This is the sixth post in a series on Katie Mann and Kasra Rafi’s paper Large-scale geometry of big mapping class groups. The purpose of this post is to discuss when a Polish group is generated by a coarsely bounded set, and give examples of mapping class groups which are locally coarsely bounded but fail this criterion.

## The general case

We have the following theorem of Rosendal.

Theorem 1.2(Rosendal). Let $G$ be a Polish group. Then $G$ is generated by a coarsely bounded set if and only if $G$ is locally coarsely bounded and not the countably infinite union of a chain of proper, open subgroups.

Let’s make sense of this. Note that countable, discrete groups are Polish. Being generated by a coarsely bounded set is the analogue of finite generation for discrete groups, so we should expect that for a countable discrete group (which is automatically locally coarsely bounded, having finite discrete neighborhoods of the identity) being finitely generated is equivalent to not being the countably infinite union of a chain of proper subgroups, which are automatically open.

Indeed, suppose $G$ is finitely generated by a set $S$ , and that $G_1 < G_2 < G_3 < \cdots$ is a strictly increasing chain of proper subgroups. Since the union of the $G_i$ is $G$ , each $s \in S$ belongs to some $G_i$ , but then at some finite stage $G_n$ , we have each $s \in S$ contained in $G_n$ , so actually $G_n = G$ .

Conversely, supposing every strictly increasing chain of proper subgroups terminates, we show that $G$ is finitely generated. Indeed, we can take a sequence of finitely generated subgroups! Begin with $G_1 = \langle s_1 \rangle$ for some $s_1 \in G$ . At each stage, add $s_{n+1} \notin G_n$ and take $G_{n+1} = \langle s_1,\ldots,s_{n+1}\rangle$ . Since this sequence terminates, we’ve proven that $G$ is finitely generated. (This proof contains the useful fact that every generating set for a finitely generated group contains a \emph{finite} generating set.)

## A non-example: limit type

Remember the “Great Wave off Kanagawa” surface from the previous post? It had genus zero and end space homeomorphic to $\omega^\omega + 1$ in the order topology. Take the connect sum of two copies of that surface; so the genus-zero surface with end space homeomorphic to $\omega^\omega \cdot 2 + 1$ . It looks a little like this:

The original “Great Wave” surface has self-similar end space and genus zero, so has coarsely bounded mapping class group. This surface $\Sigma$ has locally coarsely bounded mapping class group, but we will show that $\operatorname{Map}(\Sigma)$ is not generated by a coarsely bounded set. Consider the index-two subgroup $G$ of $\operatorname{Map}(\Sigma)$ comprising those mapping classes that fix pointwise the two maximal ends. We’ll show that $G$ is a countable union of proper open subgroups $G_0 < G_1 < \cdots$ . Since $G$ has index two in $\operatorname{Map}(\Sigma)$ , this will show that $\operatorname{Map}(\Sigma)$ is also a countable union of proper open subgroups $G'_0 < G'_1 < \cdots$ , where each $G'_i$ is obtained from $G_i$ by adding a fixed mapping class $\phi$ that swaps the two maximal ends of $\Sigma$ . (Recall that a subgroup is open if and only if it contains an open neighborhood of the identity, so the $G'_i$ are open since $G$ is open in $\operatorname{Map}(\Sigma)$ and the $G'_i$ are open in $G$ .)

To start, consider a simple closed curve $\alpha$ separating $\Sigma$ into two pieces, each containing exactly one maximal end. The identity neighborhood we consider is $U_A$ , where $A$ is an annular subsurface with core curve $\alpha$ . Since $U_A$ is a subgroup of $G$ , we’ll let $G_0 = U_A$ . Since $\operatorname{Map}(\Sigma)$ and hence $G$ is Polish, there is a countable dense subset $\{\phi_i : i \in \mathbb{N}\}$ of $G$ . Any open subgroup containing the $\phi_i$ is in fact all of $G$ , so consider the sequence of open subgroups $G_1 \le G_2 \le \cdots$ , where

$G_i = \langle G_0, \phi_1,\ldots, \phi_i \rangle.$If we can show that each $G_i$
is a proper subgroup of $G$
, we will be done,
even though *a priori* this chain may not be strictly increasing.
Consider a maximal end $\xi$
and a neighborhood basis of $\xi$
comprising nested clopen neighborhoods $U_j$
with $U_{j+1} \subset U_j$
,
beginning with $U_0$
being the end set of the component of $\Sigma - A$
containing $\xi$
.
In the figure we can think of the clopen neighborhoods as coming from the “fluting” process
described in the previous post.
Thus, $U_0 - U_j$
contains points homeomorphic to $\omega^{j-1} + 1$
but not points homeomorphic to $\omega^j + 1$
.
In plainer words, $U_0 - U_1$
contains isolated planar ends,
$U_0 - U_2$
contains ends accumulated by isolated planar ends,
$U_0 - U_3$
contains ends accumulated by ends accumulated by isolated planar ends,
so on and so forth.

