Epstein-Penner decomposition and thrice punctured spheres in hyperbolic 3-manifolds

Han YOSHIDA ^{∗}

Abstract

Let M be a cusped hyperbolic 3-manifold containing a thrice punctured sphere S. Suppose M is not the Whitehead link complement. We prove that an arc on S is isotopic to an edge of a Euclidean decomposition of M. By using the above result, we relate alternating knot diagrams and the canonical decompositions. Let K be an alternating hyperbolic knot. On a reduced alternating knot diagram of K, if we replace a crossing with a large number of half twists, the polar axis of the crossing is isotopic to an edge of canonical decomposition for the resulting knot.

Mathematics Subject Classifications: Primary 57M50, 57M25, 57N10; Secondary 51M10 Key words: hyperbolic 3-manifold, Euclidean decomposition, thrice punctured sphere, alternating knot

1 Introduction

In [9], Epstein and Penner defined Euclidean decompositions for cusped hyperbolic manifolds, namely, complete non-compact orientable hyperbolic manifolds of finite volume. A Euclidean decomposition is a kind of ideal cell decomposition. In this paper, we prove the following theorem.

Theorem 1.1. Let M be a cusped hyperbolic 3-manifold of finite volume with two cusps C_{1} and C_{2}. Suppose S is an incompressible thrice punctured sphere in M such that two of the punctures belong to C_{1} and the other puncture belongs to C_{2}. Let τ be the simple curve on S whose ends belong to the two punctures corresponding to C_{1}. Then M is the Whitehead link complement, or τ is ambient isotopic to an edge of Euclidean decomposition D([1, ϵ]) where ϵ is sufficiently small number.

∗Nara Women’s University, Kita Uoya Nishimachi Nara 630-8506, Japan.

e-mail: han@vivaldi.ics.nara-wu.ac.jp

This research is partially supported by Research Fellowships of the Japan Society for Young Scientists.

Here, D([1, ϵ]) is the Euclidean decomposition with weight 1 and ϵ on C_{1} and C_{2} respectively. (See Section 2.)

Making a suitable normalization of the definition of “Euclidean decomposition”, Weeks defined the canonical decomposition for cusped hyperbolic manifolds in [13]. In [10], Sakuma and Weeks determined canonical decompositions for many alternating hyperbolic link complements and posed the following conjecture.

Conjecture 1.2. The arc arising from the crossing of a reduced alternating link diagram of hyperbolic link L is isotopic to an edge of canonical decomposition for the link L.

Figure 1

Figure 1: The arc arising from the crossing.

For the hyperbolic 2-bridge links, the above conjecture has been proved in some particular cases [3], [5], [6].

In section 5, we prove the following corollary.

Corollary 1.3. Let K be an alternating hyperbolic knot and K′ a hyperbolic knot obtained by replacing a part of a reduced alternating knot diagram of K as in Figure 2. Let τ be an arc as specified in Figure 2. If |n| is sufficiently large, then τ is ambient isotopic to an edge of the canonical decomposition of S^{3} − K′.

Figure 2

Figure 2: An edge of canonical decomposition of S^{3} − K′.

Acknowledgment : The author would like to express her thanks to and Dr. H. Akiyoshi for their helpful comments. The author also would like to express thanks to referee for valuable suggestions.

2 Euclidean decomposition

Let M be a cusped hyperbolic 3-manifold with cusps C_{1}, …, C_{p}. Put

W = {[w_{1}, …, w_{p}] ∈ RP^{p−1}|w_{j} > 0 (j = 1, …, p)}.

For W = [w_{1}, …, w_{p}] ∈ W, Epstein and Penner defined the Euclidean decomposition of M with weight W, denoted by D(W). In [6], we have shown that the following definition of Euclidean decomposition is equivalent to the original one.

Let c be a small positive number. For each j (j = 1, …, p), take the j-th cusp horospherical cross section of C_{j}, so that vol(C_{j}) = cw_{j}. The preimages of this cusp horospherical cross sections in the universal cover H^{3} form a set of horoballs, which is denoted by H(M). Let π : H^{3} → M the covering projection.

