Generating twist subgroup of mapping class group of non-orientable surface by involutions

TL;DR

This paper proves that the twist subgroup of the mapping class group of a non-orientable surface can be generated by six involutions for certain genera, advancing understanding of the algebraic structure of these groups.

Contribution

It establishes that the twist subgroup is generated by six involutions for non-orientable surfaces of genus at least 14 or 16, providing a new generating set.

Findings

Twist subgroup generated by six involutions for g ≥ 16 or g = 14

Improves understanding of algebraic structure of mapping class groups

Provides explicit generating set for specific genera

Abstract

Let $N_{g}$ denote the closed non-orientable surface of genus $g$ and let ${\mathcal M} _g$ denote the mapping class group of $N_{g}$. Let ${\mathcal T} _g$ denote the twist subgroup of ${\mathcal M} _g$ which is the subgroup of ${\mathcal M} _g$ is generated by all Dehn twists. In this thesis, we proved that ${\mathcal T} _g$ is generated by six involutions for $g \geq 16$ or $g = 14$.

Equations

$$t_{f(a)}^{ϵ}=f{t}_a{f}^{-1},$$

$$Y_{m^{-1},a}$$

$$Y_{m,a^{-1}}$$

$$f{Y}_{m,a}{f}^{-1}$$

Full text

1. Introduction

For $g\geq 1$, let $N_{g}$ denote the closed connected non-orientable surface of genus $g$. The mapping class group $\mathcal{M}_{g}$ is the group of isotopy classes of diffeomorphisms of $N_{g}$. A simple closed curve $c$ in $N_{g}$ is two-sided (resp. one-sided) if a regular neighborhood of $c$, denoted by $N_{c}$, is an annulus (resp. a Möbius band). Let $a$ be a two-sided simple closed curve on $N_{g}$. By the definition, the regular neighborhood of $a$ is an annulus, and it has two possible orientation. Now, we fix one of its two possible orientations. For two sided simple closed curve $a$, we can define the Dehn twist $t_{a}$. We indicate the direction of a Dehn twist by an arrow beside the curve $a$ as shown in Fig. 1.

For $g\geq 2$, Lickorish ([Li1], [Li2]) first proved that $\mathcal{M}_{g}$ is not generated by Dehn twists, and $\mathcal{M}_{g}$ is generated by Dehn twists and Y-homeomorphisms. Chillingworth [C] found a finite set of generators of this group. The number of Chillingworth’s generators is improved to $g+1$ by Szepietowski [S3]. Hirose [Hi] proved that his generating set is the minimal generating set by Dehn twists and Y-homemorphisms. Szepietowski show that $\mathcal{M}_{g}$ is generated by four involutions for $g\geq 4$ [S2]. Denote by $\mathcal{T}_{g}$ the subgroup of $\mathcal{M}_{g}$ is generated by all Dehn twists. We call $\mathcal{T}_{g}$ the twist subgroup of $\mathcal{M}_{g}$. The group $\mathcal{T}_{g}$ is an index $2$ subgroup of $\mathcal{M}_{g}$ (see [Li1]). In particular, $\mathcal{T}_{g}$ is finitely generated. Chillingworth [C] proved that $\mathcal{T}_{g}$ is generated by a Dehn twist if $g=2$, two Dehn twists if $g=3$, $(3g-1)/2$ Dehn twists if $g\geq 5$ is odd, $(3g)/2$ Dehn twists if $g\geq 4$ is even. Stukow [St2] proved that $\mathcal{T}_{g}$ admits a finite presentation. Omori [O] showed that $\mathcal{T}_{g}$ is generated by $g+1$ Dehn twists for $g\geq 4$. Recently, Du [Du] showed that $\mathcal{T}_{g}$ is generated by three elements of finite order if $g\geq 5$ is odd. One of his generator is of order $2g$ and the other two are of order $2$.

On the other hand, it is not known the generating set for $\mathcal{T}_{g}$ consisting only of involutions. In this paper, we showed that $\mathcal{T}_{g}$ is generated by involutions.

Theorem 1.1**.**

For $g\geq 16$ or $g=14$, $\mathcal{T}_{g}$ is generated by six involutions.

The paper is organized as follows. In Section $2$ we recall the properties of Dehn twists and Y-homeomorphisms. In Section $3$ we construct involutions of $\mathcal{T}_{g}$ and prove the theorem 1.1.

2. Preliminaries

We represent the surface $N_{g}$ as a connected sum of $g$ projective planes as in Fig. 2. In this Figure, each encircled cross mark represents a crosscap: the interior of the encircled disk is to be removed and each pair of antipodal points on the boundary are to be identified.

Let $a_{1},a_{2},\ldots,a_{g-1},b,$ and $e$ be two-sided simple closed curves on $N_{g}$ as in Fig. 2 and Fig. 3.

Let $m_{1},m_{2},\ldots,m_{g}$ be one-sided simple closed curves on $N_{g}$ as in Fig. 4.

