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Feb 11, 2018 at 14:21 comment added მამუკა ჯიბლაძე It is believable but not entirely obvious: since $f(p)=d_pf=0$, then I believe also $df(Y)=0$ at $p$ for any $Y$. I understand this does not imply there is no $X$ with $X\cdot(Y\cdot f)$ nonzero at $p$ but still - to have an explicit example would be better I believe
Feb 9, 2018 at 15:16 comment added M.G. @Sebastian: thanks for the nice counterexample! I guess I was too enthusiastic at first.
Feb 9, 2018 at 13:04 vote accept M.G.
Feb 9, 2018 at 5:59 comment added Sebastian I mean $Y\cdot f=df(Y)$ and $X\cdot Y\cdot f=X\cdot(Y\cdot f)$.
Feb 9, 2018 at 2:09 comment added Qfwfq @მამუკა ჯიბლაძე: I presume it's $Y$ acting on $f$ as a derivation (but let's Sebastian confirm or not)
Feb 8, 2018 at 21:01 comment added მამუკა ჯიბლაძე And what is $Y\cdot f$?
Feb 8, 2018 at 19:52 history edited Sebastian CC BY-SA 3.0
added 2 characters in body
Feb 8, 2018 at 19:47 comment added Qfwfq Does $X\cdot Y$ denote composition as differential operators or the product in the OP?
Feb 8, 2018 at 19:26 history answered Sebastian CC BY-SA 3.0