The universal surface bundle over the Torelli space has no sections

Abstract

For $g>3$, we give two proofs of the fact that the Birman exact sequence for the Torelli group

$$\begin{aligned} 1\rightarrow \pi _1(S_g)\rightarrow {\mathcal I}_{g,1}\rightarrow {\mathcal I}_g\rightarrow 1 \end{aligned}$$

does not split. This result was claimed by Mess (Unit tangent bundle subgroups of the mapping class groups, MSRI Pre-print, 1990), but his proof has a critical and unrepairable error which will be discussed in the introduction. Let ${\mathcal U\mathcal I}_{g,n}\xrightarrow {Tu'_{g,n}} {\mathcal B} {\mathcal I}_{g,n}$ (resp. ${\mathcal U}{\mathcal P}{\mathcal I}_{g,n}\xrightarrow {Tu_{g,n}}{\mathcal B}{\mathcal P}{\mathcal I}_{g,n}$) denote the universal surface bundle over the Torelli space fixing n points as a set (resp. pointwise). We also deduce that $Tu'_{g,n}$ has no sections when $n>1$ and that $Tu_{g,n}$ has precisely n distinct sections for $n\ge 0$ up to homotopy.

1 Introduction

It is a basic problem to understand when bundles have continuous sections, and the corresponding group theory problem as to when short exact sequences have splittings. These are equivalent problems when the fiber, the base and the total space are all $K(\pi ,1)$-spaces. In this article, we will discuss the “section problems” and the “splitting problems” in the setting of surface bundles. Here by section we mean continuous section.

Let $S_{g,n}$ be a closed orientable surface of genus g with n marked points. Let Mod$(S_{g,n})$ (resp. PMod$(S_{g,n})$) be the mapping class group of $S_{g,n}$, i.e., the group of isotopy classes of orientation-preserving diffeomorphisms of $S_g$ fixing n points as a set (resp. pointwise). Both Mod$(S_{g,n})$ and PMod$(S_{g,n})$ act on $H^1(S_g;\mathbb {Z})$ leaving invariant the algebraic intersection numbers. Let ${\mathcal I}_{g,n}$ (resp. ${\mathcal P\mathcal I}_{g,n}$) be the Torelli group (resp. pure Torelli group) of $S_{g,n}$, i.e., the subgroup of Mod$(S_{g,n})$ (resp. PMod$(S_{g,n})$) that acts trivially on $H^1(S_g;\mathbb {Z})$. We omit n when $n=0$. The following Birman exact sequence for the Torelli group provides a relationship between ${\mathcal I}_{g,1}$ and ${\mathcal I}_{g}$; e.g., see Farb and Margalit [7, Chapter 4.2]

$$\begin{aligned} 1\rightarrow \pi _1(S_g)\xrightarrow {\text {point pushing}} {\mathcal I}_{g,1}\xrightarrow {T\pi _{g,1}} {\mathcal I}_{g}\rightarrow 1. \end{aligned}$$

(1)

The main theorem of this paper is the following:

Theorem 1.1

(Nonsplitting of the Birman exact sequence for the Torelli group) For $g>3$, the Birman exact sequence for the Torelli group (1) does not split.

Remark 1

Our proof needs the condition $g>3$. By Mess [14, Proposition 4], ${\mathcal I}_2$ is a free group. So the Birman exact sequence for ${\mathcal I}_2$ splits. The case $g=3$ is open.

Let ${\mathcal B}{\mathcal P}{\mathcal I}_{g,n}:=K({\mathcal P}{\mathcal I}_{g,n},1)$ be the pure universal Torelli space fixing n marked points pointwise and let

$$\begin{aligned} S_g\rightarrow {\mathcal U}{\mathcal P}{\mathcal I}_{g,n}\xrightarrow {Tu_{g,n}}{\mathcal B}{\mathcal P}{\mathcal I}_{g,n} \end{aligned}$$

(2)

be the pure universal Torelli bundle. Surface bundle (2) classifies smooth $S_g$-bundle equipped with a basis of $H^1(S_g;{\mathbb {Z}})$ and n ordered points on each fiber. Since ${\mathcal P}{\mathcal I}_{g,n}$ fixes n points, there are n distinct sections $\{Ts_i|1\le i\le n\}$ of the universal Torelli bundle (2). Let ${\mathcal B}{\mathcal I}_{g,n}:=K({\mathcal I}_{g,n},1)$ be the universal Torelli space fixing n marked points as a set and let

$$\begin{aligned} S_g\rightarrow {\mathcal U}{\mathcal I}_{g,n}\xrightarrow {Tu'_{g,n}} {\mathcal B}{\mathcal I}_{g,n} \end{aligned}$$

(3)

be the universal Torelli bundle. This bundle classifies smooth $S_g$-bundles equipped with a basis of $H^1(S_g;{\mathbb {Z}})$ and n unordered points on each fiber. Theorem 1.1 says that $Tu_{g,0}$ has no sections. For $n\ge 0$, we have the following complete answer for sections of (refTUB2) and (3).

Theorem 1.2

(Classification of sections for the n-pointed Torelli bundles) The following holds:

• (1) For $n\ge 0$ and $g>3$, every section of the universal Torelli bundle (2) is homotopic to $Ts_i$ for some $i\in \{1,2,...,n\}$.

• (2) For $n>1$ and $g>3$, the universal Torelli bundle (3) has no continuous sections.

Let ${\mathcal M}_g:=K(\text {Mod}(S_{g}),1)$. As is known, the universal bundle

$$\begin{aligned} S_g\rightarrow {\mathcal U}{\mathcal M}_g\xrightarrow {} {\mathcal M}_g \end{aligned}$$

has no sections. This can be seen from the corresponding algebraic problem of finding splittings of the Birman exact sequence

$$\begin{aligned} 1\rightarrow \pi _1(S_g)\rightarrow \text {Mod}(S_{g,1})\rightarrow \text {Mod}(S_{g})\rightarrow 1. \end{aligned}$$

An analysis of the finite subgroups ${{\,\mathrm{Mod}\,}}(S_g)$ and ${{\,\mathrm{Mod}\,}}(S_{g,1})$ shows that the sequence does not split: every finite subgroup of ${{\,\mathrm{Mod}\,}}(S_{g,1})$ is cyclic, but there are noncyclic finite subgroups of ${{\,\mathrm{Mod}\,}}(S_g)$; e.g., see Farb and Margalit [7, Corollary 5.11]. However, this method does not work for torsion-free subgroups of Mod$(S_g)$ like Torelli groups. For any subgroup $G<{{\,\mathrm{Mod}\,}}(S_g)$, there is an extension $\Gamma _G$ of G by $\pi _1(S_g)$ as the following short exact sequence

$$\begin{aligned} 1\rightarrow \pi _1(S_g)\rightarrow \Gamma _G\rightarrow G\rightarrow 1. \end{aligned}$$

(4)

We call (4) the Birman exact sequence for G since it is induced from the Birman exact sequence. In a later paper of Chen and Salter [5], they establish the virtual non-splitting of Birman exact sequence for finite index subgroup of Torelli groups. This gives another proof of Theorem 1.1 and strengthens Theorem 1.1 as well.

Define $\pi ^0=\pi _1(S_g)$ and $\pi ^{n+1}=[\pi ^n,\pi ^0]$. The kth term of the Johnson filtration subgroup of ${{\,\mathrm{Mod}\,}}(S_{g})$, denoted by ${{\,\mathrm{Mod}\,}}_g(k)$, is the kernel of the action of ${{\,\mathrm{Mod}\,}}(S_g)$ on $\pi ^0/\pi ^k$. For example, ${{\,\mathrm{Mod}\,}}_g(1)={\mathcal I}_g$. The Johnson filtration was defined by Johnson [11]. We pose the following open problem.

Problem

(Splitting of Birman exact sequence for Johnson filtration subgroups) Do the Birman exact sequence split for Johnson filtration subgroups Mod$_g(k)$ for $k\ge 2$?

