Positive factorizations of symmetric mapping classes

# Positive factorizations of symmetric mapping classes

Tetsuya Ito Department of Mathematics, Graduate School of Science, Osaka University
1-1 Machikaneyama Toyonaka, Osaka 560-0043, JAPAN
and  Keiko Kawamuro Department of Mathematics, The University of Iowa, Iowa City, IA 52242, USA
August 21, 2019
###### Abstract.

We study Question 7.9 in [9] raised by Etnyre and Van Horn-Morris; whether a symmetric mapping class admitting a positive factorization is a lift of a quasi-positive braid. We answer the question affirmatively for mapping classes satisfying certain cyclic conditions.

## 1. Introduction

In [9, Question 7.9] Etnyre and Van Horn-Morris ask the following question:

###### Question 1.1.

[9, Question 7.9] If a symmetric mapping class admits a positive factorization, then is a lift of a quasi-positive braid ?

This is a profound question connecting three important objects in topology; (1) symmetric mapping classes, (2) positive factorizations, and (3) quasipositive brads. We describe each here.

Let be a compact, oriented surface with non-empty boundary, which is a special (see Section 2) cyclic branched covering of a disk branched at -points. A mapping class is called symmetric (fiber-preserving) [4, p.65] if is a lift of an element of the braid group . Symmetric mapping class groups were introduced and studied by Birman and Hilden in a series of papers culminating in [3]. As Margalit and Winarski say in [23], the Birman-Hilden theory has had influence on many areas of mathematics, from low-dimensional topology, to geometric group theory, to representation theory, to algebraic geometry and more.

A positive factorization of a mapping class is a factorization of into positive (right-handed) Dehn twists about simple closed curves. In contact and symplectic geometry, positive factorizations of mapping classes play an important role due to the following fact: A contact -manifold is Stein fillable if and only if it is supported by an open book whose monodromy admits a positive factorization [2, 13, 22].

A quasipositive braid in is a braid which factorizes into positive half twists about proper simple arcs in the -punctured disk. Quasipositive knots and links are introduced and studied by Rudolph in a series of papers. Rudolph showed [25] that a quasipositive knot can be realized as an intersection (transverse -link) of the unit sphere in with an algebraic complex curve in . Boileau and Orevkov [6] proved the converse that any knot arising as such an intersection must be quasipositive.

In this paper, we give partial answers to the question of Etnyre and Van Horn-Morris.

Let be a compact, oriented surface with non-empty boundary. Let be a disk. Suppose is a special -fold cyclic branched covering of the disk branched at points. The meaning of “special” will be made clear in Section 2. In [3] Birman and Hilden show that there is a well-defined injective homomorphism whose image is the symmetric mapping class group which is defined in Section 2.

Using the homomorphism , Etnyre and Van Horn-Morris in [9] consider various submonoids in .

 P(n) := {b∈Bn | b is a positive braid } QP(n) := {b∈Bn | b is a quasi-positive braid } Dehn+(n,k) := Ψ−1(Dehn+(S)) Tight+(n,k) := Ψ−1(Tight+(S)) RV(n) := {b∈Bn | b is a right-veering braid } Veer+(n,k) := Ψ−1(Veer+(S))

Here, a braid is positive if it is a product of standard generators , and is quasi-positive if it is a product of conjugates of . We have and for . The set is a monoid generated by positive Dehn twists, is a monoid consisting of monodromies supporting tight contact structures, and is a monoid consisting of right-veering mapping classes. One can see that is right-veering if and only if is right-veering (see [19, Section 7] for the definition(s) of right-veering braids).

###### Proposition 1.2.

We have for all and .

Etnyre and Van Horn-Morris observe that [17, Lemma 3.1] implies the following:

###### Proposition 1.3.

[9, p.355] For all and we have .

We prove Proposition 1.3 in Section 2.

In summary, we have;

 P(n)⊂QP(n)⊂Dehn+(n,k)⊂Tight(n,k)⊂Veer+(n,k)=RV(n)⊊Bn.

In [18, Example 2.9], the strictness of the inclusion is shown.

###### Proposition 1.4.

In general, both the inclusions are strict as we show below:

###### Proof.

Let . By [18, Theorem 1.2] the braid is in since the fractional Dehn twist coefficient of is one. However, is not in since its exponent sum is negative which means , where the equality will be proved in Theorem 3.2. (See [18, Corollary 3.6] for a better criterion for than a naive exponent sum argument.)

