We prove that the tunnel number of the sum of n knots is at least n.Comment: 8 pages. To appear in Topolog

Scharlemann, Martin

Schultens, Jennifer

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Martin Scharlemann

Jennifer Schultens

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Topology

THE TUNNEL NUMBER OF THE SUM OF n KNOTS IS AT LEAST n

AbstractWe prove that the tunnel number of the sum of n knots is at least n

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Topology Vol. 38, No. 2, pp. 265—270, 1999( 1998 Elsevier Science LtdAll rights reserved. Printed in Great Britain0040-9383/98 $— see front matterPII: S0040-9383(98)00002-0THE TUNNEL NUMBER OF THE SUM OF n KNOTSIS AT LEAST nMARTIN SCHARLEMANN and JENNIFER SCHULTENS(Received 10 August 1997; in revised form 20 January 1998)We prove that the tunnel number of the sum of n knots is at least n. ( 1998 Elsevier Science Ltd. All rightsreserved.1. INTRODUCTIONIn [5], Norwood showed that tunnel number 1 knots are prime. This led to the moregeneral conjecture, see for instance [4, Problem 1.70B], that the tunnel number of a sum ofn knots is at least n. Here we prove this conjecture. The idea is to show that the splittigsurface of a Heegard splitting corresponding to a tunnel system realizing the tunnel numberof the sum of n knots intersects each individual knot complement essentially. Thena sophisticated Euler characteristic argument, based on the idea of untelescoping theHeegaard splitting, yields the result.2. PRELIMINARIESFor standard deÞnitions concerning knots, see [1] or [6] and for those concerning3-manifolds, see [2] or [3].DeÞnition 1. Let N be a submanifold of M, we denote an open regular neighborhood ofN in M by g (N).DeÞnition 2. Let K be a knot in S3. Denote the complement of K, S3!g(K), by C(K).Remark 1. Let K"K1dK2be the sum of two knots. Then the decomposing spheregives rise to a decomposing annulus A properly embedded in C(K) such thatC(K)"C(K1)XAC(K2). If K"K1d2dKn , then we may assume that the decomposingspheres are nested, so that C (K)"C(K1) XA1 2XAn~1C(Kn ).DeÞnition 3. A „unnel system for a knot K is a collection of disjoint arcsT"t1X2Xtn properly embedded in C(K) such that C(K)!g(T) is a handlebody. Thetunnel number of K, denoted by t (K), is the least number of arcs required in a tunnel systemfor K.DeÞnition 4. A compression body is a 3-manifold … obtained from a connected closedorientable surface S by attaching 2-handles to S]M0NLS]I and capping o⁄ any resulting2-sphere boundary components. We denote S]M1N by L‘… and L…!L‘… by L~….265DeÞnition 5. A set of deÞning disks for a compression body … is a set of disksMD1,2, DnN properly embedded in … with LDiLL‘… for i"1,2, n such that the resultof cutting … along D1X2XDn is homeomorphic to L~…]I.DeÞnition 6. A Heegaard splitting of a 3-manifold M is a decomposition M"»XS… inwhich », … are compression bodies such that »W…"L‘»"L‘…"S and M"»X….We call S the splitting surface or Heegaard surface.DeÞnition 7. Let M"»XS… be an irreducible Heegaard splitting. We may think ofM as being obtained from L~»]I by attaching all 1-handles dual to 2-handles in» followed by all 2-handles in …, followed, perhaps, by 3-handles. An untelescoping ofM"»XS… is a rearrangement of the order in which the 1-handles (of ») and the 2-handles(dual to the 1-handles of …) are attached. This rearrangement is chosen so that M isdecomposed into submanifolds M1,2, Mm, such that M1WMi‘1"Fiand Fiis anincompressible surface in M, and such that the Miinherit, from a subcollection of theoriginal 1-handles and 2-handles, strongly irreducible Heegaard splittings M1"»1XS1…1,2, Mm"»mXSm…m. Unless M is a lens space or S1]S2, no S1,2, Smisa torus. For details see [8,7]. We denote the untelescoping of M"»XS… byM"(»1XS1…1) XF12XFm~1 (»mXSm …m). For convenience, we will occasionally denoteL~»"L~»1by F0.LEMMA 2. s(S)"+mi/1s (Si)!