Definition 2.1. : A cell of the Euclidean decomposition D(W) is the projection image of an ideal cell Δ ⊂ H^{3} whose vertices are the centers of the nearest horoballs in H(M) to some point p ∈ H^{3}. In particular, the Euclidean decomposition D([1, …, 1]) is called the canonical decomposition of M.

Remark 2.2. We can expand the horospherical sections of cusps keeping the ratio w_{1}, …, w_{p}. Let us consider the situation where two horoballs H_{1} and H_{2} collide at the point p and no horoballs contain p in the universal cover H^{3}. The projection image of the geodesic which connects the center of H_{1} and H_{2} is an edge of the Euclidean decomposition D([w_{1}, …, w_{p}]).

Let M be a cusped hyperbolic 3-manifold with two cusps. We denote these cusps by C_{1}, C_{2}. Put

D(ϵ) = D([1, ϵ])

and E(ϵ) = {edges of D(ϵ) connecting the cusp C_{1}to itself}.

We remark that if ϵ_{1} and ϵ_{2} are sufficiently small, E(ϵ_{1}) = E(ϵ_{2}). For a pair of coprime integers P = (r, s), let M(P) be the manifold obtained from M by r/s-Dehn filling on the second cusp C_{2}. By the hyperbolic Dehn surgery theorem, if |r| + |s| is sufficiently large, M(P) is also a cusped hyperbolic manifold. Our aim is to find edges of the canonical decomposition of M(P). Let ι_{P} : M ↪ M(P) be the natural embedding. The following proposition is a particular case of Proposition 2.4 in [6].

Proposition 2.3. Let P = (r, s) be a pair of coprime integers and e an edge of the Euclidean decomposition D(ϵ) for sufficiently small ϵ > 0. Then ι_{P} (e) is ambient isotopic to an edge of the canonical decomposition of M(P) if |r| + |s| is sufficiently large.

3 Proof of Main Theorem

We will work in the upper-half-space model for the hyperbolic space H^{3}. We identify the boundary of H^{3} with C ∪ {∞}, and the group of orientation preserving isometries of H^{3} with P SL(2, C).

Adams showed that an incompressible thrice punctured sphere in any cusped hyperbolic 3-manifold is isotopic to a totally geodesic surface. (See [2] section 3.) On the thrice punctured sphere, complete hyperbolic structure of finite area is uniquely determined, which is obtained by glueing the edges of the ideal quadrangle whose vertices are at −1, 0, 1, ∞ in the upper half space model.

Figure 3

Figure 3: The identification of quadrangle.

Let Γ be a Kleinian group such that H^{3}/Γ = M, π : H^{3} → M the covering projection, S̃_{R} the hyperplane whose boundary consists of the real line and ∞. We may assume that π(S̃_{R}) = S, γ_{1} = ( \begin{matrix}1 & 2 \\ 0 & 1 \end{matrix} ) , γ_{2} = ( \begin{matrix}1 & 0 \\ 2 & 1 \end{matrix} ) ∈ Γ, and two ideal points 0 and ∞ correspond to the cusp C_{1}. We remark that π^{−1}(S) is a union of disjoint hyperplanes.

Put c = vol(C_{1}). We denote the horoball centered at x ∈ C ∪ {∞} which projects to C_{1} by H_{x}(c).

Lemma 3.1. We expand the horospherical cross section of the cusp C_{1} until H_{∞}(c) collides H_{0}(c). Then the Euclidean height of H_{∞}(c) and H_{0}(c) is 1.

(Proof.) Suppose c = vol(C_{1}) is a sufficiently small number. The intersection of S with ∂C_{1} consists of two disjoint curves s and s′, which are straight and parallel in the Euclidean geometry on ∂C_{1}. Therefore the two curves s and s′ have the same hyperbolic length. Thus π(∂(S̃_{R} ∩ H_{∞}(c))) and π(∂(S̃_{R} ∩ H_{0}(c))) have the same length for any c. Then the reflection along the geodesic connecting −1 and 1 takes H_{∞}(c) to H_{0}(c). Hence when the Euclidean height of H_{∞}(c) is 1, that of H_{0}(c) is also 1.

Therefore the horoballs H_{0}(c) and H_{∞}(c) collide at a point p with Euclidean height 1. After here, we fix c = vol(C_{1}) such that the Euclidean height of H_{∞}(c) is 1. We denote H_{x}(c) by H_{x}. Thus for γ ∈ Γ, γ(H_{∞}) = H_{γ(∞)} is centered at γ(∞). Put

H(M) = {γ(H_{∞})|γ ∈ Γ}.

The following fact about the Euclidean height of the horoball H_{∞} and P SL(2, C) is well known (See Section 4 of [8]).

Lemma 3.2. For γ ∈ P SL(2, C), the Euclidean height of γ(H_{∞}) is \frac{1}{|a|^{2}} where a is the first entry in the second row of γ.

Our plan is to show that if some horoball H_{x} contains p, then x must be 0, ∞, i or −i. To prove this, we consider separately the cases where x is on the real line (Lemma 3.3) and off the real line (Lemma 3.4). We will see that if four horoballs contain the point p, M is the Whitehead link complement (Proposition 4.1).

Lemma 3.3. If H_{x} ∈ H(M) contains p and x ∈ ∂S̃_{R}, then H_{x} is H_{0} or H_{∞}.

(Proof.) Let Γ_{0} be the subgroup of Γ generated by γ_{1} and γ_{2}, that is Γ ∩ P SL(2, R) and γ_{x} an element of Γ such that γ_{x}(∞) = x. As γ_{x}(∞) = x ∈ ∂S̃_{R}, γ_{x}^{−1}(S̃_{R}) is a vertical plane. Because S is totally geodesic, γ_{x}^{−1}(S̃_{R}) ∩ S̃_{R} = ∅. Then γ_{1}(γ_{x}^{−1}(S̃_{R})) = γ_{x}^{−1}(S̃_{R}). We have γ_{x}γ_{1}γ_{x}^{−1}(S̃_{R}) = S̃_{R}. Thus γ_{x}γ_{1}γ_{x}^{−1} ∈ Γ_{0} and γ_{x}γ_{1}γ_{x}^{−1} is parabolic centered at γ_{x}(∞) = x. Therefore H_{x} = γ_{x}(H_{∞}) = γ′(H_{∞}) or γ′(H_{0}) for some γ′ ∈ Γ_{0}.

First we consider γ′(H_{∞}). Since Γ_{0} is generated by γ_{1} and γ_{2}, the first entry in the second row of γ′ is an even integer. Thus γ′(H_{∞}) = H_{∞}, or the Euclidean height of γ′(H_{∞}) is less or equal to \frac{1}{2^{2}} by Lemma 3.2. As the Euclidean height of p is 1, γ′(H_{∞}) = H_{∞} or p ∉ γ′(H_{∞}).

We take a conjugate of Γ_{0} by ϕ = ( \begin{matrix}0 & i \\ i & 0 \end{matrix} ). Then ϕΓ_{0}ϕ^{−1} = Γ_{0}. Next we consider ϕγ′ϕ^{−1}(H_{∞}). By the same argument as above, ϕγ′ϕ^{−1}(H_{∞}) = H_{∞} or p ∉ ϕγ′ϕ^{−1}(H_{∞}). As ϕ^{−1}(H_{∞}) = H_{0} and ϕ(p) = p, p ∉ γ′(H_{0}) or γ′(H_{0}) = H_{0}.

This completes the proof of Lemma 3.3.

Lemma 3.4. For H_{x} ∈ H(M), if H_{x} ∩ S̃_{R} ≠ ∅ and x ∉ ∂S̃_{R}, then H_{x} ∩ S̃_{R} is only one point.

(Proof.) Let γ_{x} be an element of Γ such that γ_{x}(∞) = x. Thus H_{∞} ∩ γ_{x}^{−1}(S̃_{R}) ≠ ∅. As γ_{x}(∞) ∉ ∂S̃_{R}, ∞ ∉ ∂γ_{x}^{−1}(S̃_{R}). The hyperplane γ_{x}^{−1}(S̃_{R}) is not vertical. Put S̃ = γ_{x}^{−1}(S̃_{R}). We consider the Euclidean height of S̃. Since S is totally geodesic, γ_{1}(S̃) ∩ S̃ = ∅. Then the Euclidean height of S̃ is less or equal to 1. Because the Euclidean height of H_{∞} is 1 and S̃∩H_{∞} ≠ ∅, S̃∩H_{∞} is only one point. Thus S̃_{R}∩γ_{x}(H_{∞}) = γ_{x}(S̃)∩γ_{x}(H_{∞}) = γ_{x}(S̃∩H_{∞}) is only one point.

