Homotopically equivalent simple loops on 2-bridge spheres in 2-bridge link complements (II)

Homotopically equivalent simple loops on 2-bridge spheres in 2-bridge link complements (II)

Abstract.

This is the second of a series of papers which give a necessary and sufficient condition for two essential simple loops on a 2-bridge sphere in a 2-bridge link complement to be homotopic in the link complement. The first paper of the series treated the case of the -bridge torus links. In this paper, we treat the case of -bridge links of slope and , where is an arbitrary integer.

2010 Mathematics Subject Classification:
Primary 20F06, 57M25
The first author was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education, Science and Technology(2012R1A1A3009996). The second author was supported by JSPS Grants-in-Aid 22340013 and 21654011.

1. Introduction

Let be a 2-bridge link in and let be a 4-punctured sphere in obtained from a -bridge sphere of . In [1], we gave a complete characterization of those essential simple loops in which are null-homotopic in . The purpose of this series of papers starting from [2] and ending with [3], including the present paper as the second one, is to give a necessary and sufficient condition for two essential simple loops on to be homotopic in . In the first paper [2] of the series, we treated the case when the 2-bridge link is a -torus link. In this paper, we treat the case of 2-bridge links of slope and , where is an arbitrary integer. These two families play special roles in our project in the sense that the treatment of these links form a base step of an inductive proof of a theorem for general -bridge links giving an answer to the problem treated in this series of papers. We note that the figure-eight knot is both a 2-bridge link of slope with and a 2-bridge link of slope with . Surprisingly, the treatment of the figure-eight knot, the simplest hyperbolic 2-bridge knot, is the most complicated. In fact, the figure-eight knot group admits various unexpected reduced annular diagrams (see Section 7). This reminds us of the phenomenon in the theory of exceptional Dehn filling that the figure-eight knot attains the maximal number of exceptional Dehn fillings.

This paper is organized as follows. In Section 2, we describe the main results of this paper (Main Theorems 2.2 and 2.3). In Section 3, we set up Hypotheses A, B and C, under which we establish technical lemmas used for the proofs in Sections 46. The special case of Main Theorem 2.2 (namely, the case of a 2-bridge link of slope ) is treated in Section 4, and the remaining case of Main Theorem 2.2 (namely, the case of a 2-bridge link of slope with ) in Section 5. The proof of Main Theorem 2.3 is contained in Section 6. In the final section, Section 7, we prove Theorems 2.5 and 2.6.

2. Main results

This paper, as a continuation of [2], uses the same notation and terminology as in [2] without specifically mentioning. We begin with the following question, providing whose answer is the purpose of this series of papers.

Question 2.1.

Consider a -bridge link with . For two distinct rational numbers , when are the unoriented loops and homotopic in ?

In the first paper [2], we treated the case when for some , and obtained a complete answer (see [2, Main Theorem 2.7]). In the present paper, we solve the above question for the -bridge links and , where is an arbitrary integer.

Main Theorem 2.2.

Suppose , where is an integer. Then, for any two distinct rational numbers , the unoriented loops and are never homotopic in .

Main Theorem 2.3.

Suppose , where is an integer. Then, for two distinct rational numbers , the unoriented loops and are homotopic in if and only if both (i.e., ) and the set equals either or .

Remark 2.4.

The exceptional pairs and have the following geometric properties in the Farey tessellation. Let be the reflection of the hyperbolic plane in the geodesic with endpoints and , which bisects the Farey edge . Then preserves the Farey tessellation and interchanges and . The members of each of the exceptional pairs are interchanged by the involution .

We prove the above main theorems by interpreting the situation in terms of combinatorial group theory. In other words, we prove that two words representing the free homotopy classes of and are conjugate in the -bridge link group if and only if and satisfy the conditions given in the statements of the theorems. The key tool used in the proofs is small cancellation theory, applied to two-generator and one-relator presentations of -bridge link groups. The proofs of the main theorems also imply the following theorems.

Theorem 2.5.

Suppose , where is an integer. Then the following hold for a rational number .

  1. The loop is peripheral if and only if one of the following holds.

    1. , i.e., , and or .

    2. .

