---
abstract: 'We obtain a simple presentation of the hyperelliptic mapping class group ${{\mathcal{M}}^h}(N)$ of a nonorientable surface $N$. As an application we compute the first homology group of ${{\mathcal{M}}^h}(N)$ with coefficients in $H_1(N;{{\mathbb{Z}}})$.'
address: ' Institute of Mathematics, University of Gdańsk, Wita Stwosza 57, 80-952 Gdańsk, Poland '
author:
- Michał Stukow
title: A finite presentation for the hyperelliptic mapping class group of a nonorientable surface
---

[^1]

Introduction
============

Let $N_{g,s}^n$ be a smooth, nonorientable, compact surface of genus $g$ with $s$ boundary components and $n$ punctures. If $s$ and/or $n$ is zero, then we omit it from the notation. If we do not want to emphasise the numbers $g,s,n$, we simply write $N$ for a surface $N_{g,s}^n$. Recall that $N_{g}$ is a connected sum of $g$ projective planes, and $N_{g,s}^n$ is obtained from $N_g$ by removing $s$ open disks and specifying the set $\Sigma=\{{{z}_1,\dotsc,{z}_{n}}\}$ of $n$ distinguished points in the interior of $N_g$.

Let ${\textrm{Diff}}(N)$ be the group of all diffeomorphisms ${h\colon N\to N}$ such that $h$ is the identity on each boundary component and $h(\Sigma)=\Sigma$. By ${{\mathcal{M}}}(N)$ we denote the quotient group of ${\textrm{Diff}}(N)$ by the subgroup consisting of maps isotopic to the identity, where we assume that isotopies fix $\Sigma$ and are the identity on each boundary component. ${{\mathcal{M}}}(N)$ is called the *mapping class group* of $N$.

The mapping class group ${{\mathcal{M}}}(S_{g,s}^n)$ of an orientable surface is defined analogously, but we consider only orientation preserving maps. If we include orientation reversing maps, we obtain the so-called *extended mapping class group* ${{\mathcal{M}}}^{\pm}(S_{g,s}^n)$.

Suppose that the closed orientable surface $S_g$ is embedded in ${{\mathbb{R}}}^3$ as shown in Figure \[r01\], in such a way that it is invariant under reflections across $xy,yz,xz$ planes. Let ${{\varrho}\colon S_g\to S_g}$ be the *hyperelliptic involution*, i.e. the half turn about the $y$-axis.

![Surface $S_g$ embedded in ${{\mathbb{R}}}^3$.[]{data-label="r01"}](R01){width="90.00000%"}

The *hyperelliptic mapping class group* ${{{\mathcal{M}}}}^h(S_g)$ is defined to be the centraliser of ${\varrho}$ in ${{{\mathcal{M}}}}(S_g)$. In a similar way we define the *extended hyperelliptic mapping class group* ${{{\mathcal{M}}}}^{h\pm}(S_g)$ to be the centraliser of ${\varrho}$ in ${{{\mathcal{M}}}}^{\pm}(S_g)$.

Background
----------

The hyperelliptic mapping class group turns out to be a very interesting and important subgroup of the mapping class group. Its algebraic properties have been studied extensively – see [@Bir-Hil; @Kawazumi1] and references there. Although ${{{\mathcal{M}}}}^h(S_g)$ is an infinite index subgroup of ${{{\mathcal{M}}}}(S_g)$ for $g\geqslant 3$, it plays surprisingly important role in studying its algebraic properties. For example Wajnryb’s simple presentation [@Wajn_pre] of the mapping class group ${{{\mathcal{M}}}}(S_g)$ differs from the presentation of the group ${{{\mathcal{M}}}}^h(S_g)$ by adding one generator and a few relations. Another important phenomenon is the fact, that every finite cyclic subgroup of maximal order in ${{{\mathcal{M}}}}(S_g)$ is conjugate to a subgroup of ${{{\mathcal{M}}}}^h(S_g)$ [@MaxHyp].

Homological computations play a prominent role in the theory of mapping class groups. Let us mention that in the case of the hyperelliptic mapping class group, Bödigheimer, Cohen and Peim [@BoedCoh] computed $H^*({{{\mathcal{M}}}}^h(S_g);{{\mathbb{K}}})$ with coefficients in any field ${{\mathbb{K}}}$. Kawazumi showed in [@Kawazumi1] that if $\textrm{ch}({{\mathbb{K}}})\neq 2$ then $H^*({{{\mathcal{M}}}}^h(S_g);H^1(S_g;{{\mathbb{K}}}))=0$. For the integral coefficients, Tanaka [@Tanaka] showed that $H_1({{{\mathcal{M}}}}^h(S_g);H_1(S_g;{{\mathbb{Z}}}))\cong{{\mathbb{Z}}}_2$. Let us also mention that Morita [@MoritaJacFou] showed that in the case of the full mapping class group, $H_1({{{\mathcal{M}}}}(S_g);H_1(S_g,{{\mathbb{Z}}}))\cong{{\mathbb{Z}}}_{2g-2}$.

Main results
------------

The purpose of this paper is to extend the notion of the hyperelliptic mapping class group to the nonorientable case. We define this group ${{{\mathcal{M}}}}^h(N)$ in Section \[sec:def:hiper\] and observe that it contains a natural subgroup ${{{\mathcal{M}}}}^{h+}(N)$ of index 2 (Remark \[rem:positiv:subg\]).

Then we obtain simple presentations of these groups (Theorems \[th:004\] and \[tw:pres:pos\]). By analogy with the orientable case, these presentations may be thought of as the first approximation of a presentation of the full mapping class group ${{{\mathcal{M}}}}(N)$. In fact, for $g=3$ the hyperelliptic mapping class group ${{{\mathcal{M}}}}^h(N)$ coincide with the full mapping class group ${{{\mathcal{M}}}}(N)$ (see Corollary \[Presen:g3:Bir\]). If $g\geq 4$, then Paris and Szepietowski [@SzepParis] obtained a simple presentation of ${{{\mathcal{M}}}}(N)$, which can be rewritten (Proposition 3.3 and Theorem 3.5 of [@StukowSimpSzepPar]) so that it has the hyperelliptic involution ${\varrho}$ as one of the generators, and the hyperelliptic relations (Theorem \[th:004\]) appear among defining relations.

As an application of obtained presentations we compute the first homology groups of ${{{\mathcal{M}}}}^h(N)$ and ${{{\mathcal{M}}}}^{h+}(N)$ with coefficients in $H_1(N;{{\mathbb{Z}}})$ (Theorems \[tw:hom:positive\] and \[tw:hom:full\]).

Definitions of ${{{\mathcal{M}}}}^h(N_g)$ and ${{{\mathcal{M}}}}^{h+}(N_g)$ {#sec:def:hiper}
===========================================================================

Let $S_{g-1}$ be a closed oriented surface of genus $g-1\geqslant 2$ embedded in ${{\mathbb{R}}}^3$ as shown in Figure \[r01\], in such a way that it is invariant under reflections across $xy,yz,xz$ planes, and let ${j\colon S_{g-1}\to S_{g-1}}$ be the symmetry defined by $j(x,y,z)=(-x,-y,-z)$. Denote by $C_{{{{\mathcal{M}}}}^{\pm}(S_{g-1})}(j)$ the centraliser of $j$ in ${{{\mathcal{M}}}}^{\pm}(S_{g-1})$. The orbit space $S_{g-1}/{\langle j \rangle}$ is a nonorientable surface $N_{g}$ of genus $g$ and it is known (Theorem 1 of [@BirChil1]) that there is an epimorphism $${\pi_j\colon C_{{{{\mathcal{M}}}}^{\pm}(S_{g-1})}(j)\to {{{\mathcal{M}}}}(N_{g})}$$ with kernel $\ker \pi_j={\langle j \rangle}$. In particular $${{{\mathcal{M}}}}(N_{g})\cong C_{{{{\mathcal{M}}}}^{\pm}(S_{g-1})}(j)/{\langle j \rangle}.$$ Observe that the hyperelliptic involution ${\varrho}$ is an element of $C_{{{{\mathcal{M}}}}^{\pm}(S_{g-1})}(j)$. Hence the following definition makes sense.

