COMPARING HEEGAARD SPLITTINGS - UCSB Math

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0 E as and 1E nu. i'otice that cx na y -- fl by weak incompressibility. Suppose a generic .... The half-saddle case is similar but easier than subcases 3a and 3b below. so we consider just t,he ... Subcase 3c. At so the saddle ..... point and, for the annuli A and Af based on c and c' in B4 and Bx respectively, neither A n Q nor A' n ...
TRANSAC:TIONS O F 'THE AXIERICAV h~IATHEh~IATICA1. SOCIETY Volume 3Z0. Vmmber 2, February 1998. Pagei 689 715 S 0002-9Sl47(98)(31824-8

COMPARING HEEGAARD SPLITTINGS -THE BOUNDED CASE FIYAM RUBINSTEIN AND LIARTIN SCI-IARLEAIANN

A B S T R A ~In~ .a recent paper we used Cerf theory to compare strongly irreducible Heegaard splittings of the same closed irreducible orientable 3-manifold. This captures all irreducible splittings of non-Haken 3-manifolds. One appliq cation is a solution to the stabilization problem for such splittings: If p are the genera of two s11littings, then there is a common stabilization of genus 51, 8q - 9. Here xve show how t o obtain similar results even ~ v h e nthe 3manifold has boundary.




I. i?lc'e,rior(T)C x .

7, each. curve djy,, 1 < I; 5 rn: i s n mcridian curve for Y , and

3. 87;) is cssentinl i'r~Q . TI2c.n. th,ti?,e is a planar surface U in Q ztselJc. with 8i7the union copies qi'sorne of the dZ:, 1 5 i < rn,

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C'OLIPAZRING FIISISGAZ~I-LT) SPJ,I I ' I I N C S ' L l - I I ; ROI1KT)E:I) CASE

Proof. Since Q is weakly incompressible, yet compresses both into X and into Y , all meridian curves of both X and Y lie on a single component Qo. Compress Qo into Y along the meridian disks L>, ,1 i in: bounded by ( d T ) , ,1 5 i m: and call the result Sy c Y. Let. X" c iCi( denote the 3-ma.nifold obtained from X by attaching the corresponding 2-handles. Somewhat similarly, let Sx c X denote the surface obtained by maximally compressing Qo into X. Then the region between Sx and Svin Xf is obtained by attaching 2-handles to both sides of a collar Qo x I of Qo. Since Qo is weakly incompressible, it follows as in [CG] that both Sx and Sy are incompressible in .Xf. The choice of D iguarantees that T extends t o a disk T ~intX' with aT+ = dTo Since Sx is incompressible in X + , T' can be isotoped to be disjoint from Sx a.nd so lies in the region between Sx and Sv.Since dTo lies in Qo and is disjoint from each dTi,1 5 i 5 m, we may consider dTo C Sy.Then [CG]implies that dTo is o inessential in Sy:as required.

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