} in VV_{-} form a basis for 2,1\mathbb{C}^{2,1} then the intersection (u,v1)(u,v2)\mathcal{B}(u,v_{1})\cap\mathcal{B}(u,v_{2}) is called a Giraud disc. This is a particularly nice type of bisector intersection.

Suppose that Γ\Gamma is a discrete group of PU(2,1). Let p0p_{0} be a point of 𝐇2\mathbf{H}^{2}_{\mathbb{C}} and write Γp0\Gamma_{p_{0}} for the stabilizer of p0p_{0} in Γ\Gamma. Then the Dirichlet domain Dp0(Γ)D_{p_{0}}(\Gamma) for Γ\Gamma with centre p0p_{0} is defined to be

Dp0(Γ)={p𝐇2:d(p,p0)<d(p,g(p0))forallgΓΓp0}D_{p_{0}}(\Gamma)=\{p\in\mathbf{H}^{2}_{\mathbb{C}}:d(p,p_{0})<d(p,g(p_{0}))\,\ {\rm for\,\ all}\,\ g\in\Gamma-\Gamma_{p_{0}}\}

We define the spinal sphere 𝒮𝐇2\mathcal{S}\in\partial\mathbf{H}^{2}_{\mathbb{C}} as the boundary of the bisector \mathcal{B} in 𝐇2\mathbf{H}^{2}_{\mathbb{C}}. Note that two spinal spheres have an intersection if and only if the corresponding bisectors have an intersection.

3. Complex hyperbolic triangle groups Δ(3,3,4)\Delta(3,3,4)

Let IiI_{i} be a reflection along the complex line CiC_{i} for i=1,2,3i=1,2,3. We assume that Ci1C_{i-1} and CiC_{i} either meet at the angle π/pi\pi/p_{i} for some integer pi3p_{i}\geqslant 3 or else Ci1C_{i-1} and CiC_{i} are asymptotic, in which case they make an angle 0 and we write pi=p_{i}=\infty, where the indices are taken mod 3. The subgroup Δ(p1,p2,p3)\Delta(p_{1},p_{2},p_{3}) of PU(2,1) generated by I1,I2I_{1},I_{2} and I3I_{3} is called a complex hyperbolic triangle group. For fixed p1,p2,p3p_{1},p_{2},p_{3}, modulo conjugacy in PU(2,1), there exists in general a 11-parameter family of complex hyperbolic triangle group Δ(p1,p2,p3)\Delta(p_{1},p_{2},p_{3}).

We consider the deformation space of complex hyperbolic triangle group Δ(3,3,4)\Delta(3,3,4), generated by three complex reflections I1,I2I_{1},I_{2} and I3I_{3}. As an abstract group, it is given by

σ1,σ2,σ3σ12=σ22=σ32=(σ1σ2)4=(σ1σ3)3=(σ2σ3)3=id.\langle\sigma_{1},\sigma_{2},\sigma_{3}\mid\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma_{3}^{2}=(\sigma_{1}\sigma_{2})^{4}=(\sigma_{1}\sigma_{3})^{3}=(\sigma_{2}\sigma_{3})^{3}=id\rangle.

We will describe a parametrization of the deformation space of Δ(3,3,4)\Delta(3,3,4), which is a little different from that in [13].

Suppose that the polar vectors n1n_{1}, n2n_{2} of the complex lines C1C_{1}, C2C_{2} are given by

n1=[010]andn2=[1/21/20].n_{1}=\left[\begin{array}[]{c}0\\ 1\\ 0\\ \end{array}\right]\quad{\rm and}\quad n_{2}=\left[\begin{array}[]{c}1/\sqrt{2}\\ 1/\sqrt{2}\\ 0\\ \end{array}\right].

Then the corresponding complex reflections I1I_{1} and I2I_{2} are given by

(3.1) I1=[001010100]andI2=[001010100].I_{1}=\left[\begin{array}[]{ccc}0&0&1\\ 0&1&0\\ 1&0&0\\ \end{array}\right]\quad{\rm and}\quad I_{2}=\left[\begin{array}[]{ccc}0&0&1\\ 0&1&0\\ 1&0&0\\ \end{array}\right].

We may also suppose that the polar vector n3n_{3} of C3C_{3} is

n3=[abeiθd].n_{3}=\left[\begin{array}[]{c}a\\ be^{i\theta}\\ d\\ \end{array}\right].

Furthermore, we can assume that aa, bb, dd are nonnegative real numbers by conjugating a diagonal map Diag(eiβ,eiβ,e2iβ){\rm Diag}(e^{i\beta},e^{i\beta},e^{-2i\beta}) if necessary. After a normalization of n3n_{3}, we have

a2+b2d2=1.a^{2}+b^{2}-d^{2}=1.

The matrix for the complex reflection I3I_{3} is given by

(3.2) I3=[a2b2+d22abeiθ2ad2abeiθa2+b2+d22bdeiθ2ad2bdeiθa2b2d2].I_{3}=\left[\begin{array}[]{ccc}a^{2}-b^{2}+d^{2}&2abe^{i\theta}&2ad\\ 2abe^{-i\theta}&-a^{2}+b^{2}+d^{2}&2bde^{-i\theta}\\ -2ad&-2bde^{i\theta}&-a^{2}-b^{2}-d^{2}\\ \end{array}\right].

One may always assume θ[0,π]\theta\in[0,\pi] by complex conjugating if necessary.

The condition that I1I3I_{1}I_{3} and I2I3I_{2}I_{3} have order 3 is equivalent to tr(I1I3)=tr(I2I3)=0tr(I_{1}I_{3})=tr(I_{2}I_{3})=0. That is,

a2+3b2+d2=0-a^{2}+3b^{2}+d^{2}=0

and

4abcosθ+a2b2+d2=0.4ab\cos{\theta}+a^{2}-b^{2}+d^{2}=0.

Since we know that a2+b2=d2+1a^{2}+b^{2}=d^{2}+1, we have

(3.3) b=1/2,2acosθ=1/22a2,d2=(4a23)/4.b=1/2,\quad 2a\cos{\theta}=1/2-2a^{2},\quad d^{2}=(4a^{2}-3)/4.

We also have that d2d^{2} is nonnegative and |1/22a2|2a\left|1/2-2a^{2}\right|\leqslant 2a if and only if 3/2a(2+1)/2\sqrt{3}/2\leqslant a\leqslant(\sqrt{2}+1)/2. In other words, our parametrization of the deformation space of Δ(3,3,4)\Delta(3,3,4) is given by

a[3/2,(2+1)/2].a\in[\sqrt{3}/2,(\sqrt{2}+1)/2].

In particular, the entries of n3n_{3} are all real when a=(2+1)/2a=(\sqrt{2}+1)/2. Thus the complex hyperbolic triangle group Δ(3,3,4)\Delta(3,3,4) lies in SO(2,1)SO(2,1) when a=(2+1)/2a=(\sqrt{2}+1)/2.

Proposition 3.1.

Let I1,I2I_{1},I_{2} and I3I_{3} be given by (3.1) and (3.2). Suppose I1I3I_{1}I_{3} and I2I2I_{2}I_{2} have order 3. Then I1I3I2I3I_{1}I_{3}I_{2}I_{3} is elliptic if and only if a<1a<1.

Proof.

We compute the trace of I1I3I2I3I_{1}I_{3}I_{2}I_{3} directly and have that

tr(I1I3I2I3)\displaystyle tr(I_{1}I_{3}I_{2}I_{3})
=\displaystyle= 8ab(a2b2d2)cos(θ)+a4+a2(2b22d2)+b4+6b2d2+d4\displaystyle-8ab(a^{2}-b^{2}-d^{2})\cos(\theta)+a^{4}+a^{2}(2b^{2}-2d^{2})+b^{4}+6b^{2}d^{2}+d^{4}
=\displaystyle= 4a21.\displaystyle 4a^{2}-1.

The condition that I1I3I2I3I_{1}I_{3}I_{2}I_{3} is elliptic is equivalent to tr(I1I3I2I3)<3tr(I_{1}I_{3}I_{2}I_{3})<3. ∎

So our parameter space for I1,I2,I3\left\langle I_{1},I_{2},I_{3}\right\rangle with I1I3I2I3I_{1}I_{3}I_{2}I_{3} nonelliptic is given by

(3.4) 1a(2+1)/2.1\leqslant a\leqslant(\sqrt{2}+1)/2.

To make the computation simpler, we write a=1212ta=\frac{1}{2\sqrt{1-2t}}. Thus the parameter space for the triangle group Δ(3,3,4)\Delta(3,3,4) becomes

(3.5) 3/8t213/8\leqslant t\leqslant\sqrt{2}-1

with the new parameter tt.

Most calculations are carried out in the Siegel model. From now on, we will work on this model.

It is convenient to introduce some notations that are used throughout the paper. We define

(3.6) a(t)=6t2,\displaystyle a(t)=\sqrt{6t-2},
b(t)=t22t+1,\displaystyle b(t)=\sqrt{-t^{2}-2t+1},
c(t)=8t34t12t6t2.\displaystyle c(t)=\sqrt{\frac{8t-3}{4t-1-2t\sqrt{6t-2}}}.

