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 $\mathfrak{so(3)}$.
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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.
Solveit
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José Carlos Santos
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Why ${exp(\lambda M)}$ is the group of rotation around an axis? – Tyrion Apr 24 '21 at 11:14
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1@Tyrion: Can you compute the action on the linear subspace of $\mathbb{R}^3$ corresponding to $M\in\mathfrak{so}(3)$? – user10354138 Apr 24 '21 at 11:37
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No but i answer my self using jordan form to compute $exp(\lambda M)$ – Tyrion Apr 24 '21 at 11:46
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1If $M\in\mathfrak{so}(3)$, then there is an orthogonal matrix $N$ such that$$N^{-1}MN=\begin{bmatrix}0&0&0\0&0&-a\0&a&0\end{bmatrix}$$for some $a\in\Bbb R$. So$$N^{-1}\exp(M)N=\begin{bmatrix}1&0&0\0&\cos a&-\sin a\0&\sin a&\cos a\end{bmatrix}.$$ – José Carlos Santos Apr 24 '21 at 11:50
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Is, there a way of doing this without exponential map? – Tyrion Apr 27 '21 at 14:45
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With Lie algebras, but without the exponential map? I don't think so. – José Carlos Santos Apr 27 '21 at 14:46