Define the *hyperelliptic mapping class group* ${{{\mathcal{M}}}}^h(N)$ of a closed nonorientable surface $N$ to be the centraliser of $\pi_j({\varrho})$ in the mapping class group ${{{\mathcal{M}}}}(N)$. We say that $\pi_j({\varrho})$ is the *hyperelliptic involution* of $N$ and by abuse of notation we write ${\varrho}$ for $\pi_j({\varrho})$.

In order to have a little more straightforward description of ${\varrho}$ observe, that the orbit space $S_{g-1}/{\langle j \rangle}$ gives the model of $N_g$, where $N_g$ is a connected sum of an orientable surface $S_r$ and a projective plane (for $g$ odd) or a Klein bottle (for $g$ even) – see Figure \[r02\].

![Nonorientable surface $N_g$.[]{data-label="r02"}](R02){width="96.00000%"}

To be more precise, $N_g$ is the left half of $S_{g-1}$ embedded in ${{\mathbb{R}}}^3$ as in Figure \[r01\] with boundary points identified by the map $(x,y,z)\mapsto (-x,-y,-z)$. Note that $g=2r+1$ for $g$ odd and $g=2r+2$ for $g$ even. In such a model, ${{\varrho}\colon N_g\to N_g}$ is the map induced by the half turn about the $y$-axis.

Observe that the set of fixed points of ${{\varrho}\colon N_g\to N_g}$ consists of $g$ points $\{p_1,p_2,\ldots,p_g\}$ and the circle $p$. Therefore ${{{\mathcal{M}}}}^h(N)$ consists of isotopy classes of maps which must fix the set $\{p_1,p_2,\ldots,p_g\}$ and map the circle $p$ to itself. Moreover, the orbit space $N_g/{\langle {\varrho}\rangle}$ is the sphere $S_{0,1}^g$ with one boundary component corresponding to $p$ and $g$ distinguished points corresponding to $\{p_1,p_2,\ldots,p_g\}$. Since elements of ${{{\mathcal{M}}}}^h(N_g)$ may not fix $p$ point–wise, it is more convenient to treat $p$ as the distinguished puncture $p_{g+1}$, hence we will identify $N_g/{\langle {\varrho}\rangle}$ with the sphere $S_0^{g,1}$ with $g+1$ punctures. The notation $S_0^{g,1}$ is meant to indicate that maps of $S_0^{g,1}$ (and their isotopies) could permute the punctures $p_1,\ldots,p_g$, but must fix $p_{g+1}$.

The main goal of this section is to prove the following theorem.

\[tw:002\] If $g\geqslant 3$ then the projection $N_g\to N_g/{\langle {\varrho}\rangle}$ induces an epimorphism $${\pi_{\varrho}\colon {{{\mathcal{M}}}}^h(N_g)\to {{{\mathcal{M}}}}^{\pm}(S_0^{g,1})}$$ with $\ker \pi_{\varrho}={\langle {\varrho}\rangle}$.

Consider the following diagram $$\xymatrix@C=3pc@R=3pc{
 C_{{{{\mathcal{M}}}}^{\pm}(S_{g-1})}({\langle j,{\varrho}\rangle})\ar^{\pi_{\varrho}}[r]\ar^{\pi_j}[d]&C_{{{{\mathcal{M}}}}^{\pm}(S_0^{2g})}(j)\ar^{\pi_j}[d]\\
 {{{\mathcal{M}}}}^h(N_g)\ar@{-->}^{\pi_{\varrho}}[r] \ar@/^1pc/@{->}[u]^{i_j} &{{{\mathcal{M}}}}^{\pm}(S_0^{g,1})
}$$ The left vertical map is the restriction of the projection $${\pi_j\colon C_{{{{\mathcal{M}}}}^{\pm}(S_{g-1})}(j)\to {{{\mathcal{M}}}}(N_g)}$$ to the subgroup consisting of elements which centralise ${\varrho}$. The nice thing about $\pi_j$ is that it has a section $${i_j\colon {{{\mathcal{M}}}}(N_g)\to C_{{{{\mathcal{M}}}}^{\pm}(S_{g-1})}(j)}.$$ In fact, for any $h\in{{{\mathcal{M}}}}(N_g)$ we can define $i_j(h)$ to be an orientation preserving lift of $h$.

The upper horizontal map is the restriction of the homomorphism $${\pi_{\varrho}\colon {{{\mathcal{M}}}}^{h\pm}(S_{g-1})\to {{{\mathcal{M}}}}^{\pm}(S_0^{2g})}$$ induced by the orbit projection $S_{g-1}\to S_{g-1}/{\langle {\varrho}\rangle}$. The fact that this map is a homomorphism was first observed by Birman and Hilden [@Bir-Hil]. The kernel of this map is equal to ${\langle {\varrho}\rangle}$.

The right vertical map is again the homomorphism induced by the orbit projection $S^{2g}_0\to
S^{2g}_0/{\langle j \rangle}$. However now ${j\colon S^{2g}_0\to S^{2g}_0}$ is a reflection with a circle of fixed points. The existence of $\pi_j$ in such a case follows from the work of Zieschang (Proposition 10.3 of [@Zies1]).

Hence there is the homomorphism $${\pi_{\varrho}\colon {{{\mathcal{M}}}}^h(N_g)\to {{{\mathcal{M}}}}^{\pm}(S_0^{g,1})}$$ defined as the composition $$\pi_{\varrho}=\pi_j\circ \pi_{\varrho}\circ i_j.$$ Moreover, $$\begin{aligned}
\ker \pi_{\varrho}&=\ker (\pi_j\circ \pi_{\varrho}\circ i_j)=(\pi_j\circ \pi_{\varrho}\circ i_j)^{-1}(id)\\
&=i_j^{-1}(\pi_{\varrho}^{-1}(\pi_j^{-1}(id)))=i_j^{-1}(\pi_{\varrho}^{-1}({\langle j \rangle}))=i_j^{-1}({\langle j,{\varrho}\rangle})={\langle {\varrho}\rangle}.
\end{aligned}$$

Theorem \[tw:002\] is not true if $N=N_2$. This corresponds to the fact that the Birman-Hilden theorem does not hold for the closed torus $S=S_1$.

\[rem:positiv:subg\] Theorem \[tw:002\] shows that the group ${{{\mathcal{M}}}}^h(N_g)$ contains a very natural subgroup of index 2, namely $${{{\mathcal{M}}}}^{h+}(N_g)=\pi_{\varrho}^{-1}({{{\mathcal{M}}}}(S_0^{g,1})).$$ Geometrically, the subgroup ${{{\mathcal{M}}}}^{h+}(N_g)$ consists of these elements, which preserve the orientation of the circle $p$ (the circle fixed by ${\varrho}$). As we will see later (see Remark \[uw:perspective\]), it seems that the group ${{{\mathcal{M}}}}^{h+}(N)$ corresponds to ${{{\mathcal{M}}}}^h(S)$, whereas ${{{\mathcal{M}}}}^{h}(N)$ corresponds to ${{{\mathcal{M}}}}^{h\pm}(S)$.

Presentations for groups ${{{\mathcal{M}}}}(S_0^{g,1})$ and ${{{\mathcal{M}}}}^{\pm}(S_0^{g,1})$
================================================================================================

Let $w_1,w_2,\ldots,w_g$ be simple arcs connecting punctures $p_1,\ldots,p_{g+1}$ on a sphere $S_0^{g+1}$ as shown in Figure \[r03\].