In the Siegel model, the polar vectors n1,n2n_{1},n_{2} and n3n_{3} are given by

n1=[010],n2=[1/22/21/2],n3=1212t[22(a(t)+1)t+ib(t)22(a(t)1)].n_{1}=\left[\begin{array}[]{c}0\\ 1\\ 0\\ \end{array}\right],\,\ n_{2}=\left[\begin{array}[]{c}1/2\\ \sqrt{2}/2\\ 1/2\\ \end{array}\right],\,\ n_{3}=\frac{1}{2\sqrt{1-2t}}\left[\begin{array}[]{c}\frac{\sqrt{2}}{2}(a(t)+1)\\ -t+ib(t)\\ \frac{\sqrt{2}}{2}(a(t)-1)\\ \end{array}\right].

The corresponding complex reflections I1,I2I_{1},I_{2} and I3I_{3} are given by the matrices

I1=[100010001],I2=[1/22/21/22/202/21/22/21/2],I_{1}=\left[\begin{array}[]{ccc}-1&0&0\\ 0&1&0\\ 0&0&-1\\ \end{array}\right],\quad I_{2}=\left[\begin{array}[]{ccc}-1/2&\sqrt{2}/2&1/2\\ \sqrt{2}/2&0&\sqrt{2}/2\\ 1/2&\sqrt{2}/2&-1/2\\ \end{array}\right],

and

I3=[142(1+a(t))(t+ib(t))8t416t2a(t)8t42(a(t)1)(t+ib(t))8t4122(1+a(t))(t+ib(t))8t416t+2a(t)8t42(a(t)1)(t+ib(t))8t414],I_{3}=\left[\begin{array}[]{ccc}-\frac{1}{4}&\frac{\sqrt{2}(1+a(t))(t+ib(t))}{8t-4}&\frac{1-6t-2a(t)}{8t-4}\\ \frac{\sqrt{2}(a(t)-1)(-t+ib(t))}{8t-4}&-\frac{1}{2}&\frac{\sqrt{2}(1+a(t))(-t+ib(t))}{8t-4}\\ \frac{1-6t+2a(t)}{8t-4}&-\frac{\sqrt{2}(a(t)-1)(t+ib(t))}{8t-4}&-\frac{1}{4}\\ \end{array}\right],

respectively.

4. The Dirichlet domain of the triangle group Δ(3,3,4)\Delta(3,3,4)

4.1. The Dirichlet domain

For the convenience of the reader we recall the construction of the Dirichlet domain of the triangle group Δ(3,3,4)\Delta(3,3,4) from [13] without proof. The notations used here differ slightly from the notations used in [13].

For k,1k8k\in{\mathbb{Z}},1\leqslant k\leqslant 8, the involution AkA_{k} is denoted by

(I2I1)(k1)/2I3(I1I2)(k1)/2(I_{2}I_{1})^{(k-1)/2}I_{3}(I_{1}I_{2})^{(k-1)/2}

if kk is an odd integer and

(I2I1)(k2)/2I2I3I2(I1I2)(k2)/2(I_{2}I_{1})^{(k-2)/2}I_{2}I_{3}I_{2}(I_{1}I_{2})^{(k-2)/2}

if kk is even. One may take the index kk mod 8. Let p0p_{0} be the fixed point of I2I1I_{2}I_{1} in 𝐇2\mathbf{H}^{2}_{\mathbb{C}}. The bisector k\mathcal{B}_{k} is defined to be the bisector equidistant from p0p_{0} and Ak(p0)A_{k}(p_{0}). We define a polyhedron DD bounded by sides contained in these eight bisectors.

The combinatorial configuration of the bisectors as tt decreases from 21\sqrt{2}-1 to 3/83/8 are described as follows.

1\mathcal{B}_{1}2\mathcal{B}_{2}3\mathcal{B}_{3}4\mathcal{B}_{4}5\mathcal{B}_{5}6\mathcal{B}_{6}7\mathcal{B}_{7}8\mathcal{B}_{8}112233
Figure 2. A schematic view of Dirichlet domain of the triangle group Δ(3,3,4)\Delta(3,3,4) in the ball model.
Proposition 4.1 (Parker-Wang-Xie [13]).

Let k\mathcal{B}_{k} be defined as above. Suppose that 3/8t213/8\leqslant t\leqslant\sqrt{2}-1. Then for each k/8k\in{\mathbb{Z}}/8{\mathbb{Z}}:

  1. (1)

    The bisector k\mathcal{B}_{k} intersects k±1\mathcal{B}_{k\pm 1} in a Giraud disc. The Giraud disc is preserved by AkAk±1A_{k}A_{k\pm 1}, which has order 3.

  2. (2)

    The intersection of k\mathcal{B}_{k} with k±2\mathcal{B}_{k\pm 2} is contained in the halfspace bounded by k±1\mathcal{B}_{k\pm 1} not containing p0p_{0}.

  3. (3)

    The bisector k\mathcal{B}_{k} does not intersect k±ι\mathcal{B}_{k\pm\iota} for 3ι43\leqslant\iota\leqslant 4. Moreover, the boundaries of these bisectors are disjoint except for ι=3\iota=3 and t=3/8t=3/8, in which case the boundaries intersect in a single point, which is a parabolic fixed point.

The symmetry for DtD_{t}.

For each kk mod 8 and each nn mod 4, we have

  1. (1)

    (I2I1)n(k)=2n+k(I_{2}I_{1})^{n}(\mathcal{B}_{k})=\mathcal{B}_{2n+k};

  2. (2)

    (I2I1)nI2(k)=2n+3k(I_{2}I_{1})^{n}I_{2}(\mathcal{B}_{k})=\mathcal{B}_{2n+3-k}.

Furthermore, one can check that the side pairing maps AkA_{k} for DtD_{t} satisfies the conditions of the Poincaré polyhedron theorem for coset decomposition. Thus we have

Theorem 4.2 (Parker-Wang-Xie [13]).

Suppose that 3/8t213/8\leqslant t\leqslant\sqrt{2}-1. Let DtD_{t} be the polyhedron in 𝐇2\mathbf{H}^{2}_{\mathbb{C}} containing p0p_{0} and bounded by the eight bisectors k\mathcal{B}_{k}. Then DtD_{t} is the fundamental polyhedron of triangle group Δ(3,3,4)\Delta(3,3,4).

Let Γt\Gamma_{t} be the even subgroup of the triangle group I1,I2,I3\left\langle I_{1},I_{2},I_{3}\right\rangle. Let

g1=I3I2I1I2,g2=I2I1,g3=I1I2I3I2=g21g1g2.g_{1}=I_{3}I_{2}I_{1}I_{2},\quad g_{2}=I_{2}I_{1},\quad g_{3}=I_{1}I_{2}I_{3}I_{2}=g_{2}^{-1}g_{1}g_{2}.

Then

Γt=g1,g2.\Gamma_{t}=\left\langle g_{1},g_{2}\right\rangle.

Note that

g3=g21g1g2,g1=g2g1(g21g3)(g2g1)1.g_{3}=g_{2}^{-1}g_{1}g_{2},g_{1}=g_{2}g_{1}(g_{2}^{-1}g_{3})(g_{2}g_{1})^{-1}.

For 1k81\leqslant k\leqslant 8, we have

  1. (1)

    k=(p0,g2(k2)/2g31(p0))\mathcal{B}_{k}=\mathcal{B}\left(p_{0},g_{2}^{(k-2)/2}g_{3}^{-1}(p_{0})\right) if kk is even;

  2. (2)

    k=(p0,g2(k1)/2g1(p0))\mathcal{B}_{k}=\mathcal{B}\left(p_{0},g_{2}^{(k-1)/2}g_{1}(p_{0})\right) if kk is odd.

The side-pairing maps.

From above, it is easy to check that g1g_{1} maps the side on 4\mathcal{B}_{4} to the side on 1\mathcal{B}_{1}. Side-pairing maps for other sides can be obtained from this one by symmetry.

The Poincaré polyhedron theorem also shows that DtD_{t} is a fundamental domain for the action of Γt\Gamma_{t} modulo the action of a cyclic group g2\left\langle g_{2}\right\rangle of order 4.

When t=3/8t=3/8, the geometry of the group Γ3/8\Gamma_{3/8} had been studied in [4]. It is the holonomy representation of a uniformizable spherical CR structure on the figure-eight knot complement.

In order to study the manifold at infinity, ie the quotient of the domain of discontinuity under the action of group. The basis idea is to consider the intersection with 𝐇2\partial\mathbf{H}^{2}_{\mathbb{C}} of a fundamental domain for the action on 𝐇2\mathbf{H}^{2}_{\mathbb{C}}.

The combinatorial structure of Dt=Dt𝐇2\partial_{\infty}D_{t}=D_{t}\cap\partial\mathbf{H}^{2}_{\mathbb{C}} is simple due to the combinatorial structure of DtD_{t}. Let 𝒮i\mathcal{S}_{i} be the spinal sphere corresponding to the bisector i\mathcal{B}_{i}. We define

𝒜i=𝒮iDt.\mathcal{A}_{i}=\mathcal{S}_{i}\cap\partial_{\infty}D_{t}.

From Proposition 4.1, it is easy to see that 𝒜i\mathcal{A}_{i} is an annulus and Dt\partial_{\infty}D_{t} is bounded by eight (pairwise isometric) annuluses.