  2. The free homotopy class is primitive with the following exceptions.

    1. , and or . In this case, is the third power of some primitive element in .

    2. and . In this case, is the second power of some primitive element in .

Theorem 2.6.

Suppose , where is an integer. Then the loop is non-peripheral and primitive for any rational number .

Here, a closed loop in is said to be peripheral if it is homotopic to a loop on a peripheral torus. A loop is said to be primitive if there is no element in the -bridge link group whose proper power is conjugate to .

3. Technical lemmas

In this section, we set up Hypotheses A, B and C, under which we establish technical lemmas used for the proofs in Sections 46.

3.1. Hypothesis A

Hypothesis A.

Let be a rational number such that and for any integer . For two distinct elements , suppose that the unoriented loops and are homotopic in . Then and are conjugate in . Let be the symmetrized subset of generated by the single relator of the upper presentation of , and let be the decomposition as in [2, Proposition 3.12]. Due to [2, Lemma 4.7], there is a reduced annular -diagram such that and are, respectively, outer and inner boundary labels of . Then we see from [2, Proposition 3.19(1)] that satisfies the three hypotheses (i), (ii) and (iii) of [2, Theorem 4.9].

Let be the outer boundary layer of (see [2, Figure 7(a)]). Also let and be, respectively, the outer and inner boundary cycles of starting from , where is a vertex lying in both the outer and inner boundaries of . Here, recall from [2, Convention 4.6] that is read clockwise and is read counterclockwise. Let and be the decompositions into oriented edges in . Then clearly for each , there is a face of such that are consecutive edges in a boundary cycle of . We denote the path by and the path by . In particular, if (see [2, Figure 7(b)]), then, for each , there is a face in such that and are two consecutive edges in . Here the indices for the -cells are considered modulo , and the indices for the edges are considered modulo .

Lemma 3.1.

Under Hypothesis A, both of the following hold for every .

  1. None of , , and contains as a subsequence.

  2. None of , , and contains a subsequence of the form , where .

Proof.

By [2, Convention 4.3], each of the words , , and is a piece of the cyclic word . So, the assertion follows from (the only if part) of [2, Corollary 3.25(1)]. ∎

Lemma 3.2.

Under Hypothesis A, only one of the following holds for each face of .

  1. Both and contain as their subsequence.

  2. Both and contain subsequences of the form , where .

Proof.

By [2, Convention 4.3], each of the words and is not a piece. So, by (the if part of) [2, Corollary 3.25(1)], each contains or () as a subsequence. On the other hand, since and since and are symmetric ([2, Proposition 3.12(1)]), is equal to . So if contains (resp., ) then cannot contain (resp., ). Hence we obtain the desired result. ∎

Lemma 3.3.

Under Hypothesis A, only one of the following holds.

  1. For every face of , contains as its subsequence.

  2. For every face of , contains a subsequence of the form , where .

Proof.

Suppose on the contrary that (1) holds for and (2) holds for . Then contains both and as subsequences. By [2, Proposition 3.19(1)], , contradicting the hypothesis of the lemma. ∎

3.2. Hypothesis B

Assuming the following Hypothesis B, we will establish several technical lemmas concerning important properties of .

Hypothesis B.

Suppose under Hypothesis A that Lemma 3.3(1) holds, namely, for every face of , suppose that contains as its subsequence. Then we can decompose the word (clearly ) into

where , and may be empty, , and where (here and are possibly empty), for every . By Lemma 3.2, we also have the decomposition of the word as follows (clearly if ):

where , and may be empty, , and where (here and are possibly empty), for every . Then for every .

Notation 3.4.

Let be a reduced word in . If is not an empty word, then, by and , respectively, we denote a beginning subword and an ending subword of such that is the first term of the sequence and is the last term of . On the other hand, if is empty, then and are also empty words. (Though similar symbols, and (), are used in different meanings in [2, Lemma 3.24], we believe this does not cause any confusion, because these symbols are not used in the remainder of this paper.)

Remark 3.5.

(1) If , then, by [2, Example 3.17(2)], , where and . So, in Hypothesis B, both and are exactly of the form , where are integers.