Error in Mess [13, Proposition 2] In the unpublished paper of Mess [13, Proposition 2], he claimed that there are no splittings of the exact sequence (1). But his proof has a fatal error. Here is how the proof goes. Let C be a curve dividing $S_g$ into 2 parts S(1) and S(2) of genus p and q, where $p,q\ge 2$. Let $US_g$ be the unit tangent bundle of a surface of genus g. The unit tangent bundle subgroups of $S_p$ and $S_q$ amalgamate along the common boundary Dehn twist to give a subgroup $A\le \text {Mod}(S_g)$ satisfying the following short exact sequence

$$\begin{aligned} 1\rightarrow {\mathbb {Z}}\rightarrow \pi _1(US_p)\times \pi _1(US_q)\rightarrow A\rightarrow 1. \end{aligned}$$

In Case a) of Mess’ proof, he tried to prove the following claim, which is the key of his proof. Without the claim, the argument appears to be unrepairable.

Mess’ Claim in the proof of Proposition 2

The Birman exact sequence for A does not split. $\square$

However, this is a wrong claim. We construct a splitting of the Birman exact sequence for A as the following. Let $\text {PConf}_2(S_p)$ be the pure configuration space of $S_p$, i.e., the space of ordered 2-tuples of distinct points on $S_p$. Define

$$\begin{aligned} \text {PConf}_{1,1}(S_p)=\{(x,y,v)|x\ne y\in S_p \text { and } v\in US_p \text { a unit vector at} x\}. \end{aligned}$$

We have the following pullback diagram

(5)

The splitting of $\pi _1(US_p)$ should lie in $\pi _1(\text {PConf}_{1,1}(S_p))$ instead of $\pi _1(US_{p,1})$ as Mess claimed in Case a) of Proposition 2. As long as we can find a section of f, we will find a section of A to ${\mathcal I}_{g,1}$. By the property of pullback diagrams, a section of g can induce a section of f in diagram (5). To negate the argument of Mess Proposition 2, we only need to construct a section of g. We simply need to find a self-map of $S_g$ that has no fixed point. For example, the composition of a retraction of $S_p$ onto a curve c and a rotation of c at any nontrivial angle does not have a fixed point. Therefore Mess’ proof is invalid and does not seem to be repairable.

Strategy of the proof of Theorem 1.1 Our first strategy is the following: assume that we have a splitting $\phi : \mathcal {I}_g\rightarrow \mathcal {I}_{g,1}$ of (1). Let $T_a$ be the Dehn twist about a simple closed curve a on $S_g$. The main result of Johnson [10] shows that all bounding pair maps, i.e., $T_aT_b^{-1}$ for a pair of non-separating curves a, b that bound a subsurface, generate ${\mathcal I}_{g}$. Firstly we need to understand $\phi (T_aT_b^{-1})$. We will show that $\phi (T_aT_b^{-1})=T_{a'}T_{b'}^{-1}$ for a bounding pair $a',b'$ on $S_{g,1}$ such that $a'$ is homotopic to a and $b'$ is homotopic to b after forgetting the puncture. Moreover, the curve $a'$ does not depend on the choice of b. Then we use the lantern relation to derive a contradiction. Our main tool is the canonical reduction system for a mapping class, which in turn uses the Thurston classification of isotopy classes of diffeomorphisms of surfaces. This idea originated from Birman et al. [1].

We also give an algebraic proof using cohomology obstruction. The key tool is the classification theorem of Chen [4, Theorem 1.5] and the Johnson homomorphism.

Outline of the paper In Sect. 2, we give an introduction to canonical reduction systems and lantern relations. Then in Sect. 3, we give our first proof using those tools. In Sect. 4, we deal with the punctured case. Then in Sect. 5, we use Sect. 4 and homology to give the second proof of Theorem 1.1.

2 Canonical reduction systems and the lantern relation

In this section, we will give some background that will be used in the proof.

2.1 Canonical reduction systems

The central tool for the proof of Theorem 1.1 is the notion of a canonical reduction system, which can be viewed as an enhancement of the Nielsen–Thurston classification. We remind the reader that a curve $c\subset S$ is said to be peripheral if c is isotopic to a boundary component or surrounds a marked point of S.

The Nielsen–Thurston classification asserts that each nontrivial element $f\in {{\,\mathrm{Mod}\,}}(S)$ is of exactly one of the following types: periodic, reducible, or pseudo-Anosov. A mapping class f is periodic if $f^n=id$ for some $n \ge 1$, and is reducible if for some $n \ge 1$, there is some nonperipheral simple closed curve $c \subset S$ such that $f^n(c)$ is isotopic to c. If neither of these conditions are satisfied, f is said to be pseudo-Anosov. In this case, f is isotopic to a homeomorphism $f'$ of a very special form. We will not need to delve into the theory of pseudo-Anosov mappings, and refer the interested reader to Farb and Margalit [7, Chapter 13] and Fathi et al. [6] for more details.

Definition 2.1

(Reduction systems) A reduction system of a reducible mapping class h in $\text {Mod}(S)$ is a set of disjoint nonperipheral curves that h fixes as a set up to isotopy. A reduction system is maximal if it is maximal with respect to inclusion of reduction systems for h. The canonical reduction system $\text { CRS}(h)$ is the intersection of all maximal reduction systems of h.

Canonical reduction systems allow for a refined version of the Nielsen–Thurston classification. For a reducible element f, there exists n such that $f^n$ fixes each element in CRS(f) and after cutting out CRS(f), the restriction of $f^n$ on each component is either identity or pseudo-Anosov. This is called the canonical form of f; e.g., see Farb and Margalit [7, Corollary 13.3]. In Propositions 2.2–2.6, we list some properties of the canonical reduction systems that will be used later.

Proposition 2.2

$\text {CRS}(h^n)=\text {CRS}(h)$ for any n.

Proof

This is classical; see Farb and Margalit [7, Chapter 13]. $\square$

For two curves a, b on a surface S, let i(a, b) be the geometric intersection number of a and b. For two sets of curves P and Q, we say that P and Q intersect if there exist $a\in P$ and $b\in Q$ such that $i(a,b)\ne 0$. We emphasize that “intersection” here refers to the intersection of curves on S, and not the abstract set-theoretic intersection of P and Q as sets.

Proposition 2.3

Let h be a reducible mapping class in $\text {Mod}(S)$. If $\{\gamma \}$ and $\text {CRS}(h)$ intersect, then no power of h fixes $\gamma$.

Proof

Suppose that $h^n$ fixes $\gamma$. Therefore $\gamma$ belongs to a maximal reduction system M. By definition, ${{\,\mathrm{CRS}\,}}(h)\subset M$. However $\gamma$ intersects some curve in CRS(f); this contradicts the fact that M is a set of disjoint curves. $\square$

Proposition 2.4

Suppose that $h,f\in \text {Mod}(S)$ and $fh=hf$. Then $\text {CRS}(h)$ and $\text {CRS}(f)$ do not intersect.

Proof

Conjugating, CRS($hfh^{-1})=h({{\,\mathrm{CRS}\,}}(f))$. Since $hfh^{-1}=f$, it follows that CRS$(f)=h({{\,\mathrm{CRS}\,}}(f))$. Therefore h fixes the whole set CRS(f). There is some $n \ge 1$ such that $h^n$ fixes all curves element-wise in CRS(f). By Proposition 2.3, curves in CRS(h) do not intersect curves in CRS(f). $\square$

For a curve a on a surface S, denote by $T_a$ the Dehn twist about a. More generally, a Dehn multitwist is any mapping class of the form

$$\begin{aligned} T : = \prod T_{a_i}^{k_i} \end{aligned}$$

for a collection of pairwise-disjoint simple closed curves $\{a_i\}$ and arbitrary integers $k_i$.

Proposition 2.5

Let

$$\begin{aligned} T : = \prod T_{a_i}^{k_i} \end{aligned}$$

be a Dehn multitwist. Then

$$\begin{aligned} {{\,\mathrm{CRS}\,}}(T) = \{a_i\}. \end{aligned}$$

Proof

Firstly T cannot contain any simple closed curves b for which $i(b, a_i) \ne 0$, since no power of T preserves b. This can be seen from the equation

$$\begin{aligned} i\left( \prod T_{a_i}^{k_i}(b),b\right) =\sum |k_i|i(a_i,b)\ne 0=i(b,b); \end{aligned}$$

see Farb and Margalit [7, Proposition 3.2]. It follows that if S is any reduction system for T, then $S \cup \{a_i\}$ is also a reduction system, and hence that $\{a_i\} \subset {{\,\mathrm{CRS}\,}}(T)$. If $\gamma$ is disjoint from each element of $\{a_i\}$ but not equal to any $a_i$, then there exists some curve $\delta$, also disjoint and distinct from each $a_i$, such that $i(\gamma ,\delta ) \ne 0$. As both $\{a_i\} \cup \{\gamma \}$ and $\{a_i\} \cup \{\delta \}$ are reduction systems for T, this shows that no such $\gamma$ can be contained in ${{\,\mathrm{CRS}\,}}(T)$ and hence that ${{\,\mathrm{CRS}\,}}(T) = \{a_i\}$ as claimed. $\square$

The final result we will require follows from the theory of pseudo-Anosov mapping. It appears in McCarthy [12, Theorem 1].

