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Knot Group Epimorphisms DANIEL S Knot Group Epimorphisms DANIEL S. SILVER and WILBUR WHITTEN Abstract: Let G be a finitely generated group, and let λ ∈ G. If there 3 exists a knot k such that πk = π1(S \k) can be mapped onto G sending the longitude to λ, then there exists infinitely many distinct prime knots with the property. Consequently, if πk is the group of any knot (possibly composite), then there exists an infinite number of prime knots k1, k2, ··· and epimorphisms · · · → πk2 → πk1 → πk each perserving peripheral structures. Properties of a related partial order on knots are discussed. 1. Introduction. Suppose that φ : G1 → G2 is an epimorphism of knot groups preserving peripheral structure (see §2). We are motivated by the following questions. Question 1.1. If G1 is the group of a prime knot, can G2 be other than G1 or Z? Question 1.2. If G2 can be something else, can it be the group of a composite knot? Since the group of a composite knot is an amalgamated product of the groups of the factor knots, one might expect the answer to Question 1.1 to be no. Surprisingly, the answer to both questions is yes, as we will see in §2. These considerations suggest a natural partial ordering on knots: k1 ≥ k2 if the group of k1 maps onto the group of k2 preserving peripheral structure. We study the relation in §3. 2. Main result. As usual a knot is the image of a smooth embedding of a circle in S3. Two knots are equivalent if they have the same knot type, that is, there exists an autohomeomorphism of S3 taking one knot to the other. 3 3 ∼ 2 Let k be a knot in S . We denote its group π1(S \ intV, ∗) by πk. Here V = k × D is a tubular neighborhood of k, and the basepoint ∗ is chosen on the boundary ∂V =∼ k × S1. The element m represented by an essential simple closed curve in ∂V that is contractible in V is called a meridian; the element l represented by an essential simple closed curve in ∂V that is nullhomologous in S3 \ intV is called a longitude. A well-known algorithm enables one to express l in terms of Wirtinger generators corresponding to a diagram for k. Details can be found on page 37 of [BZ85]. The inclusion map ∂V ,→ S3 \ intV induces an injection of fundamental groups. Its image is the subgroup hm, li generated by m and l. First author is partially supported by NSF grant DMS-0304971. 2000 Mathematics Subject Classification. Primary 57M25; secondary 37B10. 1 Let ki, i = 1, 2, be knots with meridian-longitude pairs mi, li. A homomorphism πk1 → πk2 preserves peripheral structure if the image of hm1, l1i is conjugate to a subgroup of hm2, l2i. Definition 2.1. (i) k1 covers k2 (or k2 supports k1) if there is an epimorphism πk1 → πk2; (ii) If G is a finitely generated group normally generated by an element µ, and if λ ∈ G, then a knot k covers (G, µ, λ) (or briefly k covers G ) if there exists an epimorphism φ :(πk, m, l) → (G, µ, λ), where (m, l) is a meridian-longitude pair. If k covers G for a given λ ∈ G, then we say (after Johnson and Livingston [JL89]) that k realizes λ. For a given group G as above and λ ∈ G, [JL89] provides necessary and sufficient conditions for the existence of a knot k that covers (G, µ, λ). We will show that k can always be chosen to be a prime knot. Theorem 2.2. Let G be a finitely generated group that is normally generated by a single element µ, and let λ ∈ G. If there exists a knot k that realizes λ, then there exists an infinite number of distinct prime knots that realize λ. Proof. As the theorem easily follows when k is trivial, we assume that k is knotted. By Proposition 2.5 of [EKT03], we can regard k as the numerator closure T N of a prime or rational tangle T . (See [EKT03], where the authors provide general terminology and prove an even stronger condition.) Set C = (1/2)N , using Conway notation (Figure 1). Figure 1: The tangle 1/2 We use a construction of [EKT03] to form a 2-component link L = T N ∪(1/2)N = k∪C (Figure 2). The components k and C are contained in disjoint solid tori V1 and V2, respectively, the cores of which form a Hopf link. Note that C is an untwisted double of the core of V2. By