Cutting and Pasting in the Torelli subgroup of $\operatorname{Out}(F_{n})$

From Laudenbach’s sequence (1), we see that $\operatorname{Out}(F_{n})\cong\operatorname{Mod}(M_{n})/\text{STwist}(M_{n})$, where $\text{STwist}(M_{n})\cong(\mathbb{Z}/2)^{n}$ is the subgroup of $\operatorname{Mod}(M_{n})$ generated by sphere twists. Now that we have related $\operatorname{Out}(F_{n})$ to a geometrically defined group, we can start introducing boundary components. Extending the relationship given by Laudenbach’s sequence, we define “$\operatorname{Out}(F_{n})$ with boundary” as

When $b=1$, Laudenbach [19] also shows that $\mathrm{Out}(F_{n,1})\cong\operatorname{Aut}(F_{n})$. Hatcher-Wahl [14] introduced a more general version of $\mathrm{Out}(F_{n,b})$, which they denoted by $A_{n,k}^{s}$. The original definition of $A_{n,k}^{s}$ has to do with classes of self-homotopy equivalences of a certain graph. However, in [14] the authors give an equivalent definition, which says that $A_{n,k}^{s}$ is the mapping class group of $M_{n}$ with $s$ spherical and $k$ toroidal boundary components, modulo sphere twists. With this definition, we see that $\mathrm{Out}(F_{n,b})=A_{n,0}^{b}$. Similar groups have been examined in the work of Jensen-Wahl [16] and Wahl [26]. Their versions, however, involve only toroidal boundary components, and thus are distinct from $\mathrm{Out}(F_{n,b})$.

An important feature of sphere twists (discussed in Section 2) is that they act trivially on homotopy classes of embedded loops, and thus act trivially on $H_{1}(M_{n})$. Therefore, the action of $\operatorname{Mod}(M_{n,b})$ on $H_{1}(M_{n,b})$ induces an action of $\mathrm{Out}(F_{n,b})$ on $H_{1}(M_{n,b})$. We can then define the Torelli subgroup $IO_{n,b}\subset\mathrm{Out}(F_{n,b})$ to be the kernel of this action. However, this definition does not capture all homological information when $b>1$, especially when $M_{n,b}$ is being embedded in $M_{m,c}$. To see why, consider the scenario depicted in Figure 1, in which $M_{2,2}$ has been embedded into $M_{4}$. This embedding induces a homomorphism $\iota_{M}:\operatorname{Mod}(M_{2,2})\to\operatorname{Mod}(M_{4})$ obtained by extending by the identity. This map sends sphere twists to sphere twists, and so we get an induced map $\iota_{*}:\mathrm{Out}(F_{2,2})\to\operatorname{Out}(F_{4})$. However, this does not restrict to a map $IO_{2,2}\to IO_{4}$ under this definition of $IO_{n,b}$ since elements of $IO_{2,2}$ are not required to fix the homology class of the subarc of $\alpha$ lying inside $M_{2,2}$. To address this issue, we will use a slightly modified homology group.

Returning to Figure 1, let $P$ be the trivial partition of the boundary components of $M_{2,2}$ with a single $P$-adjacency class. With this choice of partition, we see that $[\alpha\cap M_{2,2}]\in H_{1}^{P}(M_{2,2})$. If $f\in IO_{2,2}^{P}$, then it follows that $\iota_{*}(f)\in\operatorname{Out}(F_{4})$ preserves the homology class of $\alpha$. Therefore, $\iota_{*}(f)\in IO_{4}$, and so $\iota_{*}$ restricts to a map $IO_{2,2}^{P}\to IO_{4}$.

As we discussed in the last paragraph, given an embedding $\iota:M_{n,b}\hookrightarrow M_{m}$, we can extend by the identity to get a map $\iota_{*}:\mathrm{Out}(F_{n,b})\to\operatorname{Out}(F_{m})$. In general, $\iota_{*}$ may not be injective. However, it is injective if no connected component of $M_{m}\setminus M_{n,b}$ is diffeomorphic to $D^{3}$ (see Appendix A). Moreover, such an embedding induces a natural partition of the boundary components of $M_{n,b}$ as follows.