Proposition 2.6

(McCarthy) Let S be a Riemann surface of finite type, and let $f \in {{\,\mathrm{Mod}\,}}(S)$ be a pseudo-Anosov element. Then the centralizer subgroup of f in ${{\,\mathrm{Mod}\,}}(S)$ is virtually cyclic.

2.2 The lantern relation

Now, we introduce a remarkable relation for $\text { Mod}(S)$ that will be used in the proof.

Proposition 2.7

(The lantern relation) Let $x,y,z,b_1,b_2,b_3,b_4$ be simple closed curves in $S=S_{0,4}$ that are arranged as the curves shown in the following figure.

In Mod(S) we have the relation

$$\begin{aligned} T_xT_yT_z=T_{b_1}T_{b_2}T_{b_3}T_{b_4}. \end{aligned}$$

Proof

This is classical; see Farb and Margalit [7, Chapter 5.1].

3 The lantern relation proof of theorem 1.1

3.1 Images of bounding pair maps

Let $\{a,b\}$ be a bounding pair, i.e., a, b are non-separating curves such that a and b bounds a subsurface. In this subsection, we will determine $\phi (T_aT_b^{-1})$. For two curves c and d, denote by i(c, d) the geometric intersection number of c and d. For a curve $c'$ on $S_{g,1}$, when we say $c'$ is isotopic to a curve c on $S_g$, we mean that $c'$ is isotopic to c on $S_g$ by forgetting the marked point.

Lemma 3.1

• (1) Let $\{a,b\}$ be a bounding pair, and fix $k>0$ such that $(T_aT_b^{-1})^k\in \Gamma$. Up to a swap of a and b, we have that $\sigma ((T_aT_b^{-1})^k)=(T_{a'}T_{b'}^{-1})^k(T_{a'}^{-1}T_{a''})^{n}$, where n is an integer and $a',a'',b'$ are three disjoint curves on $\Sigma _{g,1}$ such that $a',a''$ are isotopic to a and $b'$ is isotopic to b. Notice that n can be zero.

• (2) Let c be a separating curve on $\Sigma _g$ that divides $\Sigma _g$ into two subsurfaces each of genus at least two. For any $k >0$ such that $(T_c)^k\in \Gamma$, we have that $\sigma ((T_c)^k)=(T_{c'})^{k}(T_{c'}^{-1}T_{c''})^{n}$ where n is an integer and $c'$ and $c''$ are a pair of curves on $\Sigma _{g,1}$ that are both isotopic to c.

Proof

Let $(T_aT_b^{-1})^k\in \Gamma$ be a power of a bounding pair map. Since the centralizer of $(T_aT_b^{-1})^k$ contains a copy of ${\mathbb {Z}}^{2g-3}$ as a subgroup of ${\mathcal I}(\Sigma _g)$, the centralizer of $(T_aT_b^{-1})^k$ as a subgroup of $\Gamma$ contains a copy of ${\mathbb {Z}}^{2g-3}$ as well. By the injectivity of $\sigma$, the centralizer of $\sigma (T_aT_b^{-1})\in {\mathcal I}(\Sigma _{g,*})$ contains a copy of ${\mathbb {Z}}^{2g-3}$. When $g>3$, we have that $2g-3>3$. Therefore $\sigma ((T_aT_b^{-1})^k)\in \mathcal I(\Sigma _{g,*})$ cannot be pseudo-Anosov because the centralizer of a pseudo-Anosov element is a virtually cyclic group by Proposition 2.6. For any curve $\gamma '$ on $\Sigma _{g,*}$, denote by $\gamma$ the same curve on $\Sigma _g$. We decompose the proof into the following three steps.

• Step 1: CRS$({\varvec{\sigma }((T_aT_b^{-1})^k))}$ only contains curves that are isotopic to ${\varvec{a}}$ or ${\varvec{b}}$. Suppose that there exists $\gamma ' \in {{\,\mathrm{CRS}\,}}(\sigma ((T_aT_b^{-1})^k))$ such that $\gamma$ is not isotopic to a or b. There are two cases.

  • Case 1: ${\varvec{\gamma }}$ intersects ${\varvec{a}}$. or ${\varvec{b}}.$ Since a power of $\sigma ((T_aT_b^{-1})^k)$ fixes $\gamma '$, a power of $(T_aT_b^{-1})^k$ fixes $\gamma$. On the other hand, CRS$((T_aT_b^{-1})^k)=\{a,b\}$. Combined with Lemma 2.3, this shows that $(T_aT_b^{-1})^k$ does not fix $\gamma$. This is a contradiction.

  • Case 2: ${\varvec{\gamma }}$ does not intersect ${\varvec{a}}$ and ${\varvec{b}}$. In this case by the change-of-coordinates principle, there exists a separating curve c on $\Sigma _g$ such that $i(a,c)=0$, $i(b,c)=0$ and $i(c,\gamma )\ne 0$. Assume that $T_c^m\in \Gamma$. Since $(T_aT_b^{-1})^k$ and $T_c^m$ commute in $\Gamma$, the two mapping classes $\sigma ((T_aT_b^{-1})^k)$ and $\sigma (T_c^m)$ commute in $\mathcal I(\Sigma _{g,*})$. Therefore a power of $\sigma (T_c^m)$ fixes CRS($\sigma (T_aT_b^{-1})$); more specifically a power of $T_c^m$ fixes $\gamma$. However by Lemma 2.3, no power of $T_c$ fixes $\gamma$. This is a contradiction.

• Step 2: CRS$({\varvec{\sigma }((T_aT_b^{-1})^k))}$ must contain curves ${\varvec{a}}'$ and ${\varvec{b}}'$ that are isotopic to ${\varvec{a}}$ and ${\varvec{b}}$, respectively. Suppose that CRS$(\sigma ((T_aT_b^{-1})^k))$ does not contain a curve $a'$ isotopic to a. Then by Step 1, we have CRS$(\sigma ((T_aT_b^{-1})^k))$ either contains one curve $b'$ isotopic to b or two curves $b'$ and $b''$ both isotopic to b. After cutting $\Sigma _{g,*}$ along CRS$(\phi ((T_aT_b^{-1})^k))$, there is some component C that is not a punctured annulus. C is homeomorphic to the complement of b in $\Sigma _g$. By the Nielsen–Thurston classification, a power of $\sigma ((T_aT_b^{-1})^k)$ is either pseudo-Anosov on C or else is the identity on C. If a power of $\sigma ((T_aT_b^{-1})^k)$ is pseudo-Anosov on C, then the centralizer of $\sigma ((T_aT_b^{-1})^{k})|_C$ is virtually cyclic by Proposition 2.6. Combining with $T_{b'}$ and $T_{b''}$, the centralizer of $\sigma ((T_aT_b^{-1})^k)$ in $\mathcal I(\Sigma _{g,*})$ is virtually an abelian group of rank at most 3. This contradicts the fact that the centralizer of $\sigma ((T_aT_b^{-1})^k)$ contains a subgroup $\mathbb {Z}^{2g-3}$, since $g \ge 4$ and hence $2g-3>3$. Therefore $\sigma ((T_aT_b^{-1})^k)$ is the identity on C. However, viewing $C = \Sigma _g {\setminus } \{b\}$ as a subsurface of $\Sigma _g$ that contains a, we see that $(T_aT_b^{-1})^k$ is actually not the identity on C; this is a contradiction.