5. CR-spherical uniformizations for the \mathbb{R}-Fuchsian representation

In this section, we just focus on the \mathbb{R}-Fuchsian representation. Let t0=21t_{0}=\sqrt{2}-1. Then Γt0PO(2,1)PU(2,1)\Gamma_{t_{0}}\subset\mbox{\rm PO(2,1)}\subset\mbox{\rm PU(2,1)}. Let u0=324u_{0}=\sqrt{3\sqrt{2}-4} and v0=221v_{0}=\sqrt{2\sqrt{2}-1}. The generators g1,g2g_{1},g_{2} and g3g_{3} are given by the matrices

g1\displaystyle g_{1} =\displaystyle= [3+42+62u0+8u0422u0+2u0+2+241422u02u02241222u0+2u02241422u0+2u0+2+243+42+62u0+8u04],\displaystyle\left[\begin{array}[]{ccc}\frac{3+4\sqrt{2}+6\sqrt{2}u_{0}+8u_{0}}{4}&\frac{2\sqrt{2}u_{0}+2u_{0}+2+\sqrt{2}}{4}&-\frac{1}{4}\\ \frac{-2\sqrt{2}u_{0}-2u_{0}-2-\sqrt{2}}{4}&\frac{1}{2}&\frac{2\sqrt{2}u_{0}+2u_{0}-2-\sqrt{2}}{4}\\ -\frac{1}{4}&\frac{2\sqrt{2}u_{0}+2u_{0}+2+\sqrt{2}}{4}&\frac{3+4\sqrt{2}+6\sqrt{2}u_{0}+8u_{0}}{4}\\ \end{array}\right],
g2\displaystyle g_{2} =\displaystyle= [1/22/21/22/202/21/22/21/2],\displaystyle\left[\begin{array}[]{ccc}1/2&\sqrt{2}/2&-1/2\\ -\sqrt{2}/2&0&-\sqrt{2}/2\\ -1/2&\sqrt{2}/2&1/2\\ \end{array}\right],
g3\displaystyle g_{3} =\displaystyle= [3+2242+2+6u0+22u041224u022u042+2+6u0+22u041+2222+2+6u0+22u041224u022u042+2+6u0+22u043+224].\displaystyle\left[\begin{array}[]{ccc}\frac{3+2\sqrt{2}}{4}&\frac{\sqrt{2}+2+6u_{0}+2\sqrt{2}u_{0}}{4}&\frac{-1-2\sqrt{2}-4u_{0}-2\sqrt{2}u_{0}}{4}\\ \frac{-\sqrt{2}+2+6u_{0}+2\sqrt{2}u_{0}}{4}&\frac{1+2\sqrt{2}}{2}&\frac{\sqrt{2}+2+6u_{0}+2\sqrt{2}u_{0}}{4}\\ \frac{-1-2\sqrt{2}-4u_{0}-2\sqrt{2}u_{0}}{4}&\frac{\sqrt{2}+2+6u_{0}+2\sqrt{2}u_{0}}{4}&\frac{3+2\sqrt{2}}{4}\\ \end{array}\right].

Now g1g_{1} is a loxodromic element in PU(2,1). Let p1p_{1} and p2p_{2} be the attractive and repulsive fixed points of g1g_{1}. We denote by α1\alpha_{1} the arc of \mathbb{C}-circle p1p2{p_{1}}\curvearrowright{p_{2}}. Then α1\alpha_{1} is the axis at infinity of g1g_{1}. Let Λt0\Lambda_{t_{0}} be the limit set of Γt0\Gamma_{t_{0}}. Then it is a round circle. The crown associated to g1g_{1} is the subset of 𝕊3\mathbb{S}^{3} defined as

CrownΓt0,g1=Λt0(gΓt0gα1).{\rm Crown}_{\Gamma_{t_{0}},g_{1}}=\Lambda_{t_{0}}\cup\Bigl{(}\bigcup_{g\in\Gamma_{t_{0}}}g\cdot\alpha_{1}\Bigr{)}.

We denote ΩΓt0,g1ΩΓt0\Omega_{\Gamma_{t_{0}},g_{1}}\subset\Omega_{\Gamma_{t_{0}}} the complement of CrownΓt0,g1{\rm Crown}_{\Gamma_{t_{0}},g_{1}} in 𝕊3\mathbb{S}^{3}. Dehornoy showed

Proposition 5.1 (Dehornoy [2]).

ΩΓt0,g1/Γt0\Omega_{\Gamma_{t_{0}},g_{1}}/\Gamma_{t_{0}} is homeomorphic to the figure-eight knot complement.

We will reinterpret Proposition 5.1 by using the fundamental domain.

Note that g21g3g_{2}^{-1}g_{3} is also a loxodromic element in PU(2,1). Let q1q_{1} and q2q_{2} be the attractive and repulsive fixed points of g21g3g_{2}^{-1}g_{3}. We denote by β1\beta_{1} the arc of \mathbb{C}-circle q1q2{q_{1}}\curvearrowright{q_{2}}. Then β1\beta_{1} is the axis at infinity of g21g3g_{2}^{-1}g_{3}. Define

αi=g2i(α1),βi=g2i(β1),\alpha_{i}=g_{2}^{i}(\alpha_{1}),\quad\beta_{i}=g_{2}^{i}(\beta_{1}),

for i=2,3,4.i=2,3,4.

𝒮5\mathcal{S}_{5}𝒮2\mathcal{S}_{2}𝒮6\mathcal{S}_{6}𝒮3\mathcal{S}_{3}𝒮7\mathcal{S}_{7}𝒮4\mathcal{S}_{4}𝒮8\mathcal{S}_{8}𝒮1\mathcal{S}_{1}α1\alpha_{1}α2\alpha_{2}α3\alpha_{3}α4\alpha_{4}β1\beta_{1}β2\beta_{2}β3\beta_{3}β4\beta_{4}
Figure 3. A schematic view of the configuration of the eight spinal spheres and the eight \mathbb{C}-arcs. Each round sphere is a spinal sphere, and Dt\partial_{\infty}D_{t} is the region outside all the spinal spheres. αi\alpha_{i}(βi\beta_{i}) is the thick red (blue) arc with end points in some of the spinal spheres.
Refer to caption
Figure 4. The \mathbb{C}-arc α1^\hat{\alpha_{1}} is the part of \mathbb{C}-arc α1\alpha_{1}, which is the axis at infinity of g1g_{1}. It lies in Dt\partial_{\infty}D_{t} with end points on the spinal spheres 𝒮5\mathcal{S}_{5}(the green one) and 𝒮2\mathcal{S}_{2}(the blue one).
Refer to caption
Figure 5. A realistic view of the configuration of the eight spinal spheres and the eight \mathbb{C}-arcs.

See Figure 3 for a schematic view of the configuration of the eight spinal spheres and the eight \mathbb{C}-arcs. Figure 3 should also be compared with Figure 5.

5.1. The configuration of the eight \mathbb{C}-arcs

We will study the intersections of arcs αi,βi\alpha_{i},\beta_{i} with the spinal spheres 𝒮i\mathcal{S}_{i}. Let Ut0=Dt0U_{t_{0}}=\partial_{\infty}D_{t_{0}}. Then Ut0U_{t_{0}} is a solid torus. We denote by αi^,βi^\hat{\alpha_{i}},\hat{\beta_{i}} the segments of the arcs αi,βi\alpha_{i},\beta_{i}, which are contained in the interior of solid torus Ut0U_{t_{0}}. We define α1^\hat{\alpha_{1}}^{-}(α1^+\hat{\alpha_{1}}^{+}) to be the end point of α1^\hat{\alpha_{1}} close to the attractive fixed point p1p_{1}(repulsive fixed point p2p_{2}) of α1\alpha_{1}, see Figure 4. The end point of β1^\hat{\beta_{1}} can be defined similarly. Note that

αi^=g2i(α1^),βi^=g2i(β1^),\hat{\alpha_{i}}=g_{2}^{i}(\hat{\alpha_{1}}),\quad\hat{\beta_{i}}=g_{2}^{i}(\hat{\beta_{1}}),

for i=2,3,4.i=2,3,4.

Proposition 5.2.

The end point α4^\hat{\alpha_{4}}^{-} of α4^\hat{\alpha_{4}} is on the spinal sphere 𝒜1\mathcal{A}_{1}, and the other end point α4^+\hat{\alpha_{4}}^{+} of α4^\hat{\alpha_{4}} is on the spinal sphere 𝒜8\mathcal{A}_{8}.

Proof.

Firstly, we claim that the \mathbb{C}-arc α4\alpha_{4} is contained in the exterior of the spinal spheres 𝒮i(1i8)\mathcal{S}_{i}(1\leqslant i\leqslant 8) except for 𝒮1\mathcal{S}_{1}, 𝒮4\mathcal{S}_{4}, 𝒮5\mathcal{S}_{5} and 𝒮8\mathcal{S}_{8}.

Note that α4\alpha_{4} is the \mathbb{C}-arc, which is the axis at infinity of g21g1g2g_{2}^{-1}g_{1}g_{2}. Let vv be the polar vector of the \mathbb{C}-circle containing the arc α4\alpha_{4}. Then

v=g21(n3I2(n1))=[9+42+(102+12)u0701].v=g_{2}^{-1}\left(n_{3}\boxtimes I_{2}(n_{1})\right)=\left[\begin{array}[]{c}\frac{9+4\sqrt{2}+(10\sqrt{2}+12)u_{0}}{7}\\ 0\\ 1\\ \end{array}\right].

