Rylee Alanza Lyman

This is the third post in a series on Katie Mann and Kasra Rafi’s paper Large-scale geometry of big mapping class groups. In the previous post we explored an obstruction to coarse boundedness, while in this one we will introduce a sufficient condition for $\operatorname{Map}(\Sigma)$ to be coarsely bounded. This post includes proofs that closely follow those in the paper, which while elementary, get a little involved. I imagine this will be one of very few moments in the series where I give full proofs.

Necessary conditions

Suppose that $\operatorname{Map}(\Sigma)$ is coarsely bounded. Then by the contrapositive of the result of the previous post, if $S$ is a finite-type subsurface of $\Sigma$ , there must be a homeomorphism $f\colon \Sigma \to \Sigma$ satisfying $f(S) \cap S = \varnothing$ . Therefore by the examples given at the end of the previous post, if $\Sigma$ has finitely many maximal ends, it has at most two, and if $\Sigma$ has nonzero genus, it has infinite genus. In fact, by a variant of the former condition, if $X \subset \operatorname{Ends}(\Sigma)$ is a finite, $\operatorname{Map}(\Sigma)$ -invariant set, it has cardinality at most two. (I got worried here a little: what if the invariant ends are isolated and planar, i.e. punctures? But then I realized: any subsurface containing all the finitely many punctures of $\Sigma$ would be nondisplaceable.)

Self-similarity

Towards the construction of sufficient conditions for $\operatorname{Map}(\Sigma)$ to be coarsely bounded, we introduce the property of self-similarity of $\operatorname{Ends}(\Sigma)$ : recall that $\operatorname{Ends}(\Sigma)$ is a closed subset of the Cantor set, so the most natural open neighborhoods to work with are clopen (closed and open) neighborhoods. Supposing we have a decomposition

$\operatorname{Ends}(\Sigma) = E_1 \sqcup E_2$

of $\operatorname{Ends}(\Sigma)$ into clopen subsets, then self-similarity of $\operatorname{Ends}(\Sigma)$ is the condition that one of the subsets, say $E_1$ , contains a clopen subset $D$ such that the pair $(D,D \cap \operatorname{Ends}^g(\Sigma))$ is homeomorphic to the pair $(\operatorname{Ends}(\Sigma),\operatorname{Ends}^g(\Sigma))$ . Note that we can further split $E_1$ into two, and so on, so this condition is clearly equivalent to the condition that whenever

$\operatorname{Ends}(\Sigma) = E_1 \sqcup \cdots \sqcup E_n$

is a decomposition into finitely many clopen subsets, one of them, say $E_1$ , contains a clopen subset $D$ as above.

Anyway, suppose that $\Sigma$ has infinite or zero genus and self-similar end space. We first show that each finite-type subsurface $S$ of $\Sigma$ is displaceable in a strong sense: after enlarging $S$ , we may suppose that each boundary component of $S$ is separating (i.e. removing it cuts $\Sigma$ into two pieces) and complementary component of $\Sigma - S$ has infinite type and infinite or zero genus; the complementary components in $\Sigma - S$ together with the finite set of punctures of $S$ partition $\operatorname{Ends}(\Sigma)$ into finitely many clopen subsets, so some complementary component, call it $C_0$ , has end space containing a clopen neighborhood homeomorphic to $\operatorname{Ends}(\Sigma)$ (respecting the subspaces of nonplanar ends).

Lemma 3.2 (Mann–Rafi). With notation as above, $C_0$ contains a finite-type subsurface $R$ disjoint from but homeomorphic to $S$ via a homeomorphism $f \colon \Sigma \to \Sigma$ satisfying $f(S) = R$ , $f(C_0) \supset S$ and such that the end set of $f(C_0) \cap C_0$ contains a homeomorphic copy of $\operatorname{Ends}(\Sigma)$ (respecting the subspaces of nonplanar ends).

