Automatic Continuity of Pure Mapping Class Groups

Let $S_{\omega}$ denote the symmetric group on a countably infinite set. This is a Polish group with the compact-open topology. Any countable index subgroup $H$ determines a homomorphism $\phi$ to $S_{\omega}$ by the left multiplication action on left cosets. The subset of $S_{\omega}$ corresponding to permutations that fix $H$ is open. The pullback of this subset via $\phi$ is $H$, which is open by automatic continuity. ∎

Identify $\Lambda$ with $\mathbb{Q}$ via a bijection, and let $\Gamma=\mathbb{R}\setminus\mathbb{Q}$. For any given $\alpha\in\Gamma$, let $\Omega_{\alpha}$ be any sequence of rational numbers converging to $\alpha$. ∎

Let $\gamma_{1},\gamma_{2},...$ be an infinite sequence of irrationals independent over $\mathbb{Q}$. Now $\{\gamma_{n}\mathbb{Q}\}_{n=1}^{\infty}$ is a collection of pairwise disjoint dense subsets of the reals. Identify $\Lambda$ via a bijection with $\mathbb{N}$, and then identify each $\{n\}\times\Lambda^{\prime}$ by a bijection with $\gamma_{n}\mathbb{Q}$. Let $\Gamma=\mathbb{R}\setminus(\bigcup_{n=1}^{\infty}\gamma_{n}\mathbb{Q})$. For any given $\alpha\in\Gamma$, choose any sequence of numbers in $\bigcup_{n=1}^{\infty}\gamma_{n}\mathbb{Q}$ converging to $\alpha$ that includes infinitely many entries of $\gamma_{n}\mathbb{Q}$ for all $n$, and then let $\Omega_{\alpha}$ be the corresponding set of tuples $(n,\gamma_{n}q_{n})$ where $q_{n}\in\mathbb{Q}$. ∎

Apply Lemma 3.7 with $\mathcal{A}$ consisting of subsurfaces of the form $A_{\Lambda}=\cup_{n\in\Lambda}M_{2n}$ for some infinite set $\Lambda\subset\mathbb{Z}$. Any collection of the $M_{i}$ is admissible since we can extend a homeomorphism on the collection via the identity map to $S$. Note $\mathcal{A}$ satisfies the hypotheses of the lemma, since

1. (i)

   We can write $\mathbb{Z}$ as a countable disjoint union of infinite sets $\Lambda_{i}$, and define $U_{i}$ to be $\bigcup_{n\in\Lambda_{i}}M_{2n}$. Each such set contains a countable union of disjoint elements of $\mathcal{A}$.

2. (ii)

   Lemma 3.8 implies the same statement for the mapping class group of a disjoint union of sliced Loch Ness monsters. Thus, any element supported in $A_{\Lambda}$ may be written as the product of two commutators.

We conclude that for some such subsurface $A_{\Lambda}\in\mathcal{A}$, we have $\operatorname{Map}(A_{\Lambda})\subset W^{16}$. Now we apply Lemma 3.5. Let $\{\Omega_{\alpha}\}$ be an uncountable collection of infinite subsets of $\Lambda$ such that $\Omega_{\alpha}\cap\Omega_{\beta}$ is finite for all $\alpha\neq\beta$. Note we may assume $\Lambda$ and each $\Omega_{\alpha}$ contain infinitely many negative and positive integers.

Since all homeomorphisms are assumed to fix the boundary pointwise, we first modify each $M_{i}$ by deleting a small regular open neighborhood of the $x$-axis so that we can move them into one another with an appropriate homeomorphism. For each $\alpha$, let $f_{\alpha}$ be a homeomorphism supported in $M$ with the following property. For each $n\in\Omega_{\alpha}$, let $f_{\alpha}(M_{2n})$ be the smallest connected subsurface containing the union of $M_{2n+1},M_{2n+2},...,M_{2k-1}$ where $k\in\Omega_{\alpha}$ is the smallest element in $\Omega_{\alpha}$ larger than $n$, so that $f_{\alpha}$ maps $A_{\Omega_{\alpha}}$ into the complementary region. Also let $f_{\alpha}$ map the union of $M_{2n+1},M_{2n+2},...,M_{2k-1}$ into $M_{2k}$. Note this homeomorphism exists by the change of coordinates principle. Since $\{\Omega_{\alpha}\}$ is uncountable, there are some $\alpha$ and $\beta$ such that $f_{\alpha}$ and $f_{\beta}$ are in the same left translate $gW$ for some $g\in\overline{\operatorname{PMap}_{c}(S)}$. Therefore, $f_{\alpha}^{-1}f_{\beta}$ and $f_{\beta}^{-1}f_{\alpha}$ are both in $W^{2}$.

