Timeline for Is every singular foliation induced by a Lie algebroid?
Current License: CC BY-SA 4.0
16 events
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| May 10, 2023 at 20:50 | history | edited | LSpice | CC BY-SA 4.0 |
Proofreading, and deleted "thanks", while this is on the front page
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| May 9, 2023 at 18:17 | answer | added | Praphulla Koushik | timeline score: 2 | |
| Apr 24, 2017 at 12:14 | answer | added | Iakovos | timeline score: 6 | |
| Mar 29, 2017 at 19:53 | history | edited | Ervin | CC BY-SA 3.0 |
deleted 127 characters in body
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| Mar 27, 2017 at 18:49 | comment | added | Qfwfq | OK, I see what you meant now. Thanks for the details | |
| Mar 27, 2017 at 17:31 | comment | added | Ervin | In the regular case, due to Frobenius' theorem, a distribution is integrable iff its sheaf of sections is a sheaf of Lie subalgebras of $(\mathfrak X _M , [,])$ | |
| Mar 27, 2017 at 17:29 | comment | added | Ervin | I'm sorry, I gave the wrong definition of integrable distribution. $D$ is integrable iff for every $p\in M$ there exists a submanifold (without boundary) $S$ containing $p$ such that for all $q\in S$ we have $T_q S = D_q $. With this definition,you can immediately see that my counterexample is not integrable. However, a set of generators of $D$ is $\partial / {\partial x},f \partial/{\partial y} $ where $f:\mathbb R ^2 \to \mathbb R $ is a smooth function which is zero for $x\leq 0$ and nonzero for $x>0$. | |
| Mar 27, 2017 at 17:24 | history | edited | Ervin | CC BY-SA 3.0 |
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| Mar 27, 2017 at 17:03 | comment | added | Qfwfq | But I don't understand your example: which are the vector fields generating $D$ (or $\Sigma$, which is the same up to a standard -at least in algebraic geometry- abuse of notation/terminology) on points of the form $(0,y)\in\mathbb{R}^2$? | |
| Mar 27, 2017 at 17:00 | comment | added | Qfwfq | From what you say, I would say yes (as the sheaf of $\mathcal{C}^\infty_M$-modules $\Sigma$ is closed under Lie bracket). | |
| Mar 27, 2017 at 15:48 | history | edited | Ervin | CC BY-SA 3.0 |
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| Mar 27, 2017 at 15:36 | comment | added | Ervin | I am not sure, but I would say no. Take the distribution $D$ on $\mathbb R ^2$ which has rank 2 on points with $x >0$ and is generated by $\partial / {\partial x} $ for $x\leq 0$. This is a smooth distribution and is not integrable (the "leaves" have boundary). However, the sheaf $\Sigma$ of smooth vector fields tangent to $D$ is a sheaf of Lie algebras, in the sense that for every open $U\subseteq M $ the module $\Sigma (U) $ is closed under Lie brackets. | |
| Mar 27, 2017 at 15:21 | comment | added | Qfwfq | A smooth (possibly singular) integrable distribution is the same as a "singular foliation" in the same sense as for holomorphic foliations, i.e. a sub sheaf-of-Lie-algebras of the tangent sheaf, right? | |
| Mar 27, 2017 at 10:01 | history | edited | Ervin | CC BY-SA 3.0 |
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| Mar 27, 2017 at 9:15 | answer | added | Tsemo Aristide | timeline score: 8 | |
| Mar 27, 2017 at 7:39 | history | asked | Ervin | CC BY-SA 3.0 |