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.


migrated from physics.stackexchange.com Sep 21 '16 at 17:54

This question came from our site for active researchers, academics and students of physics.

  • 2
    $\begingroup$ How is this a physics question? $\endgroup$ – ACuriousMind Sep 21 '16 at 16:18
  • $\begingroup$ Symmetries are quite useful for particle physics. However, I bear your comment in mind and will post this to math stack exchange. $\endgroup$ – SAMCRO Sep 21 '16 at 16:22

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 isomoprhism.

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 theoritical 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 automorphis 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 auotomorphism 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.)

  • $\begingroup$ 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. $\endgroup$ – SAMCRO Sep 21 '16 at 20:29
  • $\begingroup$ 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.) $\endgroup$ – Jason DeVito Sep 22 '16 at 14:35
  • $\begingroup$ OK, now I understand, that was clear and yes the former was typo on my part. Thanks again. $\endgroup$ – SAMCRO Sep 22 '16 at 15:00
  • $\begingroup$ the outer automorphism group of U(1) is $\mathbb{R}$? $\endgroup$ – wonderich Aug 20 '18 at 16:41
  • $\begingroup$ @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. $\endgroup$ – Jason DeVito Aug 20 '18 at 16:54

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 $f(x)=e^{iax}$.

  • $\begingroup$ So what is the outer automorphism group of U(1)? $\endgroup$ – wonderich Aug 20 '18 at 16:38
  • $\begingroup$ Do you mean $f(e^{ix})=e^{iax}$ in your last sentence? $\endgroup$ – wonderich Aug 20 '18 at 16:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.