Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Lie group homo = continuous group homo? Or what do you need to show? –  Matt N. 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 –  Une Femme Douce Nov 1 '12 at 9:04
5  
How can you attempt to prove something when you don't even know what it is you have to show? –  Matt N. Nov 1 '12 at 9:05

2 Answers 2

up vote 4 down vote accepted

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.

share|improve this answer
2  
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. –  Matt N. 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.

share|improve this answer
2  
"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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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