From here, we move on to exploring the parallels between these Torelli subgroups and those of mapping class groups. There is a well-known relationship between the mapping class groups of surfaces with a different number of boundary components called the Birman exact sequence (see [12]):

Here, $UT(\Sigma_{n,b-1})$ is the unit tangent bundle of $\Sigma_{n,b-1}$, the map $\pi_{1}(UT(\Sigma_{n,b-1}))\to\operatorname{Mod}(\Sigma_{g,b})$ is given by pushing a boundary component around a loop, and the map $\operatorname{Mod}(\Sigma_{g,b})\to\operatorname{Mod}(\Sigma_{g,b-1})$ is given by attaching a disk onto this boundary component. In Section 4, we will prove versions of the Birman exact sequence for $\operatorname{Mod}(M_{n,b})$ and $\mathrm{Out}(F_{n,b})$, all culminating in the following sequence for $IO_{n,b}^{P}$.

Once we have established this version of the Birman exact sequence, in Section 5, we will define a generating set for $IO_{n,b}^{P}$. This generating set will be inspired by the generating set for $IO_{n}$ found by Magnus [21] in 1935.

Throughout this paper, we will use the convention $[a,b]=aba^{-1}b^{-1}$. Since we defined $IO_{n,b}^{P}$ to be a subgroup of $\operatorname{Mod}(M_{n,b})/\text{STwist}(M_{n,b})$, our generators will be defined geometrically rather than algebraically. However, in the case of $b=0$, they will reduce directly to Magnus’s generators. In Section 6, we will show that these elements do indeed generate $IO_{n,b}^{P}$.

This is rather striking because the analogous result for Torelli subgroups of mapping class groups with multiple boundary components is still open. We will prove this theorem by using the Birman exact sequence to reduce to $b=0$ and applying Magnus’s theorem. Unfortunately, the tools we have constructed do not seem strong enough to give a novel proof of Magnus’s theorem. We will, however, prove a weaker version in Section 7. The original proof of Magnus’s Theorem 1.1 comes in two steps: showing that the given automorphisms $\operatorname{Out}(F_{n})$-normally generate $IO_{n}$, and then showing that the subgroup they generate is normal in $\operatorname{Out}(F_{n})$. We will give a proof of the first step in our setting. For alternative proofs of the first step, as well as more information on the second step in this context, see [5] and [10].

Once we have a finite generating set for $IO_{n,b}^{P}$, a natural question arises: how does the cardinality of this generating set compare to the rank of $H_{1}(IO_{n,b}^{P})$? For $b\leq 1$, this question is answered by a result of Andreadakis [1] and Bachmuth [3].

I would like to thank my advisor Andy Putman for directing me to $\operatorname{Out}(F_{n})$ and its Torelli subgroup, and for his input during the revision process. I would also like to thank Patrick Heslin and Aaron Tyrrell for helpful conversations regarding diffeomorphism groups, as well as Dan Margalit for an enlightening question which resulted in the addition of Section 8.

In section 2, we will give a short overview of sphere twists. We then move on to proving Theorem A in Section 3. We will establish all of our versions of the Birman exact sequence (including Theorem B) in Section 4. In Section 5, we will define our candidate generators for $IO_{n,b}^{P}$, and we will prove that they generate (Theorem C) in Section 6 using the Birman exact sequence and Magnus’s Theorem 1.1. In Section 7, we will prove Theorem D. We then move on to Section 8, in which we recall the definition of the Johnson homomorphism for $IA_{n}$, and use it to compute the rank of the abelianization of $IO_{n,b}^{P}$, proving Theorem E. Finally, we conclude with two appendices. In Appendix A, we provide conditions for a map $\mathrm{Out}(F_{n,b})\to\operatorname{Out}(F_{m})$ induced by an inclusion to be injective, and in Appendix B we prove a lemma which allows us to realize bases of $H_{2}(M_{m})$ as collections of disjoint oriented spheres.

We will frequently direct the reader to figures which are intended to give some geometric intuition for the manifold $M_{n,b}$. In order to assemble $M_{n,b}$, we begin with one or more copies of $S^{3}$, remove a collection of open balls, and then glue the resulting boundary components together in pairs. These gluings will be indicated by double-sided arrows connecting the boundary spheres being glued. As an example, see Figure 2.

