Timeline for Isometric embedding of SO(3) into an euclidean space
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 25, 2014 at 11:27 | comment | added | Vít Tuček | @RobertBryant: Thank you for clearing this up. | |
| Mar 25, 2014 at 11:23 | comment | added | Robert Bryant | @VítTuček: Yes, you are right; equations (29–31). The others are for an embedding into an indefinite space; I was too hasty when I was looking back at the file. | |
| Mar 25, 2014 at 11:18 | comment | added | Vít Tuček | @RobertBryant: I see. Don't you mean equations (29) - (31)? | |
| Mar 25, 2014 at 11:10 | comment | added | Robert Bryant | @VítTuček: However, they compute the induced metric embedding parameters explicitly and relate them to the $r_\mu$ in their equations $(55)$ and $(56)$, and it's obvious from these formulae that they don't get the bi-invariant metric when $r_3=0$. That's what I mean by 'explicitly point out'. | |
| Mar 25, 2014 at 10:58 | comment | added | Vít Tuček | @RobertBryant: I don't see that statement anywhere in the paper. Rather it seems that they obtain embedding of $\mathrm{SO}(3)$ only for special choice $r_\mu = 1$, $\mu = 1, 2, 3$. For $r_3 = 0$ one basically gets the embedding you've described in mathoverflow.net/questions/161295/… | |
| Mar 25, 2014 at 8:47 | comment | added | Robert Bryant | @VítTuček: No. In their paper, they explicitly point out that the embedding into $\mathbb{R}^6$ that one gets for $r_3=0$ is not isometric for the bi-invariant metric. (Of course, it is isometric for a different left-invariant metric, just not the bi-invariant one.) | |
| Mar 25, 2014 at 6:52 | comment | added | Robert Bryant | @CarloBeenakker: Section 8 of the paper you cite only proves that the matrix embedding of $\mathrm{SO}(3)$ into $\mathbb{R}^9$ does not lie in a hyperplane (which was obvious from the beginning anyway). It says nothing about whether or not there is an isometric embedding of $\mathrm{SO}(3)$ into a lower dimensional space. | |
| Mar 25, 2014 at 0:31 | comment | added | David E Speyer | It it easy to see that we can't do better if we want the action of $SO(3) \times SO(3)$ to extend linearly to $\mathbb{R}^n$: Irreducible representations of $SO(3)$ have dimension $2m+1$, so irreps of $SO(3) \times SO(3)$ have dimension $(2m+1) \times (2n+1)$, so the smallest representation of the product on which neither $SO(3)$ acts trivially is dimension $3 \times 3=9$. But it isn't clear to me that the symmetries have to extend to the ambient space. | |
| Mar 25, 2014 at 0:21 | comment | added | Vít Tuček | @CarloBeenakker: Doesn't they state that for $r_3 = 0$ one gets isometric embedding into $\mathbb{R}^6$? | |
| Mar 25, 2014 at 0:11 | comment | added | Anton Petrunin | This is the same as the Veronese embedding $\mathbb{R}\mathrm{P}^3\hookrightarrow \mathbb{R}^9$. | |
| Mar 24, 2014 at 22:09 | comment | added | Carlo Beenakker | @PaulSiegel --- yes, 9 seems large, but unless I have misunderstood section 8 of the cited paper it is minimal. | |
| Mar 24, 2014 at 21:42 | comment | added | Paul Siegel | But of course there are lots of relations - elements above the diagonal determine the elements below the diagonal, determinant is $1$, etc. So $9$ is certainly not minimal. | |
| Mar 24, 2014 at 21:26 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |