# THE SECOND RATIONAL HOMOLOGY OF THE TORELLI GROUP IS FINITELY GENERATED

Daniel Minahan  Skiles Classroom Building, Georgia Institute of Technology,
Atlanta, GA 30332<EMAIL_ADDRESS>
###### Abstract.

We prove that second rational homology of the Torelli group of an orientable
closed surface of genus $g$ is finite dimensional for $g\geq 51$. This rules
out the simplest obstruction to the Torelli group being finitely presented and
provides a partial answer to a question of Bestvina.

###### Key words and phrases:

Torelli group

###### 2000 Mathematics Subject Classification:

Primary 54C40, 14E20; Secondary 49E25, 20C20

##  1\. Introduction

Let $S_{g}^{b}$ be a compact, orientable surface of genus $g$ with $b$
boundary components. The mapping class group $\operatorname{Mod}(S_{g})$ is
$\pi_{0}(\operatorname{Diff}^{+}(S_{g}))$. The action of
$\operatorname{Mod}(S_{g})$ on $H_{1}(S_{g};\mathbb{Z})$ induces a
representation
$\operatorname{Mod}(S_{g})\rightarrow\operatorname{Sp}(2g,\mathbb{Z})$ called
the symplectic representation, where $\operatorname{Sp}(2g,\mathbb{Z})$ is the
group of invertible linear transformations of $H_{1}(S_{g};\mathbb{Z})$ that
respect the algebraic intersection form $\langle\cdot,\cdot\rangle$. The
kernel of this representation is called the Torelli group and is denoted
$\mathcal{I}_{g}$. There is a short exact sequence

$1\rightarrow\mathcal{I}_{g}\rightarrow\operatorname{Mod}(S_{g})\rightarrow\operatorname{Sp}(2g,\mathbb{Z})\rightarrow
1.$

By the work of McCullough–Miller [19], Mess [20] and Johnson [12], the Torelli
group is finitely generated if and only if $g\neq 2$. Birman posed the
following question [4, Problem 29]:

######  Question 1.

Is $\mathcal{I}_{g}$ finitely presented for any sufficiently large $g$?

Similar questions were asked by Mess [16, Page 90] and Morita [22, Problem
2.1].

_Obstructions to finite presentability_. If a group $G$ is not finitely
presented, then the vector space $H_{2}(G;\mathbb{Q})$ is infinite
dimensional. The main result of the paper is the following theorem, which
shows that $H_{2}(\mathcal{I}_{g};\mathbb{Q})$ does not obstruct finite
presentability of $\mathcal{I}_{g}$ for $g\gg 0$.

######  Theorem A.

Let $g\geq 51$. The vector space $H_{2}(\mathcal{I}_{g};\mathbb{Q})$ is finite
dimensional.

Theorem A partially answers the following question of Bestvina [7, Page 5].
Margalit also asked a similar question [18, Question 5.12]).

######  Question 2.

For which choices of $k,g$, and commutative ring $R$ is
$H_{k}(\mathcal{I}_{g};R)$ finitely generated?

_Prior partial answers to Question 2._ Mess showed that $\mathcal{I}_{2}$ is
an infinitely generated free group [20], which implies that
$H_{1}(\mathcal{I}_{2})$ is infinitely generated and
$H_{i}(\mathcal{I}_{2})=0$ for $i\geq 2$. Johnson [13] computed
$H_{1}(\mathcal{I}_{g};\mathbb{Z})$ for $g\geq 3$ and proved that

$H_{1}(\mathcal{I}_{g};\mathbb{Q})\cong\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q}).$

In fact, Johnson defined a map
$\tau_{g}:\mathcal{I}_{g}\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Z})/H_{1}(S_{g};\mathbb{Z})$
now called the Johnson homomorphism, and showed that $(\tau_{g})_{*}$ induces
an isomorphism in first rational homology [13]. Akita showed that
$H_{*}(\mathcal{I}_{g};\mathbb{Z})$ is infinitely generated as an abelian
group for $g\geq 7$ [1]. Hain showed that $H_{3}(\mathcal{I}_{3};\mathbb{Z})$
is infinitely generated [10]. Bestvina, Bux, and Margalit showed that the
cohomological dimension of $\mathcal{I}_{g}$ is $3g-5$ [2, Theorem A], which
implies that $H_{k}(\mathcal{I}_{g};\mathbb{Z})=0$ for $k\geq 3g-4$. In this
same paper, they proved that $H_{3g-5}(\mathcal{I}_{g};\mathbb{Z})$ is
infinitely generated for $g\geq 2$ [2, Theorem C]. Gaifullin later showed that
$H_{k}(\mathcal{I}_{g};\mathbb{Z})$ is infinitely generated for $2g-3\leq
k\leq 3g-5$ [9], confirming a conjecture of Bestvina–Bux–Margalit [2].

###  1.1. The outline of the proof of Theorem A

Theorem A will follow from the following result.

######  Theorem B.

Let $g\geq 51$. Let $a\subseteq S_{g}$ be a nonseparating simple closed curve.
The cokernel of the pushforward map
$\iota_{*}:H_{2}(\operatorname{Stab}_{\mathcal{I}_{g}}(a);\mathbb{Q})\rightarrow
H_{2}(\mathcal{I}_{g};\mathbb{Q})$ is finite dimensional.

Most of the paper will be devoted to proving Theorem B.

_The connection between Theorem A and Theorem B_. Let $\mathcal{B}$ be a
finite collection of nonseparating simple closed curves such that the set of
Dehn twists

$\\{T_{c}:c\in\mathcal{B}\\}$

generates $\operatorname{Mod}(S_{g})$, e.g., the curves corresponding to the
Humphries generators [8]. Consider the map

$\varphi:H_{2}(\mathcal{I}_{g};\mathbb{Q})\rightarrow\bigoplus_{c\in\mathcal{B}}\operatorname{cok}(H_{2}(\operatorname{Stab}_{\mathcal{I}_{g}}(c);\mathbb{Q})\rightarrow
H_{2}(\mathcal{I}_{g};\mathbb{Q})).$

Now, for any $c\in\mathcal{B}$, the quotient map
$\rho_{c}:H_{2}(\mathcal{I}_{g};\mathbb{Q})\rightarrow\operatorname{cok}(H_{2}(\operatorname{Stab}_{\mathcal{I}_{g}}(c);\mathbb{Q})\rightarrow
H_{2}(\mathcal{I}_{g};\mathbb{Q}))$ satisfies

$\ker(\rho_{c})=\operatorname{im}(H_{2}(\operatorname{Stab}_{\mathcal{I}_{g}}(c);\mathbb{Q})\rightarrow
H_{2}(\mathcal{I}_{g};\mathbb{Q})).$

Hence we have

$\ker(\varphi)=\bigcap_{c\in\mathcal{B}}\operatorname{im}(H_{2}(\operatorname{Stab}_{\mathcal{I}_{g}}(c);\mathbb{Q})\rightarrow
H_{2}(\mathcal{I}_{g};\mathbb{Q})).$

Then for any $c\in\mathcal{B}$, $T_{c}$ acts trivially
$\operatorname{im}(H_{2}(\operatorname{Stab}_{\mathcal{I}_{g}}(c);\mathbb{Q})\rightarrow
H_{2}(\mathcal{I}_{g};\mathbb{Q}))$. Therefore every element of
$\\{T_{c}:c\in\mathcal{B}\\}$ acts trivially on $\ker(\varphi)$. Since
$\\{T_{c}:c\in\mathcal{B}\\}$ generates $\operatorname{Mod}(S_{g})$, the
vector space $\ker(\varphi)$ is a trivial
$\operatorname{Sp}(2g,\mathbb{Z})$–module. We will show in Lemma 9.1 that
$H_{0}(\operatorname{Sp}(2g,\mathbb{Z});H_{2}(\mathcal{I}_{g};\mathbb{Q}))$ is
finite dimensional. This is a result of Kassabov and Putman [14], for which we
provide an alternative proof. Theorem B says that each vector space
$\operatorname{cok}(H_{2}(\operatorname{Stab}_{\mathcal{I}_{g}}(c);\mathbb{Q})\rightarrow
H_{2}(\mathcal{I}_{g};\mathbb{Q}))$ is finite dimensional. Since $\mathcal{B}$
is a finite set, the vector space $\operatorname{im}(\varphi)$ is finite
dimensional. Consider the short exact sequence

$0\rightarrow\ker(\varphi)\rightarrow
H_{2}(\mathcal{I}_{g};\mathbb{Q})\rightarrow\operatorname{im}(\varphi)\rightarrow
0.$

A portion of the corresponding long exact sequence in
$\operatorname{Sp}(2g,\mathbb{Z})$–homology is given by

$\ldots\rightarrow
H_{1}(\operatorname{Sp}(2g,\mathbb{Z});\operatorname{im}(\varphi))\rightarrow
H_{0}(\operatorname{Sp}(2g,\mathbb{Z});\ker(\varphi))\rightarrow
H_{0}(\operatorname{Sp}(2g,\mathbb{Z});H_{2}(\mathcal{I}_{g};\mathbb{Q}))\rightarrow\ldots$

Since $\operatorname{Sp}(2g,\mathbb{Z})$ is finitely generated and
$\operatorname{im}(\varphi)$ is finite dimensional,
$H_{1}(\operatorname{Sp}(2g,\mathbb{Z});\operatorname{im}(\varphi))$ is finite
dimensional as well. By Lemma 9.1,
$H_{0}(\operatorname{Sp}(2g,\mathbb{Z});H_{2}(\mathcal{I}_{g};\mathbb{Q}))$ is
finite dimensional. Hence the vector space
$H_{0}(\operatorname{Sp}(2g,\mathbb{Z}),\ker(\varphi))$ is finite dimensional.
Then $H_{0}(\operatorname{Sp}(2g,\mathbb{Z});\ker(\varphi))$ is isomorphic to
$\ker(\varphi)$ because $\ker(\varphi)$ is a trivial
$\operatorname{Sp}(2g,\mathbb{Z})$–module, so $\ker(\varphi)$ is finite
dimensional. Since $\operatorname{im}(\varphi)$ is finite dimensional as well,
we have $H_{2}(\mathcal{I}_{g};\mathbb{Q})$ finite dimensional.

###  1.2. The proof of Theorem B

The proof of Theorem B proceeds by considering the action of $\mathcal{I}_{g}$
on a complex called the complex of homologous curves, and then studying the
equivariant homology spectral sequence for this group action.

_The complex of homologous curves_. In this paper, a curve on a surface $S$ is
an isotopy class of oriented essential embedded copies of $S^{1}$. Two curves
$c$ and $d$ are disjoint if they have disjoint representatives. The geometric
intersection number, denoted $|c\cap d|$, is the minimum number of
intersection points over all representatives of $c$ and $d$. Let $a\subseteq
S_{g}$ be a nonseparating curve and let $[a]\in H_{1}(S_{g};\mathbb{Z})$
denote the corresponding first homology class. The complex of homologous
curves $\mathcal{C}_{[a]}(S_{g})$ is the complex where the vertices are curves
$c$ such that $c$ represents $[a]$ in $H_{1}(S_{g};\mathbb{Z})$. A $k$–cell in
$\mathcal{C}_{[a]}(S_{g})$ is a collection of $k+1$ pairwise disjoint vertices
of $\mathcal{C}_{[a]}(S_{g})$. A theorem of the author says that
$\widetilde{H}_{k}(C_{[a]}(S_{g});\mathbb{Z})=0$ for $0\leq k\leq g-3$ [21,
Theorem A].

_The equivariant homology spectral sequences_. We consider the equivariant
homology spectral sequence $\mathbb{E}_{*,*}^{*}$ for the action of
$\mathcal{I}_{g}$ on $\mathcal{C}_{[a]}(S_{g})$ [5, Section VII]. Let
$a\subseteq S_{g}$ be a nonseparating curve and let
$X_{g}=\mathcal{C}_{[a]}(S_{g})/\mathcal{I}_{g}$. We will use the notation
$X_{g}$ throughout the paper to refer to this quotient. If $\sigma$ is a cell
of $X_{g}$, let $(\mathcal{I}_{g})_{\sigma}$ denote the stabilizer
$\operatorname{Stab}_{\mathcal{I}_{g}}(\widehat{\sigma})$ for some arbitrary
lift of $\sigma$ to $\mathcal{C}_{[a]}(S_{g})$. It is known that if $x$ and
$y$ are nonseparating homologous curves in $S_{g}$, then there is an
$f\in\mathcal{I}_{g}$ such that $fx=y$ (see, e.g., [23, Lemma 6.2]). Hence the
set $X_{g}^{(0)}$ of vertices is a singleton. It follows that page 1 of
$\mathbb{E}_{*,*}^{*}$ is as in Figure 1.

[name = homologous bcl simple, homological Serre grading, y axis gap = 1.5cm,
y axis gap = 1.5cm , x axis extend end = 2cm, classes = draw = none, xscale =
3.2, yscale = 1] [”H_2((I_g)_a)”](0,2) [”H_1((I_g)_a)”](0,1) [”e
∈X^(1)⨁H_2((I_g)_e)”](1,2) [”e ∈X^(1)⨁H_1((I_g)_e)”](1,1)
[”σ∈X^(2)⨁H_1((I_g)_σ)”](2,1) [”e∈X^(1)⨁H_0((I_g)_e)”](1,0)
[”σ∈X^(2)⨁H_0((I_g)_σ)”](2,0) [”ρ∈X^(3)⨁H_0((I_g)_ρ)”](3,0) [̣source anchor =
180, target anchor = 0, shorten ¡ = 0.5]1(1,2) [̣source anchor = 180, target
anchor = 0, shorten ¡ = 0.5]1(1,1) [̣source anchor = 174, target anchor = 6,
shorten ¡ = 0.5]1(2,1) [̣source anchor = 174, target anchor = 6, shorten ¡ =
0.5]1(3,0) [̣source anchor = 174, target anchor = 6, shorten ¡ = 0.5]1(2,0)

[name = homologous bcl simple,page = 1]

Figure 1. Page 1 of $\mathbb{E}_{*,*}^{*}$ for the action of $\mathcal{I}_{g}$
on $\mathcal{C}_{[a]}(S_{g})$. All coefficients are in $\mathbb{Q}$.

Since $\mathcal{C}_{[a]}(S_{g})$ is 2–acyclic for $g\geq 5$ [21, Theorem A],
the spectral sequence $\mathbb{E}_{*,*}^{*}$ converges to
$H_{2}(\mathcal{I}_{g})$ [5, Section VII]. Examining Figure 1, we see that the
cokernel of the map

$\iota_{*}:H_{2}(\operatorname{Stab}_{\mathcal{I}_{g}}(a);\mathbb{Q})\rightarrow
H_{2}(\mathcal{I}_{g};\mathbb{Q})$

is noncanonically identified with the direct sum

$\mathbb{E}_{1,1}^{\infty}\oplus\mathbb{E}_{2,0}^{\infty}.$

Hence the cokernel of $\iota_{*}$ is noncanonically identified with a
subquotient of

$\mathbb{E}_{1,1}^{2}\oplus\mathbb{E}_{2,0}^{2}.$

The proof of Theorem B will proceed in two steps:

  1. I.

the vector space $\mathbb{E}_{2,0}^{2}$ is finite dimensional, and

  2. II.

the vector space $\mathbb{E}_{1,1}^{2}$ is finite dimensional.

By definition, $\mathbb{E}_{2,0}^{2}$ is canonically identified with
$H_{2}(X_{g};\mathbb{Q})$. Step I is recorded as the following proposition.

######  Proposition 1.1.

Let $g\geq 51$ and $a\subseteq S_{g}$ be a nonseparating simple closed curve.
Let $X_{g}=\mathcal{C}_{[a]}(S_{g})/\mathcal{I}_{g}$. The vector space
$H_{2}(X_{g};\mathbb{Q})$ is finite dimensional.

Proposition 1.1 will be proven in Sections 3–6. Step II of the proof of
Theorem B is recorded as the following result.

######  Proposition 1.2.

Let $g\geq 51$ and let $a\subseteq S_{g}$ be a nonseparating simple closed
curve. Let $\mathbb{E}_{*,*}^{*}$ denote the equivariant homology spectral
sequence in rational coefficients for the action of $\mathcal{I}_{g}$ on
$\mathcal{C}_{[a]}(S_{g})$. The vector space $\mathbb{E}_{1,1}^{2}$ is finite
dimensional.

Proposition 1.2 will be proven in Sections 7–8. Before discussing the proofs
of Propositions 1.1 and 1.2, we will discuss the main technical tool of the
paper.

_Representations of transvective type_. In Section 2, we will introduce the
notion of a representative of transvective type. A representation of
transvective type is a group $G$, a set of elements $\mathcal{F}\subseteq G$,
and a $G$–representation $V$ that satisfies certain hypotheses. The output of
Proposition 2.1 is that if $(G,\mathcal{F},V)$ is a representation of
transvective type, then $V$ is finite dimensional. We will apply Proposition
2.1 to the following representations:

  * •

$G=\operatorname{Stab}_{\operatorname{Sp}(2g,\mathbb{Z})}([a])$,
$V=\mathbb{E}_{2,0}^{2}$, and $\mathcal{F}$ is a set of transvections along
nine linearly indpendent elements in $[a]^{\perp}$ (Proposition 1.1).

  * •

$G=\operatorname{Stab}_{\operatorname{Sp}(2g,\mathbb{Z})}([a])$,
$V=\mathbb{E}_{1,1}^{2}$, and $\mathcal{F}$ is a set of transvections along
nine linearly indpendent elements in $[a]^{\perp}$ (Proposition 1.2).

  * •

$G=\operatorname{Sp}(2g,\mathbb{Z})$, $V=H_{2}(\mathcal{I}_{g};\mathbb{Q})$,
and $\mathcal{F}$ is singleton consisting of a transvection along a primitive
element (Theorem A).

The term representative of transvective type refers to the fact that
$\mathcal{F}$ a set of transvections. A tuple $(G,\mathcal{F},V)$ is a
reprsentation of transvective type if it satisfies four hypotheses. We will
state the hypotheses explicitly in Section 2, but roughly they are as follows:

  1. (1)

a normal generating condition,

  2. (2)

a finitness of coinvariants condition,

  3. (3)

a finiteness of quotient spaces condition, and

  4. (4)

a finite generating condition.

In Section 2, we prove Lemma 2.2, which will allow us to verify that
hypotheses (1) and (4) hold for the representations mentioned above. The
content of Sections 3–9 is verifying that hypotheses (2) and (3) hold for the
representations corresponding to Proposition 1.1, Proposition 1.2, and Theorem
A.

###  1.3. The outline of the paper

The paper is divided into four parts:

  * •

Section 2,

  * •

Sections 3–6,

  * •

Sections 7–8, and

  * •

Section 9.

_Section 2_. As above, this section contains the proofs of Proposition 2.1 and
Lemma 2.2.

_Sections 3–6_. These sections comprise the proof of Proposition 1.1. The main
output of each section is as follows:

  * •

Section 3 contains preliminary work about the structure of the quotient space
$\mathcal{C}_{[a]}(S_{g})/\mathcal{I}_{g}$. In particular, we describe certain
tori in the quotient space $\mathcal{C}_{[a]}/\mathcal{I}_{g}$ called
Bestvina–Margalit tori. This content is also relevant for the proof of
Proposition 1.2.

  * •

Section 4 contains Proposition 4.1, which is used in Section 6 to verify that
hypothesis (3) of representation of transvective type holds for the
representation associated to Proposition 1.1.

  * •

Section 5 contains Lemma 5.1, which is used to verify that hypothesis (2) of
the definition of representation of transvective type holds for the
representation associated to Proposition 1.1.

  * •

Section 6, which contains the proof of Proposition 1.1. The bulk of the
section is devoted to proving Lemma 6.1, which is a lemma that allows us to
apply Proposition 4.1.

_Sections 7–8_. These sections comprise the proof of Proposition 1.2. The main
output of each section is as follows:

  * •

Section 7 contains Proposition 1.2, which is used to verify that hypothesis
(3) of the definition of representation of transvective type holds for the
representation associated to Proposition 1.2.

  * •

Section 8 contains Lemma 8.1, which is used to verify that hypothesis (2) of
the definition of representation of transvective type holds for the
representation associated to Proposition 1.2. Additionally, the section
contains the proof of Proposition 1.2 and the proof of Theorem B.

_Section 9_. This section contains Lemma 9.1, which verifies that hypothesis
(2) of the definition of represenation of transvective type holds for the
representation associated to Theorem A. Theorem B is used to verify that
hypothesis (3) holds for this representation. Additionally, this section
contains the proofs of Theorem A.

_Acknowledgments_. We would like to thank our advisor Dan Margalit for many
helpful math conversations and his continued support and encouragement. We
would also like to thank Benson Farb and Andy Putman for providing comments on
earlier drafts of this paper. We would like to thank Peter Patzt for helping
to clarify the exposition and for pointing out some confusing notational
choices. We would like to thank Igor Belegradek, Wade Bloomquist, Katherine
Williams Booth, John Etnyre, Annie Holden, Thang Le, Nick Salter, Roberta
Shapiro, Cindy Tan, and Bena Tshishiku for helpful conversations.

##  2\. A finiteness result about groups acting on vector spaces

The goal in this section is to prove Proposition 2.1. We first explicitly
define the term representation of transvective type as mentioned in the
introduction. We also prove Lemma 2.2, which is an auxiliary result about
generating sets of subgroups of the mapping class group that stabilize
nonseparating multicurves.

