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Topology Vol. 26. No. I, pp. 4 I-u. 1987. w-9383.87 ~3.00~ M) Pnnrcdin Great Bntun. Pcrgamon Journals Ltd.

KNOT COMPLEMENTS AND GROUPS

WILBUR WHITEN

In memory of R. H. Fox

(Receiced13 March 1985)

Two BASIC questions of classical theory arose after Seifert [ 163 showed, in 1933, that the group of a (composite) knot in S3 does not necessarily determine the complement.

QUESTION 1. Does the group of a (in S3) determine the complement among all (tame) knot complements?

Our principal result is the following theorem, which implies an affirmative answer, since the group of a prime knot cannot be isomorphic to that of a composite knot [7, lemma 2, p. 12861.

RIGIDITY THEOREM. Prime (in S’) with isomorphic groups hare homeomorphic complements.

The group of a prime knot does not, however, necessarily determine the topological type of the exterior. Dehn hips [12, p. 2431 on certain “essential” solid tori in the exteriors of torus knots and of cable knots produce Haken manifolds that are homotopically equivalent but not homeomorphic to the original exteriors and that, in fact, cannot be imbedded in S3.

QUESTION 2. Does the group of a prime knot determine the type of the knot?

COROLLARY 1. There exist at most two distinct prime knots with isomorphic groups.

Remark. The Rigidity Theorem and Corollary 1 improve earlier results of Simon [17, 181, Swarup [19], and Feustel and Whitten [7] on Questions 1 and 2. Moreover, the comments of Gordon [lo] and Simon [13, problem 1.13, p. 2781 put these earlier developments into perspective. The next corollary of the Rigidity Theorem is merely an observation, since composite knots have property P.

COROLLARY 2. Prime knots with isomorphic groups are equiaalent if and only if knot complements determine knot types.

The proof of the Rigidity Theorem follows immediately from Proposition 3 and Theorem 4[4, corollary 2, p. 43; 5, corollary 21. To state these results, let Q denote the rationals, let r E Q u { co >, and let K(r) denote the closed, orientable 3-manifold produced by r-surgery on a tame knot K c S3. _

41 42 Wilbur Whittsn

PROPOSITION 3. if there exist prime knots M!ithisomorphic groups and non-homeomorphic complements, then there exist a non-tricial knot K and an integer m such that

(1) K(l/m) z S3, and (2) Irnl # 0, 1, or 2.

THEOREM 4 [4, 5-J. If K is a non-trivial knot and r E Q is not equal to 1 or - 1, then K(r) is not simply connected. Moreover, K (1) and K( - 1) cannot both be simply connected.

In Section 1, we prove Proposition 3 and Corollary 1. In Section 2, we prove two consequences of Theorem 4.

5 1. PROOFS OF PROPOSITION 3 AND COROLLARY 1

We shall work in the PL-category; let K be a knot in S3. Let E(K) denote the exterior of K, and let N(K) denote a tubular neighbourhood of K. Set XK = n, E (K) = 7c1(S3 -8 (K)). If the knots K1 and K2 belong to the same knot type, then we write K, = Kz; we write K = 0 to mean that K is trivial. Now fix an orientation of S3 and orient K. A meridian-longitude pair (p, I) for K on dE (K) will be oriented in the standard way with i. homologous to K in N(K) and with intersection number p. i. = + 1. Recall that if K # 0 and if p and q are relatively prime integers, then a simple closed curve, J (p, q; K), homologous on d E (K) to pp + qL, is a (p, q)-cable knot about K, provided that 141 2 2. Recall also that if q 2 2, then Schubert’s characterization [15] of cable knots says that the isotopy type { J } of J determines and is determined by the triple (p, q, {K 1).

Proof of Proposition 3. We assume that J1 and J2 are prime knots of which nJ1 z R J2 and E ( J1) $ E ( J2). Now both E ( J1 ) and E ( J2) must contain an essential annulus, for otherwise, E (J1) z E(J,) [6, theorem 10, p. 42; 123. (We note that S3 - J1 E S3 - Jz if and only if E(J,) z E(J2) [9, proposition 10.1, p. 743.) If either J1 or Jz is a , then so is the other [2], and it follows that J1 = J2 [14, p. 611. Hence, J1 and Jz are cable knots, J(p,, ql; K,) and J(p,, q2; K2), respectively, about non-trivial knots, K, and K2.

NOW there exist annuli A1 and Az and solid tori V, and V, such that E (Ji) E E (Ki) u K

(i = 1,2). By an application of Johannson’s deformation theorem [12; 11, theorem ?.21, p. 2123 or by [7, proof of theorem 2, p. 12891, we can find a homotopy equivalence f:E(J,)-+E(Jl)suchthatfIE(K,):E(K,)-+ E(K,) isa homeomorphism,f(A,) = A,,and .f 1VI : VI + V2 is a homotopy equivalence. Let &, &) be a meridian-longitude pair for Ki on dE(K,), and note that a (pi, ql)-curve on d E (K,) goes to a f (p2, q,)-curve on d E (K2). For homological reasons, we must have jpll = lpzl and lqll = lq21 [7; proof of corollary 2, p. 12921. Now we can change the orientation of K1 (or K,) along with the orientations of both p1 and i.l (or pz and &), if need be, to guarantee that q1 = q2 = q 2 2. Set p1 = p; note that p2 = ep, with EE { - 1, I}. For convenience, we replacefl E(K,) byf. Thus (homologically) we havef,(j.,) = + AZ, f,(~~)= f ~~+rni., (for some mEZ), andf,(pp,+ql,) = &- (eppZ+qj.2). Thus,

f*(~ +9&) = + ppz+(mp + 9)&, and it follows that either m = 0 or mp = f 2q. If m = 0, then we can extend f:E(K,) + E(K,) to a homeomorphism (S3, K,) + (S3, K,). Since a boundary component of Ai KNOT COMPLEMENTS AND GROUPS 43 belongs to the same knot type as .Ji(i = 1. 2) and since_/ (A,) = A2, it follows that J1 = J2. Thus. rnp = + 2q, and so m # 0. Also, 1 I Ip I I 2 and

and so lm 1 # 0, 1, or 2, since q 2 2. Nowf: E(K,)-+ E(K,) is a homeomorphism andf, (pI) = k~~+rnj.~, Hence, either Kz(l/m) Z (S3, K,) or K,( - l/m) 2 (S’, K,), and conclusion (1) of the proposition follows. Cl

Proof of Corollary 1. Let {K,, K2, . . . > be any collection of prime knots with rcKi ~ n Kj, for all i and j. By the Rigidity Theorem, we have E (Ki) z E (Kj), for all i and j. By [4, corollary 3, p. 431 or [S, corollary 31, the collection {K,, K,, . . > contains representa- tives from, at most, two distinct knot types. 0

52. TWO COROLLARIES OF THEORE\ 4

COROLLARY 5. If K is a knot with non-trivial Arf invariant, rhen K (1 /m) $ S3 when m # 0. In particular, K is determined by its complement.

Proof: Let x (K) denote the Arf invariant of K. By [9, theorem 4, p. 611, the p-invariant of K(l/m) is mX(K)(mod 2). Hence, if Irnj = 1 and if x(K) = 1, then K(l/m) $ S3. Since K(l/m) $ S3 when jrnl 2 2 (by Theorem 4), it follows that K(l/m) $ S3 when m # 0. 0 Since non-trivial surgery on a strongly never produces a fake 3-sphere (as pointed out in [S, p. 179]), our last conclusion follows immediately from Corollary 5.

COROLLARY 6. A strongly invertible knot with non-trivial Arf invariant has property P.

Remark. Casson [3] has recently shown that knots with non-trivial Arf invariant have property P, and Bleiler and Scharlemann [l] have shown the same for strongly invertible knots. Also, Culler et al. proved a result similar to our Corollary 5; see [4, corollary 5, p. 431 or [S, corollary 51.

Acknowledgements-The author thanks the Institute for Advanced Study and the University of Iowa for their hospitality while part of this work was done. The ONR grant NOO14-85-K-0099 provided partial support. The author is grateful to M. Boileau, F. Gonzllez-Acuiia, C. Gordon, K. IMurasugi. and J. Simon for helpful comments, and to F. GonzBlez-Acuiia in particular for Corollary 5.

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Department of Mathematics Unicersity of Southwestern Louisiana Lafayette, LA 70504 U.S.A.