1
$\begingroup$

Let $M$ be a closed smooth submanifold of $N$. It is well known that tubular neigbourhoods of $M$ are diffeomorphic to the normal bundle of $M$ in $N$ and therefore to each other. Are they smoothly isotopic? I think I know how to prove that they are, but having a reference would be nice.

If they are not, how is it enough to just assign a framing to a knot to obtain a surgery? Is the pair (framing, sphere embedding) enough for high dimensional handle attachments/surgery?

Definition: for a Riemannian metric $g$ on $N$, let $\nu_{g,\epsilon}(M)$ be the $\epsilon$-neighbourhood of $M$ in $N$. For $\epsilon$ small enough, $\nu_{g,\epsilon}(M)$ is a tubular neighbourhood.

$\endgroup$
7
  • 1
    $\begingroup$ Tubular neighborhoods are unique up to isotopy. Search for "tubular neighborhod" in the index of M. Hirsch's "Differential topology". $\endgroup$ Nov 11, 2019 at 12:23
  • $\begingroup$ @IgorBelegradek Thank you! This reference additionally obtains a bundle map at the end of the isotopy. $\endgroup$
    – mathquest
    Nov 11, 2019 at 14:19
  • $\begingroup$ I forgot to add that the isotopy can be made ambient by the isotopy extension lemma (chapter 8 of Hirsch's text). $\endgroup$ Nov 11, 2019 at 14:25
  • $\begingroup$ I think the best treatment of tubular neighborhoods is given in Wall's recent book on differential topology, in particular he gives an intrinsic definition of a tubular neighborhood (linear disk bundle) instead of the very awkward definition via Riemannian metric. See Theorem 2.5.5 of Wall. $\endgroup$ Nov 16, 2019 at 10:05
  • $\begingroup$ @StefanFriedl Thank you, that does seem like a nice textbook. It's the same definition as in Hirsch. Wall, though, also shows that this definition is equivalent to the Riemannian one. $\endgroup$
    – mathquest
    Nov 18, 2019 at 12:09

0

Your Answer

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