Timeline for 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)?
Current License: CC BY-SA 3.0
17 events
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| Sep 28, 2011 at 20:21 | comment | added | Ryan Budney | $\mathbb RP^3$ 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. | |
| Sep 28, 2011 at 18:18 | comment | added | Kurt | Excellent answers. Thank you. But you say the manifold $SO(3)$ is orientable. Does it not have the structure of the projective space $RP^3$, which is not orientable? | |
| Sep 28, 2011 at 18:14 | vote | accept | Kurt | ||
| Sep 28, 2011 at 17:13 | comment | added | Ryan Budney | 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. | |
| Sep 28, 2011 at 17:07 | answer | added | Ryan Budney | timeline score: 3 | |
| Sep 28, 2011 at 16:58 | answer | added | Sai | timeline score: 0 | |
| Sep 28, 2011 at 12:28 | comment | added | Kurt | @Ryan: Yes that is what I would like to understand. How would you prove that $SO(3)$ partitions $R \times SO(3)$ into disjoint pieces? Similarly does $R \times SO(3)$ partition $R^2 \times SO(3)$, and does $R^2 \times SO(3)$ partition $SE(3)$? Thank you! | |
| Sep 28, 2011 at 9:03 | comment | added | Sam Nead | Hmmm. Perhaps you mean that the topological space $\mathbb R \times SO(3)$ nicely embeds into the isometry group of $\mathbb R^3$. Ok. | |
| Sep 28, 2011 at 9:01 | comment | added | Sam Nead | @Ryan - What is $\mathbb R \rtimes SO(3)$? It looks like you are saying that there is a special line in $\mathbb R^3$ preserved by all rotations? | |
| Sep 28, 2011 at 6:36 | comment | added | Ryan Budney | In $SE(3)$ there is $\mathbb R \rtimes SO(3)$. This is a 4-dimensional subspace of $\mathbb R^3 \rtimes SO(3)$, and $SO(3)$ partitions it into two disjoint subspaces. Is that what you're asking about? | |
| Sep 28, 2011 at 6:28 | comment | added | Ryan Budney | What does "separated by three dimensional $SO(3)$" mean? | |
| Sep 28, 2011 at 6:24 | comment | added | Kurt | 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. | |
| Sep 28, 2011 at 6:18 | comment | added | Ryan Budney | 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. | |
| Sep 28, 2011 at 6:07 | history | edited | Kurt | CC BY-SA 3.0 |
added 500 characters in body
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| Sep 28, 2011 at 5:57 | comment | added | Theo Johnson-Freyd | 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. | |
| Sep 28, 2011 at 5:43 | answer | added | John Galt | timeline score: -1 | |
| Sep 28, 2011 at 5:24 | history | asked | Kurt | CC BY-SA 3.0 |