Anyway, consider $\phi_1,\ldots,\phi_n$ . Since each $\phi_i$ leaves $\xi$ invariant, we claim that there exists $M$ large such that for all $m \ge M$ , ends homeomorphic to $\omega^m + 1$ contained in $U_m$ actually remain inside $U_m$ under each $\phi_i$ . To see this, note that if there was a sequence of ends $\{\xi_m\}$ with each $\xi_m$ homeomorphic to $\omega^m + 1$ such that each $\xi_m$ was moved outside of $U_m$ by some $\phi_i$ , then since the sequence $\{\xi_m\}$ necessarily converges to $\xi$ , by the pigeonhole principle some $\phi_i$ would have to move $\xi$ . This already shows us that $G_i$ is a proper subgroup of $G$ , since by the classification of surfaces we can move some $\xi_m$ with $m > M$ outside of $U_m$ (and into a neighborhood of the other maximal end).

This surface $\Sigma$ is an example of the general phenomenon Mannâ€“Rafi term having end space of “limit type”. The argument we just gave generalizes to show that if $\Sigma$ has limit type (see Definition 6.2 of their paper for a precise definition) then $\operatorname{Map}(\Sigma)$ cannot be generated by a coarsely bounded set.

## A non-example: infinite rank

If $G$
is a finitely generated group,
notice that all (*a fortiori* continuous) quotients of $G$
are finitely generated,
and that conversely if $G$
has a quotient that is not finitely generated,
then $G$
cannot be finitely generated.
The same is true of Polish groups and coarsely bounded generation:
if $G$
is a Polish group that has a continuous quotient
which is not coarsely boundedly generated, then $G$
is not either.
A prime example of such a group as a quotient
is the countably infinite group $\bigoplus_{n = 1}^\infty \mathbb{Z}$
.

It is possible to build continuous maps to $\bigoplus_{n=1}^\infty \mathbb{Z}$ by using the topology of the end space. Here is one example. Consider the ends $\xi_n$ constructed in the previous post which are pairwise noncomparable. The $\xi_n$ are maximal ends of self-similar surfaces, so have stable neighborhoods. We form a surface by “fluting” together the union of countably infinitely many copies of each $\xi_n$ . Necessarily each collection of ends locally homeomorphic to $\xi_n$ converges to the maximal end of the flute. Now, this surface is self-similar, with genus zero or infinity, hence has coarsely bounded mapping class group. So take the connect sum of two copies of this surface, and call the connect sum $\Sigma$ .

As before, take a simple closed curve $\alpha$ that separates $\Sigma$ into two pieces, each one containing a single one of the two maximal ends. Pick one of the maximal ends, call it $\xi$ , and let $U$ be the neighborhood of $\xi$ determined by $\alpha$ . We claim that for each end $\xi_n$ and any mapping class $\phi$ belonging to the index-two subgroup of $\operatorname{Map}(\Sigma)$ fixing $\xi$ , the number of ends locally homeomorphic to $\xi_n$ mapped into and out of $U$ is finite. Indeed, were either quantity infinite, the same argument as in the previous example shows that $\phi$ would have to move $\xi$ .

Anyway, count up the number of ends of type $\xi_n$ moved into $U$ by $\phi$ and subtract the number of ends of type $\xi_n$ moved out of $U$ by $\phi$ . This defines a homomorphism $\ell_n \colon G \to \mathbb{Z}$ . By “shifting a strip of ends locally homeomorphic to $\xi_n$ ”, we can show that $\bigoplus_{n=1}^\infty \ell_n\colon G \to \bigoplus_{n=1}^\infty \mathbb{Z}$ is surjective, and continuous, since if $A$ is an annular subsurface with core curve $\alpha$ , the open set $U_A$ is contained in the kernel of $\bigoplus_{n=1}^\infty \ell_n$ .

This surface $\Sigma$ is an example of the general phenomenon Mannâ€“Rafi term having end space of “infinite rank”. The argument we just gave generalizes to show that if $\Sigma$ has infinite rank (see Definition 6.5 for a precise definition) then $\operatorname{Map}(\Sigma)$ cannot be generated by a coarsely bounded set.