If $n\notin\Omega_{\alpha}$, then $f_{\alpha}(M_{2n})\subseteq M_{2m}$ for some $m\in\Omega_{\alpha}$. If $m\notin\Omega_{\beta}$, then $f_{\beta}^{-1}f_{\alpha}(M_{2n})$ is contained in some $M_{2k}$ where $k\in\Omega_{\beta}$. Since $\Omega_{\alpha}\cap\Omega_{\beta}$ is finite, we conclude that, with the exception of finitely many values of $n\notin\Omega_{\alpha}$, the map $f_{\beta}^{-1}f_{\alpha}$ takes $M_{2n}$ into $A_{\Omega_{\beta}}\subset A_{\Lambda}$.

Reversing the role of $\alpha$ and $\beta$, the same argument shows that with only finitely many exceptions of $n\notin\Omega_{\beta}$, $f^{-1}_{\alpha}f_{\beta}$ takes every $M_{2n}$ into $A_{\Lambda}$. Let $F^{\prime}$ be the union of these two exceptional sets of integers. Now write $\bigcup_{n\in\mathbb{Z}}M_{2n}$ as the union of $X_{1}=\bigcup_{n\notin(\Omega_{\alpha}\cup F^{\prime})}M_{2n}$, $X_{2}=\bigcup_{n\notin(\Omega_{\beta}\cup F^{\prime})}M_{2n}$, and $X_{3}=\bigcup_{n\in F^{\prime}}M_{2n}$.

It follows that $\operatorname{Map}(X_{1})$, $\operatorname{Map}(X_{2})\subset W^{20}$, so $\operatorname{Map}(X_{1}\cup X_{2})\subset W^{40}$. Now we can complete the proof by repeating the above argument to the union of the odd $M_{n}$. ∎

We show the case where $\{S_{n}\}_{n\in\mathbb{N}}$ is infinite since the finite case is similar. Apply Lemma 3.7 with $\mathcal{A}$ consisting of subsurfaces of the form $A_{\Lambda}=\bigcup_{n\in\mathbb{N},m\in\Lambda}M_{n,2m}$ to show $\operatorname{Map}(A_{\Lambda})\subset W^{16}$ for some infinite $\Lambda\subset\mathbb{Z}$. Apply Lemma 3.6 to $\mathbb{N}\times\Lambda$, so that $\{\Omega_{\alpha}\}_{\alpha\in\mathbb{R}}$ are infinite subsets of $\mathbb{N}\times\Lambda$ with the properties listed in the lemma. Note we may assume $\mathbb{N}\times\Lambda$ and each $\Omega_{\alpha}$ contain infinitely many positive and negative integers in the second coordinate.

Modify all of the standard pieces slightly as before, then for each $\alpha$, let $f_{\alpha}$ be a map supported in $\bigcup_{i\in\mathbb{N}}S_{i}$ with the following property. For all $(n,m)\in\Omega_{\alpha}$, let $f_{\alpha}(M_{n,2m})$ be the smallest connected subsurface containing the union of $M_{n,2m+1},M_{n,2m+2}$,…, $M_{n,2k-1}$ where $k$ is the smallest second component among the elements of $\Omega_{\alpha}\cap(\{n\}\times\Lambda)$ larger than $m$. Now the proof is completed as before. ∎

First, we explain the details for the sliced Loch Ness monster and then discuss how to extend the argument to the other cases.

Let $f\in\operatorname{Map}(S_{K})$ be any element. Let $g$ be one of the maps produced by applying Lemma 3.11 to $f$, and assume the conditions of the second part of Lemma 3.11 hold for $g$. Note here we are using Lemma 3.1 which implies $\operatorname{Map}(S_{K})=\overline{\operatorname{PMap}_{c}(S_{K})}$. Let $\{K_{i}\}$ be the sequence of compact subsurfaces containing the support of $g$. Let $\{M_{i}\}$ be the collection of standard pieces for $S_{K}$. We can assume by applying change of coordinates that each $M_{i}$ contains exactly one of the $K_{i}$, and each $K_{i}$ appears in some $M_{i}$. By Lemma 3.9, we have some cofinite union $T=\bigcup_{i\in\mathbb{Z}\setminus F}M_{i}$ with $\operatorname{Map}(T)\subset W^{80}$. Therefore, we can find some $g^{\prime}\in\operatorname{Map}(T)$ such that $g^{\prime}g\in\operatorname{PMap}_{c}(S_{K})$. Now let $K^{\prime}\subset S_{K}$ be a compact subsurface bounded by a single curve that contains the support of $g^{\prime}g$. Now using the density of $W^{2}$ in $\operatorname{Map}(S_{K})$, find some element $\phi\in W^{2}$ such that $\phi(K^{\prime})\subset T$. It follows that $\phi g^{\prime}g\phi^{-1}\in W^{80}$, and therefore $g^{\prime}g\in W^{84}$. Finally, this gives $g\in W^{164}$, and since the above argument also applies to the other element from fragmentation, $f\in W^{328}$.