(2) If with , then, again by [2, Example 3.17(2)], , where and . So, in Hypothesis B, both and are exactly of the form , where and are integers such that if then is necessarily for . In particular, or unless is an empty word. The same is true for , and .

(3) If with , then, by [2, Example 3.17(1)], , where and . So, in Hypothesis B, both and are exactly of the form , where are integers. In particular, and .

Lemma 3.6.

Let , where is an integer. Under Hypothesis B, suppose that is a subword of the cyclic word represented by such that corresponds to a term of . Then, after a cyclic shift of indices, is equal to one of the following subwords:

where in the first three cases and in the last case. In each of the above, the “intermediate subwords” are empty; to be precise, when we say that , for example, is a subword of , we assume that , and are empty words.

Proof.

Recall from Hypothesis B that , where and are possibly empty and . In other words, if (resp., ) is not an empty word, then there is a sign change between and (resp., between and ). The desired result follows from this observation. ∎

Throughout the remainder of this paper, we will assume the following convention.

Convention 3.7.

In Figures 123, the change of directions of consecutive arrowheads represents the change from positive (negative, resp.) words to negative (positive, resp.) words, and a dot represents a vertex whose position is clearly identified. Also an Arabic number represents the length of the corresponding positive (or negative) word. In Figures 112, the upper complementary region is regarded as the unbounded region of . Thus the outer boundary cycles runs the upper boundary from left to right.

The following Lemmas 3.8 and 3.11 show that there are strong restrictions for the shape of the word in Lemma 3.6.

Lemma 3.8.

Let . Under Hypothesis B, the following hold for every .

  1. .

  2. and .

  3. If , then .

  4. .

  5. If , then and .

  6. .

Proof.

(1) Suppose on the contrary that for some . Without loss of generality, we may assume . Then, since , we have either or . We now assume . (The other case can be treated similarly.) Then by using the fact that , we see that is locally as illustrated in Figure 1(a) which follows Convention 3.7. The Arabic numbers , , and near the upper boundary represent the lengths of the words , , and , respectively, whereas the Arabic numbers , and near the lower boundary represent the lengths of the words , and respectively (in particular ), and the change of directions of consecutive arrowheads represents the change from positive (negative, resp.) words to negative (positive, resp.) words.

Suppose first that . Then we see from Figure 1(a) that involves both a term and a term of the form with , contradicting [2, Lemma 3.8] which says that either is equal to or consists of and for some . Suppose next that . Then by Lemma 3.1(1), none of and contains as a subsequence. This means that the initial vertex of lies in the interior of the segment of with weight . (See Figure 1(b), where the initial vertex of is the left-most vertex.) Similarly, the terminal vertex of lies in the interior of the segment of with weight ; in particular, it does not lie in the segment of with weight . Hence, we see from Figure 1(b) that is of the form with . This yields that a term occurs in , which is obviously a contradiction.

Figure 1. Lemma 3.8(1) where

(2) Suppose on the contrary that . (The other case is treated similarly.) Then, since and have different signs when is non-empty and since and , the only possibility is that and . If , then we see from Figure 2(a) that involves both a term and a term of the form with , contradicting [2, Lemma 3.8]. On the other hand, if , then we see, by using Lemma 3.1(1) as in the proof of Lemma 3.8(1), that is of the form with (see Figure 2(b)). This implies that has a term , a contradiction.

Figure 2. Lemma 3.8(2) where

(3) Suppose and suppose on the contrary that . Then and we see, by using Lemma 3.1(1) as in the proof of Lemma 3.8(1), that is of the form with , as illustrated in Figure 3. This implies that contains a term , a contradiction.

Figure 3. Lemma 3.8(3) where

(4) Suppose on the contrary that . Then we have and . If , then we see from Figure 4(a) that includes both a term and a term of the form with , contradicting [2, Lemma 3.8]. So, we must have . Note that and that the terminal vertex of lies in the interior of the segment of with weight , by Lemma 3.1(1). We may assume the vertex divides the segment into segments with weights and as in Figure 4(b). (The other case where the weights of and are interchanged is treated similarly.) Then, as illustrated in Figure 4(b), includes a subsequence , where is an inner boundary cycle of the outer boundary layer, say , of . If , then contains both a term and a term , contradicting [2, Lemma 3.8]. On the other hand, if , then, by using the same argument as in the case of (1) replacing the consideration of -cells and in Figure 1(b) with the consideration of -cells and in Figure 4(b), respectively, we obtain a contradiction.