The lift of the \mathbb{C}-arc α4\alpha_{4} can be written as

(5.4) V1=[942(102+12)u07(1402+168)u0+562+1267(x+yi)1],V_{1}=\left[\begin{array}[]{c}\frac{-9-4\sqrt{2}-(10\sqrt{2}+12)u_{0}}{7}\\ \frac{\sqrt{(140\sqrt{2}+168)u_{0}+56\sqrt{2}+126}}{7}(x+yi)\\ 1\\ \end{array}\right],

where y=1x2,1x1y=\sqrt{1-x^{2}},-1\leqslant x\leqslant 1.

We claim that the intersection of α4\alpha_{4} with 𝒮2\mathcal{S}_{2} is empty. Substituting (5.4) to the equation of the bisector 2\mathcal{B}_{2}

|V1,q0|=|V1,g2g1(q0)|.|\langle V_{1},q_{0}\rangle|=|\langle V_{1},g_{2}g_{1}(q_{0})\rangle|.

We get the equation

(2607u0x+14414u0x+767x+542)9+42+12u0+102u0\displaystyle(260\sqrt{7}u_{0}x+144\sqrt{14}u_{0}x+76\sqrt{7}x+54\sqrt{2})\sqrt{9+4\sqrt{2}+12u_{0}+10\sqrt{2}u_{0}}
115422888u020242u01592=0.\displaystyle-1154\sqrt{2}-2888u_{0}-2024\sqrt{2}u_{0}-1592=0.

Solve the equation, we have

x=48162675571.0676.x=\sqrt{\frac{4816\sqrt{2}-6755}{7}}\approx-1.0676.

Therefore, α4\alpha_{4} does not intersect with 𝒮2\mathcal{S}_{2}, which is the boundary at infinity of 2\mathcal{B}_{2}. With the same argument, one can also prove that α4\alpha_{4} does not intersect with 𝒮3\mathcal{S}_{3}, 𝒮6\mathcal{S}_{6} and 𝒮7\mathcal{S}_{7}.

Next, we study the intersection of α4\alpha_{4} with 𝒮1\mathcal{S}_{1}, 𝒮4\mathcal{S}_{4}, 𝒮5\mathcal{S}_{5} and 𝒮8\mathcal{S}_{8}. The intersection point divide the arc α4\alpha_{4} into several segments. We will determine which segment is α4^\hat{\alpha_{4}}.

Substituting (5.4) to the equation of the bisector 1\mathcal{B}_{1}

|V1,q0|2=|V1,g1(q0)|2.|\langle V_{1},q_{0}\rangle|^{2}=|\langle V_{1},g_{1}(q_{0})\rangle|^{2}.

We get

(447u0x+3214u0x+127x+1014x)9+42+12u0+102u0\displaystyle(44\sqrt{7}u_{0}x+32\sqrt{14}u_{0}x+12\sqrt{7}x+10\sqrt{14}x)\sqrt{9+4\sqrt{2}+12u_{0}+10\sqrt{2}u_{0}}
1262280u01682u0112=0.\displaystyle-126\sqrt{2}-280u_{0}-168\sqrt{2}u_{0}-112=0.

The intersection point corresponds to the solution

x=8211,y=222.x=\sqrt{8\sqrt{2}-11},\quad y=2\sqrt{2}-2.

Substituting (5.4) to the equation of the bisector 4\mathcal{B}_{4}

|V1,q0|2=|V1,g23g1(q0)|2.|\langle V_{1},q_{0}\rangle|^{2}=|\langle V_{1},g_{2}^{3}g_{1}(q_{0})\rangle|^{2}.

We get

(1727u0x+12014u0x527x3414x)9+42+12u0+102u0\displaystyle(172\sqrt{7}u_{0}x+120\sqrt{14}u_{0}x-52\sqrt{7}x-34\sqrt{14}x)\sqrt{9+4\sqrt{2}+12u_{0}+10\sqrt{2}u_{0}}
1512u010642u0840=0.\displaystyle-1512u_{0}-1064\sqrt{2}u_{0}-840=0.

The intersection point corresponds to the solution

x=162+137,y=4227.x=-\frac{\sqrt{16\sqrt{2}+13}}{7},\quad y=\frac{4\sqrt{2}-2}{7}.

Substituting (5.4) to the equation of the bisector 5\mathcal{B}_{5}

|V1,q0|2=|V1,g31(q0)|2.|\langle V_{1},q_{0}\rangle|^{2}=|\langle V_{1},g_{3}^{-1}(q_{0})\rangle|^{2}.

We get

(1727u0x+12014u0x+527x+3414x)9+42+12u0+102u0\displaystyle(172\sqrt{7}u_{0}x+120\sqrt{14}u_{0}x+52\sqrt{7}x+34\sqrt{14}x)\sqrt{9+4\sqrt{2}+12u_{0}+10\sqrt{2}u_{0}}
60221512u010642u0840=0.\displaystyle-602\sqrt{2}-1512u_{0}-1064\sqrt{2}u_{0}-840=0.

The intersection point corresponds to the solution

x=162+137,y=4227.x=\frac{\sqrt{16\sqrt{2}+13}}{7},\quad y=\frac{4\sqrt{2}-2}{7}.

Substituting (5.4) to the equation of the bisector 8\mathcal{B}_{8}

|V1,q0|2=|V1,g21g31(q0)|2.|\langle V_{1},q_{0}\rangle|^{2}=|\langle V_{1},g_{2}^{-1}g_{3}^{-1}(q_{0})\rangle|^{2}.

We get

(447u0x+3214u0x127x1014x)9+42+12u0+102u0\displaystyle(44\sqrt{7}u_{0}x+32\sqrt{14}u_{0}x-12\sqrt{7}x-10\sqrt{14}x)\sqrt{9+4\sqrt{2}+12u_{0}+10\sqrt{2}u_{0}}
280u01682u0112=0.\displaystyle-280u_{0}-168\sqrt{2}u_{0}-112=0.

The intersection point corresponds to the solution

x=8211,y=222.x=\sqrt{8\sqrt{2}-11},\quad y=2\sqrt{2}-2.

By simple calculation, it find that one end point of α4\alpha_{4} lies inside 𝒮1\mathcal{S}_{1} and 𝒮5\mathcal{S}_{5} and the other end point of α4\alpha_{4} lies inside 𝒮4\mathcal{S}_{4} and 𝒮8\mathcal{S}_{8}. It is also easy to check that the intersection point of α4\alpha_{4} with 𝒮5\mathcal{S}_{5} lies in 𝒮1\mathcal{S}_{1} and the intersection point of α4\alpha_{4} with 𝒮1\mathcal{S}_{1} does not lie in any spinal sphere. So this intersection point is on 𝒜1\mathcal{A}_{1}.

we also see that the intersection point of α4\alpha_{4} with 𝒮4\mathcal{S}_{4} lies in 𝒮8\mathcal{S}_{8} and the intersection point of α4\alpha_{4} with 𝒮8\mathcal{S}_{8} does not lie in any spinal sphere. So the intersection point is on 𝒜8\mathcal{A}_{8}.

From the configuration of spinal spheres, we can see that the segment on α4\alpha_{4} between the intersect points of α4\alpha_{4} with 𝒜1\mathcal{A}_{1} and 𝒜8\mathcal{A}_{8} is the \mathbb{C}-arc α4^\hat{\alpha_{4}} that we are looking for, see Figure 7. ∎

Similarly, we have

Proposition 5.3.

The end point β1^\hat{\beta_{1}}^{-} of β1^\hat{\beta_{1}} is on the spinal sphere 𝒜2\mathcal{A}_{2}, and the other end point β1^+\hat{\beta_{1}}^{+} of β1^\hat{\beta_{1}} is on the spinal sphere 𝒜6\mathcal{A}_{6}.

From the calculations in Proposition 5.2 and Proposition 5.3, we have

α4^=[94212u0102u0722+2u0+4v062v04u0v082u0v07i1],\hat{\alpha_{4}}^{-}=\left[\begin{array}[]{c}\frac{-9-4\sqrt{2}-12u_{0}-10\sqrt{2}u_{0}}{7}\\ 2-\sqrt{2}+2u_{0}+\frac{4v_{0}-6\sqrt{2}v_{0}-4u_{0}v_{0}-8\sqrt{2}u_{0}v_{0}}{7}i\\ 1\\ \end{array}\right],
α4^+=[94212u0102u072+22u0+4v062v04u0v082u0v07i1],\hat{\alpha_{4}}^{+}=\left[\begin{array}[]{c}\frac{-9-4\sqrt{2}-12u_{0}-10\sqrt{2}u_{0}}{7}\\ -2+\sqrt{2}-2u_{0}+\frac{4v_{0}-6\sqrt{2}v_{0}-4u_{0}v_{0}-8\sqrt{2}u_{0}v_{0}}{7}i\\ 1\\ \end{array}\right],
β1^=[182u0+26u09213(32530+246312)u012172157312i((9826+66642)u0+28842+5208)(12u0)(2+32)66891],\hat{\beta_{1}}^{-}=\left[\begin{array}[]{c}18\sqrt{2}u_{0}+26u_{0}-9\sqrt{2}-13\\ \frac{(32530+24631\sqrt{2})u_{0}-12172-15731\sqrt{2}i\left((9826+6664\sqrt{2})u_{0}+2884\sqrt{2}+5208\right)\sqrt{(1-2u_{0})(2+3\sqrt{2})}}{6689}\\ 1\\ \end{array}\right],
β1^=[182u0+26u09213(21246+161052)u09540111812i((3006+23722)u0+10762+1128)(12u0)(2+32)46571].\hat{\beta_{1}}^{-}=\left[\begin{array}[]{c}18\sqrt{2}u_{0}+26u_{0}-9\sqrt{2}-13\\ \frac{(21246+16105\sqrt{2})u_{0}-9540-11181\sqrt{2}i\left((3006+2372\sqrt{2})u_{0}+1076\sqrt{2}+1128\right)\sqrt{(1-2u_{0})(2+3\sqrt{2})}}{4657}\\ 1\\ \end{array}\right].

