In Bott & Tu, a tubular neighborhood of a submanifold $S\subset M$ is defined as an open neighborhood $T$ of $S$ in $M$ such that $T$ is diffeomorphic to a vector bundle of rank $\mathrm{codim}\,S$ such that $S$ is diffeomorphic to the zero section. They then claim such tubular neighborhoods always exist and that the normal bundle $NS$ of $S$ in $M$ is the required bundle.

However, the tubular neighborhood states that $T$ is diffeomorphic to a neighborhood of the zero section in $NS$, not to the whole thing. Is it possible to "stretch" the image of $T$ in $NS$ so that it is diffeomorphic to all of $NS$? Spivak seems to have a proof in the compact case, but how does one show this in general, using the tubular neighborhood quoted above?

  • $\begingroup$ I suppose that you may choose your TN so that it works. If e.g. it is locally a product of a coordinate patch and a ball. In the non-compact case I suppose you have to worry about 'loose' ends accumulating at your submanifold (unless you exclude this explicitly). $\endgroup$
    – H. H. Rugh
    Jul 29, 2016 at 22:46

1 Answer 1


Let $E$ be a vector bundle and denote by $Z$ the image of the zero section. Any open neighbourhood $U$ of $Z$ contains an open neighbourhood $V$ of $Z$ such that there is a diffeomorphism $V \to E$ which is the identity on $Z$.

The proof of this fact is easier if the manifold is compact, but it is also true in the non-compact case. See Proposition $2.8$ of these notes from Michael Usher.

Apply the above result to $E = NS$.

As Mike said, the neighbourhood $V$ is a disc bundle of fixed radius (with respect to some Riemannian metric) if the base is compact, or of smoothly decreasing radius if the base is non-compact. You then use the diffeomorphism between a disc and $\mathbb{R}^n $ in each fibre to construct the desired diffeomorphism.

  • 2
    $\begingroup$ It's also not substantially more difficult in the noncompact case. It's just that instead of being a disc bundle of fixed radius, $V$ is now a disc bundle of smoothing varying radius. $\endgroup$
    – user98602
    Jul 29, 2016 at 22:49
  • 1
    $\begingroup$ Sure. Not conceptually more difficult, but more work is needed. $\endgroup$ Jul 29, 2016 at 22:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .