Mapping class groups (also called modular groups) are of central importance in many fields of geometry. If is an oriented surface (i.e. a -manifold), the group of orientation-preserving self-homeomorphisms of is a *topological group* with the compact-open topology. The *mapping class group* of , denoted (or by some people) is the group of path-components of , i.e. , or equivalently where is the subgroup of homeomorphisms isotopic to the identity.

When is a surface of finite type (i.e. a closed surface minus finitely many points), the group 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 when is of *infinite* type. However, such groups do arise naturally in dynamics.

**Example:** Let be a group of (orientation-preserving) homeomorphisms of the plane, and suppose that has a bounded orbit (i.e. there is some point for which the orbit is contained in a compact subset of the plane). The closure of such an orbit is compact and -invariant. Let be the union of the closure of with the set of bounded open complementary regions. Then is compact, -invariant, and has connected complement. Define an equivalence relation on the plane whose equivalence classes are the points in the complement of , and the connected components of . 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 is a totally disconnected set . The original group admits a natural homomorphism to the mapping class group of . After passing to a -invariant closed subset of if necessary, we may assume that is minimal (i.e. every orbit is dense). Since is compact, it is either a finite discrete set, or it is a Cantor set.

The mapping class group of contains a subgroup of finite index fixing the end of ; this subgroup is the quotient of a *braid group* by its center. There are many tools that show that certain groups cannot have a big image in such a mapping class group.

Much less studied is the case that is a Cantor set. In the remainder of this post, we will abbreviate by . Notice that any homeomorphism of extends in a unique way to a homeomorphism of , 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 denote the mapping class group of . Then there is a natural surjection whose kernel is (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 of orientation-preserving diffeomorphisms of the plane with a bounded orbit is circularly orderable:

**Proposition:** There is an injective homomorphism .

*Sketch of Proof:* Choose a complete hyperbolic structure on . The Birman exact sequence exhibits 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 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 . Recall that for a surface of finite type, the group acts simplicially on the *complex of curves* , a simplicial complex whose simplices are the sets of isotopy classes of essential simple closed curves in that can be realized mutually disjointly. A fundamental theorem of Masur-Minsky says that (with its natural simplicial path metric) is -hyperbolic (though it is not locally finite). Bestvina-Fujiwara show that any reasonably big subgroup of contains lots of elements that act on 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 is uniformly perfect.

*Proof:* Remember that denotes the mapping class group of . We denote the Cantor set in the sequel by .

A closed disk is a *dividing disk* if its boundary is disjoint from , and separates into two components (both necessarily Cantor sets). An element 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 , there is a homeomorphism of the sphere permuting , that takes off itself, and so that the family of disks are pairwise disjoint, and converge to a limiting point . Define to be the infinite product . Notice that is a well-defined homeomorphism of the plane permuting . Moreover, there is an identity , thereby exhibiting as a commutator. The theorem will therefore be proved if we can exhibit any element of as a bounded product of local elements.

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

The same argument just barely fails to work with in place of . One can also define dividing disks and local homeomorphisms in , with the following important difference. One can show by the same argument that local homeomorphisms in are commutators, and that for an arbitrary element there are local elements so that 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: .

The so-called “point-pushing” subgroup can be understood geometrically by tracking the image of a proper ray from to infinity. We are therefore motivated to consider the following object:

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

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

**Lemma:** Let and suppose there is a vertex such that share an edge. Then is a product of at most two local homeomorphisms.

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

It follows that if has a bounded orbit in , then the commutator lengths of the powers of are bounded, and therefore vanishes. If this is true for every , then Bavard duality implies that admits no nontrivial homogeneous quasimorphisms. This motivates the following questions:

**Question:** Is the diameter of infinite? (Exercise: show )

**Question:** Does any element of act on with positive translation length?

**Question:** Can one use this action to construct nontrivial quasimorphisms on ?

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