• Step 3:${\varvec{\sigma }((T_aT_b^{-1})^k)=(T_{a'}T_{b'}^{-1})^k(T_{a'}^{-1}T_{a''})^{n}}$, where ${\varvec{n}}$ is an integer and ${\varvec{a}',a'',b'}$ are three disjoint curves on ${\varvec{\Sigma }_{g,*}}$ such that ${\varvec{a}',a''}$ are isotopic to ${\varvec{a}}$ and ${\varvec{b}'}$ is isotopic to ${\varvec{b}}$. Suppose that $\sigma ((T_aT_b^{-1})^k)$ is pseudo-Anosov on some component C of

  $$\begin{aligned} \Sigma _{g,*} {\setminus } {{\,\mathrm{CRS}\,}}(\sigma ((T_aT_b^{-1})^k)) \end{aligned}$$

  Since the genus $g(C)\ge 1$, there exists a separating curve s on C such that $\sigma (T_s^m)$ commutes with $\sigma ((T_aT_b^{-1})^k)$ in $\sigma (\Gamma )$. Therefore, some power of $\sigma ((T_aT_b^{-1})^k)$ fixes CRS($\sigma (T_s^m))$, which is either one curve or two curves isotopic to s. Thus a power of $\sigma ((T_aT_b^{-1})^k)$ fixes some curve on C, which means that $\sigma ((T_aT_b^{-1})^k)$ is not pseudo-Anosov on C. It follows that a power of $\sigma ((T_aT_b^{-1})^k)$ must be a product of Dehn twists about the curves in CRS$(\sigma ((T_aT_b^{-1})^k))$. Since $\sigma ((T_aT_b^{-1})^k)$ is a lift of $(T_aT_b^{-1})^k$, the lemma holds.

The same argument works for $T_c^m\in \Gamma$ the Dehn twist about a separating curve c as long as both components of $\Sigma _g{\setminus } \{c\}$ have genus two or greater. $\square$

In $\phi (T_aT_b^{-1})=(T_{a'})^n(T_{a''})^{1-n}T_{b'}$, denote $(T_{a'})^n(T_{a''})^{1-n}$ by the a component of $\phi (T_aT_b^{-1})$. Notice that without loss of generality, the b component of $\phi (T_aT_b^{-1})$ could also be a product of Dehn twists. In the following lemma, we will prove that the a component of $\phi (T_aT_b^{-1})$ does not depend on the choice of b.

Lemma 3.2

For two bounding pairs $\{a,b\}$ and $\{a,c\}$, the a component of $\phi (T_aT_b^{-1})$ is the same as the a component of $\phi (T_aT_c^{-1})$.

Proof

For b, c disjoint, let us first work on the case when $\phi (T_aT_b^{-1})=(T_{a'})^n(T_{a''})^{1-n}T_{b'}^{-1}$ and $n\ne 0,1$. Then by Proposition 2.5, we know that CRS$(\phi (T_aT_b^{-1}))=\{a',a'',b'\}$. Since $T_aT_b^{-1}$ and $T_aT_c^{-1}$ commutes, we know that CRS$(\phi (T_aT_b^{-1}))$ and CRS$(\phi (T_aT_c^{-1}))$ should disjoint by Proposition 2.4. Therefore $\phi (T_aT_c^{-1})=(T_{a'})^k(T_{a''})^{1-k}T_{c'}^{-1}$ since $a',a''$ are the only curves that are isotopic to a after forgetting and disjoint from $a',a''$. However then

$$\begin{aligned} \phi (T_cT_b^{-1})=\phi (T_aT_b^{-1}))\phi (T_aT_c^{-1})^{-1}=T_{c'}T_{b'}^{-1}(T_{a'})^{n-k}(T_{a''})^{n-k}. \end{aligned}$$

This contradicts Lemma 3.1. We can do a similar argument for case $n=0$.

When b, c intersect, there are a series of curves $\{b_1=b,b_2,...,b_n=c\}$ such that $b_i, b_{i+1}$ are disjoint for $1 \le i \le n-1$ and $a,b_i$ form a bounding pair for $1 \le i \le n$; see Putman [15, Theorem 1.9]. Therefore, the a components of $\phi (T_aT_b^{-1})$ and $\phi (T_aT_c^{-1})$ are the same by applying the above argument $n-1$ times. $\square$

We denote by the capital letter A the subset of curves in CRS$(\phi (T_aT_b^{-1}))$ that are isotopic to a. By Lemma 3.2, A only depends on the curve a. It can be a one-element set or a two-element set.

Lemma 3.3

For $g\ge 4$ and two curves a, b on $S_g$ such that $i(a,b)=0$, we have that A is disjoint from B.

Proof

Suppose that a, b are non-separating. The case of separating curves are the same. If a, b bound a subsurface, then by Lemma 3.1, we know that A and B are disjoint. If a, b do not bound, then there are curves c, d such that they form the following configuration.

Notice that $g\ge 4$ is needed here. Since $\phi (T_aT_c^{-1})$ and $\phi (T_bT_d^{-1})$ commute, their canonical reduction systems do not intersect by Corollary 2.4. Therefore A and B are disjoint. $\square$

3.2 A nonsplitting lemma for the braid groups

Let $\mathbb {D}_n$ be a 2-disk with n marked points. The n-strand pure braid group is denoted by $PB_n$, i.e., the pure mapping class group of ${\mathbb {D}}_n$ fixing the n marked points pointwise. The center $Z_n$ of $PB_n$ is the Dehn twist about the curve surrounding all marked points. Let ${{\mathcal F}_n}: PB_{n+1}\rightarrow PB_n$ be the natural forgetful map forgetting a marked point. In this subsection, we prove a nonsplitting lemma for ${{\mathcal F}_3}$ that will be used in the proof of Theorem 1.1.

Lemma 3.4

There is no homomorphism ${\mathcal S}:PB_3\rightarrow PB_4$ such that Dehn twists are mapped to Dehn twists, ${\mathcal S}(T_{Z_3})=T_{Z_4}$ and ${\mathcal F_3}\circ {\mathcal S}=id$.

Proof

Suppose the opposite that we have ${\mathcal S}:PB_3\rightarrow PB_4$ such that Dehn twists map to Dehn twists, ${\mathcal S}(Z_3)=Z_4$ and ${\mathcal F_3}\circ {\mathcal S}=id$. Let c be a simple closed curve on $\mathbb {D}_3$ and we call $c'$ the curve on $\mathbb {D}_4$ such that ${\mathcal S}(T_c)=T_{c'}$. In Fig. 1, the lantern relation gives $T_aT_bT_c=T_{Z_3}\in PB_3$. Since ${\mathcal S}(T_{Z_3})=T_{Z_4}$, we have $T_{a'}T_{b'}T_{c'}=T_{Z_4}\in PB_4$. If $i(a',b')>2$, then $T_{a'}T_{b'}$ is pseudo-Anosov on a subsurface of $\mathbb {D}_4$ by Thurston’s construction, which contradicts that $T_{a'}T_{b'}=T_{Z_4}T_{c'}^{-1}$ a multitwist; e.g., see Chen [3, Proposition 2.13] for Thurston’s construction. So $i(a',b')=2$. There are several cases we need to concern about the number of points $a'$ and $b'$ surround.

Fig. 1

$T_aT_bT_c=T_{Z_3}$

Fig. 2

Case 1

Fig. 3

Case 2

• Case 1: ${\varvec{a}'}$ bounds 2 points and ${\varvec{b}'}$ bounds 2 points. Then we have $T_{a'}T_{b'}=T_{d'}T_{c''}^{-1}$ for some curve $c',d'$ by lantern relation as in Fig. 2 such that $c''$ surrounds 2 points and $d'$ surrounds 3 points. By applying $\phi$ to the original lantern relation $T_aT_bT_c=T_{Z_3}$, we obtain $T_{a'}T_{b'}T_{c'}=T_{Z_4}$. This gives us that $T_{d'}T_{c''}^{-1}=T_{Z_4}T_{c'}^{-1}$. Thus we have $c''=c'$ and $d'=Z_4$, which contradicts the fact that $d'$ only surrounds 3 points.

• Case 2: ${\varvec{a}'}$ bounds 2 points and ${\varvec{b}'}$ bounds 3 points or ${\varvec{a}'}$ bounds 3 points and. ${\varvec{b}'}$ bounds 3 points By symmetry, we only consider the case that $a'$ bounds 2 points and $b'$ bounds 3 points. Then we have $T_{a'}T_{b'}=T_{Z_4}T_{e'}T_{c''}^{-1}$ by the lantern relation as is shown in Fig. 3 where $c''$ surrounds 3 points and $e'$ surrounds 2 points. By applying $\phi$ to the original lantern relation $T_aT_bT_c=T_{Z_3}$, we obtain $T_{a'}T_{b'}T_{c'}=T_{Z_4}$. Therefore $T_{Z_4}T_{e'}T_{c''}^{-1}=T_{Z_4}T_{c'}^{-1}$, which is not possible.

