reducibility of Heegaard splittings

# Manifolds admitting both strongly irreducible and weakly reducible minimal genus Heegaard splittings

Tsuyoshi Kobayashi  and  Yo’av Rieck
July 12, 2019July 12, 2019
July 12, 2019July 12, 2019
###### Abstract.

We construct infinitely many manifolds admitting both strongly irreducible and weakly reducible minimal genus Heegaard splittings. Both closed manifolds and manifolds with boundary tori are constructed.

###### Key words and phrases:
3-manifold, Heegaard splittings, strong irreducibility, weak reducibility
###### 1991 Mathematics Subject Classification:
57M99, 57M25
The first names author was supported by Grant-in-Aid for scientific research, JSPS grant number 19540083. The second named author was supported in part JSPS (fellow number P00024) and by the 21st century COE program “Constitution for wide-angle mathematical basis focused on knots” (Osaka City University); leader: Akio Kawauchi.

The pioneering work of Casson and Gordon [1] shows that a minimal genus Heegaard splitting of an irreducible, non-Haken 3-manifold is necessarily strongly irreducible; by contrast, Haken [2] showed that a minimal genus (indeed, any) Heegaard splitting of a composite 3-manifold is necessarily reducible, and hence weakly reducible. The following question of Moriah [8] is therefore quite natural:

###### Question 1 ([8], Question 1.2).

Can a 3-manifold have both weakly reducible and strongly irreducible minimal genus Heegaard splittings?

###### Theorem 2.

There exist infinitely many closed, orientable 3-manifolds of Heegaard genus 3, each admitting both strongly irreducible and weakly reducible minimal genus Heegaard splittings.

Theorem 2 is proved in Section 2. In Remark 7 we offer a strategy to generalize Theorem 2 to construct examples of genus , for each ; it is easy to see that no such examples can exist if . In Section 3 we give examples of manifolds with one, two or three torus boundary components, each admitting both strongly irreducible and weakly reducible minimal genus Heegaard splittings. Moreover, for each manifold with two boundary components, we construct four minimal genus Heegaard surfaces, two weakly reducible, one separating the boundary components and one that does not, and similarly two strongly irreducible minimal genus Heegaard surfaces. For a precise statement, see Theorem 14.

In an effort to keep this article short we refer the reader to Section 2 of [6] for definitions and background material. Unless otherwise stated we follow the notation of that paper.

## 1. Preliminaries

### 1.1. Constructing strongly irreducible Heegaard splittings

In this subsection we introduce a method for constructing strongly irreducible Heegaard splittings using 2-bridge link exteriors; this is taken out of [5].

###### Definitions 3.
1. A 2-string tangle is a pair of 3-ball and two disjoint arcs , properly embedded in .

2. A tangle is called 2-string trivial tangle if it is homeomorphic (as a triple) to , where is a 2-disk and are two distinct points in .

Let be a 2-string trivial tangle. Let , and , . Note that is a genus 2 handlebody, , are annuli in and the pair is primitive in (see Figure 1), i.e., there exist pairwise disjoint meridian disks , so that:

1. is an essential arc in (), and

2. .

A link is called a 2-bridge link if it can be expressed as the union of two 2-string trivial tangles; more precisely, if , where and are 2-string trivial tangles, and . Note that in this paper by a 2-bridge link we always mean a two component link, and not a 2-bridge knot.

Let be as above and be a copy of , and similarly . Let be a 2-bridge link. Then we see from the above that there exists a homeomorphism such that , the exterior of , is homeomorphic to and , so that , are meridian curves (). The image of in is called a bridge sphere.

Let be a (possibly disconnected) orientable, irreducible, -irreducible 3-manifold such that consists of two tori and and each component of has non empty boundary (hence, consists of at most two components). Suppose that there exists a 3-dimensional submanifold such that:

1. Each component of is a handlebody, and is incompressible in .

2. () consists of an annulus, say , such that

1. is incompressible in , and—

2. is -incompressible in (i.e., there does not exist a disk properly embedded in that intersects in an essential arc).

3. Each component of is a handlebody such that () consists of an annulus, say satisfying:

1. is incompressible in , and—

2. is -incompressible in .

With notation as above, let be the 3-manifold obtained from and by identifying their boundary by an orientation reversing homeomorphism such that ( resp.) is mapped to ( resp.). Let and similarly . Since ( resp.) is primitive in ( resp.), we see that ( resp.) is a handlebody obtained from ( resp.) by attaching a 1-handle (see Figure 2), and therefore is a Heegaard splitting of .

For this Heegaard splitting the following holds:

###### Proposition 4.

With notation as above, if is not the trivial link or the Hopf link, then the Heegaard splitting is strongly irreducible.

