Free commuting involutions on closed two-dimensional surfaces

TL;DR

This paper investigates the maximum number of free commuting involutions on orientable surfaces, identifying minimal genus surfaces with a given number of involutions as specific real moment-angle complexes.

Contribution

It establishes a precise relationship between the number of involutions and the minimal genus of surfaces, characterizing these surfaces as real moment-angle complexes derived from polygons.

Findings

Minimal genus surfaces with n involutions are real moment-angle complexes.

The genus of these surfaces is given by g = 1 + 2^{n-1}(n-2).

Identifies the boundary of an (n+2)-gon as key in the surface structure.

Abstract

We consider the function $f(g)$ that assigns to an orientable surface $M$ of genus $g$ the maximal number of free commuting independent involutions on $M$. We show that the surface of minimal genus $g$ with $f(g)=n$ is a real moment-angle complex $R_K$, where $K$ is the boundary of an $(n+2)$-gon. The genus is given by the formula $g = 1 + 2^{n-1}(n-2)$.

Equations

$$(\text{X},\text{A})=\{(X_1,A_1),\dots ,(X_m,A_m)\}$$

$$({\textbf{{X}}},{\textbf{{A}}})^{I}=\bigl{\{}(x_{1},\ldots,x_{m})\in\prod_{j=1}^{m}X_{j}\colon\;x_{j}\in A_{j}\quad\text{for }j\notin I\bigl{\}}$$

$$({\textbf{{X}}},{\textbf{{A}}})^{\mathcal{K}}=\bigcup_{I\in\mathcal{K}}({\textbf{{X}}},{\textbf{{A}}})^{I}=\bigcup_{I\in\mathcal{K}}\Bigl{(}\prod_{i\in I}X_{i}\times\prod_{i\notin I}A_{i}\Bigl{)}.$$

$${\mathcal{R}}_{\mathcal{K}}=([-1,1],\{-1,1\}{)}^{\mathcal{K}},$$

$$g=1+2^{m-3}(m-4).$$

$$\begin{gathered} \{({\epsilon }_1,\dots ,{\epsilon }_{i-1},x_i,{\epsilon }_{i+1},\dots ,{\epsilon }_{j-1},x_j,{\epsilon }_{j+1},\dots ,{\epsilon }_m):x_i,x_j\in [-1,1]\}, \\ \text{where}{\epsilon }_1,\dots ,{\epsilon }_m=\pm 1. \end{gathered}$$

$$\{({\epsilon }_1,\dots ,{\epsilon }_{i-1},x_i,x_{i+1},{\epsilon }_{i+2},\dots ,{\epsilon }_m):x_i,x_{i+1}\in [-1,1]\},\text{where}{\epsilon }_1,\dots ,{\epsilon }_m=\pm 1$$

$$\chi ({\mathcal{R}}_{\mathcal{K}})=V-E+F=V-2F+F=V-F=2^{m-2}(4-m),$$

$$f(h)=\max \{n:\text{there is an action of Z2n on a surface of genus }h\}.$$

$$G={\pi }_1(B)=⟨a_1,b_1,\dots ,a_g,b_g|a_1{b}_1{a}_1^{-1}{b}_1^{-1}\dots a_g{b}_g{a}_g^{-1}{b}_g^{-1}=1⟩.$$

