Detecting inequivalent of the unknotting tunnels for a two-bridge knot.

Yoshiaki Uchida

1. Introduction

Let k be a knot in the 3-sphere S^{3}. An exterior of k is the closure of the complement of a regular neighborhood of k, and is denoted by E(k). An unknotting tunnel for k is an embedded arc τ in S^{3} such that τ ∩ k = ∂τ and S^{3} − IntN(k ∪ τ, S^{3}) is a genus two handlebody. N(H, K) denotes a regular neighborhood of H in K. Then we denote τ ∩E(k) by τ̂. Let τ_{1}, τ_{2} be unknotting tunnels for k. We say that τ_{1} and τ_{2} are homeomorphic if there is a self homeomorphism f of E(k) such that f(τ̂_{1}) = τ̂_{2}. τ_{1} and τ_{2} are isotoic if τ̂_{1} is ambient isotopic to τ̂_{2} in E(k).

Suppose that k is a two-bridge knot. Then by [K. Prop. 3.1], τ_{1} and τ_{2} shown in Figure 1.1 are unknoting tunnels.

Figure 1.1, 1.2

For two-bridge knots, we refer to [HLM]. To each rational number a/b with b odd, there is associated the two-bridge knot K(a/b) shown in Figure 1.2. In Figure 1.2, the central tangle consists of lines of slope ±b/a which are drawn on a square “pillowcase”. In Figure 1.2 we also describe τ_{1} and τ_{2}. By π-twist of \overline{CD} about x-axis, τ_{i} (i = 1, 2) for K(a/b) is isotopic to τ_{i} for K(a/b ± a) and K(a/ − b) is the mirror image of K(a/b). Then we can assume that 0 < b ≤ \frac{a}{2}.

Theorem. For a two-bridge knot K(a/b) with b odd and 0 < b ≤ \frac{a}{2}, τ_{1} and τ_{2} are isotopic if and only if (a, b) = (2n + 1, n) or (2n + 1, 1) (n ∈ N).

Remark. π-rotation about y-axis, shows that τ_{1} and τ_{2} are homeomorphic. M. Sakuma[MS] has proved this theorem by the argument of algebraic topology.

2. Proof of Theorem

Since the “if” part is easily seen (cf. [MS, Theorem 5.2]), we prove “only if” part. Assume that τ_{1} and τ_{2} are isotopic for K(a/b) with (a, b) ≠ (2n + 1, n) and (2n + 1, 1). Then, there exists an ambient isotopy f_{t}(0 ≤ t ≤ 1) of E(K(a/b)) such that f_{1}(τ̂_{1}) = τ̂_{2}. Let p : M ⟶ E(K(a/b)) be a covering of E(K(a/b)), f̃_{t} : M ⟶ M a lift of f_{t}, τ̃_{i}(i = 1, 2) a component of p^{−1}(τ̂_{i}) such that f̃_{1}(τ̃_{1}) = τ̃_{2}. By restrincting f̃_{t} : M × I ⟶ M to τ̃_{1} × I (≅ I × I), we get a continuous map g : I × I ⟶ M with the following properties.

(1) g |_{I×{0}} is an embedding whose image is τ̃_{1},
(2) g |_{I×{1}} is an embedding whose image is τ̃_{2}, and
(3) g({0, 1} × I) ⊂ a component of ∂M.
Then, τ̃_{1} is homotopic (rel.∂) to τ̃_{2} in M.

According to [HLM], the a-fold dihedral irregular branched covering of S^{3} branched over K(a/b) is S^{3}. And the preimage of K(a/b) (= K(11/3)), τ_{1} and τ_{2} is described as in Figure 2.1. The link p^{−1}(K(a/b)) has 1 + [\frac{a}{2}] components C_{0}, C_{1}, ..., C_{[ \frac{a}{2} ]}. In Figure 2.1, ∂τ̃_{1} is containd in ∂N(C_{b}) and only τ̃_{2} in p^{−1}(τ_{2}) whose boundary is contained in ∂N(C_{b}). Then f̃_{1}(τ̃_{1}) = τ̃_{2}. Since τ̃_{1} is homotopic (rel.∂) to τ̃_{2} in S^{3} − Int(⋃_{i=2}^{1+[ \frac{a}{2} ]} N(C_{i})), τ̃_{1} is to τ̃_{2} in S^{3} − Int(N(C_{0}) ∪ N(C_{1}) ∪ N(C_{a−[ \frac{a}{2} ]b}) ∪ N(C_{[ \frac{a}{2} ]})). Figure 2.2 shows the preimage after puncturing the “pillowcase”, twisting one of its ends and flatting out onto the plane.

Figure 2.1, 2.2

Remark. Since 0 < b ≤ \frac{a}{2}, b ≠ 1, and (a, b) ≠ (2n + 1, 1), then 0 < a − [\frac{a}{b}]b < b < [\frac{a}{2}].

According to [HLM. pp.502 - 503], the link C_{0} ∪C_{b} ∪C_{a−[ \frac{a}{b}]b} is unique. And there are only four cases for C_{[ \frac{a}{2} ]} (Figure 2.3). In the case of Figure 2.3(I), the two-fold branched covering branched over C_{a−[ \frac{a}{b}]b} is described in Figure 2.4(I). Then, in Figure 2.3(I), τ̃′_{1} is homotopic (rel.∂) to τ̃′_{2} in S^{3} − Int(⋃_{i=1}^{6} N(D_{i})). So τ̃′_{1} is homotopic to τ̃′_{2} in S^{3} − Int(N(D_{1}) ∪ N(D_{2}) ∪ N(D_{3})). Now, H_{1}(S^{3} −Int(N(D_{1})∪N(D_{2})∪N(D_{3}), ∂N(D_{1}))) ≅ Z ≅< d_{2} |>, where d_{2} is a meridian of D_{2}, and τ̃′_{2} ≃ d_{2} and τ̃′_{1} ≃ 0 in H_{1}(S^{3} − Int(N(D_{1}) ∪ N(D_{2}) ∪ N(D_{3}), ∂N(D_{1}))). Then τ̃′_{1} is not homotopic (rel.∂) to τ̃′_{2}, so τ_{1} and τ_{2} are not isotopic.

Figuere 2.3, 2.4

In the other cases, Figure 2.4 (II) ((III), (IV), resp.) describes the two-fold branched cover branched over C_{a−[ \frac{a}{b}]b} in Figure 2.3(II) ((III),(IV), resp.). Then we do the similar operation for the bold lines. Hence, we can show that τ_{1} and τ_{2} are not isotopic. This completes the proof.

References

[HLM] H. Hilden, M. Lozano, and J. Montesins, On knots that are universal, Topology 24 (1985), 499-504

[K] T. Kobayashi, A criterion for detecting inequivalent tunnels for a knot, to appear in Math. Proc. Camb. Phil. Soc..

[MS] K. Morimoto and M. Sakuma, On unknoting tunnels for knots, preprint.

Department of mathematics, Kobe University, Nada-ku, Kobe, 657 Japan