Figure 4. Lemma 3.8(4) where

(5) Suppose on the contrary that and . (The other case is treated similarly.) Then , , and we see from Figure 5 that includes both a term and a term of the form with , contradicting [2, Lemma 3.8].

Figure 5. Lemma 3.8(5) where

(6) Suppose on the contrary that . Then . If (see Figure 6(a)), then contains both a term and a term of the form with . Here, if , then by [2, Proposition 3.19(1)], contradicting the hypothesis of the theorem, while if , then we have a contradiction to [2, Lemma 3.8]. On the other hand, if (see Figure 6(b)), then we see, by using Lemma 3.1(1) as in the proof of Lemma 3.8(1), that is of the form with . This implies that a subsequence occurs in , which is a contradiction. ∎

Figure 6. Lemma 3.8(6) where
Lemma 3.9.

Let . Under Hypothesis B, the following hold, where .

  1. No two consecutive terms of can be .

  2. No two consecutive terms of can be .

  3. No two consecutive terms of can be of the form .

  4. If , then no term of can be of the form .

  5. No term of can be of the form .

  6. No term of can be of the form .

Proof.

(1) Suppose on the contrary that contains as a subsequence. Let be a subword of the cyclic word corresponding to a subsequence , where . By using Lemma 3.6 and the facts that and , we see that one of the following holds after a shift of indices.

  1. , where .

  2. , where .

  3. .

However, (i) and (ii) are impossible by Lemma 3.8(1), and (iii) is impossible by Lemma 3.8(6).

(2) Suppose on the contrary that contains as a subsequence. Let be a subword of the cyclic word corresponding to a subsequence , where . If contains for some , then we see, by using Lemma 3.6 and the identity , that either with or with . However both cases are impossible by Lemma 3.8(2). Thus cannot contain . Since is a term of , this implies that is disjoint from for every . The same conclusion also holds for , and hence for . Thus is a subword of for some . This is a contradiction, because whereas .

(3a) Suppose on the contrary that contains as a subsequence. Let be a subword of the cyclic word corresponding to a subsequence , where . By using Lemma 3.6 and the facts that and , we see that one of the following holds after a cyclic shift of indices.

  1. with .

  2. with , where or .

If (ii) holds, then either or , a contradiction to Lemma 3.8(2). So (i) holds. By applying the same argument to and by using the fact that is a subword of , we see that , where is empty. Hence we see . This contradicts Lemma 3.8(6).

(3b) Suppose and suppose on the contrary that contains a term . First suppose that contains a term . Let be a subword of the cyclic word corresponding to a term . By arguing as in the proof of Lemma 3.9(3a), we see that one of the following holds.

  1. .

  2. , where or .

However, (i) is impossible by Lemma 3.8(3), and (ii) is impossible by Lemma 3.8(2), as in the proof of Lemma 3.9(3a).

Next suppose that involves a term . Let be a subword of the cyclic word corresponding to a term . Arguing as in the proof of Lemma 3.9(3a), we may assume that one of the following holds.

  1. with .

  2. with .

  3. , where .

However, (i) and (ii) are impossible by Lemma 3.8(2), and (iii) is impossible by Lemma 3.8(4).

Finally suppose that contains a term of the form . Let be a subword of the cyclic word corresponding to a term . By using Lemma 3.6 and the facts that and , we see that must contain a subword for some . This contradicts Lemma 3.8(3).

(4) By Lemma 3.9(3b), it remains to prove the assertion for . Suppose and suppose on the contrary that contains a term of the form . Let be a subword of the cyclic word corresponding to a term . Arguing as in the proof of Lemma 3.9(3a), we may assume that one of the following holds.

  1. where and .

  2. where and .

  3. , where and .

However, (i) and (ii) are impossible by Lemma 3.8(2), and (iii) is impossible by Lemma 3.8(4) and (5).

(5) By Lemma 3.9(3b), it remains to prove the assertion for . Suppose