![Sphere $S_0^{p+1}$ and elementary braid ${\sigma}_i$.[]{data-label="r03"}](R03){width="80.00000%"}

Recall that to each such arc $w_i$ we can associate the elementary braid ${\sigma}_i$ which interchanges punctures $p_i$ and $p_{i+1}$ – see Figure \[r03\]. The following theorem is due to Magnus [@Mag_sfer]. It is also proved in Chapter 4 of [@Bir-BLMCG].

\[th:00\] If $g\geqslant 1$, then ${{{\mathcal{M}}}}(S_0^{g+1})$ has the presentation with generators ${\sigma}_1,\ldots,{\sigma}_g$ and defining relations: $$\begin{aligned}
   &{\sigma}_k{\sigma}_j={\sigma}_j{\sigma}_k\quad\text{for $|k-j|>1$},\\
&{\sigma}_j{\sigma}_{j+1}{\sigma}_j={\sigma}_{j+1}{\sigma}_j{\sigma}_{j+1}\quad\text{for $j=1,\ldots,g-1$},\\
&{\sigma}_1\cdots {\sigma}_{g-1}{\sigma}_g^2{\sigma}_{g-1}\cdots {\sigma}_1=1,\\
&({\sigma}_1{\sigma}_2\cdots {\sigma}_{g})^{g+1}=1.
  \end{aligned}$$

In order to avoid unnecessary complications, from now on assume that $g\geqslant 3$. Recall that we denote by ${{{\mathcal{M}}}}(S_0^{g,1})$ the subgroup of ${{{\mathcal{M}}}}(S_0^{g+1})$ consisting of maps which fix $p_{g+1}$.

\[th:001\] If $g\geqslant 3$, then ${{{\mathcal{M}}}}(S_0^{g,1})$ has the presentation with generators ${\sigma}_1,\ldots,{\sigma}_{g-1}$ and defining relations:

1.  ${\sigma}_k{\sigma}_j={\sigma}_j{\sigma}_k$ for $|k-j|>1$ and $k,j<g$,

2.  ${\sigma}_j{\sigma}_{j+1}{\sigma}_j={\sigma}_{j+1}{\sigma}_j{\sigma}_{j+1}$ for $j=1,2,\ldots,g-2$,

3.  $({\sigma}_1\cdots {\sigma}_{g-1})^g=1$.

By Lemma 2.2 of [@Big-lin], $${{{\mathcal{M}}}}(S_0^{g,1})\cong B_{g}/{\langle \Delta^2 \rangle},$$ where $B_{g}={{{\mathcal{M}}}}(S_{0,1}^{g})$ is the braid group on $g$ strands, and $$\Delta^2=({\sigma}_1\cdots {\sigma}_{g-1})^g$$ is the generator of the center of $B_g$. Since $B_g$ has the presentation with generators ${\sigma}_1,\ldots,{\sigma}_{g-1}$ and defining relations (A1), (A2), this completes the proof.

Theorem \[th:001\] can be also algebraically deduced from Theorem \[th:00\]. Since ${{{\mathcal{M}}}}(S_0^{g,1})$ is a subgroup of index $g+1$ in ${{{\mathcal{M}}}}(S_0^{g+1})$, for the Schreier transversal we can take $$(1,{\sigma}_{g},{\sigma}_{g}{\sigma}_{g-1},\ldots,{\sigma}_{g}{\sigma}_{g-1}\cdots {\sigma}_1).$$ If we now apply Reidemeister-Schreier process, as generators for ${{{\mathcal{M}}}}(S_0^{g,1})$ we get ${\sigma}_1,\ldots,{\sigma}_{g-1}$ and additionally $\tau_1,\ldots,\tau_g$ where $$\tau_k=\begin{cases}
          {\sigma}_g\cdots{\sigma}_{k+1}{\sigma}_k^2{\sigma}_{k+1}^{-1}\cdots{\sigma}_g^{-1}&\text{for $k=1,\ldots,g-1$}\\
{\sigma}_g^2&\text{for $k=g$}.
         \end{cases}$$ As defining relations we get

1.  ${\sigma}_k{\sigma}_j={\sigma}_j{\sigma}_k$ for $|k-j|>1$ and $k,j<g$,

2.  ${\sigma}_k\tau_j=\tau_j{\sigma}_k$ for $j\neq k,k+1$,

3.  ${\sigma}_j{\sigma}_{j+1}{\sigma}_j={\sigma}_{j+1}{\sigma}_j{\sigma}_{j+1}$ for $j=1,2,\ldots,g-2$,

4.  ${\sigma}_k\tau_{k+1}{\sigma}_{k}^{-1}=\tau_{k+1}^{-1}\tau_k\tau_{k+1}$ for $k=1,2,\ldots,g-1$,

5.  ${\sigma}_k\tau_k{\sigma}_{k}^{-1}=\tau_{k+1}$ for $k=1,2,\ldots,g-1$,

6.  $\tau_1\tau_2\cdots \tau_g=1$,

7.  ${\sigma}_{g-1}\cdots{\sigma}_2{\sigma}_1\tau_1{\sigma}_1{\sigma}_2\cdots {\sigma}_{g-1}=1$,

8.  $\left({\sigma}_{g-1}{\sigma}_{g-2}\cdots{\sigma}_{1}\tau_1\right)^g=1$.

If we now remove generators $\tau_1,\ldots,\tau_g$ from the above presentation, we obtain the presentation given by Theorem \[th:001\]. The computations are lengthy, but completely straightforward.

Recall that by ${{{\mathcal{M}}}}^{\pm}(S_0^{g,1})$ we denote the extended mapping class group of the sphere $S_0^{g,1}$, that is the extension of degree 2 of ${{{\mathcal{M}}}}(S_0^{g,1})$. Suppose that the sphere $S_0^{g,1}$ is the metric sphere in ${{\mathbb{R}}}^3$ with origin $(0,0,0)$ and that punctures $p_1,\ldots,p_g$ are contained in the $xy$-plane. Let ${{\sigma}\colon {S_0^{g,1}}\to {S_0^{g,1}}}$ be the map induced by the reflection across the $xy$-plane. We have the short exact sequence. $$\xymatrix{
1\ar[r]&{{{\mathcal{M}}}}(S_0^{g,1})\ar[r]&{{{\mathcal{M}}}}^{\pm}(S_0^{g,1})\ar[r]&{\langle {\sigma}\rangle}\ar[r]&1.
}$$ Moreover, ${\sigma}{\sigma}_i{\sigma}^{-1}={\sigma}_i^{-1}$ for $i=1,\ldots,g-1$. Therefore Theorem \[th:001\] implies the following.

\[th:003\] If $g\geqslant 3$, then ${{{\mathcal{M}}}}^{\pm}(S_0^{g,1})$ has the presentation with generators ${\sigma}_1,\ldots,{\sigma}_{g-1},{\sigma}$ and defining relations:

1.  ${\sigma}_k{\sigma}_j={\sigma}_j{\sigma}_k$ for $|k-j|>1$ and $k,j<g$,

2.  ${\sigma}_j{\sigma}_{j+1}{\sigma}_j={\sigma}_{j+1}{\sigma}_j{\sigma}_{j+1}$ for $j=1,2,\ldots,g-2$,

3.  $({\sigma}_1\cdots {\sigma}_{g-1})^g=1$,

4.  ${\sigma}^2=1$,

5.  ${\sigma}{\sigma}_i{\sigma}={\sigma}_i^{-1}$ for $i=1,2,\ldots,g-1$.

Presentations for groups ${{{\mathcal{M}}}}^h(N_g)$ and ${{{\mathcal{M}}}}^{h+}(N_g)$
=====================================================================================

By Theorem \[tw:002\] there is a short exact sequence. $$\xymatrix{
1\ar[r]&{\langle {\varrho}\rangle}\ar[r]&{{{\mathcal{M}}}}^h(N_g)\ar[r]^{\pi_{\varrho}}&{{{\mathcal{M}}}}^{\pm}(S_0^{g,1})\ar[r]&1.
}$$ Moreover, it is known that as lifts of braids ${\sigma}_1,\ldots,{\sigma}_{g-1}\in{{{\mathcal{M}}}}^{\pm}(S_0^{g,1})$ we can take Dehn twists $t_{a_1},\ldots,t_{a_{g-1}}\in {{{\mathcal{M}}}}^h(N_g)$ about circles $a_1,\ldots,a_{g-1}$ – cf Figure \[r02\] (small arrows in this picture indicate directions of twists). As a lift of ${\sigma}$ we take the symmetry $s$ across the $xy$ plane (the second lift of ${\sigma}$ is the symmetry ${\varrho}s$, that is the symmetry across the $yz$ plane).

