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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.

wonderich
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SAMCRO
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2 Answers2

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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.)

Tomáš Bzdušek
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Jason DeVito - on hiatus
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  • I am not really understanding your solution: I did math as an undergraduate, including algebraic topology quite some time ago. When you mention there are two continuous automorphisms of $$S^1$$ do you mean the automorphism group is $$\mathcal{Z}{2}$$? The outer automorphism of $$SU(3)$$ os $S{3}$, the permutation symmetries of three objects. I was hoping for an analogue of this. Apologises for my lack of understanding. – SAMCRO Sep 21 '16 at 20:29
  • I'm claiming that the outer (continuous) automorphism group of $S^1$ is $\mathbb{Z}/2\mathbb{Z}$, generated by the inversion map. Further, since you noted the center is all of $S^1$, this is the same as the full automorphism group. (Also, I have always seen Out = Aut/Inn, not Out = Aut/Z as you wrote.) – Jason DeVito - on hiatus Sep 22 '16 at 14:35
  • OK, now I understand, that was clear and yes the former was typo on my part. Thanks again. – SAMCRO Sep 22 '16 at 15:00
  • @wonderich: The continuous automorphism group of $U(1)$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, with generator given by $z\mapsto \overline{z}$. The automorphism group with no assumptions on continuity has more than $|\mathbb{R}|$ elements in it (if you believe the axiom of choice), so is not isomorphic to $\mathbb{R}$ either. – Jason DeVito - on hiatus Aug 20 '18 at 16:54
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Write $U(1)=\{e^{ix},x\in R\}$. Let $f$ be an automorphism of $U(1)$, $f(e^{ix})=e^{ig(x)}$,

Suppose $f$ is continue, $f(e^{ix}e^{iy})=f(e^{i(x+y)})=$ $f(e^{ix})f(e^{iy})$ $=e^{ig(x+y)}=e^{ig(x)+g(y)}$, implies that $h(x,y)=e^{i(g(x+y)-g(x)-g(y))}=1$. Since $h$ is continue, we deduce that there exists such that $g(x+y)-g(x)-g(y)=2\pi n$. Write $l(x)=g(x)+2\pi n$,you deduce that $l$ is linear, $l(x)=ax$ and $f(x)=e^{iax}$.

Tsemo Aristide
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