Unstabilized Self-amalgamation of a Heegaard Splitting along Disks

2016-10-13 08:43LIANGLIANGLEIFENGCHUNANDLIFENGLING

LIANG LIANG,LEI FENG-CHUN AND LI FENG-LING

(School of Mathematical Sciences,Dalian University of Technology,Dalian,Liaoning,116024)



Unstabilized Self-amalgamation of a Heegaard Splitting along Disks

LIANG LIANG,LEI FENG-CHUN AND LI FENG-LING*

(School of Mathematical Sciences,Dalian University of Technology,Dalian,Liaoning,116024)

In this paper,we prove that a self-amalgamation of a strongly irreducible Heegaard splitting along disks is unstabilized.

Heegaard splitting,self-amalgamation,unstabilized

2010 MR subject classification:57N10,57M50

Document code:A

Article ID:1674-5647(2016)02-0117-05

1 Introduction

So a natural question is when is an amalgamation of two unstabilized Heegaard splittings unstabilized?A well-known result is the Gordon conjecture:The connected sum ofunstabilized Heegaard splittings is never stabilized(see[2],Problem 3.91).This was proved independently by Bachman[3]and by Qiu and Scharlemannin[4].When the genus of the amalgamated surface is positive,there exist many counterexamples showing that an amalgamation of two unstabilized Heegaard splittings might be stabilized(see[5]–[8]).On the other hand,many sufficient conditions for an amalgamation of two unstabilized Heegaard splittings to be unstabilized are given,see[9]–[12],where the gluing maps are required to be complicated enough,and[13]–[15],where the the factor Heegaard splitings are of“high”distance.

Let M be a self-amalgamation ofalong boundary components F1and F2ofGiven a Heegaard splittingforthere is amalgamated Heegaard splittiing for M, obtained by an analogous construction to that of Schulten’s.It has been proved by Du and Qiu[16]that the self-amalgamation of a“high”distance Heegaard splitting is unstabilized. Recently,Zou et al.[17]proved that the self-amalgamation of a Heegaard splitting of distance at least 3 is unstabilized.In[18],we generalize the self-amalgamation of a Heegaard splitting to the case where the amalgamated surface could be with nonempty boundaries.And we proved that if the Heegaard splitting is strongly irreducible and annulus-busting,then any self-amalgamation of the Heegaard splitting along any essential subsurfaces is unstabilized.

In this paper,we consider a special case when the amalgamated surface is a disk.We prove that the self-amalgamation of a strongly irreducible Heegaard splitting along two disjoint disks is unstabilized.The article is organized as follows:in Section 2,we review some necessary preliminaries.The statement and proof of the main result is given in Section 3.

2 Preliminary

Let M be a compact orientable 3-manifold and F be a properly embedded surface in M. F is said to be compressible if either F is a 2-sphere which bounds a 3-ball or there is an essential simple closed curve on F which bounds a disk in M;otherwise,F is said to be incompressible.F is said to be essential if F is incompressible and no component of F is∂-parallel in M.A simple closed curve in F is said to be essential if it is not contractible or∂-parallel in F.

3 Main Result and Proof

Proof.By the definition of self-amalgamation,the self-amalgamated surfaces lie in the same side of the Heegaard surface.So we assume that F1and F2are two disjoint disks in∂−W and f is a homeomorphism from F1to F2just as above.

A compression body is called simple if there is only one essential separating or nonseparating disk up to isotopy.Then we have the following corollary:

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10.13447/j.1674-5647.2016.02.04

date:Nov.24,2014.

The NSF(11101058,11329101 and 11471151)of China and the Fundamental Research Funds(DUT14ZD208 and DUT14LK12)for the Central Universities.

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E-mail address:liang liang@aliyun.com(Liang L),dutlfl@163.com(Li F L).