Without loss of generality, we have exhausted all possibilities. $\square$

3.3 Proof of Theorem 1.1

In this proof, we do a case study on the possibilities of $\phi (T_aT_e^{-1})$ for a bounding pair map $T_aT_e^{-1}$. Case 1 and 2 are when the a component is not a single Dehn twist. Case 3 is when the component of every curve is a single Dehn twist.

Proof

We break our discussion into the following two cases.

Case 1: There is a bounding pair map ${\varvec{T}_aT_e^{-1}}$ such that the ${\varvec{a}}$ component is a multi-twist about ${\varvec{a}',a''}.$ Then there exist curves b, c, d such that a, b, c, d bound a sphere with 4 boundary components as Fig. 4. We need $g\ge 4$ here.

Fig. 4

On $S_g$

Fig. 5

On $S_{g,1}$

There are curves x, y, z such that we have the lantern relation $T_aT_x^{-1}T_bT_y^{-1}T_cT_z^{-1}T_d=1$ as in Fig. 4. The curves $\{b',c',d',x',y',z'\}$ of $\{b,c,d,x,y,z\}$ do not intersect $a',a''$ as in Fig. 5 by Lemma 3.1. The complement of a in $S_g$ and the complement of the annulus formed by $a',a''$ in $S_{g,1}$ are the same surfaces. Therefore there is an identification of curves on those subsurfaces. We call the correspondent curves $b',c',d',x',y',z'$. By observation, they should satisfy lantern relation

$$\begin{aligned} T_{a'}T_{x'}^{-1}T_{b'}T_{y'}^{-1}T_{c'}T_{z'}^{-1}T_{d'}=1. \end{aligned}$$

After applying $\phi$, we have

$$\begin{aligned} (T_{a'})^n(T_{a''})^{1-n}T_{x'}^{-1}T_{b'}T_{y'}^{-1}T_{c'}T_{z'}^{-1}T_{d'}=1. \end{aligned}$$

By computation, we have $(T_{a'})^n(T_{a''})^{1-n}=T_{a'}$, which contradicts the assumption on the components of a.

Case 2: There is a bounding curve ${\varvec{c}}$ such that the ${\varvec{c}}$ component has ${\varvec{2}}$ curves ${\varvec{c}',c''}.$ The proof of Case 1 works for this case as well.

Case 3: For any bounding pair map ${\varvec{T}_aT_e^{-1}}$, we have ${\varvec{\phi }(T_aT_e^{-1})=T_{a'}T_{e'}^{-1}}$ and for any Dehn twist ${\varvec{T}_s}$ about a separating curve ${\varvec{s}}$, we have ${\varvec{\phi }(T_s)=T_{s'}}.$ Let $S_{g,p}^b$ be a genus g surface with p marked points and b boundary components. In this case, firstly we need to locate $*$. Let us decompose the surface into pair of pants as the following Fig. 6.

Fig. 6

A decomposition

With the pants decomposition in Fig. 6, the point $*$ lies in a pair of pants P that is either like A where all three boundary curves are non-separating or like B where one of the boundary curve c is separating and the subsurface cutting out by c containing P is of genus at least 2. We will use the following procedure to find a subsurface S in two different situations.

• When the boundary curves of P are all non-separating, there is a separating curve d bounding a subsurface $S_1$ of genus 2 containing P. Find a curve c on $S_1-P$ such that c, t form a bounding pair. Then a, b, c, d bound a subsurface S which is a sphere with 4 boundary components.

• When one of the boundary curve d of P is separating and the subsurface $S_1$ cutting out by d containing P is of genus at least 2, the other two boundary curves a, t of P form a bounding pair. Find two other non-separating curves b, c on $S_1-P$ such that t, b, c form a pair of pants. Then a, b, c, d bound a subsurface S which is a sphere with 4 boundary components.

Then S satisfies the following:

• $\partial S=a\cup b\cup c\cup d$ such that d is separating and any two of a, b, c does not separate $S_g$

• The lifts $a',b',c',d'$ are 4 disjoint simple closed curves on $S_{g,1}$ such that $d'$ is separating and $a'\cup b'\cup c'\cup d'$ bounds a sphere with 4-boundary components and a puncture $*$ in $S'\approx S_{0,1}^4\subset S_{g,1}$.

Let W be the subgroup of ${\mathcal I}_g$ generated by bounding pair maps about curves inside S. Let $W'$ be the subgroup of ${\mathcal I}_{g,1}$ generated by bounding pair maps about curves inside $S'$. After gluing punctured disks to the boundaries a, b, c and $a',b',c'$ there are forgetful homomorphisms $\mu : W\rightarrow PB_3$ and $\mu ': W'\rightarrow PB_4$. The following claim and Lemma 3.4 concludes the proof of Theorem 1.1.

Claim 3.5

$\mu$ and $\mu '$ are isomorphisms and $\phi :W\cong PB_3\rightarrow W'\cong PB_4$ is a section of the forgetful map ${{\mathcal F}_3}:PB_4\rightarrow PB_3$ satisfying that Dehn twists are mapped to Dehn twists, ${\mathcal S}(T_{Z_3})=T_{Z_4}$ and ${{\mathcal F}_3}\circ {\mathcal S}=id$.

We need to show that $\mu$ is injective and surjective. Firstly, $PB_3$ is generated by Dehn twist about simple closed curves. Every simple closed curves surrounds 2 boundary components in $\mathbb {D}_3$, so forming a bounding pair with the 3rd boundary component. This shows surjectivity.

For injectivity, if $f\in \text {ker}(\mu )$ as a mapping class on S, then f is either trivial or equal to a product of Dehn twists on a, b, c because $\mu$ is forgetful map. However, we claim that a nontrivial product of Dehn twists on a, b, c is never in the Torelli group, which shows that $\mu$ is injective. Assume the opposite and without loss of generality assume that $f=T_a^{m}T_b^nT_c^l\in {\mathcal I}_g$ for $l\ne 0$. Let x, y be two curves or two elements in $H_1(S_g;\mathbb {Z})$ and denote by I(x, y) the algebraic intersection number of x and y. For $x\in H_1(S_g;\mathbb {Z})$, we have that

$$\begin{aligned} T_a^{m}T_b^nT_c^l(x)=mI(a,x)a+nI(b,x)b+lI(c,x)c+x. \end{aligned}$$

The fact that $T_a^{m}T_b^nT_c^l$ is in ${\mathcal I}_g$ implies that $mI(a,x)a+nI(b,x)b+lI(c,x)c=0$ for any x. Since [a], [b] are independent in $H_1(S_g;\mathbb {Z})$, there exists an element x such that $I(a,x)=0$ and $I(b,x)=1$. Since $a\cup b\cup c$ separates the surface, we have that $a+b+c=0$ and $I(c,x)=-1$. Then

$$\begin{aligned} mI(a,x)a+nI(b,x)b+lI(c,x)c=nb-lc=0 \end{aligned}$$

However this contradicts the fact that $[b],[c]\in H_1(S_g;\mathbb {Z})$ are independent and $l\ne 0$.

For the same proof, we can show that $\mu '$ is injective. For surjectivity, any simple closed curve $\gamma$ on $\mathbb {D}_4$ that surrounds 2 punctures has 2 possibilities without loss of generality: (1) $\gamma$ surrounds a, b; (2) $\gamma$ surrounds $a,*$. For (1), we have that $T_{\gamma }T_c^{-1}\in {\mathcal I}_{g,1}$; for (2), we have that $T_{\gamma }T_a^{-1}\in {\mathcal I}_{g,1}$. For both cases, $T_{\gamma }$ is in the image of $\mu '$. Since $PB_4$ is generated by Dehn twists about simple closed curves surrounding 2 punctures, we know that $\mu '$ is surjective. The condition on map $\phi$ is given by the assumption of Case 3. $\square$ $\square$

4 The n-pointed torelli spaces

In this section, we will prove Theorem 4.1. The main tool is Chen [4, Theorem 1.5] and a technical lemma about Torelli action on $\pi _1(S_g)$, which is proved by Johnson homomorphism.