&m~1i/1s (Fi).Proof. Let M"»XS… be a Heegaard splitting, thens(S)"s(L~»)!2(d(1-handles attached in » )!d(0-handles attached in » ) )and in an untelescoping,s(Si)"s (L~»i)!2(d(1-handles attached in »i)!d (0-handles attached in »i) )"s (Fi~1)!2(d(1-handles attached in »i)!d (0-handles attached in »i) ).So, since 1-handles are merely reordered in an untelescoping,s(S )"s (L~»i)!2 m+i/1d((1-handles attached in »i)!d (0-handles attached in »i))"s (L~»)! m+i/1s (Fi~1)# m+i/1s(Si). KLEMMA 3. ‚et P be a properly embedded incompressible surface in an irreducible 3-manifold M and let M"(»1XS1…1)XF12XFm~1 (»mXSm…m ) be an untelescoping ofa Heegaard splitting M"»XS…. „hen (Zm~1i/1Fi)X(Zmi/1Si) can be isotoped to intersectP only in curves essential in P.Remark 4. This lemma demonstrates the advantage of working with untelescopings ofHeegaard splittings rather than Heegaard splittings. It is a deep fact that the splittingsurface of a strongly irreducible Heegaard splitting can be isotoped to intersect a properlyembedded incompressible surface only in curves essential in this surface. This fact is provenfor instance in [9, Lemma 6].266 M. Scharlemann and J. SchultensProof. Here (Zm~1i/1Fi) may be isotoped to intersect P only in curves essential in P bya standard innermost disk argument, since both are incompressible. Then Pi"PWMiisa properly embedded incompressible surface in Mi. It follows that each Simay be isotopedin Mito intersect Pionly in curves essential in Pi, by [9, Lemma 6]. Note that the latterisotopies Þx (Zm~1i/1Fi). KLEMMA 5. ‚et K be a prime knot and let A be an annulus properly embedded in C(K) suchthat the components of LA are meridians. „hen A is boundary parallel.Proof. In S3, A can be extended to a sphere by adding two meridian disks. This sphereintersects K in two points. Since K is prime, one side of the sphere contains a singleunknotted arc. KLEMMA 6. ‚et P be an incompressible surface in a compression body …. „hen the result ofcutting … along P is a collection of compression bodies.Proof. This is [9, Lemma 2]. KRemark 7. In the above lemma, P need not be connected.LEMMA 8. ‚et ifA is a collection of incompressible annuli in a compression body …, thenin any component X of …!A, s(L‘…WX) s(L~…WX).Proof. Let D be a set of deÞning disks for …. We argue by induction on the pair(Ds(L~… )!s (L‘… ) D, DAWD D). If Ds (L~…)!s (L‘… ) D"0, then (D"0 and) all annuli arespanning annuli and the result follows.To complete the inductive step, suppose there is a disk D in D such that DWA"0. Theresult of cutting … along D is a compression body … @ with Ds(L~… @ )!s (L‘… @ ) D(Ds(L~…)!s(L‘… ) D, or two compression bodies … @ and …A withDs(L~… @)!s(L‘… @) D(Ds(L~…)!s (L‘…) D and Ds(L~…)!s (L‘…A ) D(Ds(L‘…) D. Thecomponents of …!A can be obtained from the components of … @!A or of … @!Aand …A!A by attaching a 1-handle either to a single component or so as to connect twocomponents. In both cases, the result follows from the inductive hypotheses.If there is no such disk, considerDWA. If there is an arc a inDWA that is inessential inA, then we may assume that a is outermost in A, and we may cut the disk D inD containing a along a and paste on two copies of the disk cut o⁄ by a inA to obtain a diskD@. Replacing D by D@ inD produces a new set of deÞning disksD@ with DAWD@ D(DAWD D.If all arcs inDWA are essential inA, let b be an arc inDWA that is outermost inD. LetA be the annulus inA that gives rise to b. Cutting and pasting A along b and the outermostdisk cut o⁄ inD yields a disk D@ disjoint fromA. If D@ is inessential, then A is inessential andcan be ignored. (Since cutting along A does not alter any components or their Eulercharacteristics.) If D@ is essential, the result follows as above. This completes the inductivestep. K3. THE COMBINATORICSIn the following, we consider a tunnel system T, realizing the tunnel numberof K1d2dKn . We also consider the Heegaard splitting C(K1d2dKn )"»XS…TUNNEL NUMBER 267corresponding to T and an untelescoping C(K1d2dKn )"(»1XS1…1)XF12XFm~1(»mXSm…m) of C(K1d2dKn )"»XS…. Set Mi"»iX…i . By Remark 1,C(K1d2dKn)"C(K1)XA1X2XAn~1C(Kn ). We