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# SOC DEVIANT BEHAVR SOC 170

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This 28 page Class Notes was uploaded by Russel Monahan on Thursday October 22, 2015. The Class Notes belongs to SOC 170 at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 26 views. For similar materials see /class/226874/soc-170-university-of-california-santa-barbara in Sociology at University of California Santa Barbara.

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GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE MARTIN SCHARLEMANN ABSTRACT There is a relation between the generalized Property R Con jecture and the Schoen ies Conjecture that suggests a new line of attack on the latter The new approach gives a quick proof of the genus 2 Schoen ies Conjecture and suf ces to prove the genus 3 case even in the absence of new progress on the generalized Property R Conjecture 1 INTRODUCTION AND PRELIMINARIES The Schoen ies Conjecture asks whether every FL or equivalently smooth 3sphere in S4 divides the 4sphere into two PL balls The appeal of the con jecture is at least 3fold o The topological version for locally at embeddings is known to be true in every dimension Both the PL and the smooth versions when properly phrased to avoid problems with exotic structures are known to be true in every other dimension 0 If the Schoen ies Conjecture is false then there is no hope for a PL prime decomposition theorem for 4manifolds for it would imply that there are 4manifolds X and Y not themselves 4spheres so that XY 2 S4 0 The Schoen ies Conjecture is weaker than the still unsolved 4 dimensional PL Poincare Conjecture and so might be more acces sible Little explicit progress has been made on the Schoen ies Conjecture for several decades a time which has nonetheless seen rapid progress in our understanding of both 3 and 4dimensional manifolds Here we outline how the Schoen ies Conjecture is connected to another important problem on the border between classical 3 and 4dimensional topology namely the generalized Property R Conjecture and we begin to outline how the last two decades of progress in combinatorial 3dimensional topology can be Date March 20 2006 Research partially supported by a National Science Foundation grant Thanks also to Catalonia s Centre Recerca Matematica for their extraordinary hospitality while this work was being completed 1 2 MARTIN SCHARLEMANN used to extend the proof of the Schoen ies Conjecture from what are called genus 2 embeddings of S3 in S4 to genus 3 embeddings We work in the PL category throughout All manifolds discussed are orientable 2 GENERALIZED PROPERTY R Recall the famous Property R theorem proven in a somewhat stronger form by DaVid Gabai Ga2 Theorem 21 Property R If O frameal surgery on a knot K C S3 yields S1 gtlt S2 then K is the unknot It is wellknown indeed it is perhaps the original motivation for the Prop erty R Conjecture that Property R has an immediate consequence for the handlebody structure of 4manifolds Corollary 22 Suppose U 4 is a homology 4 sphere anal has a handle struc ture containing exactly one 2 hanalle anal no 3 hanalles Then U is the 4 sphere Proof Since U has a handle structure with no 3handles dually it has a handle structure with no lhandles In order for U to be connected this dual handle structure must then have exactly one 0handle so the original handle structure has a single 4handle Let U C U be the union of all 0 and lhandles of U andM 9U U can be thought of as the regular neighborhood of a graph or collapsing a maximal tree in that graph as the regular neighborhood of a bouquet of circles The 4dimensional regular neighborhood of a circle in an orientable 4manifold is S1 gtlt D3 so U is the boundary connected sum hS1 gtlt D3 some n 2 0 Explicitly the number of summands n is one more than the difference between the number of lhandles and 0handles ie 1 7 x where x is the Euler characteristic of the graph It follows thatM S1 gtlt S2 and in particular H2 M E Zquot Now consider the closed complement U of U in U Via the dual handle structure U is obtained by attaching a single 2handle to B4 so it deformation retracts to a 2sphere and in particular H2 U E Z Since U is a homology 4sphere and H2U 0 it follows from the MayerVietoris sequence H3U 0gtH2MgtH2U H2UgtH2U 0 that Z 21mm 2 Zquot so n 1 andM S1gtlt 32 On the other hand U whose handle structure dual to that from U consists of a 0handle and a 2handle is Visibly the trace of surgery on a knot in S3 namely the attaching map of the 2handle The framing of the surgery is triVial since the generator of H2U is represented by gtk gtlt S2 C GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE 3 S1 gtlt S2 gM and this class visibly has trivial selfintersection Since the result of 0framed surgery on the knot is M S1 gtlt S2 the knot is trivial by Property R Theorem 21 so U is simply S2 gtltD2 Hence U is the boundary union S1 gtlt D3 Ua S2 gtlt D2 Of course the same is true of S4 since the closed complement of a neighborhood of the stan dard 2sphere in S4 is S1 gtlt D3 So we see that U can be obtained from S4 by removing the standard S1 gtlt D3 and pasting it back in perhaps differently But it is wellknown and is usefully extended to all 4dimensional handle bodies by Laudenbach and Poenaru LP that any automorphism of S 1 gtlt S2 extends to an automorphism of S1 gtlt D3 so the gluing homeomorphism ex tends across S1 gtlt D3 to give a homeomorphism of U with S4 The generalized Property R conjecture cf Kirby Problem 182 says this Conjecture 1 Generalized Property R SupposeL is a framed link of n 2 1 components in S3 and surgery onL via the speci ed framing yields n S 1 gtlt S2 Then there is a sequence of handle slides on L cfKi that converts L into a 0 framed unlink In the case n 1 no slides are possible so Conjecture 1 does indeed di rectly generalize Theorem 21 On the other hand for n gt 1 it is certainly necessary to include the possibility of handle slides For if one starts with the 0framed unlink of ncomponents and does a series of possibly compli cated handleslides the result will be a possibly complicated framed link L of ncomponents The result of doing the speci ed framed surgery on L will necessarily be the same cf Ki as for the original unlink namely S1 gtlt S2 but L itself is no longer the unlink The example L is still consistent with Conjecture 1 since simply reversing the sequence of handle slides will convertL back to the framed unlink So in some sense Conjecture l is the broadest plausible generalization of Theorem 21 The generalized Property R Conjecture naturally leads to a generalized Corollary 22 Proposition 23 Suppose Conjecture 1 is true Then any homology 4 sphere U with a handle structure containing no 3 handles is S Proof Again focus on the 3manifold M that separates U the manifold after the 0 and lhandles are attached from its closed complement U in U The dual handle structure on U shows that U is constructed by attaching some 2handles to B4 On the other hand the original handle structure shows that U is the regular neighborhood of a graph so as before for some n U g hS1 gtltD3 andM g S1 gtlt S2 In particular H2M g Zquot Since U is a homology 4sphere and H2 UL 0 it follows as before from the MayerVietoris sequence that H2 U Zquot Hence U must be 4 MARTIN SCHARLEMANN obtained from B4 by attaching exactly n 2handles Then the generalized property R conjecture would imply that U hS2 gtlt D2 It is shown in LP that any automorphism of S1 gtlt S2 annsl gtlt D3 extends to an automorphism of hS1 gtltD3 This is not quite stated explicitly in LP beyond the observation on p 342 mark that no diffeomorphism of X P was needed here Hence the only manifold that can be obtained by gluing U to U alongMis S4 D The Proposition suggests this possibly weaker conjecture Conjecture 2 Weak generalized Property R conjecture Suppose attach ing n 2 handles to B4 gives a manifold W whose boundary is S1 gtlt S2 Then W 2 M32 gtlt D2 We have then Proposition 24 The weak generalized PropertyR conjecture Conjecture 2 is equivalent to the conjecture that any homology 4 sphere U with a handle structure containing no 3 handles is S4 Proof The proof of Proposition 23 really required only Conjecture 2 so only the converse needs to be proved Suppose we know that any homology 4sphere with a handle structure containing no 3handles is S4 Suppose W is a 4manifold constructed by attaching n 2handles to B4 and BW is S1 gtlt S2 Consider the exact sequence ofthe pair W9W 0 H3W9WgtH29WgtH2WgtH2W9WgtH19WgtH1W 0 Since the last two nontrivial terms are both Zquot the inclusion induces an isomorphism of the rst two nontrivial terms H2 8WaH2 Attach V hS1 gtltD3 to W by a homeomorphism of their boundaries and call the result U There is an obvious homeomorphism of boundaries and any other one will give the same 4manifold per LP Then the MayerVietoris sequence for the pair WVshows that U is a homology 4sphere hence under our assumption U S4 V C U is just a regular neighborhood ofthe wedge of n circles F Since U is simply connected F is homotopic to a standard ie planar wedge of circles in U whose complement is hS2 gtltD2 In dimension 4 homotopy of lcomplexes implies isotopy apply general position to the levelpreserving mapl gtltaUgtltI so infacthhnS gtltD asrequired D Setting aside conjecture here is a concrete extension of Property R Proposition 25 Suppose a 2 handle is attached to a genus n 4 dimensional handlebodyN hS1 gtltD3 and the resulting 4 manifoldN has boundary n1sl gtlt 32 Then N 2 in1sl gtltD3 GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE 5 Proof The proof is by induction on n when n 1 this is Property R Sup pose then that n gt 1 and letK C 9hnS1gtltD3 nS1gtlt S2 be the attach ing map for the 4dimensional 2handle The hypothesis is then that surgery on K yields n1S1 gtlt S2 a reducible manifold But examining the possi bilities in Sch2 we see that this is possible only if n 31 gtlt 32 7K is itself reducible so in particular one of the nonseparating 2spheres gtlt S2 is disjoint from K Following LP this 2sphere bounds a 3ball in N SplitN along this 3ball convertingN to n1S1 gtlt D3 and 9N to rkz S1 gtlt S2 By inductive hypothesis the split open N is mpg S1 gtltD3 so originally N 2 in1sl XD3 D Remark Experts will note that rather than use Sch2 one can substi tute the somewhat simpler Gal Ifn gt 1 then H2 nS1 gtlt S2 7 n 31 0 Since both 00 and 0framed surgery on K yield reducible hence nontaut 3manifolds from Gal it follows that n S1 gtlt S2 7 17K is itselfnot taut hence is reducible 3 APPLICATION HEEGAARD UNIONS Let H quot hnS1 gtlt D2 denote a 3dimensional genus n orientable handle body and Jquot hnS1 gtlt D3 denote a 4dimensional genus n orientable han dlebody H quot and Jquot can also be thought of as regular neighborhoods in re spectively R3 and R4 of any graph F with Euler characteristic 950 l 7 n De nition 31 Suppose for some p0p1p2 E N Hquot0 is embedded into both 9191 and 9192 so that its complement in each aJp J 12 is also a handlebody Then the 4 manifold W Jpl UHpo Jquot2 is called the Heegaard union ofthe 191 along Hp See Figure 1 copies of H H Hp 0 FIGURE 1 The term Heegaard union comes from the fact that H 90 is half of a Hee gaard splitting of both 91 and 91192 Moreover if W is such a Heegaard 6 MARTIN SCHARLEMANN union then 9191 7H UaHpO 9192 7H9 is a Heegaard splitting of 9W The construction here is tangentially related to the construction in BC 24 of a 4dimensional cobordism between three Heegaardsplit 3manifolds Indeed if two of the three 3manifolds in the BirmanCraggs construction are ofthe form iS1 gtlt S2 and are then lled in with copies of S1 gtlt D3 the result is a Heegaard union Lemma 32 Ifa Heegaard union W Jpl UJPZ is a rational homology ball then 0 p1 p2 Proof The rst and second homology groups rational coef cients of W are trivial so the result follows from the MayerVietoris sequence of W Jpl UHpo Jpzi H2ltWgt 0gtH1Hp gtH1Jpl H1ltJp2gtH1W 0 D Proposition 33 Suppose aHeegaaral union W Jpl UHpO Jquot2 is a homol ogy ball anal 9W 2 S3 I f the weak generalized propertyR conjecture Con jecture 2 is true for minp1p2 components then W B4 Proof Suppose with no loss of generality that pl 3 p2 Let Ji denote Jp i 12 and H0 denote H90 Consider the genus p0 Heegaard split ting of 912 given by H0 UaHo 8J2 7H0 According to Waldhausen Wa there is only one such Heegaard splitting of 9 up to homeomorphism obtained as follows Regard J2 as the product of the interval with a genus p2 3dimensional handlebody H Then H gtlt 0 C 9H gtlt I 912 and 912 7 H gtlt are both 3dimensional handlebodies The resulting Hee gaard splitting of 912 is called the product splitting It can be regarded as the natural Heegaard splitting of 912 2 pz S1 gtlt S2 Any other Heegaard splitting eg the genus p0 splitting at hand is homeomorphic to a stabiliza tion of this standard splitting As proven in LP and noted above any automorphism of 912 extends over J2 itself so we may as well assume that the Heegaard splitting H 0 UaHo 8J2 7H0 actually is a stabilization of the product splitting In particular and most dramatically if p0 p2 then no stabilization is required so J2 is justHO gtlt I and W 211 Much the same is true ifpo p2 1 most ofHO is just H so its attachment to J1 has no effect on the topology ole The single stabilization changes the picture slightly and is best conveyed by consider ing what the effect would be of attaching a 4ball to J1 not along one side of the minimal genus splitting of BB4 ie along B3 C S3 which clearly leaves J1 unchanged but rather along one side of the oncestabilized splitting of BB4 That is B4 is attached to J1 along a solid torus unknotted in BB4 But this is exactly a description of attaching a 2handle to J1 So W can be GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE 7 viewed as J1 with a single 2handle attached In the general situation in which the product splitting is stabilized po 7 p2 times W is homeomorphic to J1 with po 7 p2 2handles attached The result now follows from Lemma 32 and Proposition 24 D Remark The link along which the 2handles are attached has p1 com ponents and viewed in S3 is part of a genus p0 Heegaard splitting So its tunnel number can be calculated pl 7 l tunnels are needed to connect the link into a genus p1 handlebody and another po 7 p1 are needed to make it half of a Heegaard splitting Hence the tunnel number is po 7 1 This fact may be useful but anyway explains why Schl could be done just knowing Property R for tunnel number one knots Corollary 34 Suppose a H eegaard union W Jpl UHpo Jquot2 is a homology ball and W 2 331pr g 3 then W 34 Proof By Lemma 32 p1 p2 S 3 hence minp1p2 S l The result then follows from Proposition 33 and Theorem 21 D 4 HANDLEBODY STRUCTURE ON 3MANIFOLD COMPLEMENTS Suppose M C S4 is a connected closed FL or smooth 3submanifold In this section we discuss the handlebody structure of each complementary component of M It is a classical result cf KL thatM can be isotoped so that it is in the form of a recti ed critical level embedding We brie y review what that means Informally the embeddingM C S4 is in the form of a critical level embed ding if it has a handlebody structure in which each handle is horizontal with respect to the natural height function on S4 andM intersects each region of S4 between handle levels in a vertical collar of the boundary of the part of M that lies below or symmetrically above More formally regard S4 as the boundary ofD4 gtlt 71 1 so S4 consists oftwo 4balls D4 gtlt i1 called the poles added to the ends of S3 gtlt 711 Let p S3 gtlt 71177ll be the natural projection For 71 lt t lt 1 denote p710 by SEE Then M C S3 