On a question of Etnyre and Van Horn-Morris
We answer Question 6.12 in  asked by Etnyre and Van Horn-Morris.
Let be a compact oriented surface with boundary. Let denote the mapping class group of , namely the group of isotopy classes of orientation preserving homeomorphisms of that fix the boundary pointwise. Let be a boundary component of and let
The following theorem answers [1, Question 6.12] of Etnyre and Van Horn-Morris: For which the set forms a monoid?
Let be a surface that is not a pair of pants and has negative Euler characteristic. Let be a boundary component of . The set and hence is a monoid if and only if .
If is a pair of pants then is a monoid if and only if .
Theorem 1.1 states that is not a monoid. But contains , the monoid of right-veering mapping classes.
2. Basic study of quasi-morphisms
As shown in [4, Corollary 4.17] the FTDC map is not a homomorphism but a quasi-morphism if the surface has negative Euler characteristic. In order to prove Theorem 1.1 we first study general quasi-morphisms and obtain a monoid criterion (Theorem 2.2).
Let be a group. A map is called a quasi-morphism if
is finite. The value is called the defect of the quasi-morphism. A quasi-morphism is homogeneous if for all and . Every quasi-morphism can be modified to a homogeneous quasi-morphism by taking the limit:
A typical example of homogeneous quasi-morphism is the translation number
Here is the group of orientation-preserving homeomorphisms of that are lifts of orientation-preserving homeomorphisms of . The limit does not depend on the choice of . An important property of we will use is that
if then for all .
Given a quasi-morphism and let
It is easy to see that:
The set forms a monoid if .
Let be a homogeneous quasi-morphism which is a pull-back of the translation number quasi-morphism , namely, there is a homomorphism such that . Then forms a monoid for .
Let . Assume to the contrary that id not a monoid. There exist such that . That is . Take an integer so that
By the definition of the defect we have
Let and . By the property we have .
On the other hand, since by the property again we have for all . Therefore and
which is a contradiction. ∎
Proof of Theorem 1.1.
Next we show that is not a monoid for . For any non-separating simple closed curve and any boundary component of we have . Therefore we have i.e.,
(Case 1) Recall that for any surface of genus the group is generated by Dehn twists along non-separating curves (see p.114 of ). If were a monoid then this fact and (3.1) imply that which is clearly absurd. Thus is not a monoid if and .
(Case 2) If and let be the boundary components and be the simple closed curves as shown in Figure 1-(1). Let and . Since are non-separating
By the lantern relation, for any positive integer with we have
thus . This shows that is not a monoid for all and .
(Case 3) If and add additional boundary components in the place of as shown in Figure 1-(2). By a similar argument using the lantern relation we can show that is not a monoid for all and any . By the symmetry of the surface we can further show that is not a monoid for all and .
(Case 4) If and the group is generated by Dehn twists about non-separating curves. Thus this case is subsumed into Case 1.
Parallel arguments show that does not form a monoid for . ∎
We close the paper by proving Corollary 1.2.
The authors thank John Etnyre for pointing out an error in an early draft of the paper. TI was partially supported by JSPS Grant-in-Aid for Young Scientists (B) 15K17540. KK was partially supported by NSF grant DMS-1206770.
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