###### Sketch of proof.

Proposition 4 is identical to Proposition 3.1 of [5] and the proof can be found there. For the convenience of the reader, we sketch it here. Let and be a pair of meridian disks. Minimize the intersection of with , and the intersection of with . By symmetry, we have the following three cases:

1. and

2. and

3. and

In the first case, (resp. ) is the meridian disks of the tangles (resp. ); since is not the trivial link or the Hopf link, intersects more than twice. In the second case, is the meridian disk of the tangles . Consider an outermost disk on , say . Note that . If the arc of on or is inessential, we can pinch it off; the proof now is the same as the first case. Else, gives a boundary compression for or . Again, since is not the trivial link or the Hopf link, we see that .

In the third case, we consider outermost disks, on , and on . If the arc of on or is inessential, or the arc of on or is inessential, then arguments similar to the above work. Suppose on or , and on or are essential. Since is not the trivial link or the Hopf link, we see that .

### 1.2. Spines of amalgamated Heegaard splittings.

A spine of a compression body is a graph embedded in so that is homeomorphic to . Let be a Heegaard splitting of a manifold ; a graph is a spine for if there exists an ambient isotopy of so that the image of after this isotopy is contained in as a spine. Simultaneous spines of are two disjointly embedded graphs , , so that after an ambient isotopy of the image of ( resp.) is contained in ( resp.) as a spine. For the definition of amalgamation of Heegaard splitting see [9].

###### Proposition 5.

Let and be manifolds so that and are connected and homeomorphic. For , let be Heegaard splittings of , where is a handlebody and a compression body. Let (resp. ) be a spine of (resp. ). Let be a manifold obtained by gluing and along their boundaries. Let be the amalgamation of and .

Then there exist simultaneous spines of so that is contained in a spine of or , and is contained in a spine of the other.

###### Proof.

We denote the image of in by , the image of in by , and the image of in by . By transversality, we assume as we may that . The Heegaard surface that gives amalgamation of and is given by tubing along into and along into , see Figure 3 (this figure is based on Schultens’ [9, Figure 3]). Note that the intersection of and the amalgamated Heegaard surface is not transverse.

We may suppose that is contained in and is contained in . By compressing along the disks we obtain two handlebodies. One handlebody is isotopic to and so we may take as its spine. The other handlebody contains and admits a deformation retract onto it; moreover, intersects each disk of in exactly one point and has no other intersections with the boundary of this handlebody. Since the two handlebodies were obtained from by compressing along the disks , it is easy to construct a spine for by connecting to . is treated similarly; the proposition follows. ∎

## 2. Proof of Theorem 2

We adopt the notation of Section 1.

Let be the trefoil knot and the figure eight knot. Let be a hyperbolic 2-bridge link. Denote by ().

We note that there exists an essential annulus in such that the closures of the components of are solid tori, say and , where wraps around longitudinally twice, and around longitudinally three times. Hence, and are incompressible and boundary incompressible. On the other hand, we note that is a genus 1 fibered knot. Hence we have the following: let be a minimal genus Seifert surface for (note that is a once punctured torus). Let and . Then ( resp.) is homeomorphic to , where ( resp.) corresponds to . Note that is homeomorphic to a genus 2 handlebody, and is incompressible and -incompressible in . Let be a bridge sphere in . Then as in Section 1, separates into two genus 2 handlebodies, called and . Finally, let be a 3-manifold obtained from and by identifying their boundaries by a homeomorphism so that satisfies the following conditions:

1. , hence .

2. , hence .

Note that the conditions of Proposition 4 are satisfied, and so we see that admits a strongly irreducible genus 3 Heegaard splitting. Explicitly, the splitting surface is obtained from the bridge sphere by attaching (that is, two once-punctured tori) in and in . Denote this splitting by , where and are the handlebodies and resp., and is the splitting surface.

The decomposition is the torus decomposition for . In [4, Theorem], a complete list of Heegaard genus 2 3-manifolds admitting non-trivial torus decomposition is given. By consulting that list, we see that . Above we constructed a strongly irreducible genus 3 Heegaard splitting for . We conclude that , and that admits a strongly irreducible minimal genus Heegaard splitting.

We claim that the submanifold admits a genus 2 Heegaard splitting. Since is primitive in and is primitive in , and are genus 2 handlebodies. Let and . Let and . It is clear that is a genus 2 handlebody. It is easy to see that is primitive in , i.e., there is a meridian disk of such that is an essential arc in . This implies that is a genus 2 compression body with . Denoting by , we see that is a genus 2 Heegaard splitting of .

