Topological groups on the circle $S^1$

Question

On the circle $S^1$ there is the usual circle group, i.e. the group isomorphic to $\{e^{i\varphi}\mid \varphi\in[0,2\pi)\}$ with complex multiplication as group operation. This group is a topological group in the sense that $S^1$ is a topological space and the group operation and the inverse are continuous.

Question: Are there other topological groups on $S^1$ essentially different from the circle group, assuming $S^1$ with the standard topology? What about other abelian topological groups?

My question was motivated by this other post. The group described there turned out to be just the usual one.

Two notes:

• I am looking for groups involving all of $S^1$, not only a subset. Especially no subgroups of the circle group.
• I am looking for groups not isomorphic to the circle group.

Answer 1

It is a theorem that any two topological group structures on $S^1$ are isomorphic, and similarly for the $n$-torus $(S^1)^n$, $n \ge 1$: any compact topological group structure on the $n$-torus is isomorphic to the standard one.

Here is a high-powered proof by quoting two theorems.

One is the theorem quoted in Francesco Pollizi's answer to this MathOverflow question: any compact abelian Lie group of dimension $n$ is isomorphic to $(S^1)^n$.

The other theorem is the solution to Hilbert's 5th problem by Gleason, Montgomery, and Zippin: every topological group which is a manifold has a smooth structure which makes it a Lie group.

Now, my guess is that for the case of $S^1$ there is probably an elementary proof of this theorem, perhaps embedded somewhere as an example in the literature of topological groups, but I do not know.

Answer 2

While Lee Mosher's answer is correct, it is an overkill. Suppose that $G$ is a topological group homeomorphic to $S^1$. Since $G$ is compact, you do not need to quote the solution of Hilbert-V: The Peter-Weyl theorem (which is much simpler) implies that each compact topological group admits a faithful finite dimensional linear representation and, hence, is a Lie group (Cartan's theorem). Since the Lie algebra ${\mathfrak g}$ of $G$ is 1-dimensional, it is commutative. Hence $\exp: {\mathfrak g}={\mathbb R}\to G$ is a homomorphism (necessarily an epimorphism), hence, $G$ is isomorphic to $U(1)$.