# Self-Similar End Spaces

14 Aug 2022

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 $$(FU)^5$$. $$\blacksquare$$