To prove Proposition 4.1 in Section 4, we prepare the following corollary.

Corollary 3.5. For H_{x} ∈ H(M) and γ ∈ Γ, if H_{x} ∩ γ(S̃_{R}) ≠ ∅ and x ∉ ∂γ(S̃_{R}), then H_{x} ∩ γ(S̃_{R}) is only one point.

By Lemma 3.3 and 3.4, we get the following lemma.

Lemma 3.6. If p is contained H ∈ H(M), then H ∈ {H_{0}, H_{∞}, H_{i}, H_{−i}} and p ∈ ∂H.

(Proof.) Suppose the center of H is on ∂S̃_{R}. As H contains p ∈ S̃_{R}, H is H_{0} or H_{∞} by Lemma 3.3.

Suppose the center of H is not on ∂S̃_{R}. As H contains p ∈ S̃_{R}, H ∩ S̃_{R} ≠ ∅. Then H ∩ S̃_{R} = {p} by Lemma 3.4. Hence H is H_{i} or H_{−i} and p ∈ ∂H.

(Proof of Theorem 1.1.) By Lemma 3.6, we have three cases.

Case 1. Suppose p ∈ ∂H_{0}, ∂H_{∞} and p ∉ H for any H ∈ H(M)\{H_{0}, H_{∞}}. Let τ̃ be the geodesic which connects the centers of the horoballs H_{0} and H_{∞}. By Remark 2.2, π(τ̃) is an edge of the Euclidean decomposition D([1, ϵ]). Then the Theorem 1.1 follows immediately.

Case 2. Suppose p ∈ ∂H_{i}(resp. ∂H_{−i}), ∂H_{0}, ∂H_{∞} and p ∉ H for any H ∈ H(M)\{H_{i}, H_{0}, H_{∞}} (resp. H(M)\{H_{−i}, H_{0}, H_{∞}}). Then the nearest horoballs of p in H(M) are H_{0}, H_{∞} and H_{i}(resp. H_{−i}). Let Δ be the ideal triangle with vertices 0, ∞ and i(resp. −i). By the definition of Euclidean decomposition, π(Δ) is a cell of the Euclidean decomposition D([1, ϵ]). Since τ̃ is an edge of Δ, π(τ̃) is an edge of the Euclidean decomposition D([1, ϵ]). Then Theorem 1.1 follows immediately.

Case 3. Suppose p ∈ ∂H_{0}, ∂H_{∞}, ∂H_{i}, ∂H_{−i} and p ∉ H for any H ∈ H(M)\{H_{i}, H_{−i}, H_{0}, H_{∞}}. We will show that M is the Whitehead link complement in the next section.

4 The case where four horoballs collide

To complete the proof of Theorem 1.1, we show the following Proposition by using the above notation.

Proposition 4.1. Suppose p ∈ ∂H_{0}, ∂H_{∞}, ∂H_{i}, ∂H_{−i} and p ∉ H for any H ∈ H(M)\{H_{i}, H_{−i}, H_{0}, H_{∞}}. Then M is the Whitehead link complement.

(Proof.) Let δ_{+} and δ_{−} be the elements of Γ such that δ_{+}(∞) = i and δ_{−}(∞) = −i. Put S̃_{+} = δ_{+}^{−1}(S̃_{R}) and S̃_{−} = δ_{−}^{−1}(S̃_{R}). As H_{i} ∩ S̃_{R} = {p},

H_{∞} ∩ S̃_{+} = δ_{+}^{−1}(H_{i}) ∩ δ_{+}^{−1}(S̃) = δ_{+}^{−1}(H_{i} ∩ S̃) = {one point}.

The Euclidean height of H_{∞} is 1. Thus the Euclidean height of S̃_{+} is 1. By the similar fashion, the Euclidean height of S̃_{−} is 1. The Euclidean diameter of these hyperplanes is 2. We denote the geodesic whose end points are S̃_{+} ∩ γ_{1}(S̃_{+}) and S̃_{+} ∩ γ_{1}^{−1}(S̃_{+}) (resp. S̃_{−} ∩ γ_{1}(S̃_{−}) and S̃_{−} ∩ γ_{1}^{−1}(S̃_{−})) by l_{+}(resp. l_{−}). We show the following Claim.

