Automorphisms of U(1)

This probably a very simple question but I would appreciate some clarification.

So $U(1)$ is an Abelian Lie group. As it is Abelian, all elements of the group commute with each other, this means the centre of the group is the group itself.

Since $$\text{Inn}(G)=G/Z\implies \text{Inn}(U(1))=\mathcal{I} $$

I would like to know the outer automorphisms

$$\text{Out}(G)=\text{Aut}(G)/\text{Inn}(G) $$

However I do not know the automorphism group in order to make this deduction.

Solution 1:

First, we need to distinguish between the continuous automorphism group and the abstract automorphism group.

I claim that there is a unique continuous non-identity automorphism of $S^1$.

First, since $S^1$ is abelian, the inversion map $i:S^1\rightarrow S^1$ is a continuous isomorphism.

Why is it the only one? Let $g:S^1\rightarrow S^1$ be any continuous automorphism. If $\pi:\mathbb{R}\rightarrow \mathbb{R}/\mathbb{Z}\cong S^1$ is the canonical projection, then the map $\pi \circ g$ lifts to a unique continuous map $\tilde{g}:\mathbb{R}\rightarrow \mathbb{R}$ with $\tilde{g}(0) = 0$. One can easily verify that $\tilde{g}$ is a homomorphism, that it $\tilde{g}$ satisfies the Cauchy functional equation. In particular, $\tilde{g}(x) = cx$ for some constant $c\in \mathbb{R}$.

Now, because $\tilde{g}(\mathbb{Z})\subseteq \mathbb{Z}$, it follows that $g(1) = c\in \mathbb{Z}$. This establishes that every continuous homomorphism $S^1\rightarrow S^1$ is of the form $z\rightarrow z^c$ with $c$ an integer. (Here, we are thinking of $S^1$ as the unit complex numbers). Then note that if $|c|\neq 1$, this map is not injective. So there are precisely two continuous automorphisms of $S^1$, the identity and the inverse.

Now, if one is simply interested in the group theoretical automorphism group of $S^1$, then there are a ton more (assuming you believe in the axiom of choice.)

To see this, note that, as shown in this surprising post, $S^1$ is, group theoretically, isomorphic to $\mathbb{C}^\times$. In particular, every automorphism of $\mathbb{C}$ as a field can be thought of as a group isomorphism of $S^1$. According to this post, there are at precisely $2^{|\mathbb{R}|}$ such automorphisms.

(Note that it's possible that an automorphism of $S^1\cong \mathbb{C}^\times$ does not extend to a field isomorphism of $\mathbb{C}$, so there may be even more automorphisms of $S^1$. This may also allow one to create discontinuous automorphisms even in the absence of choice, but I'm not sure what happens in that case.)