Homotopy $_0()$ and surfaces of transitivity of a generic Lie group $$

Question

Given a Lie group $G$,

• the path connected pieces are counted via the homotopy $_0()$.

• the distinct number of surfaces of transitivity --- According to surfaces of transitivity (of $G$): If a group $G$ acts on a space $V$, then a surface $S ⊂ V$ is a surface of transitivity if $S$ is invariant under $G$, i.e., $∀ g ∈ G$, $∀s ∈ S : gs ∈ S$, and for any two points $ s_1 , s_2 ∈ S$ there is a $g ∈ G$ such that $gs_1 = s_2$.

My question is that how the distinct number of surfaces of transitivity relates to the homotopy $_0()$? Is it possible that we have only a kind of surfaces of transitivity, but with a nonzero $_0() \neq 0$? Alternatively, is it possible that we have many ditinct kind of surfaces of transitivity, but with a zero $_0()= 0$? Are there interesting relations between Homotopy $_0()$ and surfaces of transitivity of Lie group $$, at all?

I will give three example.

1. Lorentz group $O(1,3)$, it preserves the quadratic form $$Q(x) = x_0^2 - x_1^2 - x_2^2 - x_3^2,$$ the surfaces of transitivity have at least 3 kinds of surfaces of transitivity: 2-sheet hyperbola, lightcone, 1-sheet hyperbola. (in a separate question we are arguing how to count such surfaces Relation between Homotopy $\pi_0(G)$ and Surfaces of transitivity --- of Lorentz group $G$)

While $$\pi_0(O(1,3))=\mathbf{Z}/2\oplus \mathbf{Z}/2.$$ We have some discussions here in Relation between Homotopy $\pi_0(G)$ and Surfaces of transitivity --- of Lorentz group $G$

2. Special orthogonal group $SO(3)$, it preserves the quadratic form $$Q(x) = x_1^2 + x_2^2 + x_3^2,$$ allows only Lie group elements with determinant 1. The surfaces of transitivity have at least 1 kind of surfaces of transitivity: $$S^2,$$ Possibly only one kind. While $$\pi_0(SO(3))=0.$$

3. Special orthogonal group $SO(3)$, it preserves the quadratic form $$Q(x) = x_1^2 + x_2^2 + x_3^2,$$ but allows only Lie group elements with determinant $\pm 1$. The surfaces of transitivity have at least 1 kind of surfaces of transitivity: $S^2$. Possibly only one kind of surface. While $$\pi_0(O(3))=\mathbf{Z}/2.$$

Answer 1

As I responded to your related question, there isn't a particular link here. The strongest result that I'm aware of here is that if $G$ is connected then all of its orbits must be connected as well (each element $g$ is path connected to the identity and you let this path descend to the space you are looking at).

Other than that, adding more components to a connected group can only reduce the number of orbits. That is, elements of these new components might exchange some of the orbits of the connected component of the identity, thus reducing them to a single orbit.