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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 $\mathbb{R}^3$.

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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
    
What does "separated by three dimensional $SO(3)$" mean? –  Ryan Budney Sep 28 '11 at 6:28
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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

3 Answers 3

up vote 3 down vote accepted

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 $SE(3)$.

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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.

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I am not entirely sure what you mean by separation of space. But, would n't it depend on the representation of SE(3) and SO(3). For example, one can take the view that SE(3) is a dual projective space R\hat{P}^3 by using dual quaternion representation for spatial rigid body displacement.

I am not a mathematician, so please ignore me if what I say does not make sense.

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