Planar-busting Curves on the Boundary of a Handlebody

SUN DONG-QI,TANG JING-YANAND LI FENG-LING

(1.School of Mathematical Sciences,Dalian University of Technology,

Dalian,Liaoning,116024) (2.College of Science,Harbin Engineering University,Harbin,150001) (3.School of Mathematical Sciences,Harbin Normal University,Harbin,150025)

Communicated by Lei Feng-chun

Planar-busting Curves on the Boundary of a Handlebody

SUN DONG-QI1,2,TANG JING-YAN1,3AND LI FENG-LING1

(1.School of Mathematical Sciences,Dalian University of Technology,

Dalian,Liaoning,116024) (2.College of Science,Harbin Engineering University,Harbin,150001) (3.School of Mathematical Sciences,Harbin Normal University,Harbin,150025)

Communicated by Lei Feng-chun

Let Hnbe an orientable handlebody of genus n.It has been proved that for n not less than 2,there exists an annulus-busting curve in?Hn.In the present paper,we prove that for n not less than 2,there exists an essential simple closed curve C in?Hnwhich intersects each essential planar surface in Hnnon-emptily. Furthermore,we show that for n not less than 3,a pants-busting curve must also be an annulus-busting curve.

handlebody,planar surface,planar-busting curve,pants-busting curve

1 Introduction

Let Hnbe an orientable handlebody of genus n.A planar surface in Hnis a 2-sphere with some holes.The relation between planar surfaces and thin positions is considered in[1–2].

Rubinstein and Scharlemann[3]considered the maximal essential annuli in H2.Lei and Tang[4]detected the maximal essential annuli in Hn(n≥2).It is a result in[5]that for each n≥2,there exists an essential simple closed curve C on the boundary of Hnsuch that C intersects every essential annulus in Hnnon-emptily.This curve C is called an annulus-busting curve.

In the present paper,we show the existence of planar-busting curves.Namely,for each n≥2,there exists an essential simple closed curve C on the boundary of Hnsuch thatC intersects every essential planar surface in Hnnon-emptily.Furthermore,we show that under some conditions,a pants-busting curve is also an annulus-busting curve.

In Section 2,we show some useful propositions,and we use them to prove the main results in Sections 3 and 4.

All the manifolds considered in the paper are assumed to be compact,orientable and connected.The def i nitions and terminologies not def i ned here are standard;see,for example, [6–7].

2 Preliminaries

Let Hnbe an orientable handlebody of genus n.A connected properly embedded surface P in Hnis essential if P is incompressible and is not boundary parallel in Hn.

Def i nition 2.1Let Hnbe a handlebody of genus n.An essential simple closed curve C in?Hnwhich intersects each essential pair of pants(annulus)in Hnnon-emptily is called a pants-busting(annulus-busting)curve.An essential simple closed curve C in?Hnwhich intersects each essential planar surface in Hnnon-emptily is called a planar-busting curve.

From a result of Schultens[8],it is easy to see the following proposition.

Proposition 2.1Let Hnbe a handlebody with genus n≥2,and P be an essential planar surface in Hn.Then the manifold obtained from cutting Hnopen along P may be a handlebody or consist of two handlebodies.

Let S be a closed orientable surface,α0,α1,···,αnbe a sequence of essential simple closed curves in S such that for each 1≤i≤n,αi-1and αican be isotoped to be disjoint. Then we say that the sequence is a path of length n.

The distance d(α,β)between a pair α,β of essential simple closed curves in S is the smallest integer n such that there is a path from α to β of length n.Le tbe a Heegaard splitting of a 3-manifold M.

Denote by d(S)the distance of the Heegaard splitting which is def i ned as

d(S)=min{d(C1,C2)|Cibounds an essential disk in Vi,i=1,2}.

The following theorem of Hempel[9]is important for proving our main theorem.

Theorem 2.1[9]For positive integers m,n≥2,there exists a Heegaard splittingof genus n for a closed orientable 3-manifold M with the distance d(S)＞m.

For convenience,we give the following def i nition and we will use it later.

Def i nition 2.2Let Hn(n≥2)be a handlebody of genus n,A be an essential annulus in Hn,and?A=a∪b.Let α be a simple arc in?Hnwith α∩A=α∩a=?α(or α∩A=α∩b=?α),and both ends of α meet a(or b)from the same sides.N(α)=α×[-1,1]is the regular neighborhood of α in?Hn,α=α×{0},and N(α)∩A=?α×[-1,1].Connect N(α)and A along?α×[-1,1]and push α×(-1,1)into the interior of Hn.Denote this new properly embedded pair of pants by P.We say that P is the band sum of A along α. This process is also called to do a band sum to A along α to get a pair of pants.See Fig. 2.1.

3 The Existence of Planar-busting Curves

The following lemma is important for the proof of the main theorem.

Lemma 3.1Let M be a compact orientable 3-manifold,andbe a Heegaard splitting of M.If d(S)≥k+1(k≥2),then there is no essential planar surface P with |?P|≤k in M.

Proof.Otherwise,assume that M contains an essential planar surface P with|?P|≤k in M.When|?P|=k,for d(S)≥k+1≥3,M=V1∪SV2is strongly irreducible.It is clear that P∩S/=?.Then P∩S are essential simple closed curves in P and S.Denote

So Piis essential in Vi.Then Piis?-compressible in Vi,and

Compress Piin Vifor ki-1 times.Then we get at least one essential disk in Vi.So

It contradicts the assumption.

When|?P|＜k,the proof is similar to that the case for|?P|=k.

Theorem 3.1For each n≥2,there exists an essential simple closed curve C on the boundary of a handlebody Hnof genus n such that C intersects every essential planar surface P non-emptily.

Proof.Let Hnbe a handlebody of genus n,and P be a properly embedded essential planar surface with|?P|=k in Hn.By Theorem 2.1,there exists a Heegaard splitting M′=V1∪S′V2of genus n≥2 for a closed orientable 3-manifold M′with d(S′)≥k+1 (k≥2).Let Hn=V1,and C be a meridian curve of?V2,that is,C be the boundary of an essential disk in V2.Let M be the 3-manifold obtained by adding a 2-handle to V1along C. Pushing S′slightly into the interior of M by isotopy,we get a surface S which is,in fact,a Heegaard surface of M.Clearly,d(S)≥d(S′)≥k+1.Suppose that P∩C=?.Then we have the following claim:

Claim 3.1P is essential in M.

Otherwise,there are two possibilities:

(1)P is compressible in M;

(2)P is boundary parallel in M.

Suppose that case(1)happens.For P is a properly embedded essential planar surface in Hn,we can?-compress P in Hn.However,?-compressing P in Hnis just the case that isotope P in M.For|?P|≤k,after?-compressing P at most k-1 times,we must get an essential disk in Hn.For P∩C=?,we have d(S)≤k,contradicting the condition of the theorem.

In case(2),P is?-parallel in M.Then P is separating both in M and Hn.Denote MP=H∪N,in which H is a handlebody of genus k-1 and N~=M.By Proposition 2.1,HnP=H1∪H2,where both H1and H2are handlebodies.Suppose that C??H1-P. Then H is obtained from adding a 2-handle to H1along C.Thus,H1is a handlebody of genus k,and C is a longitude on?H1.Note C∩P=?.Thus d(S)≤k,contradicting the assumption d(S)≥k+1.

Thus,P is essential in M.On the other hand,by Lemma 3.1,M contains no essential planar surface P with|?P|≤k,a contradiction.

4 Annulus-busting Curves and Pants-busting Curves

Lemma 4.1Let Hn(n≥2)be a handlebody of genus n,A be an essential annulus in Hn,and?A=a∪b.Then both a and b are non-separating in?Hn.

Proof.Let S=?Hn(a∪b).Denote the two cutting sections of a(b)by a+,a-(b+,b-, respectively).If A is separating in Hn,then S will have two components S1and S2.Suppose that

Then

So a is non-separating in?Hn.Similarly,b is non-separating in?Hn.

If A is non-separating in Hn,then S is connected and

?Hna can be obtained from S by gluing b+and b-together.So?Hna is connected,and a is non-separating in?Hn.Similarly,b is non-separating in?Hn.

