# Lie algebra homomorphism $\varphi: \frak{g} \to h$ not of the form $D_0\phi$

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?

• 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 $G$. – Jason DeVito May 5 at 23:21
• @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? – Mariah May 6 at 12:24
• For your map $\varphi$, use an isomorphism (which exists, since, as you said, the Lie algebras are isomorphic). – Jason DeVito May 6 at 16:13
• @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 $(\mathbb{R}, + )$? – Mariah May 6 at 16:37
• Perfect. If you'd like, write up your own answer to this question. (And include a proof that $S^1$ can't cover $\mathbb{R}$. – Jason DeVito May 6 at 19:37