Timeline for Lie subalgebras of $\chi^{\infty}(M)$ of codimension $n = \dim M$
Current License: CC BY-SA 3.0
24 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 4, 2017 at 19:28 | vote | accept | Ali Taghavi | ||
| Apr 4, 2017 at 19:28 | comment | added | Ali Taghavi | @PeterMichor Prof. Michor thank you again for your very interesting answer and very interesting reference by Grabowski. My sincere apoloigy for my late "accept". | |
| Apr 4, 2017 at 19:24 | vote | accept | Ali Taghavi | ||
| Apr 4, 2017 at 19:28 | |||||
| S Aug 9, 2016 at 11:54 | history | suggested | Ali Taghavi | CC BY-SA 3.0 |
I add a link
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| Aug 9, 2016 at 10:59 | comment | added | Ali Taghavi | @PeterMichor Thank you for this revised version and very helpful information. | |
| Aug 9, 2016 at 10:57 | review | Suggested edits | |||
| S Aug 9, 2016 at 11:54 | |||||
| Aug 8, 2016 at 18:38 | history | bounty ended | CommunityBot | ||
| Aug 7, 2016 at 12:48 | history | edited | Peter Michor | CC BY-SA 3.0 |
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| Aug 7, 2016 at 11:52 | history | edited | Peter Michor | CC BY-SA 3.0 |
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| Aug 7, 2016 at 7:40 | history | edited | Peter Michor | CC BY-SA 3.0 |
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| Aug 6, 2016 at 18:12 | comment | added | David E Speyer | I just found a paper of Hurtado's which seems relevant; see my answer to your other question. | |
| Aug 6, 2016 at 18:11 | comment | added | Ali Taghavi | @DavidSpeyer Thank you very much for your attention to my question. | |
| Aug 6, 2016 at 17:25 | comment | added | David E Speyer | I can't think of any finite codimension subalgebra $L$ which is not contained in some $L_p$. One wants to prove this by reducing to the analogous result for $C^{\infty}(M)$. A subgoal I've been thinking about is to take $M = \mathbb{R}^n$, in which case $\mathrm{Vect}(M) \cong C^{\infty}(M)^{\oplus n}$, and try to show that finite codimension sub-Lie-algebras of $\mathrm{Vect}(M)$ must be $C^{\infty}(M)$ submodules. But no success so far, and I wouldn't be amazed if the reason is that there really are some very unusual sub-Lie-algebras which I haven't thought of. | |
| Aug 6, 2016 at 16:11 | comment | added | Ali Taghavi | @DavidSpeyer Yes. Me too I think the space is not closed under lie bracket. So, according to your previous comments, what do you guess about the minimum codimension of Lie subalgebras of $\chi^{\infty}(M)$? | |
| Aug 6, 2016 at 16:03 | comment | added | David E Speyer | Actually, I think the computation must be wrong somehow. In $\mathbb{R}^2$, let $X = \partial/\partial x$ and $Y = x \partial/\partial y$. Then $X$ and $Y$ both have coefficient of $\partial/\partial y$ equal to $0$ at the origin, and both have divergence $0$, but $[X,Y] = \partial/\partial y$, for which the coefficient of $\partial/\partial y$ at the origin is nonzero. | |
| Aug 6, 2016 at 15:30 | comment | added | David E Speyer | Here is a more concrete question, although it may simply be an ignorance of how currents work: Choose $p$ and $\alpha$ as you say, and let $H$ be the hypersurface $u^1=0$ near $p$. If $X$ is a vector field tangent to $H$, then you show that $i_X(\alpha \otimes \delta_p)=0$. But I think it makes sense to flow $\alpha \otimes \delta_p$ along $X$ for a positive amount of time and, if I do, I think I get a different current, supported at some $p' \in H$ other than $p$. How can the current be killed by the Lie algebra action, but moved by its exponential? | |
| Aug 6, 2016 at 15:24 | comment | added | David E Speyer | While I agree with the verification, I am confused by the result. I would have thought that a codimension $k$ subalgebra of $\mathrm{Vect}(M)$ would correspond to a codimension $k$ subgroup of $\mathrm{Diff}(M)$, and thus to a $k$ dimensional space on which $\mathrm{Diff}(M)$ acts. But, if we allow $\mathrm{Diff}(M)$ to act on pairs $(p, a)$ where $a$ is in $T^{\ast}_p M$, we expect an orbit of dimension $2 \dim M$, not $2$. Is there some way to correct my intuition, beyond just saying "infinite dimensional Lie groups are hard."? | |
| S Aug 6, 2016 at 14:38 | history | suggested | Ali Taghavi | CC BY-SA 3.0 |
I replace dimension by codimension
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| Aug 6, 2016 at 13:56 | review | Suggested edits | |||
| S Aug 6, 2016 at 14:38 | |||||
| Aug 5, 2016 at 9:57 | comment | added | Ali Taghavi | My apology for this second message. Am I missing some thing if I think that may the space you mentioned is not closed under lie bracket? | |
| Aug 3, 2016 at 18:41 | history | edited | Peter Michor | CC BY-SA 3.0 |
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| Aug 3, 2016 at 14:29 | comment | added | Ali Taghavi | Prof. Michor Thank you very much for your answer. May be I am missing some thing to understand your answer: In the usual metric of $\mathbb{R}^{2}$ is it obvious that the space of all vector field $X$ with $\{X: X^1(p)=0, \text{div}(X)(p)=0\}$ is a Lie algebra? | |
| Aug 3, 2016 at 11:21 | history | answered | Peter Michor | CC BY-SA 3.0 |