Homotopy $\pi_0(G)$ of Lorentz group $G$
We know the Lorentz group is an indefinite orthogonal group $O(1,3)$ with $\mathbf{R}$ coefficients, which Lie group topology has four connected pieces due to $$ \pi_0(O(1,3))=\mathbf{Z}/2\oplus \mathbf{Z}/2 $$ The generators are called $P$ and $T$ in Wiki
Surfaces of transitivity of Lorentz group $G$
According to surfaces of transitivity (of Lorentz group $G$): 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 $gs_1 = s_2$. By definition of the Lorentz group, it preserves the quadratic form $$Q(x) = x_0^2 - x_1^2 - x_2^2 - x_3^2.$$
The surfaces of transitivity of the orthochronous Lorentz group $O_+(1, 3)$, $Q(x) {{=}} const.$ of spacetime are the following:
$Q(x) > 0, x_0 > 0$ is the upper branch of a hyperboloid of two sheets. Points on this sheet are separated from the origin by a future time-like vector.
$Q(x) > 0, x_0 < 0$ is the lower branch of this hyperboloid. Points on this sheet are the past time-like vectors.
$Q(x)= 0, x_0 > 0$ is the upper branch of the light cone, the future light cone.
$Q(x)= 0, x_0 <0$ is the lower branch of the light cone, the past light cone.
$Q(x) <0 is a hyperboloid of one sheet. Points on this sheet are space-like separated from the origin.
The origin $x_\mu = 0$.
I am not sure how to count the number of surfaces of transitivity for the orthochronous Lorentz group $O_+(1, 3)$.
But according to Wiki - surfaces of transitivity (of Lorentz group $G$), for the full Lorentz group $O(1, 3)$, the surfaces of transitivity are only four since the transformation T takes an upper branch of a hyperboloid (cone) to a lower one and vice versa.
$$ \text{So the number of surfaces of transitivity for the full Lorentz group $O(1, 3)$ is 4.} $$ But how is this 4 counted? The transformation T can also bring the future light cone to the past light cone. So according to this logic, $$ \text{we may only have 3 kinds of surfaces of transitivity? see below} $$
$Q(x) > 0, x_0 > 0$ and $Q(x) > 0, x_0 < 0$ are the upper and lower branch of a hyperboloid of two sheets. Points on this sheet are separated from the origin by a future time-like vector. But the two sheets are related by the $T$ transformation.
$Q(x)= 0, x_0 > 0$ and $Q(x)= 0, x_0 <0$: the upper and lower branches of the light cone, the future light cone. But the two sheets are related by the $T$ transformation.
$Q(x) <0 is a hyperboloid of one sheet. Points on this sheet are space-like separated from the origin.
Question: Relation between $\pi_0(O(1,3))=\mathbf{Z}/2\oplus \mathbf{Z}/2$ and $O(1,3)'s$ Surfaces of transitivity.
My question is that how the four connected pieces of Lorentz group $O(1,3)$ is related or established from the surfaces of transitivity of Lorentz group? https://en.wikipedia.org/wiki/Lorentz_group#Surfaces_of_transitivity
The missing piece of the puzzle there is the origin. I'm not sure if you can really call that a surface since it is 0-dimensional but it is present in the list on wikipedia so I assume that is the 4th kind of surface to which they refer.
On to how this relates to the connected components of $O(1,3)$. The best way to understand these components is by dividing the group up based on two properties. Firstly, as with the definite orthogonal group $O(n)$ some elements preserve orientation and some reverse it (i.e. they have determinant $\pm 1$) and the subgroup of those that preserve it, the special orthogonal group, is labelled $SO(1,3)$ and this is disconnected from the rest of the group. On the other hand we also have some elements which preserve the forward time direction (equally they preserve the forward light cone) and some which exchange the forward direction with the backward direction. The subgroup which preserves the time direction is the orthochronous group $O_+(1,3)$.
How do these interact? Well the intersection of these subgroups which we shall label $SO_+(1,3)$ is the elements which preserve both orientation and the forward light cone. This subgroup is connected and indeed it is the connected component of the identity. We could write all the components now as cosets of this component. If $A \in SO(1,3)$ which doesn't preserve the forward light cone and $B \in O_+(1,3)$ which doesn't preserve the orientation then the components are $SO_+(1,3)$, $A\cdot SO_+(1,3)$, $B\cdot SO_+(1,3)$, $AB\cdot SO_+(1,3)$ and for example $SO(1,3) = SO_+(1,3)\cup A\cdot SO_+(1,3)$.
So what have we seen through all this. Well firstly, the orientation preserving/not-preserving distinction doesn't make any difference to the orbits. $O(1,3)$, $SO(1,3)$ have the same "surfaces of transitivity". However, if we restrict to the orthochronous group some of the orbits are split into $2$. Specifically the light cone and the two-sheeted hyperboloids are split into their forward and backward components.