gtlt 71 l C S4 is a critical level embedding ifthere are a collection t1 lt Q lt lt in of values in 711 and a collection of closed surfaces F1F C S3 sothat 1 itlztnl C 7171 2 for each 1 g i g n 71M S3 gtlttt1F gtlt tun1 3 M S2 B3 with boundary F1 4 For each 2 S i S n F is obtained from E71 by a 39surgery some 0 S j S 3 That is there is a 3ball Dj gtltD3 j C S3 incident to E71 8 MARTIN SCHARLEMANN in BB gtltD3 j andE is obtained from E71 by replacing aDj gtltD3 j withD gtlt aDH 5 for each 2 S i S nM S2 is the trace ofthe surgery above That is it is the union ofFFl F andDj gtltD3 j Such an embedding gives rise to a handle structure onM with n handles added successively at levels t1 tn j is the indeX ofthe handle DJ gtltD3 j A critical level embedding is called recti ed if for 0 S j S 2 each handle of index j occurs at a lower level than each handle of index j 1 Furthermore all 0 and 1handles lie below S8 and all 2 and 3handles lie above S8 See Figure 2 D4 north pole all 2 amp 3handles attached above D s3x0 all 0 amp 1handles attached below 4 D south pole FIGURE 2 Note that the surface M S8 is a Heegaard surface for M since all 0 and 1handles lie on one side namely in S3 gtlt 710 and all 2 and 3handles lie on the other S3 gtlt 0 1 In particularM S8 is connected It is easy to see Sch1 Lemma 14 though not completely obvious that if the rst 1 handle attached to the boundary of a 0handle is incident to the 0handle at only one end then the handles cancel and there is a recti ed embedding of M in which neither handle appears So minimizing the number of handles we will henceforth assume that the rst 1handle incident to each 0handle is incident to it in both ends Equally important is the dual to this remark the boundary of the core of any 2handle is essential in the surface to which the 2handle is attached To summarize Lemma 41 Any recti ed critical level embedding of M may be isotoped rel M S8 to a recti ed critical level embedding with no more but perhaps fewer handles of any index such that o the rst 1 handle incident to each 0 handle is incident to it in both ends and GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE 9 o the core of any 2 handle attached in S is a compressing disk for M O Sig We will henceforth consider only recti ed critical level embeddings with these two properties De nition 42 The genus of the embedding of M in S4 is the genus of the H eegaard surface M O S8 It will be important to understand how a recti ed critical level embed ding induces a handlebody structure on each of its closed complementary components X and Y LetX denote the component of S4 7M that contains the south pole D4 gtlt 71 For each generic t 6 711 let Yf resp Xf Mf be the part of Y resp X M lying below level I or more formally the 4manifold with boundary Y O S3 gtlt 71tj rest S3 gtlt 71tj 3 manifold with boundary M O S3 gtlt 71tj Symmetrically let Y resp X M be the part on resp X M lying above level I that is the 4 manifold with boundary Y O S3 gtlt a 1 rest S3 gtlt L 1 3manifold with boundaryM S3 gtlt L l Finally let Yf resp XM be the part of Y resp X M lying at level I that is the 3manifold with boundary Y O S resp X OS closed surface M S Thus BY is the union ofo and Yi Ifti lt t lt n1 then 9M 9Mr M E C S and Y consists ofa col lection of closed complementary components of F in SEE Each component of F in S is incident to Y on exactly one side and to X on the other Clearly as long as no t lies between the values I lt t then Yi g Y since the region between them is just a collar on part of the boundary On the other hand for each ti consider the relation between YEg and We know thatE is obtained from E71 by doing j surgery along a 39disk in S3 iii1 If that 39handle lies on the Y side of E71 in the sphere S278 then Yale is homeomorphic to just Ytlig with that 39handle removed So Ytljrg is still just Yrg with a collar added to part of its boundary but only to the complement of the 39handle in Y I 8 Hence it is still true that e 2 Ytl g On the other hand if the 39handle lies on the X side of Fi1 then Ytljrg is homeomorphic to Yrg but with a 4dimensional j handle added namely the product of the interval thti s with the 3dimensional j handle added to Maia in S2 We have then the general rule sometimes called the rising water rule cf Figure 3 2 Lemma 43 1 If the 39 surgery at level I has its core in Y then Yale 198 10 MARTIN SCHARLEMANN 2 If the j surgery at level t has its core in X then Ytljrg Yrg with a j handle attached X handle added to X X handle on Y side FIGURE 3 Of course the symmetric statements hold for X Note that since X con tains the south pole Xtrg 234 whereas Ytrg 0 Just as MS is a Hee gaard surface for M X07 and Y0 are connected 4manifolds constructed from just 0 and lhandles In other words there are integers many 2 0 so thath M51 gtltD3 and Y0 W31 XD3 Each handle in M07 corresponds to a handle of the same index in exactly one of X07 or Y0 so there is a connection between nxmy and the genus g of M3 The critical level embedding de nes a handlebody structure on M with a 0handles and b lhandles where b 7 a I l g If a gt g then there would be at least one 0handle in the critical level em bedding which is rst incident to a lhandle on a single one of its ends violating the Handle Cancellation Lemma 41 So a S g Let away resp bmby denote the number of 0 resp 1 handles in the critical level embedding whose cores lie inX and Y We have from above that axay abxby bnx byiay and my bxiaxl l The asym metry is explained by noting that the south pole is a 0handle for X It follows that nx my g Another way of counting nx and my is this Suppose a lhandle at critical level t has its core lying in X say If the ends of the lhandle lie in distinct components of Fi1 then the lhandle adds a lhandle to Y but nothing to its genus In contrast if the ends of the lhandle lie on the same component of E71 then it adds 1 to the genus of Y A count of the total number of the latter sort of lhandles lying inX resp Y gives ny resp nX GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE 11 For everything that has been said about X 7 and Y 7 there is a dual state ment for X and Y easily obtained by just inverting the height function The result is that beyond the standard 4dimensional duality of handle structures onX and on Y there is a kind of 3dimensional duality between handles in X and handles in Y induced by the 3dimensional duality of handles in M See Figure 4 To be concrete X04r is also a solid 4dimensional handlebody To deter mine its genus consider the core of each 2handle say at critical height I gt 0 If the core of the 2handle lies on the X side of E71 then the co core lies on the Y side ofE so it corresponds to a lhandle in XOT This lhandle adds genus to X04r if and only if the boundary of the 2handle is nonseparating in E71 FIGURE 4 To see how this occurs consider the dual rule to Lemma 43 That is suppose again that F is obtained from E71 by doing j surgery along a j disk in S3 iii1 and ask how 8 and g dilTer If the 39surgery at level I has its core in Y then viewed from above instead of below there is a corresponding 3 7 j surgery with its core in X So following the argument of Lemma 43 Yt Ytthg with a 3 7 jhandle attached On the other hand if the core of the 39handle lies in X Y is unchanged This might be called the descending hydrogen rule cf Figure 5 To summarize all possibilities 12 MARTIN SCHARLEMANN Lemma 44 Suppose F is obtained from E71 by doing j surgery along a 39 disk in S3 711771 1 If the 39 surgery at level t has its core in Y then 18 1 i 0 thg XII with a 39 handle attached Yntg g Ytlt g with a 3 ij handle attached g t 8 t 8 39 2 If the 39 surgery at level t has its core in X then 0 Y e 2 18 with a 39 handle attached 0 Y 2 his 18 0 X518 gthtre with a 3 7 j handle attached handle on X side Y X handle added to Y FIGURE 5 Here is a simple