4.1 Translation to a group-theoretic problem

We first translate the “section problem” of the universal Torelli surface bundle into a group-theoretic statement. As is discussed in Chen [4, Section 2.1], we have the following correspondence when $g>1$:

$$\begin{aligned} \left\{ \begin{matrix} \hbox {Conjugacy classes of}\\ \hbox {representations}\\ \rho : \pi _1(B)\rightarrow \text {Mod}(S_g) \end{matrix} \right\} \Longleftrightarrow \left\{ \begin{matrix} \hbox {Isomorphism classes}\\ \hbox {of oriented}\\ S_g-\hbox {bundles over} B \end{matrix} \right\} . \end{aligned}$$

(6)

Let $f:E\rightarrow B$ be a surface bundle determined by $\rho : \pi _1(B)\rightarrow \text {Mod}(S_g)$. Let $f_*:\pi _1(E)\rightarrow \pi _1(B)$ be the map on the fundamental groups. By the property of the pullback diagram, finding a splitting of $f_*$ is the same as finding a homomorphism p that makes the following diagram commute, i.e., $\pi _{g,1}\circ p=\rho$.

(7)

Now we have a new correspondence:

$$\begin{aligned} \left\{ \begin{matrix} \hbox {Homotopy classes of}\\ \hbox {continuous sections of}\\ S_g\rightarrow E\xrightarrow {f} B \end{matrix} \right\} \Longleftrightarrow \left\{ \begin{matrix} \hbox {Homomorphisms} p \hbox {satisfying diagram} ~(7)~ \hbox {up}\\ \hbox {to conjugacy by an element in}\\ \hbox {Ker}(\pi _{g,1})\cong \pi _1(S_g) \end{matrix} \right\} . \end{aligned}$$

(8)

Let ${\mathcal P}{\mathcal I}_{g,n} \xrightarrow {{T\pi _{g,n}}} {\mathcal I}_g$ and ${\mathcal I}_{g,n} \xrightarrow {{T\pi '_{g,n}}} {\mathcal I}_g$ be the forgetful maps forgetting the marked points. Let ${\mathcal I}_{g,n}\xrightarrow {Tp_{g,n,i}} \text {Mod}(S_{g,1})$ be the forgetful homomorphism forgetting the fixed points $\{x_1,...,{\hat{x_i}},...,x_n\}$. Let $PB_n(S_g)$ (resp. $B_n(S_g)$) be the n-strand surface braid group, i.e., $PB_n:={{\,\mathrm{PConf}\,}}_n(S_g)$ (resp. $B_n:={{\,\mathrm{Conf}\,}}_n(S_g)$), where ${{\,\mathrm{PConf}\,}}_n(S_g)$ (resp. ${{\,\mathrm{Conf}\,}}_n(S_g)$) denotes the space of ordered (resp. unordered) n-tuples of distinct points on $S_g$. By the generalized Birman exact sequence (e.g., see Farb and Margalit [7, Theorem 9.1]), we have that $Ker(T\pi _{g,n})\cong PB_n(S_g)$ and $Ker(T\pi '_{g,n})\cong B_n(S_g)$. By the correspondence (8), we can translate Theorem 1.2 into the following group-theoretic statement.

Proposition 4.1

For $g>1$ and $n\ge 0$. The following holds:

(1) Every homomorphism p satisfying the following diagram is either conjugate to a forgetful homomorphism $Tp_{g,n,i}$ by an element in ${\mathcal P}{\mathcal I}_{g,n}$, or else factors through $T\pi _{g,n}$, i.e., there exists $f: {\mathcal I}_g\rightarrow {{\,\mathrm{Mod}\,}}(S_{g,1})$ such that $p=f\circ T\pi _{g,n}$;

(9)

(2) For $n>1$, every homomorphism $p'$ satisfying the following diagram factors through $T\pi _{g,n}'$, i.e., there exists $f':{\mathcal I}_g\rightarrow {{\,\mathrm{Mod}\,}}(S_{g,1})$ such that $p'=f'\circ T\pi _{g,n}'$

(10)

We will prove Proposition 4.1 in the next subsection. Now we finish the proof of Theorem 1.2 using Proposition 4.1.

Proof of Theorem 1.2 assuming Proposition 4.1

The monodromies of bundles

$$\begin{aligned} Tu_{g,n}:{\mathcal U}{\mathcal P}{\mathcal I}_{g,n}\rightarrow {\mathcal P}{\mathcal I}_{g,n}\text { and }Tu'_{g,n}:{\mathcal U}{\mathcal I}_{g,n}\rightarrow {\mathcal B}{\mathcal I}_{g,n} \end{aligned}$$

are the natural projections

$$\begin{aligned} {\mathcal P}{\mathcal I}_{g,n}\rightarrow {\mathcal I}_g\rightarrow {{\,\mathrm{Mod}\,}}(S_g)\text { and }{\mathcal I}_{g,n}\rightarrow {\mathcal I}_g\rightarrow {{\,\mathrm{Mod}\,}}(S_g). \end{aligned}$$

By correspondence (8) and Proposition 4.1, every section of $Tu_{g,n}$ is either a pullback of a section of $Tu_g$ or one of the canonical sections; every section of $Tu'_{g,n}$ for $n\ge 2$ is a pullback of a section of $Tu_g$. However, by Theorem 1.1, the short exact sequence

$$\begin{aligned} 1\rightarrow \pi _1(S_g)\rightarrow {\mathcal I}_{g,1}\xrightarrow {\pi _{g,1}} {\mathcal I}_{g}\rightarrow 1 \end{aligned}$$

has no section, which implies that $Tu_g$ has no section. This concludes the proof of Theorem 1.2$\square$

4.2 The proof of Proposition 4.1

The top exact sequence of diagram (9) gives us a representation

$$\begin{aligned} \rho _T:{\mathcal I}_g\rightarrow \text {Out}(PB_n(S_g)). \end{aligned}$$

The following lemma describes a property of $\rho _T$. Let $p_i: PB_n(S_g)\rightarrow \pi _1(S_g)$ be the induced map on the fundamental groups of the forgetful map forgetting all points except the ith point. The following lemma says that even though ${\mathcal I}_g$ preserves all surjections $\pi _1(S_g)\rightarrow \mathbb {Z}$ but ${\mathcal I}_g$ does not preserve any surjection $\pi _1(S_g)\rightarrow F_h$ for $h>1$, where $F_h$ denotes the free group with h generators.

Lemma 4.2

Let $h>1$. For any surjective homomorphism $\phi : PB_n(S_g)\rightarrow F_h$, there exists an element $t\in {\mathcal I}_g$ such that $t(\text {Ker}(\phi ))\ne \text {Ker}(\phi )$.

Proof

By Chen [4, Lemma 3.5], any surjective homomorphism $\phi :PB_n(S_g)\rightarrow F_h$ factors through some $p_i$. Thus we only need to deal with the case $n=1$. We will prove the lemma by contradiction.

Suppose the opposite that there exists a surjective homomorphism $\phi :\pi _1(S_g)\rightarrow F_h$ such that for any element $e\in {\mathcal I}_g$, we have $e(\text {Ker}(\phi ))= \text {Ker}(\phi )$. Since $\phi$ is surjective, the induced map on $H_1(\_\_,\mathbb {Z})$ is also surjective. Let $a_1,a_2,...,a_h\in \pi _1(S_g)$ be group elements such that $\phi (a_1),...,\phi (a_h)$ generate $F_h$. Since the cup product $H^1(F_h,\mathbb {Z})\otimes H^1(F_h,\mathbb {Z})\xrightarrow {cup} H^2(F_h,\mathbb {Z})$ is trivial, the image of $\phi ^*:H^1(F_h;\mathbb {Z})\rightarrow H^1(S_g;\mathbb {Z})$ is an isotropic subspace with dimension at most g. Thus we can find $b\in \pi _1(S_g)$ such that $\phi (b)=1$ and $[b]\ne 0\in H_1(S_g;\mathbb {Z})$. It is clear that [b] and $\{[a_1],...,[a_h]\}$ are linearly independent in $H_1(S_g;\mathbb {Z})$. Setting $\pi ^0=\pi _1(S_g)$, and $\pi ^{n+1}=[\pi ^n,\pi ^0]$, we have the following exact sequence.

$$\begin{aligned} 1\rightarrow \pi ^1/\pi ^2\rightarrow \pi ^0/\pi ^2\rightarrow \pi ^0/\pi ^1\rightarrow 1. \end{aligned}$$

Define $H:=H_1(S_g;\mathbb {Z})$ and $\omega =\sum _{j=1}^g a_j\wedge b_j$. For the following discussion, we refer the reader to Johnson [9] and Farb and Margalit [7, Section 6.6] for more details. We know that $\pi ^1/\pi ^2\cong \wedge ^2 H/\mathbb {Z}\omega$, where the identification is given by $[x,y]\rightarrow x \wedge y$. Notice that ${\mathcal I}_g$ acts trivially on both $\pi ^1/\pi ^2$ and $\pi ^0/\pi ^1$ but nontrivially on $\pi ^0/\pi ^2$, which is measured by the Johnson homomorphism $\tau :{\mathcal I}_g\rightarrow \text {Hom}(H,\wedge ^2 H/\mathbb {Z}\omega )$. For $t\in {\mathcal I}_g$ and $x\in H$, the Johnson homomorphism is defined by $\tau (t)(x)=t(\tilde{x})\tilde{x}^{-1}\in \pi ^1/\pi ^2$, where $\tilde{x}\in \pi ^0$ is any element such that $[\tilde{x}]=x$. It is standard to check that $\tau (t)$ does not depend on the choice of $\tilde{x}$.

