Is an action of $\mathbb{S^1}$ on itself effective if and only if it is free?

Question

Obviously if the action is free it is effective. In the case of $\mathbb{S^1}$ acting on itself by rotations/complex multiplication this is clearly the case, but are there other possible actions of $\mathbb{S^1}$ on itself which I haven't considered?

Answer 1

Let $f: \mathbb{S}^1 \to \mathbb{S}^1$ be any homemorphism. It induces a $\mathbb{Z}$ action on $\mathbb{S}^1$ in an obvious way: for $x \in \mathbb{S}^1$, $n \cdot x$ is defined as $f^n(x)$. Let $n$ be smallest such that $f^n$ is identity. Then induced action of $\mathbb{Z}/n\mathbb{Z}$ on $\mathbb{S}^1$ is effective. So, the question boils down to finding a self-homeomorphism of $\mathbb{S}^1$ that is not identity, but has a fixed point.

It is pretty easy to come up with such -- map $1$ to $1$, and then as you go counter-clockwise around domain $\mathbb{S}^1$, wrap it around the codomain slightly faster than you walk along the domain, and once you are halfway on the domain, slow down to make up for it.

Answer 2

Short version: If you assume the action of $S^1$ on itself is continuous, then effective implies free. But there are discontinuous actions which are effective but not free.

Let's first assume the action is continuous and effective. We will show it is free.

The action is given by a continuous injective homomorphism $\phi:S^1\rightarrow Homeo(S^1)$. Since $S^1$ is connected, the image acutally lands in $Homeo^+(S^1)$, the group of orientation preserving homeomorphisms of $S^1$. Further, since $S^1$ is compact, the image is also compact.

Now, according to

Ghys, Etienne: Groups acting on the circle, Enseign. Math. (2) 47 (2001), no. 3-4, 329–407

up to conjugacy, there is a unique maximal compact subgroup of $Homeo^+(S^1)$: $SO(2)$. So, there is a $g\in Homeo^+(S^1)$ with $g \phi(S^1)g^{-1}\subseteq SO(2)$.

Now, pick $h\in S^1$ and assume $\phi(h)\in Homeo^+(S^1)$ fixes a point $p\in S^1$. First note that $g(p)$ is fixed by $g\phi(h)g^{-1}$: $g\phi(h)g^{-1}g(p) = g\phi(h)(p)= g(p)$. Since $g\phi(h)g^{-1}\in SO(2)$, it's a regular good old fashion rotation. Since it has a fixed point, it must be the identity rotation: $g\phi(h)g^{-1} = Id$. This easily implies that $\phi(h) = Id$. Since $\phi$ is injective, $h = Id$, so the action is free.

Now, what if the action isn't continuous? Well, according to this MSE question, $S^1$ is isomorphic as a group to $S^1\times \mathbb{R}$. From this question, we know that $\mathbb{R}$ is isomorphic to $\mathbb{R}^2$ as a group. So, if we can find an effective non-free action of $S^1\times \mathbb{R}$ on $S^1\times \mathbb{R}^2$, we can use the isomorphisms to create an effective non-free action of $S^1$ on itself. To do this, it's enough to find an effective non-free action of $\mathbb{R}$ on $\mathbb{R}^2$ because we can then "product" this action with the usual effective, free action of $S^1$ on itself to get an effective non-free action of $S^1\times \mathbb{R}$ on $S^1\times \mathbb{R}^2$.

So, let's create a smooth action of $\mathbb{R}$ on $\mathbb{R}^2$ which is effective but not free. To do so, let $f:\mathbb{R}\rightarrow \mathbb{R}$ be defined by $f(t) = \begin{cases} 0 & t\leq 0\\ e^{-1/t} & t > 0\end{cases}$. Then let $\mathbb{R}$ act on $\mathbb{R}^2$ by $t\ast(x,y) = (x+\phi(y)t,y)$.

We first note that this is an action: if $t = 0$, then $t\ast(x,y) = (x,y)$ and \begin{align*} s\ast(t\ast(x,y)) &= s\ast(x+\phi(y)t, y)\\ &= (x+\phi(y)t + \phi(y)s, y)\\ &= (x+\phi(y)(s+t), y)\\ &= (s+t)\ast(x,y).\end{align*}

This action is effective: If $y > 0$, then $\phi(y) > 0$, so if $t\neq 0$, then $\phi(y)t\neq 0$, so the $x$-coordinates of $(x,y)$ and of $t\ast(x,y) = x+\phi(y)t, y)$ are different.

This action is not free: if $y\leq 0$, then $\phi(y) = 0$ so $t\ast(x,y) = (x,y)$.