Timeline for Isometric embedding of SO(3) into an euclidean space
Current License: CC BY-SA 3.0
13 events
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| Mar 25, 2014 at 14:37 | comment | added | Danny Ruberman | @RobertBryant: Well, I'd say that it's straightforward that the normal bundle of $S^2$ is trivial; there is a series of embeddings $S^2 \subset \mathbb{R^3} \subset \mathbb{R^4} \subset \mathbb{R^5}$, each of which has trivial normal bundle. But everyone has their own idea of `obvious'. | |
| Mar 25, 2014 at 11:19 | comment | added | Robert Bryant | @DannyRuberman: OK. However, it's already obvious that $\mathrm{SO}(3)$ embeds into $S^2\times S^2$ as order pairs of orthonormal vectors, so you don't need to mention the disk bundle. After that, any way you get $S^2\times S^2$ embedded into $\mathbb{R}^5$ does the job. Your way, you do need to know that the normal bundle of $S^2$ in $\mathbb{R}^5$ is trivial, which is not completely obvious (to me, anyway; I had to think about it). | |
| Mar 25, 2014 at 8:38 | comment | added | Danny Ruberman | Here's an easier (to me anyway) non-isometric embedding: SO(3) is the unit sphere bundle of $S^2$, and hence bounds the Euler class 2 disk bundle over the 2-sphere, say $W$. Doubling $W$ gives $S^2 \times S^2$, which bounds $S^2 \times B^3$, which embeds in $\mathbb{R}^5$ as a neighborhood of $S^2$. Essentially the same argument shows that every orientable $3$-manifold embeds in $\mathbb{R}^5$, a result of Hirsch. | |
| Mar 25, 2014 at 6:30 | comment | added | Robert Bryant | If you want to embed $\mathrm{SO}(3)$ into $\mathbb{R}^5$ (non-isometrically), the simplest way I know is to first embed it into $\mathbb{R}^6$ by projecting onto the first two columns of the matrix element in $\mathrm{SO}(3)$. Then, note that the image of this embedding lies in a hypersphere $S^5\subset \mathbb{R}^6$. Remove a point of this $S^5$ that does not lie in the image and then stereographically project this complement onto $\mathbb{R}^5$. This gives a non-isometric, non-equivariant embedding of $\mathrm{SO}(3)$ into $\mathbb{R}^5$. | |
| Mar 25, 2014 at 1:03 | comment | added | Robert Bryant | @DavidSpeyer: Actually, no. The generic orbit in this $\mathbb{R}^5$ is not $\mathrm{SO}(3)$; it is a quotient $\mathrm{SO}(3)/(\mathbb{Z}_2\times\mathbb{Z}_2)$. This is because the generic $3$-by-$3$ symmetric matrix with trace zero and distinct eigenvalues is stabilized by the subgroup of order $4$ that fixes the eigenspaces. Moreover, none of the metrics induced on these orbits is bi-invariant. | |
| Mar 25, 2014 at 0:34 | comment | added | David E Speyer | The embedding isn't hard to see. $SO(3)$ acts (by conjugation) on the space of $3 \times 3$ symmetric matrices of trace $0$, which is isomorphic to $\mathbb{R}^5$. A generic $SO(3)$ orbit in this space is isomorphic to $SO(3)$. But that isn't isometric, at least not for a general orbit. | |
| Mar 25, 2014 at 0:18 | comment | added | Vít Tuček | Right. :) I've changed my answer accordingly. | |
| Mar 25, 2014 at 0:18 | history | edited | Vít Tuček | CC BY-SA 3.0 |
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| Mar 25, 2014 at 0:05 | comment | added | Mariano Suárez-Álvarez | @AntonPetrunin, yeah, sure. We all default to $C^\infty$, though :-) | |
| Mar 25, 2014 at 0:04 | comment | added | Anton Petrunin | @MarianoSuárez-Alvarez OP did not specify the class of embeddings, if it is $C^1$ then by Nash--Kuiper there is one, for $C^\infty$-embeddings 9 might be the best. | |
| Mar 25, 2014 at 0:00 | history | edited | Vít Tuček | CC BY-SA 3.0 |
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| Mar 24, 2014 at 23:50 | comment | added | Mariano Suárez-Álvarez | But the question is about isometric embedding for a specific metric. A flat torus embeds as a submanifold in 3-space but not isometrically, for example. | |
| Mar 24, 2014 at 23:44 | history | answered | Vít Tuček | CC BY-SA 3.0 |