Timeline for Do the germs of local vector fields which are generated by a Lie pseudogroup form a Lie algebra?
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| Feb 1, 2016 at 12:58 | comment | added | Sebastian Goette | I don't know what the "Lie" in "Lie pseudogroup" stands for. Nevertheless, the local flow of the Lie bracket will be contained in the closure of $\mathcal G$ in an appropriate sense, by arguments similar to my comments. So if you demand or prove that $\mathcal G$ is closed, you can get those commutators. | |
| Jan 30, 2016 at 10:50 | comment | added | hase_olaf | @Sebastian: How can a curve in $\mathcal G$ be differentiable? As $\mathcal G$ is merely a Lie pseudogroup, it does not possess a differentiable structure. Thanks for helping! | |
| Jan 29, 2016 at 13:30 | comment | added | Sebastian Goette | Let $\Phi_t$, $\Psi_t$ be the local flows. I think the curve $\Phi_{\sqrt t}\Psi_{\sqrt t}\Phi_{-\sqrt t}\Psi_{-\sqrt t}$ in $\mathcal G$ is differentiable from above in $t$ at $t=0$ and has the Lie bracket as differential. | |
| Jan 29, 2016 at 10:34 | comment | added | hase_olaf | Hi @Sebastian. I know how to calculate the commutator pointwise in terms of local flows. But I need the local flow of the commutator vector field to be contained in $\mathcal{G}$. I do not see why this is true, although I guess it is. | |
| Jan 28, 2016 at 15:17 | comment | added | Sebastian Goette | Have you tried to use the definition of a Lie bracket using (local) flows? It seems the answer is yes because all steps needed to set up the Lie bracket this way can be performed in $\mathcal G$. | |
| Jan 28, 2016 at 10:36 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Language editing.
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| Jan 28, 2016 at 10:26 | review | First posts | |||
| Jan 28, 2016 at 10:36 | |||||
| Jan 28, 2016 at 10:24 | history | asked | hase_olaf | CC BY-SA 3.0 |