To obtain a presentation for the group ${{{\mathcal{M}}}}^h(N_g)$ we need to lift relations (B1)–(B5) of Theorem \[th:003\]. Each relation of the form $$w({\sigma}_1,\ldots,{\sigma}_{g-1},{\sigma})=1$$ lifts either to $w(t_{a_1},\ldots,t_{a_{g-1}},s)=1$ or to $w(t_{a_1},\ldots,t_{a_{g-1}},s)={\varrho}$. In order to determine which of these two cases does occur it is enough to check whether the homeomorphism $w(t_{a_1},\ldots,t_{a_{g-1}},s)$ changes the orientation of the circle $a_1$ or not. This can be easily done and as a result we obtain the following theorem.

\[th:004\] If $g\geqslant 3$, then ${{{\mathcal{M}}}}^h(N_g)$ has the presentation with generators $t_{a_1},\ldots,t_{a_{g-1}},s,{\varrho}$ and defining relations:

1.  $t_{a_k}t_{a_j}=t_{a_j}t_{a_k}$ for $|k-j|>1$ and $k,j<g$,

2.  $t_{a_j}t_{a_{j+1}}t_{a_j}=t_{a_{j+1}}t_{a_j}t_{a_{j+1}}$ for $j=1,2,\ldots,g-2$,

3.  $(t_{a_{1}}\cdots t_{a_{g-1}})^g=\begin{cases}
                                                  1&\text{for $g$ even}\\
    {\varrho}&\text{for $g$ odd,}\\\end{cases}$

4.  $s^2=1$,

5.  $st_{a_j}s=t_{a_j}^{-1}$ for $j=1,2,\ldots,g-1$,

6.  ${\varrho}^2=1$,

7.  ${\varrho}t_{a_j}{\varrho}=t_{a_j}$ for $j=1,2,\ldots,g-1$,

8.  ${\varrho}s{\varrho}=s$.

\[cr:001\] If $g\geqslant 3$, then $$H_1({{{\mathcal{M}}}}^h(N_g))=\begin{cases}
                   {{\mathbb{Z}}}_2\oplus{{\mathbb{Z}}}_2&\text{for $g$ odd}\\
{{\mathbb{Z}}}_2\oplus{{\mathbb{Z}}}_2\oplus{{\mathbb{Z}}}_2&\text{for $g$ even.}
                  \end{cases}$$

Relation (C2) implies that the abelianization of the group ${{{\mathcal{M}}}}^h(N_g)$ is an abelian group generated by $t_{a_1},s,{\varrho}$. Defining relations take form $$\begin{aligned}
   &t_{a_1}^{(g-1)g}=\begin{cases}
            1&\text{for $g$ even}\\
{\varrho}&\text{for $g$ odd}\\\end{cases}\\
&s^2=1,\quad t_{a_1}^2=1,\quad {\varrho}^2=1.
  \end{aligned}$$ Hence $H_1({{{\mathcal{M}}}}^h(N_g))={\langle t_{a_1},s \rangle}\simeq {{\mathbb{Z}}}_2\oplus{{\mathbb{Z}}}_2$ for $g$ odd and $H_1({{{\mathcal{M}}}}^h(N_g))={\langle t_{a_1},s,{\varrho}\rangle}\simeq
{{\mathbb{Z}}}_2\oplus{{\mathbb{Z}}}_2\oplus{{\mathbb{Z}}}_2$ for $g$ even.

The main theorem of [@Chil] implies that the group ${{{\mathcal{M}}}}(N_3)$ is generated by $a_1,a_2$ and a crosscap slide which commutes with ${\varrho}$. Hence ${{{\mathcal{M}}}}^h(N_3)={{{\mathcal{M}}}}(N_3)$ and Theorem \[th:004\] implies the following.

\[Presen:g3:Bir\] The group ${{{\mathcal{M}}}}(N_3)$ has the presentation with generators $t_{a_1},t_{a_2},s$ and defining relations:

1.  $t_{a_1}t_{a_2}t_{a_1}=t_{a_2}t_{a_1}t_{a_2}$,

2.  $(t_{a_1}t_{a_2}t_{a_1})^4=1$,

3.  $s^2=1$,

4.  $st_{a_j}s=t_{a_j}^{-1}$ for $j=1,2$.

By Theorem \[th:004\], the group ${{{\mathcal{M}}}}(N_3)$ is generated by $t_{a_1},t_{a_2},{\varrho},s$ with defining relations:

1.  $t_{a_1}t_{a_{2}}t_{a_1}=t_{a_{2}}t_{a_1}t_{a_{2}}$,

2.  $(t_{a_{1}}t_{a_2})^3={\varrho}$,

3.  $s^2=1$,

4.  $st_{a_j}s=t_{a_j}^{-1}$ for $j=1,2$,

5.  ${\varrho}^2=1$,

6.  ${\varrho}t_{a_j}{\varrho}=t_{a_j}$ for $j=1,2$,

7.  ${\varrho}s{\varrho}=s$.

Using (C2), we can rewrite (C3) in the form $$\begin{aligned}
 {\varrho}=t_{a_{1}}t_{a_2}t_{a_{1}}(t_{a_2}t_{a_{1}}t_{a_2})=t_{a_{1}}t_{a_2}t_{a_{1}}(t_{a_1}t_{a_{2}}t_{a_1
})=(t_{a_1}t_{a_{2}}t_{a_1})^2.
  \end{aligned}$$ Hence we can remove ${\varrho}$ from the generating set and then (C6) will transform into (D2). It remains to check that relations (C7) and (C8) are superfluous. Let start with (C7). $$\begin{aligned}
t_{a_1}{\varrho}t_{a_1}^{-1}&=t_{a_1}(t_{a_1}t_{a_{2}}t_{a_1})(t_{a_1}t_{a_{2}}t_{a_1})t_{a_1}^{-1}\\
&=t_{a_1}(t_{a_2}t_{a_{1}}t_{a_2})(t_{a_1}t_{a_{2}}t_{a_1})t_{a_1}^{-1}=
(t_{a_1}t_{a_2}t_{a_{1}})(t_{a_1}t_{a_2}t_{a_{1}})={\varrho}\\
t_{a_2}{\varrho}t_{a_2}^{-1}&=t_{a_2}(t_{a_1}t_{a_{2}}t_{a_1})(t_{a_1}t_{a_{2}}t_{a_1})t_{a_2}^{-1}\\
&=t_{a_2}(t_{a_1}t_{a_{2}}t_{a_1})(t_{a_2}t_{a_{1}}t_{a_2})t_{a_2}^{-1}=
(t_{a_1}t_{a_2}t_{a_{1}})(t_{a_1}t_{a_2}t_{a_{1}})={\varrho}\end{aligned}$$ Now we check (C8). $$\begin{aligned}  
s{\varrho}s&=
s(t_{a_1}t_{a_{2}}t_{a_1})^2s=(t_{a_1}^{-1}t_{a_{2}}^{-1}t_{a_1}^{-1})^2=(t_{a_1}t_{a_2}t_{a_1})^{-2
} = (t_{a_1}t_{a_2}t_{a_1})^{2}={\varrho}.
  \end{aligned}$$

By restricting homomorphism ${\pi_{\varrho}\colon {{{\mathcal{M}}}}^{h}(N_g)\to {{{\mathcal{M}}}}^{\pm}(S_0^{g,1})}$ to the subgroup ${{{\mathcal{M}}}}^{h+}(N_g)$ we obtain the exact sequence $$\xymatrix{
1\ar[r]&{\langle {\varrho}\rangle}\ar[r]&{{{\mathcal{M}}}}^{h+}(N_g)\ar[r]^{\pi_{\varrho}}&{{{\mathcal{M}}}}(S_0^{g,1})\ar[r]&1.
}$$ Now if we lift the presentation from Theorem \[th:001\], we get the following.

\[tw:pres:pos\] If $g\geqslant 3$, then ${{{\mathcal{M}}}}^{h+}(N_g)$ has the presentation with generators $t_{a_1},\ldots,t_{a_{g-1}},{\varrho}$ and defining relations:

1.  $t_{a_k}t_{a_j}=t_{a_j}t_{a_k}$ for $|k-j|>1$ and $k,j<g$,

2.  $t_{a_j}t_{a_{j+1}}t_{a_j}=t_{a_{j+1}}t_{a_j}t_{a_{j+1}}$ for $j=1,2,\ldots,g-2$,

3.  $(t_{a_{1}}\cdots t_{a_{g-1}})^g=\begin{cases}
                                                  1&\text{for $g$ even}\\
    {\varrho}&\text{for $g$ odd,}\\\end{cases}$

4.  ${\varrho}^2=1$,

5.  ${\varrho}t_{a_j}{\varrho}=t_{a_j}$ for $j=1,2,\ldots,g-1$.

\[cr:002\] If $g\geqslant 3$, then $$H_1({{{\mathcal{M}}}}^{h+}(N_g))=\begin{cases}
                   {{\mathbb{Z}}}_{2(g-1)g}&\text{for $g$ odd}\\
{{\mathbb{Z}}}_{(g-1)g}\oplus{{\mathbb{Z}}}_2&\text{for $g$ even.}
                  \end{cases}$$

Relation (E2) implies that the abelianization of the group ${{{\mathcal{M}}}}^{h+}(N_g)$ is an abelian group generated by $t_{a_1},{\varrho}$. Defining relations take form: $$\begin{aligned}
   &t_{a_1}^{(g-1)g}=\begin{cases}
            1&\text{for $g$ even}\\
{\varrho}&\text{for $g$ odd,}\\\end{cases}\\
&{\varrho}^2=1.
  \end{aligned}$$ Hence $H_1({{{\mathcal{M}}}}^{h+}(N_g))={\langle t_{a_1} \rangle}\simeq {{\mathbb{Z}}}_{2(g-1)g}$ for $g$ odd and $H_1({{{\mathcal{M}}}}^{h+}(N_g))={\langle t_{a_1},{\varrho}\rangle}\simeq
{{\mathbb{Z}}}_{(g-1)g}\oplus{{\mathbb{Z}}}_2$ for $g$ even.

\[uw:perspective\] To put Corollaries \[cr:001\] and \[cr:002\] into perspective, recall that in the oriented case (Theorem 8 of [@Bir-Hil]), $$\begin{aligned}
   {{{\mathcal{M}}}}^{h}(S_g)=\langle& t_{a_1},\ldots,t_{a_{2g+1}},{\varrho}{\;|\;}t_{a_k}t_{a_j}=t_{a_j}t_{a_k},\ t_{a_j}t_{a_{j+1}}t_{a_j}=t_{a_{j+1}}t_{a_j}t_{a_{j+1}},\\
 &(t_{a_1}t_{a_2}\cdots t_{a_{2g+1}})^{2g+2}=1,\ {\varrho}=t_{a_1}t_{a_2}\cdots t_{a_{2g+1}}t_{a_{2g+1}}\cdots t_{a_2}t_{a_1},\\ &{\varrho}^2=1, {\varrho}t_{a_1}{\varrho}=t_{a_1}\rangle,\quad
\text{where $j=1,2,\ldots,2g$, $|k-j|>1$}.
  \end{aligned}$$ The presentation for the group ${{{\mathcal{M}}}}^{h\pm}(S_g)$ is obtained from the above presentation by adding one generator $s$ and three relations: $$s^2=1,\ st_{a_1}s=t_{a_1}^{-1},\ {\varrho}s{\varrho}=s.$$ Consequently, $H_1({{{\mathcal{M}}}}^{h\pm}(S_g))={{\mathbb{Z}}}_2\oplus{{\mathbb{Z}}}_2$ and $$H_1({{{\mathcal{M}}}}^{h}(S_g))=\begin{cases}
            {{\mathbb{Z}}}_{4g+2}&\text{for $g$ even}\\
{{\mathbb{Z}}}_{8g+4}&\text{for $g$ odd.}\\\end{cases}$$ This suggests that algebraically the group ${{{\mathcal{M}}}}^{h+}(N)$ corresponds to ${{{\mathcal{M}}}}^h(S)$, whereas ${{{\mathcal{M}}}}^{h}(N)$ corresponds to ${{{\mathcal{M}}}}^{h\pm}(S)$.

Computing $H_1({{{\mathcal{M}}}}^{h+}(N_g); H_1(N_g;{{\mathbb{Z}}}))$ and $H_1({{{\mathcal{M}}}}^{h}(N_g); H_1(N_g;{{\mathbb{Z}}}))$
====================================================================================================================================

Homology of groups
------------------

Let us briefly review how to compute the first homology of a group with twisted coefficients. Our exposition follows [@Brown; @Tanaka].

For a given group $G$ and $G$-module $M$ (that is ${{\mathbb{Z}}}G$-module) we define the *bar resolution* which is a chain complex $(C_n(G))$ of $G$-modules, where $C_n(G)$ is the free $G$-module generated by symbols $[h_1|\cdots|h_n]$, $h_i\in G$. For $n=0$, $C_0(G)$ is the free module generated by the empty bracket $[\cdot]$. Our interest will restrict to groups $C_2(G),C_1(G),C_0(G)$ for which the boundary operator ${\partial_n\colon C_n(G)\to C_{n-1}(G)}$ is defined by formulas: $$\begin{aligned}
\partial_2([h_1|h_2])&=h_1[h_2]-[h_1h_2]+[h_1],\\
   \partial_1([h])&=h[\cdot]-[\cdot].
  \end{aligned}$$

The homology of $G$ with coefficients in $M$ is defined as the homology groups of the chain complex $(C_n(G)\otimes M)$, where the chain complexes are tensored over ${{\mathbb{Z}}}G$. In particular, $H_1(G;M)$ is the first homology group of the complex $$\xymatrix@C=3pc@R=3pc{C_2(G)\otimes M\ar[r]^{\ \partial_2\otimes {\rm id}}&C_1(G)\otimes M
\ar[r]^{\ \partial_1\otimes {\rm id}}&C_0(G)\otimes M.}$$ For simplicity, we denote $\partial\otimes {\rm id}={\overline{\partial}}$ henceforth.

If the group $G$ has a presentation $G=\langle X\,|\,R\rangle$, denote by $${\langle {\overline{X}} \rangle}={\langle [x]\otimes m{\;|\;}x\in X, m\in M \rangle}{\subseteq}C_1(G)\otimes M.$$ Then, using the formula for $\partial_2$, one can show that $H_1(G;M)$ is a quotient of ${\langle {\overline{X}} \rangle}\cap \ker{\overline{\partial}}_1$.

The kernel of this quotient corresponds to relations in $G$ (that is elements of $R$). To be more precise, if $r\in R$ has the form $x_1\cdots x_k=y_1\cdots y_n$ and $m\in M$, then $r$ gives the relation (in $H_1(G;M)$) $${\overline{r}}\otimes m\!:\ \sum_{i=1}^{k}x_1\cdots x_{i-1}[x_i]\otimes m=\sum_{i=1}^{n}y_1\cdots y_{i-1}[y_i]\otimes m.\label{eq_rew_rel}$$ Then $$H_1(G;M)={\langle {\overline{X}} \rangle}\cap \ker{\overline{\partial}}_1/{\langle {\overline{R}} \rangle},$$ where $${\overline{R}}=\{{\overline{r}}\otimes m{\;|\;}r\in R,m\in M\}.$$

Action of ${{{\mathcal{M}}}}^{h}(N_g)$ on $H_1(N_g;{{\mathbb{Z}}})$
-------------------------------------------------------------------

Let $c_1,\ldots,c_g$ be one-sided circles indicated in Figure \[r04\]. In this figure surface $N_g$ is represented as the sphere with $g$ crosscaps (the shaded disks represent crosscaps, hence their interiors are to be removed and then the antipodal points on each boundary component are to be identified).

![Surface $N_g$ as a sphere with crosscaps.[]{data-label="r04"}](R04){width="80.00000%"}

The same set of circles is also indicated in Figure \[r02\] – for a method of transferring circles between two models of $N_g$ see Section 3 of [@Stukow_HomTw].