will always assume thatL~»1"LC(K1d2dKn) and that Zm~1i/1 Fi and Zm~1i/1 Si intersect Zn~1j/1Ai only in curvesessential in Zn~1j/1Ai. We will, furthermore, assume that, subject to these constraints, thenumber of intersections of Zm~1i/1Fiand Zmi/1Siwith Zn~1j/1Ajis minimal.DeÞnition 8. Set Sij"SiWC(Kj), Fij"FiWC(Kj) and Aij"MiWAj.LEMMA 9. For all i, j, s (Sij) and s (Fij) are even.Proof. Here Fiis separating, so FiWAj~1is separating. Since LAj~1LLC(K1d2dKn) which is a torus, hence connected, both components lie on one sideof Fi, hence DFiWAj~1D is even. The same is true for DFiWAjD. Thus s (FiWC(Kj) )"2!2(genus(FiWC(Kj) ))!DFiW(Aj~1XAj) D is even. Similarly for Si. KDeÞnition 9. Set xij"!1/2s(Fij) and yij"!1/2s (Sij).LEMMA 10. ”nder the assumptions above, yij*maxMxi~1j, xijN.Proof. This follows from Lemma 8. KLEMMA 11. For all j, there is an i, such that yij’0.Proof. Suppose yij"0 for i"1,2, m. Then xij"0 for i"1,2, m!1. SoGj"(Zm~1i/1Fij)X(Zmi/1Sij)LC(Kj)is a collection of annuli and tori. Since the tori arise only in Zm~1i/1Fij, they are incompress-ible separating tori. Thus if a torus component „ of Fiis in C(Kj), then so is a component ofSi{, which cannot be a torus, for some i@. But this would contradict yi{j"0. HenceGjconsists entirely of annuli. By Lemma 5, the annuli are all boundary parallel. Hencecutting C(Kj) along the annular components of Gjyields a copy of C(Kj). By Lemma 6, allcomponents of C(Kj) cut along G are compression bodies, a contradiction. KLEMMA 12. For all j,m+i/1yij’m~1+i/1xij.Proof. This follows by comparing the tables in Fig. 1. By Lemma 10, the largest valueencountered in a given column of the table in Fig. 1(a) occurs one time more often in thecorresponding columns of the table in Fig. 1(b). If the largest value encountered in a columnin the table in Fig. 1(a) is zero, then by Lemma 11, there must be nonzero entries in thecorresponding column of the table in Fig. 1b. KRemark 13. Since all numbers involved are integers, it follows that +mi/1yij*1#+m~1i/1xij, for all j.THEOREM 14. t(K1d2dKn)*n.268 M. Scharlemann and J. SchultensFig. 1.Proof. Heren+j/1Am+i/1yijB*n+j/1A1#m~1+i/1xijB"n#n+j/1m~1+i/1xijHence,n+j/1m+i/1yij! n+j/1m~1+i/1xij*n.Thus,n+j/1m+i/1!2(yij)! n+j/1m~1+i/1!2(xij) !2nand by deÞnitionn+j/1m+i/1s(SiWC(Kj))! n+j/1m~1+i/1s(FiWC(Kj) ) !2n.So,s(S)" m+i/1s (Si)!m~1+i/1s(Fi) !2n.Whencegenus(S)*n#1andt(K1d2dKn)*n.KTUNNEL NUMBER 269Acknowledgements—The Þrst author is supported by an NSF grant. The second author is supported by an NSFpostdoctoral fellowship. We wish to thank MSRI, where part of this work was carried out.REFERENCES1. Burde, G., Zieschang, H., Knots, Studies in Mathematics, Vol. 5. de Gruyter, Berlin, 1985.2. Hempel, J., Annals of Mathematial Studies, 1976, 86.3. Jaco, W., Lectures on Three-manifold Topology, Regional Conference Series in Mathematics, Vol. 43. Amer.Math. Soc., Providence, RI, 1981.4. Kirby, R., ProblemÕs in low-dimensional topology. In Geometric „opology—Proceedings of the 1993 GeorgiaInternational „opology Conference, ed. W. H. Kazez, AMS/IP Studies in Advanced Mathematics, InternationalPress, Cambridge, MA, 1997, Vol. 2, Part 2, pp. 35—473.5. Norwood, F.H., Proceedings of the American Mathematical Society, 1982, 86, 143—147.6. Rolfsen, D., Knots and Links, Publish or Perish, Houston, Texas.7. Scharlemann, M. and Thompson, A., AMS Contemporary Mathematics, 1994, 164, 231—238.8. Scharlemann, Heegaard splittings of compact 3-manifolds. Handbook of Geometric „opology, eds R. Davermanand R. Sherr. Elsevier, Amsterdam (to appear).9. Schultens, J., Additivity of tunnel number. Preprint.Department of Mathematics”CSBSanta Barbara, CA 93106”.S.A.Department of Mathematics and Computer ScienceEmory ”niversityAtlanta, GA 30322”.S.A.270 M. Scharlemann and J. Schultens

THE TUNNEL NUMBER OF THE SUM OF n KNOTS IS AT LEAST n 

The tunnel number of the sum of n knots is at least n

The tunnel number of the sum of n knots is at least n

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