All connected closed subgroups in $\rm{SO}(3)$ I want to find all connected closed subgroups in $\rm{SO}(3)$. Is there a direct classification without attracting Lie algebras?
Thank you very much!
 A: Let $G\subseteq SO(3)$ be a connected closed subgroup.
There is a continuous map $A:SO(3)\to [0,\pi]$ which takes a rotation to its (unsigned) angle of rotation.  In particular, $A(G)$ must be a connected subset of $[0,\pi]$.  Since the identity is the unique rotation of angle $0$, if $A(G)$ is nontrivial, it must contain a nontrivial interval $[0,\epsilon]$.
In particular, if $G$ is nontrivial, it must contain a rotation by some irrational angle.  Since $G$ is closed, this implies $G$ contains the entire subgroup of rotations around the same axis, because powers of the irrational rotation are dense in all rotations about the same axis.
Now, say $v\in S^2$ is such that $G$ contains all rotations which fix $v$.  Suppose there exists $g\in G$ which does not fix $v$.  Let us now consider the orbit $Gv$, which must be connected and have more than one point.  Note that for every $w\in Gv$, $G$ contains all rotations which fix $w$ (by conjugating the rotations that fix $v$), and so the set $Gv$ must be closed under rotation around any element of $Gv$.
I now claim that $Gv$ contains an element which is orthogonal to $v$.  If $Gv$ contains a point whose inner product with $v$ is nonpositive, then this is immediate by the connectedness of $Gv$.  Otherwise, given $w\neq v\in G$, note that if we rotate $v$ around $w$ by $\pi$, we get a vector whose angle with $v$ is twice the angle formed by $w$ and $v$.  Iterating this, we will eventually reach a vector whose angle with $v$ is at least $\pi$, thus reducing to the first case.
So, $Gv$ contains a point which is orthogonal to $v$.  This means there are two orthogonal axes with respect to which $G$ contains all rotations.  But such rotations act transitively on $S^2$ (use one of them to set the "latitude" and then use the other to set the "longitude"), so $Gv=S^2$.  This means $G$ contains all rotations around all axes, so $G=SO(3)$.
So, to sum up, we have shown that if $G$ is nontrivial, it must contain all rotations about some axis, and if $G$ also contains another rotation which is not about the same axis, then $G$ must be all of $SO(3)$.  So the only connected closed subgroups of $SO(3)$ are the trivial group, the group of rotations about some fixed axis, and $SO(3)$ itself.