Under the action of g2g_{2}, we can obtain the end points of all \mathbb{C}-arcs αi^\hat{\alpha_{i}}^{-} and βi^\hat{\beta_{i}}^{-}. We summarize these in Table 1.

Table 1. The positions and the coordinates of the end points of the eight arcs.
\mathbb{C}-arc End points
α1^\hat{\alpha_{1}} α1^𝒜2\hat{\alpha_{1}}^{-}\in\mathcal{A}_{2}, α1^+𝒜5\hat{\alpha_{1}}^{+}\in\mathcal{A}_{5}
α2^\hat{\alpha_{2}} α2^𝒜3\hat{\alpha_{2}}^{-}\in\mathcal{A}_{3}, α2^+𝒜6\hat{\alpha_{2}}^{+}\in\mathcal{A}_{6}
α3^\hat{\alpha_{3}} α3^𝒜4\hat{\alpha_{3}}^{-}\in\mathcal{A}_{4}, α3^+𝒜7\hat{\alpha_{3}}^{+}\in\mathcal{A}_{7}
α4^\hat{\alpha_{4}} α4^𝒜1\hat{\alpha_{4}}^{-}\in\mathcal{A}_{1}, α4^+𝒜8\hat{\alpha_{4}}^{+}\in\mathcal{A}_{8}
β1^\hat{\beta_{1}} β1^𝒜2\hat{\beta_{1}}^{-}\in\mathcal{A}_{2}, β1^+𝒜6\hat{\beta_{1}}^{+}\in\mathcal{A}_{6}
β2^\hat{\beta_{2}} β2^𝒜3\hat{\beta_{2}}^{-}\in\mathcal{A}_{3}, β2^+𝒜7\hat{\beta_{2}}^{+}\in\mathcal{A}_{7}
β3^\hat{\beta_{3}} β3^𝒜4\hat{\beta_{3}}^{-}\in\mathcal{A}_{4}, β3^+𝒜8\hat{\beta_{3}}^{+}\in\mathcal{A}_{8}
β4^\hat{\beta_{4}} β4^𝒜1\hat{\beta_{4}}^{-}\in\mathcal{A}_{1}, β4^+𝒜5\hat{\beta_{4}}^{+}\in\mathcal{A}_{5}

5.2. The configuration of the eight cutting disks

Recall the affine disk bounded by a \mathbb{C}-circle in Definition 2.4.

Definition 5.4.

From Propositions 5.2 and 5.3, for each \mathbb{C}-arc αi^\hat{\alpha_{i}} (or βi^\hat{\beta_{i}}), there is a affine disk bounded by the \mathbb{C}-circle containing this \mathbb{C}-arc. We define the cutting disk to be the part of the affine disk bounded by the \mathbb{C}-arc and two spinal spheres containing the end points of the \mathbb{C}-arc.

See Figure 6 for a realistic view of the cutting disk corresponding to β1^\hat{\beta_{1}}.

Refer to caption
Figure 6. The embedded cutting disk (the black one) corresponding to β1^\hat{\beta_{1}} is the region of the affine disk that lies outside the spinal spheres 𝒮2\mathcal{S}_{2} (the yellow one), 𝒮6\mathcal{S}_{6} (the cyan one) and is constrained by the \mathbb{C}-arc β1^\hat{\beta_{1}}.

From the definition, we know that each cutting disk properly embeds in the fundamental domain at infinity. There are eight embedded cutting disks corresponding to the eight \mathbb{C}-arcs. We have

Proposition 5.5.

The eight cutting disks are disjoint.

The proof of this proposition for the deformation from the \mathbb{R}-Fuchsian case to the degenerate case will be given in Section 6 (see Proposition 6.2). We include this proposition in the \mathbb{R}-Fuchsian case here just for completeness.

From some routine calculations and the relation

g2g1(g21g3)(g2g1)1=g1,g_{2}g_{1}(g_{2}^{-1}g_{3})(g_{2}g_{1})^{-1}=g_{1},

we can check that g2g1(β1^)α1^g_{2}g_{1}(\hat{\beta_{1}})\cup\hat{\alpha_{1}} is the fundamental domain of the g1g_{1} acting on its axis at infinity α1{\alpha_{1}}. This allow us to get the following proposition.

Proposition 5.6.
CrownΓt0,g1=Λt0(gΓt0gα1)=Λt0(gΓt0g(α1^β1^)).{\rm Crown}_{\Gamma_{t_{0}},g_{1}}=\Lambda_{t_{0}}\cup\Bigl{(}\bigcup_{g\in\Gamma_{t_{0}}}g\cdot\alpha_{1}\Bigr{)}=\Lambda_{t_{0}}\cup\Bigl{(}\bigcup_{g\in\Gamma_{t_{0}}}g\cdot(\hat{\alpha_{1}}\cup\hat{\beta_{1}})\Bigr{)}.

The principal significance of Proposition 5.6 is that it allows us to get the figure-eight knot complement from Dehornoy’s result by applying techniques using the fundamental domain. Since Dt0(i=14(αi^βi^))\partial_{\infty}D_{t_{0}}-(\cup^{4}_{i=1}(\hat{\alpha_{i}}\cup\hat{\beta_{i}})) is a subset of ΩΓt0,g1\Omega_{\Gamma_{t_{0}},g_{1}}, the restriction of the quotient map on ΩΓt0,g1ΩΓt0,g1/Γt0\Omega_{\Gamma_{t_{0}},g_{1}}\rightarrow\Omega_{\Gamma_{t_{0}},g_{1}}/\Gamma_{t_{0}} to Dt0(i=14(αi^βi^))\partial_{\infty}D_{t_{0}}-(\cup^{4}_{i=1}(\hat{\alpha_{i}}\cup\hat{\beta_{i}})) also gives a quotient space, which can be viewed as the quotient of Dt0(i=14(αi^βi^))\partial_{\infty}D_{t_{0}}-(\cup^{4}_{i=1}(\hat{\alpha_{i}}\cup\hat{\beta_{i}})) by side-pairings on

j=18𝒜i(i=14(αi^βi^)).\cup^{8}_{j=1}\mathcal{A}_{i}\cup(\cup^{4}_{i=1}(\hat{\alpha_{i}}\cup\hat{\beta_{i}})).
Proposition 5.7.

The quotient space of

Dt0(i=14(αi^βi^))\partial_{\infty}D_{t_{0}}-(\cup^{4}_{i=1}(\hat{\alpha_{i}}\cup\hat{\beta_{i}}))

by side-pairings and the quotient space ΩΓt0,g1/Γt0\Omega_{\Gamma_{t_{0}},g_{1}}/\Gamma_{t_{0}} are homeomorphic. So it is the figure-eight knot complement.

Proof.

We denote by MM the quotient space of Dt0i=14(αi^βi^)\partial_{\infty}D_{t_{0}}-\cup^{4}_{i=1}(\hat{\alpha_{i}}\cup\hat{\beta_{i}}) by side-pairings. Then it is trivial that MM is a subspace of ΩΓt0,g1/Γt0\Omega_{\Gamma_{t_{0}},g_{1}}/\Gamma_{t_{0}}. Conversely, from the side-pairings on j=18𝒜i(i=14(αi^βi^))\cup^{8}_{j=1}\mathcal{A}_{i}\cup(\cup^{4}_{i=1}(\hat{\alpha_{i}}\cup\hat{\beta_{i}})), we have MM contains ΩΓt0,g1/Γt0\Omega_{\Gamma_{t_{0}},g_{1}}/\Gamma_{t_{0}}. So they are homeomorphic. Then by Proposition 5.1, MM is the figure-eight knot complement.

6. Geometric stability in the deformation

In this section we focus on the group Γt\Gamma_{t} for t(3/8,21]t\in(3/8,\sqrt{2}-1]. The combinatorics of the fundamental domain of Γt\Gamma_{t} does not change for t(3/8,21]t\in(3/8,\sqrt{2}-1] due to the work of [13]. Therefore, we only need to show that the configurations of the eight arcs and the eight cutting disks are the same as the \mathbb{R}-Fuchsian representation.

Proposition 6.1.

The spinal spheres where the end points of eight arcs are located do not change during the deformation.

Proof.

We will show that one end point of \mathbb{C}-arc α4,t^\hat{\alpha_{4,t}} is always located on the spinal sphere 𝒮1,t\mathcal{S}_{1,t} during the deformation. The proofs in other cases are similar, and we omit them.

Our proof is in three steps:

Step 1:

To show that the end point p4,2p_{4,2} of the \mathbb{C}-arc α4,t\alpha_{4,t} lies in the spinal spheres 𝒮1,t\mathcal{S}_{1,t} and 𝒮5,t\mathcal{S}_{5,t} and the end point p4,1p_{4,1} of the \mathbb{C}-arc α4,t\alpha_{4,t} lies in the spinal spheres 𝒮4,t\mathcal{S}_{4,t} and 𝒮8,t\mathcal{S}_{8,t}.

