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Let $M$ be smooth manifold and $S^s \hookrightarrow M^m$ a smooth embedding. Denote the normal bundle of $S$ in $M$ via $N_{S/M}$.

A tubular neighbourhood (of $S$ in $M$) is a disc bundle over $S$ with smoothly varying radius; it is embedded in the normal bundle $N_{S/M}$ induced by an embedding $S \hookrightarrow M$. While the tubular neighbourhood is only locally resembling the normal bundle (in the embedded setting), abstractly it is isomorphic to $N_{S/M}$. Indeed,

  • if $S$ is compact, the radius is fixed and we can use the same diffeomorphism from the disc to $\mathbb{R}^{m-s}$ in each fibre.
  • for non-compact $S$, a more detailed argument is needed, I assume, but the statement remains true, nonetheless.

So, tubular neighbourhoods are exactly normal bundles of $S$ in $M$ (over varying embeddings)? Now, since every vector bundle $E$ over $S$ can be realized as $E \cong N_{S/E}$, does this mean:

Tubular neighbourhoods = Normal bundles = Vector bundles over $S$?

This seems wrong, but I keep confusing myself.

Edit: With "this equals that equals that", I mean the following: Let

  • $\tau(S,M)$ be the set of tubular neighbourhoods of $S$ in $M$
  • $\mathcal{N}(S,M)$ the set of normal bundles of $S$ in $M$, one for each embedding of $S$,
  • $\operatorname{Vect}(S)$, the space of vector bundles over $S$

Are there bijections between $\tau(S,M)$, $\mathcal{N}(S,M)$ and $\operatorname{Vect}(S)$?

Edit 2: I guess $\tau(S,M) \to \mathcal{N}(S,M)$ is a bijection, but the bijection to the space of vector bundles over $S$ is false the reason being that $M$ may not embedded in $E$ although $S$ does. So, $N_{S/E} \not \cong N_{S/M}$. For example, let $S = [0,1]$ and $E = \mathbb{R} \times [0,1]$ and $M$ be the Mobius band.

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    $\begingroup$ No, it does not mean "this equals that equals the other thing". "Equality" is not the appropriate concept here, if you want to express this precisely and correctly. A tubular neighborhood of $S$ in $M$ is a certain subset of $M$. You can express everything you are trying to say in the language of continuous functions and homeomorphisms and so on; replacing that language with "this equals that equals the other thing" is certainly lazy and moreover is wrong. $\endgroup$
    – Lee Mosher
    Commented Oct 30, 2023 at 15:52
  • $\begingroup$ See my comment. I $\textit{am}$ looking for a precise statement, indeed. $\endgroup$
    – ferhenk
    Commented Oct 30, 2023 at 15:54
  • $\begingroup$ An oblique but perhaps helpful anecdote: As a student I confused myself one afternoon with a comparable thought. "The total space of a (smooth) vector bundle over a (smooth) manifold is a manifold. But every manifold is a rank-0 vector bundle over itself. Does that mean manifolds are vector bundles? That doesn't sound right." <> The moral: What may help here is to ponder (i) exactly what structure each type of object entails, (ii) whether or not an instance of one structure uniquely induces another structure, and (iii) whether this "inducement" is in some sense bijective. $\endgroup$ Commented Oct 30, 2023 at 16:31

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