Fix a smoothly embedded 2-sphere $S\subset M_{n,b}$, and let $U\cong S\times[0,1]$ be a tubular neighborhood of $S$. Recall that $\pi_{1}(\operatorname{SO}(3),\operatorname{id})\cong\mathbb{Z}/2\mathbb{Z}$, and the nontrivial element $\gamma:[0,1]\to\operatorname{SO}(3)$ is given by rotating $\mathbb{R}^{3}$ one full revolution about any fixed axis through the origin. Fix an identification $S=S^{2}\subset\mathbb{R}^{3}$. Then, we define the sphere twist about $S$, denoted $T_{S}\in\operatorname{Mod}(M_{n,b})$, to be the class of the diffeomorphism which is the identity on $M_{n,b}\setminus U$ and is given by $(x,t)\mapsto(\gamma(t)\cdot x,t)$ on $U\cong S\times[0,1]$. The isotopy class of $T_{S}$ depends only on the isotopy class of $S$. In fact, more is true: Laudenbach [19] showed that the class of $T_{S}$ depends only on the homotopy class of $S$.

Since $\pi_{1}(\operatorname{SO}(3),\operatorname{id})\cong\mathbb{Z}/2\mathbb{Z}$, we see that sphere twists have order at most two. However, it is tricky to show that sphere twists are actually nontrivial because they act trivially on homotopy classes of embedded arcs and surfaces. To see why this is true, let $S\subset M_{n,b}$ be an embedded 2-sphere, and let $U=S\times[0,1]$ be a tubular neighborhood of $S$. Suppose that $\alpha$ is an arc or surface embedded in $M_{n,b}$. We can homotope $\alpha$ such that it is either disjoint from $U$ or intersects $U$ transversely. Let $p\in S$ be one of points in $S$ which lies on the axis of rotation used to construct $T_{S}$. We can homotope $\alpha$ such that $\alpha\cap U$ collapses into $p\in[0,1]$. Note that this process is not an isotopy, and $\alpha$ is no longer embedded in $M_{n,b}$. This is not an issue because a result of Laudenbach [19] shows that if $\alpha$ is fixed up to homotopy, then it is fixed up to isotopy. Since $T_{S}$ fixes $p\times[0,1]$ pointwise, it follows that $T_{S}$ fixes $\alpha$ up to homotopy. The upshot of this is that a more sophisticated invariant must be constructed to detect the nontriviality of $T_{S}$. In [18, 19], Laudenbach uses framed cobordisms to show that for $b=0,1$, the sphere twist $T_{S}$ is trivial if and only if $S$ is separating. In the case of no boundary components, Brendle, Broaddus, and Putman [6] give another proof of this fact by showing that sphere twists act nontrivially on a trivialization of the tangent bundle of $M_{n}$ up to isotopy.

Let $\text{STwist}(M_{n,b})\subset\operatorname{Mod}(M_{n,b})$ be the subgroup generated by sphere twists. Given $\mathfrak{f}\in\operatorname{Mod}(M_{n,b})$ and a sphere twists $T_{S}$, we have the “change of coordinates” formula

This shows that $\text{STwist}(M_{n,b})$ is a normal subgroup of $\operatorname{Mod}(M_{n,b})$. In fact, even more is true. Letting $\mathfrak{f}=T_{S^{\prime}}$ in the above formula and using the fact that sphere twists act trivially on embedded surfaces up to isotopy, we find that

which implies $\text{STwist}(M_{n,b})$ is actually abelian. Since nontrivial sphere twists have order two, it follows that $\text{STwist}(M_{n,b})$ is isomorphic to a product of copies of $\mathbb{Z}/2\mathbb{Z}$. For $b=0,1$, another result of Laudenbach shows that $\text{STwist}(M_{n,b})\cong(\mathbb{Z}/2\mathbb{Z})^{n}$ and is generated by the sphere twists about the $n$ core spheres $*\times S^{2}$ in each $S^{1}\times S^{2}$ summand. For $b>1$, one can show that $\text{STwist}(M_{n,b})\cong(\mathbb{Z}/2\mathbb{Z})^{n+b-1}$. The $-1$ in the exponent reflects the fact that the product of all the sphere twists about boundary components is trivial. Since we will need this fact later, we include a proof here.

If $b=1$, this shows that the sphere twist about the boundary component is trivial. However, if $b>1$, then the sphere twists about boundary components are nontrivial. We will also need this fact, so we prove it here.