Claim 4.2. The end points of δ_{+}(l_{+}) and δ_{−}(l_{−}) are 0 and ∞.

We prove the endpoints of δ_{+}(l_{+}) are 0 and ∞. We can prove the endpoints of δ_{−}(l_{−}) are 0 and ∞ by the similar fashion. To prove simplify, we put δ = δ_{+}, l = l_{+}, S̃ = S̃_{+} respectively. Let u and v be the ideal endpoints of δ(l).

Suppose {u, v} ≠ {0, ∞}. The geodesic δ(l) contains the point p whose Euclidean height is 1. Then we have uv < 0 and |u − v| > 2. We assume {u} = S̃_{R} ∩ δγ_{1}(S̃), {v} = S̃_{R} ∩ δγ_{1}^{−1}(S̃), u > 0 and v < 0. The other cases are proved similarly.

The Euclidean height of γ_{1}(S̃) is 1. Since H_{∞} touches γ_{1}(S̃), δ(H_{∞}) = H_{i} touches δγ_{1}(S̃). Then as u becomes bigger, the Euclidean height of δγ_{1}(S̃) becomes bigger. The horoballs H_{i} and γ_{1}(H_{i}) = H_{2+i} are symmetric about the hyperplane {(t, z) ∈ H^{3}|z = 1}. If u = 1, δγ_{1}(S̃) touches H_{i} and H_{2+i}. Therefore if u > 1, ∂δγ_{1}(S̃) must contain the center of H_{2+i} by Corollary 3.5.

Suppose u > 1. We will show that γ_{1}^{−1}δγ_{1}(S̃) intersects δγ_{1}^{−1}(S̃). The center of H_{2+i}, which is 2 + i, is contained ∂δγ_{1}(S̃) and δ(H_{∞}) = H_{i} touches δγ_{1}(S̃). Then the Euclidean radius of δγ_{1}(S̃) is 6 − 2\sqrt{7} = 0.703 … and u = 4 − \sqrt{7} = 1.354 . . ..

Figure 4

Figure 4: Horoballs and a hyperplane and δ(l).

First we consider the hyperplane γ_{1}^{−1}δγ_{1}(S̃). As γ_{1} is parabolic centered at ∞, the Euclidean radius of γ_{1}^{−1}δγ_{1}(S̃) is 6 − 2\sqrt{7} = 0.703 … and

S̃_{R} ∩ γ_{1}^{−1}δγ_{1}(S̃) = γ_{1}^{−1}(S̃_{R} ∩ δγ_{1}(S̃)) = {u − 2} = {−0.645 …}.

Next we consider δγ_{1}^{−1}(S̃). The geodesic δ(l) contains p = (1, 0) ∈ R^{+} × C = H^{3} and u = 4 − \sqrt{7} = 1.354 . . .. Then it is not hard to see that \{v\} = S̃_{R} ∩ δγ_{1}^{−1}(S̃) = \{−(4 + \sqrt{7})/9\} = \{−0.738 …\}. As H_{∞} touches γ_{1}^{−1}(S̃), δ(H_{∞}) = H_{i} touches δγ_{1}^{−1}(S̃). We can show that δγ_{1}^{−1}(S̃) is a hyperplane with Euclidean radius (26−4\sqrt{7})/81 = 0.386 … and the Euclidean radius of γ_{1}^{−1}δγ_{1}(S̃) is 6 − 2\sqrt{7} = 0.703 . . .. Thus δγ_{1}^{−1}(S̃) intersects γ_{1}^{−1}δγ_{1}(S̃) (See Figure 5). It contradicts the assumption that π(S̃) = S is totally geodesic.

Figure 5

Figure 5: Boundaries of hyperplanes.

By the same argument as above, v = −4+\sqrt{7} = −1.354 … or v ≥ −1. If v = −4+\sqrt{7}, δγ_{1}^{−1}(S̃) is a hyperplane with Euclidean radius 6−2\sqrt{7} and u = (4+\sqrt{7})/9 = 0.738 … and the Euclidean radius of δγ_{1}^{−1}(S̃) is (26 − 4\sqrt{7})/81 = 0.386 . . .. The hyperplane γ_{1}δγ_{1}^{−1}(S̃) intersects δγ_{1}(S̃). It is contradiction.