_Representations of transvective type_. Let $R$ be a commutative unital
Noetherian ring. After this section, we will always take $R=\mathbb{Q}$. Let
$G$ be a group and $V$ an $R$–module equipped with a linear $G$–action. If
$\mathcal{S}\subseteq G$ is a set, we will let
$\operatorname{C}_{G}(\mathcal{S})$ denote the $G$–centralizer of
$\mathcal{S}$. Let $\mathcal{F}\subseteq G$ be a set of pairwise commuting
elements. For each subset $\mathcal{F}^{\prime}\subsetneq\mathcal{F}$, let
$\operatorname{C}^{\prime}_{G}(\mathcal{F}^{\prime})\subseteq\operatorname{C}_{G}(\mathcal{F})$
be a subgroup. Assume that for
$\mathcal{F}^{\prime\prime}\subseteq\mathcal{F}^{\prime}$, we have
$\operatorname{C}^{\prime}_{G}(\mathcal{F}^{\prime\prime})\supseteq\operatorname{C}^{\prime}_{G}(\mathcal{F}^{\prime})$.
For $\mathcal{F}=\emptyset$, we let
$\operatorname{C}^{\prime}_{G}(\emptyset)=G$. Assume that the following hold:

  1. (1)

For any $\mathcal{F}^{\prime}\subsetneq\mathcal{F}$ and any
$f\in\mathcal{F}\setminus\mathcal{F}^{\prime}$,
$f\in\operatorname{C}^{\prime}_{G}(\mathcal{F}^{\prime})$ normally generates
$\operatorname{C}^{\prime}_{G}(\mathcal{F}^{\prime})$.

  2. (2)

For any $\mathcal{F}^{\prime}\subsetneq\mathcal{F}$, the homology group

$H_{0}\left(\operatorname{C}_{G}^{\prime}(\mathcal{F}^{\prime});V\right)$

is finitely generated as an $R$–module.

  3. (3)

The cokernel

$\operatorname{cok}\left(\bigoplus_{f\in\mathcal{F}}V^{f}\rightarrow V\right)$

is finitely generated as an $R$–module, where $V^{f}$ denotes the subspace of
$V$ fixed by $f$.

  4. (4)

For any subset $\mathcal{F}^{\prime}\subsetneq\mathcal{F}$, the subgroup
$\operatorname{C}_{G}^{\prime}(\mathcal{F}^{\prime})\subseteq\operatorname{C}_{G}(\mathcal{F}^{\prime})$
has a finite generating set contained in

$\bigcup_{h\in\operatorname{C}_{G}^{\prime}(\mathcal{F}^{\prime})}h(\mathcal{F}\setminus\mathcal{F}^{\prime})h^{-1}.$

We say that such a tuple $(G,\mathcal{F},V)$ is a representation of
transvective type. The terminology comes from the fact that Proposition 2.1
will be applied when:

  * •

$G=\operatorname{Sp}(L;\mathbb{Z})$ for $L$ a free abelian group equipped with
an alternating bilinear form,

  * •

$\mathcal{F}\subseteq G$ is a set of transvections along a set $\mathcal{V}$
of primitive, linearly independent elements of $L$,

  * •

$V$ is a $\mathbb{Q}$–vector space, and

  * •

the subgroups $\operatorname{C}^{\prime}_{G}(\mathcal{F}^{\prime})$ for
$\mathcal{F}^{\prime}\subseteq\mathcal{F}$ corresponding to
$\mathcal{V}^{\prime}\subseteq\mathcal{V}$ are subgroups of
$(\mathcal{V}^{\prime})^{G}$.

_Remark_. The choice to use a subgroup
$\operatorname{C}^{\prime}_{G}(\mathcal{F}^{\prime})\subseteq\operatorname{C}_{G}(\mathcal{F}^{\prime})$
instead of the whole centralizer $\operatorname{C}_{G}(\mathcal{F}^{\prime})$
in the definition of representation of transvective type hinges on the fact
that it is more convenient to deal with stabilizers of homology classes rather
than actual centralizers of transvections.

We are now ready to state the main result of Section 2.

######  Proposition 2.1.

Let $(G,\mathcal{F},V)$ be a representation of transvective type over a
commutative unital ring $R$. Then $V$ is finitely generated as an $R$–module.

###### Proof.

We induct on $|\mathcal{F}|$. Note that in the proof, “finitely generated”
means “finitely generated as an $R$–module”.

_The base case: $|\mathcal{F}|=1$_. By hypothesis (4) of the definition of
representation of transvective type, $G$ has a finite generating set
$\mathcal{H}$ consisting of conjugates of the unique $f\in\mathcal{F}$. Let
$\varphi$ be the map

$\varphi:V\rightarrow\bigoplus_{g\in\mathcal{H}}V/V^{g}.$

For $g\in\mathcal{H}$, there is an $h_{g}\in G$ such that
$h_{g}fh_{g}^{-1}=g$. Hence conjugating by $h_{g}$ induces an isomorphism
$V/V^{f}\rightarrow V/V^{g}$. By hypothesis (2), $V/V^{f}$ is finitely
generated, so $V/V^{g}$ is finitely generated. Then since $\mathcal{H}$ is
finite, the $R$–module $\bigoplus_{g\in\mathcal{H}}V/V^{g}$ is finitely
generated as well. Then since $R$ is Noetherian, the $R$–module
$\operatorname{im}(\varphi)$ is finitely generated. Consider now the short
exact sequence

$0\rightarrow\ker(\varphi)\rightarrow
V\rightarrow\operatorname{im}(\varphi)\rightarrow 0$

Since $\operatorname{im}(\varphi)$ is finitely generated, it suffices to show
that $\ker(\varphi)$ is finitely generated. The rightmost portion of the
corresponding long exact sequence in $G$–homology is given by

$\ldots\rightarrow H_{1}(G;\operatorname{im}(\varphi))\rightarrow
H_{0}(G;\ker(\varphi))\rightarrow H_{0}(G;V)\rightarrow
H_{0}(G;\operatorname{im}(\varphi))\rightarrow 0.$

Since $G=\operatorname{C}_{\emptyset}(G)$ is finitely generated by hypothesis
(4) and $\operatorname{im}(\varphi)$ is finitely generated,
$H_{1}(G;\operatorname{im}(\varphi))$ is finitely generated as well.
Furthermore, $G=\operatorname{C}_{\emptyset}(G)$, so $H_{0}(G;V)$ is finitely
generated by hypothesis (2). Hence $H_{0}(G;\ker(\varphi))$ is finitely
generated. By construction, $\ker(\varphi)=\bigcap_{g\in\mathcal{H}}V^{g}$.
Hence $\ker(\varphi)$ is a trivial $G$–module, since any $v\in\ker(\varphi)$
is fixed by every element of $\mathcal{H}$ and $G$ is generated by
$\mathcal{H}$. Therefore $\ker(\varphi)=H_{0}(G;\ker(\varphi))$, so
$\ker(\varphi)$ is finitely generated. Since $\ker(\varphi)$ and
$\operatorname{im}(\varphi)$ are finitely generated, $V$ is finitely generated
as well.

_The inductive step: $\left|\mathcal{F}\right|>1$_. Choose an element
$f\in\mathcal{F}$. Consider the right exact sequence

$\bigoplus_{g\in\mathcal{F}\setminus\\{f\\}}V^{g}\rightarrow V\rightarrow
V^{\prime}\rightarrow 0.$

Let $G_{f}=\operatorname{C}^{\prime}_{G}(\mathcal{F}\setminus\\{f\\})$. We
have the following claim.

_Claim_. The tuple $(G_{f},\\{f\\},V^{\prime})$ is a representation of
transvective type.

_Proof of claim_. We will verify in turn that each hypothesis in the
definition of representation of transvective type holds for
$(G_{f},\\{f\\},V^{\prime})$.

  1. (1)

By hypothesis (1) applied to $(G,\mathcal{F},V)$, the element $f$ normally
generates $G_{f}$. Since the only proper subset of $\\{f\\}$ is $\emptyset$,
hypothesis (1) holds.

  2. (2)

Hypothesis (2) implies that $H_{0}(G_{f};V)$ is finitely generated. Since the
pushforward map $H_{0}(G_{f};V)\rightarrow H_{0}(G_{f};V^{\prime})$ is
surjective, hypothesis (2) holds for $(G_{f},\\{f\\},V^{\prime})$ as well.

  3. (3)

By construction, there is a surjection

$\operatorname{cok}\left(\bigoplus_{g\in\mathcal{F}}V^{g}\rightarrow
V\right)\twoheadrightarrow\operatorname{cok}\left((V^{\prime})^{f}\rightarrow
V^{\prime}\right).$

Hypothesis (3) for $(G,\mathcal{F},V)$ therefore implies that
$\operatorname{cok}((V^{\prime})^{f}\rightarrow V^{\prime})$ is finitely
generated, so hypothesis (3) holds for $(G_{f},\\{f\\},V^{\prime})$.

  4. (4)

Hypothesis (4) follows since the only proper
$\mathcal{F}^{\prime}\subsetneq\\{f\\}$ is $\mathcal{F}^{\prime}=\emptyset$.
Hence hypothesis (4) for $(G,\mathcal{F},V)$ with
$\mathcal{F}^{\prime}=\mathcal{F}\setminus\\{f\\}$ says that $G_{f}$ has a
finite generating set consisting of $G_{f}$–conjugates of $f$.

Therefore $(G_{f},\\{f\\},V^{\prime})$ is a representation of transvective
type.

Given the claim, the inductive hypothesis says that $V^{\prime}$ is finitely
generated. We now show that

$(G,\mathcal{F}\setminus\\{f\\},V)$

is a representation of transvective type. Hypotheses (1), (2), and (4) for
$(G,\mathcal{F}\setminus\\{f\\},V)$ follow from the corresponding hypotheses
for $(G,\mathcal{F},V)$, since
$\mathcal{F}\setminus\\{f\\}\subseteq\mathcal{F}$. Hypothesis (3) follows from
the above proof that $(G,\\{f\\},V^{\prime})$ is a representation of
transvective type, since the base case of the induction verifies that

$V^{\prime}=\operatorname{cok}\left(\bigoplus_{g\in\mathcal{F}\setminus\\{f\\}}V^{g}\rightarrow
V\right)$

is finitely generated. Hence the inductive hypothesis applied to
$(G,\mathcal{F}\setminus\\{f\\},V)$ says that $V$ is finitely generated. ∎

In order to use Proposition 2.1 in the paper, we will need a general result
that allows us to verify that hypotheses (1) and (4) in the defintion of
representation of transvective type hold. A quasi–unimodular lattice is a
finitely generated free abelian group $L$ equipped with an alternating form
$\langle\cdot,\cdot\rangle:L\times L\rightarrow\mathbb{Z}$. The lattice $L$ is
unimodular if the form $\langle\cdot,\cdot\rangle$ is nondegenerate. The genus
of $L$, denoted $g(L)$, is $\frac{1}{2}\operatorname{rk}(W)$, where
$W\subseteq L$ is a maximal free abelian subgroup of $L$ such that the
restriction of $\langle\cdot,\cdot\rangle$ to $W$ is nondegenerate. If $v\in
L$ is some element, then $v$ is primitive if there is no integer $m\geq 2$ and
nonzero $w\in L$ with $mw=v$. For any $v\in L$, the transvection along $v$ is
the homomorphism $T_{v}:L\rightarrow L$ given by

$T_{v}(w)=w+\langle v,w\rangle v.$

We will say that the transvection $T_{v}$ is primitive if $v$ is primitive. We
will say that a subgroup $L^{\prime}\subseteq L$ is primitive if $L$ is
generated by primitive elements. The set of degenerate elements in $L$, i.e.,
elements $v\in L$ such that $\langle v,\cdot\rangle:L\rightarrow\mathbb{Z}$ is
the zero map, will be denoted $L^{\operatorname{degen}}$. If
$\mathcal{V}\subseteq L$ is a set of elements, we will denote by
$\mathcal{V}^{\perp}$ the subgroup $\\{w\in L:\langle w,v\rangle=0\text{ for
all }v\in\mathcal{V}\\}$. We let

$\operatorname{Sp}(L,\mathbb{Z})=\\{g\in\operatorname{GL}(L,\mathbb{Z}):\langle
gv,gw\rangle=\langle v,w\rangle\text{ for all }v,w\in L\\}.$

If $L^{\prime}\subseteq L$ is a primitive, unimodular subgroup, let
$\operatorname{proj}_{L^{\prime}}:L\rightarrow L^{\prime}$ denote the
projection map induced by the form $\langle\cdot,\cdot\rangle$.

_Multicurves_. We will say that a multicurve $M\subseteq S_{g}$ is an isotopy
class of pairwise disjoint oriented curves. We will use the notation
$\pi_{0}(M)$ to denote the set of isotopy classes of curves in $M$. The
stabilizer group $\operatorname{Stab}_{\operatorname{Mod}(S_{g})}(M)$ will
denote the subgroup of $\operatorname{Mod}(S_{g})$ consisting of elements that
stabilize each curve $c\in\pi_{0}(M)$. Note that since our multicurves are
oriented, the stabilizer $\operatorname{Stab}_{\operatorname{Mod}(S_{g})}(M)$
does not change the orientation of the curves of $M$.

_Notation_. For the remainder of the paper, we will use ${\ssearrow}$ to
denote Farb and Margalit’s notion of cutting along curves on surfaces [8,
Section 3.6].

_Mapping class groups of cut open surfaces_. Let $M\subseteq S_{g}$ be a
nonseparating multicurve. The mapping class group
$\operatorname{Mod}(S_{g}{\ssearrow}M)$ is the mapping class group of the cut
open surface. There is a natural inclusion $S_{g}{\ssearrow}M\hookrightarrow
M$, and this induces a map
$\operatorname{Mod}(S_{g}{\ssearrow}M)\rightarrow\operatorname{Mod}(S_{g})$
given by extending $\varphi\in\operatorname{Mod}(S_{g}{\ssearrow}M)$ by the
identity along $S_{g}\setminus\operatorname{im}(S_{g}{\ssearrow}M\rightarrow
S_{g})$.

We conclude Section 2 by proving the showing that hypotheses (1) and (4) of
the definition of representation of transvective type hold for subgroups
$G\subseteq\operatorname{Sp}(2g,\mathbb{Z})$, where $G$ will be the image of
$\operatorname{Mod}(S_{g}{\ssearrow}M)\rightarrow\operatorname{Mod}(S_{g})\rightarrow\operatorname{Sp}(2g,\mathbb{Z})$
for $M\subseteq S_{g}$ some nonseparating multicurve. This lemma will be used
in the proofs of Proposition 1.1, Proposition 1.2, and Theorem A to verify
that hypotheses (1) and (4) of the definition of representation of
transvective type hold.

######  Lemma 2.2.

Let $g\geq 2$ and $b\geq 0$, and let $S=S_{g}^{b}$ be an oriented, compact
surface of genus $g$ with $b$ boundary components. Let $M\subseteq S$ be a
nonseparating multicurve such that $g(S{\ssearrow}M)\geq 1$. Let $G$ be the
image of the composition

$\operatorname{Mod}(S{\ssearrow}M)\rightarrow\operatorname{Mod}(S)\rightarrow\operatorname{Aut}(H_{1}(S_{g}^{b};\mathbb{Z})).$

Then the following hold:

  1. (1)

$G$ has a finite generating set consisting of transvections, and

  2. (2)

the $G$–normal closure of any primitive transvection $T_{v}\in G$ is all of
$G$.

###### Proof.

We prove each statement in turn.

_Statement (1)_. Korkmaz [17, Theorem 3.1] has shown that if
$S_{g^{\prime}}^{b^{\prime}}$ is a compact, oriented surface with genus
$g^{\prime}\geq 1$, then $\operatorname{Mod}(S_{g^{\prime}}^{b^{\prime}})$ has
a finite generating set
$\mathcal{D}\subseteq\operatorname{Mod}(S_{g^{\prime}}^{b^{\prime}})$
consisting of Dehn twists along nonseparating curves. Then $S{\ssearrow}M$ is
a compact, oriented surface, and has genus at least one by hypothesis. Hence
there is a finite set of Dehn twists along nonseparating curves
$\mathcal{D}\subseteq\operatorname{Mod}(S{\ssearrow}M)$ such that
$\mathcal{D}$ generates $\operatorname{Mod}(S{\ssearrow}M)$. Then the image of
$T_{d}\in\mathcal{D}$ under the symplectic representation is a transvection
along $[d]$. Hence $G$ has a finite generating set consisting of primitive
transvections, as desired.

_Statement (2)_. By the argument in statement (1),
$\operatorname{Mod}(S{\ssearrow}M)$ is generated by Dehn twists along
nonseparating curves. By the change of coordinates principle [8, Section 1.3],
any Dehn twist $T_{d}\in\operatorname{Mod}(S{\ssearrow}M)$ along a
nonseparating curve $d\subseteq\operatorname{Mod}(S{\ssearrow}M)$ is
$\operatorname{Mod}(S{\ssearrow}M)$–conjugate to any other such Dehn twist
$T_{e}$. Hence $\operatorname{Mod}(S{\ssearrow}M)$ is normally generated by
any Dehn twist along a nonseparating curve $T_{d}$. Since the image of $T_{d}$
is the transvection $T_{[d]}$, we have that $T_{[d]}$ normally generates $G$
as well, which completes the proof. ∎

##  3\. Bestvina–Margalit tori

For the remainder of this section, fix a $g\geq 3$ and $a\subseteq S_{g}$ a
nonseparating curve. Let $X_{g}=\mathcal{C}_{[a]}(S_{g})/\mathcal{I}_{g}$. A
Bestvina–Margalit torus (defined below), is a certain type of 2–torus embedded
in $X_{g}$. There is a subspace
$\operatorname{BM}_{2}(X_{g};\mathbb{Q})\subseteq H_{2}(X_{g};\mathbb{Q})$
generated by the fundamental classes of the Bestvina–Margalit tori. The goal
in this section is to prove basic structural results about $X_{g}$ and
Bestvina–Margalit tori.

_Bestvina–Margalit tori, by example_. In unpublished work [3], Bestvina and
Margalit associate to each 2-cell $\sigma$ in $X_{g}$ a corresponding dual
cell. The cells $\sigma$ and $\sigma^{\prime}$ form a torus which we refer to
as a Bestvina–Margalit torus. The idea is that $\sigma$ and $\sigma^{\prime}$
have the same edges, except the edges are in a different cyclic order. Let
$x,y,z,w\in\mathcal{C}_{[a]}(S_{g})$ be as in Figure 2. Let $\sigma\subseteq
X_{g}$ be the 2-cell such that there is a lift $\widehat{\sigma}\subseteq
C_{[a]}(S_{g})$ with vertices $x,y,$ and $w$, and let
$\sigma^{\prime}\subseteq X_{g}$ be the 2–cell with a lift
$\widehat{\sigma}^{\prime}$ with vertices $x,z,$ and $w$. Then the edges
$e_{x,y}$ and $e_{y,w}$ are in the same orbit under the action of
$\mathcal{I}_{g}$ to $e_{z,w}$ and $e_{x,z}$, respectively [24, Proposition
3.29]. However, the cells $\widehat{\sigma}$ and $\widehat{\sigma}^{\prime}$
are not in the same orbit under the action of $\mathcal{I}_{g}$, since they
induce decompositions of $H_{1}(S_{g};\mathbb{Z})$ with different cyclic
orders. Hence the cells $\sigma$ and $\sigma^{\prime}$ form a torus in
$X_{g}$.

$x$$y$$w$$z$ Figure 2. The construction of dual cells.

We will denote this torus $\operatorname{BM}_{\sigma}$ and its fundamental
class by $[\operatorname{BM}_{\sigma}]$. In general, a Bestvina–Margalit torus
is any subcomplex of $X_{g}$ given by the union of a pair of 2–cells $\sigma$
and $\sigma^{\prime}$ such that each edge of $\sigma$ is in the same
$\mathcal{I}_{g}$–orbit as an edge of $\sigma^{\prime}$, but
$\sigma\neq\sigma^{\prime}$. Let $\operatorname{BM}_{2}(X_{g};\mathbb{Q})$
denote the subspace of $H_{2}(X_{g};\mathbb{Q})$ generated by the set of
fundamental classes $\\{[\operatorname{BM}_{\sigma}]\\}_{\sigma\in
X_{g}^{(2)}}$. This subspace plays a crucial role in the proof of Proposition
1.1.

###  3.1. Structural results about $\mathcal{C}_{[a]}(S_{g})/\mathcal{I}_{g}$

In this section we will prove Lemma 3.1, which will tell us when two $k$–cells
$\sigma,\sigma^{\prime}\subseteq\mathcal{C}_{[a]}(S_{g})$ are in the same
$\mathcal{I}_{g}$–orbit for the action of $\mathcal{I}_{g}$ on
$\mathcal{C}_{[a]}(S_{g}).$ It is known that two nonseparating curves $c$ and
$d$ in $S_{g}$ are in the same orbit under the action of the Torelli group if
and only if $c$ and $d$ are homologous (see, e.g., Putman [23, Lemma 6.2]).
Hence the complex $X_{g}$ has one vertex. Since $X_{g}$ has one vertex, there
is a group generated by the edges of $X_{g}$ with relations given by 2–cells.
We will use this fact to denote the composition of two edges $y$ and $z$ that
share a 2–cell by $yz$, where this edge is the third edge of the 2–cell. This
specifies an orientation on the edges, since there is one orientation of the
edges $y$ and $z$ such that $y*z=yz$, where $*$ is the composition in the
group generated by the edges on $X_{g}$. If
$e\subseteq\mathcal{C}_{[a]}(S_{g})$ is an edge, there is a bounding pair
$T_{c,c^{\prime}}$ taking one endpoint of $e$ to the other endpoint. See
Figure 3 for an example.