To see , recall [24, Proposition 3.1] (for the open book ) and [19, Proposition 3.14] (for general open books) that every braid with respect to an open book is transversely isotopic to a right-veering braid after suitable positive stabilizations. Let be a right-veering braid that is a stabilization of . Then for each , is in but not in . ∎

With these terminologies, Question 1.1 of Etnyre and Van Horn-Morris is equivalent to the following:

###### Question 1.5.

[9, Question 7.9] Do we have ?

In [9] they say “the answer is almost certainly no”. Thus our goal can be set to find sufficient conditions for .

### 1.1. Motivation

Our particular branched covering is closely related to the cyclic branched covering of the standard contact 3-sphere . Let be a transverse knot in represented by the closure of an -braid with respect to the open book . Let be the -fold cyclic branched covering, branched along . Then is equipped with a contact structure that is a perturbation of the kernel of the pull-back . Such a contact structure is supported by the open book . Thus, .

Let be the unit complex ball giving a Stein filling of . If the braid is quasi-positive, a factorization of as a product of positive half twists gives rise to an immersed Seifert surface of with ribbon intersections as shown in [9, Figure 9]. Pushing this surface into the interior of we have a properly embedded symplectic surface in such that . Let be the -fold cyclic branched cover of branched along . Then gives a Stein filling of

On the other hand, a factorization of into positive half twists induces a factorization of into positive Dehn twists (Proposition 1.3), to which one can associate a Legendrian surgery diagram (see Section 2 and [17, Fig. 12]). Let be the 4-dimensional handlebody obtained by attaching 2-handles to according to the Legendrian surgery diagram. By [7], [14, Proposition 2.3] the manifold is a Stein filling of .

In fact, these two constructions of Stein fillings give the same manifold. Namely, the two 4-manifolds and are diffeomorphic, which follows from the proof of [20, Claim 2.1]. Thus, the branched covering construction behaves nicely not only for contact structures but also for Stein fillings.

With the above discussion in mind, we may extend Question 1.1 to the following question about Stein filling and -action:

Assume that . Then is Stein fillable because and is contained in the monoid of Stein fillable open books [7, 13]. The contact manifold also admits an -action as a contactomorphism with the quotient space . Our question is:

Can we find a Stein filling coming from the above construction; namely, can the -action on extend to a holomorphic -action on some with the quotient space ?

This new question suggests that, even if Question 1.1 may have negative answer in general as Etnyre and Van Horn-Morris expect in [9], it is worth trying to find sufficient conditions for to be hold.

### 1.2. Main results

The following are our main results that give sufficient conditions for .

Theorem 3.2. For , .

Theorem 5.1. Let be a deck transformation. Let be a simple closed geodesic curve in such that are pairwise disjoint and for some that divides . Let . Suppose that with

 Ψ(bd)=(TC∘Tι(C)∘Tι2(C)∘⋯∘Tιe−1(C))j.

Then and so

This theorem states that Question 1.1 has an affirmative answer for a symmetric mapping class which is a root of symmetric product of positive Dehn twists (see also Corollary 5.2).

Theorem 5.4. Suppose that the subsurfaces are pairwise non-isotopic. Assume that is either

• a non-negative power of a single Dehn twist (i.e., is an annulus which is neighborhood of a simple closed geodesic curve), or

• a pseudo-Anosov map (i.e., is a hyperbolic surface).

Suppose that satisfies

 Ψ(b)=[f1f2⋯fk]

then .

We say that a simple closed curve is symmetric if it is invariant under some deck translation. Question 1.1 can be understood as a question of the existence of factorizations of elements of into positive Dehn twists about symmetric simple closed curves.

A well-known example where a positive factorization coincides with a product of Dehn twists about symmetric simple closed curves may be the daisy relation [8]. In Example 5.7 we see that the technical looking assumptions in Theorem 5.4 can be understood as a generalization of the setting of the daisy relation, and view Theorem 5.4 as a generalization of the daisy relation.

### 1.3. Organization of the paper

In Section 2 we review results of Birman and Hilden that we need in this paper.

In Section 3, we show that the answer to the question is affirmative when the number of branch points is two (Theorem 3.1) or the degree of the branched covering is two with branch points (Theorem 3.2).