$$G/H\approx {\mathbb{Z}}_2^n.$$

$$[a_i{]}^2=[b_j{]}^2=1,$$

$$G={\pi }_1(B)=⟨a_1,\dots ,a_g|a_1^2\dots a_g^2=1⟩$$

$$P=⟨g^2,g\in G⟩,$$

$$\chi (M_g)=a\cdot 2^n\text{with }n⩽2-a,a⩽1.$$

$$H(g)=\frac{W\left(\frac{1}{2}(g-1)\ln 2\right)}{\ln 2}+2,$$

$$g=1+2^{n-1}(b-2),\text{where}n⩽b,b⩾1.$$

$$g(n)=1+2^{n-1}(n-2)⩽1+2^{n-1}(b-2)=g,$$

Full text

1. Introduction

One of the main objects of study in toric topology is the moment-angle-complex $\mathcal{Z}_{\mathcal{K}}$, which is a cell complex with a torus action constructed from a simplicial complex $\mathcal{K}$, see [1]. Along with the moment-angle complex $\mathcal{Z}_{\mathcal{K}}$, its real analog $\mathcal{R}_{\mathcal{K}}$ is considered and has many interesting properties. If $\mathcal{K}$ is a simplicial subdivision of an $(n-1)$-dimensional sphere with $m$ vertices, then $\mathcal{Z}_{\mathcal{K}}$ is an $(m+n)$-dimensional (closed) manifold, and $\mathcal{R}_{\mathcal{K}}$ is an $n$-dimensional manifold. In particular, if $\mathcal{K}$ is the boundary of an $m$-gon, then $\mathcal{R}_{\mathcal{K}}$ is an orientable closed two-dimensional manifold (a surface). Given a positive integer $g$, let $f(g)$ be the maximal number of free commuting independent involutions on a closed orientable surface $M_{g}$ of genus $g$ (that is, $f(g)$ is the largest possible $n$ such that $(\mathbb{Z}_{2})^{n}$ acts on $M_{g}$ freely). In this paper, it is proved that the function $f(g)$ attains a local maximum on the surfaces $\mathcal{R}_{\mathcal{K}}$ see Fig. 1. In other words, for any integer $n$, the surface of minimum genus $g$ that supports a free action of $\mathbb{Z}_{2}^{n}$ has the form $\mathcal{R}_{\mathcal{K}}$, where $\mathcal{K}$ is an $(n+2)$-gon. The genus is given by the formula $g=1+2^{n-1}(n-2)$.

The author is grateful to her supervisor Taras Panov for suggesting the problem and attention to this work.

2. Basic notions and preliminary statements

A moment-angle complex is a special case of the following construction:

Construction 2.1** (polyhedral product).**

Let $\mathcal{K}$ be a simplicial complex on the set $[m]={1,2,\ldots,m}$ and let

[TABLE]

be a collection of $m$ pairs of topological spaces, $A_{i}\subset X_{i}$. For each subset $I\subset[m]$ denote

[TABLE]

and define the polyhedral product of $({\textbf{{X}}},{\textbf{{A}}})$ corresponding to a simplicial complex $\mathcal{K}$ by

[TABLE]

Here the union is a subset of $\prod_{j=1}^{m}X_{j}$.

In the case when all pairs $(X_{i},A_{i})$ are the same, i.e. $X_{i}=X$ and $A_{i}=A$ for $i=1,\ldots,m$, we use the notation $(X,A)^{\mathcal{K}}$ for $({\textbf{{X}}},{\textbf{{A}}})^{\mathcal{K}}$.

If $(X,A)=(D^{2},S^{1})$ then $\mathcal{Z}_{\mathcal{K}}=(D^{2},S^{1})^{\mathcal{K}}$ is called the moment-angle complex. We will consider its real analogue:

[TABLE]

where $[-1,1]=D^{1}$  is the a line segment, and $\{-1,1\}=\partial[-1,1]$  is the pair of points.

If $\mathcal{K}$  is a simplicial decomposition of $(n-1)$-dimensional sphere, then $\mathcal{R}_{\mathcal{K}}$ is an $n$-dimensional manifold with the action of group $(\mathbb{Z}_{2})^{m}$ (see [1, Theorem 4.1.7]). In other words, we have $m$ commuting involutions on $\mathcal{R}_{\mathcal{K}}$. Of these $m$ involutions no more than $m-n$ act freely. For completeness, below we present a proof of these facts for $\mathcal{K}$ being the boundary of an $m$-gon. The surfaces $\mathcal{R}_{\mathcal{K}}$ corresponding to $m$-gons appear as ‘‘regular topological skew polyhedra’’ in the 1937 work of Coxeter  [2].

