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?

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

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.

  • $\begingroup$ Thank you very much for the explanation. What you have said makes perfect sense! $\endgroup$ – Anon Apr 15 '12 at 0:39

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 $\alpha \in \Bbb{R}$.


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.