A Y-homeomorphism is defined as follow. For a one-sided simple closed curve $m$ and a two-sided oriented simple closed curve $a$ which intersects $m$ transversely in one point, the regular neighborhood $K$ of $m\cup a$ is homomeomorphic to the Klein bottle with one hole. Let $M$ be the regular neighborhood of $m$. Then the Y-homeomorphism $Y_{m,a}$ is the isotopy class of the diffeomorphism obtained by pushing $M$ once along $a$ keeping the boundary of $K$ fixed (see Fig. 5).

Dehn twists and Y-homeomorphisms have the following properties.

Lemma 2.1**.**

For any element $f$ in $\mathcal{M}_{g}$ and a two-sided simple closed curve $a$, we have

[TABLE]

where if $f\mid_{N_{a}}$ is an orientation preserving diffeomorphism (resp. orientation reversing diffeomorphism), then $\epsilon=1$ (resp. $\epsilon=-1$).

Lemma 2.2**.**

For a one-sided simple closed curve $m$, a two-sided simple closed curve $a$, and diffeomorphism $f$ on $N_{g}$, we have the following.

[TABLE]

Omori [O] reduced the number of Dehn twist generators for $\mathcal{T}_{g}$ and showed:

Theorem 2.3**.**

For $g\geq 4$, $\mathcal{T}_{g}$ is generated by $g+1$ Dehn twists $t_{a_{1}},t_{a_{2}},\ldots,t_{a_{g-1}},t_{b},$ and $t_{e}$.

Remark 2.4**.**

The minimum number of generators for $\mathcal{T}_{g}$ by Dehn twists is at least $g$ for $g\geq 4$ (see [O]). Omori asked the following problem: Which of $g$ and $g+1$ is the minimum number of generators for $\mathcal{T}_{g}$ by Dehn twists when $g\geq 4$?

We set a basis of ${\mathrm{H}_{1}}(N_{g},\mathbb{R})$ as in Fig. 6.

It is known that a homology group ${\mathrm{H}_{1}}(N_{g},\mathbb{R})$ has a basis in which every linear map $f_{*}:{\mathrm{H}_{1}}(N_{g},\mathbb{R})\rightarrow{\mathrm{H}_{1}}(N_{g},\mathbb{R})$, induced by a diffeomorphism $f:N_{g}\rightarrow N_{g}$, has a matrix with integral coefficients. Therefore we can define the homomorphism $D:\mathcal{M}_{g}\rightarrow\mathbb{Z}_{2}$ as follows: $D(f)=det(f_{*})$. We recall the properties of this homomorphism $D$ (see [St1]).

Lemma 2.5**.**

Let $f$ be an element of $\mathcal{M}_{g}$. Then $D(f)=1$ (resp. $D(f)=-1$) if $f$ is (resp. is not) in $\mathcal{T}_{g}$.

Remark 2.6**.**

Let $y$ be any Y-homeomorphism. Then $D(y)=-1$.

3. Proof of Theorem 1.1

In this section, we prove Theorem 1.1. For $r\geq 7$, We assume that $g=2r$.

We deform $N_{g}$ in Fig. 2 to $N_{g}$ in Fig.7 by diffeomorphism $\psi$

so that $N_{g}$ in Fig .7 is symmetrical with respect to the plane, denoted by mirror, illustrated in the middle of this surface. Let $\widetilde{\sigma}$ be a reflection in the mirror and let $\sigma$ be a composition $\psi^{-1}\widetilde{\sigma}\psi$. Then $\sigma$ is involution of $\mathcal{M}_{g}$. By an easy calculation, we found $D(\sigma)=-1$ if $r$ is odd and $D(\sigma)=1$ if $r$ is even. When $g$ is odd, $\sigma$ is not element in $\mathcal{T}_{g}$ by lemma 2.5. So, we consider a product $\sigma Y_{m_{g-1},a_{g-1}}$. This element $\sigma Y_{m_{g-1},a_{g-1}}$ is in $\mathcal{T}_{g}$. Since $\sigma(a_{g-1})=a_{g-1}$ and since $\sigma\mid_{N_{a_{g-1}}}$ is orientation reversing, we found that $\sigma Y_{m_{g-1},a_{g-1}}$ is an involution by lemma 2.2.We rewrite $\sigma Y_{m_{g-1},a_{g-1}}$ as $\sigma$. Therefore, for $r\geq 7$, $\sigma$ is involution in $\mathcal{T}_{g}$.

In the same way, we define $\tau$ as reflection in the mirror in Fig. 8.

We found $D(\tau)=1$ if $r$ is odd and $D(\tau)=-1$ if $r$ is even. When $g$ is even, $\tau$ is not element in $\mathcal{T}_{g}$ by lemma 2.5. So, we consider a product $\tau Y_{m_{g-1},a_{g-1}}$. This element $\tau Y_{m_{g-1},a_{g-1}}$ is in $\mathcal{T}_{g}$. By lemma 2.2, $\tau Y_{m_{g-1},a_{g-1}}$ is an involution. We rewrite $\tau Y_{m_{g-1},a_{g-1}}$ as $\tau$. Therefore, for $r\geq 7$, $\tau$ is involution in $\mathcal{T}_{g}$.

