Connected lie subgroup of $SO(3)$

Question

In this post All connected closed subgroups in $\rm{SO}(3)$ there is a classification of all connected lie subgroup of $SO(3)$. I would like to know if is possibile to arrive to the same conclusion using the one to one correspondence between connected lie subgroup and subalgebra of $\mathfrak{so(3)}$.

Answer 1

The Lie algebra $\mathfrak{so}(3)$ consists of the matrices of the form$$\begin{bmatrix}0&a&c\\-a&0&b\\-c&-b&0\end{bmatrix},$$with $a,b,c\in\Bbb R$. It is isomorphic to $(\Bbb R^3,\times)$, where $\times$ is the cross product. But it follows from this that it has no two-dimensional Lie subalgebras, since this would be the same thing as asserting that there are two linearly independent vectors $v,w\in\Bbb R$ such that $v\times w$ is a linear combination of $v$ and $w$.

So, if $\mathfrak g$ is a Lie subalgebra of $\mathfrak{so}(3)$ then either it is $\{0\}$, or it is the set $\{\lambda M\mid \lambda\in\Bbb R\}$ for some $M\in\mathfrak{so}(3)$, or it is the whole of $\mathfrak{so}(3)$.

If $\mathfrak g=\{0\}$, then the only connected Lie subgroup of $SO(3,\Bbb R)$ whose Lie algebra is $\mathfrak g$ is $\{\operatorname{Id}\}$.

If $\mathfrak g=\{\lambda M\mid \lambda\in\Bbb R\}$ for some $M\in\mathfrak{so}(3)$, then the only connected Lie subgroup of $SO(3,\Bbb R)$ for some $M\in\mathfrak{so}(3)$ whose Lie algebra is $\mathfrak g$ is $\{\exp(\lambda M)\mid\lambda\in\Bbb R\}$, which is the group of all rotations around an axis.

And if $\mathfrak g=\mathfrak{so}(3)$, then the only Lie subgroup of $SO(3,\Bbb R)$ whose Lie algebra is $\mathfrak g$ is $SO(3,\Bbb R)$ itself.