How to prove the group of automorphisms of $S^1$ as a topological group is $\mathbb Z_2$?

Question

The title basically says it all. How does one prove the group of automorphisms of $S^1$ (the unit circle in $\mathbb C$), as a topological group, is $\mathbb Z_2$? I was surprised not to find the answer here already, maybe I just didn't use the right search terms but I looked pretty far and wide and didn't find it. Nor did a general google search turn up the proof. I should add that this is an exercise in a "basic" topology book so I'm looking for an elementary self-contained proof that does not refer to any major theorems.

Answer 1

It is well-known that $\text{End}(S^1) \cong \mathbb Z$ (cf. Pontryagin dual of the unit circle), hence $\text{Aut}(S^1) \cong \mathbb Z^{\times} \cong \mathbb Z/2$.

Answer 2

I managed to prove it using basic principles. It seems there may be a much easier way. This way is actually pretty easy but it's not that easy to write down precisely. Anyway, here is an answer.

Let $f:S^1\rightarrow S^1$ be an automorphism. Let $a=e^{i\theta}$ and $b=e^{i\phi}$ in $S^1$ where $0\leq\theta\leq\phi\leq2\pi$. Let $I_{a,b}=\{e^{ix}\mid \theta\leq x\leq\phi\}$. In words $I_{a,b}$ is the closed segment of the circle going counter-clockwise from $a$ to $b$. Then for example $I_{1,-1}$ is the upper half of the circle $\{z\in S^1\mid\text{im }z\geq0\}$, and $I_{-1,1}$ is the lower half of the circle $\{z\in S^1\mid\text{im }z\leq0\}$. Note that $\overline{I_{a,b}}=I_{\overline b,\overline a}$

By the intermediate value theorem and the fact that $f$ is one-to-one, it must be that $$ f(I_{a,b})=I_{f(a),f(b)}\text{ or }f(I_{a,b})=I_{f(b),f(a)} \ \ \ \ \ \ \ \ \ \ (1) $$ Let $n\in\mathbb N$. There are exactly $n$ solutions to the equation $x^n-1=0$ in $\mathbb C$, called $n$-th roots of unity, evenly spaced around the circle. If $z^n=1$ then $1=\overline{z^n}=\overline{z}^n$. Thus the conjugate of an $n$-th root of unity is another $n$-th root of unity. We will prove by induction on $n$ that for every root of unity $z$, $f(z)=z$ or for every root of unity $f(z)=\overline{z}$.

Precisely let the statement $S(n)$ be:

$f(z)=z$ $\forall$ $n$-th roots of unity $z$ and for consecutive $n$-th roots of unity $a$ and $b$, $f(I_{a,b})=I_{a,b}$

or

$f(z)=\overline z$ $\forall$ $n$-th roots of unity $z$ and for consecutive $n$-th roots of unity $a$ and $b$, $f(I_{a,b})=\overline{I_{a,b}}$.

Base case $n=2$. The square roots of unity are $\pm1$. Since $f$ is a homomorphism of groups $f(1)=1$. Since $(-1)^2=1$, $1=f(1)=f((-1)^2)=(f(-1))^2$ Thus $f(-1)=\pm1$. Since $f(1)=1$ it cannot be that also $f(-1)=1$, thus $f(-1)=-1$. Thus $f$ fixes 1 and $-1$. Then by (1) $f(I_{1,-1})=I_{1,-1}$ or $f(I_{1,-1})=I_{-1,1}$. The base case $n=2$ therefore is true.

Assume $S(m)$ is true $\forall$ $m<n$. Let $\zeta$ be an $n$-th root of unity. Then $\zeta^n=1$, so $1=f(\zeta^n)=(f(\zeta))^n$. Thus $f(\zeta)$ is also an $n$-th root of unity. Now let $a,b\in S^1$ be consective $(n-1)$-st roots of unity. We will use (repeatedly) the following key point: there is one and only one $n$-th root of unity $z\in I_{a,b}$ (the $n$th roots of unity are equally spaced around the circle, so this is geometrically evident). By the induction hypothesis $f$ either fixes $a$ and $b$ or sends them to their (respective) conjugates. If it fixes $a$ and $b$ then, by the induction hypothesis, $f(I_{a,b})=I_{a,b}$. Thus the only $n$-th root of unity in $f(I_{a,b})$ is $z$. Since $f(z)$ must be an $n$-th root of unity, it must be that $f(z)=z$. On the other hand, if $f$ sends $a$ and $b$ to their conjugates, then by the induction hypothesis $f(I_{a,b})=\overline{I_{a,b}}$, which equals $I_{\overline b,\overline a}$ and $\overline b, \overline a$ are consecutive $(n-1)$-st roots of unity, thus as before $f(I_{a,b})$ contains one and only one $n$-th root of unity, $\overline{z}$. Since $f(z)\in I_{\overline b,\overline a}$ must be an $n$-th root of unity, it follows that $f(z)=\overline{z}$. Thus we have shown that $f$ either fixes all of the $n$-th roots of unity or sends them all to their conjugates. The required behavior of $f$ on the intervals between consecutive $n$-th roots of unity follows from (1).

Thus we have shown that $f$ fixes all of the roots of unity. The roots of unity correspond to the rational numbers in $[0,1]$ under the map $x\mapsto e^{2\pi ix}$. Thus the roots of unity are dense in $S^1$. Therefore if $f$ fixes all of the roots of unity, and $f$ is continuous, then $f$ must be the identity function. And if $f$ sends all roots of unity to their conjugates, then $f$ must send all elements of $S^1$ to their conjugates. Thus the identity function and conjugation are the only two possibilities for $f$.

Answer 3

The elementary, clear way to do this problem is to show that the only continuous group homomorphisms $\mathbf R\rightarrow\mathbf R$ are $x\mapsto \lambda x$ for some $\lambda\in\mathbf R$. Using this and unique path lifting one shows that the only continuous group homomorphisms $\mathbf R\rightarrow S^1$ are $x\mapsto e^{i\lambda x}$ for fixed $\lambda\in\mathbf R$ as before. Finally, $\mathbf R/\mathbf Z\cong S^1$ as top gps, so any continuous gp hom $\mathbf R/\mathbf Z\cong S^1\rightarrow S^1$ is the same as a $\mathbf Z$-periodic continuous gp hom $\mathbf R\rightarrow S^1$; these are just the involutions corresponding to $\lambda=\pm2\pi$. $\blacksquare$