In Section 4, we discuss roots of quasi-positive braids and find conditions that a root of a quasi-positive braid is also quasi-positive. We obtain results that are used in Section 5.

In Section 5, we prove our main results (Theorems 5.1 and 5.4).

## 2. Birman-Hilden theory

Throughout the paper, unless otherwise stated, we always assume that the boundary of a surface is non-empty, any homeomorphism of a surface with marked points (punctures) fixes the boundary pointwise and permutes the marked points, and any isotopy of homeomorphisms pointwise fixes the boundary and the marked points. We denote by the isotopy class of a homeomorphism . For a simple closed curve in a surface , we denote by a right-handed Dehn twist about which is supported in a neighborhood of , and denote its isotopy class by . A simple closed curve in a surface is called essential if it is not homotopic to a point, a puncture, or a boundary component of the surface.

Let be points in the interior of a 2-disk . We may identify the braid group with the mapping class group of the -punctured disk . The fundamental group is the free group of rank and generated by , where is a loop winding once around counter-clockwise. For let be a homomorphism defined by for all . For and , let

 π:S≈Sn,k→D

be the -fold cyclic branched covering branched at such that the associated regular covering is the -fold cyclic cover corresponding to the subgroup of . We denote by the set of branch points in . Let be a generator of the deck transformation group .

We visualize and as follows. The covering can be viewed as the canonical Seifert surface of the -torus link represented as the closure of the -braid (see Figure 1). The deck transformation is the rotation of the surface about the braid axis through the branch points , and is not pointwise fixed by .

Let and be a homeomorphism representing the braid . Since via the action of on , there is a lift of (see [3, Lemma 5.1] and [12, Theorem 1.1]). Among the possible lifts which are related to each other by deck transformations, is the unique lift that fixes pointwise.

When two homeomorphisms and represent the same braid we note that an isotopy between and lifts to an isotopy between their lifts and . Thus, we have a well-defined homomorphism

 Ψ:Bn→Mod(S)

defined by . Let

 Θ=Θk:Mod(S)→Mod(S)

be an automorphism defined by .

###### Definition 2.1.

Define

 SMod(S):={[f]∈Mod(S) ∣∣ f∈Homeo+(S) and Θ([f])=[f]}

and call it the symmetric mapping class group. An element of is called symmetric mapping class.

The following fact is attributed to Birman and Hilden.

###### Proposition 2.2.

The homomorphism is injective and onto .

For the sake of completeness, we include a proof of Proposition 2.2.

###### Proof of Proposition 2.2.

The injectivity of follows from [3]. A homeomorphism is called fiber-preserving if for any with . Suppose that satisfy . According to [3, Theorem 1], since the lifts represent the same element of and are fiber-preserving, there exists a fiber-preserving isotopy between and . Since is fiber-preserving the composition is a well-defined homeomorphism of and it gives an isotopy between and ; hence, .

To see , assume that satisfies . The same argument as in the proof of [3, Theorem 3] with Teichmüller’s theorem for bordered surfaces [1, page 59] shows that can be represented by a homeomorphism such that .

For let be a pre-image of under the branched covering map . Define a homeomorphism by

 b(x)=π(f′(~x)).

Since , the image is independent of the choice of the pre-image .

We observe that acts on the branch set as a permutation. Suppose that and satisfy . Since fixes the branch set point-wise we have . That is, is a fixed point of and . We get which shows .

Since the map is the unique lift of . We obtain .

To see , let be an simple closed curve that goes through th and th sheets and th and st twisted bands as depicted in Figure 1. Let be a positive Dehn twist about . It is shown in [17, Lemma 3.1] that for the standard braid generators of the Artin braid group we have

 Ψ(σi)=[ti,1∘ti,2∘⋯∘ti,k−1]∈Mod(S).

For and let (see Figure 1) be properly embedded arcs on the th band between th and st sheets such that . Let denote the isotopy class of relative to . We get

 [ι−1∘(ti,1∘ti,2∘⋯∘ti,k−1)∘ι][λjl]=[ti,1∘ti,2∘⋯∘ti,k−1][λjl]

for all and . Knowing that the surface cut along the union of arcs yields disjoint disks, the Alexander trick [10, Proposition 2.8] implies that . This shows that . ∎

## 3. Positive answers for simple cases

###### Theorem 3.1.

We have for all .