Recall that $H_1(N_g;{{\mathbb{Z}}})$ as a ${{\mathbb{Z}}}$-module is generated by $\gamma_1=[c_1],\ldots,\gamma_g=[c_g]$ with respect to the single relation $$2(\gamma_1+\gamma_2+\cdots+\gamma_g)=0.$$ There is a ${{\mathbb{Z}}}_2$-valued intersection paring $\langle\,,\rangle$ on $H_1(N_g;{{\mathbb{Z}}})$ defined as the symmetric bilinear form (with values in ${{\mathbb{Z}}}_2$) satisfying $\langle\gamma_i,\gamma_j\rangle=\delta_{ij}$ for $1\leqslant i,j\leqslant g$. The mapping class group ${{{\mathcal{M}}}}(N_g)$ acts on $H_1(N_g;{{\mathbb{Z}}})$ via automorphisms which preserve $\langle\,,\rangle$, hence there is a representation $${\psi\colon {{{\mathcal{M}}}}(N_g)\to {\rm Iso}(H_1(N_g;{{\mathbb{Z}}}))}.$$ In fact it is known that this representation is surjective – see [@Pinkal; @Gadgil].

Since we have very simple geometric definitions of $t_{a_i},s,{\varrho}\in {{{\mathcal{M}}}}^h(N_g)$ it is straightforward to check that $$\begin{aligned}
   \psi(t_{a_i})&=I_{i-1}\oplus \begin{bmatrix}
0&1\\
-1&2\end{bmatrix}\oplus I_{g-i-1}\\
\psi(t_{a_i}^{-1})&=I_{i-1}\oplus \begin{bmatrix}
2&-1\\
1&0
    \end{bmatrix}\oplus I_{g-i-1}\\
\psi(s)&=\begin{bmatrix}
          -1&2&-2&2&\ldots&(-1)^{g}\cdot 2\\
0&1&-2&2&\ldots&(-1)^{g}\cdot 2\\
0&0&-1&2&\ldots&(-1)^{g}\cdot 2\\
0&0&0&1&\ldots&(-1)^{g}\cdot 2\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&0&\ldots& (-1)^{g}\cdot 1
         \end{bmatrix}
\\
\psi({\varrho})&=-I_g
  \end{aligned}$$ where $I_k$ is the identity matrix of rank $k$.

The above matrices are written with respect to the generating set $(\gamma_1,\gamma_2,\ldots,\gamma_g)$. Note that $H_1(N_g;{{\mathbb{Z}}})$ is not free, hence one has to be careful with matrices – two different matrices may represent the same element.

Computing ${\langle {\overline{X}} \rangle}\cap \ker {\overline{\partial}}_1$
-----------------------------------------------------------------------------

Observe that if $G={{{\mathcal{M}}}}^h(N_g)$, $M=H_1(N_g;{{\mathbb{Z}}})$ and $h\in G$ then $${\overline{\partial}}_1([h]\otimes \gamma_j)=(h-1)[\cdot]\otimes \gamma_j=[\cdot]\otimes(\psi(h)^{-1}-I_g)\gamma_j.$$ If we identify $C_0(G)\otimes M$ with $M$ by the map $[\cdot]\otimes m\mapsto m$, this formula takes form $${\overline{\partial}}_1([h]\otimes \gamma_j)=(\psi(h)^{-1}-I_g)\gamma_j.$$ Let us denote $[{\varrho}]\otimes \gamma_j, [s]\otimes \gamma_j,
[t_{a_i}]\otimes \gamma_j$ respectively by ${\varrho}_j,s_j$ and $t_{i,j}$. Using the above formula, we obtain $$\begin{aligned}
{\overline{\partial}}_1({\varrho}_j)&=-2\gamma_j\\
{\overline{\partial}}_1(s_j)&=\begin{cases}
-2\sum_{k=1}^{j}\gamma_k &\text{for $j$ odd}\\
-{\overline{\partial}}_1(s_{j-1}) &\text{for $j$ even}
                       \end{cases}
\\
   {\overline{\partial}}_1(t_{i,j})&=\begin{cases}
                              \gamma_i+\gamma_{i+1}&\text{for $j=i$}\\
-\gamma_i-\gamma_{i+1}&\text{for $j=i+1$}\\
0&\text{otherwise.}
                             \end{cases}
  \end{aligned}$$

Let $g\geqslant 3$ and $G={{{\mathcal{M}}}}^{h+}(N_g)$ then ${\langle {\overline{X}} \rangle}\cap\ker{\overline{\partial}}_1$ is the abelian group which admits the presentation with generators:

1.  $t_{i,j}$, where $i=1,\ldots,g-1$ and $j=1,\ldots,i-1,i+2,\ldots,g$

2.  $t_{j,j}+t_{j,j+1}$, where $j=1,\ldots,g-1$

3.  $2t_{j,j}+{\varrho}_j+{\varrho}_{j+1}$ , where $j=1,\ldots,g-1$

4.  $\begin{cases}
     2t_{1,1}+2t_{3,3}+\cdots+2t_{g-2,g-2}-{\varrho}_g&\text{for $g$ odd}\\
    2t_{1,1}+2t_{3,3}+\cdots+2t_{g-1,g-1}&\text{for $g$ even}
          \end{cases}$

and relations $$\begin{aligned}
r_{t_j}\!&:0=2t_{j,1}+\cdots +2(t_{j,j}+t_{j,j+1})+\cdots+2t_{j,g}\quad\text{for $j=1,\ldots,g-1$}\\
r_{{\varrho}}\!&:\begin{cases}
              2(2t_{1,1}+{\varrho}_1+{\varrho}_2)+\cdots+2(2t_{g-2,g-2}+{\varrho}_{g-2}+{\varrho}_{g-1})\\
\qquad\qquad =2(2t_{1,1}+2t_{3,3}+\cdots+2t_{g-2,g-2}-{\varrho}_g)\quad \text{for $g$ odd}\\
              2(2t_{1,1}+{\varrho}_1+{\varrho}_2)+\cdots+2(2t_{g-1,g-1}+{\varrho}_{g-1}+{\varrho}_{g})\\
\qquad\qquad\qquad =2(2t_{1,1}+2t_{3,3}+\cdots+2t_{g-1,g-1})\quad \text{for $g$ even.}
             \end{cases}
\end{aligned}$$

By Theorem \[tw:pres:pos\], ${\langle {\overline{X}} \rangle}$ is generated by $t_{i,j}$ and ${\varrho}_j$. Using formulas for ${\overline{\partial}}_1(t_{i,j})$ and ${\overline{\partial}}_1({\varrho}_j)$ it is straightforward to check that elements (F1)–(F4) are elements of $\ker{\overline{\partial}}_1$. Moreover, $$2t_{j,1}+2t_{j,2}+\cdots+2t_{j,g}=[t_{a_j}]\otimes 2(\gamma_1+\cdots +\gamma_g)=0,$$ hence $r_{t_j}$ is indeed a relation. Similarly we check that $r_{{\varrho}}$ is a relation.

Observe that using relations $r_{t_j}$ and $r_{{\varrho}}$ we can substitute for $2t_{j,g}$ and $2{\varrho}_1$ respectively, hence each element in ${\langle {\overline{X}} \rangle}$ can be written as a linear combination of $t_{i,j},{\varrho}_j$, where each of $t_{1,g},t_{2,g},\ldots,t_{g-1,g},{\varrho}_1$ has the coefficient 0 or 1. Moreover, for a given $x\in {\langle {\overline{X}} \rangle}\subset C_1(G)\otimes H_1(N_g;{{\mathbb{Z}}})$ such a combination is unique. Hence for the rest of the proof we assume that linear combinations of $t_{i,j},{\varrho}_j$ satisfy this condition.

Suppose that $h\in{\langle {\overline{X}} \rangle}\cap \ker{\overline{\partial}}_1$. We will show that $h$ can be uniquely expressed as a linear combination of generators (F1)–(F4).

First observe that $h=h_1+h_2$, where $h_1$ is a combination of generators (F1)–(F2), and $h_2$ does not contain generators of type (F1) nor elements $t_{j,j+1}$. Moreover, $h_1$ and $h_2$ are uniquely determined by $h$.

