The group of three dimensional rotations $SO(3)$ is a subgroup of the Special Euclidean Group $SE(3) = \mathbb{R}^3 \rtimes SO(3)$. The manifold of $SO(3)$ is the three dimensional real projective space $RP^3$. Does $RP^3$ cause a separation of space in the manifold of $SE(3)$? 


(edit) Sorry about lack of clarity. My question should be worded as 'does $SO(3)$ partition any four dimensional subspace of $SE(3)$ into exactly two disjoint pieces?'


I am basically interested in understanding whether a generalization of the Jordan curve separation theorem works in such non Euclidean spaces. In particular, I want to know if (non) orientability of $SO(3)$ affects the generalization, especially since it is used to construct $SE(3)$ as a product space with $\mathbb{R}^3$.



Does the manifold of the three dimensional group of rotations SO(3) cause a separation of space in the group of rigid motions SE(3)?
Okay, now I think I understand your question. This is the question I will answer:


* Question: Let $X$ be a connected $4$-dimensional subspace of $SE(3)$ that contains $SO(3)$. Is it possible for $X \setminus SO(3)$ to be connected? Disconnected?


The answer to both questions is yes. So there is no Jordan separation theorem for $4$-dimensional subspaces of $SE(3)$ containing $SO(3)$. 


Observation 1: As a space, $SE(3)$ is just the cartesian product of $SO(3)$ with $\mathbb R^3$. Explicitly, we will think of $SE(3)$ as the set $SO(3) \times \mathbb R^3$. 


Observation 2: If $X := SO(3) \times \mathbb R$ embeds in $SE(3)$, therefore $SO(3) \times \{0\}$ disconnects it. 


Observation 3: If $X := SO(3) \times S^1$, where $S^1 = \{ x \in \mathbb R^2 : |x|=1\}$, then the map $X \to SO(3) \times \mathbb R^3$ given by $(p,x) \longmapsto (p,x,0)$ is an embedding. In particular, $X \setminus (SO(3) \times \{1\})$ is connected. 


So the answer to both your questions is yes. 


I'd like to suggest looking at the proof of the generalized Jordan-Brouwer theorem in Guillemin and Pollack, or perhaps in an algebraic topology textbook like Bredon's. This will give you a very flexible set of tools that will let you know quite generally when you can expect a separation theorem, and when you can't. 


Notice: my answer had nothing to do with the fact that $SO(3)$ has a non-trivial fundamental group, or whether or not it is orientable. The key part of the construction is that $SO(3)$ has co-dimension *at least* $2$ (And actually co-dimension $3$) in $SE(3)$. 


