Is a vector bundle which is trivial on a smooth submanifold trivial in a neighbourhood?

Question

Let $E$ be a (real or complex) vector bundle on a smooth (possibly compact) manifold $M$. Suppose $N \subset M$ is a closed submanifold, such that $E|_N$ is trivial. Does there exist an open neighbourhood $N \subset U \subset M$ such that $E|_U$ is trivial? If not, what is a counter example?

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This clearly holds if $N$ is a point. I also think this should be true if $E$ is a complex line bundle: Let $U \subset M$ be a tubular neighbourhood, diffeomorphic to the normal bundle of $N$ in $M$. Then the inclusion $N \to U$ is a homotopy equivalence, so it induces an isomorphism $$H^2(U, \mathbb Z) \xrightarrow{\cong} H^2(N, \mathbb Z).$$ As $c_1(E|_N) = 0$, we see that also $c_1(E|_U) = 0$, and since line bundles are classified by their first Chern class, $E|_U$ is trivial.

As far as I understand, Pontryagin classes of real line bundles are always zero, so a similar argument cannot apply to the real case? Also, vector bundles of higher rank are not classified by their Chern classes.

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I have the following argument in mind, but I'm not sure if it works out, or if I made a mistake: Choose a tubular neighbourhood $U \subset M$ of $N$. As $U$ is diffeomorphic to the normal bundle of $N$ in $M$, there is a map $\pi: U \to N$. Is it possible to arrange $(U, \pi)$ in such a way that, for each $x \in N$, there is a neighbourhood $V_x \subset N$, such that $E$ is trivial on $U_x = \pi^{-1}(V_x)$? I guess this should be possible if we make $U$ small enough, certainly if $N$ is compact.

If we have this, choose trivialisations $\varphi_x: E_{U_x} \to U_x \times \mathbb C^n$. We can then extend our section $s$ to $U$ by mapping each point $y \in U$ to the same vector as $\pi(y)$, under the trivialisation $\varphi_{\pi(y)}$. In formulas, $$s_U: y \mapsto (y, \operatorname{pr}_2 \varphi_{\pi(y)}(s(\pi(y)))).$$

Does that work?

Answer 1

Yes, if $M$ is paracompact there exists such an open subset $N\subset U\subset M$ on which $E$ is trivial. Here is why:
Let $\theta$ be the trivial bundle of the same rank as $E$.
There exists a vector bundle $HOM(E, \theta)$ defined on $M$ whose fiber $HOM(E, \theta)(m) $ at $m\in M$ is the set of linear maps $L (E(m), \theta(m))$ and whose set of sections over an open (resp. closed) submanifold $V\subset M$ is $$\Gamma(V,HOM(E, \theta))=Hom (E\vert V,\theta \vert V)$$ Your isomorphism is a section $s\in \Gamma(N, HOM(E,\theta))=Hom(E\vert N,\theta\vert N)$ and that section can be extended to a section $S_0\in \Gamma(M, HOM(E,\theta))=Hom (E,\theta)$ (a standard extension property valid for any vector bundle $F$ on $M$ and proved with partitions of unity).
Now the set of points $m\in M$ where $S_0(m):E(m)\to \theta(m)$ is an isomorphism is (by a local computation) an open subset $N\subset U\subset M$ and $S_0\vert U\in \Gamma(U,HOM(E, \theta))$ is an isomorphism in $Hom (E\vert U,\theta\vert U)$ solving your problem in the affirmative.
Reference
Karoubi's K-Theory, Chapter 1, Corollary 5.11, page 24.
Complement (September 5, 2023)
The above result is no longer true in algebraic geometry: here is a counterexample.
Let $C$ be a smooth projective curve of positive genus, $Pic^0(C)$ be its Jacobian and $$p:Pic^0(C)\times C \to Pic^0(C), q:Pic^0(C)\times C \to C$$ be the projections.
On $Pic^0(C)\times C$ there is a Poincaré line bundle $\mathcal P$ which has the property that for any line bundle $L$ of degree zero on $C$ $$\mathcal P\vert p^{-1}[L]=q^*(L)\vert p^{-1}[L] $$ Now, denoting by $\theta$ the trivial line bundle on $C$, the Poincaré bundle $\mathcal P$ is clearly trivial on the subvariety $p^{-1}([\theta])\subset Pic^0(C)\times C$, but is trivial on no neighbourhood of that subvariety since it is non-trivial on every fiber $p^{-1}([L])$ with $[L]\neq [\theta]$.

Answer 2

Motivated by the beautiful answer provided by @Georges Elencwajg, I will add here more details in his proof for the convenience of further readers.

Detail: (partition of unity) Given a vector bundle $W$ (corresponding to $HOM(E,\theta)$ in his answer), then the restriction map $\Gamma(M, W)\rightarrow \Gamma(N,W|_N)$ is surjective. Here, a section $s\in\Gamma(M,W)$ is supposed to be continuous.

Proof: As $M$ is paracompact, one can find a locally finite cover $\left\{U_i\right\}$ such that $W|_{U_i}$ is trivial. Let $\left\{V_i\right\}$ be another cover such that $\overline {V_i}\subset U_i$. Set $T_i:=N\cap\overline {V_i}$. For any $t\in \Gamma(N,W|_N)$, let $t|_{T_i}:=t_i$. Then one can apply Tietez extension theorem to see that $t_i$ have a local extension $s_i\in \Gamma(\overline{V_i},W)$. Now let $(\alpha_i)$ be a partition of unity associated with the cover $\left\{V_i\right\}$ and for any $i$, define $\widetilde{s_i}(x)=\alpha_i(x)s_i(x)$ if $x\in\overline{V_i}$ and $\widetilde{s_i}(x)=0$ otherwise. It implies that $\widetilde{s_i}$ is a continuous section which is zero over all but only a finite number of $V_i$. We then set $s=\sum_i\widetilde{s_i}$, then for any $x\in X$, the sum $\sum_i\widetilde{s_i}$ is actually a finite sum over a neighborhood of $x$. The finiteness can then guarantee that $s$ is a continuous section over the whole manifold. One can see that the ``paracompact" assumption is really important! Finally, for any $x\in N$, $$s(x)=\sum_i \alpha_i(x)t(x)=t(x).$$ The desired result is obtained.