Skip to main content
MathOverflow

Return to Question

Language editing.
Source Link
Stefan Kohl ♦
Stefan Kohl ♦
  • 19.6k
  • 21
  • 75
  • 137

Do the germs of local vectorfields, thatvector fields which are generated by a Lie pseudogroup form a lieLie algebra?

Let $(M,g)$ a semi Riemannian-Riemannian manifold, and let $\mathcal{G}$ a be Lie pseudogroup thatwhich acts by local isometries on $(M,g)$. Clearly the germs of all local killing vector fields at $p \in M$ formsform a lieLie 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 lieLie algebra as well? If so, why?

Do the germs of local vectorfields, that are generated by a Lie pseudogroup form a lie algebra?

Let $(M,g)$ a semi Riemannian manifold and let $\mathcal{G}$ a Lie pseudogroup that acts by local isometries on $(M,g)$. Clearly the germs of all local killing vector fields at $p \in M$ forms 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?

Do the germs of local vector fields which are generated by a Lie pseudogroup form a Lie algebra?

Let $(M,g)$ a semi-Riemannian 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?

Source Link
hase_olaf
hase_olaf
  • 121
  • 2

Do the germs of local vectorfields, that are generated by a Lie pseudogroup form a lie algebra?

Let $(M,g)$ a semi Riemannian manifold and let $\mathcal{G}$ a Lie pseudogroup that acts by local isometries on $(M,g)$. Clearly the germs of all local killing vector fields at $p \in M$ forms 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?