Timeline for A question on involutions on the Lie algebra of vector fields
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| S Jul 29, 2014 at 6:51 | history | bounty ended | Ali Taghavi | ||
| S Jul 29, 2014 at 6:51 | history | notice removed | Ali Taghavi | ||
| Jul 29, 2014 at 6:44 | vote | accept | Ali Taghavi | ||
| Jul 29, 2014 at 6:06 | answer | added | Peter Michor | timeline score: 3 | |
| Jul 26, 2014 at 12:19 | comment | added | Peter Michor | @Ali Taghavi: $H^\infty=\bigcap_k H^k$ is the intersection of all Sobolev spaces. On a compact manifold it is $C^\infty$. In any case it is a Frechet space. | |
| Jul 26, 2014 at 11:31 | comment | added | Ali Taghavi | @PeterMichor Thanks for your comment. What is the exact definition of $H^{\infty}$? Is It a Hilbert space? I think such space soves the problem which you said | |
| Jul 25, 2014 at 14:01 | comment | added | Peter Michor | @Ali Taghavi: If $X$ is a smooth vector field then $\text{ad}_X$ math the Sobolev space of $H^k$-vector fields to the Sobolev space of $H^{k-1}$-vector fields. It is an unbounded operator on $H^{k-1}$ and its domain of definition consists of all $H^{k-1}$ vector fields which are $k$-times weakly differentiable along the flow lines of $X$. So there is no change to search for trace class operators among them. | |
| Jul 22, 2014 at 21:38 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| S Jul 22, 2014 at 21:32 | history | bounty started | Ali Taghavi | ||
| S Jul 22, 2014 at 21:32 | history | notice added | Ali Taghavi | Draw attention | |
| Jul 22, 2014 at 21:29 | comment | added | Ali Taghavi | @IanAgol This gives us a Lie subalgebra of $\chi^{\infty}(M)$, consists vector fields for which $ad$ is a trace class operator. Now a good question: To what extent this Lie algebra can be represented? What type of vector fields belong to this Lie algebra? Obviously the Killing form can be defined on this Lie algebra.Can you help to modify this idea? | |
| Jul 22, 2014 at 21:24 | comment | added | Ali Taghavi | @IanAgol your very essntial comment about the killing form is a motivation to consider the following:The problem is that we do not have a "trace" for defining the killing form. From this obstraction, we extract the following construction. For a compact Riemannian manifold $M$, one can construct a sobolov hilbert space which contains $\chi^{\infty}(M)$ as a dense space(ex:$H^{\infty}$) such that for each smooth vector field $X$, the operator $ad(X)$ is a bounded operator. Now we search for "trace class" operators, those $X$ for which $ad(X)$ is a trace class operator. | |
| Jul 21, 2014 at 10:09 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Jul 21, 2014 at 4:16 | answer | added | Victor Protsak | timeline score: 2 | |
| Jul 20, 2014 at 23:14 | comment | added | Ian Agol | On odd dimensional spheres, the antipodal map works. | |
| Jul 20, 2014 at 20:06 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Jul 20, 2014 at 20:03 | comment | added | Ali Taghavi | @IanAgol I apologize for my mistake in the question, I revised it. Thank you for your comment | |
| Jul 20, 2014 at 17:29 | comment | added | Ian Agol | What do you have in mind for the Killing form for the space of vector fields? | |
| Jul 20, 2014 at 10:13 | history | edited | Ali Taghavi |
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| Jul 20, 2014 at 9:58 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Jul 20, 2014 at 9:41 | history | asked | Ali Taghavi | CC BY-SA 3.0 |