# 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

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

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

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

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