We construct third involution. We define $\upsilon$ as reflection in the mirror in Fig. 9 and we found $D(\upsilon)=1$. Hence, the element $\upsilon$ is involution in $\mathcal{T}_{g}$.

Let $\rho_{1}$, $\rho_{2}$, and $\rho_{3}$ be $\sigma t_{a_{1}}$, $\sigma t_{b}$, and $\tau t_{e}$, respectively. By lemma 2.1, $\rho_{i}$ is involution in $\mathcal{T}_{g}$ for $i=1,2,3$. Let $G$ be a subgroup of $\mathcal{T}_{g}$ generated by $\sigma$, $\tau$, $\upsilon$, $\rho_{1}$, $\rho_{2}$, and $\rho_{3}$. Since $t_{a_{1}}$ is a product $\sigma\rho_{1}$, $t_{a_{1}}$ is a product of two involutions. In the same way, $t_{b}$ and $t_{e}$ are products of two involutions. We show that $t_{a_{2}},t_{a_{3}},\ldots,t_{a_{g-1}}$ are products of involutions.

Let $a$ and $b$ be two-sided simple closed curves on $N_{g}$. For $f\in G$, the symbol $a\stackrel{{\scriptstyle f}}{{\longleftrightarrow}}b$ means that $f(a)=b$ or $f^{-1}(a)=b$. Since $a_{1}$ is mapped to $a_{2},a_{3},\ldots,a_{g-1}$ by $\sigma$, $\tau$, and $\upsilon$ (see Fig. 10), $t_{a_{i}}$ is a product of involutions by lemma 2.1 for $i=1,2,\ldots,g-1$. Since all Omori’s generators for $\mathcal{T}_{g}$ is in $G$, we prove that $G$ is equal to $\mathcal{T}_{g}$.

In the case of $g=2r+1(r\geq 7)$, we can prove Theorem 1.1 in same way. Involutions $\sigma$, $\tau$, and $\upsilon^{\prime}$ are reflection in the mirror in Fig. 11 , Fig. 12 , and Fig. 13 , respectively.

If $\sigma$ (resp. $\tau$) is not in $\mathcal{T}_{g}$, we rewrite $\sigma Y_{m_{g-1},a_{g-1}}$ (resp.  $\tau Y_{m_{g-1},a_{g-1}}$) as $\sigma$ (resp. $\tau$). Since the element $\upsilon^{\prime}$ is not in $\mathcal{T}_{g}$, we define $\upsilon$ as a product $\upsilon^{\prime}Y_{m_{9},a_{9}}$. This element $\upsilon$ is in $\mathcal{T}_{g}$ and this element is an involution by lemma 2.2. Involution $\rho_{1}$, $\rho_{2}$, and $\rho_{3}$ are the same as the case of $g=2r$. In the same way as the case of $g=2r$, all Omori’s generators for $\mathcal{T}_{g}$ is in $G$.

So we proved Theorem 3.1:

Theorem 3.1**.**

For $g\geq 16$ or $g=14$, $\mathcal{T}_{g}$ is generated by involutions $\sigma$, $\tau$, $\upsilon$, $\rho_{1}$, $\rho_{2}$, and $\rho_{3}$.

4. Concluding Remarks

We found the construction of $\upsilon$ needs $g\geq 16$ or $g=14$. By constructing of three other involutions instead of $\upsilon$, $\mathcal{T}_{g}$ can be generated by eight involutions in case of $g\geq 8$. In this case, $\mathcal{T}_{g}$ is generated by $\sigma,\tau,\rho_{1},\rho_{2},\rho_{3}$ in Section $3$ and additionally $\eta$, $\theta$, and $\rho_{4}$. Let $\eta^{\prime}$ and $\theta^{\prime}$ are involutions of $\mathcal{T}_{g}$ which are reflections in the mirrors in Fig. 14 and Fig. 15, respectively. We define $\eta=\eta^{\prime}Y_{m_{7},a_{7}}$ and $\theta=\theta^{\prime}Y_{m_{1},a_{1}}$. A element $\rho_{4}$ is a product of $\eta t_{a_{3}}$. Then, $\eta$, $\theta$, and $\rho_{4}$ are involutions in $\mathcal{T}_{g}$ by lemma 2.2 and 2.5. In the same way as Section 3, We can proved the following:

Theorem 4.1**.**

For $g\geq 8$, $\mathcal{T}_{g}$ is generated by eight involutions.

In the case of $\mathcal{M}_{g}$, Szepietowski show that $\mathcal{M}_{g}$ is generated by $4$ involutions for $g\geq 4$.

We can consider following problem:

Problem 4.2**.**

For $g\geq 4$, can the twist subgroup $\mathcal{T}_{g}$ be generated by $4$ involutions?

Acknowledgments

The author would like to thank Susumu Hirose and Naoyuki Monden for many advices. He would also like to thank Genki Omori for many helpful suggestions and comments of the twist subgroup.