###### Proof.

The first equality is obvious.

By Proposition 1.3 we only need prove . Take a braid . There exists a positive integer such that . Let . By Lemma 3.1 of [17] we have . Clearly . Since we have ; hence,

###### Theorem 3.2.

For , .

The equality has been known and used in the literature. However for sake of completeness, we give a proof of this case along with the case of .

###### Proof.

Since the case is shown in Theorem 3.1 we may assume or .

Let (for the case) and (for the case) be simple closed curves on as depicted in Figure 2. The Dehn twists about these curves generate , and holds.

If then there exist non-boundary parallel simple closed curves (possibly for ) and such that

 Ψ(b)={TΓ1∘⋯∘TΓm∘Txδ(n=3)TΓ1∘⋯∘TΓm∘Txδ1∘Tyδ2(n=4).

We may assume that all of the curves are non-separating since the positive Dehn twist about a separating curve can be written as a product of the positive Dehn twists about non-separating curves. Since is non-separating for some . We can write as

 fi=Tεlajl∘⋯∘Tε1aj1

for some and . Since the Dehn twist along the boundary does not affect simple closed curves in , the words , and are not needed in the expression. Note that for . Therefore, we have

 TΓi=Tf(a1)=fi∘Ta1∘f−1i=Ψ(bi)Ψ(σ1)Ψ(b−1i)=Ψ(biσ1b−1i)

and for all . In particular, by Proposition 2.2.

For the case , since , and which give

 TΓ1∘⋯∘TΓm∘Txδ1∘Tyδ2 = Θ(TΓ1∘⋯∘TΓm∘Txδ1∘Tyδ2) = Θ(TΓ1∘⋯∘TΓm)∘Θ(Txδ1)∘Θ(Tyδ2) = TΓ1∘⋯∘TΓm∘Txδ2∘Tyδ1.

Therefore, .

With the chain relations [10, Proposition 4.12] and , we can conclude . ∎

## 4. Roots in quasi-positive braids

In this section we address the following question. Among the results, Corollary 4.6 plays an important role to prove our main theorems in Section 5.

###### Question 4.1.

Are and closed under taking roots? That is,

(1) Does for an integer imply ?

(2) Does for an integer imply ?

###### Definition 4.2 (Property (QP-root)).

We say that a quasi-positive braid has the property (QP-root) if the following condition is satisfied.

 If b=xd for x∈Bn and d≥2 then x∈QP(n).

It is shown in [15, Theorem 1.1] that the -th root of a braid (if it exists) is unique up to conjugacy; namely, implies that and are conjugate to each other. This leads to the following observation.

###### Proposition 4.3.

For assume that . Then if and only if .

Let be a non-periodic reducible braid. According to [16, Definition 5.1], up to conjugacy, is in the following regular form.

Let be an essential multi-curve in the -punctured disk such that

• is isotopic to , and

• any simple closed curve which has non-zero geometric intersection with must not be preserved by any power of .

Such a multi-curve is called a canonical reduction system for [5] and it is unique up to conjugacy. It always exists for a non-periodic reducible braid [21].

Let be the set of outermost curves of . There could exist punctures in not enclosed by any circle in . Define a set of curves to be the union of and one circle around each such puncture of . We may enumerate the elements of as

 A={a1,1,…,a1,r1,a2,1,…,a2,r2,…,ac,1,…,ac,rc}

so that and . This action of on gives disjoint braided tubes (see Figure 3 (i)). Identifying each tube with a string, we get an -braid which we call the tubular braid associated to . By the nature of the canonical reduction system, this braid is unique up to conjugacy.

The braid closure of gives a -component link. The braid closure of the original braid can be viewed as a satellite of the -component link. Let denote the braiding inside the tube which starts at and ends at (where the indices are considered up to modulo ), which we call the interior braids.

We say that is in regular form if

 (4.1) the only non-trivial interior braids are b1,r1,…,bc,rc.

We denote these non-trivial interior braids by . The condition (4.1) can be realized by shifting non-trivial interior braid along the closure of the tubular braid , which is equivalent to taking a conjugate (see Figure 3 (ii)). In this process, the tubular braid remains the same and

 (4.2) b[i]=bi,1bi,2⋯bi,ri (up to conjugacy).
###### Remark 1.

In [16] a regular form requires one more property that if and are conjugate then . But we do not use this property in this paper.