example of how this 3dimensional duality can be useful Proposition 45 Suppose there is a recti ed critical level embedding of M S3 in S4 so that the 0 and l handles as they are successively attached all lie on the X side ThenX 2 B4 Proof Following Lemma 44 X has no 0 or lhandles so it only has 2 and 3handles Dually in the standard 4dimensional handle duality of X X can be constructed with only 1 and 2handles Neither of these statements in itself is enough to show thatX is a 4ball Consider however what the given information tells us about Y following Lemma 44 applied to the construction of Y from above The possible 2 and 3handles in the construction of X from below correspond respectively to l and 0 handles in the construction of Y from above Similarly the lack of 0 and lhandles beyond the south pole for X constructed from below corresponds to a lack of 3 and 2handles for Y when constructed from above Hence Y has only 0 and lhandles ie it is a 4dimensional handlebody On the other hand because it is the complement of S3 in S4 it is a homotopy 4ball so the handlebody must be of genus 0 ie Y is a 4ball Then its complementX is also a 4ball D GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE 13 5 TWO PROOFS OF THE GENUS TWO SCHOENFLIES CONJECTURE Informed by the ideas above we present two proofs of the genus 2 Schoen ies Conjecture The rst is similar in spirit though dilTerent in detail to the original proof of Schl The second uses a different approach one that aims to simplify the picture by reimbeddingX or Y Here is a more general statement relevant to the classical approach Proposition 51 Suppose a 3 sphereM has a genus g recti ed critical level embedding in S4 with at most two 0 handles or at most two 3 handles If the generalized propertyR conjecture is true for links of g 7 1 components thenM divides S4 into two PL 4 balls Proof Perhaps inverting the height function assume without loss of gen erality that M has at most two 3handles The roles of X and Y can be interchanged by passing the lowest 0handle over the south pole so we can also assume without loss of generality that the rst that is the lowest 3 handle forM lies in Y and so represents the addition of a 3handle to X The second 3handle and so the last handle of M either lies in X or in Y but these options are isotopic by passing the handle over the north pole So via an isotopy of this handle we can choose whether both 3handles of M lie in Y and so represent attaching of 3handles to X and not Y or one each lies inX and Y See Figure 6 FIGURE 6 Now consider the genera nx and ny of the 4dimensional handlebodies X0 Y0 with g nx ny If nx 0 then X07 is a 4ball X is obtained from this 4ball by attaching some number of 2 and 3handles and also a 4handle if the north pole of S4 lies in X There are as many total of 2 and 4 handles as there are 3handles since X is a homotopy 4ball and the argument of the previous paragraph ensures that we can arrange it so that X contains at most one 3handle Viewed dually this means thatX can be constructed from 9X S3 with no 3handles and at most one each of l and 2handles The result then follows from Corollary 22 14 MARTIN SCHARLEMANN If nx gt 1 then ny S g 7 l and rst arranging as above so that Y has no 3handles the result again follows from the proof of Proposition 23 D Corollary 52 Sch Each complementary component of a genus 2 embed ding ofM S3 in S4 is a 4 ball Proof As noted above we can assume that the number a of 0handles in the recti ed embedding of M is no larger than g 2 Proposition 51 then shows the result follows from Property R via Corollary 22 D The reimbedding proof of the genus 2 Schoen ies Conjecture begins with a more general claim that follows from our results above for Heegaard unions Proposition 53 Suppose M E S3 has a recti ed critical level embedding in S4 so that Y resp X3 is a handlebody of genus p0 If the generalized propertyR conjecture is true for pg 2 components then Y g B4 rest g B4 Proof It was noted above that Y0 is a 4dimensional handlebody andM0 is a 3dimensional handlebody The latter fact and the hypothesis imply that M U Y is a Heegaard splitting of BYOT Viewing the critical level embed ding from the top down we symmetrically see that YOJr is a 4dimensional handlebody and Y is a Heegaard union of Y0 and YOJr along Y Let phpg denote the genera of Y0 and YOJr respectively Since M is a 3sphere each complementary component of M is a homotopy 4ball In particular following lemma 32 p1 p2 p0 The result now follows from Proposition 33 Proposition 53 suggests a clear strategy for a proof of the general Schoen ies Conjecture assuming the generalized Property R Conjecture Given a recti ed critical level embedding of M S3 in S4 try to reimbed X or Y still a recti ed critcal level embedding so that afterwards either the 3 manifolng or its complement Y is a handlebody Or at least more closely resembles a handlebody For even if a series of reimbeddings rst of X then of its new complement Y then of the new complement of Y etc eventually leads to a handlebody crosssection at height 0 we are nished For once one of the complementary components of the multiply reimbed dedM is a 4ball we have that both are hence the previous complementary components in succession leading back to the original X and Y are all 4balls This is more formally explained in the proof of Corollary 82 What follows is a proof of the genus 2 Schoen ies Conjecture built on this strategy In order to be as exible as possible in reimbeddingX or Y we rst prove atechnical lemma which roughly shows that at the expense of some vertical GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE 15 rearrangement of the 3handles or dually the 0handles the core of a 2handle resp the cocore of a lhandle can be moved rel its boundary to another position without affecting the isotopy class of M or even the embedding of M below the speci ed 2handle resp above the speci ed I handle Suppose as above M is a recti ed critical level embedded S3 in S4 Lemma 54 PrairieDog Lemma Let E C S2 iii1 be the core of the 2 handle added to E71 at critical level t gt 0 and lett be a generic height such that ti1 lt t lt t LetE C S iii1 be another disk with BE isotopic to BE in E71 Then there is aproper isotopy ofMr in S3 gtlt t 1 so that afterwards o the new embedding M ofS3 is still a recti ed critical level embed ding o the critical levels and their indices are the same forM and M o the core of the 2 handle at critical level t is E and o for any generic level t below the level of the rst 3 handle M E Mf Proof With no loss we take aE parallel hence disjoint from 9E Let k be the number of 2handles above level t and n lE E The proof is by induction on the pair k n leXicographically ordered Case1kn0 In this case 2handle attached at level t is the last 2handle attached and E is disjoint from E Then the union ofE and E and a collar between their boundaries is a 2sphere in S3 iii1 if it bounds a 3ball in S3 iii1 then the disks are isotopic and there is nothing to prove If it does not bound a 3ball let S be the parallel reducing sphere for S3 7F andB C S3 be the ball it bounds on the side that does not contain the component to which E and E are attached Since t is the highest 2handle each component of F B is a sphere and each is eventually capped off above t by a 3handle If all are capped off by 3handles that lie within B push all of M B gtlt t 1 vertically down to a height just above t so that afterwards E is isotopic to E in S278 7M Perform the isotopy then pushM B gtlt t 1 back up so that the 3handles are attached above height t The number of 3handles attached namely the number of components of F is the same so although perhaps rearranged in order the critical heights at which the 3handles are attached can be restored to the original set of critical heights See Figure 7 The picture is