D. Johnson showed that $\tau ({\mathcal I}_g)$ is a subspace of $\text {Hom}(H,\wedge ^2 H/\mathbb {Z}\omega )$ that is isomorphic to $\wedge ^3 H/H$. Under this isomorphism, an element $t\in {\mathcal I}_g$ such that $\tau (t)=b\wedge a_1 \wedge a_2\in \wedge ^3 H/H$ corresponds to an element $\tau (t)\in \text {Hom}(H,\wedge ^2 H/\mathbb {Z}\omega )$ satisfying $\tau (t)(b)=a_1\wedge a_2$. By the definition of the Johnson homomorphism, we have that $t(b)b^{-1}=[a_1,a_2]T$, where $T\in \pi ^2$. Since $\phi (b)=1$, we have that $\phi (t(b))=1$ by the assumption that $t(\text {Ker}(\phi ))= \text {Ker}(\phi )$. As a result, $\phi ([a_1,a_2])\phi (T)=1$. Let $F_h^1=[F_h,F_h]$ and $F_h^{n+1}=[F_h^n,F_h]$. We have that $\phi (\pi ^n)\subset F_h^n$, which implies that $\phi (T)\in F_h^2$. However $\phi ([a_1,a_2])\ne 1\in F_h^1/F_h^2$, which contradicts the fact that $\phi ([a_1,a_2])=\phi (T)^{-1}$. $\square$

A natural question follows Lemma 4.2:

Problem

Does any of the Johnson filtration subgroups ${{\,\mathrm{Mod}\,}}_g(k)$ (as defined in the introduction) preserve a surjective homomorphism $\phi : \pi _1(S_g)\rightarrow F_h$ for $h>1$?

We need the following lemma from Handel and Thurston [8, Lemma 2.2].

Lemma 4.3

(Handel and Thurston [8]) For $g>1$, a pseudo-Anosov element of Mod$(S_{g,n})$ does not fix any nonperipheral isotopy class of curves including nonsimple curves. Equivalently, viewing a mapping class as an outer automorphism of $\pi _1(S_{g,n})$, a pseudo-Anosov mapping class does not preserve any nontrivial conjugacy class in $\pi _1(S_{g,n})$.

To prove statement (2) in Proposition 4.1, we need the following lemma.

Lemma 4.4

For $n>1$, the image of any homomorphism $B_n(S_g)\rightarrow \pi _1(S_g)$ is a free group.

Proof

Suppose that there exists a homomorphism $\Phi :B_n(S_g)\rightarrow \pi _1(S_g)$ such that the image is not a free group. By the classification of subgroups of $\pi _1(S_g)$, we have Image$(\Phi )\cong \pi _1(S_h)$ is a finite index subgroup of $\pi _1(S_g)$. Since $PB_n(S_g)$ is a finite index subgroup of $B_n(S_g)$, we know that $\Phi '=\Phi |_{PB_n(S_g)}:PB_n(S_g)\rightarrow \pi _1(S_g)$ has image a nontrivial finite index subgroup in $\pi _1(S_g)$. By Chen [4, Theorem 1.5], the map $\Phi '$ factors through some $p_i$. Since the betti number of a finite index subgroup of $\pi _1(S_g)$ is bigger than the betti number of $\pi _1(S_g)$, there is no surjection from $\pi _1(S_g)$ to a nontrivial finite index subgroup of $\pi _1(S_g)$. This is a contradiction. $\square$

Now we start the proof of Proposition 4.1.

Proof of Proposition 4.1

For any $p:{\mathcal P} {\mathcal I}_{g,n}\rightarrow {{\,\mathrm{Mod}\,}}(S_{g,1})$ satisfying diagram (9) and for $e\in {\mathcal P}{\mathcal I}_{g,n},x\in PB_n(S_g)$, the restriction map R satisfies

$$\begin{aligned} R(exe^{-1})=p(exe^{-1})=p(e)R(x)p(e)^{-1}. \end{aligned}$$

Denote by $C_e$ the conjugation by e in any group. This induces the following diagram:

(11)

By Chen [4, Theorem 1.5], a homomorphism $R:PB_n(S_g)\rightarrow \pi _1(S_g)$ either factors through a forgetful homomorphism $p_i$ or has cyclic image. By Lemma 4.4, we know that $R': B_n(S_g)\rightarrow \pi _1(S_g)$ is not a surjection. Therefore, the image of $R'$ is either cyclic or a noncyclic free group. We will only discuss R in the following, but exact same reasoning works for $R'$. We break our discussion into the following four cases.

• Case 1: $\mathbf{Im} {\varvec{(}R)=1}$ In this case, p factors through $T\pi _{g,n}$.

• Case 2: Image${\varvec{(}R)\cong \mathbb {Z}}$ In diagram (11), the conjugation map $C_{p(e)}$ corresponds to the outer automorphism induced by e. Then $C_{p(e)}$ preserves Image(R) for any e. However, it is known that ${\mathcal I}_g$ contains pseudo-Anosov elements, which does not preserve Image(R) by Lemma 4.3; e.g., see Farb and Margalit [7, Corollary 14.3] for the fact that the Torelli group contains pseudo-Anosov maps.

• Case 3: $\mathbf{Im} {\varvec{(}R)=F_h}$ for ${\varvec{h}>1}$ We have the following diagram.

  

  (12)

  Therefore p(e) has to preserves the kernel of the surjection R for any $e\in {\mathcal I}_g$, which contradicts Lemma 4.2.

• Case 4: $\mathbf{Im} {\varvec{(}R)=\pi _1(S_g)}$ Therefore $R=A\circ p_i$ for $A\in \text {Aut}(\pi _1(S_g))\cong {{\,\mathrm{Mod}\,}}(S_{g,1})$ by Dehn–Nielsen–Baer Theorem (e.g., see Farb and Margalit [7, Theorem 8.8]). We claim that then $p= A\circ p_i$ and $A\in \pi _1(S_g)$ for $g>2$, which concludes this case. For $e\in {\mathcal I}_{g,n}$ and $f\in PB_n(S_g)$,

  $$\begin{aligned} A\circ p_i(efe^{-1})=p(efe^{-1})=p(e)A(p_i(f))p(e)^{-1}. \end{aligned}$$

  We know that

  $$\begin{aligned} A\circ p_i(efe^{-1})=A(p_i(e)p_i(f)p_i(e)^{-1})=A(p_i(e))A(p_i(f))A(p_i(e)^{-1}) \end{aligned}$$

  Therefore $A(p_i(e))p(e)^{-1}$ is in the centralizer of $A(p_i(f))$ in ${{\,\mathrm{Mod}\,}}(S_{g,1})$, which is trivial for $g\ge 3$. So $A(p_i(e))p(e)^{-1}=i$ the trivial element. So $A(p_i(e))=p(e)$. As an element in ${{\,\mathrm{Mod}\,}}(S_{g,1})$, we have that $A(p_i(e))=Ap_i(e)A^{-1}$. For $x\in {{\,\mathrm{Mod}\,}}(S_{g,1})$, let $\overline{x}$ be its image in ${{\,\mathrm{Mod}\,}}(S_g)$ under the forgetful map. Therefore we have that

  $$\begin{aligned} \overline{A} \overline{p_i(e)} \overline{A}^{-1}=\overline{p_i(e)}\in {{\,\mathrm{Mod}\,}}(S_g) \end{aligned}$$

  because $A(p_i(e))$ and $p_i(e)$ have to be equal in ${{\,\mathrm{Mod}\,}}(S_g)$ for any e. Therefore, we have

  $$\begin{aligned} \overline{A}\in \text {Center} (\text {Mod}(S_g)). \end{aligned}$$

  For $g>2$, $\text {Center}(\text {Mod} (S_g))=1$, therefore we have $A\in \pi _1(S_g)$; e.g., see Farb and Margalit [7, Chapter 3.4].