$v$$w$$c$$c^{\prime}$ Figure 3. The bounding pair map
$T_{c}T_{c^{\prime}}^{-1}$ and edge $e=e_{v,w}$.

Lemma 3.1 describes the cells of the complex $X_{g}$. Before stating and
proving the lemma, we introduce some notation. Let $\sigma\subseteq X_{g}$ be
a $k$–cell with a lift $\widehat{\sigma}\subset\mathcal{C}{[a]}(S_{g})$, where
$\widehat{\sigma}$ corresponds to a multicurve $a_{0}\sqcup\ldots\sqcup
a_{k}$. Let $S_{0},\ldots,S_{k}$ be the connected components of
$S_{g}{\ssearrow}\widehat{\sigma}$. Assume that $S_{i}\cap S_{j}\neq\emptyset$
if and only if $i-j\equiv\pm 1\operatorname{mod}k+1$. For each $S_{i}$, let
$\mathcal{H}_{i}^{\sigma}$ denote the image in $H_{1}(S_{g};\mathbb{Z})$ of
the pushforward map $H_{1}(S_{i};\mathbb{Z})\rightarrow
H_{1}(S_{g};\mathbb{Z})$. Let $\mathcal{H}(\sigma)$ denote the set of free
abelian groups $\mathcal{H}_{0}^{\sigma},\ldots,\mathcal{H}_{k}^{\sigma}$.
Associated to $\mathcal{H}(\sigma)$ is a directed cycle $C$. The vertices of
$C$ are the connected components $S_{i}$ and there is a directed edge
$S_{i}\rightarrow S_{j}$ if the following hold:

  * •

$a_{\ell}\subseteq S_{i}\cap S_{j}$ for some $0\leq\ell\leq k$, and

  * •

$a_{\ell}$ is oriented so that $S_{i}$ is on the left of $a_{\ell}$.

This directed cycle induces a cyclic ordering on the set
$\mathcal{H}(\sigma)$. For convenience, we will index the $\mathcal{H}_{i}$ so
that the cyclic order is
$\mathcal{H}_{0}<\mathcal{H}_{1}<\ldots<\mathcal{H}_{k}<\mathcal{H}_{0}$. We
will refer to this cyclically ordered set $\mathcal{H}(\sigma)$ as the
$\textit{cyclic decomposition of
}H_{1}(S_{g};\mathbb{Z}){inducedby}\widehat{\sigma}$. We will say that the
$\textit{genus of }\widehat{\sigma}$ is the cyclically ordered tuple
$\left(g(\mathcal{H}_{0}),\ldots,g(\mathcal{H}_{k})\right)$, and we will
denote this tuple by $g(\widehat{\sigma})$. We will prove the following
result.

######  Lemma 3.1.

Let $\sigma$ and $\sigma^{\prime}$ be two $k$–cells of
$\mathcal{C}_{[a]}(S_{g})$. Then there is an $f\in\mathcal{I}_{g}$ such that
$f\sigma=\sigma^{\prime}$ if and only if $\sigma$ and $\sigma^{\prime}$ induce
the same decomposition of $H_{1}(S_{g};\mathbb{Z})$.

###### Proof.

We first prove the forwards implication and then we prove the backwards
implication.

_The forward implication_. Let $\sigma,\sigma^{\prime}\subseteq
C_{[a]}(S_{g})$ be 2–cells in the same $\mathcal{I}_{g}$ orbit. Let
$f\in\mathcal{I}_{g}$ be an element with $f\sigma=\sigma^{\prime}$. Since
$f\in\operatorname{Mod}(S_{g})$, $f$ takes any
$\mathcal{H}\in\mathcal{H}(\sigma)$ to some
$\mathcal{H}^{\prime}\in\mathcal{H}(\sigma^{\prime})$, and since
$f\in\mathcal{I}_{g}$, we must have $\mathcal{H}=\mathcal{H}^{\prime}$.
Therefore $\mathcal{H}(\sigma)=\mathcal{H}(\sigma^{\prime})$ as unordered
sets, so it remains to show that $f$ preserves the cyclic order on
$\mathcal{H}(\sigma)$ and $\mathcal{H}(\sigma^{\prime})$. Let $b\subseteq
S_{g}$ be an oriented curve such that for any vertex $d\in\sigma$, $b$
intersects $d$ once with signed intersection $-1$. Then since
$f(\sigma)=\sigma^{\prime}$, the image $f(b)$ satisfies the property that for
any $d^{\prime}\in\sigma^{\prime}$, $f(b)$ intersects $cd$ once with signed
intersection number $-1$. Now, for each
$\mathcal{H}_{i}\in\mathcal{H}(\sigma)$, let $c_{i}$ be a nonseparating simple
closed curve disjoint from $\sigma$ and not homologous to $[c]$ such that
$c_{i}$ intersects $b$ once with signed intersection $-1$. An example with
$\sigma$ a 2–cell of the curves $\sigma,b$, and $c_{i}$ can be found in Figure
4, if we denote the vertices of $\sigma$ by $d_{0},d_{1},d_{2}$.

$d_{0}$$c_{0}$$d_{1}$$c_{1}$$d_{2}$$c_{1}$$b$ Figure 4. The curves
$a_{0},a_{1},a_{2}$, $b$, and $c_{i}$

Let $c^{\prime}_{i}=f(c_{i})$. Then $[c^{\prime}_{i}]=[c_{i}]$ since
$f\in\mathcal{I}_{g}$, and
$[c^{\prime}_{i}]\in\mathcal{H}_{i}(\sigma^{\prime})$. Then the cyclic order
of the elements of $\mathcal{H}(\sigma)$ is the same as the cyclic order of
the intersections of $c_{i}$ with $b$, where the cyclic order on the $c_{i}$
is induced by the orientation of $b$, so the lemma follows.

_The backwards implication_. Suppose now that $\sigma$ and $\sigma^{\prime}$
induce the same decomposition of $H_{1}(S_{g};\mathbb{Z})$. Reindex
$\mathcal{H}(\sigma^{\prime})$ so that
$\mathcal{H}_{i}^{\sigma}=\mathcal{H}_{i}^{\sigma^{\prime}}$ for all $0\leq
i\leq\dim(\sigma)$, and let $S_{i}^{\sigma}$, $S_{i}^{\sigma^{\prime}}$ denote
the connected components of $S_{g}{\ssearrow}\sigma$ and
$S_{g}{\ssearrow}\sigma^{\prime}$. Let
$\alpha_{1},\beta_{1},\ldots,\alpha_{g},\beta_{g}$ be a symplectic basis for
$H_{1}(S_{g};\mathbb{Z})$ such that:

  * •

$[a]=\alpha_{1}$ and

  * •

there is a partition $\nu\vdash\\{2,\ldots,g\\}$ where $\nu$ has $k+1$ blocks
such that, for each $P_{i}\in\nu$, the set $\\{\alpha_{j},\beta_{j}\\}_{j\in
P_{i}}$ is a symplectic basis for a maximal unimodular subgroup of
$\mathcal{H}_{i}^{\sigma}\in\mathcal{H}(\sigma)$.

For each $\mathcal{H}_{i}^{\sigma}$ and $\mathcal{H}_{i}^{\sigma^{\prime}}$,
let $\\{a_{j},b_{j}\\}_{j\in P_{i}}$ and
$\\{a_{j}^{\prime},b_{j}^{\prime}\\}_{P_{i}}$ be representatives in
$S_{i}^{\sigma}$ and $S_{i}^{\sigma^{\prime}}$ respectively for
$\\{\alpha_{j},\beta_{j}\\}_{j\in P_{i}}$ such that $|a_{i}\cap
b_{j}|=|a^{\prime}_{i}\cap
b^{\prime}_{j}|=\left|\langle\alpha_{i},\beta_{j}\rangle\right|$ for any
$1\leq i,j\leq g$. By the change of coordinates principle, there is an
$f\in\operatorname{Mod}(S_{g})$ taking $\sigma$ to $\sigma^{\prime}$ that also
takes $a_{j}\rightarrow a^{\prime}_{j},b_{j}\rightarrow b_{j}^{\prime}$ for
all $2\leq j\leq g$. Then $f_{*}(\alpha_{i})=\alpha_{i}$ and
$f_{*}(\beta_{i})=\beta_{i}$ for $2\leq i\leq g$. Then $f$ takes a vertex of
$\sigma$ to a vertex of $\sigma^{\prime}$, so $f_{*}(\alpha_{1})=\alpha_{1}$.
This implies that $\langle f_{*}\beta_{1},\alpha_{i}\rangle=\langle
f_{*}\beta_{1},f_{*}(\beta_{i})\rangle=0$ for any $2\leq i\leq g$, and
$\langle\alpha_{1},f_{*}(\beta_{1})\rangle=1$. Hence
$f(\beta_{1})=\beta_{1}+n\alpha_{1}$ for some $n\in\mathbb{Z}$, so $f$ is not
a priori in the Torelli group, since it may act nontrivially on $\beta_{1}$.
We now find another element $h\in\operatorname{Mod}(S_{g})$ such that
$hf\sigma=\sigma^{\prime}$ and $hf\in\mathcal{I}_{g}$.

Let $a^{\prime}\in\sigma^{\prime}$ be a curve in the $k$–cell
$\sigma^{\prime}$. Then
$T_{a^{\prime}}^{-n}f\sigma=T_{a^{\prime}}^{-n}\sigma^{\prime}=\sigma^{\prime}$
since $f\sigma=\sigma^{\prime}$ and $T_{a^{\prime}}$ acts trivially on
$\sigma^{\prime}$ by construction. Furthermore, for any $m\in\mathbb{Z}$, we
have

$T_{[a^{\prime}]}(\beta_{1}+m\alpha_{1})=\beta_{1}+m\alpha_{1}+\langle[a^{\prime}],\beta_{1}+m\alpha_{1}\rangle[a^{\prime}].$

Since $[a^{\prime}]=\alpha_{1}$, we therefore have
$T_{[a^{\prime}]}(\beta_{1}+m\alpha_{1})=\beta_{1}+m\alpha_{1}+1\alpha_{1}=\beta_{1}+(m+1)\alpha_{1}$.
Therefore
$T_{[a^{\prime}]}^{-n}(\beta_{1}+m\alpha_{1})=\beta_{1}+(m-n)\alpha_{1}$.
Therefore we have we have
$T_{[a^{\prime}]}^{-n}f_{*}(\beta_{1})=T_{[a^{\prime}]}^{-n}(\beta_{1}+n\alpha_{1})=\beta_{1}+n\alpha_{1}-n\alpha_{1}=\beta_{1}$.
Then $T_{a^{\prime}}$ acts trivially on the set
$\\{\alpha_{1},\alpha_{2},\beta_{2},\ldots,\alpha_{g},\beta_{g}\\}$ since
$[a^{\prime}]=\alpha_{1}$ trivially intersects every element in this set.
Hence $(T_{a}^{-n}f)_{*}$ fixes the set
$\\{\alpha_{1},\beta_{1},\ldots,\alpha_{g},\beta_{g}\\}$, so
$T_{a}^{-n}f\in\mathcal{I}_{g}$. Hence $T_{a}^{-n}f$ is an element of the
Torelli group taking $\sigma$ to $\sigma^{\prime}$, as desired. ∎

Given Lemma 3.1, if $\sigma\subseteq X_{g}$ is a cell, the cyclically ordered
decomposition $\mathcal{H}(\widehat{\sigma})$ for $\widehat{\sigma}$ a lift of
$\sigma$ to $\mathcal{C}_{[a]}(S_{g})$ depends only on $\sigma$, so we will
use the notation $\mathcal{H}(\sigma)$ to refer to the decomposition of
$H_{1}(S_{g};\mathbb{Z})$ induced by $\widehat{\sigma}$, and refer to
$\mathcal{H}(\sigma)$ as the decomposition of $H_{1}(S_{g};\mathbb{Z})$
induced by $\sigma$.

###  3.2. Bestvina–Margalit tori

We will now give a formal definition of a Bestvina–Margalit torus and prove a
collection of results about fundamental classes of Bestvina–Margalit tori.

_Bestvina–Margalit tori_. Let $\sigma\subseteq X_{g}$ be a 2–cell. There is a
unique cell $\sigma^{\prime}$ called the dual cell to $\sigma$ such that
$\mathcal{H}(\sigma)=\mathcal{H}(\sigma^{\prime})$ as unordered sets, but with
$\mathcal{H}(\sigma)\neq\mathcal{H}(\sigma^{\prime})$ as cyclically ordered
sets. This $\sigma^{\prime}$ is unique by Lemma 3.1 since there are exactly
two cyclic orders on the set of three elements. The Bestvina–Margalit torus
$\operatorname{BM}_{\sigma}$ is the union $\sigma\cup\sigma^{\prime}$, and the
fundamental class of $\operatorname{BM}_{\sigma}$ is denoted
$[\operatorname{BM}_{\sigma}]$. We have the following lemma that allows us to
add fundamental classes of Bestvina–Margalit tori.

######  Lemma 3.2.

Let $x,y,z$ be three edges in $X_{g}$ such that $\operatorname{BM}_{x,y}$,
$\operatorname{BM}_{x,z}$ and $\operatorname{BM}_{y,z}$ are all
Bestvina–Margalit tori. Let $yz$ denote the third edge of a 2–cell $\sigma$
with $y,z\subseteq\sigma$ oriented so that in the group on the edges of
$X_{g}$, we have $y*z=yz$. A lift of a 3–cell $\tau\subseteq X_{g}$ containing
these edges can be seen in Figure 5. Then in $H_{2}(X_{g};\mathbb{Q})$, the
relation
$[\operatorname{BM}_{x,y}]+[\operatorname{BM}_{x,z}]=[\operatorname{BM}_{x,yz}]$
is satisfied.

$a$$a_{1}$$a_{2}$$a_{3}$ Figure 5. A 3–cell containing lifts of the edges
$x,y,z,yz$ as in Lemma 3.2. $x$ lifts to $(a,a_{1})$, $y$ lifts to
$(a_{1},a_{2})$, $z$ lifts to $(a_{2},a_{3})$ and $yz$ lifts to
$(a_{1},a_{3})$.

Before proving the lemma, we will explain how to compute differentials in
$X_{g}$. If $\tau\subseteq\mathcal{C}_{[a]}(S_{g})$ is a $k$–cell with
vertices $\\{c_{0},\ldots,c_{k}\\}$, then the differential $\partial\tau$ is
given by

$\partial\tau=\sum_{i=0}^{k}(-1)^{i}(c_{0},\ldots,\widehat{c_{i}},\ldots,c_{k})$

where $(c_{0},\ldots,\widehat{c_{i}},\ldots,c_{k})$ denotes the $(k-1)$–cell
with vertices given by $c_{0},\ldots,c_{i-1},c_{i+1},\ldots,c_{k}$. This
differential descends to $X_{g}$ by definition. Now, if $\tau\subseteq X_{g}$
is a $k$–cell, Lemma 3.1 tells us that $\tau$ is determined by the
decomposition

$\mathcal{H}(\tau)=\\{\mathcal{H}_{0},\ldots,\mathcal{H}_{k}\\}.$

Then, if $\overline{\tau}$ is a lift of $\tau$ to $\mathcal{C}_{[a]}(S_{g})$,
forgetting a vertex $c\subseteq\overline{\tau}$, which yields a $k$–cell
$\overline{\sigma}$, corresponds to replacing
$\mathcal{H}_{i},\mathcal{H}_{i+1}$ in $\mathcal{H}(\tau)$ with
$\mathcal{H}_{i}+\mathcal{H}_{i+1}$. Therefore the differential $\partial\tau$
is given by

$\partial\tau=\sum_{i=0}^{k}(-1)^{i}\sigma_{i}$

where for $0\leq i\leq k-1$, each $\sigma_{i}$ satisfies

$\mathcal{H}(\sigma_{i})=\\{\mathcal{H}_{0},\ldots,\mathcal{H}_{i-1},\mathcal{H}_{i}+\mathcal{H}_{i+1},\mathcal{H}_{i+2},\ldots,\mathcal{H}_{k}\\}.$

We have that
$\mathcal{H}(\sigma_{k})=\\{\mathcal{H}_{1},\ldots,\mathcal{H}_{k-1},\mathcal{H}_{k}+\mathcal{H}_{0}\\}$.

###### Proof of Lemma 3.2.

We will find 3–dimensional subcomplex $\overline{L}\subseteq X_{g}$ such that
the boundary realizes the relation in the lemma. Let $\tau$ be a 3–cell with
$\mathcal{H}(\tau)=\\{\mathcal{H}_{0},\mathcal{H}_{1},\mathcal{H}_{2},\mathcal{H}_{3}\\}$
such that $\mathcal{H}_{0}\in\mathcal{H}(x)$,
$\mathcal{H}_{1}\in\mathcal{H}(x)$, and $\mathcal{H}_{2}\in\mathcal{H}(x)$.
The space $\overline{L}$ is given as the union of the following 3–cells:

  * •

$\tau_{0}=\tau$,

  * •

$\tau_{1}$ that satisfies
$\mathcal{H}(\tau_{1})=\\{\mathcal{H}_{1},\mathcal{H}_{0},\mathcal{H}_{2},\mathcal{H}_{3}\\}$,
and

  * •

$\tau_{2}$ that satisfies
$\mathcal{H}(\tau_{2})=\\{\mathcal{H}_{1},\mathcal{H}_{2},\mathcal{H}_{0},\mathcal{H}_{3}\\}$.

We can compute the differentials of each 3–cell above as follows, signed
appropriately to yield the desired relation in the lemma. For notational
convenience, we will conflate $\sigma$ and $\mathcal{H}(\sigma)$ for
$\sigma\subseteq X_{g}$.

  * •

$\partial\tau_{0}=\\{\mathcal{H}_{0}+\mathcal{H}_{1},\mathcal{H}_{2},\mathcal{H}_{3}\\}-\\{\mathcal{H}_{0},\mathcal{H}_{1}+\mathcal{H}_{2},\mathcal{H}_{3}\\}+\\{\mathcal{H}_{0},\mathcal{H}_{1},\mathcal{H}_{2}+\mathcal{H}_{3}\\}-\\{\mathcal{H}_{0}+\mathcal{H}_{3},\mathcal{H}_{1},\mathcal{H}_{2}\\}$,

  * •

$-\partial\tau_{1}=-\\{\mathcal{H}_{0}+\mathcal{H}_{1},\mathcal{H}_{2},\mathcal{H}_{3}\\}+\\{\mathcal{H}_{1},\mathcal{H}_{0}+\mathcal{H}_{2},\mathcal{H}_{3}\\}-\\{\mathcal{H}_{1},\mathcal{H}_{0},\mathcal{H}_{2}+\mathcal{H}_{3}\\}+\\{\mathcal{H}_{1}+\mathcal{H}_{3},\mathcal{H}_{0},\mathcal{H}_{2}\\}$,
and

  * •

$\partial\tau_{2}=\\{\mathcal{H}_{1}+\mathcal{H}_{2},\mathcal{H}_{0},\mathcal{H}_{3}\\}-\\{\mathcal{H}_{1},\mathcal{H}_{2}+\mathcal{H}_{0},\mathcal{H}_{3}\\}+\\{\mathcal{H}_{1},\mathcal{H}_{2},\mathcal{H}_{0}+\mathcal{H}_{3}\\}-\\{\mathcal{H}_{1}+\mathcal{H}_{3},\mathcal{H}_{2},\mathcal{H}_{0}\\}.$

Now, we see that in the sum
$\partial\tau_{0}-\partial\tau_{1}+\partial\tau_{2}$, the term
$\\{\mathcal{H}_{0}+\mathcal{H}_{1},\mathcal{H}_{2},\mathcal{H}_{3}\\}$ in
$\partial\tau_{0}$ cancels with
$-\\{\mathcal{H}_{0}+\mathcal{H}_{1},\mathcal{H}_{2},\mathcal{H}_{3}\\}$ in
$-\partial\tau_{1}$, the term
$-\\{\mathcal{H}_{0}+\mathcal{H}_{3},\mathcal{H}_{1},\mathcal{H}_{2}\\}$ in
$\partial\tau_{0}$ cancels with
$\\{\mathcal{H}_{1},\mathcal{H}_{2},\mathcal{H}_{0}+\mathcal{H}_{3}\\}$ in
$\partial\tau_{2}$, and
$\\{\mathcal{H}_{1},\mathcal{H}_{0}+\mathcal{H}_{2},\mathcal{H}_{3}\\}$ in
$-\partial\tau_{1}$ cancels with
$-\\{\mathcal{H}_{1},\mathcal{H}_{2}+\mathcal{H}_{0},\mathcal{H}_{3}\\}$ in
$\partial\tau_{2}$. This means that

$\displaystyle\partial\tau_{0}-\partial\tau_{1}+\partial\tau_{2}=$
$\displaystyle-\\{\mathcal{H}_{0},\mathcal{H}_{1}+\mathcal{H}_{2},\mathcal{H}_{3}\\}+\\{H_{0},\mathcal{H}_{1},\mathcal{H}_{2}+\mathcal{H}_{3}\\}$
$\displaystyle-\\{\mathcal{H}_{1},\mathcal{H}_{0},\mathcal{H}_{2}+\mathcal{H}_{3}\\}+\\{\mathcal{H}_{1}+\mathcal{H}_{3},\mathcal{H}_{0},\mathcal{H}_{2}\\}$
$\displaystyle+\\{\mathcal{H}_{1}+\mathcal{H}_{2},\mathcal{H}_{0},\mathcal{H}_{3}\\}-\\{\mathcal{H}_{1}+\mathcal{H}_{3},\mathcal{H}_{2},\mathcal{H}_{0}\\}.$

Now, by rearranging terms, we have

$\displaystyle\partial\tau_{0}-\partial\tau_{1}+\partial\tau_{2}=$
$\displaystyle-\\{\mathcal{H}_{0},\mathcal{H}_{1}+\mathcal{H}_{2},\mathcal{H}_{3}\\}+\\{\mathcal{H}_{1}+\mathcal{H}_{2},\mathcal{H}_{0},\mathcal{H}_{3}\\}$
$\displaystyle+\\{H_{0},\mathcal{H}_{1},\mathcal{H}_{2}+\mathcal{H}_{3}\\}-\\{\mathcal{H}_{1},\mathcal{H}_{0},\mathcal{H}_{2}+\mathcal{H}_{3}\\}$
$\displaystyle+\\{\mathcal{H}_{1}+\mathcal{H}_{3},\mathcal{H}_{0},\mathcal{H}_{2}\\}-\\{\mathcal{H}_{1}+\mathcal{H}_{3},\mathcal{H}_{2},\mathcal{H}_{0}\\}.$

Then each pair of two cells term on the right side of the above equation is a
Bestvina–Margalit torus, so we have

$\partial\tau_{0}-\partial\tau_{1}+\partial\tau_{2}=-[\operatorname{BM}_{x,yz}]+[\operatorname{BM}_{x,y}]+[\operatorname{BM}_{x,z}]$

and therefore the relationship given in the lemma holds. ∎

We now prove Lemma 3.3, which says when two Bestvina–Margalit tori are equal
in $X_{g}$.