only a little changed if one of the components of F lying in B is eventually capped olf by a 3handle not in B The proof is by induction on the number of such handles Consider the highest such handle say at 16 MARTIN SCHARLEMANN FIGURE 7 level If capping off a sphere component S of B in S2 Let B be the 3ball component of Sj 7S that does lie in B ie the complement of the 3handle in S If there are no components of in the interior of B then the 3handle is isotopic to B via passing over the north pole This isotopy decreases by one the number of 3handles not lying in B completing the inductive step If some components of do lie in B note that eventually they are capped off by 3handles lying in B by choice of If Simply push M B gtlt 01 down below level If and do the pole pass described above CaseZ k0ngt0 Consider an innermost disk E6 C E cut off by E in E Then the union of E6 and the subdisk E0 of E bounded by 9E6 is a sphere bounding a ball B whose interior is disjoint from E If no component ofE lies in B E0 can be isotoped past E6 reducing n by at least one and maybe more thereby completing the inductive step If some components of F lie in B then follow the recipe given in Case 1 For example if all components of F B eventually bound 3handles that lie in B push M B gtlt L 1 vertically down to just above level L do the isotopy then raiseM B gtlt L 1 back up aga1n Case 3 kgt 0 Depending on whether 71 0 or n 2 1 let S and B be the reducing sphere and 3balls described in cases 1 and 2 above The inductive hypothesis and a standard innermost disk argument tells us that any 2handle attached above level I can starting from highest to lowest be replaced by a 2handle disjoint from the sphere S Suppose If is the level of the rst 3handle after the replacement the entire product S gtlt Ltj is disjoint from M In fact following the argument of Case 2 we can isotope the 3handles of M possibly rearranging the ordering of the 3handles so that all of S gtlt L 1 is disjoint from M Then pushM B gtlt L 1 down tojust above level L do GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE 17 the isotopy across B as described in Case 1 or 2 then precisely restore the height ome S3 gtlt i1 D Note that since all moves are by isotopy X and Y don t change Lemma 55 Torus Unknotting Lemma Suppose that Y or symmet rically X3 lies in a knotted solid torus W C S8 Let ho W S1 gtltD2 be an orientation preserving homeomorphism to the unknotted solid torus S1 gtlt D2 C S3 Then there is a reimbedding h YaS4 to a recti ed criti cal level embedding so that hY Sg ho Y3 and the number of handles of each index is unchanged For t any generic height between the highest 0 handle and lowest 3 handle both ofMi are unchanged Proof If 9W compresses in X6 there is nothing to prove areduce W to get a 3ball which can be isotoped into S1 gtlt D2 C S8 and that same isotopy can be applied at every level ofS3 gtlt I C S4 So we assume that 9W is incompressible in X6 In each successively in creasing critical level t gt 0 ask whether the 2handle attached at t can be replaced as in the PrairieDog Lemma 54 by a 2handle that lies in W If it can be done then do so This may alter the critical level embedding of M but only above level ti71 If success is possible at the critical levels of all 2handles the same can be accomplished for the 3handles as described in Case 2 of the proof of Lemma 54 Similarly at each successively de creasing critical level t J lt 0 try to replace cocores of lhandles by disks that lie in W If this can be done for all lhandles then also replace cocores of 0handles by 3balls in W If success is possible for all 1 and 2 handles hence at all levels thenM 8W gtlt 71 1 0 and so Y C W gtlt 711 Thenthe function ho gtlt 711 on W gtlt 711 restricted to Y C W gtlt 711 is the required reimbedding We are left with the case where successful replacement of the core of a 2handle or cocore of a lhandle is not always possible Suppose without loss of generality that t gt 0 is the lowest critical level for which the core of the associated 2handle cannot be replaced by one that lies in W C S2 Without loss we assume that the replacements of lower 2handles have been done so Y S3 gtlt 0ti 7 8 C W gtlt 0ti 7 e In particular the core ofthe 2handle must lie in X578 Choose a diskD C thig so that its boundary is the same as that of the core of the 2handle and among all such disks D intersects 9W in as few components as possible An innermost circle ofD 9W then cuts off a subdisk of D whose boundary is essential in 9W by choice of D so since W is knotted the subdisk must be a meridian disk for the solid torus W 18 MARTIN SCHARLEMANN So at the generic level I 7 8 BW compresses in thig H W In particu lar Ytlig lies in a 3ball B C W It is a classical result that simple con ing extends the homeomorphism holB B h0B C S1 gtlt D2 C S3 to an orientationpreserving homeomorphism H S3gtS3 De ne then the em bedding on0 into S3 gtlt 01 so that W m S3 gtlt an i 5 ho gtlt an i e and fort 2 t 78 MY HlYf The same argument applies symmetrically to construct le0 Either all cocores of lhandles can be replaced by disks in W in which case we afterwards simply use kg at every level I 6 710 or there is a highest crit ical level I for which the cocore of the associated handle on M cannot be replaced by a disk in W and we apply the symmetric version of the con struction above Here then is an alternative proof of Corollary 52 Proof Like any surface in S3 the genus 2 surface Mg compresses in S3 and so it compresses into either X6 or Y0 say the former Maximally compress M3 in X6 IfXg is a handlebody then Proposition 53 says X 2 B4 hence its complement Y 2 B4 If X6 is not a handlebody then the surface F resulting from maximally compressing the surface M3 into X6 consists of one or two tori Like any surface in S3 F compresses in S3 The torus component ofF that com presses in the complement of F bounds a solid torus W on the side on which the compressing disk lies That side can t lie in XS since F is maximally compressed in that direction so W must contain M3 and so indeed all of Y0 The solid torus W is knotted else F would still compress further into X6 Now apply the Torus Unknotting Lemma 55 to reimbed Y in S4 in a level preserving way so that afterwards W is unknotted in particular afterwards F does compress further into the new complement of Y0 After perhaps a further iteration of the argument when F originally consisted of two tori we have a levelpreserving reembedding of Y in S4 so that afterwards its complement is a handlebody It follows from Proposition 53 then that after such a reimbedding S4 7 Y 2 B4 hence also Y 2 B4 6 STRAIGHTENING CONNECTING TUBES BETWEEN TORI IN S3 Enlightened by Corollary 34 observe that there is no generalized Prop erty R obstacle to applying Proposition 53 to the proof of the genus 3 Schoen ies Conjecture All that is needed is a suf ciently powerful version of the Torus Unknotting Lemma 55 that would instruct us how to reimbed some complementary component Y of a genus three S3 in S4 so that its new complement in S8 more closely resembles a handlebody eg it areduces to a surface of lower genus than X6 or Y did originally GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE l9 Fuelling excitement in this direction is the classic theorem of Fox Fo that any compact connected 3dimensional submanifold of S3 can be re imbedded as the complement of handlebodies What seems di icult to nd is a way to extend such a reimbedding of Y to all of Y as is done in Lemma 55 It is crucial in the proof of Lemma 55 that a solid torus has a unique meridian whereas of course higher genus handlebodies have in nitely many meridians For genus 3 embeddings there is indeed enough information to make such a reimbedding strategy work The key is a genus 2 analogue of the Torus Unknotting Lemma called the Tube Straightening