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My student Steven Frankel has just posted his paper Quasigeodesic flows and Mobius-like groups on the arXiv. This heartbreaking work of staggering genius interesting paper makes a deep connection between dynamics, hyperbolic geometry, and group theory, and represents the first significant progress that I know of on a conjectural program I formulated a few years ago.

One of the main results of the paper is to show that every quasigeodesic flow on a closed hyperbolic 3-manifold either has a closed orbit, or the fundamental group of the manifold admits an action on a circle with some very peculiar properties, namely that it is Mobius-like but not Mobius. The problem of giving necessary and sufficient conditions on a vector field on a 3-manifold to guarantee the existence of a closed orbit is a long and interesting one, and the introduction to the paper gives a brief sketch of this history as follows:

Mapping class groups (also called modular groups) are of central importance in many fields of geometry. If $S$ is an oriented surface (i.e. a $2$-manifold), the group $\text{Homeo}^+(S)$ of orientation-preserving self-homeomorphisms of $S$ is a topological group with the compact-open topology. The mapping class group of $S$, denoted $\text{MCG}(S)$ (or $\text{Mod}(S)$ by some people) is the group of path-components of $\text{Homeo}^+(S)$, i.e. $\pi_0(\text{Homeo}^+(S))$, or equivalently $\text{Homeo}^+(S)/\text{Homeo}_0(S)$ where $\text{Homeo}_0(S)$ is the subgroup of homeomorphisms isotopic to the identity.

When $S$ is a surface of finite type (i.e. a closed surface minus finitely many points), the group $\text{MCG}(S)$ is finitely presented, and one knows a great deal about the algebra and geometry of this group. Less well-studied are groups of the form $\text{MCG}(S)$ when $S$ is of infinite type. However, such groups do arise naturally in dynamics.

Example: Let $G$ be a group of (orientation-preserving) homeomorphisms of the plane, and suppose that $G$ has a bounded orbit (i.e. there is some point $p$ for which the orbit $Gp$ is contained in a compact subset of the plane). The closure of such an orbit $Gp$ is compact and $G$-invariant. Let $K$ be the union of the closure of $Gp$ with the set of bounded open complementary regions. Then $K$ is compact, $G$-invariant, and has connected complement. Define an equivalence relation $\sim$ on the plane whose equivalence classes are the points in the complement of $K$, and the connected components of $K$. The quotient of the plane by this equivalence relation is again homeomorphic to the plane (by a theorem of R. L. Moore), and the image of $K$ is a totally disconnected set $k$. The original group $G$ admits a natural homomorphism to the mapping class group of $\mathbb{R}^2 - k$. After passing to a $G$-invariant closed subset of $k$ if necessary, we may assume that $k$ is minimal (i.e. every orbit is dense). Since $k$ is compact, it is either a finite discrete set, or it is a Cantor set.

The mapping class group of $\mathbb{R}^2 - \text{finite set}$ contains a subgroup of finite index fixing the end of $\mathbb{R}^2$; this subgroup is the quotient of a braid group by its center. There are many tools that show that certain groups $G$ cannot have a big image in such a mapping class group.

Much less studied is the case that $k$ is a Cantor set. In the remainder of this post, we will abbreviate $\text{MCG}(\mathbb{R}^2 - \text{Cantor set})$ by $\Gamma$. Notice that any homeomorphism of $\mathbb{R}^2 - \text{Cantor set}$ extends in a unique way to a homeomorphism of $S^2$, fixing the point at infinity, and permuting the points of the Cantor set (this can be seen by thinking of the “missing points” intrinsically as the space of ends of the surface). Let $\Gamma'$ denote the mapping class group of $S^2 - \text{Cantor set}$. Then there is a natural surjection $\Gamma \to \Gamma'$ whose kernel is $\pi_1(S^2 - \text{Cantor set})$ (this is just the familiar Birman exact sequence).

The following is proved in the first section of my paper “Circular groups, planar groups and the Euler class”. This is the first step to showing that any group $G$ of orientation-preserving diffeomorphisms of the plane with a bounded orbit is circularly orderable:

Proposition: There is an injective homomorphism $\Gamma \to \text{Homeo}^+(S^1)$.

Sketch of Proof: Choose a complete hyperbolic structure on $S^2 - \text{Cantor set}$. The Birman exact sequence exhibits $\Gamma$ as a group of (equivalence classes) of homeomorphisms of the universal cover of this hyperbolic surface which commute with the deck group. Each such homeomorphism extends in a unique way to a homeomorphism of the circle at infinity. This extension does not depend on the choice of a representative in an equivalence class, and one can check that the extension of a nontrivial mapping class is nontrivial at infinity. qed.

This property of the mapping class group $\Gamma$ does not distinguish it from mapping class groups of surfaces of finite type (with punctures); in fact, the argument is barely sensitive to the topology of the surface at all. By contrast, the next theorem demonstrates a significant difference between mapping class groups of surfaces of finite type, and $\Gamma$. Recall that for a surface $S$ of finite type, the group $\text{MCG}(S)$ acts simplicially on the complex of curves $\mathcal{C}(S)$, a simplicial complex whose simplices are the sets of isotopy classes of essential simple closed curves in $S$ that can be realized mutually disjointly. A fundamental theorem of Masur-Minsky says that $\mathcal{C}(S)$ (with its natural simplicial path metric) is $\delta$-hyperbolic (though it is not locally finite). Bestvina-Fujiwara show that any reasonably big subgroup of $\text{MCG}(S)$ contains lots of elements that act on $\mathcal{C}(S)$ weakly properly, and therefore such groups admit many nontrivial quasimorphisms. This has many important consequences, and shows that for many interesting classes of groups, every homomorphism to a mapping class group (of finite type) factors through a finite group. In view of the potential applications to dynamics as above, one would like to be able to construct quasimorphisms on mapping class groups of infinite type.

