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Jan 27, 2017 at 23:08 vote accept Totentanz
Sep 28, 2016 at 14:55 comment added Michael Bächtold You may also find a proof in the book: Nestruev, Smooth Manifolds and Observables, p. 127. springer.com/la/book/9780387955438
Sep 28, 2016 at 8:06 answer added Sebastian timeline score: 5
Sep 27, 2016 at 21:31 comment added Simon Henry A suggestion : If you already now how to prove that any derivation (i.e. such that $\delta(ab)= \delta(a)b+a \delta(b)$ ) on the space of functions is the derivation along a vector fields then you can observe that, (at least when $k=1$) the condition with the commutator is very similar to the property of being a derivation and try the same argument.
Sep 27, 2016 at 20:43 history asked Totentanz CC BY-SA 3.0