Classify the embedding of Lie groups: $U(1)$ in $SU(2)$ versus $U(1)$ in $SO(3)$ I am interested in knowing the way to embed a Lie group to another Lie group.
For example, we can
embed  $$U(1) \subset SU(2) \tag{1}$$ also
$$U(1) \subset SO(3). \tag{2}$$
Here they are all regarded as some Lie groups. But in terms of differentiable manifolds, $SU(2) \cong S^3$ as a 3-sphere and $SO(3)  \cong RP^3$ as a real protective space both in real 3-dimensions.
Now, my question is that how do we classify the way of their embedding? Here I must specify a way to define what exactly is a classification. But I am not certain what is the correct math definition. What I can prescribe is that there should be some identification of deformations between different embedding. As long as the embedding map can be continuous deformed to each other (maybe in the sense of homeomorphic), then the embedding maps are identified.

Question: Is there a concrete math definition of such classification of the Lie group embedding identified via continuous deformations or homeomorphic)?


*

*My take is that to classify the embedding of Lie groups: $U(1)  \cong S^1 \subset SU(2) \cong S^3$, since they are all Lie groups so their identities must be the same identity group element $1$. Different embedding of $U(1)$ must intersect at the identity point on $SU(2)$. Next we can continuously deform such $U(1)$ on $SU(2)$. For those maps which cannot be continuous deformed, I expect that they are classified by the homotopy class
$$
[U(1), SU(2)]=[S^1,S^3]=\pi_1(S^3)=0.
$$
Thus there is only one way of embedding of $U(1) \subset SU(2)$ in terms of homotopy or homeomorphic.


*To classify the embedding of Lie groups: $U(1)  \cong S^1 \subset SO(3) \cong RP^3$, since they are all Lie groups so their identities must be the same identity group element $1$. Different embedding of $U(1)$ must intersect at the identity point on $SO(3)$. Next we can continuously deform such $U(1)$ on $SO(3)$. For those maps which cannot be continuous deformed, I expect that they are classified by the homotopy class
$$
[U(1), SO(3)]=[S^1,RP^3]=\pi_1(RP^3)=\mathbf{Z}/2.
$$
Thus there are two ways of embedding of $U(1) \subset SO(3)$ in terms of homotopy or homeomorphic.
In terms of the lift map, we can see that the nontrivial $\mathbf{Z}/2$ class of the embedding $[U(1), SO(3)]$ is the obstruction to lift the $U(1) \to SO(3)$ to $U(1) \to SU(2)$.
And a lift diagram is like:
$$
\begin{array}{ccc}
  &  & SU(2)\\
          &\nearrow &           \downarrow\\
  U(1) & \longrightarrow & SO(3)
\end{array}.
$$
Please correct me and point toward the good way to think about the classifications of the embedding of Lie groups.
 A: Further to my comment. The fact that any two maximal tori are conjugate can be found in any textbook on Lie groups (or the wikipedia page on maximal torus https://en.wikipedia.org/wiki/Maximal_torus). It means that If $T_1,T_2 \subset G$ are maximal tori, then there is $g \in G $ such that $$g T_1 g^{-1} = T_2 (*)$$
It proves if $G$ is a connected Lie group then image of the maximal torus in $\pi_1(G)$ is the same for all of the maximal tori. Since, if (*) holds, then pick a continous path $\phi: [0,1] \rightarrow G$, such that $\phi(0)=Id$, $\phi(1)=g$. Then the isotopy of $G$. $$(h,t) \mapsto \phi(t)h \phi(t)^{-1} ,$$ deforms $T_{1}$ to $T_{2}$ continously.
It remains to see whether the class of the maximal torus $U(1) \subset SO(3)$ is trivial or not. The maximal torus in $SO(3)$ indeed is the non-trivial element of $\pi_1(SO(3)) = \mathbb{Z}_{2}$.
For a brief sketch of this, lets recall that the set of unit quaterions $\{a+bi+cj+dk: a^2+b^2+c^2+d^2=1\}$is the Lie group $SU(2) = S^3$, and a $U(1)$ subgroup is given by $\{a+bi : a^2+b^2=1\}$.Geometrically this is the intersection of a $2$-dimensional plane through the origin with $S^3$. Next, we use the identification $S^{3}/{\pm 1} = SO(3)$.
When we quotient by $\pm 1$, the maximal torus in $S^3$ double covers the maximal torus in $SO(3)$. Hence the maximal torus in $SO(3)$ has a lift which is a path in $S^3$ between antipodal points. Such curves are well-known to generate $\pi_{1}(SO(3))$.