Next we decompose $h_2=h_3+h_4$, where $h_3$ is a combination of generators (F3) and $h_4$ does not contain ${\varrho}_j$ for $j<g$. As before, $h_3$ and $h_4$ are uniquely determined by $h_2$.

Element $h_4$ has the form $$h_4=\sum_{j=1}^{g-1} \alpha_jt_{j,j}+\alpha {\varrho}_g,$$ for some integers $\alpha,\alpha_1,\ldots,\alpha_{g-1}$. Hence $$0={\overline{\partial}}_1(h_4)=\alpha_1\gamma_1+(\alpha_1+\alpha_2)\gamma_2+\cdots+(\alpha_{g-2}+
\alpha_{g-1})\gamma_{g-1}+(\alpha_{g-1}-2\alpha)\gamma_g.$$ If $g$ is odd this implies that $$\alpha_1=\alpha_3=\cdots=\alpha_{g-2}=2k,\ \alpha_2=\alpha_4=\cdots=\alpha_{g-1}=0,\ \alpha=-k,$$ for some $k\in{{\mathbb{Z}}}$. For $g$ even we get $$\alpha_1=\alpha_3=\cdots=\alpha_{g-1}=2k,\ \alpha=\alpha_2=\alpha_4=\cdots=\alpha_{g-2}=0.$$ In each of these cases $h_4$ is a multiple of the generator (F4).

By an analogous argument we get

\[gen\_ker\_roz\] Let $g\geqslant 3$ and $G={{{\mathcal{M}}}}^h(N_g)$ then ${\langle {\overline{X}} \rangle}\cap\ker{\overline{\partial}}_1$ is the abelian group which admits the presentation with generators: (F1)–(F4),

1.  $s_j+s_{j-1}$, where $j$ is even,

2.  $s_j-{\varrho}_1-{\varrho}_2-\cdots-{\varrho}_j$, where $j$ is odd.

The defining relations are $r_{t_j}$, $r_{{\varrho}}$ and $$r_{s}\!:\begin{cases}
0=2(s_2+s_1)+2(s_4+s_3)+\cdots+2(s_{g-1}+s_{g-2})\\
\qquad\qquad\qquad\qquad\  +2(s_{g}-{\varrho}_1-{\varrho}_2-\cdots-{\varrho}_g)\quad\text{for $g$ odd}\\
            0=2(s_2+s_1)+2(s_4+s_3)+\cdots+2(s_g+s_{g-1})\quad\text{for $g$ even.}
           \end{cases}$$

Rewriting relations
-------------------

Using formula we rewrite relations (E1)–(E5) as relations in $H_1({{{\mathcal{M}}}}^{h+}(N_g); H_1(N_g;{{\mathbb{Z}}}))$.

Relation (E1) is symmetric with respect to $k$ and $j$, hence we can assume that $j+1<k$. This relation gives $$\begin{aligned}r^{(E1)}_{k,j:i}\!:0&=([t_{a_k}]+t_{a_k}[t_{a_j}]-[t_{a_j}]-t_{a_j}[t_{a_k}])\otimes \gamma_i\\
&=t_{k,i}+[t_{a_j}]\otimes \psi(t_{a_k}^{-1})\gamma_i-t_{j,i}-[t_{a_k}]\otimes \psi(t_{a_j}^{-1})\gamma_i\\
&=\pm\begin{cases}
   0&\text{if $i\neq k,k+1,j,j+1$}\\
t_{j,k}+t_{j,k+1}&\text{if $i=k$ or $i=k+1$}\\
t_{k,j}+t_{k,j+1}&\text{if $i=j$ or $i=j+1$.}
  \end{cases}
  \end{aligned}$$ Relation (E2) gives $$\begin{aligned}r^{(E2)}_{j:i}\!:0=&
([t_{a_j}]+t_{a_{j}}[t_{a_{j+1}}]+t_{a_{j}}t_{a_{j+1}}[t_{a_{j}}]\\
&-[t_{a_{j+1}}]-t_{a_{j+1}}[t_{a_{j}}]-t_{a_{j+1}}t_{a_{j}}[t_{a_{j+1}}])\otimes \gamma_i\\
=&\begin{cases}
                 t_{j,i}-t_{j+1,i}&\text{\hspace{-1.5cm}if $i\neq j,j+1,j+2$}\\
 t_{j,j+2}-t_{j+1,j}&\text{\hspace{-1.5cm}if $i=j+2$}\\
(*)+2(t_{j,j}+t_{j,j+1})&\text{\hspace{-1.5cm}if $i=j$}\\
(*)-(t_{j,j}+t_{j,j+1})-(t_{j+1,j+1}+t_{j+1,j+2})&\text{if $i=j+1$.}
                 \end{cases}
\end{aligned}$$ In the above formula $(*)$ denotes some expression homologous to 0 by previously obtained relations. Carefully checking relations $r^{(E1)}_{k,j:i}$ and $r^{(E2)}_{j:i}$ we conclude that generators (F1) generate a cyclic group, and generators (F2) generate a cyclic group of order at most 2.

We next turn to the relation (E5). It gives $$\begin{aligned}r^{(E5)}_{j:i}\!:0=&([{\varrho}]+{\varrho}[t_{a_j}]-[t_{a_j}]-t_{a_j}[{\varrho}])\otimes \gamma_i\\
   &=\begin{cases}
        -2t_{j,i}&\text{if $i\neq j,j+1$}\\
 -2t_{j,j}-{\varrho}_j-{\varrho}_{j+1}&\text{if $i=j$}\\
({\varrho}_j+{\varrho}_{j+1}+2t_{j,j})-2(t_{j,j}+t_{j,j+1})&\text{if $i=j+1$.}
       \end{cases}
\end{aligned}$$ These relations imply that generators (F3) are homologically trivial, and generators (F1) generate at most ${{\mathbb{Z}}}_2$.

