As we know the figure-8 ($4_1$) complement can be obtained by quotienting $\mathbb{H}^3$ with an arithmetic Kleinian group, which has index 12 inside $PSL(2,\mathcal{O}_3)$. The resulting complete hyperbolic manifold has a cusp corresponding to the torus boundary.

On the other hand, the "conformal boundary" (boundary at infinity) of $\mathbb{H}^3$ is the sphere at infinity $S^2_{\infty}$. For a Kleinian group $\Gamma$, the hyperbolic manifold $\mathbb{H}^3/\Gamma$ inherits a conformal boundary $\Lambda/\Gamma$, where $\Lambda$ is the domain of discontinuity ($S^2_{\infty}$ with the limit set of $\Gamma$ excluded).

Then how is the empty "conformal boundary" of $4_1$ complement related to its cusp? (I know that the conformal structure of the torus boundary is called "cusp shape"); i.e., should both of them correspond to the $z=0$ "end" in Poincare half-space model? (since the torus boundary results from the truncated vertices on $S^2_{\infty}$ of 2 ideal tetrahedra under developing map upon gluing).

Update: the limit set of $\Gamma$ here is the entire $S^2_{\infty}$ (Corollary 11.8 in Bonahon), so domain of discontinuity as well as the conformal boundary is $\emptyset$.

Confusion: according to section 5 of Thurston, the entire boundary of a Kleinian manifold $(\mathbb{H}^3\cup\Lambda)/\Gamma$ is $\Lambda/\Gamma$, which is empty for $4_1$ complement. Then what is the role of the torus boundary (cusp)?

P.S.: As a physicist, I may not have stated the definition of "conformal boundary" rigorously, but in my mind it is the boundary in the sense of AdS/CFT correspondence (in Euclidean signature, $AdS_3$ is $\mathbb{H}^3$).


1 Answer 1


The group of parabolic isometries fixing a point at infinity is isomorphic to ${\mathbb C}$. (Because it acts simply transitively on a horosphere $H$.) The discrete group $\Gamma$ intersects this stabilizer of the parabolic fixed point (i.e., the stabilizer of $H$) in a lattice $\Lambda\subset{\mathbb C}$. The quotient $H/\Lambda={\mathbb C}/\Lambda$ is a torus with a conformal structure.

  • $\begingroup$ Thank you ThiKu! So if I understand correctly, the triangles at truncated vertices of ideal tetrahedra are on the horospherical "plane", and the torus $H/\Gamma$ is the torus boundary of figure-8 complement as well as its conformal boundary. Am I right? $\endgroup$
    – David Sun
    May 11, 2018 at 7:02
  • $\begingroup$ Yes. The identification of the horosphere with the complex plane is not unique, but only canonical up to conformal automorphims of the plane. This is why you „only“ get a canonical conformal structure. Choosing some identification of the horosphere with the Euclidean plane would give you an Euclidean structure on the torus, but this is not unique. $\endgroup$
    – ThiKu
    May 11, 2018 at 10:49
  • $\begingroup$ Sorry I am confused again. According to Corollary 11.8 in the book "Low-Dimensional Geometry" by Bonahon, the limit set of the Kleinian group $\Gamma_8$ of figure-8 is the entire $S^2_{\infty}$, meaning that the domain of continuity is $\emptyset$, then the conformal boundary should be $\emptyset$ as well, instead of the torus boundary. I am perplexed by this paradox. $\endgroup$
    – David Sun
    May 12, 2018 at 20:55
  • $\begingroup$ Yes, the conformal structure is not coming from the domain of discontinuity, which is empty. $\endgroup$
    – ThiKu
    May 13, 2018 at 6:06
  • $\begingroup$ I see. Then how does the torus metric behave near the cusp? Does it goes to 0 or blow up? I have the confusion due to the two paragraphs below Eq.(2.7) in this 3d gravity paper by Maloney-Witten, where they mention that this Kleinian manifold has two ends, including the cusp where metric goes to 0. $\endgroup$
    – David Sun
    May 13, 2018 at 7:19

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