######  Lemma 3.3.

Let $\operatorname{BM}_{\sigma}$ and $\operatorname{BM}_{\tau}$ be two
Bestvina–Margalit tori. Then
$\operatorname{BM}_{\sigma}=\operatorname{BM}_{\tau}$ if and only if
$\mathcal{H}(\sigma)=\mathcal{H}(\tau)$ as unordered sets.

###### Proof.

We prove both directions.

_Backwards direction_. We prove the contrapositive. If
$\operatorname{BM}_{\sigma}\neq\operatorname{BM}_{\tau}$, then
$\sigma\neq\tau$ since every cell $\sigma$ has a unique dual 2–cell
$\sigma^{\prime}$ such that $\sigma\cup\sigma^{\prime}$ is a Bestvina–Margalit
torus. This implies that $\mathcal{H}(\sigma)\neq\mathcal{H}(\tau)$ as
decompositions of $[a]^{\perp}$, which implies in particular that
$\mathcal{H}(\sigma)\neq\mathcal{H}(\tau)$ as unordered sets.

_Forwards direction_. Suppose that
$\operatorname{BM}_{\sigma}=\operatorname{BM}_{\tau}$. By Lemma 3.1, $\sigma$
and $\tau$ are identified under the action of $\mathcal{I}_{g}$ if and only if
$\mathcal{H}(\sigma)=\mathcal{H}(\tau)$ as decompositions of $[a]^{\perp}$.
But then if $\mathcal{H}(\sigma)\neq\mathcal{H}(\tau)$, we must have
$\mathcal{H}(\sigma)=\mathcal{H}(\tau^{\prime})$, where $\tau^{\prime}$ is the
2–cell dual to $\tau$, since there are only two cyclic orders on a set of
three elements. ∎

In particular, if $\operatorname{BM}_{\sigma}$ is a Bestvina–Margalit torus,
we will use the notation $\mathcal{H}(\operatorname{BM}_{\sigma})$ to denote
the decomposition $\mathcal{H}(\sigma)$ without the cyclic order, and we will
refer to this decomposition as the decomposition of  $H_{1}(S_{g};\mathbb{Z})$
induced by $\operatorname{BM}_{\sigma}$. If $\sigma$ is a 2–cell with edges
$x$ and $y$, we will denote the splitting
$\mathcal{H}(\operatorname{BM}_{\sigma})$ by
$\mathcal{H}(\operatorname{BM}_{x,y})$. We describe how $\mathcal{H}$
interacts with the addition of tori in the following lemma. The idea is that
if we have a 3–cell $\tau$ as in Figure 5, then we can describe the splitting
induced by one torus corresponding to a 2–cell $\sigma\subseteq\tau$ in terms
of the splitting induced by tori corresponding to other 2–cells contained in
$\tau$.

######  Lemma 3.4.

Let $g\geq 51$ and let $a\subseteq S_{g}$ be a nonseparating simple closed
curve. Let $x,y,z\subseteq X_{g}$ be three edges such that
$\operatorname{BM}_{x,y}$, $\operatorname{BM}_{x,z}$ and
$\operatorname{BM}_{y,z}$ are all Bestvina–Margalit tori. Suppose that we have

$\mathcal{H}(\operatorname{BM}_{x,y})=\\{\mathcal{H}_{0},\mathcal{H}_{1},\mathcal{H}_{2}\\}\text{
and
}\mathcal{H}(\operatorname{BM}_{x,z})=\\{\mathcal{H}_{0},\mathcal{H}^{\prime}_{1},\mathcal{H}^{\prime}_{2}\\}$

with $\mathcal{H}_{1}\cap\mathcal{H}^{\prime}_{1}=\mathbb{Z}[a]$. Suppose
additionally that we have $\mathcal{H}_{0}\in\mathcal{H}(x)$,
$\mathcal{H}_{1}\in\mathcal{H}(y)$ and
$\mathcal{H}^{\prime}_{1}\in\mathcal{H}(z)$. Let $yz$ denote the third edge of
a 2–cell $\sigma$ with $y,z\subseteq\sigma$ with $yz$ oriented so that
$y*z=yz$ in the group generated by edges of $X_{g}$. Then we have

$\mathcal{H}(\operatorname{BM}_{x,yz})=\\{\mathcal{H}_{0},\mathcal{H}_{1}+\mathcal{H}_{1}^{\prime},\mathcal{H}_{2}\cap\mathcal{H}^{\prime}_{2}\\}.$

###### Proof.

We first want to compute the decomposition $\mathcal{H}(yz)$. The 2–cell
$\sigma$ containing $y,z$ and $yz$ has
$\mathcal{H}_{1},\mathcal{H}^{\prime}_{1}\in\mathcal{H}(\sigma)$. This implies
that the third element of $\mathcal{H}(\sigma)$ is
$\mathcal{H}_{1}^{\perp}\cap(\mathcal{H}^{\prime}_{1})^{\perp}$. Hence

$\mathcal{H}(yz)=\\{\mathcal{H}_{1}^{\perp}\cap(\mathcal{H}^{\prime}_{1})^{\perp},(\mathcal{H}_{1}^{\perp}\cap(\mathcal{H}^{\prime}_{1})^{\perp})^{\perp}\\}=\\{\mathcal{H}_{1}^{\perp}\cap(\mathcal{H}^{\prime}_{1})^{\perp},\mathcal{H}_{1}+\mathcal{H}_{1}^{\prime}\\}$

and in particular, we have
$\mathcal{H}_{1}+\mathcal{H}^{\prime}_{1}\in\mathcal{H}(yz)$. Now, the
splitting $\mathcal{H}(\operatorname{BM}_{x,yz})$ contains $\mathcal{H}_{0}$
and $\mathcal{H}_{1}+\mathcal{H}_{1}^{\prime}$, so the third element of
$\mathcal{H}(\operatorname{BM}_{x,yz})$ is given by

$\mathcal{H}_{0}^{\perp}\cap(\mathcal{H}_{1}+\mathcal{H}_{1}^{\prime})^{\perp}=(\mathcal{H}_{0}+\mathcal{H}_{1})^{\perp}\cap(\mathcal{H}_{0}+\mathcal{H}_{1}^{\prime})^{\perp}=\mathcal{H}_{2}\cap\mathcal{H}_{2}^{\prime},$

so the lemma holds. ∎

Before moving forward with the paper, we will briefly discuss a notation for
cells in $X_{g}$, as well as the action of subgroups of
$\operatorname{Sp}(2g,\mathbb{Z})$ on $X_{g}$.

_Graphical representations for Bestvina–Margalit tori_. Let $x,y,z,yz\subseteq
X_{g}$ be a set of edges as in Lemma 3.4. Let
$\widehat{\mathcal{H}}=\\{\mathcal{H}_{0},\mathcal{H}_{1},\mathcal{H}_{1}^{\prime},\mathcal{H}_{2}\cap\mathcal{H}_{2}^{\prime}\\}$
be an unordered set. Then $\widehat{\mathcal{H}}$ corresponds to the labeled
graph as in Figure 6.

$\mathcal{H}_{0}$$\mathcal{H}_{1}$$\mathcal{H}_{1}^{\prime}$$\mathcal{H}_{2}\cap\mathcal{H}_{2}^{\prime}$
Figure 6. The graphical representation of $\widehat{\mathcal{H}}$

The advantage of this notation is that the decompositions corresponding to
$\mathcal{H}(x,y)$, $\mathcal{H}(x,z)$ and $\mathcal{H}(x,yz)$ can be read off
from the graph. Each of these Bestvina–Margalit tori corresponds to the
decomposition given by taking the span of the union of two subgroups not equal
to $\mathcal{H}_{0}$. We will use this notation in Section 5 to keep track of
certain decompositions of $[a]^{\perp}\subseteq H_{1}(S_{g};\mathbb{Z})$.

_The action of $\operatorname{Sp}([a]^{\perp},\mathbb{Z})$ on $X_{g}$_.
Throughout the rest of the paper, we will consider the action of
$\operatorname{Sp}([a]^{\perp},\mathbb{Z})$ on $X_{g}$. The subgroup
$\Gamma_{a}\subseteq\operatorname{Mod}(S_{g})$ consisting of all elements that
act trivially on the class $[a]\in H_{1}(S_{g};\mathbb{Z})$ acts on
$\mathcal{C}_{[a]}(S_{g})$. We have $\mathcal{I}_{g}\subseteq\Gamma_{a}$ by
definition, and so the quotient
$\Gamma_{a}/\mathcal{I}_{g}=\operatorname{Sp}([a]^{\perp},\mathbb{Z})$ acts on
$X_{g}$.

##  4\. Abelian cycles in $H_{2}(\mathcal{I}_{g};\mathbb{Z})$

The goal of this section is to prove Proposition 4.1, which says that certain
subspaces of $H_{2}(\mathcal{I}_{g};\mathbb{Q})$ generated by abelian cycles
of bounding pair maps can be generated by abelian cycles supported on finitely
many subsurfaces.

_Abelian cycles_. Let $G$ be a group and ${c_{1},\ldots,c_{k}}\subseteq G$ be
a collection of pairwise commuting elements of infinite order. Let $A=\langle
c_{1},\ldots,c_{k}\rangle\subseteq G$. The abelian cycle
$\operatorname{ab}(c_{1},\ldots,c_{k})\in H_{k}(G;\mathbb{Q})$ is the image of
the class $[c_{1}]\wedge\ldots\wedge[c_{k}]\in H_{k}(A;\mathbb{Q})$ under the
pushforward map $H_{k}(A;\mathbb{Q})\rightarrow H_{k}(G;\mathbb{Q})$. If
$k=2$, we will denote the abelian cycle $\operatorname{ab}(c_{1},c_{2})$ by
$[c_{1},c_{2}]$. If $G$ is a group, we let
$H_{k}^{\operatorname{ab}}(G;\mathbb{Q})$ denote the subspace of
$H_{k}(G;\mathbb{Q})$ generated by abelian cycles.

_Torelli group of a subsurface_. Let $S$ be a compact surface and
$\iota:S\hookrightarrow S_{g}$ be an embedding. We will denote the genus of
$S$ by $g(S)$. We say that $\iota$ is clean if no connected component of
$S_{g}\setminus\iota(S)$ is a disk. Let $\mathcal{I}_{g}(S,S_{g})$ denote the
intersection $\iota_{*}(\operatorname{Mod}(S))\cap\mathcal{I}_{g}$, where the
pushforward map $\iota_{*}$ is given by extending
$\varphi\in\operatorname{Mod}(S)$ by the identity over
$S_{g}\setminus\iota(S)$. Let $\mathcal{I}(\iota)$ denote the inclusion map
$\mathcal{I}(S,S_{g})\rightarrow\mathcal{I}_{g}$.

_Bounding pair maps_. If $c\subseteq S_{g}$ is a curve, let $T_{c}$ denote the
Dehn twist along $c$. We say that $f\in\mathcal{I}_{g}$ is a bounding pair map
if $f=T_{c}T_{c^{\prime}}^{-1}$ for $c$ and $c^{\prime}$ two disjoint
homologous curves. We will denote the bounding pair map
$T_{c}T_{c^{\prime}}^{-1}$ by $T_{c,c^{\prime}}$. Let
$H_{2}^{\operatorname{ab},\operatorname{bp}}(\mathcal{I}(S,S_{g});\mathbb{Q})$
denote the subspace of $H_{2}(\mathcal{I}(S,S_{g});\mathbb{Q})$ generated by
abelian cycles $[T_{c,c^{\prime}},T_{d,d^{\prime}}]$ such that the bounding
pairs $c\cup c^{\prime}$ and $d\cup d^{\prime}$ are disjoint.

The main goal of the section is to prove the following result.

######  Proposition 4.1.

Let $g\geq 51$ and let $M\subseteq S_{g}$ be a nonseparating multicurve with
$\left|\pi_{0}(M)\right|=9$. Let $\mathcal{V}=\\{[c]:c\in\pi_{0}(M)\\}$. The
natural map

$\bigoplus_{v\in\mathcal{V}}H_{2}^{\operatorname{ab},\operatorname{bp}}(\mathcal{I}_{g};\mathbb{Q})^{T_{v}}\rightarrow
H_{2}^{\operatorname{ab},\operatorname{bp}}(\mathcal{I}_{g};\mathbb{Q})$

is surjective.

The idea of Proposition 4.1 is that if there are enough curves on a surface
$S_{g}$, then any abelian cycle of bounding pairs is a linear combination of
abelian cycles that are each disjoint from at least one of the curves.

_Outline of the proof of Proposition 4.1_. If $G$ is a group acting on a
vector space $V$ and $\mathcal{F}\subseteq G$ is a subset, we let
$V^{\mathcal{F}}$ denote the subspace of $V$ fixed by all elements of
$\mathcal{F}$. If $\mathcal{W}\subseteq H_{1}(S_{g};\mathbb{Z})$, let
$T_{\mathcal{W}}$ denote the set of transvections along elements of
$\mathcal{W}$. If $F\in\mathcal{I}_{g}$, let $[F]$ denote the corresponding
class in $H_{1}(\mathcal{I}_{g};\mathbb{Q})$. Let
$[T_{d,d^{\prime}},T_{e,e^{\prime}}]\in
H_{2}^{\operatorname{ab},\operatorname{bp}}(\mathcal{I}_{g};\mathbb{Q})$ be an
abelian cycle. The proof of the proposition proceeds in two steps.

  1. (1)

There is a relation

$[T_{d,d^{\prime}},T_{e,e^{\prime}}]=\sum_{i=1}^{k}\lambda_{i}[T_{d,d^{\prime}},T_{f_{i},f^{\prime}_{i}}]$

with $[T_{f_{i},f^{\prime}_{i}}]\in
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{\mathcal{V}^{\prime}}}$ for some
$\mathcal{V}^{\prime}\subseteq\mathcal{V}$ and
$\left|\mathcal{V}^{\prime}\right|=4$ and for all $1\leq i\leq k$.

  2. (2)

If $[T_{d,d^{\prime}},T_{f_{i},f^{\prime}_{i}}]$ is an abelian cycle with
$T_{f_{i},f^{\prime}_{i}}\in
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{\mathcal{V}^{\prime}}}$ for some
$\mathcal{V}^{\prime}\subseteq V$ and $\left|\mathcal{V}^{\prime}\right|=4$,
then there is a relation

$[T_{d,d^{\prime}},T_{f_{i},f^{\prime}_{i}}]=\sum_{j=1}^{m}\theta(\mathcal{V})_{j}[T_{h_{j},h^{\prime}_{j}},T_{f_{i},f^{\prime}_{i}}]$

with each $[T_{h_{j},h^{\prime}_{j}}]\in
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{v}}$ for some
$v\in\mathcal{V}^{\prime}.$

Step (1) is the content of Lemma 4.11 and Step (2) is the content of Lemma
4.9.

_Outline of the section_. We will begin by describing the Johnson
homomorphism. We will then prove Lemma 4.5, which describes the vector space
$H_{1}(\mathcal{I}(S,S_{g});\mathbb{Q})$ for $S\hookrightarrow S_{g}$ a clean
embedding with $g(S)\geq 3$. We then prove Lemma 4.8 which is a result about
generating sets of exterior powers of quasi–unimodular lattices, and then use
Lemma 4.8 to prove Lemma 4.9. We also use Lemma 4.8 to prove Lemma 4.10, which
is another result about generating sets of exterior products of vector spaces.
We use Lemma 4.10 to prove Lemma 4.11. We then combine these lemmas to prove
Proposition 4.1.

_The Johnson homomorphism_. Let $a_{1},b_{1},\ldots,a_{g},b_{g}$ be a
symplectic basis for $H_{1}(S_{g};\mathbb{Z})$ and let $\omega=a_{1}\wedge
b_{1}+\ldots+a_{g}\wedge
b_{g}\in\scaleobj{1.5}{\wedge}^{2}H_{1}(S_{g};\mathbb{Z})$. There is an
inclusion
$H_{1}(S_{g};\mathbb{Z})\hookrightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Z})$
given by $[c]\rightarrow[c]\wedge\omega$. Johnson [11] constructed a map

$\tau_{g}:\mathcal{I}_{g}\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Z})/H_{1}(S_{g};\mathbb{Z})$

and showed that the pushforward map

$(\tau_{g})_{*}:H_{1}(\mathcal{I}_{g};\mathbb{Q})\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q})$

is an isomorphism [13]. We will make use of Lemma 4.2, which describes
$\tau_{g}(T_{c,c^{\prime}})$ for $T_{c,c^{\prime}}$ a bounding pair map. The
lemma is due to Johnson [11, Lemma 4B].

######  Lemma 4.2.

Let $g\geq 3$. Let $T_{c,c^{\prime}}$ be a bounding pair map and let $S$ be a
connected component of $S_{g}{\ssearrow}\left(c\cup c^{\prime}\right)$. Let
$a_{1},b_{1},\ldots,a_{h},b_{h}$ be a symplectic basis for a maximal
nondegenerate subspace of $H_{1}(S;\mathbb{Z})$. We have

$\tau_{g}(T_{c,c^{\prime}})=[c]\wedge\left(\sum_{i=1}^{h}a_{i}\wedge
b_{i}\right).$

We use Lemma 4.2 to prove Lemma 4.5, which describes
$H_{1}(\mathcal{I}(S;S_{g});\mathbb{Z})$ for clean embeddings
$S\hookrightarrow S_{g}$. We first prove an auxiliary lemma, which is
essentially just a corollary of a result of Putman [23, Theorem 1.3].

######  Lemma 4.3.

Let $g\geq 4$ and let $\iota:S\hookrightarrow S_{g}$ be a clean embedding such
that $g(S)\geq 3$. Then $\mathcal{I}(S,S_{g})$ is generated by bounding pair
maps.

###### Proof.

A result of Putman [23, Theorem 1.3] tells us that $\mathcal{I}(S,S_{g})$ is
generated by bounding pair maps and Dehn twists along separating curves. Hence
it suffices to show that if $\delta\subseteq\iota(S)$ is a separating curve in
$S_{g}$, then $T_{\delta}$ is a product of bounding pair maps. Since $g(S)\geq
3$, there is an embedded copy of $S_{1}^{2}\subseteq S$ such that the image of
one boundary component of $S_{1}^{2}$ is the curve $\delta$. Then by embedding
a copy of $S_{0}^{4}\hookrightarrow S_{1}^{2}$, we can use the lantern
relation [8, Section 5.1] to rewrite $T_{\delta}$ as a product of three
bounding pair maps, so the lemma follows. ∎

We also need one more fact about quasi–unimodular lattices.

######  Lemma 4.4.

Let $L$ be a quasi–unimodular lattice such that $g(L)\geq 1$. The free abelian
group $\scaleobj{1.5}{\wedge}^{2}L$ is spanned by elements of the form
$\gamma\wedge\delta$ with $\gamma,\delta$ primitive and
$\langle\gamma,\delta\rangle=1$.

###### Proof.

Let $\mathcal{S}\subseteq\scaleobj{1.5}{\wedge}^{2}L$ be the set of all
elements $\gamma\wedge\delta$ as in the statement of the lemma. Let

$\mathcal{B}=\\{\gamma_{1},\delta_{1},\gamma_{2},\delta_{2},\ldots,\gamma_{g(L)},\delta_{g(L)},\gamma_{g(L)+1},\gamma_{g(L)+2},\ldots,\gamma_{\ell}\\}$

be a partial symplectic basis for $L$. The group $\scaleobj{1.5}{\wedge}^{2}L$
is generated by all elements of the form $r\wedge s$ for $r,s\in\mathcal{B}$,
so it suffices to show that any element $r\wedge s$ with $r,s\in\mathcal{B}$
is a $\mathbb{Z}$–linear combination of elements as in the statement of the
lemma.

