Tukukkk

Lie's second theorem says that if $G$ is a simply connected Lie group, then every isomorphism $Phi$ of its Lie algebra $mathfrak{g}$ lifts to an isomorphism $phi$ of $G$, i.e. such that $dphi_e = Phi$ where we identify $mathfrak{g}$ with $T_e G$, the tangent space to $G$ at the identity.

Now consider $G = mathrm{SO}(3)$, which is not simply connected. Is there an explicit counterexample to Lie's second theorem in this case? That is, can we write down a Lie algebra isomorphism $Phi$ of $mathfrak{so}(3)$ which is not the differential of any isomorphism $phi$ of $mathrm{SO}(3)$?

I feel like, if $(eta_1, eta_2, eta_3)$ is the usual basis of $mathfrak{so}(3)$, where $[eta_i, eta_j] = eta_k$ cyclically, then a map like $Phi(eta_1)=eta_2$, $Phi(eta_2)=eta_1$, $Phi(eta_3)=-eta_3$ ought to work, but I can't figure out how to prove there is no such group isomorphism $phi$. I've been trying to find $t_1, t_2, t_3 in mathbb{R}$ and a relation involving $(e^{t_1 eta_1}, e^{t_2 eta_2}, e^{t_3 eta_3})$ that is not satisfied by $(e^{t_1 eta_2}, e^{t_2 eta_1}, e^{-t_3 eta_3})$ but I can't come up with one.

There is an isomorphism $r: text{Aut}(SU(2)) to text{Aut}(SO(3))$, because any automorphism of $SU(2)$ necessarily preserves the center and $SO(3) = SU(2)/Z(SU(2))$. To see that $r$ is injective, observe that if $r(rho) = text{Id}$, then necessarily $rho(g) = pm g$ for each $g$. By continuity, the sign is independent of $g$, and because $rho$ is a homomorphism, the sign is 1, as desired.

To see that $r$ is surjective, observe that taking the derivative at the identity gives an embedding $text{Aut}(G) hookrightarrow text{Aut}(mathfrak g)$. By Lie's theorem, the map $text{Aut}(SU(2)) to text{Aut}(mathfrak{su}(2))$ is an isomorphism, and in particular we have factored this isomorphism as the composite of two injective maps. Both maps must therefore be isomorphisms.

In fact, $$text{Aut}(SU(2)) = text{Inn}(SU(2)) cong text{Inn}(SO(3)) = SO(3).$$

José's answer gives an example of a map of Lie algebras which may not be lifted to a map of Lie groups, but we see here that if you're looking to find an automorphism of $mathfrak{so}(3)$ which does not lift to an automorphism of $SO(3)$, you are out of luck. You'll have to try something more complicated.

See $SU(2)$ as the group of quaterninons with norm $1$. Let$$operatorname{Im}mathbb{H}={ai+bj+ckinmathbb{H},|,a,b,cinmathbb{R}}$$and, for each $qin SU(2)$, let $varphi(q)colonoperatorname{Im}mathbb{H}longrightarrowoperatorname{Im}mathbb{H}$ be the map defined by $varphi(q)(v)=q^{-1}vq$. Then $operatorname{Im}mathbb{H}simeqmathbb{R}^3$ and $varphi$ is a surjective group homomorphism from $SU(2)$ onto $SO(3,mathbb{R})$, whose kernel is $pm1$. Then $Dvarphi_1colonmathfrak{su}(2)longrightarrowmathfrak{so}(3,mathbb{R})$ is a Lie algebra isomorphism and its inverse is a representation of $mathfrak{so}(3,mathbb{R})$ in $mathbb{C}^2$. But you cannot lift it to $SO(3,mathbb{R})$, since $SU(2)$ and $SO(3,mathbb{R})$ are not isomorphic.