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Sep 16, 2022 at 22:33 vote accept Audrey Rosevear
Sep 16, 2022 at 18:45 answer added Andy Putman timeline score: 3
Sep 16, 2022 at 7:21 comment added Sam Nead I think that you mean "coordinates where $f$ looks like $(x, y, z) \mapsto (x^2 - y^2, 2xy)$". In your version $f$ is not surjective near the origin. Also, I think that $D = \mathrm{ker} f$ should be $D = \mathrm{ker} \, d\!f$.
Sep 16, 2022 at 5:44 comment added Audrey Rosevear The leaves here are precisely the preimages of points under $f$, so they should be compact.
Sep 16, 2022 at 3:18 comment added Andy Putman I haven’t read that paper, so this might be totally misguided. But why are the leaves of the 1-dimensional distribution closed? All closed oriented 3-manifolds support 1-dimensional foliations (since their Euler characteristic is 0, you can find a nonvanishing vector field on them).
Sep 16, 2022 at 2:04 history asked Audrey Rosevear CC BY-SA 4.0