We begin by proving this in the case $r=\gamma_{i},s=\delta_{j}$. We have
$\gamma_{i}\wedge\delta_{j}=(\gamma_{i}+\gamma_{j})\wedge\delta_{j}-(\gamma_{j}\wedge\delta_{j})$.
Each of these latter elements is a wedge of primitive elements with algebraic
intersection number one, so $r\wedge s$ is in the subgroup of
$\scaleobj{1.5}{\wedge}^{2}L$ generated by $\mathcal{S}$.

Now, suppose that $r\in\mathcal{B},s=\delta_{i}$ with $r\neq\gamma_{i}$. We
have $r\wedge s=(r+\gamma_{i})\wedge\delta_{i}-\gamma_{i}\wedge\delta_{i}$,
and each of these latter elements are in $\mathcal{S}$, since the only element
of $\mathcal{B}$ that $\delta_{i}$ intersects nontrivially is $\gamma_{i}$.

Finally, suppose that $r=\gamma_{i},s=\gamma_{j}$. If $i$ or $j$ are less than
$g(L)$, then without loss of generality we have $j\leq g(L)$. Then we have
$r\wedge s=gamma_{i}+\delta_{j}\wedge\gamma_{j}-\delta_{j}\wedge\gamma_{j}$,
and these latter elements are all in $\mathcal{S}$. Otherwise, $i,j>g(L)$. In
this case, we have

$r\wedge
s=(\gamma_{i}+\delta_{g(L)})\wedge(\gamma_{j}+\gamma_{g(L)})-\delta_{g(L)}\wedge\gamma_{j}-\delta_{g(L)}\wedge\gamma_{j}-\delta_{g(L)}-\gamma_{g(L)}.$

The first and less elements are in $\mathcal{S}$, and the middle two elements
are $\mathbb{Z}$–linear combinations of elements in $\mathcal{S}$ by the
previous cases, so $r\wedge s\in\operatorname{Span}_{\mathbb{Z}}(\mathcal{S})$
and thus the lemma holds. ∎

We now describe $H(\mathcal{I}(S,S_{g});\mathbb{Q})$ for certain clean
embeddings $\iota:S\hookrightarrow S_{g}$.

######  Lemma 4.5.

Let $S$ be a compact surface with boundary with $\left|\pi_{0}(\partial
S)\right|\leq 2$. Let $\iota:S\hookrightarrow S_{g}$ be a clean embedding such
that $S_{g}{\ssearrow}S$ is connected. Suppose additionally that
$g(S_{g}{\ssearrow}S)\geq 1$. Then there is a surjection

$f:H_{1}(\mathcal{I}_{g}(S,S_{g});\mathbb{Q})\twoheadrightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})$

such that the following diagram commutes:

${H_{1}(\mathcal{I}(S,S_{g});\mathbb{Q})}$${\wedge^{3}H_{1}(S;\mathbb{Q})}$${H_{1}(\mathcal{I}_{g};\mathbb{Q})}$${\wedge^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q}).}$$\scriptstyle{f}$$\scriptstyle{\mathcal{I}(\iota)_{*}}$$\scriptstyle{\wedge^{3}\iota_{*}}$$\scriptstyle{\tau_{g}\otimes\mathbb{Q}}$

Furthermore, suppose that $g(S)\geq 3$. Then $f$ is an isomorphism, and the
map $\wedge^{3}\iota_{*}\circ f$ is injective.

###### Proof.

We begin with the first part of the lemma. We show that the image of the
composition the composition

$H_{1}(\mathcal{I}_{g}(S,S_{g});\mathbb{Q})\xrightarrow{\mathcal{I}(\iota)_{*}}H_{1}(\mathcal{I}_{g};\mathbb{Q})\xrightarrow{\tau_{g}\otimes\mathbb{Q}}\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q})$

is $\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}\iota_{*})$, and that
$\scaleobj{1.5}{\wedge}^{3}\iota_{*}$ is an injection. If these two statements
hold, then $f$ is given by

$f=(\scaleobj{1.5}{\wedge}^{3}\iota_{*})^{-1}|_{\operatorname{im}(\tau_{g}\otimes\mathbb{Q}\circ\mathcal{I}(\iota)_{*}}\circ\tau_{g}\otimes\mathbb{Q}\circ\mathcal{I}(\iota)_{*}.$

We prove each containment in turn, and then prove injectivity.

_Showing that
$\operatorname{im}(\tau_{g}\otimes\mathbb{Q}\circ\mathcal{I}(\iota)_{*})\subseteq\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q}))$_.
Lemma 4.3 says that $\mathcal{I}_{g}(S,S_{g})$ is generated by bounding pair
maps, so it suffices to show that if
$T_{c,c^{\prime}}\in\mathcal{I}_{g}(S,S_{g})$ is a bounding pair map, then
$\tau_{g}\otimes\mathbb{Q}\circ\mathcal{I}(\iota)_{*}(T_{c,c^{\prime}})\in\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q}))$.
Since the bounding pair $c\cup c^{\prime}$ is contained in $S$, we have
$[c]\in\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})$. Then there must be a
connected component $S^{\prime}$ of $S_{g}{\ssearrow}\left(c\cup
c^{\prime}\right)$ such that $S^{\prime}\subseteq S$. Let
$\alpha_{1},\beta_{1},\ldots,\alpha_{h},\beta_{h}$ be a symplectic basis for a
maximal unimodular subgroup of $H_{1}(S^{\prime};\mathbb{Z})$. Lemma 4.2 says
that
$\tau_{g}\circ\mathcal{I}(\iota)_{*}([T_{c,c^{\prime}}])=[c]\wedge\left(\alpha_{1}\wedge\beta_{1}+\ldots+\alpha_{h}\wedge\beta_{h}\right)\in\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q}))$,
so the $\subseteq$ containment holds.

_Showing that
$\operatorname{im}(\tau_{g}\circ\mathcal{I}(\iota)_{*})\supseteq\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q}))$_.
Observe that if $v_{1},v_{2},v_{3}\in H_{1}(S;\mathbb{Z})$ are primitive
classes with $\langle v_{1},v_{2}\rangle=1$ and $\langle
v_{1},v_{3}\rangle=\langle v_{2},v_{3}\rangle=0$, then Lemma 4.2 says that if
$c,c^{\prime}$ is a bounding pair with $[c]=v_{3}$ and $v_{1},v_{2}$ a
symplectic basis for a maximal nondegenerate subgroup of the first homology of
a connected component of $S{\ssearrow}\left(c\cup c^{\prime}\right)$, then
$\tau_{g}(T_{c,c^{\prime}})=v_{1}\wedge v_{2}\wedge
v_{3}\in\operatorname{im}(\tau_{g}\circ\mathcal{I}(\iota)_{*})$. Since the set
of such triples of elements spans
$\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Z})$ by Lemma 4.4, we have
$\operatorname{im}(\tau_{g}\circ\mathcal{I}(\iota)_{*})\supseteq\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q}))$.

_Showing that $\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})$ injects into
$\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q})$_.
We first show that the exterior power of pushforwards
$\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})$
is injective. Since $S_{g}{\ssearrow}S$ is connected, the pushforward
$H_{1}(S;\mathbb{Q})\rightarrow H_{1}(S_{g};\mathbb{Q})$ is injective, so
$\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})$
is injective as well. Hence it suffices to show that
$\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q}))\cap
H_{1}(S_{g};\mathbb{Q})\wedge\omega=0$, where
$\omega\in\scaleobj{1.5}{\wedge}^{2}H_{1}(S_{g};\mathbb{Q})$ is the
characteristic element. Suppose otherwise, so there is some nonzero
$\alpha\wedge\omega\in\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q}))$.
Since $\omega$ does not depend on the choice of symplectic basis, we can
choose a symplectic basis $\alpha_{1},\beta_{1},\ldots,\alpha_{g},\beta_{g}$
with
$\alpha\wedge\omega=m\cdot\alpha_{1}\wedge(\alpha_{1}\wedge\beta_{1}+\alpha_{2}\wedge\beta_{2}+\ldots+\alpha_{g}\wedge\beta_{g})=m\cdot\alpha_{1}\wedge(\alpha_{2}\wedge\beta_{2}+\ldots+\alpha_{g}\wedge\beta_{g})$
for $m$ some positive integer. But this implies that there are $(2g-1)$
linearly independent elements
$\alpha_{1},\alpha_{2},\beta_{2},\ldots,\alpha_{g},\beta_{g}\in\operatorname{im}(H_{1}(S;\mathbb{Q})\rightarrow
H_{1}(S_{g};\mathbb{Q}))$. We have assumed that $g(S_{g}{\ssearrow}S)\geq 1$,
so $\dim(H_{1}(S;\mathbb{Q}))\leq 2g-2$, which is a contradiction.

We now prove the second part of the lemma, where we assume that $g(S)\geq 3$.
In this case, since $g(S)\geq 3$ and $\iota:S\hookrightarrow S_{g}$ is clean,
a theorem of Putman [25, Theorem B] states that the pushforward map
$\mathcal{I}(\iota)_{*}$ is an injection. Therefore the composition
$(\scaleobj{1.5}{\wedge}^{3}\iota_{*})^{-1}|_{\operatorname{im}(\tau_{g}\otimes\mathbb{Q}\circ\mathcal{I}(\iota)_{*}}\circ\tau_{g}\otimes\mathbb{Q}\circ\mathcal{I}(\iota)_{*}$
is an injection. By definition this is $f$, so $f$ is injective as well, and
is therefore an isomorphism. ∎

We will also prove Lemma 4.7, which is a result that describes how stabilizers
of transvections interact with the abelianization of certain
$\mathcal{I}(S,S_{g})$. We first prove the following result.

######  Lemma 4.6.

Let $g\geq 3$ and let $w\in H_{1}(S_{g};\mathbb{Z})$ be a nonzero primitive
element. The fixed space $H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{w}}$ is sent
to
$\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}[w]^{\perp}\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q}))$
under the Johnson homomorphism.

###### Proof.

Since the Johnson homomorphism is a map of
$\operatorname{Sp}(2g,\mathbb{Z})$–representations, it suffices to show that
$\left(\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q})\right)^{T_{w}}=\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}[w]^{\perp}\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q}))$.
Since $T_{w}$ acts trivially on $[w]^{\perp}$ by the definition of
transvection, the transvection $T_{w}$ acts trivially on
$\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}[w]^{\perp}\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q}))$,
and thus
$\tau_{g}(H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{w}})\supseteq\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}[w]^{\perp}\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q}))$.
It therefore remains to prove the other containment.

Let $\mathcal{B}=\\{\alpha_{1},\beta_{1},\ldots,\alpha_{g},\beta_{g}\\}\in
H_{1}(S_{g};\mathbb{Z})$ be a symplectic basis so that $\alpha_{1}=w$. The
vector space $\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})$ has a basis
consisting of all pure wedge products of elements in $\mathcal{B}$. Then after
modding out by
$H_{1}(S_{g};\mathbb{Q})\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})$,
we see that any pure tensor of the form
$\alpha_{1}\wedge\beta_{1}\wedge\gamma$ is a linear combination

$\alpha_{2}\wedge\beta_{2}\wedge\gamma+\ldots+\alpha_{g}\wedge\beta_{g}\wedge\gamma.$

Hence
$\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q})$
has a basis consisting of all pure tensors of elements in $\mathcal{B}$ except
those containing $\alpha_{1}\wedge\beta_{1}$. This set is spanning in
$\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q})$,
and is therefore a basis by counting dimensions. Now, $T_{w}$ acts trivially
on any such wedge products except for those containing $\beta_{1}$. For such a
basis element $\beta_{1}\wedge\gamma\wedge\delta$ for
$\gamma,\delta\in\mathcal{B}\setminus\\{\alpha_{1}\\}$, we have
$T_{w}(\beta_{1}\wedge\gamma\wedge\delta)=\beta_{1}\wedge\gamma\wedge\delta+\alpha_{1}\wedge\gamma\wedge\delta$.
Now, if we have a $\mathbb{Q}$–linear combination of such elements

$\sum_{i=1}^{n}\lambda_{i}\beta_{1}\wedge\gamma_{i}\wedge\delta_{i},$

then we have

$T_{w}\left(\sum_{i=1}^{n}\lambda_{i}\beta_{1}\wedge\gamma_{i}\wedge\delta_{i}\right)=\sum_{i=1}^{n}\lambda_{i}(\beta_{1}+\alpha_{1})\wedge\gamma_{i}\wedge\delta_{i}=(\beta_{1}+\alpha_{1})\wedge\sum_{i=1}^{n}\lambda_{i}\gamma_{i}\wedge\delta_{i}.$

Then to have
$\sum_{i=1}^{n}\lambda_{i}\beta_{1}\wedge\gamma_{i}\wedge\delta_{i}$ fixed by
$T_{w}$, we must have

$\sum_{i=1}^{n}\lambda_{i}\beta_{1}\wedge\gamma_{i}\wedge\delta_{i}=(\beta_{1}+\alpha_{1})\wedge\sum_{i=1}^{n}\lambda_{i}\gamma_{i}\wedge\delta_{i},$

or equivalently we must have

$\alpha_{1}\wedge\sum_{i=1}^{n}\lambda_{i}\gamma_{i}\wedge\delta_{i}=0$

Since the span of the set $\\{\gamma_{i},\delta_{i}:1\leq i\leq n\\}$ does not
contain $\alpha_{1}$, we conclude that
$\sum_{i=1}^{n}\lambda_{i}\gamma_{i}\wedge\delta_{i}=0$. In particular, this
implies that only elements of the form
$\scaleobj{1.5}{\wedge}^{3}[w]^{\perp}\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q})$
are fixed by $T_{w}$, as desired. ∎

We apply Lemma 4.6 in the following context.

######  Lemma 4.7.

Let $g\geq 4$, and let $\iota:S\hookrightarrow S_{g}$ be a clean embedding
satisfying the hypotheses of Lemma 4.5. Let $M\subseteq S_{g}$ be a
nonseparating multicurve, and let $\mathcal{W}\subseteq
H_{1}(S_{g};\mathbb{Z})$ be given by $\mathcal{W}=\\{[c]:c\in\pi_{0}(M)\\}.$
Then, after passing to $\mathbb{Q}$–coefficients, the Johnson homomorphism
$\tau_{g}$ restricts to an isomorphism

$\operatorname{im}\left(H_{1}(\mathcal{I}(S,S_{g});\mathbb{Q})\rightarrow
H_{1}(\mathcal{I}_{g};\mathbb{Q})\right)\cap
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{\mathcal{W}}}\rightarrow\scaleobj{1.5}{\wedge}^{3}\operatorname{im}(H_{1}(S;\mathbb{Q})\rightarrow
H_{1}(S_{g};\mathbb{Q}))\cap\scaleobj{1.5}{\wedge}^{3}(\mathcal{W}^{\perp})$

###### Proof.

By Lemma 4.5, we have
$\operatorname{im}\left(H_{1}(\mathcal{I}(S,S_{g});\mathbb{Q})\rightarrow
H_{1}(\mathcal{I}_{g};\mathbb{Q})\right)\cong\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}\iota_{*})\cong\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})$.
Now, the Johnson homomorphism is
$\operatorname{Sp}(2g,\mathbb{Z})$–equivariant, so if $w\in\mathcal{W}$, the
fixed set $H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{w}}$ is isomorphic to
$\operatorname{im}(\tau_{g}\otimes\mathbb{Q})^{T_{w}}$. Then by Lemma 4.6, we
have

$\tau_{g}(H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{w}})=\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}[w]^{\perp}\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q})).$

Then we have
$H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{\mathcal{W}}}=\bigcap_{w\in\mathcal{W}}H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{w}}$
and
$\scaleobj{1.5}{\wedge}^{3}\mathcal{W}^{\perp}=\bigcap_{w\in\mathcal{W}}\scaleobj{1.5}{\wedge}^{3}[w]^{\perp}$.
Therefore we have
$\tau_{g}(H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{\mathcal{W}}})=\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}\mathcal{W}^{\perp}\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S_{g};\mathbb{Q})/H_{1}(S_{g};\mathbb{Q}))$.
Hence the intersection

$\operatorname{im}\left(H_{1}(\mathcal{I}(S,S_{g});\mathbb{Q})\rightarrow
H_{1}(\mathcal{I}_{g};\mathbb{Q})\right)\cap
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{\mathcal{W}}}$

is sent to
$\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}H_{1}(S;\mathbb{Q})\rightarrow\operatorname{im}(\tau_{g}))\cap\operatorname{im}(\scaleobj{1.5}{\wedge}^{3}\mathcal{W}^{\perp}\rightarrow\operatorname{im}(\tau_{g}))$
by the Johnson homomorphism, and this last intersection is equal to

$\scaleobj{1.5}{\wedge}^{3}(\operatorname{im}(H_{1}(S;\mathbb{Q})\rightarrow
H_{1}(S_{g};\mathbb{Q})))\cap\scaleobj{1.5}{\wedge}^{3}\mathcal{W}^{\perp}$

as desired. ∎

The remaining steps to do now before proving Proposition 4.1 are to prove
Lemma 4.9 and Lemma 4.11, which will allow us to rewrite classes
$[T_{d,d^{\prime}},T_{e,e^{\prime}}]$ as linear combinations of abelian cycles
of bounding pairs maps contained in
$H_{2}(\mathcal{I}_{g};\mathbb{Q})^{T_{v}}$ for various primitive $v\in
H_{1}(S_{g};\mathbb{Z})$.

###  4.1. The proof of Lemma 4.9

We now recall a basic fact from linear algebra.

_Inclusion–exclusion for vector spaces_. Let $V$ be a finite dimensional
vector space, and let $V_{1},\ldots,V_{n}\subseteq V$ be a set of subspaces
such that the natural map

$\bigoplus_{1\leq i\leq n}V_{i}\rightarrow V$

is a surjection. Then the following chain complex

$\ldots\rightarrow\bigoplus_{\mathcal{S}\subseteq\\{1,\ldots,n\\}:\left|\mathcal{S}\right|=k}\bigcap_{j\in\mathcal{S}}V_{j}\rightarrow\bigoplus_{\mathcal{S}\subseteq\\{1,\ldots,n\\}:\left|\mathcal{S}\right|=k-1}\bigcap_{j\in\mathcal{S}}V_{j}\rightarrow\ldots\rightarrow\bigoplus_{i=1}^{n}V_{i}\rightarrow
V\rightarrow 0$

is exact, where the differentials are given by direct sums of inclusion maps
with appropriate signs. Assume that for any
$\mathcal{S},\mathcal{S}^{\prime}\subseteq\\{1,\ldots,n\\}$ with
$\left|\mathcal{S}\right|=\left|\mathcal{S}^{\prime}\right|$, we have
$\dim\left(\bigcap_{j\in\mathcal{S}}V_{j}\right)=\dim\left(\bigcap_{j\in\mathcal{S}^{\prime}}V_{j}\right)$.
Let $d_{k}=\dim\left(\bigcap_{j\in\mathcal{S}}V_{j}\right)$ for any
$\mathcal{S}\subseteq\\{1,\ldots,n\\}$ with $\left|\mathcal{S}\right|=k$.
Therefore, by taking the alternating sum of the dimensions of each term of
this chain complex and applying the fact that the complex is exact, the
integers $d_{k}$ satsify the relation

$\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}d_{k}=0,$

and hence the relation

$d_{0}=\sum_{k=1}^{n}(-1)^{k+1}\binom{n}{k}d_{k}.$

We will refer to this as the inclusion–exclusion principle for vector spaces.
We will use this to prove the following.

######  Lemma 4.8.

Let $(L,\langle\cdot,\cdot\rangle)$ be quasi–unimodular lattice. Let
$\mathcal{W}=\\{w_{1},\ldots,w_{4}\\}\subseteq L$ be a set of primitive
elements such that the image of $\mathcal{W}$ under the adjoint map
$L\rightarrow\operatorname{Hom}_{\mathbb{Z}}(L,\mathbb{Z})$ is a linearly
independent set. Then the natural map

$\varphi:\bigoplus_{w\in\mathcal{W}}\scaleobj{1.5}{\wedge}^{3}\left(w^{\perp}\otimes\mathbb{Q}\right)\rightarrow\scaleobj{1.5}{\wedge}^{3}\left(L\otimes\mathbb{Q}\right)$

is surjective.

###### Proof.

Let $n=\dim(L\otimes\mathbb{Q})$. We will show that
$\operatorname{im}(\varphi)=\scaleobj{1.5}{\wedge}^{3}(L\otimes\mathbb{Q})$.
If $\mathcal{W}^{\prime},\mathcal{W}^{\prime\prime}\subseteq\mathcal{W}$ are
two subsets with
$\left|\mathcal{W}^{\prime}\right|=\left|\mathcal{W}^{\prime\prime}\right|$,
then we have
$\dim\left(\mathcal{W}^{\prime}\right)^{\perp}=\dim\left(\mathcal{W}^{\prime\prime}\right)^{\perp}$,
and therefore
$\dim(\scaleobj{1.5}{\wedge}^{3}\left(\mathcal{W}^{\prime}\right)^{\perp})=\dim(\scaleobj{1.5}{\wedge}^{3}\left(\mathcal{W}^{\prime\prime}\right)^{\perp})$.
Furthermore, since
$\dim(\mathcal{W}^{\prime})^{\perp}=n-\left|\mathcal{W}^{\prime}\right|$, we
have
$\dim(\scaleobj{1.5}{\wedge}^{3}\left(\mathcal{W}^{\prime}\right)^{\perp})=\binom{n-\left|\mathcal{W}^{\prime}\right|}{3}$.
Therefore the inclusion–exclusion principle for vector spaces applied to
$V=\operatorname{im}(\varphi)$ and
$V_{i}=\scaleobj{1.5}{\wedge}^{3}w_{i}^{\perp}$ for $1\leq i\leq 4$ says that

$\dim(\operatorname{im}(\varphi))=\sum_{k=1}^{4}(-1)^{k+1}\binom{4}{k}\binom{n-k}{3}=\binom{n}{3},$

where the last equality is a computation. But then
$\dim\scaleobj{1.5}{\wedge}^{3}L\otimes\mathbb{Q}=\binom{n}{3}$, so
$\operatorname{im}(\varphi)=\scaleobj{1.5}{\wedge}^{3}L\otimes\mathbb{Q}$ and
thus $\varphi$ is surjective, as desired. ∎

For our purpose, the important takeaway from Lemma 4.8 is the following
result.