Therefore 0 < u ≤ 1 and −1 ≤ v < 0. Since the Euclidean distance of the endpoints of δ(l) is greater than or equal to 2, u = 1 and v = −1. We have δ(S̃ ∩ γ_{1}(S̃)) = 1, δ(S̃ ∩ γ_{1}^{−1}(S̃)) = −1. Thus S̃ ∩ γ_{1}(S̃) = δ^{−1}(1) and S̃ ∩ γ_{1}^{−1}(S̃) = δ^{−1}(−1). As γ_{1}(S̃ ∩ γ_{1}^{−1}(S̃)) = S̃ ∩ γ_{1}(S̃), δ^{−1}(1) = γ_{1}δ^{−1}(−1). We have δγ_{1}δ^{−1}(−1) = 1. Since δγ_{1}δ^{−1} is parabolic centered at δ(∞) = i, δγ_{1}δ^{−1} = ( \begin{matrix}1 + i & 1 \\ 1 & 1 − i \end{matrix} ) . Thus

γ_{1}^{−1}δγ_{1}δ^{−1} = ( \begin{matrix}1 & −2 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 + i & 1 \\ 1 & 1 − i \end{matrix} ) = ( \begin{matrix}−1 + i & −1 + 2i \\ 1 & 1 − i \end{matrix} ) ∈ Γ.

As trace(γ_{1}^{−1}δγ_{1}δ^{−1}) = 0, γ_{1}^{−1}δγ_{1}δ^{−1} is elliptic. It is contradiction.

Claim 4.3. We may assume the center of S̃_{+} is 1 − i

The endpoints of the geodesic δ_{+}(l_{+}) are 0 and ∞, and those of l_{+} are S̃_{+} ∩γ_{1}(S̃_{+}) and S̃_{+} ∩ γ_{1}^{−1}(S̃_{+}). Then δ_{+}γ_{1}δ_{+}^{−1} ∈ Γ is parabolic centered at δ_{+}(∞) = i and δ_{+}γ_{1}δ_{+}^{−1}(0) = ∞

or δ_{+}γ_{1}^{−1}δ_{+}^{−1}(0) = ∞. Thus δ_{+}γ_{1}δ_{+}^{−1} or δ_{+}γ_{1}^{−1}δ_{+}^{−1} = ( \begin{matrix}2 & −i \\ −i & 0 \end{matrix} ) . Put γ_{+} = ( \begin{matrix}2 & −i \\ −i & 0 \end{matrix} ) ∈

Γ. By the similar fashion, δ_{−}γ_{1}δ_{−}^{−1} or δ_{−}γ_{1}^{−1}δ_{−}^{−1} = ( \begin{matrix}2 & i \\ i & 0 \end{matrix} ) . Put γ_{−} = ( \begin{matrix}2 & i \\ i & 0 \end{matrix} ). Thus we get

γ_{+}γ_{−}^{−1} = ( \begin{matrix}2 & −i \\ −i & 0 \end{matrix} ) ( \begin{matrix}0 & −i \\ −i & 2 \end{matrix} ) = ( \begin{matrix}−1 & −4i \\ 0 & −1 \end{matrix} ) ∈ Γ.