Propositions 2.3 and 2.4 of [EKT03], L = k ∪ C is a prime link. 2 C k T Figure 2: The link L Let Kq denote the knot obtained from k by 1/q-surgery on the trivial knot C. We take q > 0 for simplicity. It is helpful to view Kq as a satellite of a (−2q)-twist knot γ2q. (The notation γ2q here means that the usual regular projection of γ2q has −2q negative half-twists.) From this perspective, it follows immediately that each Kq is nontrivial. If k is composite, then it follows from Theorem 4.1 of [G-S97] that Kq is prime for q ≥ 3, since k ∪ C is prime. On the other hand, if k is prime and Kq0 is composite for some 0 0 q , then Kq will be prime when q ≥ q + 3. Thus there exists d ≥ 3 such that Kq is prime whenever q ≥ d; cf. Remarks 3.5 (ii). For a fixed meridian-longitude pair (m, l) of k, we have an epimorphism φ : πk → G for which φ(m) = µ and φ(l) = λ. The elements m and l are represented by oriented simple closed curves m1 and l1 on the boundary of a tubular neighborhood of k in V1. Assume that m1 ∩ l1 is a single point ∗, which we take as the basepoint for π(k ∪ C), πKq and πk, as previously mentioned. The curve m1 certainly represents a meridian of Kq, and since the linking number of k and C is zero, the curve l1 represents a longitude (see pages 266–267 of [R76]). Furthermore, if (w, c) is a meridian-longitude pair for C (based q at ∗), we have πk = π(k ∪ C)/hhwii and πKq = π(k ∪ C)/hhwc ii. (Here and throughout hh ii denotes normal closure.) Thus q q πKq/hhwii = π(k ∪ C)/hhw, wc ii = πk/hhc ii. But c = 1 in πk, since C represents the identity element of πk, and so πKq/hhwii = πk. The induced homomorphism η : πKq → πk is an epimorphism taking meridian to meridian and longitude to longitude, preserving orientations; cf. Remarks 3.5 (ii). Therefore, φq = φ ◦ η : πKq → G is an epimorphism that maps the classes of m1 and l1 to µ and λ, respectively. 3 Recall that Kq (q ≥ d from now on) is a satellite of a (−2q)-twist knot γ2q. Since γ2q is the double of the unknot with twisting number −2q, it follows from the form of the Alexander polynomial for doubled knots (see p. 136 of [BZ85], for example) that the exteriors Ext(γ2q) and Ext(γ2¯q) are not homeomorphic if q 6=q ¯. Finally, consider the canonical splitting of the exterior Eq of Kq into a union Σq of Seifert pieces (the characteristic submanifold of Eq) and a union Λq of atoroidal pieces (Λq = cl(Eq \ Σq)) [JS79], [J76]. Each of Σq and Λq has a finite number of components, and ∼ cl[(Σq ∪ Λq) \ Ext(γ2q)] = cl[(Σr ∪ Λr) \ Ext(γ2r)], for q, r ≥ d. Now choose p ∈ {d, d + 1,...} so large that a copy of Ext(γ2q) is not a component of Λq \ Ext(γ2q), for any q ≥ p. Thus if q 6= r and q, r ≥ p, then Eq is not homeomorphic to Er, since Ext(γ2q) is not homeomorphic to Ext(γ2r). Therefore, there exist infinitely many distinct prime knots realizing λ. Remarks 2.3. (i) Notice the seemingly large number of choices we have in the above construction of prime knots that realize λ. For example, we might double and redouble C itself. (ii) To see that the answer to each of Questions 1.1 and 1.2 is yes, see Example 2.6. Less specifically, let G be the group of a composite knot k with meridian-longitude pair (µ, λ), and let k1 and k2 be ambient isotopic copies of k. Assume that k is oriented and that each of k1 and k2 inherits this orientation. Let (mi, li) be a meridian-longitude pair for ki (i = 1, 2), and let φi : πki → πk be an isomorphism such that φ(mi) = µ and φi(li) = λ. Then φ1 and φ2 induce an epimorphism φ : π(k1]k2) → G such that φ(m) = µ 2 and φ(l) = λ , for some choice of meridian-longitude pair (m, l) for k1]k2. Thus k1]k2 2 2 covers (G, µ, λ ), and so there exist prime knots Kq covering (G, µ, λ ) by Theorem 2.2. (iii) In the proof of Theorem 2.2, we have the epimorphism η : πKq → πk, which pre- serves both meridian and longitude. By construction [EKT03], k is geometrically essential in the solid torus V1, and meets a meridianal disk of V1 transversely in two points. Hence the wrapping number of k in V1 must be 2 and the winding number of k (after it is given an orientation) in V1 is either 0 or ±2.
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