# Lie group homomorphism from $\mathbb{R}\rightarrow S^1$

I need to prove that every Lie group homomorphism from $\mathbb{R}\rightarrow S^1$ is of the form $x\mapsto e^{iax}$ for some $a\in\mathbb{R}$.

Here is my attempt: As it is group homomorphism so it must satisfies $\phi(x+y)=\phi(x).\phi(y),\forall x,y\in\mathbb{R}$, I know one result if some continous function satisfies this rule, then it is of the form $e^x$, is this the same trick here we need to apply?

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Lie group homo = continuous group homo? Or what do you need to show? –  Rudy the Reindeer Nov 1 '12 at 8:59
I dont know that a lie group homo is continous but I guess it will be smooth map so it must be continuous –  El Angel Exterminador Nov 1 '12 at 9:04
How can you attempt to prove something when you don't even know what it is you have to show? –  Rudy the Reindeer Nov 1 '12 at 9:05

This is from an answer I posted to one of my own questions:

Let $f: \mathbb R \to S^1$ be a continuous group homomorphism. Then $f$ are of the form $e^{i \lambda x}$ for $\lambda \in \mathbb{R}$. To see this note that $(e^{ix}, \mathbb{R})$ is a covering space of $S^1$. Then by the unique lifting property we get that for a continuous homomorphism $f: \mathbb{R} \to S^1$ there is a unique continuous homomorphism $\alpha : \mathbb{R} \to \mathbb{R}$ such that $f = g \circ \alpha$ where $g (x) = e^{ix}$ is the covering map. By (i) we get that $f$ has to be of the form $x \mapsto e^{i\lambda x}$.

Here (i) is the following: If $\alpha : \mathbb{R} \to \mathbb{R}$ is a continuous homomorphism then $\alpha$ is of the form $x \mapsto \lambda x$ for some $\lambda \in \mathbb{R}$. This follows from the fact that $\alpha$ is a linear map and one dimensional matrices are multiplication by scalars.

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To answer your query in the comments to the question: Usually a homomorphism of Lie groups is defined to be a homomorphism which is also a smooth map. However, it is a theorem that a continuous homomorphism between Lie groups is automatically smooth (and in this instance it is clear that the map is smooth). –  commenter Nov 1 '12 at 10:28
Dear @commenter, thank you for the comment. –  Rudy the Reindeer Nov 1 '12 at 10:30

Here is a proof that does not use as much advanced material. A Lie group homomorphism is a continuous group homomorphism between any two Lie groups. Any continuous group homomorphism $f : \Bbb{R} \rightarrow S^1$ is of the form $t \mapsto e^{i\theta(t)}$ for some function $\theta(t)$. Now observe that because $f$ is a group homomorphism we get that for all rational numbers $q$,

$$f(q) = e^{i\theta(1)q}.$$

Because $\Bbb{Q}$ is dense in $\Bbb{R}$, it follows that for all $t \in \Bbb{R}$ we have $f(t) = e^{i\theta(1) t}$. Since $\theta(1)$ can be any real number we conclude that any continuous group homomorphism from $\Bbb{R}$ to $S^1$ is of the form $e^{iat}$ for $a$ some real number.

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"Any continuous group homomorphism $f\colon \mathbb{R} \to S^1$ is of the form $t \mapsto e^{i\theta(t)}$ for some continuous function $\theta(t)$." Yes, and how do you propose to prove the existence of $\theta$? I agree that once you know that $\theta$ exists then there's not much left to do, but its existence requires an argument. –  commenter Nov 1 '12 at 10:24
I didn't mean to imply that there is no elementary proof of the existence of $\theta$, but if you look at the argument you'll see that you prove the path-lifting property of the covering map $\exp\colon \mathbb{R} \to S^1$, and your subsequent argument justifies the sentence that a homomorphism lifts to a homomorphism in Matt N.'s answer. –  commenter Nov 1 '12 at 10:41
@commenter I did not do any magic here: (1) Do you agree that every element in $S^1$ is of the form $e^{it}$ and (2) $t$ will map to $e^{i\theta(t)}$ for some $\theta(t)$. We don't actually need to assume that $\theta$ is continuous but from the fact that $f$ is a group homomorphism we will get that $\theta$ is linear. –  user38268 Nov 1 '12 at 10:52
What you will get (from knowing that $e^{ix} = e^{iy}$ iff $x - y \in 2\pi \mathbb{Z}$) is that $\theta(s+t) - \theta(s) - \theta(t) \in 2\pi \mathbb{Z}$, which is good enough to conclude. I agree with that. –  commenter Nov 1 '12 at 11:01