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I am revising for my Lie Groups exam and am stuck on the following question.

Find all Lie Group homomorphisms

a) $ \ F : \mathbb{R} \longrightarrow S^1 \ $ (Hint: Consider the corresponding homomorphisms of Lie algebras $F_{\ast}$)

b) $ \ F : S^1 \longrightarrow \mathbb{R} $

where $S^1$ is the unit circle.

I don't really know where to begin with this so any help would be very much appreciated?

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For the second, at least, you shouldn't have much trouble ... are there any compact Lie subgroups of $\mathbb{R}$? –  Neal Apr 14 '12 at 2:50
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2 Answers 2

up vote 2 down vote accepted

For a), follow the hint and consider $F_* : \mathbb R \to \mathbb R = Lie(S^1)$. Since $F_*$ is linear, it is of the form $t\mapsto ct$ for some constant $c$. For $\mathbb R$ you can think of the exponential map as being the identity map and for $S^1$ you can think of the exponential map $\mathbb R \to S^1$ as $t \mapsto e^{it}$. Using the general fact about the exponential map that $F \circ \exp = \exp \circ F_*$, we then have $$ F(t) = e^{ict}. $$ Now it is straightforward to check that for any $c$, $F$ is a homomorphism.

For b) use Neal's hint: $S^1$ is compact and connected so (since $F$ is continuous) the image of $F$ must also be compact and connected. But the only compact and connected Lie subgroup of $\mathbb R$ is just $\{0\}$. This should be pretty clear but one way to see it is that there is a one to one correspondence between Lie subalgebras of $\mathbb R$ and connected Lie subgroups. But the only subalgebras of $\mathbb R$ are itself and the 0 vector space which correspond to the Lie subgroups $\mathbb R$ and $\{0\}$. Now only the latter is compact. So the only such $F$ is the trivial homomorphism.

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Thank you very much for the explanation. What you have said makes perfect sense! –  Alex Kite Apr 15 '12 at 0:39
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Here is an alternative argument that I like using elementary analysis. Suppose you have $\phi : \Bbb{R} \to S^1$ a Lie group homomorphism. Then this means that $t \in \Bbb{R}$ is mapped to $e^{if(t)}$ for some function $f(t)$. Our goal now is to show that $f(t)$ is a linear function. Now it is elementary to show that

$$e^{if(t)} = e^{iat}$$

for all $t \in \Bbb{Q}$ where $a = f(1)$. We now have two continuous functions that are equal on a dense subset and hence are equal on all of $\Bbb{R}$. This completes the proof that any Lie group homomorphism $\phi : \Bbb{R} \to S^1$ is of the form $e^{i\alpha t}$ with $\alpha \in \Bbb{R}$.

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