return

Celebratio Mathematica

Joel Hass

Hass, Thompson and Thurston on stabilizations of Heegaard splittings

by Rob Kirby

A Hee­gaard split­ting of a closed 3-di­men­sion­al man­i­fold M3 is a cen­tur­ies-old [e1] de­scrip­tion of M as a uni­on of two handle­bod­ies with their bound­ar­ies (sur­faces of genus g) glued to­geth­er by a dif­feo­morph­ism. The 3-sphere is the uni­on of two 3-balls, or, more in­ter­est­ingly, the uni­on of two donuts (neigh­bor­hoods of the unit circle in the xy-plane and the z-ax­is uni­on in­fin­ity). The lat­ter split­ting is used to sta­bil­ize a Hee­gaard split­ting of M, in­creas­ing its genus to g+1. One can ask if two Hee­gaard split­ting of the same man­i­fold are equi­val­ent; this was proved by Re­idemeister [e3] and Sing­er [e2] to be true after sta­bil­iz­ing.

A Hee­gaard split­ting of M also arises from a Morse-like func­tion f:M[0,3] where crit­ic­al points of in­dex i are mapped to i. Then the Hee­gaard sur­face is S=f1(3/2). Cerf the­ory then gives an­oth­er proof of sta­bil­iz­a­tion be­cause sta­bil­iz­ing amounts to a birth of a can­cel­ling 1–2-pair.

A key ques­tion arose: Is one sta­bil­iz­a­tion suf­fi­cient or might more sta­bil­iz­a­tions be ne­ces­sary to trans­form one Hee­gaard genus g split­ting to an­oth­er? All ex­amples known needed only one sta­bil­iz­a­tion un­til the Hass–Thompson–Thur­ston pa­per [1], which gave ex­amples where g sta­bil­iz­a­tions are ne­ces­sary. These are ob­tained by switch­ing the two handle­bod­ies, or equi­val­ently turn­ing the Morse func­tion up­side down by chan­ging f to f.

The main the­or­em is:

For each g>2 there is a 3-man­i­fold Mg with two genus g Hee­gaard split­tings that re­quire g sta­bil­iz­a­tions to be­come equi­val­ent.

It is al­ways true that g sta­bil­iz­a­tions are enough to turn any Hee­gaard split­ting of any closed 3-man­i­fold up­side down. A sketch of the ar­gu­ment is giv­en in a re­mark at the end of the In­tro­duc­tion (Sec­tion 2) of [1].

The proof that g sta­bil­iz­a­tions are needed em­bod­ies a beau­ti­ful to­po­lo­gic­al idea which re­quires subtle geo­metry in­volving curvature, volume, area and har­mon­ic maps.

Let Σ be the ori­gin­al genus g Hee­gaard sur­face. Then we sta­bil­ize Σ k times to ob­tain a Hee­gaard sur­face S of genus g+k, and we want to show that k must at least equal g.

To turn the Hee­gaard sur­face S up­side down, we need an iso­topy fs:MM,s[0,1] which takes S to it­self but with the op­pos­ite ori­ent­a­tion.

M can be ex­pressed as Σ×(1,1) uni­on two spines which are the wedges of g circles. As S flips up­side down, we have a two para­met­er fam­ily of sur­faces, Ss,t,s[0,1],t(1,1), where the sur­face Ss,t is the im­age of the slice S×t un­der the iso­topy fs at time s. These sur­faces di­vide M in­to two parts, and we col­or one red and one blue, giv­ing the sur­face a red and a blue side.

Figure 1.

Now choose s and t so that the sur­face Ss,t di­vides the volume of M in half. There is a 1-para­met­er fam­ily λ of such sur­faces. Con­sider the two halves, left and right, of M when cut by Σ×0. Then (0,0) and (1,0) both be­long to λ but with the left side red in the first case, and the right side red in the second case. Thus there must be an (s,t) for which Ss,t di­vides the left side in equal parts and sim­il­arly the right side (see Fig­ure 1).

Met­rics are chosen so that Σ×(1,1) is very long and thin. Then be­cause of area and volume bounds, each Ss,t will in­ter­sect a high per­cent­age of the Σ×t in very small circles, each one smal­ler than the in­jectiv­ity ra­di­us of Σ×t and thus bound­ing a disk in Σ×t. All disks will be the same col­or be­cause their com­ple­ment is con­nec­ted (this is only true if the circles are not nes­ted, in which case the ar­gu­ment is slightly more subtle).

Since Ss,t is so thin most of the time, we can as­sume it is thin at Σ×0. Then we can sur­ger each of the small circles at time 0, thus cap­ping off the tubes with disks. As­sume these disks cap off red tubes. Now we fo­cus on the left side of M and find a slice Σ×τ, at which the tubes are thin and mostly blue (oth­er­wise the volume of the left side would not be di­vided in half by red and blue re­gions). Now cap off these disks to get a closed sur­face F in­side Σ×[τ,0].

The key ob­ser­va­tion now is that we can run a green arc from Σ×0 through a red cap, thus in­ter­sect­ing F in a point and go­ing from a blue re­gion in­to a red re­gion, and then stay­ing in the red re­gion to its end point at Σ×τ. The green arc may in­ter­sect F in two more points whenev­er in­ter­sect­ing a com­pon­ent of F which en­closes a blue re­gion. This green arc in­ter­sects F an odd num­ber of times and hence is ho­mo­lo­gic­ally dual to F. This means that F rep­res­ents H2(Σ×[τ,0],Z/2), and thus must have genus great­er than or equal to g be­cause there is no de­gree one map of a sur­face of lower genus to a sur­face of high­er genus.

Now we do the same for the right side, again get­ting a closed sur­face of genus great­er than or equal to g. It fol­lows that our split­ting Ss,t has genus equal to at least 2g.

The bulk of this pa­per is de­voted to del­ic­ate geo­met­ric­al ar­gu­ments. For ex­ample, the sur­faces Ss,t need to be de­formed to a fam­ily of har­mon­ic or en­ergy min­im­iz­ing maps which may res­ult in 2-chains (not ne­ces­sar­ily em­bed­ded) which still al­low the ho­mo­lo­gic­al min­im­iz­ing ar­gu­ments above.

The pa­per in­tro­duces a new tech­nique in its use of har­mon­ic maps to un­der­stand the to­po­logy of 3-man­i­folds. This is re­min­is­cent of the im­port­ant role played by min­im­al sur­faces and seems prom­ising for fu­ture ap­plic­a­tions.

Works

[1] J. Hass, A. Thompson, and W. Thur­ston: “Sta­bil­iz­a­tion of Hee­gaard split­tings,” Geom. To­pol. 13 : 4 (2009), pp. 2029–​2050. MR 2507114 Zbl 1177.​57018 ArXiv 0802.​2145 article