$\square$

5 Cohomological proof of theorem 1.1

5.1 A nonsplitting statement

In this subsection, we will prove the following. Notice that the proof only depends on Proposition 4.1 and Lefschetz fixed point theorem, which does not use Theorem 1.1. Let $F_i:{\mathcal P}{\mathcal I}_{g,2}\rightarrow {\mathcal P}{\mathcal I}_{g,1}$ be the forgetful map forgetting the ith point.

Lemma 5.1

For $g>1$, the forgetful map $F_2$ does not has a section.

Proof

Assume that $s:{\mathcal P}{\mathcal I}_{g,1}\rightarrow {\mathcal P}{\mathcal I}_{g,2}$ is a section of $F_2$. After composing with $F_1$, we obtain a map $F_1\circ s:{\mathcal P\mathcal I}_{g,1}\rightarrow {\mathcal P}{\mathcal I}_{g,1}$. This map $F_1\circ s$ satisfies the condition of Proposition 4.1 for $n=1$. Then either $F_1\circ s$ factors through ${\mathcal I}_g$ or $F_1\circ s$ is conjugate to a forgetful homomorphism.

Since $S_g$ is a $K(\pi ,1)$-space, any group homomorphism of $\eta : \pi _1(S_g)\rightarrow \pi _1(S_g)$ will induce a geometric map $\widetilde{\eta }:S_g\rightarrow S_g$. Since the map $s({\pi _1(S_g)})\subset PB_2(S_g)$, then some geometric map $\widetilde{F_1\circ s|_{\pi _1(S_g)}}:S_g\rightarrow S_g$ of $F_1\circ s|_{\pi _1(S_g)}$ has no fixed point. If $F_1\circ s$ factors through ${\mathcal I}_g$, then $F_1\circ s|_{\pi _1(S_g)}=i$ the trivial map. The Lefschetz number of the trivial map is 1, which contradicts the fact that $\widetilde{F_1\circ s|_{\pi _1(S_g)}}$ is fixed point free by Lefschetz fixed point theorem. When $F_1\circ s$ is conjugate to identity by an element in $\pi _1(S_g)$, the map $F_1\circ s$ is conjugate to identity by an element in $\pi _1(S_g)$. This means that a geometric map $\widetilde{F_1\circ s|_{\pi _1(S_g)}}:S_g\rightarrow S_g$ is the identity map. The Lefschetz number of the trivial map is $2-2g\ne 0$, which contradicts the fact that $\widetilde{F_1\circ s|_{\pi _1(S_g)}}$ is fixed point free by Lefschetz fixed point theorem. $\square$

5.2 Second proof of Theorem 1.1

We want to point out here that the marked pointed case can help us with the case of no marked point, i.e., Lemma 5.1 can give us another proof of Theorem 1.1. Notice that the proof of Lemma 5.1 does not depend on Theorem 1.1. Let ${\mathcal I}_{g,p}^b$ be the Torelli group of $S_{g,p}^b$, i.e., the subgroup of Mod$(S_{g,p}^b)$ that acts trivially on $H^1(S_g;\mathbb {Z})$.

Second proof of Theorem 1.1

Again assume $g>3$. Assume that the exact sequence (1) has a splitting which is denoted by $\phi$ such that $F\circ \phi =id$. By Lemma 3.1, the image $\phi (T_s)$ of $T_s$ the Dehn twist about a separating curve s is $T_{s'}^nT_{s''}^{1-n}$ where $s'$ and $s''$ are curves on $S_{g,1}$ that are isotopic to s. Let $US_g$ be the unit tangent bundle of genus g surface. Let s be a separating curve that separates $S_g$ into two parts $C_1\cong S_p^1$ and $C_2\cong S_q^1$ such that $p,q \ge 2$. We denote by $C(s)\subset {\mathcal I}_g$ the stabilizer of s by the action of ${\mathcal I}_g$ on curves, which satisfies the following:

$$\begin{aligned} 1\rightarrow \mathbb {Z}\xrightarrow {(T_s,T_s^{-1})} {\mathcal I}_p^1\times {\mathcal I}_q^1\rightarrow C(s) \rightarrow 1. \end{aligned}$$

The disk pushing subgroup is $\pi _1(US_p)\rightarrow {\mathcal I}_p^1$; e.g., see Farb and Margalit [7, Page 118]. The amalgamation of disk pushing subgroups of $C_1$ and $C_2$ gives us a subgroup $A\subset C(s)$ satisfying the following short exact sequence, which has already been encountered when we describe the work of Mess.

$$\begin{aligned} 1\rightarrow \mathbb {Z}\xrightarrow {(T_s,T_s^{-1})} \pi _1(US_p)\times \pi _1(US_q) \rightarrow A\rightarrow 1 \end{aligned}$$

(13)

Claim 5.2

$\phi (T_s)=T_{s'}$ for a curve $s'$ on $S_{g,1}$ that is isotopic to s.

Proof

We have already proved this result in the proof of Case 1 of Theorem 1.1. Here we give another proof. If the s component contains 2 curves, then sections on A in Birman exact sequence will induces a section of (13). We will prove that (13) does not split using Euler class and group cohomology. For a $\mathbb {Z}$-central extension of a group T

$$\begin{aligned} 1\rightarrow \mathbb {Z}\rightarrow \widetilde{T}\xrightarrow {\alpha } T\rightarrow 1, \end{aligned}$$

(14)

there is an associated Euler class $Eu(\alpha )\in H^2(T;\mathbb {Z})$. The extension $\alpha$ splits if and only if $Eu(\alpha )$ vanishes; see Brown [2, Chapter 4.3]. $Eu(\alpha )$ can be constructed using the Lyndon-Hochschild-Serre spectral sequence of (14), by taking $Eu(\alpha )=d_2(1)$. Here $d_2$ is the differential $d_2:\mathbb {Z}\rightarrow H^2(T;\mathbb {Z})$ on the $E_2$ page. The (rational) Betti number $b_1(\tilde{T})$ can be computed from the spectral sequence as

$$\begin{aligned} b_1(\widetilde{T})=b_1(T)+\dim (\ker ( d_2)). \end{aligned}$$

Therefore $Eu(\alpha )\ne 0$ is nonvanishing if and only if $b_1(\widetilde{T})=b_1(T)$. By the above discussion, we only need to show that $b_1(A)=b_1(\pi _1(UT\Sigma _p)\times \pi _1(UT\Sigma _q))$. However, since $p \ge 2$ and $q \ge 2$ by assumption,

$$\begin{aligned} b_1(\pi _1(UT\Sigma _p)\times \pi _1(UT\Sigma _q))=b_1(\pi _1(\Sigma _p)\times \pi _1(\Sigma _q)). \end{aligned}$$

Since $A\rightarrow \pi _1(\Sigma _p)\times \pi _1(\Sigma _q)$ is surjective, it follows that $b_1(A)\ge b_1(\pi _1(\Sigma _p)\times \pi _1(\Sigma _q))$, and so $b_1(A)=b_1(\pi _1(UT\Sigma _p)\times \pi _1(UT\Sigma _q))$ as desired. $\square$

We denote by $C(s')\subset {\mathcal I}_{g,1}$ the stabilizer of $s'$ by the action of ${\mathcal I}_{g,1}$ on curves. Since ${\mathcal I}_{g,1}$ fixes homology, $C(s')$ fixes the two components of $S_{g,1}-s'$. Therefore $C(s')$ satisfies the following exact sequence

$$\begin{aligned} 1\rightarrow \mathbb {Z}\xrightarrow {} {\mathcal I}_p^1\times {\mathcal I}_{q,1}^1\rightarrow C(s')\rightarrow 1. \end{aligned}$$

So we have a section of ${\mathcal F}:{\mathcal I}_{q,1}^1\rightarrow {\mathcal I}_q^1$ which maps $T_{s}$ to $T_{s'}$, which implies that this section gives a section of ${\mathcal F}_{q,2,1}$ in the following commutative diagram.

(15)

However, we already prove that ${\mathcal F}_{q,2,1}$ does not have a section in Lemma 5.1, which implies that ${\mathcal F}$ does not have a section. The statement follows. $\square$