Proposition 2.2**.**

Let $\mathcal{K}$  be the boundary of an $m$-gon. Then $\mathcal{R}_{\mathcal{K}}$  is an orientable closed surface of genus

[TABLE]

Proof.

Consider the cube $I^{m}=[-1,1]^{m}$ with the standard structure of cubical complex. Its two-dimensional skeleton consists of faces of the form

[TABLE]

By definition, $\mathcal{R}_{\mathcal{K}}$ is a cubical subcomplex of $I^{m}$. Indeed, $\mathcal{R}_{\mathcal{K}}$ is a union of two-dimensional faces of cube of the following form:

[TABLE]

(here and below we consider subscripts modulo $m$, i.e., $m+1\equiv 1$). Each one-dimensional face $\{(\varepsilon_{1},\dots,x_{i},\dots,\varepsilon_{m})\colon x_{i}\in[-1,1]\}$ is the boundary of precisely two squares, namely $(x_{i},x_{i+1})\in[-1,1]^{m}$ and $(x_{i-1},x_{i})\in[-1,1]^{m}$. Moreover, the intersection of two squares is either empty, or a vertex, or a one-dimensional face (since this is valid for the whole cube and one-dimensional faces of $\mathcal{R}_{\mathcal{K}}$ form the whole one-dimensional skeleton of the cube). Each vertex is contained in precisely $m$ squares. This implies that $\mathcal{R}_{\mathcal{K}}$ is a closed two-dimensional orientable manifold.

The surface $\mathcal{R}_{\mathcal{K}}$ is glued of $F=m2^{m-2}$ squares. Each square has $4$ edges, and each edge is contained in two $2$ squares. Therefore the number of edges is $E=4F/2=2F$. All vertices of the cube are vertices of our surface, therefore the number of vertices is $V=2^{m}$. Thus, the Euler characteristic is

[TABLE]

which immedially implies the formula for genus of the surface. ∎

Example 2.3**.**

Let $\mathcal{K}$ be the boundary of a triangle. Then $\mathcal{R}_{\mathcal{K}}$ is the boundary of a $3$-dimensional cube and therefore is homeomorphic to a sphere, i.e., the genus is [math].

Lemma 2.4**.**

If $\mathcal{K}$ is the boundary of an $m$-gon, then there ia a free action of $\mathbb{Z}_{2}^{m-2}$ on $\mathcal{R}_{\mathcal{K}}$.

Proof.

We describe a freely acting subgroup $\mathbb{Z}_{2}^{m-2}\subset\mathbb{Z}_{2}^{m}$ by explicitly defining its generators. Let $\psi_{i}$ be the involution sending $x_{i}$ to $-x_{i}$ and fixing the other coordinates. The involutions $\psi_{i}$, $i=1,\ldots,m$, commute and therefore generate a $\mathbb{Z}_{2}^{m}$-action. However, this action is not free as $\psi_{i}$ fixes the points whose $i$-th coordinate is zero. Consider the composition $\psi_{i}\circ\psi_{j}$, where $|i-j|>1$ (it corresponds to a diagonal in the polygon $\mathcal{K}$). The set of fixed points of the involution $\psi_{i}\circ\psi_{j}$ is the set with coordinates $x_{i}=x_{j}=0$, but $\mathcal{R}_{\mathcal{K}}$ does not contain such points, since $x_{i},x_{j}$ are not consecutive. If $m=2k$, then consider the $m-2$ involutions $\psi_{1}\circ\psi_{3}$, $\psi_{3}\circ\psi_{5}$, $\dots$, $\psi_{2k-3}\circ\psi_{2k-1}$ and $\psi_{2}\circ\psi_{4}$, $\psi_{4}\circ\psi_{6}$, $\dots$, $\psi_{2k-2}\circ\psi_{2k}$. These involutions commute pairwise and generate a free action of $\mathbb{Z}_{2}^{m-2}$. Similarly, for an odd $m=2k+1$ consider the $m-2$ involutions $\psi_{1}\circ\psi_{3}$, $\psi_{3}\circ\psi_{5}$, $\dots$, $\psi_{2k-3}\circ\psi_{2k-1}$, $\psi_{2}\circ\psi_{4}$, $\psi_{4}\circ\psi_{6}$, $\dots$, $\psi_{2k-2}\circ\psi_{2k}$ and $\psi_{1}\circ\psi_{2k}\circ\psi_{2k+1}$. ∎