######  Lemma 4.9.

Let $M^{\prime}\subseteq S_{g}$ be a nonseparating multicurve with
$\left|\pi_{0}(M^{\prime})\right|\geq 4$, and let
$\mathcal{V}^{\prime}=\\{[c]:c\in\pi_{0}(M^{\prime})\\}$. Let
$d,d^{\prime}\subseteq S_{g}$ be a bounding pair such that
$[T_{d,d^{\prime}}]\in
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{\mathcal{V}^{\prime}}}$. Let
$[T_{d,d^{\prime}},T_{f,f^{\prime}}]\in
H_{2}^{\operatorname{ab},\operatorname{bp}}(\mathcal{I}_{g};\mathbb{Q})$ be an
abelian cycle. There is a $\mathbb{Q}$–linear relation

$[T_{d,d^{\prime}},T_{f,f^{\prime}}]=\sum_{\ell=1}^{m}\lambda_{\ell}[T_{d,d^{\prime}},T_{f_{\ell},f_{\ell}^{\prime}}]$

in $H_{2}^{\operatorname{ab},\operatorname{bp}}(\mathcal{I}_{g};\mathbb{Q})$
such that, for each $1\leq\ell\leq m$, there is a $v\in\mathcal{V}^{\prime}$
such that $[T_{f_{\ell},f_{\ell}^{\prime}}]\in
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{v}}$.

###### Proof.

Let $S^{\prime}$ and $S^{\prime\prime}$ be the two connected components of
$S_{g}{\ssearrow}\left(d\cup d^{\prime}\right)$. Without loss of generality,
we may assume that $f,f^{\prime}\subseteq S^{\prime}$. Indeed, if
$f,f^{\prime}$ are in distinct connected components of
$S_{g}{\ssearrow}\left(d\cup d^{\prime}\right)$, then $f$ and $f^{\prime}$ are
homologous to $d$, so we have a relation

$[T_{d,d^{\prime}},T_{f,f^{\prime}}]=[T_{d,d^{\prime}},T_{f,d}]+[T_{d,d^{\prime}},T_{d,f^{\prime}}]$

We now have two cases.

_Case 1:
$\mathcal{V}^{\prime}\not\subseteq\operatorname{im}(H_{1}(S^{\prime};\mathbb{Z})\rightarrow
H_{1}(S_{g};\mathbb{Z}))$_. In this case, there is a
$v\in\mathcal{V}^{\prime}$ such that
$v\in\operatorname{im}(H_{1}(S^{\prime\prime};\mathbb{Z})\rightarrow
H_{1}(S_{g};\mathbb{Z}))$. Then $T_{f,f^{\prime}}\in
H_{1}(\mathcal{I}_{g};\mathbb{Z})^{T_{v}}$, so we are done.

_Case 2:
$\mathcal{V}^{\prime}\subseteq\operatorname{im}(H_{1}(S^{\prime};\mathbb{Q}))$_.
We see that since $\mathcal{V}^{\prime}$ is linearly independent and the
elements of $\mathcal{V}^{\prime}$ have pairwise trivial algebraic
intersection, we have $g(S^{\prime})\geq 4$. Lemma 4.5 says that
$H_{1}(\mathcal{I}(S^{\prime},S_{g});\mathbb{Q})\cong\scaleobj{1.5}{\wedge}^{3}H_{1}(S^{\prime};\mathbb{Q})$.
Lemma 4.8 implies that there is a surjection

$\bigoplus_{v\in\mathcal{V}^{\prime}}\scaleobj{1.5}{\wedge}^{3}v^{\perp}\cap\scaleobj{1.5}{\wedge}^{3}H_{1}(S^{\prime};\mathbb{Q})\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S^{\prime};\mathbb{Q}).$

Then Lemma 4.7 applied to each term on the summand on the left implies there
is a surjection

$\bigoplus_{v\in\mathcal{V}^{\prime}}H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{v}}\cap
H_{1}(\mathcal{I}(S^{\prime},S_{g});\mathbb{Q})\rightarrow
H_{1}(\mathcal{I}(S^{\prime},S_{g});\mathbb{Q}).$

Hence there is a relation in $H_{1}(\mathcal{I}(S^{\prime},S_{g});\mathbb{Q})$
given by

$[T_{f,f^{\prime}}]=\sum_{r=1}^{n}\lambda_{r}[F_{r}]$

where each $F_{r}\in H_{1}(\mathcal{I}(S^{\prime},S_{g});\mathbb{Q})^{T_{v}}$
for some $v\in\mathcal{V}^{\prime}$. Then each $F_{r}$ is supported on
$S^{\prime}$, so each $F_{r}$ commutes with $T_{d,d^{\prime}}$. Therefore,
there is a relation in $H_{2}(\mathcal{I}_{g};\mathbb{Q})$ given by

$[T_{d,d^{\prime}},T_{f,f^{\prime}}]=\sum_{r=1}^{n}\lambda_{r}[T_{d,d^{\prime}},F_{r}]$

where each $[F_{r}]\in
H_{1}(\mathcal{I}(S^{\prime},S_{g});\mathbb{Q})^{T_{v}}$ for some
$v\in\mathcal{V}^{\prime}$.

We now show that each $[F_{r}]$ can be written as a sum of classes represented
by bounding pair maps. Choose an $r$ with $1\leq r\leq n$ and let
$v\in\mathcal{V}^{\prime}$ such that $[F_{r}]\in
H_{1}(S^{\prime};\mathbb{Q})^{T_{v}}$. Lemma 4.7 says that
$\scaleobj{1.5}{\wedge}^{3}H_{1}(\mathcal{I}(S^{\prime},S_{g});\mathbb{Q})^{T_{v}}\cong\scaleobj{1.5}{\wedge}^{3}v^{\perp}\cap\scaleobj{1.5}{\wedge}^{3}H_{1}(S^{\prime};\mathbb{Q})$,
and this isomorphism is induced by restricting the Johnson homomorphism. Then
Lemma 4.2 and Lemma 4.4 combine to tell us that
$H_{1}(\mathcal{I}(S^{\prime}{\ssearrow}c,S_{g});\mathbb{Q})^{T_{v}}$ is
generated by classes represented by bounding pair maps, since each element in
the basis for $\scaleobj{1.5}{\wedge}^{3}v^{\perp}\cap
H_{1}(S^{\prime};\mathbb{Q})$ in Lemma 4.4 is the image under $\tau_{g}$ of a
bounding pair map. Therefore we may rewrite $[F_{r}]$ as a linear combination
of classes
$[T_{f_{1,r},f_{1,r}^{\prime}}],\ldots,[T_{f_{\ell_{r},r},f_{m_{r},r}^{\prime}}]\in
H_{1}(\mathcal{I}(S^{\prime},S_{g});\mathbb{Q})^{T_{v}}$. Since each
$T_{f_{\ell,r},f^{\prime}_{\ell,r}}$ is supported on $S^{\prime}$, each
$T_{f_{\ell,r},f^{\prime}_{\ell,r}}$ commutes with $T_{d,d^{\prime}}$.
Therefore we have a relation

$[T_{d,d^{\prime}},T_{f,f^{\prime}}]=\sum_{r=1}^{n}\lambda_{r}\sum_{\ell=1}^{m_{r}}[T_{d,d^{\prime}},T_{f_{\ell,r},f^{\prime}_{\ell,r}}]$

which by forgetting the $r$ index yields a relation

$[T_{d,d^{\prime}},T_{f,f^{\prime}}]=\sum_{\ell=1}^{m}\lambda_{\ell}[T_{d,d^{\prime}},T_{f_{\ell},f_{\ell}^{\prime}}].$

Now, since each $T_{f_{\ell},f_{\ell}^{\prime}}$ is stabilized by some
$T_{v}\in T_{\mathcal{V}^{\prime}}$, we have each
$[T_{f_{\ell},f^{\prime}_{\ell}}]\in
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{v}}$ for some
$v\in\mathcal{V}^{\prime}$, as desired. ∎

###  4.2. The proof of Lemma 4.11

We begin by extending Lemma 4.8 into the following.

######  Lemma 4.10.

Let $L$ be a quasi–unimodular lattice. Let
$\mathcal{W}=\\{w_{1},\ldots,w_{7}\\}$ be a set of primitive elements in $L$
such that the image of $\mathcal{W}$ under the adjoint map
$L\rightarrow\operatorname{Hom}_{\mathbb{Z}}(L,\mathbb{Z})$ is
$\mathbb{Q}$–linearly independent. Then the natural map

$\varphi:\bigoplus_{\mathcal{W}^{\prime}\subseteq\mathcal{W}:\left|\mathcal{W}^{\prime}\right|=4}\scaleobj{1.5}{\wedge}^{3}\left(\left(\mathcal{W}^{\prime}\right)^{\perp}\otimes\mathbb{Q}\right)\rightarrow\scaleobj{1.5}{\wedge}^{3}\left(L\otimes\mathbb{Q}\right)$

is surjective.

###### Proof.

We will prove inductively that the natural maps

$\varphi_{k}:\bigoplus_{\mathcal{W}^{\prime}\subseteq\mathcal{W}:\left|\mathcal{W}^{\prime}\right|=k}\scaleobj{1.5}{\wedge}^{3}\left(\left(\mathcal{W}^{\prime}\right)^{\perp}\otimes\mathbb{Q}\right)\rightarrow\scaleobj{1.5}{\wedge}^{3}\left(L\otimes\mathbb{Q}\right)$

are surjective for $k\leq 4$.

_Base case: $k=1$_. This follows by applying Lemma 4.8 with any subset
$\mathcal{W}^{\prime}\subseteq\mathcal{W}$ with
$\left|\mathcal{W}^{\prime}\right|=4$.

_Inductive step: $2\leq k\leq 4$_. Assume that the map $\varphi_{k^{\prime}}$
is surjective for all $1\leq k^{\prime}<k$. We will show that $\varphi_{k}$ is
surjective as well. Since $\varphi_{k^{\prime}}$ is surjective, it suffices to
show that, for any $\mathcal{W}^{\prime\prime}\subseteq\mathcal{W}$ with
$\left|\mathcal{W}^{\prime\prime}\right|=k-1$, we have

$\operatorname{im}(\varphi_{k})\supseteq\operatorname{im}\left(\scaleobj{1.5}{\wedge}^{3}\left(\mathcal{W}^{\prime\prime}\right)^{\perp}\otimes\mathbb{Q}\rightarrow\scaleobj{1.5}{\wedge}^{3}L\otimes\mathbb{Q}\right).$

Since $k\leq 4$ and $\left|\mathcal{W}^{\prime\prime}\right|<k$, we have
$\left|\mathcal{W}\setminus\mathcal{W}^{\prime\prime}\right|\geq 7-k>7-4=3$,
so $\left|\mathcal{W}\setminus\mathcal{W}^{\prime\prime}\right|\geq 4$. Hence
the natural map

$\varphi_{\mathcal{W}^{\prime\prime}}:\bigoplus_{w\in\mathcal{W}\setminus\mathcal{W}^{\prime\prime}}\scaleobj{1.5}{\wedge}^{3}\left(\mathcal{W}^{\prime\prime}\cup\\{w\\}\right)^{\perp}\otimes\mathbb{Q}\rightarrow\scaleobj{1.5}{\wedge}^{3}\left(\mathcal{W}^{\prime\prime}\right)^{\perp}\otimes\mathbb{Q}$

is surjective by Lemma 4.8. But then
$\operatorname{im}(\varphi_{k^{\prime}}\circ\varphi_{\mathcal{W}^{\prime\prime}})\subseteq\operatorname{im}(\varphi_{k})$
for any $\mathcal{W}^{\prime\prime}\subseteq\mathcal{W}$, so
$\operatorname{im}(\varphi_{k^{\prime}})\subseteq\operatorname{im}(\varphi_{k})$.
Since
$\operatorname{im}(\varphi_{k^{\prime}})=\scaleobj{1.5}{\wedge}^{3}L\otimes\mathbb{Q}$
by the inductive hypothesis, the proof is complete. ∎

We use Lemma 4.10 to prove the following result about
$H_{2}^{\operatorname{ab},\operatorname{bp}}(\mathcal{I}_{g};\mathbb{Q})$.

######  Lemma 4.11.

Let $g\geq 51$ and let $[T_{d,d^{\prime}},T_{e,e^{\prime}}]\in
H_{2}^{\operatorname{ab},\operatorname{bp}}(\mathcal{I}_{g};\mathbb{Q})$ be an
abelian cycle. Assume that one connected component of
$S_{g}{\ssearrow}\left(d\cup d^{\prime}\right)$ has genus one. Let $M\subseteq
S_{g}$ be a nonseparating multicurve with $\left|\pi_{0}(M)\right|=9$ and let
$\mathcal{V}=\\{[c]:c\in\pi_{0}(M)\\}$. Then there is a linear relation

$[T_{d,d^{\prime}},T_{e,e^{\prime}}]=\sum_{i=1}^{k}\lambda_{i}[T_{d,d^{\prime}},T_{f_{i},f^{\prime}_{i}}]$

with each $T_{f_{i},f^{\prime}_{i}}\in
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{\mathcal{V}^{\prime}}}$ for some
$\mathcal{V}^{\prime}\subseteq\mathcal{V}$ and
$\left|\mathcal{V}^{\prime}\right|=4.$

###### Proof.

The proof follows the same approach as Lemma 4.9. Let $S\sqcup
S=S_{g}{\ssearrow}\left(d\cup d^{\prime}\right)$ with $g(S)=1$. Since $e\cup
e^{\prime}$ is disjoint from $d\cup d^{\prime}$, we have
$e,e^{\prime}\subseteq S_{1}$. Since $g(S)=1$, we must have
$e,e^{\prime}\subseteq S^{\prime}$. Now, choose a primitive element $[b]\in
H_{1}(S_{g};\mathbb{Z})$ such that $[b]\in[e]^{\perp}$ and
$\langle[b],[d]\rangle=1$. Since $g(S)=1$ and $\langle[b],[d]\rangle=1$, we
have $g(H_{1}(S^{\prime};\mathbb{Z})\cap[b]^{\perp})=g-2$. Then since the
elements of $\mathcal{V}$ have pairwise trivial algebraic intersection, the
projection of $\mathcal{V}$ to $H_{1}(S^{\prime};\mathbb{Z})\cap[b]^{\perp}$
must contain a set of at least 7 elements whose image under the adjoint map
$H_{1}(S^{\prime};\mathbb{Z})\cap[b]^{\perp}\rightarrow\operatorname{Hom}_{\mathbb{Z}}(H_{1}(S^{\prime};\mathbb{Z}),\mathbb{Z})$
is $\mathbb{Q}$–linearly independent. Let $\overline{\mathcal{V}}$ be the
image of the projection of $\mathcal{V}$ to
$H_{1}(S^{\prime};\mathbb{Z})\cap[b]^{\perp}$. Let
$\mathcal{B}\subseteq\overline{\mathcal{V}}$ be a set of at least 7 linearly
independent elements. By applying Lemma 4.10 to $H_{1}(S^{\prime};\mathbb{Q})$
and $\mathcal{B}$, there is a surjection

$\bigoplus_{\mathcal{B}^{\prime}\subseteq\mathcal{B}:\left|\mathcal{B}^{\prime}\right|=4}\scaleobj{1.5}{\wedge}^{3}\left(\mathcal{B}^{\prime}\right)^{\perp}\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S^{\prime};\mathbb{Q}).$

Let $\mathcal{V}_{\mathcal{B}}$ be the preimage of $\mathcal{B}$ under the
projection map $\mathcal{V}\rightarrow\overline{\mathcal{V}}$. Since for any
$v\in\mathcal{V}$, we have $v^{\perp}\cap
H_{1}(S^{\prime};\mathbb{Q})=\left(\operatorname{proj}_{H_{1}(S^{\prime};\mathbb{Q})\cap[b]^{\perp}}v\right)^{\perp}\cap
H_{1}(S^{\prime};\mathbb{Q})$, we therefore have a surjection

$\bigoplus_{\mathcal{V}^{\prime}\subseteq\mathcal{V}_{\mathcal{B}}:\left|\mathcal{V}^{\prime}\right|=4}\scaleobj{1.5}{\wedge}^{3}\left(\mathcal{V}^{\prime}\right)^{\perp}\cap
H_{1}(S^{\prime};\mathbb{Q})\rightarrow\scaleobj{1.5}{\wedge}^{3}H_{1}(S^{\prime};\mathbb{Q}).$

By construction, the complement $S_{g}{\ssearrow}S^{\prime}$ is connected and
has $g(S_{g}{\ssearrow}S^{\prime})\geq 1$. Hence by Lemma 4.5, there is an
isomorphism
$H_{1}(\mathcal{I}(S^{\prime},S_{g});\mathbb{Q})\cong\scaleobj{1.5}{\wedge}^{3}H_{1}(S^{\prime};\mathbb{Q})$.
Furthermore, by applying Lemma 4.7, we have
$\scaleobj{1.5}{\wedge}^{3}\left(\mathcal{V}^{\prime}\right)^{\perp}\cap
H_{1}(S^{\prime};\mathbb{Q})\cong\scaleobj{1.5}{\wedge}^{3}H_{1}(S^{\prime};\mathbb{Q})^{T_{\mathcal{V}^{\prime}}}$.
Then, since $T_{e,e^{\prime}}\in\mathcal{I}(S^{\prime},S_{g})$, we have a
relation in $H_{1}(\mathcal{I}(S^{\prime},S_{g});\mathbb{Q})$ given by

$[T_{e,e^{\prime}}]=\sum_{i=1}^{k}\lambda_{i}[F_{i}]$

where each $[F_{i}]\in
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{\mathcal{V}^{\prime}}}$ for some
$\mathcal{V}^{\prime}\subseteq\mathcal{V}_{\mathcal{B}}$ with
$\left|\mathcal{V}^{\prime}\right|=4$. Since any
$f\in\mathcal{I}(S^{\prime},S_{g})$ commutes with $T_{d,d^{\prime}}$, we have
a relation in $H_{2}(\mathcal{I}_{g};\mathbb{Q})$ given by

$[T_{d,d^{\prime}},T_{e,e^{\prime}}]=\sum_{i=1}^{k}\lambda_{i}[T_{d,d^{\prime}},F_{i}]$

where each $[F_{i}]\in
H_{1}(\mathcal{I}(S^{\prime},S_{g});\mathbb{Q})^{T_{\mathcal{V}^{\prime}}}$
for some $\mathcal{V}^{\prime}\subseteq\mathcal{B}$ with
$\left|\mathcal{V}^{\prime}\right|=4$. By Lemma 4.7, we have an isomorphism
$H_{1}(\mathcal{I}(S^{\prime},S_{g});\mathbb{Q})^{T_{\mathcal{V}^{\prime}}}\cong\scaleobj{1.5}{\wedge}^{3}\operatorname{im}(H_{1}(S^{\prime};\mathbb{Q})\rightarrow
H_{1}(S_{g};\mathbb{Q}))\cap\scaleobj{1.5}{\wedge}^{3}(\mathcal{V}^{\prime})^{\perp}$.
Then Lemma 4.4 gives us a spanning set for the vector space
$\scaleobj{1.5}{\wedge}^{3}\operatorname{im}(H_{1}(S^{\prime};\mathbb{Q})\rightarrow
H_{1}(S_{g};\mathbb{Q}))\cap\scaleobj{1.5}{\wedge}^{3}(\mathcal{V}^{\prime})^{\perp}$,
and by Lemma 4.2 each element of this spanning set is given by applying
$\tau_{g}$ to a bounding pair map supported on a bounding pair $f\cup
f^{\prime}\subseteq S^{\prime}$. Hence there is a relation in
$H_{2}(\mathcal{I}_{g};\mathbb{Q})$ given by

$[T_{d,d^{\prime}},T_{e,e^{\prime}}]=\sum_{i=1}^{k}\lambda_{i}[T_{d,d^{\prime}},T_{f_{i},f^{\prime}_{i}}]$

such that each $T_{f_{i},f^{\prime}_{i}}\in
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{\mathcal{V}^{\prime}}}$ for some
$\mathcal{V}^{\prime}\subseteq\mathcal{V}$ with
$\left|\mathcal{V}^{\prime}\right|=4$, so the lemma holds. ∎

We will need one more auxiliary result, which well help us apply Lemma 4.11.