Hence (γ_{+}γ_{−}^{−1})^{k}(S̃_{R}) is the vertical plane whose boundary is {a + 4ki | a ∈ R}. Furthermore γ_{+}(S̃_{R}) is the vertical plane whose boundary is {a + 2i | a ∈ R}. Thus (γ_{+}γ_{−}^{−1})^{k}γ_{+}(S̃_{R}) is a vertical plane whose boundary is {a + (4k + 2)i | a ∈ R}. The hyperplane γ_{+}^{−1}(S̃_{R}) (resp. γ_{−}^{−1}(S̃_{R})) is centered at i/4 (resp. −i/4) and its Euclidean diameter is 1/2. Then γ_{1}^{m}(γ_{+}γ_{−}^{−1})^{k}γ_{+}^{−1}(S̃_{R}) (resp. γ_{1}^{m}(γ_{+}γ_{−}^{−1})^{k}γ_{−}^{−1}(S̃_{R})) is the hyperplane centered at m+(1/4+4k)i (resp. m+(−1/4+4k)i) with Euclidean diameter 1/2. Now we consider the hyperplanes S̃_{+} with diameter 2 which project to the thrice punctured sphere S in M. Since S is totally geodesic, the hyperplane S̃_{+} does not intersect (γ_{+}γ_{−}^{−1})^{k}(S̃_{R}), (γ_{+}γ_{−}^{−1})^{k}γ_{+}(S̃_{R}), γ_{1}^{m}(γ_{+}γ_{−}^{−1})^{k}γ_{+}^{−1}(S̃_{R}) and γ_{1}^{m}(γ_{+}γ_{−}^{−1})^{k}γ_{−}^{−1}(S̃_{R}) for any k, m ∈ Z. Then by taking suitable integers p and q, γ_{1}^{p}(γ_{+}γ_{−})^{q}(S̃_{+}) is the hyperplane centered at 1 + i or 1 − i with diameter 2.

Figure 6

Figure 6: Hyperplanes.

We have δ_{+}γ_{1}^{p}(γ_{+}γ_{−}^{−1})^{q}(∞) = δ_{+}(∞) = i. We denote δ_{+}γ_{1}^{p}(γ_{+}γ_{−}^{−1})^{q} by δ_{+} again. The hyperplane S̃_{+} = δ_{+}(S̃_{R}) is centered at 1 + i or 1 − i with Euclidean diameter 2.

We will show that the center of S̃_{+} is not 1 + i. We have shown that δ_{+}(l_{+}) is the geodesic with endpoints 0 and ∞. If the center of S̃_{+} is 1 + i, the end points of l_{+} are i and 2 + i. Thus δ_{+}(i) = 0 or ∞. As δ_{+}(∞) = i and δ_{+} is not elliptic, δ_{+}(i) = 0. Thus δ_{+}(2 + i) = ∞. By δ_{+}(i) = 0, δ_{+}(2 + i) = ∞ and δ_{+}(∞) = i, we have δ_{+} =

( \begin{matrix}1 & −i \\ −i & −1 + 2i \end{matrix} ). Thus

δ_{+}γ_{+} = ( \begin{matrix}1 & −i \\ −i & −1 + 2i \end{matrix} ) ( \begin{matrix}2 & −i \\ −i & 0 \end{matrix} ) = ( \begin{matrix}1 & −i \\ 2 − i & −1 \end{matrix} )

As trace(δ_{+}γ_{+}) = 0, δ_{+}γ_{+} is elliptic. It is contradiction. Hence the center of S̃_{+} is 1 − i.

Claim 4.4. δ_{+} = \frac{1+i}{2} ( \begin{matrix}i & −1 \\ 1 & −2 + i \end{matrix} ) or \frac{1+i}{2} ( \begin{matrix}1 & −2 + i \\ −i & 1 \end{matrix} ).

The hyperplane S̃_{+} is centered at 1 − i and δ_{+}(l_{+}) is the geodesic with endpoints 0 and ∞. Then we have

δ_{+}(−i) = 0, δ_{+}(2 − i) = ∞

or

δ_{+}(−i) = ∞, δ_{+}(2 − i) = 0.

By δ_{+}(∞) = i, we get δ_{+} = \frac{1+i}{2} ( \begin{matrix}i & −1 \\ 1 & −2 + i \end{matrix} ) or \frac{1+i}{2} ( \begin{matrix}1 & −2 + i \\ −i & 1 \end{matrix} ). This completes the

proof of Claim 4.4.

Let O be the ideal octahedron with the vertices 1, −1, i, −i, 0 and ∞. It is well known that the Whitehead link complement is obtained by glueing the faces of O so that the arrows much.(See Figure 7.)

Figure 7

Figure 7: The identification of ideal octahedron.

Denote by a the orientation preserving isometry in H^{3} which translates across the face A taking A′ to A. Similarly we denote by b, c, d the translation across B, C, D respectively. Then they are determined by the following.

a : (∞, 1, −i) ⟶ (i, 1, 0)
b : (∞, −1, −i) ⟶ (i, 1, ∞)
c : (0, 1, −i) ⟶ (i, −1, 0)
d : (0, −1, −i) ⟶ (i, −1, ∞)

Thus the fundamental group of the Whitehead link complement is generated by a, b, c, d. The Whitehead link complement is not a covering space of any hyperbolic 3-manifold. Then it is enough to show a, b, c, d ∈ Γ.

