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Let $M$ be a submanifold of $N$, i.e. there is a smooth map $i:M\rightarrow N$ which is a topological embedding and whose differential is everywhere invective.

I call tubular neighborhood of M in N an open neighborhood of $i(M)$ that has the structure of a $(dim(N)-dim(M))$-vector bundle over $i(M)$, with $i(M)$ as the zero section.

For the past days I’ve been reading about this on every Differentiable Manifolds book I can find (e.g. Kosinski, Lee, Bredon, ...) and also online, and I am now really confused as to when such a tubolar neighborhood exists.

Does it always exist, or only when $i(M)$ is closed, as Kosinski and some resources online (such as this) seem to suggest?

I tried to look at the proofs, but everyone uses results based on other results based on other results and I get lost in every book’s conventions.

Angelo Brillante Romeo
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1 Answers1

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First, please correct the spelling to 'tubular', this will make it much easier for searching.

Second, you don't get a vector bundle structure because that would imply you have a whole copy of $\mathbb{R}^{n-m}$ at each point, you only get a disk/ ball at each point.

Finally, for you actual question. I think the difference is compactness. Some books assume manifolds are compact. In this case $i(M)$ is automatically closed and you always get existence of a tubular neighborhood. If you do not assume your manifolds to be compact, you need to require that $i(M)$ is closed to get a tubular neighborhood.

Edit: As per the comments, an open disk is homeomorphic to $\mathbb{R}^n$, so you can get a homeomorphism from a vector bundle to the tubular neighborhood. However, the tubular neighborhood in $N$ is not itself a vector bundle. For example if $M$ and $N$ are compact, it has compact closure in the topology of $N$.

quarague
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  • Sorry for the spelling, in Italian it is "Tubolare" so I got confused. Why wouldn't I get a full bundle structure? A small open disk in a Manifold is diffeomorphic to $\mathbb{R}^n$, so an open disk should be fine. – Angelo Brillante Romeo Jun 28 '19 at 07:33
  • Small enough open disk of an n-dimensional manifold I mean, of course. – Angelo Brillante Romeo Jun 28 '19 at 08:12
  • Here, for example (p.112) Manifolds don’t seem to be compact and the statement is existence of tubular nhd for every embedded submanifold. http://people.dm.unipi.it/martelli/didattica/matematica/2019/ist_geo.html – Angelo Brillante Romeo Jun 28 '19 at 09:25
  • In response to your edit: I still do not understand why you are claiming that I cannot give the tubular neighborhood the structure of a vector bundle. It has a well defined vector space structure on each fiber, it has a smooth projection on $i(M)$, etc. If it is “just diffeomorphic” to a vector bundle, then you can just use that diffeomorphism to give the tubular neighborhood the structure of a v. bundle. P.S. the definition I wrote comes from Kosinski (except that he does not require it to be open) – Angelo Brillante Romeo Jun 28 '19 at 09:34
  • I guess it is a difference of convention. I would not say that the open unit disk 'has the structure of $\mathbb{R}^2$' although it is diffeomorphic to it but of course you can give it that structure through the diffeomorphism. – quarague Jun 28 '19 at 09:40