1
$\begingroup$

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 $\bar F \colon \cal E \to \cal E'$?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ Closely related: math.stackexchange.com/questions/1893383/… $\endgroup$ – Eric Wofsey Jan 25 at 6:31
2
$\begingroup$

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 $M'$).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.