Let $f\in\operatorname{Map}(S_{K})$ be any element. Let $g$ be one of the maps produced by applying fragmentation to $f$, and assume the conditions of the second part of Lemma 3.11 hold for $g$ when restricted to each component of $S_{K}$. In this case, we need to apply Lemma 3.11 to each component separately and then combine. Let $\{K_{i}\}$ be the collection of compact subsurfaces containing the support of $g$. Using Lemma 2.2, we can cut $S_{K}$ along a collection of disjoint arcs into sliced Loch Ness monsters $\{S_{n}\}$. We can also assume these arcs are chosen to miss the $K_{i}$.

Let $\{M_{n,m}\}$ be the collection of standard pieces for $S_{n}$. By change of coordinates, we can assume each $M_{n,m}$ contains exactly one $K_{i}$, and each $K_{i}$ appears in some $M_{n,m}$. By Lemma 3.10, there is a finite set $F$ such that $T=\bigcup_{(n,m)\notin F}M_{n,m}$ and $\operatorname{Map}(T)\subset W^{80}$. Proceed as before.

Let $W$ be a countably syndetic symmetric subset of $\operatorname{PMap}(S)$. Let $U$ be an open neighborhood of the identity such that $W^{2}$ is dense in $U$, and find some compact subsurface $K$ such that $\operatorname{Stab}(K)\subseteq U$. As before we can enlarge $K$ if needed so that $S\searrow K$ is homeomorphic to a disjoint union of finitely many disks with handles. Since there are finitely many ends accumulated by genus, we can further assume that each component of $S\searrow K$ is a sliced Loch Ness monster. This ensures that $\operatorname{Stab}(K)\subset\overline{\operatorname{PMap}_{c}(S)}$ so we can use fragmentation as in the proof of Proposition 4.1 to show $\operatorname{Stab}(K)\subset W^{328}$. The proof for $\operatorname{Map}(S)$ is identical. Note that we must also use a minor adaptation of Lemma 3.7 where $W$ is a countably syndetic symmetric subset of $\operatorname{PMap}(S)$ or $\operatorname{Map}(S)$ instead of $\overline{\operatorname{PMap}_{c}(S)}$. ∎

Recall the reverse direction was shown in Proposition 4.1. When $S$ either has infinitely many interior planar ends, infinitely many compact boundary components, infinitely many boundary chains, or at least one interior end accumulated by genus we will show one of the conditions of Lemma 4.3 is satisfied so that $\overline{\operatorname{PMap}_{c}(S)}$ does not have automatic continuity. After possibly filling in the finite number of punctures and capping the finite number of compact boundary components, we apply Lemma 2.3 to conclude $S$ is a connected sum of finitely many disks with handles. The original $S$ can then be obtained by connect summing with a finite-type surface with punctures and boundary components.

Recall the reverse direction was shown in Proposition 4.2. Now we consider two cases, and then we are done by using Lemma 2.3 as in Theorem B.

Recall the Loch Ness monster case was shown by Domat and the author [6], so we can assume the interior of $S$ has at least two ends. By Lemma 4.3, $\overline{\operatorname{PMap}_{c}(S)}$ has a nonopen countable index subgroup. Since $\operatorname{PMap}(S)$ is finite index in $\operatorname{Map}(S)$ by the assumption of finitely many ends, we can apply the same proof as the second case of Theorem A to show $\operatorname{Map}(S)$ does not have automatic continuity. ∎

Suppose $S$ is a disk with handles. Proceeding by contradiction, let $H$ be the kernel of a nontrivial map from $\overline{\operatorname{PMap}_{c}(S)}$ to a countable group with the discrete topology. By automatic continuity, $H$ is open and closed. Since $H$ is closed, it suffices to show that it contains every compactly supported mapping class since then $H$ is dense in $\overline{\operatorname{PMap}_{c}(S)}$, and in fact $H=\overline{\operatorname{PMap}_{c}(S)}$. Since $H$ is open, it contains $\operatorname{Stab}(K)$ for some compact subsurface $K$. Now let $\phi$ be any compactly supported mapping class. Since $\phi$ fixes the boundary pointwise, we can isotope it so that it is supported in a subsurface $K^{\prime}$ that does not intersect the boundary. Since $S$ is a disk with handles, its interior is a Loch Ness monster, and there exists some homeomorphism supported in the interior that takes $K^{\prime}$ into the complement of $K$. Therefore, a conjugate of $\phi$ lies in $\operatorname{Stab}(K)\subset H$, and we are done since $H$ is normal. There are no proper finite index subgroups in $\overline{\operatorname{PMap}_{c}(S)}$ since any index $n$ subgroup determines a nontrivial homomorphism to the symmetric group on $n$ elements via the left multiplication action on the left cosets. ∎