Lemma 4.2Let Hn(n≥2)be a handlebody of genus n,A be an essential annulus in Hn,and?A=a∪b.Let α be a simple arc in?Hnwith α∩A=?α∩a=?α,and P be the pair of pants which is the band sum of A along α.If P is boundary parallel in Hn,then A cuts Hninto a handlebody H1of genus two and a handlebody H2of genus n-1 with α∈?H1.At this time,the cutting section of A in H1and the arc α are as illustrated in Fig.4.1,where A1denotes a cutting section of A.

Fig.4.1Arc α and a cutting section of A

Proof.Let D be the essential disk in Hnwhich comes from?-compressing A in Hn.Since P is boundary parallel in Hnand P is separating in Hn,A is separating in Hn.Denote

Then

Denote

Then

Suppose that D?H1A.There are two cases for the arc α: Case 1.α??H2A.

One of H1Pand H2P,say H1P,is obtained from H1Aby adding a 1-handle to it.Thus,H2Pis isotopic to H2A,g(H1P)=g(H1A)+1 and g(H2P)=g(H2A).

For this case,we prove the conclusion by three steps as follows.

Step 1.We prove g(H2P)=g(H2A)≥2.

In fact,if g(H2P)=g(H2A)=1,then

which implies that one boundary component of P is trivial in?Hn,a contradiction.

Step 2.We prove g(H1A)≥2.

In fact,if g(H1A)=1,then A is twisted on?H1Aand D must be boundary parallel in Hn,and this is a contradiction.

Step 3.We prove that the conclusion holds.

Since g(H1A)≥2,

By the assumption,P is boundary parallel,so we can only push P into?Hntowards H2P. Thus,

and the cutting section of A and the arc α in H1Amust be the cases illustrated in Fig.4.1.

Case 2.α??H1A.

One of H1Pand H2P,say H2P,is obtained from H2Aby adding a 1-handle to it.Thus,H1Pis isotopic to H1A,g(H2P)=g(H2A)+1,and g(H1P)=g(H1A).

Similarly to Case 1,we prove the conclusion by three steps as follows.

Step 1.We prove g(H1P)=g(H1A)=2.

In fact,if g(H1P)=g(H1A)/=2,then either

or

When g(H1P)=g(H1A)=1,A is twisted in?H1A,so D is trivial in Hn,and this is a contradiction.

When g(H1P)=g(H1A)＞2,we can only push P into?Hntowards H2P.Thus,

and A is twisted in?H2A,which implies that at least one boundary component of P is twisted in?H2P,and P cannot be boundary parallel in Hn,a contradiction.

Step 2.We prove g(H2A)≥2.

In fact,if g(H2A)=1,then A is twisted in?H2A,which implies that at least one boundary component of P is twisted in?H2P,and P cannot be boundary parallel in Hn,contradicting the assumption.

Step 3.We prove that the conclusion holds.

Since g(H2A)≥2,one hasand we can only push P into?Hntowards H1P.By the assumption,D is contained in H1A. Then the cutting section of A,say A1,in H1Ais illustrated in Fig.4.1(1),and the arc α must be the case illustrated in Fig.4.1(1).

Theorem 4.1Let Hn(n≥3)be a handlebody of genus n,and C be an essential simple closed curve in?Hn.If C is pants-busting,then C is annulus-busting.

Proof.Otherwise,there would exist essential annuli in Hndisjoint from C.Let A be a maximal collection of all essential annuli in Hnwhich disjoint from C.By the assumption, A/=?.Let A1be some component of A,?A1=α1∪β1,S=?Hn(C∪?A).Denote the cutting sections of C,α1,β1by C+,C-,α+1,α-1,β+1,β-1,respectively.There exists some component of S,say S1,which contains a cutting section of C,say C+,on its boundary.

For the curve C,we have the proposition as follows:

Proposition 4.1C does not bound an essential disk in Hn.

