Finite subgroup of a group of homeomorphisms of closed unit disc

Question

Example of a non-trivial finitely generated subgroup of finite order of a group of homeomorphisms (which fix boundary point wise) of closed unit disc?

This question is related to this one (Finitely generated subgroup of infinite order of a group of homeomorphisms of closed unit disc) where now I need a finite group. I actually made a mistake in my previous question but the answers were very useful to learn more.

Answer 1

No such example exists, and this can be deduced from the theorem that braid groups are torsion free.

From that theorem, one can deduce that for every homeomorphism $f : D^2 \to D^2$, if $f$ is the identity on the boundary circle $S^1$ and if $f$ has finite order then $f$ is the identity map.

For the proof, suppose that $f$ is not the identity map, and so there exists $x \in \text{interior}(D^2) = D^2 - S^1$ such that $f(x) \ne x$. Since $f$ has finite order, there is a least integer $n \ge 2$ such that $f^n(x)=x$. We have a finite set of points in $D^2$, of cardinality $n \ge 2$, given by $$P = \{p_1,...,p_n\}, \quad x = p_1, \quad p_{k+1}=f(p_k) \quad\text{for $k=1,...,n-1$} $$ This set has the property that $f(P)=P$.

I'll define the braid group of the set $P$, denoted $B_P$ (also known as the $n$-strand braid group and denoted $B_n$). I'm going to give you the topologists definition which is quick and dirty, although there are more intuitive definitions to be found such as in the link above.

Start with the group $\cal H_P$ of all homeomorphisms of $D^2$ which fix each point of $S^1$ and which take $P$ to itself. There is a normal subgroup which I will denote $\cal H_{P;0}$ consisting of all $h \in \cal H_P$ that are isotopic to the identity relative to $S^1 \cup P$. This means that there exists a continuous function $H : D^2 \times [0,1] \to D^2$ such that $H(x,0)=h(x)$, $H(x,1)=x$, for each $x \in S^1 \cup P$ and each $t \in [0,1]$ we have $H(x,t)=x$, and for each $t \in [0,1]$ the function $D^2 \mapsto D^2$ defined by $x \mapsto H(x,t)$ is a homeomorphism.

In particular, it follows directly from the description of $H$ that:

If $h \in \cal H_{P;0}$ then for each $x_i \in P$ we have $h(x_i)=x_i$.

Define the braid group $B_P$ to be the quotient group $\cal H_P / \cal H_{P;0}$. Given an element $h \in \cal H_P$ let $[h] \in B_P$ denote the corresponding element of $B_P$. Formally $[h]$ is coset of $\cal H_{P;0}$ that contains $h$. What I need is that $[h]$ is the image of $h$ under the natural quotient homomorphism $\cal H_P \mapsto B_P$ whose kernel is $\cal H_{P;0}$.

Now there's the theorem which I mentioned earlier: the group $B_P$ is torsion free. Chapter 2 of the link I gave above gives three different proofs of this theorem.

So, why does this imply that $f$ does not have finite order?

First, $f \in \cal H_P$.

Second, since $f$ has finite order in the group $\cal H_P$ then its image $[f]$ has finite order in the group $B_P$.

But $B_P$ is torsion free, and so $[f]$ is the identity on $B_{P}$.

This means that $f \in \cal H_{P;0}$.

It follows that $f(x_i)=x_i$ for each $i$. Taking $i=1$ we get $f(x)=x$.

But that's a contradiction, because $f(x) \ne x$.

Answer 2

The proof of the theorem in the above answer by Lee Mosher:

Every periodic homeomorphism of a 2-disk fixed on the boundary is the identity

can also be found in the paper

A. Constantin, B. Kolev. The theorem of Kerékjártó on periodic homeomorphisms of the disc and the sphere. Enseign. Math. (2) 40 (1994), no. 3-4, 193-204.

https://www.e-periodica.ch/cntmng?pid=ens-001%3A1994%3A40%3A%3A81