Can one always extend a smooth triangulation from a smooth submanifold $S$ to the ambient manifold $M$? (For simplicity both $S$ and $M$ are compact without boundary). Is the extension possible when $S$ is a sphere?

What I know is summarized in this answer. Namely, in the PL category the answer is yes, by a result of Armstrong, i.e., one can always extend a triangulation from a locally flat PL submanifold. Also one can always extend a smooth triangulation from a codimension one separating hypersurface (this is due to Munkres). I think, the same is true for non-separating hypersurfaces. Thus it is enough to extend the smooth triangulation of $S$ to a tubular neighborhood of $S$.

Naively, I would expect that the extension is always possible. Otherwise, one can probably use it to define a potentially interesting invariant of smooth manifolds, and I have never heard of such an invariant.

  • $\begingroup$ Is this the same as this question mathoverflow.net/q/206212/8103? $\endgroup$
    – Mark Grant
    Aug 3 '20 at 7:53
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    $\begingroup$ @MarkGrant: it is not the same. An exercise in Munkres' "Elementary Differential topology", and probably, a theorem in Verona's book in your linked answer says that there is a smooth triangulation of $M$ in which $S$ is a subcomplex. What I ask for is to extend a smooth triangulation of $S$ to a smooth triangulation of $M$. I don't want to change the given triangulation of $S$. I do not want to subdivide it. $\endgroup$ Aug 3 '20 at 11:20
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    $\begingroup$ A good example is PL knotted ball pair $(B^4, B^2)$ that is the cone on a nontrivial knot in $S^3$. There is of course a triangulation of $B^4$ that restricts to a triangulation of $B^2$ but if we triangulate $B^2$ as a simplex, this triangulation does not extend to a triangulation of $B^4$ because it does not capture the local knottedness at the cone point. $\endgroup$ Aug 3 '20 at 12:01

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