GEOMETRIC PRESENTATIONS OF CLASSICAL KNOT GROUPS

The question addressed by thls paper is, how close is the tunnel number of a knot to the minimum number of relators in a presentation of the knot group? A dubious, but useful conjecture, is that these two Invarlants are equal. (The analogous assertion applied to 3-manifolds Is known to be false. [I]) It has been shown recently [2] that not all presentations of a knot group are "geometric". The maln result in this paper asserts that the tunnel number is equal to the nflnimum number of relators among presentations satisfying a somewhat restrictive condition, that Is, that such presentations are always geometric. I. DEFINITIONS AND CONVENTIONS. Let K be any non-trlvlal knot in $3; V CI(N(K) solid torus containing K; X CI(S3 V), the knot exterior. G the knot group I(S 3 K). If G has the deflclency-I presentation Let < gl,...,gn+ll rl,...,rn >, we will abuse notation and use gl and rj to denote loops representing the corresponding group elements. We restrict consideration to presentations having representative loops gl and rj such that there are disks Dj X wlth Dj rj and int(Dj) gl for all i,J. Let s s(K) nKnlmum number of relators in such a presentation. 0 CI(N( gl)), a neighborhood of a "bouquet" of generators, Is a handlebody of genus s+l. Let t t(K) be the tunnel number of K, which Is the minimum number of arcs that need to be attached to K so that the complement is an open handlebody. Equivalently, t(K) is the minimum number of l-handles (tunnels) that need to be removed from the knot exterior to obtain a Heegard decomposition of S 3. t(K) is also called the "Heegard" genus of K [3,2]. 290 J. ERBLAND AND M. GUTERRIEZ 2. THEOREM: s(K) L(K) for any tame knot K. The direction s(K) t(K) is easy: The fundamental group of the complement of K together with the t t(K) arcs is free on t+l generators. For each arc, a loop around that arc provides a relation. Such a presentation Is called geometric. Outline of proof, of t(K) g sg): For each rj, the corresponding disk Dj can be 3 moved so that 3Dj _3X0, int (Dj)_S X0. We build up a space Xs from X0 by stages using 2-handle surgeries along these disks; Xj is the result of the jth surgery. Xs fs the result of the last surgery, with S3 Xs N’(K), a solid torus neighborhood of K. Thus a Heegard decomposition of S3 can be obtained by drilling tunnels through the dlsks used in the 2-handle surgeries. The rest of this paper is concerned with filling in the details. 3. THE CONSTRUCTION. Let X XoUCI(N(DI)) by 2-handle surgery along DI; DI’ DI. For each J > I, for each I < J, if N(Di Dj then modify Dj to remove intersections. (Fig. la,b. Of course, there may be more intersections, but they must all be simple arcs or circles. In the latter case, the Intersections can simply be "capped off". In the case of multiple arcs, start with the outermost arc of the new disk and continue until all of the Intersections have been removed. The modification may result in an increase in the number of disks. If any disk is "trivial", simply "throw it out". We consider a disk to be trivial If the union of the disk with Xj_ and previous disks contalns an Incompressible 2-sphere. The determination of which disks are trivial is not unlque: after Dj has been modified, If extra dlsks result, then select one at a tlme. If you get an Incompressible sphere, discard that disk from consideration and move on to the next. Let Dj be the union of disks obtained from Dj after modification then removal of trivial disks. For j > I, let Xj be the space obtained from Xj_ by 2-handle surgeries along D’j. By mlnlmality of s, at least s 2-handle surgeries are performed. The boundary of Xj differs from that of Xj_ elther,by a decrease in the genus of one component or by an increase in the number of components. At the same time none of the boundary components can be a 2-sphere. Thus the end result of the construction Xs is bounded by torl only: X TO Tk. We need to show that k 0, where TO bounds a solid torus N’(K). 