Suppose δ_{+} = \frac{1+i}{2} ( \begin{matrix}i & −1 \\ 1 & −2 + i \end{matrix} ). We have

δ_{+}(∞, 1, −i) = (i, 1, 0)

and

δ_{+}γ_{1}(∞, −1, −i) = δ_{+}(∞, 1, 2 − i)
= (i, 1, ∞).

In the proof of Claim 2, we have shown that γ_{−} = ( \begin{matrix}2 & i \\ i & 0 \end{matrix} ) ∈ Γ. We get

δ_{+}γ_{−}(0, 1, −i) = δ_{+}(∞, 1 − 2i, −i)
= (i, −1, 0)

and

γ_{1}^{−1}δ_{+}γ_{1}(0, −1, −i) = γ_{1}^{−1}δ_{+}(2, 1, 2 − i)
= γ_{1}^{−1}(2 + i, 1, ∞)
= (i, −1, ∞).

Then a = δ_{+}, b = δ_{+}γ_{1}, c = δ_{+}γ_{−} and d = γ_{1}^{−1}δ_{+}γ_{1}. Hence a, b, c, d ∈ Γ.

Suppose δ_{+} = \frac{1+i}{2} ( \begin{matrix}1 & −2 + i \\ −i & 1 \end{matrix} ). Let α be an isometry of H^{3} such that α(z) = 1/z̄ on ∂H^{3} = C ∪ {∞}. It is enough to show that a, b, c, d ∈ αΓα^{−1} to prove H^{3}/Γ is the Whitehead link complement. We have

αδ_{+}γ_{−}α^{−1}(∞, 1, −i) = αδ_{+}γ_{−}(0, 1, −i)
= αδ_{+}(∞, 1 − 2i, −i)
= α(i, 1, ∞)
= (i, 1, 0),
αγ_{2}δ_{+}γ_{1}α^{−1}(∞, −1, −i) = αγ_{2}δ_{+}γ_{1}(0, −1, −i)
= αγ_{2}δ_{+}(2, 1, 2 − i)
= αγ_{2}(\frac{i}{1 − 2i}, −1, 0)
= α(i, 1, 0)
= (i, 1, ∞),
αδ_{+}α^{−1}(0, 1, −i) = αδ_{+}(∞, 1, −i)
= α(i, −1, ∞)
= (i, −1, 0)

and

αδ_{+}γ_{1}α^{−1}(0, −1, −i) = αδ_{+}γ_{1}(∞, −1, −i)
= αδ_{+}(∞, 1, 2 − i)
= α(i, −1, 0)
= (i, −1, ∞).

Then a = αδ_{+}γ_{−}α^{−1}, b = αγ_{2}δ_{+}γ_{1}α^{−1}, c = αδ_{+}γ_{−}α^{−1} and d = αδ_{+}γ_{1}α^{−1}. Hence a, b, c, d ∈ αΓα^{−1}.

5 Alternating knot and canonical decomposition

In this section, we prove Corollary 1.3. For a given alternating hyperbolic knot diagram, we replace a part of a reduced knot diagram of K as in Figure 8. Then the resulting link L is still hyperbolic [1].

Figure 8

Figure 8: The new link diagram obtained from K.

Let C_{1} be the component of L corresponding to the original knot K. We perform (1, n) Dehn surgery on the added trivial component. Then we get a knot K′ which is obtained by replacing the knot diagram of K as in Figure 2.

Case 1. Suppose S^{3}\L is not the Whitehead link complement. Then the arc as in Figure 8 is an edge of Euclidean decomposition D([1, ϵ]) by Theorem 1.1. Corollary 1.3 follows immediately.

Case 2. Suppose S^{3}\L is the Whitehead link complement. The resulting knot K′ is a twist knot. In [3], Akiyoshi has proved that the arc as in Figure 9 is an edge of canonical decomposition for the twist knot K.

Figure 9

Figure 9: An edge of canonical decomposition for twist knot.

This completes the proof of Corollary 1.3.

References

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