Let $(M,g)$ a semiRiemannian manifold, and let $\mathcal{G}$ a be Lie pseudogroup which acts by local isometries on $(M,g)$. Clearly the germs of all local killing vector fields at $p \in M$ form a Lie algebra. What if I restrict myself to germs of those local killing fields whose local flow is contained in $\mathcal{G}$?  Do they form a Lie algebra as well? If so, why?
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1$\begingroup$ 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 $\mathcal G$. $\endgroup$ – Sebastian Goette Jan 28 '16 at 15:17

$\begingroup$ 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. $\endgroup$ – hase_olaf Jan 29 '16 at 10:34

$\begingroup$ 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. $\endgroup$ – Sebastian Goette Jan 29 '16 at 13:30

$\begingroup$ @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! $\endgroup$ – hase_olaf Jan 30 '16 at 10:50

$\begingroup$ 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. $\endgroup$ – Sebastian Goette Feb 1 '16 at 12:58