Step 2:

To show that the \mathbb{C}-arc α4,t\alpha_{4,t}(p4,1p4,2{p_{4,1}}\curvearrowright{p_{4,2}}) intersects with the spinal spheres 𝒮1,t\mathcal{S}_{1,t}, 𝒮4,t\mathcal{S}_{4,t}, 𝒮5,t\mathcal{S}_{5,t} and 𝒮8,t\mathcal{S}_{8,t} only once and does not intersect other spinal spheres.

Step 3:

Note that the spinal sphere 𝒮1,t\mathcal{S}_{1,t} only intersects with two spinal spheres, 𝒮5,t\mathcal{S}_{5,t} and 𝒮8,t\mathcal{S}_{8,t}. In the beginning of the deformation, one end point of α4,t^\hat{\alpha_{4,t}} is on the spinal sphere 𝒜1,t\mathcal{A}_{1,t}, see Figure 7. If the configuration in Figure 7 turns into the configurations in Figure 8, then the \mathbb{C}-arc α4,t\alpha_{4,t} will pass through 𝒜1,t𝒜5,t\mathcal{A}_{1,t}\cap\mathcal{A}_{5,t} or 𝒜1,t𝒜8,t\mathcal{A}_{1,t}\cap\mathcal{A}_{8,t} at some time during the deformation by a geometric continuity argument. We will show that this is impossible.

We begin with the Step 1. By a simple calculation, we have

(6.1) p4,1=[2(t6t2+12tt2i)26t28t33t12(t+6t2+12tt2i)26t2],p4,2=[2(t6t2+12tt2i)26t28t33t12(t+6t2+12tt2i)26t2].p_{4,1}=\left[\begin{array}[]{c}-\frac{\sqrt{2}(-t-\sqrt{6t-2}+\sqrt{1-2t-t^{2}}i)}{2\sqrt{6t-2}}\\ \sqrt{\frac{8t-3}{3t-1}}\\ -\frac{\sqrt{2}(-t+\sqrt{6t-2}+\sqrt{1-2t-t^{2}}i)}{2\sqrt{6t-2}}\\ \end{array}\right],\quad p_{4,2}=\left[\begin{array}[]{c}-\frac{\sqrt{2}(-t-\sqrt{6t-2}+\sqrt{1-2t-t^{2}}i)}{2\sqrt{6t-2}}\\ -\sqrt{\frac{8t-3}{3t-1}}\\ -\frac{\sqrt{2}(-t+\sqrt{6t-2}+\sqrt{1-2t-t^{2}}i)}{2\sqrt{6t-2}}\\ \end{array}\right].

Substitute (6.1) to the equations of the bisectors of 1,t\mathcal{B}_{1,t} and 5,t\mathcal{B}_{5,t}. Then we get

|p4,2,q0|2|p4,2,g2g1(q0)|2=8t38t34t2,\displaystyle|\langle p_{4,2},q_{0}\rangle|^{2}-|\langle p_{4,2},g_{2}g_{1}(q_{0})\rangle|^{2}=\frac{8t-3-\sqrt{8t-3}}{4t-2},
|p4,2,q0|2|p4,2,g31(q0)|2=(4t1)8t38t+32(2t1)2.\displaystyle|\langle p_{4,2},q_{0}\rangle|^{2}-|\langle p_{4,2},g_{3}^{-1}(q_{0})\rangle|^{2}=\frac{(4t-1)\sqrt{8t-3}-8t+3}{2(2t-1)^{2}}.

It is easy to check that

8t38t34t2>0,(4t1)8t38t+32(2t1)2>0\frac{8t-3-\sqrt{8t-3}}{4t-2}>0,\quad\frac{(4t-1)\sqrt{8t-3}-8t+3}{2(2t-1)^{2}}>0

for t(3/8,21]t\in(3/8,\sqrt{2}-1].

That is, the point p4,2p_{4,2} lies inside the spinal sphere 𝒮1,t\mathcal{S}_{1,t} and 𝒮5,t\mathcal{S}_{5,t}.

Substitute (6.1) to the equations of the bisectors of 4,t\mathcal{B}_{4,t} and 8,t\mathcal{B}_{8,t}. Then we get

|p4,2,q0|2|p4,2,g21g31(q0)|2=8t3+8t34t2,\displaystyle|\langle p_{4,2},q_{0}\rangle|^{2}-|\langle p_{4,2},g_{2}^{-1}g_{3}^{-1}(q_{0})\rangle|^{2}=\frac{8t-3+\sqrt{8t-3}}{4t-2},
|p4,2,q0|2|p4,2,g21g1(q0)|2=(14t)8t38t+32(2t1)2.\displaystyle|\langle p_{4,2},q_{0}\rangle|^{2}-|\langle p_{4,2},g_{2}^{-1}g_{1}(q_{0})\rangle|^{2}=\frac{(1-4t)\sqrt{8t-3}-8t+3}{2(2t-1)^{2}}.

Both equations’ right sides are  negative for t(3/8,21]t\in(3/8,\sqrt{2}-1]. So the point p4,2p_{4,2} lie outside the spinal sphere 𝒮4,t\mathcal{S}_{4,t} and 𝒮8,t\mathcal{S}_{8,t}.

Similarly, we can show that the point p4,1p_{4,1} lies inside the spinal sphere 𝒮4,t\mathcal{S}_{4,t} and 𝒮8,t\mathcal{S}_{8,t} and lies outside the spinal sphere 𝒮1,t\mathcal{S}_{1,t} and 𝒮5,t\mathcal{S}_{5,t}.

Next, we will complete the Step 2.

The polar vector of α4,t\alpha_{4,t} is given by

[6t2+t+i12tt26t2ti12tt201].\left[\begin{array}[]{c}\frac{\sqrt{6t-2}+t+i\sqrt{1-2t-t^{2}}}{\sqrt{6t-2}-t-i\sqrt{1-2t-t^{2}}}\\ 0\\ 1\\ \end{array}\right].

To make the calculation simpler, we apply the following transformation

T1=[4t12t6t2026t2t22t+1i4t12t6t20100014t12t6t2].T_{1}=\left[\begin{array}[]{ccc}\sqrt{4t-1-2t\sqrt{6t-2}}&0&-\frac{2\sqrt{6t-2}\sqrt{-t^{2}-2t+1}i}{\sqrt{4t-1-2t\sqrt{6t-2}}}\\ 0&1&0\\ 0&0&\frac{1}{\sqrt{4t-1-2t\sqrt{6t-2}}}\\ \end{array}\right].

Then the polar vector of α4,t\alpha_{4,t} is given by

[8t301].\left[\begin{array}[]{c}8t-3\\ 0\\ 1\\ \end{array}\right].

The lift of the \mathbb{C}-arc T1(α4,t)T_{1}(\alpha_{4,t}) can be written as

(6.2) Vt=[38t28t3(x+yi)1],V_{t}=\left[\begin{array}[]{c}3-8t\\ \sqrt{2}\sqrt{8t-3}(x+yi)\\ 1\\ \end{array}\right],

where y=1x2,t12txt12ty=-\sqrt{1-x^{2}},-\frac{t}{\sqrt{1-2t}}\leqslant x\leqslant\frac{t}{\sqrt{1-2t}}.

We claim that the intersection of T1(α4,t)T_{1}(\alpha_{4,t}) with T1(𝒮3,t)T_{1}(\mathcal{S}_{3,t}) is empty. Substituting (6.2) to the equation of the bisector T1(3,t)T_{1}(\mathcal{B}_{3,t})

|Vt,T1(q0)|=|Vt,T1g22g1(q0)|.|\langle V_{t},T_{1}(q_{0})\rangle|=|\langle V_{t},T_{1}g_{2}^{2}g_{1}(q_{0})\rangle|.

We get the equation

k3,1x+k3,2y+k3,0=0,k_{3,1}x+k_{3,2}y+k_{3,0}=0,

where

k3,1=24c(t)(t1/3)(26t(a2+a/21/4)3/2t(t1/2))(2t1)2,\displaystyle k_{3,1}=-\frac{24c(t)(t-1/3)\left(\sqrt{2-6t}(a^{2}+a/2-1/4)-3/2t(t-1/2)\right)}{(2t-1)^{2}},
k3,2=24c(t)(t1/3)((t/21/4)t2+12t1+2t(3t1)(a2+2t1))(2t1)2,\displaystyle k_{3,2}=\frac{24c(t)(t-1/3)\left((t/2-1/4)\sqrt{t^{2}+12t-1}+\sqrt{2}t\sqrt{(3t-1)(a^{2}+2t-1)}\right)}{(2t-1)^{2}},
k3,0=24(t1/3)(3t27/2t+3/4)(2t1)2.\displaystyle k_{3,0}=\frac{24(t-1/3)(3t^{2}-7/2t+3/4)}{(2t-1)^{2}}.

By using some computer algebra software, we find that the minimum of the expression

k3,02k3,12+k3,22\frac{k_{3,0}^{2}}{k_{3,1}^{2}+k_{3,2}^{2}}

is given approximately by 6.5907 for t[3/8,21]t\in[3/8,\sqrt{2}-1]. So the family of lines does not intersect the circle x2+y2=1x^{2}+y^{2}=1. Thus α4,t\alpha_{4,t} does not intersect with the spinal sphere 𝒮3,t\mathcal{S}_{3,t}. With the same argument, one can also prove that α4,t\alpha_{4,t} does not intersect with 𝒮2,t\mathcal{S}_{2,t}, 𝒮6,t\mathcal{S}_{6,t}, 𝒮7,t\mathcal{S}_{7,t}.

