Timeline for Topological structure of SO(n) as a product
Current License: CC BY-SA 3.0
15 events
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| Oct 30, 2015 at 9:27 | history | edited | Cepu | CC BY-SA 3.0 |
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| Oct 29, 2015 at 7:13 | history | edited | Cepu | CC BY-SA 3.0 |
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| Oct 9, 2015 at 13:29 | comment | added | მამუკა ჯიბლაძე | @BertramArnold Thanks for the clarification. Can something similar be said about $SO(8)$? | |
| Oct 9, 2015 at 12:00 | comment | added | Bertram Arnold | @მამუკაჯიბლაძე Identify $S^3$ with the unit quaternions. Then the map $S^3\times S^3\to SO(4), (v,w)\mapsto v\cdot w^{-1}$ is a two-fold covering. In this picture, the projection $SO(4)\to S^3$ sends $(v,w)$ to $vw^{-1}$, so it is not a group homomorphism. Also, the two Lie groups $SO(4)$ and $SO(3)\times S^3$ are not isomorphic, since the first has an outer automorphism of order 2 (orientation reversal) and the second doesn't. | |
| Oct 8, 2015 at 21:49 | comment | added | David Roberts♦ | Also for $n=1$ if one uses $O(n)$ instead of $SO(n)$ :-) | |
| Oct 8, 2015 at 20:36 | comment | added | მამუკა ჯიბლაძე | Also in b), are you sure it is product of groups and not something more complicated? Because if it is just product or even semidirect product or even a nontrivial extension by a nonabelian cocycle, it would imply that $SO(3)$ is a normal subgroup of $SO(4)$ which I doubt. | |
| Oct 8, 2015 at 20:34 | comment | added | მამუკა ჯიბლაძე | I became quite excited about c): does it mean that one can somehow express the group structure of $SO(8)$ via group structure of $SO(7)$ and the quasigroup structure of $S^7$, possibly using some kind of cocycle? | |
| Oct 8, 2015 at 20:32 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Oct 8, 2015 at 13:19 | history | edited | Cepu | CC BY-SA 3.0 |
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| Oct 7, 2015 at 9:04 | comment | added | Bertram Arnold | @PVAL Yes - it is the sum of the pullback of the tangent bundle of $S^{n-1}$ and a trivial bundle, and the first becomes trivial when you add a trivial line bundle to it. user51223, in general you do not get a trivialization of $E/F$ from a trivialization of $E$ and $F$. | |
| Oct 6, 2015 at 23:32 | comment | added | PVAL | Isn't $S^{n-1}\times SO(n-1)$ always parallelizable? | |
| Oct 6, 2015 at 22:43 | comment | added | Bertram Arnold | I don't understand how it follows that $S^{n-1}$ is parallelizable if $S^{n-1}\times SO(n-1)$ is - don't you just get that the tangent bundle is stably trivial? For instance, $S^{n-1}\times\mathbb{R}$ is always parallelizable since it is diffeomorphic to $\mathbb{R}^n\setminus\{0\}$. | |
| Oct 6, 2015 at 14:23 | vote | accept | Florian Oppermann | ||
| Oct 6, 2015 at 13:55 | history | edited | Cepu | CC BY-SA 3.0 |
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| Oct 6, 2015 at 13:50 | history | answered | Cepu | CC BY-SA 3.0 |