Before we sketch the proof, here’s an example. The surface $\Sigma$ has a Cantor space of ends, of which exactly one is accumulated by genus. This end space is self-similar, because any clopen subset of a Cantor set is itself a Cantor set, and after chopping the end space into finitely many pieces, one piece contains the nonplanar end, and that piece is therefore homeomorphic to the whole end space. Suppose $S$ is a pair of pants (a genus-zero subsurface with three boundary components) cutting the end space into three pieces. The claim is that within the piece that contains the nonplanar end, there is a pair of pants $R$ disjoint from $S$ , a homeomorphism of $\Sigma$ taking $S$ to $R$ and the “big” complementary component of $S$ to the “big” complementary component of $R$ (i.e. the ones containing the nonplanar end). In this case, the intersection of the “big” components is again homeomorphic to a Cantor set of ends with one nonplanar end. By the way, Ian Biringer suggested the name “Cantor tree in early spring” for the surface $\Sigma$ ; I’m a fan.

Proof. The proof is ultimately an appeal to the Kerékjártó–Richards classification of surfaces. Let $C_0,\ldots,C_n$ be the components of $\Sigma - S$ , where $C_0$ is the indicated “big” component. Write $E_i$ for the end set of $C_i$ , and $P_S$ for the set of punctures of $S$ . We have

$\operatorname{Ends}(\Sigma) = E_0 \sqcup \cdots \sqcup E_n \sqcup P_S.$

Since the end set $E_0$ of $C_0$ contains a clopen subset $D$ homeomorphic to $\operatorname{Ends}(\Sigma)$ respecting the subset of nonplanar ends, we may find a decomposition of $D$ as

$D = E'_0 \sqcup \cdots \sqcup E'_n \sqcup P_R,$

where each $E'_i$ is homeomorphic (respecting nonplanar ends) to $E_i$ , and $P_R$ is a finite collection of isolated planar ends of the same cardinality of $P_S$ . There are two things to claim. The first is that we can find a finite-type subsurface $R$ disjoint from $S$ with genus equal to that of $S$ , puncture set $P_R$ and $n+1$ boundary components partitioning the end space of $\Sigma$ into $E'_1,\ldots, E'_n$ , along with $P_R$ and the remaining ends, namely

$E^R_0 = E'_0 \sqcup (E_0 - D) \sqcup E_1 \sqcup \cdots \sqcup E_n \sqcup P_S.$

It is clear (replace $P_S$ with $P_R$ , erase the primes and use the decomposition of $D$ above) that this latter set of ends $E^R_0$ is actually homeomorphic to $E_0$ respecting the subspace of nonplanar ends. Depending on your comfort level with arguing about subsurfaces, this may take some thought. I further claim that we can do this so that each complementary component of $\Sigma - R$ also has infinite or zero genus (necessarily according to whether the “corresponding” component of $\Sigma - S$ does). For me the key observation in both claims was that I really could shuffle finite amounts of genus around to arrange this.

Since $R$ and $S$ have the same number of punctures, boundary components and the same genus, there is a homeomorphism from $S$ to $R$ ; we may take it to match up the boundary components according to the labeling of the complementary regions, so for example the boundary component corresponding to $E_0$ is sent to the component corresponding to $E^R_0$ . Similarly, for each complementary component of $\Sigma - S$ , the corresponding complementary component of $\Sigma - R$ has the same genus and homeomorphic end space, so by the classification of surfaces, the given homeomorphism of the end spaces extends to a homeomorphism sending the complementary component of $\Sigma - S$ to the corresponding component of $\Sigma - R$ . These $n + 2$ homeomorphisms glue together to yield the desired homeomorphism $f \colon \Sigma \to \Sigma$ . By construction, we have $f(C_0) \supset S$ and since $f(C_0) \cap C_0$ contains $E'_0$ in its end set, it contains a homeomorphic copy of $\operatorname{Ends}(\Sigma)$ respecting the subspace of nonplanar ends. $\blacksquare$

Note that since we only ever enlarged $S$ to get it to satisfy all the technical things we desired, the lemma is actually true for smaller finite-type subsurfaces as well. Anyway, the lemma is the main tool to prove the following theorem.

Proposition 3.1 (Mann–Rafi). Suppose $\Sigma$ has self-similar end space and infinite or zero genus. Then $\operatorname{Map}(\Sigma)$ is coarsely bounded.

Proof. We aim to show that for any open neighborhood $U$ of the identity in $\operatorname{Map}(\Sigma)$ , there exists a finite set $F$ and $k \ge 1$ such that $\operatorname{Map}(\Sigma) \subset (FU)^k$ . Without loss of generality, we may assume that $U$ is

$U_S = \{[g] \in \operatorname{Map}(\Sigma) : g|_S = 1\}$

for some finite-type subsurface $S$ satisfying the hypotheses and hence the conclusions of Lemma 3.2 above (note that enlarging $S$ shrinks $U_S$ , which is good for our purposes). That is, we have a subsurface $R$ disjoint from $S$ and a homeomorphism $f\colon \Sigma \to \Sigma$ taking $S$ to $R$ . The finite set $F$ will be the mapping classes of the identity, $f$ and $f^{-1}$ . It suffices to take $k = 5$ .

Take $[g] \in \operatorname{Map}(\Sigma)$ . If $g(S) = S$ and $g|_S = 1$ , we are done, since $g \in U$ , so certainly in $(FU)^5$ . If $g(R) = R$ and $g|_R = 1$ , then since $f(S) = R$ , we have $f^{-1}gf(S) = f^{-1}g(R) = f^{-1}(R) = S$ and $f^{-1}gf|_S = 1$ . In this case we see that $g \in FUF \subset (FU)^2$ . This illustrates the strategy: if we can, using finitely many elements of $F$ and $U$ , reduce $g$ to the case where it is the identity on $R$ or $S$ , we win. Now, in the notation of Lemma 3.2, the end set $E_0$ of $C_0$ satisfies the condition that $E_0 \cap f(E_0)$ contains a homeomorphic copy $E'$ of $\operatorname{Ends}(\Sigma)$ . Since $f(E_0) \cup E_0 = \operatorname{Ends}(\Sigma)$ , we have a decomposition

$\operatorname{Ends}(\Sigma) = (E_0 - f(E_0)) \sqcup (E_0 \cap f(E_0)) \sqcup (f(E_0) - E_0),$

so the intersection of one of these sets with the self-similar set $g(E') \cong \operatorname{Ends}(\Sigma)$ contains a clopen subset $E''$ homeomorphic to $\operatorname{Ends}(\Sigma)$ . At the cost of one application of $f^{-1}$ , we may assume that $E''$ is contained in $E_0$ , the end set of the “big” complementary component $C_0$ of $\Sigma - S$ .

Now, notice that $C_0$ and $R$ taken by themselves satisfy the hypotheses and hence the conclusions of Lemma 3.2. This is key: arguing using the “big” collection of ends $E''$ , we can find a subsurface $R'$ homeomorphic to $R$ in $C_0 \cap g(C_0)$ and a homeomorphism $s\colon C_0 \to C_0$ satisfying $s(R) = R'$ and $R \subset s(C_0 \cap f(C_0))$ . By choosing $s$ carefully, we may extend it to a homeomorphism $\tilde s\colon \Sigma \to \Sigma$ by declaring it to be the identity on $\Sigma - C_0$ . We have $\tilde s \in U$ .

The advantage of having this $R'$ is that we may “undo” $g$ on $S$ while fixing $R'$ via a homeomorphism $u\colon \Sigma \to \Sigma$ satisfying $ug(S) = S$ and $ug|_S = 1$ . This is again a classification of surfaces argument: the point is that the decomposition of $\operatorname{Ends}(\Sigma)$ given by $S$ is equivalent to that provided by $g(S)$ (the homeomorphism $g$ sends one to the other) and both $C_0$ and $g(C_0)$ contain $R'$ , so after cutting away $R'$ , the end set decompositions are still equivalent, so there is a homeomorphism returning $g(S)$ to $S$ while fixing $R'$ , and postcomposing with a homeomorphism supported on $S$ , we may suppose this homeomorphism has the desired properties of $u$ .

So let’s put it together. We have $ug \in U$ , so we win if we can write $u$ using elements of $F$ and $U$ . But since $u$ is the identity on $R'$ , the mapping class of $(\tilde s f)^{-1}u(\tilde sf)$ is the identity on $(\tilde s f)^{-1}(R') = S$ . We have $(\tilde s f)^{-1}u(\tilde s f) \in U$ , so $u \in UFUFU \subset (FU)^3$ , which implies that $g \in (FU)^4$ . But remembering that we had to pay an application of $f^{-1}$ a few paragraphs back, we see that the original $g$ is contained in $(FU)^5$ . $\blacksquare$