######  Lemma 4.12.

Let $g\geq 51$ and let $[T_{d,d^{\prime}},T_{e,e^{\prime}}]\in
H_{2}^{\operatorname{ab},\operatorname{bp}}(\mathcal{I}_{g};\mathbb{Q})$. Then
there is a linear relation

$[T_{d,d^{\prime}},T_{e,e^{\prime}}]=\sum_{j=1}^{r}[T_{d_{j},d^{\prime}_{j}},T_{e,e^{\prime}}]$

in $H_{2}^{\operatorname{ab},\operatorname{bp}}(\mathcal{I}_{g};\mathbb{Q})$
such that , for each $1\leq j\leq r$, at least one connected component of
$S_{g}{\ssearrow}\left(d_{j}\cup d^{\prime}_{j}\right)$ has genus one.

###### Proof.

Let $S\cup S^{\prime}$ be the connected components of $S_{g}{\ssearrow}(d\cup
d^{\prime})$. If $e,e^{\prime}$ are contained in different connected
components of $S_{g}{\ssearrow}\left(d\cup d^{\prime}\right)$, we have a
relation $T_{e,e^{\prime}}=T_{e,d}T_{d,e^{\prime}}$. Hence there is a relation

$[T_{d,d^{\prime}},T_{e,e^{\prime}}]=[T_{d,d^{\prime}},T_{e,d}]+[T_{d,d^{\prime}},T_{d,e^{\prime}}]$

in $H_{2}^{\operatorname{ab},\operatorname{bp}}(\mathcal{I}_{g};\mathbb{Q})$.
Therefore without loss of generality, we may assume that
$e,e^{\prime}\subseteq S$. Now, let $d_{0},\ldots,d_{k}\subseteq S^{\prime}$
be curves such that the following hold:

  * •

$d_{0}=d$,

  * •

$d_{k}=d^{\prime}$,

  * •

$d_{i}$ and $d_{j}$ are disjoint for all $0\leq i<j\leq k$,

  * •

$d_{i}$ is homologous to $d$ for all $0\leq i\leq k$, and

  * •

the connected components of $S^{\prime}{\ssearrow}\left(\bigcup_{0\leq i\leq
k}d_{i}\right)$ all have genus one.

By construction, we have $T_{d,d^{\prime}}=\prod_{0\leq i\leq
k-1}T_{d_{i},d_{i+1}}$. Furthermore, each bounding pair $d_{i}\cup d_{i+1}$ is
contained in $S^{\prime}$, and hence each $T_{d_{i},d_{i+1}}$ commutes with
$T_{e,e^{\prime}}$. Therefore there is a relation in
$H_{2}^{\operatorname{ab},\operatorname{bp}}(\mathcal{I}_{g};\mathbb{Q})$
given by

$[T_{d,d^{\prime}},T_{e,e^{\prime}}]=\sum_{i=0}^{k-1}[T_{d_{i},d_{i+1}},T_{e,e^{\prime}}],$

so the lemma holds. ∎

We now conclude Section 4.

###### Proof of Proposition 4.1.

Let $[T_{d,d^{\prime}},T_{e,e^{\prime}}]\in
H_{2}^{\operatorname{ab},\operatorname{bp}}(\mathcal{I}_{g};\mathbb{Q}))$ be
an abelian cycle. The proof of the proposition proceeds in two steps.

  1. (1)

There is a relation

$[T_{d,d^{\prime}},T_{e,e^{\prime}}]=\sum_{i=1}^{k}\lambda_{i}[T_{d,d^{\prime}},T_{f_{i},f^{\prime}_{i}}]$

with each $[T_{f_{i},f^{\prime}_{i}}]\in
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{\mathcal{V}^{\prime}}}$ with
$\mathcal{V}^{\prime}\subseteq\mathcal{V}$ and
$\left|\mathcal{V}^{\prime}\right|=4.$

  2. (2)

If $[T_{d,d^{\prime}},T_{f_{i},f^{\prime}_{i}}]$ is an abelian cycle with
$T_{f_{i},f^{\prime}_{i}}\in
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{\mathcal{V}^{\prime}}}$ with
$\mathcal{V}^{\prime}\subseteq V$ and $\left|\mathcal{V}^{\prime}\right|=4$,
then there is a relation

$[T_{d,d^{\prime}},T_{f_{i},f^{\prime}_{i}}]=\sum_{j=1}^{m}\lambda_{j}[T_{h_{j},h^{\prime}_{j}},T_{f_{i},f^{\prime}_{i}}]$

with each $[T_{h_{j},h^{\prime}_{j}}]\in
H_{1}(\mathcal{I}_{g};\mathbb{Q})^{T_{v}}$ for some
$v\in\mathcal{V}^{\prime}.$

Lemma 4.12 says that we may assume without loss of generality that at least
one connected component of $S_{g}{\ssearrow}(d\cup d^{\prime})$ has genus one.
Then Step (1) is the content of Lemma 4.11 and Step (2) is the content of
Lemma 4.9. ∎

##  5\. Finiteness of coinvariants in
$\operatorname{BM}_{2}(X_{g};\mathbb{Q})$

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

######  Lemma 5.1.

Let $g\geq 51$ and $a\subseteq S_{g}$ a nonseparating curve. Let $[b]\in
H_{1}(S_{g};\mathbb{Z})$ be a homology class with $\langle[a],[b]\rangle=1$.
Let $b\subseteq S_{g}$ be a representative of $[b]$ with geometric
intersection $\left|a\cap b\right|=1$. Let $M\subseteq S_{g}{\ssearrow}(a\cup
b)$ be a nonseparating multicurve with $\left|\pi_{0}(M)\right|=9$. Let $G$ be
the image of the map $\operatorname{Mod}(S_{g}{\ssearrow}\left(a\cup b\cup
M\right))\rightarrow\operatorname{Sp}(2g,\mathbb{Z})$. The vector space

$H_{0}(G;\operatorname{BM}_{2}(X_{g};\mathbb{Q}))$

is finite dimensional.

_The outline of Section 5_. Let $g\geq 51$ and $a,b,M$ be as in the statement
of Lemma 5.1 and let $\mathcal{V}=\\{[c]:c\in\pi_{0}(M)\\}$. We will begin by
describing some algebraic invariants of Bestvina–Margalit tori, which record
how the elements in $\mathcal{V}$ project onto the different subgroups in
$\mathcal{H}(\operatorname{BM}_{\sigma})$. We will then prove Lemma 5.2, which
allows us to use these algebraic invariants to determine when two
Bestvina–Margalit tori $\operatorname{BM}_{\sigma}$ and
$\operatorname{BM}_{\tau}$ are in the same $G$–orbit for certain subgroups
$G\subseteq\operatorname{Sp}(2g,\mathbb{Z})$. We will then prove Lemma 5.3,
which describes how these algebraic invariants interact with addition of
fundamental classes. We will then prove Lemma 5.5 and Lemma 5.7. These two
results will allow decompose a class $[\operatorname{BM}_{\sigma}]$ into
“simpler” classes, where “simpler” means roughly that the algebraic invariants
of the torus $\operatorname{BM}_{\sigma}$ are bounded. These last two lemmas
make up the main work of the section.

_Algebraic invariants of Bestvina–Margalit tori_. We begin by describing the
invariants of Bestvina–Margalit tori that we will use to prove Lemma 5.1. Let
$\mathcal{H}(\operatorname{BM}_{\sigma})=\\{\mathcal{H}_{0}^{\sigma},\mathcal{H}_{1}^{\sigma},\mathcal{H}_{2}^{\sigma}\\}$
as in Section 3. For each $v_{i}\in\mathcal{V}$, let
$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\sigma})$ denote the maximal
$m\in\mathbb{Z}$ such that that
$\operatorname{proj}_{\mathcal{H}_{k}^{\sigma}\cap[b]^{\perp}}(v_{i})=mw$ for
some nonzero $w\in\mathcal{H}_{k}^{\sigma}$. For each
$v_{i},v_{j}\in\mathcal{V}$, let

$\theta(\mathcal{V})_{i,j,k}(\sigma)=\langle\operatorname{proj}_{\mathcal{H}_{k}^{\sigma}\cap[b]^{\perp}}(v_{i}),\operatorname{proj}_{\mathcal{H}_{k}^{\sigma}\cap[b]^{\perp}}(v_{j})\rangle.$

We now prove a result that describes the orbits of Bestvina–Margalit tori
under the action of stabilizer subgroups of
$\operatorname{Sp}(2g,\mathbb{Z})$.

######  Lemma 5.2.

Let $M\subseteq S_{g}$ be a nonseparating multicurve disjoint from $a$ and
$b$. Let $\mathcal{V}=\\{[c]:c\in\pi_{0}(M)\\}$, and assume that the elements
of $\mathcal{V}$ are indexed as $\mathcal{V}=\\{v_{1},\ldots,v_{n}\\}$. Let
$G=\operatorname{im}(\operatorname{Mod}(S_{g}{\ssearrow}\left(a\cup b\cup
M\right))\rightarrow\operatorname{Sp}(2g,\mathbb{Z}))$. Let
$\sigma,\tau\subseteq X_{g}$ be 2–cells and let
$\operatorname{BM}_{\sigma},\operatorname{BM}_{\tau}$ be the corresponding
Bestvina–Margalit tori. The tori $\operatorname{BM}_{\sigma}$ and
$\operatorname{BM}_{\tau}$ are in the same $G$–orbit if, after possibly
reindexing $\mathcal{H}(\operatorname{BM}_{\tau})$, we have:

  * •

$g(\sigma)=g(\tau)$,

  * •

$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\sigma})=\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\tau})$
for $1\leq i\leq n,0\leq k\leq 2,$ and

  * •

$\theta(\mathcal{V})_{i,j,k}(\sigma)=\theta(\mathcal{V})_{i,j,k}(\tau)$ for
$1\leq i,j\leq n,0\leq k\leq 2.$

###### Proof.

Let $\operatorname{BM}_{\sigma},\operatorname{BM}_{\tau}$ be two
Bestvina–Margalit tori that satisfy the hypotheses of the lemma. Because
$g(\sigma)=g(\tau)$, there is an
$f\in\operatorname{Stab}_{\operatorname{Sp}(2g,\mathbb{Z})}([a],[b])$
satisfying $f\operatorname{BM}_{\sigma}=\operatorname{BM}_{\tau}$. Therefore,
after possibly reindexing, we have
$f(\mathcal{H}_{k}^{\sigma})=\mathcal{H}_{k}^{\tau}$ for every $0\leq k\leq
2$.

We claim that $f$ sends
$\operatorname{proj}_{\mathcal{H}_{k}^{\sigma}\cap[b]^{\perp}}(v_{i})$ to
$\operatorname{proj}_{\mathcal{H}_{k}^{\tau}\cap[b]^{\perp}}(v_{i})$ for every
$i,k$. Indeed, such an element exists since
$g(\mathcal{H}(\sigma))=g(\mathcal{H}(\tau))$ and $\operatorname{BM}_{\sigma}$
and $\operatorname{BM}_{\tau}$ have
$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\sigma})=\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\tau})$
and $\theta(\mathcal{V})_{i,j,k}(\sigma)=\theta(\mathcal{V})_{i,j,k}(\tau)$.
Hence for each $\mathcal{H}_{k}(\tau)$ there is an
$f_{k}\in\operatorname{Stab}_{\operatorname{Sp}(2g,\mathbb{Z})}([a],[b])$
which is the identity on $\mathcal{H}_{k^{\prime}}(\tau)$ for $k^{\prime}\neq
k$ and which takes $f(v_{i,k}^{\sigma})$ to $v_{i,k}^{\tau}$. Then we can
replace $f$ with $f_{0}f_{1}f_{2}f$, which satisfies
$f\operatorname{proj}_{\mathcal{H}_{k}^{\sigma}\cap[b]^{\perp}}(v_{i})=\operatorname{proj}_{\mathcal{H}_{k}^{\tau}\cap[b]^{\perp}}(v_{i})$.
But then this new $f$ fixes every $v_{i}$, since

$f(v_{i})=\sum_{k\in\\{1,2,3\\}}f(\operatorname{proj}_{\mathcal{H}_{k}^{\sigma}\cap[b]^{\perp}}(v_{i}))=\sum_{k\in\\{1,2,3\\}}\operatorname{proj}_{\mathcal{H}_{k}^{\tau}\cap[b]^{\perp}}(v_{i})=v_{i}.$

It now remains to show that $f\in G$, i.e., that there is some
$F\in\operatorname{Mod}(S_{g}{\ssearrow}(a\cup b\cup M))$ such that the image
of $F$ under the symplectic representation is $f$. Since the symplectic
representation
$\operatorname{Mod}(S_{g})\rightarrow\operatorname{Sp}(2g,\mathbb{Z})$ is
surjective, there is some $F^{\prime}\in\operatorname{Mod}(S_{g})$ such that
$F^{\prime}$ is sent to $f$ by the symplectic representation. Since
$\mathcal{V}$ can be represented by $M$ disjoint from $a$ and $b$, we can
extend $\mathcal{V}\cup\\{a,b\\}$ to a symplectic basis
$\mathcal{B}=\\{\alpha_{1},\beta_{1},\ldots,\alpha_{g},\beta_{g}\\}$. Choose a
set of curves $\widehat{\mathcal{B}}$ such that:

  * •

$\pi_{0}(M),\\{a,b\\}\subseteq\widehat{\mathcal{B}}$,

  * •

$\mathcal{B}=\\{[c]:c\in\widehat{\mathcal{B}}\\}$, and

  * •

$\left|c\cap c^{\prime}\right|=\left|\langle[c],[c^{\prime}]\rangle\right|$
for all $c,c^{\prime}\in\widehat{\mathcal{B}}$.

Let $\widehat{\mathcal{B}}^{\prime}$ be another set of representatives for
$\mathcal{B}$ satisfying the above conditions, except we have

$\pi_{0}(F^{\prime}(M)),\\{F^{\prime}(a),F^{\prime}(b)\\}\subseteq\widehat{\mathcal{B}}^{\prime}.$

But now by the change of coordinates principle, there is some
$F^{\prime\prime}\in\operatorname{Mod}(S_{g})$ taking
$\widehat{\mathcal{B}}^{\prime}\rightarrow\widehat{\mathcal{B}}$ in such a way
that $F^{\prime\prime}_{*}$ acts trivially on $\mathcal{B}$. But then
$F^{\prime\prime}\in\mathcal{I}_{g}$, so the image of $F^{\prime\prime}\cdot
F^{\prime}$ under the symplectic representation is the same as $F^{\prime}$,
which is $f$. Then we have $F^{\prime\prime}\cdot F^{\prime}(M)=M$,
$F^{\prime\prime}\cdot F^{\prime}(a)=a$, and $F^{\prime\prime}\cdot
F^{\prime}(b)=b$. Hence $F^{\prime\prime}\cdot
F^{\prime}\in\operatorname{Mod}(S_{g}{\ssearrow}(M\cup a\cup b))$ and
$F^{\prime\prime}\cdot F^{\prime}$ maps to $f$ under the symplectic
representation, so $f\in G$, as desired. ∎

We now prove a lemma that describes how the $\theta(\mathcal{V})$ and
$\operatorname{rk}^{\mathcal{V}}$ interact with the Bestvina–Margalit tori
configured as in Lemma 5.3.

######  Lemma 5.3.

Let $g\geq 51$, $a,b,M\subseteq S_{g}$ and $\mathcal{V}\subseteq
H_{1}(S_{g};\mathbb{Z})$ be as in Lemma 5.2. Let $x,y,z\subseteq X_{g}$ and
$\mathcal{H}_{0},\mathcal{H}_{1},\mathcal{H}_{2},\mathcal{H}^{\prime}_{1},\mathcal{H}^{\prime}_{2}$
be as in Lemma 3.4. For any choice of $*=y,z$, and $yz$, let $\sigma_{*}$
denote the 2–cell containing $x$ and $*$. Let $v_{i,k}^{*}$ denote
$\operatorname{proj}_{\mathcal{H}_{i}\in\mathcal{H}(\sigma_{*})\cap[b]^{\perp}}(v_{i})$,
and let $\theta(\mathcal{V})_{i,j,k}(*)$ denote
$\theta(\mathcal{V})_{i,j,k}(\sigma_{*})$. Then the following equalities hold
for all $1\leq i,j\leq 9$:

  1. (1)

$\operatorname{rk}^{\mathcal{V}}(v_{i,0}^{y})=\operatorname{rk}^{\mathcal{V}}(v_{i,0}^{z})=\operatorname{rk}^{\mathcal{V}}(v_{i,0}^{yz})$,

  2. (2)

$\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{yz})=\gcd(\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{y}),\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{z}))$,

  3. (3)

$\operatorname{rk}^{\mathcal{V}}(v_{i,2}^{y})=\gcd(\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{z}),\operatorname{rk}^{\mathcal{V}}(v_{i,2}^{yz}))$,

  4. (4)

$\operatorname{rk}^{\mathcal{V}}(v_{i,2}^{z})=\gcd(\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{y}),\operatorname{rk}^{\mathcal{V}}(v_{i,2}^{yz}))$,

  5. (5)

$\theta(\mathcal{V})_{i,j,0}(y)=\theta(\mathcal{V})_{i,j,0}(z)=\theta(\mathcal{V})_{i,j,0}(yz)$
,

  6. (6)

$\theta(\mathcal{V})_{i,j,1}(y)+\theta(\mathcal{V})_{i,j,1}(z)=\theta(\mathcal{V})_{i,j,1}(yz)$,

  7. (7)

$\theta(\mathcal{V})_{i,j,1}(z)+\theta(\mathcal{V})_{i,j,2}(yz)=\theta(\mathcal{V})_{i,j,2}(y)$,
and

  8. (8)

$\theta(\mathcal{V})_{i,j,1}(y)+\theta(\mathcal{V})_{i,j,2}(yz)=\theta(\mathcal{V})_{i,j,2}(z)$.

###### Proof.

Since we have chosen the multicurve $M$ to be disjoint from $a$ and $b$, we
have $\mathcal{V}\subseteq[b]^{\perp}$. Then, if $X,Y\subseteq[a]^{\perp}$ are
two quasi–unimodular lattices with $X\cap Y=\mathbb{Z}[a]$ and
$\operatorname{rk}^{\mathcal{V}}(X)=2g(X)+1$,
$\operatorname{rk}^{\mathcal{V}}(Y)=2g(Y)+1$, we see that for any
$v\in[a]^{\perp}\cap[b]^{\perp}$, we have

$\operatorname{proj}_{(X+Y)\cap[b]^{\perp}}(v)=\operatorname{proj}_{X\cap[b]^{\perp}}(v)+\operatorname{proj}_{Y\cap[b]^{\perp}}(v).$

Now, Lemma 3.4 says that the following hold:

  1. (1)

$\mathcal{H}_{0}^{y}=\mathcal{H}_{0}^{z}=\mathcal{H}_{0}^{yz}$,

  2. (2)

$\mathcal{H}_{1}^{y}+\mathcal{H}_{1}^{z}=\mathcal{H}_{1}^{yz}$,

  3. (3)

$\mathcal{H}_{1}^{y}+\mathcal{H}_{2}^{yz}=\mathcal{H}_{2}^{z}$, and

  4. (4)

$\mathcal{H}_{1}^{z}+\mathcal{H}_{2}^{yz}=\mathcal{H}_{2}^{y}$.

Hence the above observation about projections tells us that the following
equalities among $v_{i,k}^{y}$, $v_{i,k}^{z}$ and $v_{i,k}^{yz}$ hold for all
$1\leq i\leq 9$:

  1. (a)

$v_{i,0}^{y}=v_{i,0}^{z}=v_{i,0}^{yz}$,

  2. (b)

$v_{i,1}^{y}+v_{i,1}^{z}=v_{i,1}^{yz}$,

  3. (c)

$v_{i,1}^{y}+v_{i,2}^{yz}=v_{i,2}^{z}$, and

  4. (d)

$v_{i,1}^{z}+v_{i,2}^{yz}=v_{i,2}^{y}$.

The relations in the statement of the lemma are derived as follows.

_Relations (1) and (5)_. Since $v_{i,0}^{y}=v_{i,0}^{z}=v_{i,0}^{yz}$ by
equality (a), we must have
$\operatorname{rk}^{\mathcal{V}}(v_{i,0}^{y})=\operatorname{rk}^{\mathcal{V}}(v_{i,0}^{z})=\operatorname{rk}^{\mathcal{V}}(v_{i,0}^{yz})$,
and similarly for
$\theta(\mathcal{V})_{i,j,0}(y)=\theta(\mathcal{V})_{i,j,0}(z)=\theta(\mathcal{V})_{i,j,0}(yz).$

_Relation (2) and (6)_. We have $v_{i,1}^{y}+v_{i,1}^{z}=v_{i,1}^{yz}$ by
relation (b). If $w_{i,1}^{*}$ is a primitive class with
$\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{*})w_{i,1}^{*}=v_{i,1}^{*}$ for some
choice of $*=y,z,yz$, then by relation (b) we have
$\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{y})w_{i,1}^{y}+\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{z})w_{i,1}^{z}=v_{i,1}^{yz}$.
By Lemma 3.4, the group $\mathcal{H}_{1}^{x,yz}\in\mathcal{H}(\sigma_{yz})$ is
given by $\mathcal{H}_{1}^{x,y}+\mathcal{H}^{x,z}_{1}$ with
$\mathcal{H}_{1}^{x,y}\cap\mathcal{H}^{x,z}_{1}=\mathbb{Z}[a]$, so we may
represent $w_{i,1}^{y}$ and $w_{i,1}^{z}$ using disjoint curves
$c_{y},c_{z}\subseteq S_{g}$. Then the homology class

$\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{y})/\gcd(\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{y}),\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{z}))\cdot[c_{y}]+\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{z})/\gcd(\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{y}),\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{z}))\cdot[c_{z}]$

is primitive, since by definition
$\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{y})/\gcd(\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{y}),\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{z}))$
and
$\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{z})/\gcd(\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{y}),\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{z}))$
are relatively prime. Therefore
$\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{yz})=\gcd(\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{y}),\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{z}))$.
For relation (6), relation (b) implies that

$\theta(\mathcal{V})_{i,j,1}^{yz}=\langle
v_{i,1}^{y}+v_{i,1}^{z},v_{j,1}^{y}+v_{j,1}^{z}\rangle.$

As above, $v_{i,1}^{y}$ and $v_{j,1}^{z}$, can be represented by multiples of
disjoint curves, and similarly for $v_{j,1}^{z}$ and $v_{i,1}^{y}$. Therefore
we have

$\theta(\mathcal{V})_{i,j,1}^{yz}=\langle
v_{i,1}^{y},v_{j,1}^{y}\rangle+\langle v_{i,1}^{y},v_{j,1}^{z}\rangle+\langle
v_{i,1}^{z},v_{j,1}^{y}\rangle+\langle v_{i,1}^{z},v_{j,1}^{z}\rangle=\langle
v_{i,1}^{y},v_{j,1}^{y}\rangle+\langle
v_{i,1}^{z},v_{j,1}^{z}\rangle=\theta(\mathcal{V})_{i,j,1}(y)+\theta(\mathcal{V})_{i,j,1}(z).$

_Relations (3) and (7)_. These two relations follow from relation (c) using an
argument similar to that for relations (2) and (6).

