Consider $\cal M$ and $\cal M'$, smooth embedded submanifolds of two linear manifolds $\cal E$ and $\cal E'$ (respectively). Let $F \colon \cal M \to \cal M'$ be a smooth map.

From Lee's textbook (2012, Intro to smooth manifolds), Lemma 5.34, we know that for the special case where $\cal M' = \mathbb{R}$ we can smoothly extend $F$, at least locally:

There exists an open neighborhood $U$ of $\cal M$ in $\cal E$ and a smooth function $\bar F \colon U \to \mathbb{R}$ such that $F$ is the restriction of $\bar F$ to $\cal M$, that is, $F = \bar F|_{\cal M}$.

My question: for the more general case where $\cal M'$ is not simply equal to $\mathbb{R}$ but can be any embedded submanifold of a linear manifold $\cal E'$, can I also have such a smooth extension with a map $\bar F \colon \cal E \to \cal E'$?

  • $\begingroup$ Do you want a map defined on all of $\mathcal E$ or will a neighborhood $U$ of a point in $M$ do? Remember that maps to $\mathcal E'$ are vector-valued functions. $\endgroup$ – Ted Shifrin Jan 24 at 23:15
  • $\begingroup$ I expect a smooth extension to all of $\cal E$ won't be possible, because already for $\cal M' = \mathbb{R}$ this requires $\cal M$ to be properly embedded (Lee, Lemma 5.34 and Exercise 5-18). So a smooth extension to a neighborhood of $\cal M$ in $\cal E$ will have to do. My difficulty is with the co-domain $\cal M' \subseteq \cal E'$. Would it suffice to simply consider $F$ as a map into $\cal E' \approxeq \mathbb{R}^d$, and to study its individual components? $\endgroup$ – Nicolas Boumal Jan 24 at 23:32
  • $\begingroup$ (Also, I suspect that if such smooth extensions exist, they would exist regardless of the linear structure of $\cal E'$.) $\endgroup$ – Nicolas Boumal Jan 24 at 23:33
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    $\begingroup$ Closely related: math.stackexchange.com/questions/1893383/… $\endgroup$ – Eric Wofsey Jan 25 at 6:31

You should be able to do this using the tubular neighborhood theorem of Riemannian geometry. Let $\pi:NM \to M$ be the basepoint map from the normal bundle of M to M, and let $V$ be a sufficiently small neighborhood of (the canonical image of) $M$ in $NM$, such that there is a diffeomorphism $\psi: V \to U$ onto some tubular neighborhood of $M$ (by the TNT). Then set $\widetilde{F} = F \circ \pi \circ \psi^{-1}: U \to M'$. This is your extension (and it takes its values in $M'$).

  • $\begingroup$ This is great, thanks Stephen! Further composing the map you propose with the inclusion map from M' into E' gives the final answer as a map from (a neighborhood of M in) E to E', and illustrates the role of having M' embedded in E' (so that the inclusion would be smooth). $\endgroup$ – Nicolas Boumal Jan 25 at 16:15
  • $\begingroup$ Yes, good catch! $\endgroup$ – Stephen M Jan 25 at 19:39

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