Substituting (6.2) to the equation of the bisector T1(1,t)T_{1}(\mathcal{B}_{1,t})

|Vt,T1q0|=|Vt,T1g1(q0)|.|\langle V_{t},T_{1}q_{0}\rangle|=|\langle V_{t},T_{1}g_{1}(q_{0})\rangle|.

We get the equation

(6.3) k1,1x+k1,2y+k1,0=0,k_{1,1}x+k_{1,2}y+k_{1,0}=0,

where

k1,1=(6t2)c(t)(t26t)2t1,\displaystyle k_{1,1}=\frac{(6t-2)c(t)\left(t-\sqrt{2-6t}\right)}{2t-1},
k1,2=(6t2)c(t)t2+12t12t1,\displaystyle k_{1,2}=\frac{(6t-2)c(t)\sqrt{t^{2}+12t-1}}{2t-1},
k1,0=(6t2)(8t3)2t1.\displaystyle k_{1,0}=\frac{(6t-2)(8t-3)}{2t-1}.

The intersection point corresponds to the solution

y\displaystyle y =k1,0k1,2k1,14+k1,12k1,22k1,02k1,12k1,12+k1,22,\displaystyle=\frac{-k_{1,0}k_{1,2}-\sqrt{k_{1,1}^{4}+k_{1,1}^{2}k_{1,2}^{2}-k_{1,0}^{2}k_{1,1}^{2}}}{k_{1,1}^{2}+k_{1,2}^{2}},
x\displaystyle x =k1,2y+k1,0k1,1.\displaystyle=-\frac{k_{1,2}y+k_{1,0}}{k_{1,1}}.

Similarly, it can be showed that the \mathbb{C}-arc α4,t\alpha_{4,t} has only one intersection with the spinal spheres 𝒮4,t\mathcal{S}_{4,t}, 𝒮5,t\mathcal{S}_{5,t} and 𝒮8,t\mathcal{S}_{8,t}.

In the last step, we show that α4,t\alpha_{4,t} can not pass through the intersection of the spinal spheres 𝒮1,t\mathcal{S}_{1,t} and 𝒮5,t\mathcal{S}_{5,t}.

Substituting (6.2) to the equation of the bisector T1(5,t)T_{1}(\mathcal{B}_{5,t})

|Vt,T1q0|=|Vt,T1g1(q0)|.|\langle V_{t},T_{1}q_{0}\rangle|=|\langle V_{t},T_{1}g_{1}(q_{0})\rangle|.

We get the equation

(6.4) k5,1x+k5,2y+k5,0=0,k_{5,1}x+k_{5,2}y+k_{5,0}=0,

where

k5,1=6(t1/3)(t1/4)c(t)(t26t)(t1/2)2,\displaystyle k_{5,1}=-\frac{6(t-1/3)(t-1/4)c(t)\left(t-\sqrt{2-6t}\right)}{(t-1/2)^{2}},
k5,2=6(t1/3)(t1/4)c(t)t2+12t1(t1/2)2,\displaystyle k_{5,2}=-\frac{6(t-1/3)(t-1/4)c(t)\sqrt{t^{2}+12t-1}}{(t-1/2)^{2}},
k5,0=6(t1/3)(2t3/4)(t1/2)2.\displaystyle k_{5,0}=\frac{6(t-1/3)(2t-3/4)}{(t-1/2)^{2}}.

Then we have

k1,1=k5,112t4t1,k1,2=k5,212t4t1,k1,0=k5,0(12t).k_{1,1}=k_{5,1}\frac{1-2t}{4t-1},\quad k_{1,2}=k_{5,2}\frac{1-2t}{4t-1},\quad k_{1,0}=k_{5,0}(1-2t).

So the equations (6.3) and (6.4) has no common solution. With the same argument, we can prove that α4,t\alpha_{4,t} can not pass through the intersection of the spinal spheres 𝒮4,t\mathcal{S}_{4,t} and 𝒮8,t\mathcal{S}_{8,t}.

Refer to captionα4,t\alpha_{4,t}𝒮1,t\mathcal{S}_{1,t}𝒮5,t\mathcal{S}_{5,t}𝒮8,t\mathcal{S}_{8,t}𝒮4,t\mathcal{S}_{4,t}
Figure 7. The configuration of arc α4,t\alpha_{4,t} and the spinal spheres 𝒮1,t\mathcal{S}_{1,t}, 𝒮4,t\mathcal{S}_{4,t}, 𝒮5,t\mathcal{S}_{5,t}, 𝒮8,t\mathcal{S}_{8,t}.
Refer to captionα4,t\alpha_{4,t}𝒮5,t\mathcal{S}_{5,t}𝒮1,t\mathcal{S}_{1,t}𝒮8,t\mathcal{S}_{8,t}𝒮4,t\mathcal{S}_{4,t}α4,t\alpha_{4,t}𝒮5,t\mathcal{S}_{5,t}𝒮8,t\mathcal{S}_{8,t}𝒮1,t\mathcal{S}_{1,t}𝒮4,t\mathcal{S}_{4,t}
Figure 8. The impossible configurations of arc α4,t\alpha_{4,t} and the spinal spheres 𝒮1,t\mathcal{S}_{1,t}, 𝒮4,t\mathcal{S}_{4,t}, 𝒮5,t\mathcal{S}_{5,t}, 𝒮8,t\mathcal{S}_{8,t} during the deformation. Compare this with Figure 7.
Proposition 6.2.

The eight cutting disks are disjointed during the deformation.

Proof.

First, we note that each pair of the eight \mathbb{C}-circles containing the \mathbb{C}-arcs is not linked for t[3/8,2/5]t\in[3/8,2/5]. Therefore, both these eight discs and their corresponding cutting discs do not intersect. These observations suggest dividing the analysis into two cases.

Case 1: t[3/8,2/5)t\in[3/8,2/5). Let v1,tv_{1,t}, v2,tv_{2,t} be the polar vectors of the \mathbb{C}-circles containing the \mathbb{C}-arc α1,t\alpha_{1,t} and \mathbb{C}-arc β1,t\beta_{1,t}. Then

v1,t=[12a(t)(t+b(t)i)22t11]andv2,t=[ta(t)1+b(t)i412t2a(t)412tt+a(t)1+b(t)i412t].v_{1,t}=\left[\begin{array}[]{c}-1\\ -\frac{\sqrt{2}a(t)(-t+b(t)i)}{2\sqrt{2t-1}}\\ 1\\ \end{array}\right]\quad{\rm and}\quad v_{2,t}=\left[\begin{array}[]{c}-\frac{t-a(t)-1+b(t)i}{4\sqrt{1-2t}}\\ \frac{\sqrt{2}a(t)}{4\sqrt{1-2t}}\\ \frac{t+a(t)-1+b(t)i}{4\sqrt{1-2t}}\\ \end{array}\right].

By the non-linking condition (2.1), we have

L(v1,t,v2,t)=2(13t)2t1.L(v_{1,t},v_{2,t})=\frac{2(1-3t)}{2t-1}.

It is easy to see that the \mathbb{C}-circles containing the \mathbb{C}-arc α1,t\alpha_{1,t} and \mathbb{C}-arc β1,t\beta_{1,t} can not be linked.

A simple calculation yield

L(v1,t,g2(v1,t))=2(15t211t+2)(2t1)2.L\left(v_{1,t},g_{2}(v_{1,t})\right)=-\frac{2(15t^{2}-11t+2)}{(2t-1)^{2}}.

So the \mathbb{C}-circles containing the \mathbb{C}-arcs α1,t\alpha_{1,t} and \mathbb{C}-arc α2,t\alpha_{2,t} can not be linked.

Similar calculations will allow us to see that each pair of the eight \mathbb{C}-circles containing the eight \mathbb{C}-arcs can not be linked for 3/8t<2/53/8\leqslant t<2/5.

Case 2: t[2/5,21]t\in[2/5,\sqrt{2}-1]. As an example, we only show that the cutting disks corresponds to the \mathbb{C}-arcs α1,t\alpha_{1,t} and \mathbb{C}-arc α2,t\alpha_{2,t} are disjointed.

In this case, the \mathbb{C}-circles containing \mathbb{C}-arcs α1,t\alpha_{1,t} and \mathbb{C}-arc α2,t\alpha_{2,t} are linked.

From the polar vector v1,tv_{1,t}, we see that the contact plane containing the \mathbb{C}-arcs α1^\hat{\alpha_{1}} based at the point with Heisenberg coordinate [x1,y1,t1][x_{1},y_{1},t_{1}], where

x1=t2a(t)22t1,y1=2a(t)b(t)22t1,t1=0.x_{1}=\frac{t\sqrt{2}a(t)}{2\sqrt{2t-1}},\,y_{1}=-\frac{\sqrt{2}a(t)b(t)}{2\sqrt{2t-1}},\,t_{1}=0.

The projection of the \mathbb{C}-circle to \mathbb{C}-plane is Euclidean circle center at (x1,y1)(x_{1},y_{1}) with radius r1=616t2t1r_{1}=\sqrt{\frac{6-16t}{2t-1}}.

