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I'm looking for two connected Lie groups $G, H$ with Lie algebras $\frak{g}, h$ and a Lie algebra homomorphism $\varphi: \frak{g} \to h$ that isn't of the form $D_0\phi$ for any Lie group homomorphism $\phi: G \to H$.

I'm thinking of looking for a locally injective $\varphi: \frak{so}(3) \to so(2)$. If I can find one then if $\varphi = D_0\phi$ for such Lie group homomorphism then we'd have:

$\phi \circ \text{exp} = \text{exp} \circ D_0\phi: \frak{}so(3) \to$ $SO(2)$. Since $\exp: \frak{so}(2) \to$ $SO(2)$ is surjective and in general exp is locally injective a known theorem states that then $\phi$ is a covering map. But this would imply that $\varphi$ is an isomorphism, which is a contradiction.

Is there such a homomorphism, or can you give a hint for another possible example?

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    $\begingroup$ This example won't work: since $\dim \mathfrak{so}(3) = 3$ and $\dim \mathfrak{so}(2) = 2$, every linear $\phi$ has kernel at least one dimensional. Thus no such $\phi$ is locally injective. Hint for another example: Pick $G$ to be non-simply connected and $H$ to be the universal cover of $G$. $\endgroup$ – Jason DeVito May 5 at 23:21
  • $\begingroup$ @JasonDeVito if we pick $G = S^1$ and $H = (\mathbb{R}, + )$ then their Lie algebras are isomorphic (right?). I'm not sure I see if this leads to a counter example. Can you elaborate please? $\endgroup$ – Mariah May 6 at 12:24
  • $\begingroup$ For your map $\varphi$, use an isomorphism (which exists, since, as you said, the Lie algebras are isomorphic). $\endgroup$ – Jason DeVito May 6 at 16:13
  • $\begingroup$ @JasonDeVito so if $\varphi$ is an isomorphism and $\varphi = D_0\phi$ this implies $\phi$ is a covering map. And $S^1$ cannot be a cover of $(\mathbb{R}, + )$? $\endgroup$ – Mariah May 6 at 16:37
  • $\begingroup$ Perfect. If you'd like, write up your own answer to this question. (And include a proof that $S^1$ can't cover $\mathbb{R}$. $\endgroup$ – Jason DeVito May 6 at 19:37

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