Lemma whose proof is reminiscent of that in Ga3 or Th We precede it with a prepara tory lemma in 3manifold topology Lemma 61 Suppose F C S3 is a genus two surface anal k C F is a sepa rating curve in F Denote the complementary components of F by U anal V anal suppose k bounds a aliskE in V so that U U n is reducible Then either 0 Any simple closeal curve in 9N that bounals a disk in S3 7N bounals a disk in V 7N o N can be isotopeal in V to be disjoint from E or o k bounals a disk in U Proof Suppose some component of 9N is a sphere S Since V is irre ducible S bounds a ball B in V Nothing is lost by adding B to N if N is incident to S on the outside of B or removing B N from N if N is in cident to S on the inside of B So we may as well assume that 9N has no sphere components Essentially the same argument shows that we may take V 7N to be irreducible For if S is a reducing sphere bounding a ball B in V then any curve in B 9N that bounds a disk in S3 7N bounds a disk in B 7N C ViN and any curve in aNiB that bounds a disk in V 7N UB also bounds a disk in V 7N so without loss we may delete B N from N The proof then will be by induction on 7958N 2 0 assuming now that V 7N is irreducible Case 1 9N does not compress in V 7N If 9N doesn t compress in S3 7N either then the rst conclusion holds vacuously Suppose then that S3 7N is a reducible A reducing sphere S for U U 17E must separate the two tori that are obtained from F by compress ing along E since otherwise one ball that S bounds in S3 would lie entirely in U U 17E This implies that S intersects V in an odd number of copies of E or put another way it intersects U in a perhaps disconnected pla nar surface with an odd number of boundary components lying on a regular 20 MARTIN SCHARLEMANN neighborhood of k inF LetP be a component of the planar surface that has an odd number of boundary components on F Consider then the result of 0framed surgery on k in the manifold S3 7N P can be capped off to give a sphere which is nonseparating in the new manifold since a meridian of 17k intersects P in an odd number of points On the other hand S3 7N itself is areducible No options ae in Sch2 Theorem 62 are consistent with this outcome in particular the manifold called M there haVing a non separating sphere so we conclude that S3 7 N U nk is either reducible or areducible In the latter case consider how a areducing disk D would intersect the surface F 7 17k We know the framing ofF 917K is a 0framing since each component of F 7 17k is a Seifert surface for k so D F if nonempty consists entirely of simple closed curves An inner most disk in D cut off by the intersection perhaps all of D lies either in U or V 7N But each component ofF 7 17K is a oncepunctured torus so if it compresses in U or V 7N so does its boundary ie a copy of k Hence we have that k bounds a disk in either U or V 7N Since V is irreducible in the latter case the disk can be isotoped to E in V thereby isotoping N in V off of E The same argument applies if S3 7 N U nk is reducible since both U and V 7N are irreducible such a reducing sphere must intersect F 7 17k and an innermost disk cut off by the intersection leads to the same contradiction Case 2 9N compresses in V 7N The proof is by contradiction Choose an essential curve X C 9N and a compressing diskD for 9N in V 7N so that X bounds a disk D in S3 7N but does not bound a disk in V 7N and among all such choices of DD ID Dgl is minimal IfD and D are disjoint then letN NU n Ifthere is then a sphere component of 8N the ball it bounds in V can without loss be deleted from N In any case perhaps after deleting the ball if a sphere appears in 9M 7 95 BN lt 7959N and the inductive hypothesis holds for N But the conclusion for N implies the conclusion for N which is contained in N eg if bounds a disk in V 7N it bounds a disk in V 7N so this is impossible If D and D are not disjoint note that all curves of intersection must be arcs else an innermost disk cut off in D could be used to surger Di and would lower ID Dzl Similarly let D denote the disk cut off fromD by an outermost arc ofD Dg in D and let DD C V 7N denote the two disks obtained by the acompression of D to 9N along D Both of these disks intersect D in fewer components than D did so by choice of X and D the boundary of each new disk must bound a disk in V 7N of course if either GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE 21 curve is inessential in 9N then it automatically bounds a disk in 9N C V A standard innermost disk argument shows that D and Di can be taken to be disjoint band them together via the band in 9N which undoes the 9 compression by D To be more explicit note that the arc aD 9N de ned a band move on 8D that split aDg into 8D and 9D Undo that band move to recover the curve BDZ now bounding a disk namely the band sum of D and Di that lies in V 7N This contradicts our original choice of E D Lemma 62 Tube Straightening Lemma Suppose thatY Sg or symmet rically X Sa lies in V C S3 with closed complement U and 9U 9V is of genus 2 Suppose V contains a separating compressing diskE so that the manifold U obtained from U by attaching a 2 handle alongE is reducible Then there is an embedding ho VgtS3 so that h09E bounds a disk in U the complement ofh0V There is also a reimbedding h YaS4 to a recti ed critical level embedding so that W msg h0Y0 o the number of handles of each index is unchanged and o for t any generic height between the highest 0 handle and lowest 3 handle both ofMi are unchanged Proof Let ho VaS3 be the reimbedding unique up to isotopy that re places the lhandle in V that is dual to E with a handle intersecting the reducing sphere for U in a single point Then after the reimbedding 9E bounds a disk in the complement of ho V namely the complement of E in the reducing sphere In each successively increasing critical level t gt 0 ask whether the 2 handle attached at t can be replaced as in the PrairieDog Lemma 54 by a 2handle that lies in V If it can then do so This may alter the critical level embedding ofM but only above level ti71 If success is possible at the critical levels of all 2handles the same can be accomplished for the 3handles as described in Case 2 of the proof of Lemma 54 Similarly at each successively decreasing critical level tj lt 0 try to replace cocores of lhandles by disks that lie in V If this can be done for all lhandles then also replace cocores of 0handles by 3balls in V If success is possible for all 1 and 2 handles hence at all levels thenM 9V gtlt 71 1 0 and so Y C V gtlt 711 Thenthe function ho gtlt 711 on V gtlt 711 restricted to Y C V gtlt 71 l is the required reimbedding We are left with the case where successful replacement of the core of a 2handle or cocore of a lhandle is not always possible Suppose without loss of generality that t gt 0 is the lowest critical level for which the core of the associated 2handle cannot be replaced by one that lies in V C S2 Without loss we assume that the replacements of lower 2handles have 22 MARTIN SCHARLEMANN been done so Y S3 gtlt 00 7 8 C V gtlt 00 7 In particular the core of the 2handle must lie in X578 Now apply Lemma 61 using Itlig for N By assumption the boundary of the core of the 2handle bounds no disk in V thig so the rst possible conclusion of Lemma 61 cannot hold If the last holds there was nothing to prove to begin with Take h identity Hence we conclude that the second conclusion holds Itlig can be isotoped to be disjoint from E But once this is true the reimbedding ho has no effect on Itlig that is Ytlig is isotopic to h0Yt Hence we can de ne hY S3 gtlt 0 to be ho gtlt 00 7 e on Y S3 gtlt 00 7 8 followed by a quick isotopy of ho Ynig to ItlieZ followed by the unaltered embedding above level 0 7 e 2 Note that this unaltered embedding is not necessarily the original embedding because of changes made while ensuring