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The question is a follow-up to this one.

  • Let $N\subset M$ be a closed smooth submanifold of codimension $k\geq 1$ of a smooth manifold $M$. Let $E_1\rightarrow M$ and $E_2\rightarrow M$ be two vector bundles over $M$. Suppose $E_1|_{M-N}\sim E_2|_{M-N}$. Is it true that $E_1|_N\sim E_2|_N$?

  • Can we impose some conditions on $k$ and $n=\dim M$ such that under the condition $E_1|_{M-N}\sim E_2|_{M-N}$, we can conclude $E_1\sim E_2$? (a counter example for this statement in general is provided by $E_1=T\mathbb{S}^1\rightarrow\mathbb{S}^1,E_2=\text{mobius}\rightarrow\mathbb{S}^1$ and $N=\{\cdot\}$) (the difference with the previous question is that $M$ and $M-N$ are generally not homotopic).

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  • $\begingroup$ The answer to the first question is No. Just think of the inclusion of the circle in the 2-dimensional sphere. $\endgroup$
    – user43326
    Commented Jun 19, 2020 at 17:17
  • $\begingroup$ @user43326: That doesn't work. Any vector bundle on $S^2$ is trivial on $S^2\setminus S^1 = D^2\sqcup D^2$ and on $S^1$. $\endgroup$
    – Michael Albanese
    Commented Jun 19, 2020 at 17:20
  • $\begingroup$ @MichaelAlbanese Thank you, yes, we need an example where the inclusion of N in M isn't null homotopic. $\endgroup$
    – user43326
    Commented Jun 19, 2020 at 17:22

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Here's an example to show that the answer to the first question is no.

Let $M = \mathbb{CP}^2$ and $N = \mathbb{CP}^1$. Note that $M - N$ is diffeomorphic to $\mathbb{C}^2$ and is therefore contractible. So for any two vector bundles $E_1, E_2 \to \mathbb{CP}^2$ of the same rank, we have $E_1|_{M - N} \cong E_2|_{M-N}$. However, we need not have $E_1|_N \cong E_2|_N$. For example, if $E_1 = \mathcal{O}_{\mathbb{CP}^2}(1)$ and $E_2 = \mathcal{O}_{\mathbb{CP}^2}$, then $E_1|_N \cong \mathcal{O}_{\mathbb{CP}^1}(1)$ and $E_2|_N \cong \mathcal{O}_{\mathbb{CP}^1}$ which are distinguished by the second Stiefel-Whitney class.

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    $\begingroup$ Am I right to say that the real version with $\mathbb{RP}^2$ and $\mathbb{RP}^1$ also provides a counter-example in the case of real vector bundles? $\endgroup$
    – anonymous67
    Commented Jun 19, 2020 at 18:39
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    $\begingroup$ @quangtu123: Yes. Although note that the counterexample I mention in my answer works for real and complex bundles: $E_1|_N$ and $E_2|_N$ are not isomorphic as real bundles and hence cannot be isomorphic as complex bundles either, although $E_1|_{M-N}$ and $E_2|_{M-N}$ are isomorphic as real bundles and as complex bundles. $\endgroup$
    – Michael Albanese
    Commented Jun 19, 2020 at 21:34

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