Timeline for Manifold_Lie algebra compatibility
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 25, 2015 at 14:22 | comment | added | Robert Bryant | @AliTaghavi: Yes, it is obvious: If such an $\alpha$ exists, let $F\to M$ be the bundle of coframes (i.e., linear isomorphisms) $u:T_xM\to L$ that are isomorphisms of $(T_xM,\alpha_x)$ with $(L,[,])$ as Lie algebras. Then $F\to M$ is a reduction of the $\mathrm{GL}(n,\mathbb{R})$ coframe bundle to a principal right $\mathrm{Aut}(L)$-subbundle. | |
| Aug 25, 2015 at 12:44 | comment | added | Ali Taghavi | I am sorry if my question is elementary: Is it obvious that the conditions in my initial question automatically implies that the structure group can be reduced to Aut(L)? | |
| Aug 25, 2015 at 11:13 | comment | added | Robert Bryant | If $\alpha$ is smooth, then, for any linearly independent vector fields $X$ and $Y$, $\alpha(X,Y)$ will be a smooth nonvanishing vector field that takes values in the subspace at each point, so the range will be a smooth subbundle. For your second comments question, the answer is 'yes'. | |
| Aug 25, 2015 at 11:13 | comment | added | Ali Taghavi | Thank you for the answer. Does the question automatically imply that "We have a Bundle of Lie algebras"? | |
| Aug 25, 2015 at 11:11 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added 471 characters in body
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| Aug 25, 2015 at 11:08 | comment | added | Ali Taghavi | Fibrewise the range is a subvector space of tangent space. But why is it a subbundle? | |
| Aug 25, 2015 at 11:04 | vote | accept | Ali Taghavi | ||
| Aug 25, 2015 at 11:01 | history | answered | Robert Bryant | CC BY-SA 3.0 |