Is the following assertion true?

Suppose $M$ is a smooth manifold of dimension $n$, and $S$ an embedded submanifold of dimension $k$. If there is an embedding $S\times \Bbb R^{n-k}\to M$ (with $S\times 0$ corresponding to $S$ in the obvious way), then the normal bundle of $S$ in $M$ is trivial.

I think this can be proved as follows, but I'm not sure about my argument: If $f$ is such an embedding, then the image of $f$ should be an open subset of $M$, so the normal bundle of $S$ in $M$ and the normal bundle of $S$ in $f(S\times \Bbb R^{n-k})$ are the same, but the latter is a trivial bundle.


1 Answer 1


The sub-assertion

... the normal bundle of $S$ in $M$ and the normal bundle of $S$ in $f(S \times \mathbb R^{n-k})$ are the same...

is false as stated literally. They aren't the "same". But it contains a grain of truth which you can then use to turn into a proof.

The key thing to keep in mind is that whenever you are tempted to use the "s" word, you should instead ask yourself: What kind of isomorphism are we talking about here?

The true statement which you should have in place of the one above is

... the derivative $Df : T(S \times \mathbb R^{n-k}) \to TM$ restricts to an isomorphism between the normal bundle of $S \times 0$ in $S \times \mathbb R^{n-k}$ and the normal bundle of $f(S \times 0)$ in $M$...

And now you can go on to prove it.

  • $\begingroup$ Is this the tubular neighborhood theorem? $\endgroup$
    – Prototank
    Nov 12, 2020 at 14:06
  • $\begingroup$ No, in this question the existence of a product tubular neighborhood is already being assumed. The tubular neighborhood theorem is about the proof of existence of tubular neighborhoods in general. $\endgroup$
    – Lee Mosher
    Nov 12, 2020 at 14:59

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