_Relations (4) and (8)_. These two relations follow from relation (d) using an
argument to that for relations (2) and (6). ∎

###  5.1. The proof of Lemma 5.5

We begin by proving an auxiliary result, which we will need to prove Lemma 5.5
and Lemma 5.7.

######  Lemma 5.4.

Let $g\geq 51$, $a,b,M\subseteq S_{g}$, and $\mathcal{V}\subseteq
H_{1}(S_{g};\mathbb{Z})$ be as in Lemma 5.2. Let $\sigma\subseteq X_{g}$ be a
2–cell. Suppose that there is a $1\leq m\leq 9$ such that the following hold:

  * •

$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\sigma})\leq 1$ for all $1\leq i<m$
and $0\leq k\leq 2$,

  * •

$\operatorname{rk}^{\mathcal{V}}(v_{m,2}^{\sigma})\geq 2$,

  * •

$\operatorname{rk}^{\mathcal{V}}(v_{m,2}^{\sigma})=\max\\{\operatorname{rk}^{\mathcal{V}}(v_{m,k}^{\sigma}):0\leq
k\leq 2\\}$, and

  * •

$\operatorname{rk}^{\mathcal{V}}(v_{m,1}^{\sigma})\neq
0,\operatorname{rk}^{\mathcal{V}}(v_{m,2}^{\sigma})$.

Then there is a relation in $\operatorname{BM}_{2}(X_{g};\mathbb{Q})$ given by

$[\operatorname{BM}_{\sigma}]=\sum_{\iota=1}^{q}\lambda_{\iota}[\operatorname{BM}_{\sigma_{\iota}}]$

such that, for all $1\leq\iota\leq q$:

  1. (1)

we have $\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\sigma_{\iota}})\leq 1$ for
all $1\leq i<m$ and $0\leq k\leq 2$,

  2. (2)

the set $\\{\operatorname{rk}^{\mathcal{V}}(v_{m,k}^{\sigma_{\iota}}):0\leq
k\leq 2\\}$ is upper bounded by
$\operatorname{rk}^{\mathcal{V}}(v_{m,2}^{\sigma})$,

  3. (3)

the set $\\{\operatorname{rk}^{\mathcal{V}}(v_{m,k}^{\sigma_{\iota}}):0\leq
k\leq 2\\}$ has no more elements equal to
$\operatorname{rk}^{\mathcal{V}}(v_{m,2}^{\sigma_{\iota}})$ than the does the
set $\\{\operatorname{rk}^{\mathcal{V}}(v_{m,k}^{\sigma}):0\leq k\leq 2\\}$,

  4. (4)

if $\max_{0\leq k\leq
2}\\{\operatorname{rk}^{\mathcal{V}}(v_{m,k}^{\sigma_{\iota}})\\}=\operatorname{rk}^{\mathcal{V}}(v_{m,2}^{\sigma})$,
we have
$0<\operatorname{rk}^{\mathcal{V}}(v_{m,1}^{\sigma_{\iota}})<\max\\{\operatorname{rk}^{\mathcal{V}}(v_{m,k}^{\sigma_{\iota}}):0\leq
k\leq 2\\}$, and

  5. (5)

we have $g(\mathcal{H}_{1}^{\sigma_{\iota}})\geq 10$.

###### Proof.

If $g(\mathcal{H}_{1}^{\sigma})\geq 10$, then we are done by taking $q=1$,
$\lambda_{1}=1$ and
$[\operatorname{BM}_{\sigma_{1}}]=[\operatorname{BM}_{\sigma}]$. Otherwise,
since $g\geq 51$, there is a $\kappa=0,2$ such that
$g(\mathcal{H}_{\kappa}^{\sigma})\geq 19$. Then since
$\left|\mathcal{V}\right|=9$, the fact that
$g(\mathcal{H}_{\kappa}^{\sigma})\geq 19$ implies that there is a primitive
$\mathcal{H}\subseteq\mathcal{H}_{\kappa}^{\sigma}$ such that the following
hold:

  * •

$a\in\mathcal{H}$,

  * •

$v_{i,\kappa}^{\sigma}\in\mathcal{H}^{\perp}$ for all $1\leq i\leq 9$,

  * •

$g(\mathcal{H})=9$,

  * •

$g(\mathcal{H})+g(\mathcal{H}_{1}^{\sigma})\geq 10$, and

  * •

$2*g(\mathcal{H})+1=\operatorname{rk}^{\mathcal{V}}(\mathcal{H})$.

Let
$\widehat{\mathcal{H}}=\\{\mathcal{H}_{0}^{\sigma},\mathcal{H},\mathcal{H}^{\perp}\cap\mathcal{H}_{1}^{\sigma},\mathcal{H}_{2}^{\sigma}\\}$.
We can describe $\widehat{\mathcal{H}}$ using the graphical notation of
Section 3 as in Figure 7.

$\mathcal{H}_{0}^{\sigma}$$\mathcal{H}$$\mathcal{H}^{\perp}\cap\mathcal{H}_{1}^{\sigma}$$\mathcal{H}_{2}^{\sigma}$
Figure 7. The graphical representation of $\widehat{\mathcal{H}}$

Let $z\subseteq X_{g}$ be the edge with $\mathcal{H}\in\mathcal{H}(z)$. Let
$y\subseteq X_{g}$ the edge with
$\mathcal{H}^{\perp}\cap\mathcal{H}_{1}^{\sigma}\in\mathcal{H}(y)$. Let
$x\subseteq X_{g}$ be the edge with
$\mathcal{H}_{\kappa^{\prime}}^{\sigma}\in\mathcal{H}(y)$, where
$0\leq\kappa^{\prime}\leq 2$ and $\kappa^{\prime}\neq 1,\kappa$. After
orienting $y$ and $z$ correctly, we let $yz$ denote the third edge of a 2–cell
in $X_{g}$ containing $y$ and $z$. We have
$\mathcal{H}+\mathcal{H}^{\perp}\cap\mathcal{H}_{1}^{\sigma}=\mathcal{H}_{1}^{\sigma}\in\mathcal{H}(yz)$,
so $x$ and $yz$ are two edges of the 2–cell $\sigma$. Now, Lemma 3.2 says that
there is a relation

$[\operatorname{BM}_{x,yz}]=[\operatorname{BM}_{x,y}]+[\operatorname{BM}_{x,z}].$

Now, let $\sigma_{y},\sigma_{z}\subseteq X_{g}$ be 2–cells containing $x$ and
then $y$ and $z$ respectively. We will now show that $\sigma_{z}$ and
$\sigma_{y}$ desired properties for $\sigma_{\iota}$.

_The 2–cell $\sigma_{z}$ satisfies the desired properties of
$\sigma_{\iota}$_. Assume that $\mathcal{H}(\sigma_{z})$ is indexed so that
$\mathcal{H}=\mathcal{H}_{\kappa}^{\sigma_{z}}$,
$\mathcal{H}_{1}^{\sigma_{z}}=\mathcal{H}^{\perp}\cap\mathcal{H}_{\kappa}^{\sigma}+\mathcal{H}_{1}^{\sigma}$,
and
$\mathcal{H}_{\kappa^{\prime}}^{\sigma_{z}}=\mathcal{H}_{\kappa^{\prime}}^{\sigma}$.
By our assumption that $v_{i,\kappa}^{\sigma}\in\mathcal{H}^{\perp}$ for all
$1\leq i\leq 9$, the following hold:

  1. (1)

$v_{i,\kappa}^{\sigma_{z}}=0$ for all $1\leq i\leq 9$,

  2. (2)

$v_{i,1}^{\sigma_{z}}=v_{i,\kappa}^{\sigma}+v_{i,1}^{\sigma}$ for all $1\leq
i\leq 9$, and

  3. (3)

$v_{i,\kappa^{\prime}}^{\sigma_{z}}=v_{i,\kappa^{\prime}}^{\sigma}$ for all
$1\leq i\leq 9$.

Then we can compute
$\operatorname{rk}^{\mathcal{V}}(v_{i,\kappa}^{\sigma_{z}})$ as follows.

  1. (1)

For $1\leq i<m$ and $0\leq k\leq 2$, we have $v_{i,k}^{\sigma_{z}}$ primitive.
The class $v_{i,k}^{\sigma}$ is primitive for all $1\leq i<m$ and $0\leq k\leq
2$ by hypothesis, so $v_{i,\kappa^{\prime}}^{\sigma_{z}}$ is primitive.
Additionally, for each $1\leq i<m$, the class $v_{i,\kappa}^{\sigma}$ can be
represented by a simple closed curve disjoint from a representative for the
class $v_{i,1}^{\sigma}$, so $v_{i,1}^{\sigma_{z}}$ is primitive. Hence
$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\sigma_{z}})\leq 1$ for $1\leq i<m$
and $0\leq k\leq 2$.

  2. (2)

For $i=m$, we have
$\operatorname{rk}^{\mathcal{V}}(v_{i,\kappa}^{\sigma_{z}})=0$,
$\operatorname{rk}^{\mathcal{V}}(v_{i,\kappa^{\prime}}^{\sigma_{z}})=\operatorname{rk}^{\mathcal{V}}(v_{i,\kappa^{\prime}}^{\sigma_{z}})$,
and
$\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{\sigma_{z}})=\gcd(\operatorname{rk}^{\mathcal{V}}(v_{i,\kappa}^{\sigma}),\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{\sigma}))$.

The computation of $\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\sigma_{z}})$
above implies that $\sigma_{z}$ satisfies property (1) of $\sigma_{\iota}$.
Then the computation of
$\operatorname{rk}^{\mathcal{V}}(v_{m,k}^{\sigma_{z}})$ implies that
$\sigma_{z}$ satisfies properties (2)–(4) of $\sigma_{\iota}$. Indeed,
properties (2) and (3) follow from the fact that
$\gcd(\operatorname{rk}^{\mathcal{V}}(v_{i,\kappa}^{\sigma}),\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{\sigma}))\leq\operatorname{rk}^{\mathcal{V}}(v_{i,\kappa}^{\sigma}),\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{\sigma})$.
Property (4) follows from the fact that
$\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{\sigma_{z}})=\gcd(\operatorname{rk}^{\mathcal{V}}(v_{i,\kappa}^{\sigma}),\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{\sigma}))\leq\operatorname{rk}^{\mathcal{V}}(v_{i,1}^{\sigma})<\operatorname{rk}^{\mathcal{V}}(v_{i,2}^{\sigma})$.
Finally, property (5) follows from the fact that
$g(\mathcal{H}_{1}^{\sigma_{z}})=g(\mathcal{H}_{1}^{\sigma})+g(\mathcal{H}_{\kappa}^{\sigma})-g(\mathcal{H})$.
Indeed, we have assumed that $g(\mathcal{H}_{\kappa}^{\sigma})\geq 19$, so
$g(\mathcal{H}_{1}^{\sigma_{z}})\geq 19-g(\mathcal{H})$. Then since
$g(\mathcal{H})=9$ by hypothesis, we have $g(\mathcal{H}_{1}^{\sigma_{z}})\geq
19-9\geq 10$, so $\sigma_{z}$ satisfies property (5) of $\sigma_{\iota}$.

_The 2–cell $\sigma_{y}$ satisfies the desired properties of
$\sigma_{\iota}$_. Assume that $\mathcal{H}(\sigma_{y})$ is indexed so that
$\mathcal{H}_{1}^{\sigma_{y}}=\mathcal{H}_{1}^{\sigma}+\mathcal{H}$,
$\mathcal{H}_{\kappa^{\prime}}^{\sigma_{y}}=\mathcal{H}_{\kappa}^{\sigma}$,
and
$\mathcal{H}_{\kappa}^{\sigma_{y}}=\mathcal{H}^{\perp}\cap\mathcal{H}_{\kappa}^{\sigma_{y}}$.
The assumption that $v_{i,1}^{\sigma}\in\mathcal{H}^{\perp}$ for all $1\leq
i\leq 9$ implies that $v_{i,k}^{\sigma}=v_{i,k}^{\sigma_{y}}$ for all $1\leq
i\leq 9$, $0\leq k\leq 2$. Therefore since $\sigma$ satisfies properties
(1)–(4) of $\sigma_{\iota}$ by hypothesis, then $\sigma_{\iota}$ does as well.
Then
$g(\mathcal{H}_{1}^{\sigma_{y}})=g(\mathcal{H}_{1}^{\sigma})+g(\mathcal{H})\geq
1+9=10$ by assumption, so $\sigma_{y}$ satisfies property (5) of
$\sigma_{\iota}$.

Now, we have shown that
$[\operatorname{BM}_{x,yz}]=[\operatorname{BM}_{x,y}]+[\operatorname{BM}_{x,z}]$.
By our choice of $x$ and $y$, we have
$\operatorname{BM}_{x,yz}=\operatorname{BM}_{\sigma}$. Therefore we have a
relation

$[\operatorname{BM}_{x,y}]=[\operatorname{BM}_{x,z}]-[\operatorname{BM}_{x,yz}].$

By taking $s=2$, $\tau_{1}=\sigma_{z}$, $\tau_{2}=\sigma_{yz}$,
$\lambda_{1}=1$, and $\lambda_{2}=-1$, the proof is complete. ∎

We now prove Lemma 5.5, which says that any fundamental class
$[\operatorname{BM}_{\sigma}]$ is a linear combination of classes
$[\operatorname{BM}_{\tau_{\ell}}]$ with
$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\tau_{\ell}})\leq 1.$

######  Lemma 5.5.

Let $g\geq 51$, $a,b,M\subseteq S_{g}$ and $\mathcal{V}\subseteq
H_{1}(S_{g};\mathbb{Z})$ be as in Lemma 5.2. Let $\operatorname{BM}_{\sigma}$
be a Bestvina–Margalit torus. Then there is a collection of 2–cells
$\tau_{1},\ldots,\tau_{p}\subseteq X_{g}$ that satisfy

$[\operatorname{BM}_{\sigma}]=\sum_{\ell=1}^{p}\lambda_{\ell}[\operatorname{BM}_{\tau_{\ell}}]$

and such that $\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\tau_{\ell}})\leq 1$
for every $1\leq\ell\leq p$, $1\leq i\leq 9$ and $k=0,1,2$.

###### Proof.

We induct on the number $m$ with $0\leq m\leq 9$ such that we can write
$[\operatorname{BM}_{\sigma}]$ as a linear combination of classes
$[\operatorname{BM}_{\tau_{\ell}}]$ that satisfy, for $1\leq\ell\leq p$ and
$1\leq i\leq m$, the inequality
$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\tau_{s}})\leq 1$.

_Base case: $m=0$_. In this case, the result holds trivially, since there are
no $i$ with $1\leq i$ and $i\leq m$.

_Inductive step: $m\geq 1$_. Inductively, assume that
$[\operatorname{BM}_{\sigma}]$ can be written as a linear combination of
classes $[\operatorname{BM}_{\tau_{\ell}}]$ that satisfy, for $1\leq\ell\leq
p$ and $1\leq i\leq m$, the inequality
$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\tau_{s}})\leq 1$. Hence without
loss of generality, we may assume that $\operatorname{BM}_{\sigma}$ satisfies
$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\sigma})\leq 1$ for $1\leq i<m$. We
will show that $[\operatorname{BM}_{\sigma}]$ is a linear combination of
fundamental classes $[\operatorname{BM}_{\tau}]$, each of which has
$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\tau})\leq 1$ for all $1\leq i\leq
m$ and $0\leq k\leq 2$. We will perform a double induction on two quantities
associated to $\sigma$:

  * •

$\operatorname{maxrk}_{m}(\sigma)=\max_{0\leq k\leq
2}\\{\operatorname{rk}^{\mathcal{V}}(v_{m,k}^{\sigma})\\}$, and

  * •

$\operatorname{nummaxrk}_{m}(\sigma)=\left|\\{k:0\leq k\leq
2,\operatorname{rk}^{\mathcal{V}}(v_{m,k}^{\sigma})=\operatorname{maxrk}_{m}(\sigma)\\}\right|$.

In particular, we suppose that $\sigma$ is a 2–cell with
$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\sigma})\leq 1$ for all $1\leq i<m$
and $0\leq k\leq 2$. We suppose that for each 2–cell $\tau$ with:

  * •

$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\tau})\leq 1$ for $1\leq i<m$ and
$0\leq k\leq 2$, and either:

    * –

$\operatorname{maxrk}_{m}(\tau)<\operatorname{maxrk}_{m}(\sigma)$, or

    * –

$\operatorname{maxrk}_{m}(\tau)\leq\operatorname{maxrk}_{m}(\sigma)$ and
$\operatorname{nummaxrk}_{m}(\tau)<\operatorname{nummaxrk}_{m}(\sigma)$,

we know that $[\operatorname{BM}_{\tau}]$ is a $\mathbb{Q}$–linear combination
of classes $[\operatorname{BM}_{\tau_{\ell}}]$ such that
$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\tau_{\ell}})\leq 1$ for all $1\leq
i\leq m$ and $0\leq k\leq 2$. We will show that this implies that the same
holds for $[\operatorname{BM}_{\sigma}]$.

_Base case: $\operatorname{maxrk}_{m}(\sigma)=1$_. In this case,
$\operatorname{BM}_{\sigma}$ satisfies
$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\sigma})\leq 1$ for all $1\leq i\leq
m$ and $0\leq k\leq 2$, so the inductive step for the induction on $m$ is
complete.

Inductive step for $\operatorname{maxrk}_{m}(\sigma)$ and
$\operatorname{nummaxrk}_{m}(\sigma)$: The inductive hypothesis for the
induction on $m$ holds for all $\tau\subseteq X_{g}$ with:

  * •

either $\operatorname{maxrk}_{m}(\tau)<\operatorname{maxrk}_{m}(\sigma)$ or

  * •

both $\operatorname{maxrk}_{m}(\tau)\leq\operatorname{maxrk}_{m}(\sigma)$ and
$\operatorname{nummaxrk}_{m}(\tau)<\operatorname{nummaxrk}_{m}(\sigma)$.

We will show that

$[\operatorname{BM}_{\sigma}]=\sum_{\ell=1}^{p}\lambda_{\ell}[\operatorname{BM}_{\tau_{\ell}}]$

such that the following hold:

  * •

at least one of the following holds:

    * –

$\operatorname{maxrk}_{m}(\sigma_{\iota})<\operatorname{maxrk}_{m}(\sigma)$ or

    * –

$\operatorname{maxrk}_{m}(\sigma_{\iota})\leq\operatorname{maxrk}_{m}(\sigma)$
and
$\operatorname{nummaxrk}_{m}(\sigma_{\iota})<\operatorname{nummaxrk}_{m}(\sigma)$,

  * •

$\lambda_{\ell}\in\mathbb{Q}$, and

  * •

$\operatorname{rk}^{\mathcal{V}}(v_{i,k}^{\tau_{\ell}})\leq 1$ for all$1\leq
i<m$ and $k=0,1,2$.

This completes the proof, since every $\tau_{\ell}$ as above satisfies the
inductive hypothesis for the induction on $m$ by the inductive hypothesis for
the induction on $\operatorname{maxrk}_{m}(\sigma)$ and
$\operatorname{nummaxrk}_{m}(\sigma)$.

If $\operatorname{rk}^{\mathcal{V}}(v_{m,k}^{\sigma})=0$ for two distinct
choices of $k$, then $\operatorname{rk}^{\mathcal{V}}(v_{m,k}^{\sigma})=1$ for
the third choice of $k$, since
$v_{m}=v_{m,0}^{\sigma}+v_{m,1}^{\sigma}+v_{m,2}^{\sigma}$ and $v_{m}$ is
primitive by assumption. Otherwise, reindex
$\mathcal{H}(\operatorname{BM}_{\sigma})$ such that
$\operatorname{rk}^{\mathcal{V}}(v_{m,2}^{\sigma})$ is maximal among all
$\operatorname{rk}^{\mathcal{V}}(v_{m,k}^{\sigma})$. Note that at least one
remaining $\operatorname{rk}^{\mathcal{V}}(v_{m,i}^{\sigma})$ must have