4. Before finishing the proof, we make some essential observations. (1) 7I(xs) 7!(X) G, the isomorphism being induced by inclusion, follows from a standard Selfert-Van Kampen argument. (ll) There is no 2-sphere SXs that separates two of these torl. Thls follows from the construction itself. (ill)For each I, l(Ti)/ l(Xs) is inJective. Otherwise there exists an essential loop in TI and a disk D in Xs bounded by . D {-I,I} together with part of the torus is a sphere that can be moved away from the torus and separates it GEOMETRIC PRESENTATIONS OF CLASSICAL KNOT GROUPS 291 from the other tort, contradicting remark (li). (iv) Each torus bounds a solid torus in S 3 on one side and hence a (possibly trivial) knot exterior on the other [4]. If > O, then the solid side of T must contain the knot; otherwise, there exists a contractible loop in the exterior of K which is not contractible in Xs, thus contradicting (i). (v) For i I, let BI= component of S3-Xs bounded by T i. Let B0 be the component of X X bounded by O N(K). By (iv) Bi is a knot exterior for s hence l(Ti)/ l(Bi) is inJective. For related reasons, I(To) I(Bo) is inJective. 5. We will now proceed to show that BI for i > 0. Let W 0 X s B0, Wj --Wj_ Bj for j > O. In particular, Wk X. For i O, we apply the Selfert-Van Kampen Theorem to the injectivlty of l(Ti)/ l(Xs) and I(TI)/ l(Bi) to get injectlvity of I(BI)/ l(Xs Bi) and l(Xs) l(Xs BI). In particular, l(Xs) I(Wo) is InJective. We apply the inductive hypothesis that l(Xs) I(Wj_I) is inJectlve, together with another application of Selfert-Van Kampen, to get InJectlvlty of I (Wj_I) I (Wj) and of "l(Xs U Bj) I(Wj)" Since Wk X, we have t(Xs l(Wo)* l(wl l(x) G. Since this composition is an isomorphism, all of the component injections must also be Isomorphisms. From this and the Selfert-Van Kampen construction, it follows that l(Xs) I(Xs Bi) and hence l(Bi) l(Ti) for i ) O, which is impossible for a knot exterior. Thus only B0 is non-empty. 6. It remains to verify that tnt(B0 V) N’(K), Just a fatter neighborhood of K, thus completing the proof. The boundary of B0 is the union of two tort, one of which ts BN(K). The inclusions in both cases must Induce isomorphisms of fundamental groups. Since I(Bo) I(T0) I(BN(K)) Z + Z is abelian, the Hurewlcz homomorphlsm is an 292 J. ERBLAND AND M. GUTERRIEZ isomorphism onto HI(B0). It follows that the merldional loops m and m2 on the bounding tori are homologous. Let a be a path from the base point of m to the base point of m2. Then mlm2 a is homogous to 0 and hence homotoplcally trivial. Thus the meridians bound an annulus and the torus TO is paraltel to N(K), which suffices to complete the proof. REFERENCES I. MONTESINOS, J.M. Note on a result of Boileau-Zieschang, in Low-dimensional Topology and Klelnlan Groups, LMS Lecture Notes Set. 112 (1986), 241-252. 2. BOILEAU, M., ROST, M., ZIESCHANG, H. On Heegard Decompositions of Torus Knot Exteriors and Related Selfert Fibre Spaces, Math. Ann. 279 (1988), 553-581. 3. CLARK, B. The Heegard genus of manifolds obtained by surgery on links and knots Int. J. Math. & Math. Scl. 3 (1980), 583-589. 4. ROLFSEN, D. Knots and Links, Publish or Perish Inc. (1976).