We now turn to the most difficult relation, namely (E3). This relation gives $$\begin{aligned}
r^{(E3)}_{i}\!:0&=\sum_{k=0}^{g-1}\sum_{n=1}^{g-1}(t_{a_1}\cdots t_{a_{g-1}})^kt_{a_1}\cdots t_{a_{n-1}}[t_{a_n}]\otimes \gamma_i-{\varepsilon}{\varrho}_i\\
&=\sum_{n=1}^{g-1}[t_{a_n}]\otimes \psi(t_{a_1}\cdots t_{a_{n-1}})^{-1} \sum_{k=0}^{g-1}\psi(t_{a_1}\cdots t_{a_{g-1}})^{-k}\gamma_i-{\varepsilon}{\varrho}_i\\
&=\sum_{n=1}^{g-1}[t_{a_n}]\otimes Y_{n} \sum_{k=0}^{g-1}Y_{g}^k\gamma_i-{\varepsilon}{\varrho}_i.
\end{aligned}$$ Where ${\varepsilon}=0$ for $g$ even, ${\varepsilon}=1$ for $g$ odd, and $Y_n=\psi(t_{a_1}\cdots t_{a_{n-1}})^{-1}$. Using the matrix formula for $\psi(t_{a_i}^{-1})$, we obtain $$Y_n\gamma_i=\begin{cases}
               -\gamma_{i-1}&\text{if $2\leq i\leq n$}\\
\gamma_{i}&\text{if $i> n$}\\
2\gamma_1+\cdots+2\gamma_{n-1}+\gamma_n&\text{if $i=1$.}
              \end{cases}$$ In particular $$Y_g^k\gamma_i=(-1)^k\gamma_{i-k},$$ where we subtract indexes modulo $g$. Therefore we have $$r^{(E3)}_{i}\!:0=\sum_{n=1}^{g-1}[t_{a_n}]\otimes Y_{n}\sum_{k=0}^{g-1}(-1)^k\gamma_{i-k}-{\varepsilon}{\varrho}_i.$$ In order to simplify computations we replace relations: $$r_1^{(E3)},\ r_2^{(E3)},\ldots,\ r_g^{(E3)}$$ with relations: $$r_1^{(E3)}+r_2^{(E3)},\ r_2^{(E3)}+r_3^{(E3)},\ldots,r_{g-1}^{(E3)}+r_g^{(E3)},\ r_g^{(E3)}.$$ Let us begin with $r_g^{(E3)}$. $$\begin{aligned}
r_g^{(E3)}\!:0=&\sum_{n=1}^{g-1}[t_{a_n}]\otimes Y_{n}\sum_{k=0}^{g-1}(-1)^k\gamma_{g-k}-{\varepsilon}{\varrho}_{g}\\
=&\sum_{n=1}^{g-1}[t_{a_n}]\otimes \left(\sum_{k=0}^{g-n-1}(-1)^k\gamma_{g-k}+\sum_{k=g-n}^{g-2}(-1)^{k+1}\gamma_{g-k-1}\right.\\
&\qquad + (-1)^{g-1}(2\gamma_1+\cdots+2\gamma_{n-1}+\gamma_n)\bigg)-{\varepsilon}{\varrho}_{g}.
  \end{aligned}$$ Since all generators of type (F1) are homologous to a single generator, say $t$, and $2t=0$, the above relation can be rewritten as $$r_g^{(E3)}\!:0=
(g-1)(g-2)t+\sum_{n=1}^{g-1}[t_{a_n}]\otimes \left((-1)^{g-n-1}\gamma_{n+1}+(-1)^{g-1}\gamma_n\right)-{\varepsilon}{\varrho}_{g}.$$ If $g$ is even, this gives the relation $$\begin{aligned}
r_g^{(E3)}\!:0=&(-t_{1,1}+ t_{1,2})+(-t_{2,2}-t_{2,3})+\cdots+(-t_{g-1,g-1}+t_{g-1,g})\\
=&(t_{1,1}+ t_{1,2})-(t_{2,2}+t_{2,3})+\cdots+(t_{g-1,g-1}+t_{g-1,g})\\
&-2(t_{1,1}+t_{3,3}+\cdots+t_{g-1,g-1}).
\end{aligned}$$ If $g$ is odd, we have $$\begin{aligned}
r_g^{(E3)}\!:0=&(t_{1,1}- t_{1,2})+(t_{2,2}+t_{2,3})+\cdots+(t_{g-1,g-1}+t_{g-1,g})-{\varrho}_g\\
=&-(t_{1,1}+ t_{1,2})+(t_{2,2}+t_{2,3})-\cdots+(t_{g-1,g-1}+t_{g-1,g})\\
&+2(t_{1,1}+t_{3,3}+\cdots+t_{g-2,g-2})-{\varrho}_g.
\end{aligned}$$ In both cases relation $r_g^{(E3)}$ implies that generator (F4) is superfluous.

Now we concentrate on the relation $r_i^{(E3)}+r_{i+1}^{(E3)}$. $$\begin{aligned}
   r_i^{(E3)}+r_{i+1}^{(E3)}&\!:0=\sum_{n=1}^{g-1}[t_{a_n}]\otimes Y_{n}
\sum_{k=0}^{g-1}(-1)^k\left(\gamma_{i-k}+\gamma_{i+1-k}\right)-{\varepsilon}({\varrho}_i+{\varrho}_{i+1})\\
&=\sum_{n=1}^{g-1}[t_{a_n}]\otimes Y_{n}
\left(\gamma_{i+1}+(-1)^{g-1}\gamma_{i+1}\right)-{\varepsilon}({\varrho}_i+{\varrho}_{i+1}).
  \end{aligned}$$ If $g$ is even, this relation is trivial, and if $g$ is odd it gives $$\begin{aligned}
   r_i^{(E3)}&+r_{i+1}^{(E3)}\!:0=2\sum_{n=1}^{g-1}[t_{a_n}]\otimes Y_{n}
\left(\gamma_{i+1}\right)-({\varrho}_i+{\varrho}_{i+1})\\
&=2(t_{1,i+1}+\cdots+t_{i,i+1}-t_{i+1,i}-\cdots-t_{g-1,i})-({\varrho}_i+{\varrho}_{i+1})\\
&=(*)+2(t_{i,i}+t_{i,i+1})-(2t_{i,i}+{\varrho}_i+{\varrho}_{i+1}).
  \end{aligned}$$ Hence this relation gives no new information.

Relation $(E4)$ gives no new information, hence we proved the following theorem.

\[tw:hom:positive\] If $g\geqslant 3$, then $$H_1({{{\mathcal{M}}}}^{h+}(N_g); H_1(N_g;{{\mathbb{Z}}}))={{\mathbb{Z}}}_2\oplus{{\mathbb{Z}}}_2.$$

Computing $H_1({{{\mathcal{M}}}}^{h}(N_g); H_1(N_g;{{\mathbb{Z}}}))$
--------------------------------------------------------------------

If $G={{{\mathcal{M}}}}^{h}(N_g)$, then by Proposition \[gen\_ker\_roz\] the kernel ${\langle {\overline{X}} \rangle}\cap\ker{\overline{\partial}}_1$ has two more types of generators: (F5), (F6), and by Theorem \[th:004\] there are three additional relations: (C4),(C5),(C8). $$\begin{aligned}
   r_i^{(C4)}\!:0&=[s]\otimes \gamma_i+s[s]\otimes\gamma_i=s_i+[s]\otimes\psi(s)\gamma_i\\
&=2(-1)^i\left(s_1+s_2+\cdots+s_{i-1}+\frac{1+(-1)^i}{2}s_i\right).
  \end{aligned}$$ This (inductively) implies that each generator of type (F5) has order at most 2. $$\begin{aligned}
   r_i^{(C8)}\!:0&=([{\varrho}]+{\varrho}[s]-[s]-s[{\varrho}])\otimes\gamma_i={\varrho}_i-2s_i-[{\varrho}]\otimes \psi(s)\gamma_i\\
&={\varrho}_i-2s_i-(-1)^i(2{\varrho}_1+2{\varrho}_2+\cdots +2{\varrho}_{i-1}+{\varrho}_i)\\
&=\begin{cases}
   -2(s_i-{\varrho}_1-\cdots-{\varrho}_i)&\text{ for $i$ odd}\\
-2(s_{i-1}+s_i)+2(s_{i-1}-{\varrho}_1-\cdots -{\varrho}_{i-1})&\text{ for $i$ even.}
  \end{cases}
  \end{aligned}$$ This implies that generator (F6) has also order at most 2. $$\begin{aligned}
   r_i^{(C5)}\!:0&=([t_{a_j}]+t_{a_j}[s]+t_{a_j}s[t_{a_j}]-[s])\otimes \gamma_i\\
&=t_{j,i}+[s]\otimes\psi(t_{a_j}^{-1})\gamma_i+[t_{a_j}]\otimes \psi(s)\psi(t_{a_j}^{-1})\gamma_i-s_i.
  \end{aligned}$$ If $i\neq j$ and $i\neq j+1$, then $$r_i^{(C5)}\!:0=(-1)^i\left(2t_{j,1}+\cdots+2t_{j,i-1}+(1+(-1)^i)t_{j,i}\right),$$ which gives no new information. If $i=j$ or $i=j+1$ and $j$ is odd, then $$r_i^{(C5)}\!:0=(*)\pm [(s_j+s_{j+1})+(t_{j,j}+t_{j,j+1})],$$ where as usual $(*)$ denotes homologically trivial element. This relation implies that generators (F5) are superfluous.

Finally, if $i=j$ or $i=j+1$ and $j$ is even then $$r_i^{(C5)}\!:0=(*)\pm[(s_{j+1}-{\varrho}_1-\cdots-{\varrho}_{j+1})-(s_{j-1}-{\varrho}_1-\cdots-{\varrho}_{j-1})].$$ This implies that all generators of type (F6) are homologous, hence we proved the following.

\[tw:hom:full\] If $g\geqslant 3$, then $$H_1({{{\mathcal{M}}}}^{h}(N_g); H_1(N_g;{{\mathbb{Z}}}))={{\mathbb{Z}}}_2\oplus{{\mathbb{Z}}}_2\oplus{{\mathbb{Z}}}_2.$$

Acknowledgements {#acknowledgements .unnumbered}
================

The author wishes to thank the referee for his/her helpful suggestions.

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[^1]: Supported by MNiSW grant N201 366436.