Unfortunately, this does not seem so easy.

Proposition: The group $\Gamma'$ is uniformly perfect.

Proof: Remember that $\Gamma'$ denotes the mapping class group of $S^2 - \text{Cantor set}$. We denote the Cantor set in the sequel by $C$.

A closed disk $D$ is a dividing disk if its boundary is disjoint from $C$, and separates $C$ into two components (both necessarily Cantor sets). An element $g \in \Gamma$ is said to be local if it has a representative whose support is contained in a dividing disk. Note that the closure of the complement of a dividing disk is also a dividing disk. Given any dividing disk $D$, there is a homeomorphism of the sphere $\varphi$ permuting $C$, that takes $D$ off itself, and so that the family of disks $\varphi^n(D)$ are pairwise disjoint, and converge to a limiting point $x \in C$. Define $h$ to be the infinite product $h = \prod_i \varphi^i g \varphi^{-i}$. Notice that $h$ is a well-defined homeomorphism of the plane permuting $C$. Moreover, there is an identity $[h^{-1},\varphi] = g$, thereby exhibiting $g$ as a commutator. The theorem will therefore be proved if we can exhibit any element of $\Gamma'$ as a bounded product of local elements.

Now, let $g$ be an arbitrary homeomorphism of the sphere permuting $C$. Pick an arbitrary $p \in C$. If $g(p)=p$ then let $h$ be a local homeomorphism taking $p$ to a disjoint point $q$, and define $g' = hg$. So without loss of generality, we can find $g' = hg$ where $h$ is local (possibly trivial), and $g'(p) = q \ne p$. Let ${}E$ be a sufficiently small dividing disk containing $p$ so that $g'(E)$ is disjoint from ${}E$, and their union does not contain every point of $C$. Join ${}E$ to $g'(E)$ by a path in the complement of $C$, and let $D$ be a regular neighborhood, which by construction is a dividing disk. Let $f$ be a local homeomorphism, supported in $D$, that interchanges ${}E$ and $g'(E)$, and so that $f g'$ is the identity on $D$. Then $fg'$ is itself local, because the complement of the interior of a dividing disk is also a dividing disk, and we have expressed $g$ as a product of at most three local homeomorphisms. This shows that the commutator length of $g$ is at most $3$, and since $g$ was arbitrary, we are done. qed.

The same argument just barely fails to work with $\Gamma$ in place of $\Gamma'$. One can also define dividing disks and local homeomorphisms in $\Gamma$, with the following important difference. One can show by the same argument that local homeomorphisms in $\Gamma$ are commutators, and that for an arbitrary element $g \in \Gamma$ there are local elements $h,f$ so that $fhg$ is the identity on a dividing disk; i.e. this composition is anti-local. However, the complement of the interior of a dividing disk in the plane is not a dividing disk; the difference can be measured by keeping track of the point at infinity. This is a restatement of the Birman exact sequence; at the level of quasimorphisms, one has the following exact sequence: $Q(\Gamma') \to Q(\Gamma) \to Q(\pi_1(S^2 - C))^{\Gamma'}$.

The so-called “point-pushing” subgroup $\pi_1(S^2 - C)$ can be understood geometrically by tracking the image of a proper ray from $C$ to infinity. We are therefore motivated to consider the following object:

Definition: The ray graph $R$ is the graph whose vertex set is the set of isotopy classes of proper rays $r$, with interior in the complement of $C$, from a point in $C$ to infinity, and whose edges are the pairs of such rays that can be realized disjointly.

One can verify that the graph $R$ is connected, and that the group $\Gamma$ acts simplicially on $R$ by automorphisms, and transitively on vertices.

Lemma: Let $g \in \Gamma$ and suppose there is a vertex $v \in R$ such that $v,g(v)$ share an edge. Then $g$ is a product of at most two local homeomorphisms.

Sketch of proof: After adjusting $g$ by an isotopy, assume that $r$ and $g(r)$ are actually disjoint. Let $E,g(E)$ be sufficiently small disjoint disks about the endpoint of $r$ and $g(r)$, and $\alpha$ an arc from ${}E$ to $g(E)$ disjoint from $r$ and $g(r)$, so that the union $r \cup E \cup \alpha \cup g(E) \cup g(r)$ does not separate the part of $C$ outside $E \cup g(E)$. Then this union can be engulfed in a punctured disk $D'$ containing infinity, whose complement contains some of $C$. There is a local $h$ supported in a neighborhood of $E \cup \alpha \cup g(E)$ such that $hg$ is supported (after isotopy) in the complement of $D'$ (i.e. it is also local). qed.

It follows that if $g \in\Gamma$ has a bounded orbit in $R$, then the commutator lengths of the powers of $g$ are bounded, and therefore $\text{scl}(g)$ vanishes. If this is true for every $g \in \Gamma$, then Bavard duality implies that $\Gamma$ admits no nontrivial homogeneous quasimorphisms. This motivates the following questions:

Question: Is the diameter of $R$ infinite? (Exercise: show $\text{diam}(R)\ge 3$)

Question: Does any element of $\Gamma$ act on $R$ with positive translation length?

Question: Can one use this action to construct nontrivial quasimorphisms on $\Gamma$?