[I]) It has been shown recently [2] that not all presentations of a knot group are "geometric".The maln result in this paper asserts that the tunnel number is equal to the nflnimum number of relators among presentations satisfying a somewhat restrictive condition, that Is, that such presentations are always geometric.

I. DEFINITIONS AND CONVENTIONS.
Let K be any non-trlvlal knot in $3; V CI(N(K) solid torus containing K; X CI(S 3 V), the knot exterior.
G the knot group I(S

K).
If G has the deflclency-I presentation Let < gl,...,gn+ll rl,...,r n >, we will abuse notation and use gl and rj to denote loops representing the corresponding group elements.
We restrict consideration to presentations having representative loops gl and rj such that there are disks Dj X wlth Dj rj and int(Dj) gl for all i,J.

Let s s(K)
nKnlmum number of relators in such a presentation.X 0 CI(N( gl)), a neighborhood of a "bouquet" of generators, Is a handlebody of genus s+l.Let t t(K) be the tunnel number of K, which Is the minimum number of arcs that need to be attached to K so that the complement is an open handlebody.Equivalently, t(K) is the minimum number of l-handles (tunnels) that need to be removed from the knot exterior to obtain a Heegard decomposition of S 3. t(K) is also called the "Heegard" genus of K [3,2].
The direction s(K) t(K) is easy: The fundamental group of the complement of K together with the t t(K) arcs is free on t+l generators.
For each arc, a loop around that arc provides a relation.Such a presentation Is called geometric.
Outline of proof, of t(K) g sg): For each rj, the corresponding disk Dj can be 3 moved so that 3Dj _3X0, int (Dj)_S X 0. We build up a space X s from X 0 by stages using 2-handle surgeries along these disks; Xj is the result of the jth surgery.X s fs the result of the last surgery, with S 3 X s N'(K), a solid torus neighborhood of K. Thus a Heegard decomposition of S 3 can be obtained by drilling tunnels through the dlsks used in the 2-handle surgeries.
The rest of this paper is concerned with filling in the details.
Let X XoUCI(N(DI)) by 2-handle surgery along DI; D I' D I. For each J > I, for each I < J, if N(D i Dj then modify Dj to remove intersections. (Fig.
la,b.Of course, there may be more intersections, but they must all be simple arcs or circles.In the latter case, the Intersections can simply be "capped off".In the case of multiple arcs, start with the outermost arc of the new disk and continue until all of the Intersections have been removed.The modification may result in an increase in the number of disks.If any disk is "trivial", simply "throw it out".We consider a disk to be trivial If the union of the disk with Xj_ and previous disks contalns an Incompressible 2-sphere.The determination of which disks are trivial is not unlque: after Dj has been modified, If extra dlsks result, then select one at a tlme.
If you get an Incompressible sphere, discard that disk from consideration and move on to the next.
Let Dj be the union of disks obtained from Dj after modification then removal of trivial disks.
For j > I, let Xj be the space obtained from Xj_ by 2-handle surgeries along D'j.By mlnlmality of s, at least s 2-handle surgeries are performed.The boundary of Xj differs from that of Xj_ elther,by a decrease in the genus of one component or by an increase in the number of components.At the same time none of the boundary components can be a 2-sphere.Thus the end result of the construction X s is bounded by torl only: X T O T k.
We need to show that k 0, where T O bounds a solid torus N'(K). 4. Before finishing the proof, we make some essential observations.
(iv) Each torus bounds a solid torus in S 3 on one side and hence a (possibly trivial) knot exterior on the other [4].If > O, then the solid side of T must contain the knot; otherwise, there exists a contractible loop in the exterior of K which is not contractible in Xs, thus contradicting (i).
(v) For i I, let BI= component of S3-Xs bounded by T i.Let B 0 be the component of X X bounded by T O N(K).By (iv) B i is a knot exterior for s hence l(Ti)/ l(Bi) is inJective.For related reasons, I(To) I(Bo) is inJective.
5. We will now proceed to show that B I for i > 0.
Let W 0 X s B0, Wj --Wj_ Bj for j > O.In particular, W k X.
For i O, we apply the Selfert-Van Kampen Theorem to the injectivlty of l(Ti)/ l(Xs) and I(TI)/ l(Bi) to get injectlvity of I(BI)/ l(Xs B i) and l(Xs) l(Xs BI).In particular, l(Xs) I(Wo) is InJective.We apply the inductive hypothesis that l(Xs) I(Wj_I) is inJectlve, together with another application of Selfert-Van Kampen, to get InJectlvlty of I (Wj_I) I (Wj) and of "l(Xs U Bj) I(Wj)" Since W k X, we have t(Xs l(Wo)* l(wl l(x) G.
Since this composition is an isomorphism, all of the component injections must also be Isomorphisms.From this and the Selfert-Van Kampen construction, it follows that l(Xs) I(Xs B i) and hence l(Bi) l(Ti) for i ) O, which is impossible for a knot exterior.Thus only B 0 is non-empty.

6.
It remains to verify that tnt(B 0 V) N'(K), Just a fatter neighborhood of K, thus completing the proof.
The boundary of B 0 is the union of two tort, one of which ts BN(K).
The inclusions in both cases must Induce isomorphisms of fundamental groups.Since I(Bo) I(T0) I(BN(K)) Z + Z is abelian, the Hurewlcz homomorphlsm is an