# Showing $\mathfrak{so}_3(\mathbb{C}) \simeq \mathfrak{sl}_2(\mathbb{C})$ does not lift to a Lie group morphism

The adjoint representation induces an isomorphism of lie algebras $$\mathfrak{so}_3(\mathbb{C}) \simeq \mathfrak{sl}_2(\mathbb{C}).$$

While the lie algebra isomorphism from $\mathfrak{sl}_2(\mathbb{C})$ to $\mathfrak{so}_3(\mathbb{C})$ does lift to a corresponding Lie group morphism (because $SL_2(\mathbb{C})$ is simply connected), the same cannot be said for the isomoprhism on the opposite direction.

This is not surprising since $SO_3(\mathbb{C})$ is not simply connected. But I don't know how to prove this morphism does indeed not lift.

• Note that $O(3)$ and $SO(3)$ have isomorphic Lie algebra, but this isomorphism does not lift to the group level, of course. – Dietrich Burde Jan 22 '17 at 15:21

Let $\psi : SL_2(\mathbb{C}) \rightarrow SO_3(\mathbb{C})$ be the map lifting the inverse of the isomorphism stated above which exists since $SL_2(\mathbb{C})$ is simply connected. Assume $\phi : SO_3(\mathbb{C}) \rightarrow SL_2(\mathbb{C})$ be a map lifting the given isomorphism.
Claim : the map $\phi$ is an isomorphism.
Note that $d(\phi \circ \psi) = d\phi \circ d\psi = Id_{\mathfrak{sl}_2(\mathbb{C})}$ since these maps on the level of lie algebras are inverse to each other by definition. Hence $\phi \circ \psi$ and $Id$ are two group homomorphism with domain as connected groups with the same derivative at identity and hence are equal. Thus $\phi \circ \psi = Id_{SL_2(\mathbb{C})}$. Note that in the argument in this paragraph we have not used that $SL_2(\mathbb{C})$ is simply connected. Hence the same argument can be applied to get $\psi \circ \phi = Id_{SO_2(\mathbb{C})}$. Thus $\phi$ as well as $\psi$ are ismomorphisms.