# 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)?

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 $\mathbb{R}^3$.

• I don't understand the question. In particular, I don't understand the phrase "cause a separation of space". Certainly a codimension-3 submanifold does not separate the manifold into disconnected pieces when removed, which was my first read, and I don't have a second-read proposal. I recommend you look at mathoverflow.net/howtoask . In particular, do please define what you mean in more detail. Some context would also be very helpful. – Theo Johnson-Freyd Sep 28 '11 at 5:57
• I think your question isn't well-formulated. In particular, no $4$-dimensional subspace of $SE(3)$ can separate, since $SE(3)$ is 6-dimensional. Your question is analogous to asking if a point in $\mathbb R^2$ separates $\mathbb R^2$. I suggest reading the section on the Jordan-Brouwer Separation Theorem in Guillemin and Pollack's "Differential Topology" text, as it should both help you formulate your question and answer it. – Ryan Budney Sep 28 '11 at 6:18
• I am not asking if a four dimensional subspace of $SE(3)$ separates $SE(3)$. Rather, I would like to know if there is a four dimensional subspace of $SE(3)$ that is separated by three dimensional $SO(3)$, especially because $SO(3)$ is non orientable. Thanks. – Kurt Sep 28 '11 at 6:24
• FYI, the manifold $SO(3)$ is orientable, but the answer to your question would be unchanged if it was non-orientable -- say if you were interested in the same question with $SE(3)$ replaced by $M \times \mathbb R^3$ where $M$ is a non-orientable $3$-manifold. – Ryan Budney Sep 28 '11 at 17:13
• $\mathbb RP^3$ is orientable. $\mathbb RP^3$ is $S^3$ modulo the antipodal map $x \longmapsto -x$. This is an orientation-preserving map of $S^3$, since it's homotopic to the identity -- think of $S^3$ as the unit sphere in $\mathbb C^2$, so $x \longmapsto -x$ is $x \longmapsto zx$ where $z = -1$. But this map also makes sense for $z$ any unit complex number -- sliding $z$ from $-1$ to $1$ is the null-homotopy of the antipodal map. – Ryan Budney Sep 28 '11 at 20:21

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.

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 $SE(3)$.
I cannot post comments yet, but I am interested in the answer to these questions. It appears $R^2 \times SO(3)$ will not partition $SE(3)$ into disconnected pieces because $R^2 \times SO(3)$ is not compact. What about the set $M \times RP^3$ where $M$ is the Mobius strip? That is a five dimensional surface. Does it partition $SE(3)$? Also the original question is unanswered, does $SO(3)$ partition $R \times SO(3)$ into disconnected pieces? Curious to know.