Here is a simple observation.

###### Lemma 4.4.

If a braid is in regular form with quasi-positive tubular braid and interior braids then is also quasi-positive.

The converse of Lemma 4.4 is not true. The reducible -braid is quasi-positive and in regular form with the tubular braid and the interior braid .

The next proposition gives a criterion of the property (QP-root).

###### Proposition 4.5.

Let be a non-periodic reducible, quasi-positive braid. If has a regular form such that all of its tubular and interior braids are quasi-positive and having the property (QP-root), then also has the property (QP-root).

###### Proof.

Let be a non-periodic reducible, quasi-positive braid. Assume that for some and . We will show that is quasi-positive. Suppose that is in regular form with quasi-positive tubular braid and quasi-positive interior braids all of which have the property (QP-root).

Since is non-periodic and reducible, the root is also non-periodic and reducible. We may assume that is in regular form with tubular braid and interior braids . Since we see that and is conjugate to . That is, there exists such that . By the property (QP-root) of , the tubular braid (hence, ) is quasi-positive.

As for the interior braids, each is conjugate to a power of some single interior braid . Moreover, for each there exists such that is conjugate to a power of . Therefore, by the property (QP-root) of , all the interior braids are quasi-positive.

By Lemma 4.4 we conclude that is quasi-positive. ∎

A classical theorem of Keréjártó and Eilenberg states that any root of a periodic -braid is conjugate to either or for some . Thus, every periodic braid has the property (QP-root). This fact and Proposition 4.5 imply the following corollary.

###### Corollary 4.6.

Let admitting a factorization where are pairwise disjoint simple closed curves in and . If for some then .

The above results show that Question 4.1, whether a quasi-positive braid has the property (QP-root), is reduced to pseudo-Anosov braids.

The next proposition gives a potential negative answer to Question 4.1.

###### Proposition 4.7.

Let be a pseudo-Anosov quasi-positive braid. Let denote the abelianization (exponent sum) map. If for some and with then does not have the property (QP-root).

###### Proof.

Since , if is quasi-positive then must be the positive half twist about a simple arc connecting two distinct punctures of . Such a mapping class is reducible. Hence, is reducible and cannot be pseudo-Anosov. ∎

At this time of writing, we do not know whether a pseudo-Anosov quasi-positive braid that satisfies the assumption of Proposition 4.7 exists or not.

## 5. Sufficient conditions for quasi-positive

In this section, we study quasi-positive braids admitting certain symmetric conditions. Throughout this section, we fix a hyperbolic structure on so that the deck transformation is an isometry (see [11, Theorem 11.6]). Let be a connected subsurface of that satisfies one of the following conditions.

• is an annular neighborhood of a geodesic simple closed curve in .

• is a hyperbolic surface with geodesic boundary and the inclusion is an isometry.

Note that the surface also satisfies the same property. For let be a homeomorphism extending such that for . For let

 fi:=ιi−1∘ˆf∘ι−i+1∈Homeo+(S).

Our goal is to study elements of the form

 ϕ=f1f2⋯fk∈Homeo+(S)

and find sufficient conditions that guarantees .

We first study the following special case.

###### Theorem 5.1.

Let be a simple closed geodesic curve in such that are pairwise disjoint with for some that divides . Let . Suppose that with

 Ψ(bd)=(TC∘Tι(C)∘Tι2(C)∘⋯∘Tιe−1(C))j.

Then and so

When we get the following.

###### Corollary 5.2.

Let . If with then .

By Proposition 2.2 it is easy to see that implies .

###### Proof of Theorem 5.1.

Since is simple and , the projection is also simple.

First, we treat an exceptional case where the projection is a simple proper arc in the -punctured disk connecting two distinct punctures. This can be realized only if , and is a non-separating simple closed curve in . Let be the braid represented by a positive half twist about the arc . We have . Thus, . Since is injective (Proposition 2.2) . Let be the abelianization map defined by . Since we get and . Proposition 4.3 implies that .

Next, suppose that the projection is a simple closed curve in the punctured disk . Let . Since the map restricted to each connected component of is a cover we have

 (5.1) Ψ((Tπ(C))k′)=TC∘Tι(C)∘Tι2(C)∘⋯∘Tιe−1(C).

Hence,

 Ψ(bd)=(TC∘Tι(C)∘Tι