###### Remark 6.

For future reference, we note the following: let be a core curve of the solid torus and a core curve of the solid torus . By construction, is contained in a spine of the handlebody and is contained in a spine of the compression body . Similarly, the decomposition , where and , gives another (possibly isotopic) genus 2 Heegaard splitting of so that is contained in a spine of the handlebody and is contained in a spine of the compression body .

It is well known that admits a genus 2 Heegaard splitting. By amalgamating a genus 2 Heegaard splitting for with a genus 2 Heegaard splitting of we obtain a weakly reducible Heegaard splitting of ; by [9] (see also [6, Lemma 2.7] for a more general statement) this Heegaard splitting has genus 3. This establishes the existence of weakly reducible minimal genus Heegaard splittings of .

This completes the proof of Theorem 2.

###### Remark 7.

The following is a suggestion for a way to generalize the results of this paper. Fix . Let (resp. ) be a genus handlebody, and , (resp. , ) be two primitive annuli. A construction similar to above gives handlebodies , of genus . The curve complex distance of a Heegaard splitting was defined by Hempel [3] and was generalized by several authors to bridge decompositions; note that is a genus , 2 bridge decomposition (for details see, for example, the proof of Proposition 2.2 of [7]). It is reasonable to expect that if the distance of is large, then is strongly irreducible and minimal genus (Tomova’s [10] should be useful here). Similar to the construction above, one obtains weakly reducible minimal genus Heegaard splittings by considering the decomposition and . This would give manifolds of genus , for arbitrary , admitting both weakly reducible and strongly irreducible minimal genus Heegaard splittings.

## 3. Further examples: the bounded case

Throughout this section, let be any of the manifolds constructed in the previous section. Let be the strongly irreducible Heegaard splitting constructed there.

Let be the simple closed curve given in Figure 4. By Figure 4 (a), is contained in a once punctured torus that is a fiber of the fibration of over . We may choose this fiber to be a component of .

###### Remark 8.

We connect to by an arc as in Figure 4 (b). It is directly observed that the exterior of a regular neighborhood of ( together with the 1-complex) is a genus 2 handlebody. This shows that is contained in a spine of a compression body (not handlebody) component of a genus 2 Heegaard splitting of .

Let , be as in Remark 6, so that and . Denote by . Denote the boundary components of by , , and .

###### Lemma 9.

admits two genus 3 weakly reducible Heegaard surfaces, denoted by and , so that:

1. separates and .

2. separates and .

###### Proof.

By applying Proposition 5 to the Heegaard splitting (recall Remark 6) and the genus 2 Heegaard splitting of given in Remark 8 we obtain a genus 3 Heegaard splitting of such that the Heegaard surface separates and , is contained in a spine of one of the handlebodies, and is contained in a spine of the other handlebody. This gives .

Analogously, by applying Proposition 5 to the Heegaard splitting (recall Remark 6) and the genus 2 Heegaard splitting of given in Remark 8 we obtain . ∎

.

###### Proof.

Since is obtained from by Dehn filling, we have that . On the other hand, is a genus 3 Heegaard surface for , showing that . ∎

###### Definition 11.

Let be a compression body, and simple closed curves. We say that are simultaneous cores if the following two conditions hold:

1. There exist mutually disjoint annuli so that one component of is and the other is on .

2. There exist mutually disjoint meridian disks so that intersects transversely in one point and for , .

###### Remark 12.

It is easy to see that are simultaneous cores if and only if is a compression body.

Recall that . Let (resp. ) be a curve obtained by pushing slightly into (resp. ).

###### Lemma 13.

The curves , and , are simultaneous cores.

###### Proof.

Recall the definition of the handlebody given in Section 1, and let , be the meridian disks of shown in Figure 1. Let be a meridian disk of the solid torus that intersects the annulus essentially. By attaching two copies of to we obtain a meridian disk for , denoted by , that intersects once and is disjoint from .

Recall that is homeomorphic to , where is a once punctured torus. We may suppose that corresponds to a curve , where is an essential curve on . Let be a vertical disk in that intersects once, that is, corresponds to a disk of the form , where is an arc properly embedded in that intersects transversely once. By attaching two copies of to we obtain a meridian disk for , denoted by , that intersects once and is disjoint from .

It is easy to see that , and that there exist a pair of disjoint annuli, say and , so that one component of is and the other is on and one component of is and the other is on . Hence, and are simultaneous cores.

The curves and are treated similarly. ∎

###### Theorem 14.

For , there exists infinitely many manifolds so that consists of exactly tori, , and each admits both strongly irreducible and weakly reducible minimal genus Heegaard splittings.

Moreover, each manifold admits four distinct minimal genus Heegaard surfaces, denoted , , , , so that the following four conditions hold.