[t2a(t)22t1,2a(t)b(t)22t1,0],[0,0,4a(t)b(t)2ta(t)+4t1]\left[\frac{t\sqrt{2}a(t)}{2\sqrt{2t-1}},-\frac{\sqrt{2}a(t)b(t)}{2\sqrt{2t-1}},0\right],\quad\left[0,0,\frac{-4a(t)b(t)}{2ta(t)+4t-1}\right]

respectively.

After normalization, the polar vector g2(v1,t)g_{2}(v_{1,t}) of the \mathbb{C}-arc α2,t\alpha_{2,t} is given by

[8t32a(t)b(t)i2ta(t)+4t101].\left[\begin{array}[]{c}\frac{8t-3-2a(t)b(t)i}{2ta(t)+4t-1}\\ 0\\ 1\\ \end{array}\right].

Then the contact plane containing the \mathbb{C}-arc α2,t^\hat{\alpha_{2,t}} and α2,t^\hat{\alpha_{2,t}} based at the points with Heisenberg coordinates [x2,y2,t2][x_{2},y_{2},t_{2}] where

x2=0,y2=0,t2=4a(t)b(t)2ta(t)+4t1.x_{2}=0,\,y_{2}=0,\,t_{2}=\frac{-4a(t)b(t)}{2ta(t)+4t-1}.

The projection of the \mathbb{C}-circle containing α2,t\alpha_{2,t} to \mathbb{C}-plane is Euclidean circle center at (x2,y2)(x_{2},y_{2}) with radius r2=16t62ta(t)+4t1r_{2}=\sqrt{\frac{16t-6}{2ta(t)+4t-1}}.

Define

k1=b(t)t,k2=2(2t1)b(t)t(2ta(t)+4t1).k_{1}=-\frac{b(t)}{t},\quad k_{2}=-\frac{\sqrt{2}(2t-1)b(t)}{t(2ta(t)+4t-1)}.

The intersection of these contact planes is an affine line given by

{[x,k1x+k2,t2]|x}.\{[x,k_{1}x+k_{2},t_{2}]|x\in\mathbb{R}\}.

By studying the intersection of the affine line with the \mathbb{C}-circles, we get that the intersection of these two affine disks is an affine segment given by

{[x,0,0]|ι1xι2},\{[x,0,0]|\iota_{1}\leqslant x\leqslant\iota_{2}\},

where

ι1=k12r22k22+r22k1k21+k12,\displaystyle\iota_{1}=\frac{\sqrt{k_{1}^{2}r_{2}^{2}-k_{2}^{2}+r_{2}^{2}}-k_{1}k_{2}}{1+k_{1}^{2}},
ι2=k1y1+x1k1k2+(k1y1+x1k1k2)2(k222y1k2+2)(1+k12)1+k12.\displaystyle\iota_{2}=\frac{k_{1}y_{1}+x_{1}-k_{1}k_{2}+\sqrt{(k_{1}y_{1}+x_{1}-k_{1}k_{2})^{2}-(k_{2}^{2}-2y_{1}k_{2}+2)(1+k_{1}^{2})}}{1+k_{1}^{2}}.

Define

vs=[(ι1s+(1s)ι2)22ι1s+(1s)ι21],v_{s}=\left[\begin{array}[]{c}\frac{-\left(\iota_{1}s+(1-s)\iota_{2}\right)^{2}}{2}\\ \iota_{1}s+(1-s)\iota_{2}\\ 1\\ \end{array}\right],

where s[0,1]s\in[0,1]. By using some computer algebra software, we find that the minimum of the expression

|vs,q0|2|vs,g2g1(q0)|2|\langle v_{s},q_{0}\rangle|^{2}-|\langle v_{s},g_{2}g_{1}(q_{0})\rangle|^{2}

is given approximately by 0.3616753 for s[0,1]s\in[0,1] and t[2/5,21]t\in[2/5,\sqrt{2}-1]. We omit writing the explicit expression, because it is a bit too complicated to fit on paper. This means that the affine segment lies inside the spinal sphere 𝒮2,t\mathcal{S}_{2,t}. So the intersection of the cut disks corresponding to the arcs α1,t^\hat{\alpha_{1,t}} and α2,t^\hat{\alpha_{2,t}} is empty, see Figure 9.

Refer to caption𝒮2,t\mathcal{S}_{2,t}
Figure 9. Two blue affine disks contain the \mathbb{C}-arcs α1,t^\hat{\alpha_{1,t}} and α2,t^\hat{\alpha_{2,t}}, respectively. The intersection of two blue affine disks is the red affine segment, which lies in spinal sphere 𝒮2,t\mathcal{S}_{2,t}.

The proof of Theorem 1.2. Similar to Proposition 5.6, for t(3/8,21]t\in(3/8,\sqrt{2}-1], the quotient space of

Dt(i=14(αi,t^βi,t^))\partial_{\infty}D_{t}-(\cup^{4}_{i=1}(\hat{\alpha_{i,t}}\cup\hat{\beta_{i,t}}))

by the natural side-pairings on

j=18𝒜i,t(i=14(αi,t^βi,t^))\cup^{8}_{j=1}\mathcal{A}_{i,t}\cup(\cup^{4}_{i=1}(\hat{\alpha_{i,t}}\cup\hat{\beta_{i,t}}))

is homeomorphic to the quotient space ΩΓt,g1/Γt\Omega_{\Gamma_{t},g_{1}}/\Gamma_{t}. By the geometric stability in the deformation, that is, Propositions 6.1 and 6.2, the topology and combinatoris of Dt(i=14(αi,t^βi,t^))\partial_{\infty}D_{t}-(\cup^{4}_{i=1}(\hat{\alpha_{i,t}}\cup\hat{\beta_{i,t}})) does not change in the deformation, and the side-pairing pattern also does not change. So the quotient space of Dt(i=14(αi,t^βi,t^))\partial_{\infty}D_{t}-(\cup^{4}_{i=1}(\hat{\alpha_{i,t}}\cup\hat{\beta_{i,t}})) is homeomorphic to the quotient space of Dt0(i=14(αi^βi^))\partial_{\infty}D_{t_{0}}-(\cup^{4}_{i=1}(\hat{\alpha_{i}}\cup\hat{\beta_{i}})) whenever t(3/8,21]t\in(3/8,\sqrt{2}-1]. By Proposition 5.6, the quotient space is the figure-eight knot complement. This ends the proof of Theorem 1.2.

References

  • [1] M. Acosta. Spherical CR uniformization of Dehn surgeries of the Whitehead link complement. Geom. Topol. 23 (2019) 2593–2664.
  • [2] P. Dehornoy. Le complémentaire du huit dans T1Σ0;3,3,4T^{1}\Sigma_{0;3,3,4}. To appear in Actes du séminaire Théorie Spectrale et Géométrie de Grenoble.
  • [3] M. Deraux, On spherical CR uniformization of 3-manifolds. Exp. Math. 24 (2015) 355–370.
  • [4] M. Deraux and E. Falbel. Complex hyperbolic geometry of the figure eight knot. Geom. Topol. 19 (2015) 237–293.
  • [5] E. Falbel, A. Guilloux and P. Will. Slim curves, limit sets and spherical CR uniformisations. arXiv: 2205.08797.
  • [6] W.M. Goldman. Complex Hyperbolic Geometry. Oxford Mathematical Monographs. Oxford University Press, 1999.
  • [7] W.M. Goldman and J.R. Parker. Complex hyperbolic ideal triangle groups. J. Reine Angew. Math. 425 (1992), 71–86.
  • [8] Y. Jiang, J. Wang and B. Xie. A uniformizable spherical CR structure on a two-cusped hyperbolic 3-manifold. arXiv: 2101.09861. To appear in Alg. Geo. Top.
  • [9] F. Kassel. Geometric structures and representations of discrete groups. In Proceedings of the international congress of mathematicians, ICM 2018, Rio de Janeiro, Brazil, August 1–9, 2018. Volume II. Invited lectures, pages 1115–1151. Hackensack, NJ: World Scientific; Rio de Janeiro: Sociedade Brasileira de Matemática (SBM), 2018.
  • [10] J. Ma and B. Xie. Three-manifolds at infinity of complex hyperbolic orbifolds. arXiv: 2205.11167[math.GT]. .
  • [11] J. Ma, B. Xie. Spherical CR uniformization of the magic 3-manifold. arXiv: 2106.06668 [math.GT]. To appear in Comm. Anal. Geo.
  • [12] J.R. Parker. Notes on complex hyperbolic geometry. Preprint.
  • [13] J.R. Parker, J. Wang and B. Xie. Complex hyperbolic (3,3,n) triangle groups. Pacific J. Math. 280 (2016) 433–453.
  • [14] J.R. Parker and P. Will. A complex hyperbolic Riley slice. Geom. Topol. 21 (2017) 3391–3451.
  • [15] R.E. Schwartz. Ideal triangle groups, dented tori, and numerical analysis. Ann. of Math. 153 (2001) 533–598.
  • [16] R.E. Schwartz. Degenerating the complex hyperbolic ideal triangle groups. Acta Math. 186 (2001) 105–154.
  • [17] R.E. Schwartz. Complex hyperbolic triangle groups. Proceedings of the International Congress of Mathematicians (2002) Volume II: Invited Lectures, 339–350.
  • [18] R.E. Schwartz. A better proof of the Goldman-Parker conjecture. Geom. Topol. 9 (2005) 1539–1602.