Otherwise,C bounded an essential disk DCin Hn.DCis either separating or nonseparating in Hn.If DCis separating in Hn,then denote HnDC=H1∪H2.For n≥3,at least one of g(H1)＞1 and g(H2)＞1 holds.Without loss of generality,assume g(H1)＞1. Then there must exist an essential pair of pants P in H1with P∩C=? and P also essential in Hn,which is a contradiction.If DCis non-separating in Hn,then denote HnDC=H. Evidently,

and there must exist an essential pair of pants P in H with P∩C=? and P also essential in Hn,a contradiction.

For the surface S1,we have the proposition as follows:

Proposition 4.2S1is not an annulus.

Otherwise,S1were an annulus.Without loss of generality,assume?S1=C+∪α+1. Then C is parallel to α1in?Hn.Now we can obtain an essential pair of pants P from A1which contains a boundary component parallel to C.Thus C∩P=?,and this is a contradiction.

Similarly to the proof of Proposition 4.2,we can prove that the component of S,which contains C-on its boundary,is not an annulus.

Now we consider the cutting section C-of C.We break the proof into two cases, depending on whether C-is contained in?S1or not.

Case 1.C-??S1.

By Lemma 4.1,|?S1|≥4.Without loss of generality,we can assume α+1∈?S1,so there exists a properly embedded simple arc γ in S1with?γ∈α+1such that the pair of pants P, obtained from doing a band sum to A1along γ,has a boundary component parallel to C. So P∩C=?.Note that C is annulus-busting,so P is inessential in Hn.Thus P is eithercompressible or?-parallel in Hn,and we will prove that there is a contradiction in either case.

If P is compressible in Hn,then at least one component of?P bounds an essential disk in Hn.Let?P=C∪β1∪δ.Since neither C nor β1bounds essential disk in Hn,δ must bound an essential disk,denoted by Dδ,in Hn.Let Aδ=P∪δDδbe an annulus in Hncoming from pasting Dδto P along δ and pushing δ slightly into the interior of Hn.?Aδ=C∪β1. By Proposition 4.2,C is not parallel to β1,so Aδis not?-parallel.Thus,Aδis essential in Hnand Aδdoes not belong to A,contradicting the maximality of A.

If P is boundary parallel in Hn,then by Lemma 4.2,A1separates Hninto a genus two handlebody H1and a genus n-1 handlebody H2,and C must be the cases illustrated in Fig.4.2.Thus,H1contains no other essential annulus in A,S1is a 4-punctured sphere with?S1=C+∪C-∪α+1∪β+1,and there must exist an essential pair of pants P′in H2, which is also essential in Hn,with P′∩C=?,contradicting the assumption.

Fig.4.2Two possibilities of curve C

Case 2.C-?S1.

Without loss of generality,we can assume|?S1|≥2.In fact,if|?S1|=1,then the component of S including C-on its boundary,denoted by S2,has more than one boundary components,so we just need to consider S2instead of S1.As in Case 1,we can assume α+1∈?S1.It is clear that either g(S1)＞0 or g(S1)=0,and we will prove that there is a contradiction in either case.

If g(S1)＞0,then there would exist a simple properly embedded arc γ in S1withsuch that the pair of pants P which is obtained from doing a band sum to A1along γ has a boundary component paralle to C.By the assumption,P is inessential in Hn,so P is either compressible or boundary parallel in Hn.In either case,there is a contradiction.In fact,if P is compressible in Hn,then similarly to the Case 1,we obtain an essential annulus which is disjoint from C and does not belong to A,contradicting the maximality of A.If P is boundary parallel in Hn,then similarly to Case 1,S1is a 4-punctured sphere,contradicting g(S1)＞0.

If g(S1)=0,then by Proposition 4.2,,and?S1would contain a component, say,of a boundary component of some Aiin A.Then there exists a simple properly embedded arc γ in S1with?γ∈α+1such that the pair of pants P obtained from doing a band sum to A1along γ has a boundary component paralle to C.By the assumption,P is inessential in Hn,so P is either compressible or boundary parallel in Hn.In either cases, as in Case 1,there is a contradiction.

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1674-5647(2013)02-0184-09

Received date:Sept.17,2012.

The grant(09XBKQ09)of Harbin Normal University,the NSF(11101058)of China,and China Postdoctoral Science Foundation(2011M500049).

E-mail address:sundq1029@yahoo.com.cn(Sun D Q).

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