that earlier 2handles lie in V Finally the argument can be applied symmetrically on Y S3 gtlt 710 D 7 WEAK FOX REIMBEDDING VIA UNKNOTTING AND STRAIGHTENING In this section we show that for a genus 3 surface in S3 the operations of Torus Unknotting and Tube Straightening described above suf ce to give a weak version of Fox reimbedding That is for a genus 3 surface F C S3 there is a sequence of such reimbeddings not necessarily all operating on the same complementary component of F so that eventually a complemen tary component is a handlebody Although the context of this section ap pears to be 3manifold theory the notation is meant to be suggestive of the eventual application to the genus 3 Schoen ies Conjecture In particular the terms torus unknottirig and tube straightening as used in this section re fer to the 3dimensional reimbedding ho given in respectively Lemma 55 and Lemma 62 Suppose F C S3 is a surface dividing S3 into two components denoted X and Y Suppose D1 is a compressing disk for F in X giving rise to a new surface F1 C S3 with complementary components X1 X 7 17D1 and Y1 XU 17D1 Suppose D2 is a compressing disk for F1 in X1 or Y1 giv ing rise to a new surface F2 C S3 with complementary components X2 and Y2 Continue to make such a series of compressions via compressing disks Dii l n so that each D lies either in Xi1 or Y1 the complementary components of Fi1 C S3 until all components of Fquot are spheres De nition 71 F can be straightened if there is a sequence of torus unknot tings and tube straightenings of F as described in Lemmas 55 and Lemma 62 so that afterwards either 0 X or Y is a handlebody or GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE 23 o the order of the compressions D1 D can be rearranged so that afterwards some 9D is inessential in Fi1 and so can be eliminated from the sequence In the sequence of compressions any diskDi can be replaced by a diskD in the same complementary component so long as 9D and 9D are isotopic in E71 The following series of lemmas assumes we are given such a sequence for F C S3 a connected genus 3 surface dividing S3 into two components denoted X and Y and D1 is a compressing disk for F in X There are of course symmetric statements if D1 C Y Lemma 72 IfD2 CX1 thenF can be straightened More speci cally there is a sequence of torus unknottings which convertX into a handlebody Proof Since X1 CX the disks D1 and D2 are disjoint and the compressions given by D1D2 can be performed simultaneously F2 consists of one two or three tori depending on how many of D1 D2 are nonseparating The proof is by induction on the number of D1 D2 that are separating disks If there are none so both D1D2 are nonseparating then F2 is a single torus if the solid torus it bounds lies in X2 then the original X was a han dlebody and we are done If the solid torus that F2 bounds lies in Y2 then all of F also lies in that solid torus After a torus unknotting F2 also bounds a solid torus in X2 andX is ahandlebody as required If D1 or D2 is separating then one of the components T of F1 is atorus If T bounds a solid torus in X1 then we may as well have used the meridian of that torus for D1 and invoked the inductive hypothesis If on the other hand T bounds a solid torus in Y1 then all of F also lies in that solid torus After perhaps a torus unknotting T bounds a solid torus in X1 as well and again we could replace D1 by a meridian of that torus and invoked the inductive hypothesis D Lemma 73 IfD1 CX and D2 C Y1 are both separating and D2 C Y Y1 ie D2 does not pass through the l handle dual to D1 then F can be straightened Proof Since both D1D2 are disjoint from F their order can be rearranged so there is symmetry between the two Since D1 is separating there is a torus component T1 C F1 and T1 bounds a component W1 of X 1 whose interior is disjoint from F2 After perhaps a torus unknotting of its complement we may as well assume that W1 is a solid torus Eventually some compression D will compress T1 to a sphere if D C W1 then we could have done D before D1 making D1 coplanar to Di redundant and thereby reduced the number of compressions Thus we may as well assume D C Yi1 so W1 24 MARTIN SCHARLEMANN is an unknotted solid torus Similarly the torus component T2 C F2 not incident to D1 bounds a solid unknotted torus W2 C Y2 See Figure 8 FIGURE 8 Assume with no loss of generality that D3 C Y2 We have already seen that D3 can t compress T2 else the number of compressions could be re duced It follows that D3 either compresses T1 or it compresses the third torus T3 created from F by the compressions D1D2 Either case implies that the component U Y2 7 W2 of Y2 is reducible Apply tube straighten ing to U U Y so that the tube dual to D1 passes through the reducing sphere of U once After the straightening the disk D that eventually com presses T1 into Y lies in Y so it can be the rst compression making D1 redundant Lemma 74 IfD1 CX andD2 C Y1 are both separating and ifD3 or both D4 and D5 are towards the X side then F can be straightened Proof In this case there is not necessarily symmetry between D1 and D2 but the argument of Lemma 73 still applies if the compression disk for T3 lies on the X side rather than the Y side or if T2 compresses before T3 This is what we now verify If D3 lies on the X side it compresses T2 or T3 into X2 and we are done On the other hand if D4 and D5 both lie on the X side then since T1 compresses on the Yside one ofD4 or D5 is the compression disk for T3 D Lemma 75 If D1 CX is separating and Y is a reducible then either D2 C Y Y1 orF can be straightened GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE 25 Proof Consider the torus component T1 of F1 and the component W1 C X1 it bounds As noted above the interior of W1 is disjoint from F1 and perhaps after atorus unknotting we can assume that W1 is a solid torus and moreover the disk D that eventually compresses T1 lies on the Y side and not in W1 Hence W1 is an unknotted solid torus Following Lemma 72 we may as well assume that D2 C Y1 By assump tion Y is areducible and after attaching a 2handle a neighborhood of D1 to get Y1 the resulting manifold still is areducible via D2 It follows from the Jaco handle addition lemma Ja that there is a areducing disk J C Y for Y whose boundary is disjoint from aDl Take J to be nonseparating if this is possible If 9 lies on T1 then it is parallel to the diskDi that eventually compresses T1 so we may as well do that compression before D1 making D1 redundant and so reducing the number of compressions So we henceforth suppose 9 lies on the other genus 2 component ofF 7 aDl and so lies on BE 7 T1 If a is inessential in F1 then it is parallel to 9D1 in F Thus 9D1 also bounds the diskJ C Y and the union of the two gives a reducing sphere for Y1 that intersects the lhandle dual to D1 in a single point It follows that the longitude 91 of T1 bounds a disk in Y and we are done as before If a is essential in F1 compress Y along J to get YJ and consider the component WJ C Y J in fact all of Y J if J is nonseparating such that 9D1 C BWJ and BWJ has genus 2 The manifold Wr obtained by attaching 17D1 to WJ has boundary the union of two tori Wr can also be viewed as a component of Y1 7 n We claim that either D2 C Y or W is reducible This is obvious if there is a disk component of D2 7J that can t be removed by an isotopy or if aDg J 0 The alternative then is that D2 J is a nonempty collection of arcs Consider an outermost disk D cut off from D2 by J We may as well assume D C WJT For if it s not then J is separating D is a meridian of the other complementary component a solid torus and we may as well have used D for J thereby eliminating the case that J is separating With then D C W compress the torus boundary component of 9VVJ along D to get a reducing sphere for WJT Finally if W is reducible apply tube