Remark*.*

For even $m$, each element of the freely acting group $\mathbb{Z}_{2}^{m-2}$ defined above preserves the orientation of $\mathcal{R}_{\mathcal{K}}$, since it is a composition of an even number of elementary involutions $\psi_{i}$. For odd $m$, the involution $\psi_{1}\circ\psi_{2k}\circ\psi_{2k+1}$ reverses the orientation of $\mathcal{R}_{\mathcal{K}}$.

Next, consider an arbitrary closed two-dimensional surface $M$. We ask the following question: find the maximal $n$ such that there is a free action of $\mathbb{Z}_{2}^{n}$ on $M$. Obviously, $n$ depends only on the genus $h$ of the surface $M$. (The Euler characteristic of an orientable surface of genus $h$ is $2-2h$, and the Euler characteristic of a nonorientable surface of genus $h$ is $2-h$.) Define the function

[TABLE]

Let $f(h)=n$. Then $B=M/\mathbb{Z}_{2}^{n}$ is a closed two-dimensional manifold with Euler characteristic $\chi(B)=\chi(M)/2^{n}$.

Proposition 2.5**.**

Let $g$ be the genus of $B=M/\mathbb{Z}_{2}^{n}$. If the surface $B$ is orientable, then $n\leqslant 2g$. If $B$ is nonorientable, then $n\leqslant g$.

Proof.

Consider the case of orientable $B$. The fundamental group of $B$ is

[TABLE]

The fundamental group $\pi_{1}(M)=H$ of the manifold $M$ is a normal subgroup of the group $G$ (since the action of $\mathbb{Z}_{2}^{n}$ is free) and

[TABLE]

Therefore the square of each coset is 1,

[TABLE]

for all $i,j=1,...,g$. Moreover, all cosets (i.e., elements of $G/H\thickapprox\mathbb{Z}_{2}^{n}$) commute. The quotient group $G/H$ is generated by the cosets $[a_{1}],[b_{1}],\dots,[a_{g}],[b_{g}]$. Since they commute and their squares equal the unit, we can compose at most $2^{2g}$ words of them. Therefore the order of the group $G/H\thickapprox\mathbb{Z}_{2}^{n}$ is at most $2^{2g}$. Hence, $n\leqslant 2g$. For a nonorientable surface $B$ the argument is similar: the fundamental group

[TABLE]

has $g$ generators and the order of the quotient $G/H$ is at most $2^{g}$. ∎

Proposition 2.6**.**

• a)

For any orientable surface $B$ of genus $g$ and an integer $n\leqslant 2g$, there exists an orientable surface $M$ with a free action of $\mathbb{Z}_{2}^{n}$ such that $B=M/\mathbb{Z}_{2}^{n}$.

• b)

For any nonorientable surface $B$ of genus $g$ and an integer $n\leqslant g$, there exists a surface $M$ with a free action of $\mathbb{Z}_{2}^{n}$ such that $B=M/\mathbb{Z}_{2}^{n}$.

Proof.

Let $B$ be orientable. Consider the subgroup $P$ in the fundamental group $\pi(B)=G$ generated by the squares of all elements

