$\mathrm{Homeo}(S^1)$ and the Mapping Class Group

Question

Is there a full description of $\mathrm{Homeo}(S^1)$ (i.e. the group of self-homeomorphisms of the circle)? By full description I mean a presentation/list of subgroups ect. Basically anything tangible. Also, what is the mapping class group $\pi_0(\mathrm{Homeo}(S^1))$?

Answer 1

$\mbox{Homeo}(S^1)$ is huge. In particular, it is not finitely generated. Some nice subgroups though include the group of orientation-preserving homeomorphisms $\mbox{Homeo}^+(S^1)$ (note that the automorphism group of $S^1$ is isomorphic to a semi-direct product of this group with the subgroup generated by the map $x\mapsto -x$).

Then there is also the subgroup of rigid automorphisms, that is the automorphism group of the circle as an object in the subcategory of metric spaces, where elements are isometries - this group is also sometimes called the circular dihedral group or the orthogonal group $O(2)$. This is still not finitely generated, but we can at least describe a generating set by noting that any isometry can be described by a rotation possibly followed by a reflection in some chosen axis. This subgroup in turn has every finite dihedral group as a subgroup by simply considering finite orbits of some rotation together with reflection.

To help give some intuition as to just how large $\mbox{Homeo}(S^1)$ really is, note that the coset space $\mbox{Homeo}(S^1)/O(2)$ is itself an uncountable space (note $O(2)$ is not a normal subgroup so this is not a quotient group, but there is an action of $O(2)$ on $\mbox{Homeo}(S^1)$ so we can consider the orbit space given by quotienting out by this action). You might like to think of this as the space of orientation preserving homeomorphisms of $S^1$ which fixes some predetermined point (say the point $(1,0)$).

There is then also the group of diffeomorphisms of $S^1$ to itself sometimes called $\mbox{Diff}(S^1)$ - this is the automorphism group of $S^1$ as an object in the subcategory of smooth manifolds.

The mapping class group $\mbox{MCG}(S^1)$ is more manageable though. Note that any two orientation preserving automorphisms of the circle are homotopic and so $\mbox{MCG}(S^1)=\pi_0(\mbox{Homeo}(S^1))$ consists of exactly two classes. The class of orientation-preserving, and the orientation-reversing automorphisms. The first class is the identity in this group which is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. Actually, we more readily get this result by noting that $\mbox{MCG}(S^1)\cong \mbox{Homeo}(S^1)/\mbox{Homeo}^+(S^1)$.