1. The Heegaard splittings given by and are strongly irreducible.

2. The Heegaard splittings given by and are weakly reducible.

3. and separate the two boundary components of .

4. and do not separate the boundary components of .

Before proving Theorem 14 we give the following definition.

###### Definition 15.

Let and be manifolds so that is obtained from by Dehn filling (equivalently, is obtained from by removing an open regular neighborhood of a link in it). Note that .

Let be any Heegaard surface. Then is a Heegaard surface of . We say that is an induced Heegaard surface (or the Heegaard surface induced by ).

Let be a Heegaard surface. Suppose that and that is a Heegaard surface of . We say that is an induced Heegaard surface (or the Heegaard surface induced by .)

The proof of the following lemma is easy and left to the reader:

###### Lemma 16.

Let and be as above. If a Heegaard surface of is weakly reducible, then so is the induced Heegaard surface. On the other hand, if is a strongly irreducible Heegaard surface that induces a Heegaard surface for , then the induced Heegaard surface is strongly irreducible.

###### Proof of Theorem 14.

We deal with the cases , and in increasing order of difficulty.

For , let . Then by Lemma 9, and admit a weakly reducible minimal genus Heegaard splitting.

Note that is isotopic to ; hence is homeomorphic to . By Lemma 13 and Remark 12, induces a genus 3 Heegaard splitting of . Since is strongly irreducible, Lemma 16 shows that the induced Heegaard splitting is strongly irreducible. The case follows.

For , let . Then . Since is obtained from by removing an open neighborhood of and , . We see that .

Note that is obtained by filling two boundary components of . Hence the genus 3 weakly reducible Heegaard splittings for given in Lemma 9 induces genus 3 weakly reducible Heegaard splittings for .

By Lemma 13 and Remark 12, induces a genus 3 Heegaard splitting for . As above, the induced Heegaard splitting is strongly irreducible. The case follows.

For , let . Similar to , it is easy to see that .

By Lemma 16, each of the two genus 3 weakly reducible Heegaard splittings given in Lemma 9 induces a genus 3 weakly reducible Heegaard splitting on , one not separating the components of (corresponding to Lemma 9 (1)), and the other separating them (corresponding to Lemma 9 (2)). These are the surfaces and in the theorem.

Note that ( resp.) is isotopic to ; hence, is homeomorphic to ( resp.). By Lemma 13, induces a Heegaard splitting for that does not separate the boundary components of . The corresponding Heegaard surface for is the surface . Similarly, by Lemma 13, induces a Heegaard splitting for that separates the boundary components of . The corresponding Heegaard surface for is the surface . By Lemma 16, and are strongly irreducible. The case follows. ∎

## References

• [1] A. J. Casson and C. McA. Gordon, Reducing Heegaard splittings, Topology Appl. 27 (1987), no. 3, 275–283. MR 89c:57020
• [2] Wolfgang Haken, Some results on surfaces in -manifolds, Studies in Modern Topology, Math. Assoc. Amer. (distributed by Prentice-Hall, Englewood Cliffs, N.J.), 1968, pp. 39–98. MR MR0224071 (36 #7118)
• [3] John Hempel, 3-manifolds as viewed from the curve complex, Topology 40 (2001), 631–657.
• [4] Tsuyoshi Kobayashi, Structures of the Haken manifolds with Heegaard splittings of genus two, Osaka J. Math. 21 (1984), no. 2, 437–455. MR 85k:57011
• [5] by same author, A construction of -manifolds whose homeomorphism classes of Heegaard splittings have polynomial growth, Osaka J. Math. 29 (1992), no. 4, 653–674. MR 93j:57007
• [6] Tsuyoshi Kobayashi and Yo’av Rieck, Heegaard genus of the connected sum of m-small knots, Communications in Analysis and Geometry 14 (2006), no. 5, 1033–1073, preprint available at arxiv.org/abs/math.GT/0503229.
• [7] by same author, Knot exteriors with additive Heegaard genus and Morimoto’s conjecture, Algebr. Geom. Topol. 8 (2008), no. 2, 953–969, preprint available at arxiv.org/abs/math.GT/0701765v2.
• [8] Yoav Moriah, Connected sums of knots and weakly reducible Heegaard splittings, Topology Appl. 141 (2004), no. 1-3, 1–20. MR 2 058 679
• [9] Jennifer Schultens, The classification of Heegaard splittings for (compact orientable surface), Proc. London Math. Soc. (3) 67 (1993), no. 2, 425–448. MR MR1226608 (94d:57043)
• [10] Maggy Tomova, Multiple bridge surfaces restrict knot distance, Algebr. Geom. Topol. 7 (2007), 957–1006, preprint available at arxiv.org/abs/math.GT/0511139.
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