[TABLE]

and consider its normalizer subgroup $H=GPG^{-1}$. Then $H$ is a normal subgroup and $H\neq G$, since $H$ contains only elements with even number of letters. The relations $[a_{i}]^{2}=[b_{j}]^{2}=[a_{i}b_{j}a_{i}b_{j}]=1$ in the quotient group $G/H$ imply that $G/H\cong\mathbb{Z}_{2}^{2g}$. Hence there exists a regular covering of $B$, which gives a free action of $G/H\cong\mathbb{Z}_{2}^{2g}$ (see [3]). Now add one generator to $P$: $P_{1}=\langle P,a_{1}\rangle$ and consider $H_{1}=GP_{1}G^{-1}$. Then $H_{1}$ is a proper normal subgroup of $G$ (since in each of its elements the generator $b_{g}$ occurs even number of times), and the quotient group is $G/H_{1}\cong\mathbb{Z}_{2}^{2g-1}$. The corresponding covering of $B$ is regular and gives a free action of $\mathbb{Z}_{2}^{2g-1}$. Continuing this process, we add to $P$ the other generators $a_{i},b_{i}$, and obtain regular coverings of $B$ corresponding to free actions of $\mathbb{Z}_{2}^{2g-k}$ for all $k=1,2,...,2g-1$.

The nonorientable case is considered similarly. ∎

3. Main results

Let $M_{g}$ be a surface of genus $g$.

Proposition 3.1**.**

Let $n$ be the maximal integer such that

[TABLE]

Then $f(g)=n$ if $a$ is even, and $n-1\leqslant f(g)\leqslant n$ if $a$ is odd.

Proof.

Let $\mathbb{Z}^{k}_{2}$ act freely on a surface $M_{g}$. Then $\chi(M_{g})=a^{\prime}\cdot 2^{k}$ where $a^{\prime}=\chi(M_{g}/~{}\mathbb{Z}^{k}_{2})$, and $k\leqslant 2-a^{\prime}$ by Proposition 2.5. By definition, $n$ is the maximum of such $k$, hence $f(g)\leqslant n$. Now we estimate $f(g)$ from below. Consider the two cases:

Case 1: $a$ is even. Consider an orientable surface $B$ with Euler characteristic $a$ and genus $\frac{2-a}{2}$. Since $n\leqslant 2\frac{2-a}{2}$, by Proposition 2.6 there exists an orientable surface $M$ such that $\mathbb{Z}^{n}_{2}$ acts freely on $M$ and $M=B/\mathbb{Z}^{n}_{2}$. The Euler characteristic is $\chi(M)=a\cdot 2^{n}=\chi(M_{g})$, therefore $M$ has genus $g$ and $f(g)\geqslant n$.

Case 2: $a$ is odd. Write $\chi(M)=2a\cdot 2^{n-1}$. Consider an orientable surface $B$ of Euler characteristic $2a$, i.e., of genus $\frac{2-2a}{2}$. We have $n-1\leqslant 2-2a$, since $n\leqslant 2-a$ and $a\leqslant 1$. By Proposition 2.6 there exists an orientable surface $M$ such that $\mathbb{Z}^{n-1}_{2}$ acts freely on $M$ and $M=B/\mathbb{Z}^{n-1}_{2}$. The Euler characteristic is $\chi(M)=2a\cdot 2^{n-1}=\chi(M_{g})$, therefore $M$ has genus $g$ and $f(g)\geqslant n-1$. ∎

Define $H(g)$ as the inverse function to $g(x)=1+2^{x-1}(x-2)$, that is,

[TABLE]

where $W$ denotes the Lambert function (i.e., the function inverse to $xe^{x}$).

Theorem 3.2**.**

Theorem $f(g)\leqslant H(g)$, and an equality is attained on real moment-angle manifolds and only on them.

Proof.

Substituting $\chi(M_{g})=2-2g$ and $a=2-b$ in (3.1) we obtain

[TABLE]

By Proposition 3.1, for the maximal $n$ satisfying the conditions above we have $f(g)\leqslant n$. On the other hand we have

[TABLE]

or equivalently $n\leqslant H(g)$. This implies the required inequality $f(g)\leqslant H(g)$. An equality is attained if and only if the genus $g$ can be written as $g=1+2^{n-1}(n-2).$ Compairing this expession with the formula from Proposition 2.2, we obtain that $M\cong\mathcal{R}_{\mathcal{K}}$, where $\mathcal{K}$ is the boundary of $(n+2)$-gon. ∎

The values of the function $f(g)$ are shown in Fig. 1, where the dashed line is the graph of $H(g)$.