# Tubular neighbourhoods are unique up to ambient isotopy?

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.

• Tubular neighborhoods are unique up to isotopy. Search for "tubular neighborhod" in the index of M. Hirsch's "Differential topology". Nov 11, 2019 at 12:23
• @IgorBelegradek Thank you! This reference additionally obtains a bundle map at the end of the isotopy. Nov 11, 2019 at 14:19
• I forgot to add that the isotopy can be made ambient by the isotopy extension lemma (chapter 8 of Hirsch's text). Nov 11, 2019 at 14:25
• 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. Nov 16, 2019 at 10:05
• @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. Nov 18, 2019 at 12:09