straightening to the surface BWJ replacing the handle dual to D1 by a handle intersecting the reducing sphere once allowing the same conclusion as before D Corollary 76 IfD1 C X anal D2 C Y1 are both separating anal Y is 9 reducible then F can be straightened Proof Combine Lemmas 75 and 73 D Corollary 77 If D1 C X is separating then either 26 MARTIN SCHARLEMANN 0 at least two of the three non separating compressing disks are on the Y side and ifD2 is separating Y is 9 irreducible or o F can be straightened Proof Following Lemmas 72 74 and 75 we may as well assume that D2 C Y Y1 and D2 is nonseparating As before let T1 be the torus com ponent of F1 bounding a component W1 of X 1 whose interior is disjoint from F1 If the compressing disk D that eventually compresses T1 lies on the X side then D C W1 and W1 is a solid torus No earlier compressing disk can be incident to T1 so in fact D could have been done before D1 making the latter redundant This reduces the number of compressions If on the other hand D lies on the Yside then both D1 and D2 are non separating disks lying on the Y side as required D Lemma 78 If D1 C X is non separating then either all succeeding disks D2D3 are non separating or F can be straightened Proof Since D1 is nonseparating F1 is a genus 2 surface Its complemen tary component Y1 contains all of Y If D2 is also nonseparating then F3 is a torus for which any compressing disk is nonseparating giVing the result So suppose D2 is separating Following Lemma 72 we may as well assume D2 C Y1 so 9Y2 consists of two tori Hence D3 is nonseparating and compresses one of the tori If D3 C Y2 then D3 could have been done before D2 making D2 redundant If D3 C X2 then the result of the compression is a sphere which could have been Viewed as a reducing sphere for Y2 Apply tube straightening using Y1 for N After that reimbedding D3 C X1 X2 does not pass through D2 so the compression along D3 could be done before the compression along D2 making D2 redundant D We note in passing though the fact will not be used that if there are nonseparating compressing disks on both sides they may be taken to be disjoint Proposition 79 If D1 C X is non separating and Y has a non separating a reducing disk E then either 9D1 9E 0 orF can be straightened Proof Following Lemma 72 we may as well assume that D2 C Y1 By assumption Y is areducible and after attaching a 2handle a neighborhood of D1 to get Y1 the resulting manifold still areducible Via D2 It follows from the Jaco handle addition lemma Ja that there is a areducing diskJ C Y for Y whose boundary is disjoint from 9D1 Take J to be nonseparating ifthis is possible Suppose rst that 9 is inessential in F1 Then the disk it bounds in F1 contains both copies of D1 resulting from the compression of F along D1 GENERALIZED PROPERTY R AND THE SCHOENFLIES CONJECTURE 27 Put another way J cuts olT a component WJ from Y that has torus boundary and whose interior is disjoint from F Following perhaps a torus unknot ting we may assume that WJ is a solid torus A standard innermost disk outermost arc argument between J and D2 ensures that they can be taken to be disjoint so D2 compresses the other genus 2 component of Y 7 n J This compression together with the compression Via the meridian of WJ areduces Y to one or two components each with a torus boundary After perhaps some further torus unknottings Y becomes then a handlebody as required So suppose henceforth that a is essential in Fl Compress Y along J to get Y J and consider the component WJ C Y J in fact all of Y J if J is non separating such that 9D1 C BWJ and BWJ has genus 2 The manifold Wf obtained by attaching 17D1 to WJ has boundary a torus Wf can also be Viewed as a component on1 7 n Consider an outermost disk E ofE cut off by J or let E E ifE is disjoint from J We may as well assume that E lies in WJ since ifJ is separating and E lies in the other component we should have taken E for J If E is inessential in WJ then E E is parallel to J since E is non separating so 9D1 9E 0 and we are done If E is essential in WJ then each component of Y 7 17J U 17E has interior disjoint from F and is bounded by atorus Following some torus unknottings we can take them to be solid tori In that case Y is a handlebody as required 8 THE GENUS 3 SCHOENFLIES CONJECTURE We now apply the results of the preVious sections to complete the proof of the genus 3 Schoen ies Conjecture Theorem 81 Suppose M is a genus 3 recti ed critical level embedding of S3 in S4 Then after a series of reimbeddings of M via other genus 3 recti ed critical level embeddings each of which changes at most one of the complementary components ofM one of those complementary components is B Proof We assume that any possible genus 3 recti ed critical level reimbed ding of M that preserves at least one complementary component and simul taneously decreases the number of handles has been done If any further such a sequence of reimbeddings Via Lemmas 62 or 55 reimbeddings that don t raise the number of handles results in X6 resp Y0 becoming a han dlebody thenX resp Y is B4 Via Proposition 53 So we assume no such further sequence eXists and use the results of the preVious section to see if there are other options With no loss of generality assume the cocore D1 of the highest lhandle lies in X6 and let E1 denote the core of the lowest 2handle 28 MARTIN SCHARLEMANN Following Lemma 72 we can assume that the cocore D2 of the previous lhandle lies on the Y side and the cocore E2 of the next lhandle lies on the side opposite the one on which E1 lies Claim Perhaps after rearranging the ordering of the handles at least one of D1 and E1 is nonseparating Suppose D1 and E1 are both separating Ifboth D1 and E1 lie inX0 then it follows from Corollary 77 that at least two of the nonseparating cores of the 2handles and at least two of the nonseparating cocores of the l handles all lie on the Y side of the surfaces to which they are attached So both Y5r and Y0 are 4dimensional handlebodies of genus at least 2 But the MayerVietoris sequence for Y5r Y0 glued along YJ then contradicts HAY HB4 cfthe proofofLemma 32 On the other hand if D1 CX0 and E1 C Y then following Corollary 76 and Lemma 75 D2 is nonseparating and lies in Y3 Then interchange D1 and D2 using the lhandle dual to D2 as the highest lhandle The new arrangement establishes the claim Following the claim we can with no loss assume thatE1 is nonseparating Then according to Lemma 78 all of the cores of 2handles are nonseparating so each surface F at or above height t 0 are connected Hence there is at most one 3handle inM and passing this 3handle over the north pole if necessary this guarantees that each of X and Y have induced handle struc tures without 3handles Following the comments preceding Lemma 44 the sum of the genera of the 4dimensional handlebodies X07 and Y0 is 3 so one of them say X07 has genus S l X is then obtained from the genus 0 or 1 handlebody X07 by attaching some 2handles but no 3handles Moreover 9X is a sphere If genusX 0 resp genusX 1 then no resp l 2 handle must be attached to ensure that H1X H2 X 0 In the rst case since no 2handles hence no handles at all are attached X 2X 234 In the second case one 2handle is attached to X07 2 S1 gtltD3 and so X 2 B4 by Corollary 22 D Corollary 82 Each complementary component of a genus three recti ed critical level embedding ofS3 in S4 is a 4 ball Proof LetM be a genus three recti ed critical level embedding of S3 Fol lowing Theorem 81 there is a sequence of such embeddings MM07M17M277Mn so that one of the complementary components of Mn is B4 and for each i 0 n 7 1 one of the complementary components of M is homeomorphic to a complementary component ofMi1 The argument is by induction on n exploiting the fact that the complement of B4 in S4 is B4

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