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Can a smooth vector bundle be trivial as a smooth fiber bundle but not as smooth vector bundle? I haven't tried much, except maybe use the global trivialization of the fiber bundle to construct a global frame, but found no way to garantee that the isomorphism would take LI vectors in LI vectors. Any help is appreciated!

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No, if a vector bundle is trivial as a smooth fiber bundle, then it is also trivial as a vector bundle. In fact, a more general result is true: if any two smooth vector bundles are isomorphic as smooth fiber bundles, then they are isomorphic as vector bundles.

[This proof is a slightly modified version of the one I originally posted, adapted to prove the more general result. For reference, my original proof is reproduced below.]

The key idea is that every smooth fiber bundle with a global section has a vector bundle associated with it, namely the pullback of the vertical tangent bundle along the section; and if two fiber bundles are isomorphic, then so are their pullback vertical bundles. On the other hand, if a fiber bundle also happens to have the structure of a smooth vector bundle, then the pullback vertical bundle is naturally isomorphic to the vector bundle itself.

In more detail, here's how it works. Suppose first that $\pi\colon E\to M$ is a smooth fiber bundle with $k$-dimensional fibers. There is a rank-$k$ vector bundle $T^V E\to E$, called the vertical tangent bundle, whose fiber at a point $p\in E$ is the tangent space to the fiber $E_{\pi(p)} = \pi^{-1}(\pi(p))$: in other words, $T^V_pE = T_p(E_{\pi(p)}) = \ker d\pi_p$.

If $E$ has a global section $\sigma\colon M\to E$, we let $E_\sigma\subset E$ be the image of the global section, which is a smooth embedded submanifold diffeomorphic to $M$. The restriction $T^V\!E|_{E_\sigma}$ is a rank-$k$ vector bundle over $E_\sigma$, which we denote by $E^V\to E_\sigma$. It can be considered as the subset of $TE$ consisting of all vertical vectors over points of $E_\sigma$.

Now suppose $\pi'\colon E'\to M$ is another smooth fiber bundle that is isomorphic over $M$ to $E$ (as a smooth fiber bundle). Thus there is a smooth diffeomorphism $\Phi\colon E\to E'$ covering the identity map of $M$. We obtain a global section $\sigma'=\Phi\circ\sigma\colon M\to E'$, and we can perform the same construction on $E'$ to yield a vector bundle $E^{\prime V}\to E'_{\sigma'}$. Because $\Phi$ is a bundle map, the global differential $d\Phi\colon TE\to TE'$ restricts to a bundle isomorphism from $E^V$ to $E^{\prime V}$ covering the diffeomorphism $\Phi|_{E_{\sigma}}\colon E_{\sigma} \to E_{\sigma'}'$.

On the other hand, if $E\to M$ is a smooth vector bundle and $\sigma\colon M\to E$ is any global section (for example, the zero section), we can construct the vector bundle $E^V\to E_{\sigma}$ as before. But in this case, for each point $q\in M$, the fiber $E_q\subseteq E$ is a vector space, and the fiber $E^V_{\sigma(q)}\subseteq E^V$ is the tangent space to $E_q$ at $\sigma(q)$. Each tangent space to the finite-dimensional vector space $E_q$ is canonically isomorphic to the vector space $E_q$ itself; the isomorphism is given by sending an element $v\in E_q$ to the derivation $D_v\colon C^\infty(E_q) \to \mathbb R$ defined by $D_v(f) = (d/dt)|_{t=0} f(\sigma(q)+tv)$.

Let $\alpha\colon E \to E^V$ be the map whose restriction to each fiber $E_q\subseteq E$ is the canonical isomorphism $E_q\to T_{\sigma(q)}(E_q) = E^V_{\sigma(q)}$. Then $\alpha$ is a vector bundle isomorphism covering the diffeomorphism $\sigma\colon M\to E_{\sigma}$, provided it is smooth. In a neighborhood $U$ of any point of $M$, there is a local vector bundle trivialization $\Psi\colon \pi^{-1}(U)\to U\times \mathbb R^k$. Its differential restricts to a smoooth local trivialization $d\Psi|_{(\pi^V)^{-1}(U)}\colon (\pi^V)^{-1}(U) \to U\times \mathbb R^k$. Unwinding the definitions shows that the map $d\Psi\circ\alpha\circ \Psi^{-1}\colon U\times \mathbb R^k\to U\times \mathbb R^k$ has the form $d\Psi\circ\alpha\circ \Psi^{-1}(q,v) =(q,v)$. Since $\Psi$ and $d\Psi|_{(\pi^V)^{-1}(U)}$ are diffeomorphisms, this shows that $\alpha$ is smooth in a neighborhood of each point.

Putting this all together, if $E\to M$ and $E'\to M$ are smooth vector bundles that are isomorphic over $M$ as smooth fiber bundles, then we have a composition of vector bundle isomorphisms $$ E\overset{\alpha}{\longrightarrow} E^V \overset{d\Phi|_{E^V}}{\longrightarrow} E^{\prime V} \overset{\alpha^{\prime-1}}{\longrightarrow} E' $$ covering the identity of $M$, thus showing the $E$ and $E'$ are isomorphic as vector bundles.

Here's the less general proof I originally posted.

Suppose first that $\pi\colon E\to M$ is a smooth fiber bundle with $k$-dimensional model fiber $F$. There is a rank-$k$ vector bundle $T^V E\to E$, called the vertical tangent bundle, whose fiber at a point $p\in E$ is the tangent space to the fiber $E_{\pi(p)} = \pi^{-1}(\pi(p))$: in other words, $T^V_pE = T_p(E_{\pi(p)}) = \ker d\pi_p$. If $E$ has a global section $\sigma\colon M\to E$, then $T^V E$ pulls back to a vector bundle over $M$, which I'll denote by $E^V = \sigma^*(T^V E)$ with projection $\pi^V\colon E^V\to M$.

Now suppose $E$ has a global trivialization (as a fiber bundle) $\Phi\colon E\to M\times F$. Thus $\Phi$ is a diffeomorphism satisfying $\pi_1\circ\Phi = \pi$ (where $\pi_1\colon M\times F\to M$ is the projection on the first factor). Because $\Phi$ is a bundle map, the global differential $d\Phi\colon TE\to T(M\times F)$ restricts to a bundle isomorphism from $T^V E$ to $T^V (M\times F)$, and therefore $T^V E$ is trivial. It follows that $E^V$ is also trivial, since it's the pullback of a trivial bundle.

Now suppose $E$ also has the structure of a smooth vector bundle. The zero section is a smooth global section, so we obtain the pullback vertical bundle $E^V$ as before, whose fiber at each point $q\in M$ is $T_0(E_q)$. In this case, since $E_q$ has the structure of a finite-dimensional vector space, the tangent space $T_0(E_q)$ is canonically isomorphic to the vector space $E_q$ itself; the isomorphism is given by sending an element $v\in E_q$ to the derivation $D_v\colon C^\infty(E_q) \to \mathbb R$ defined by $D_v(f) = (d/dt)|_{t=0} f(tv)$. Putting together these isomorphisms for all $q\in M$ shows that the vector bundle $E$ is canonically isomorphic to $E^V$, provided the map $\alpha\colon E\to E^V$ so obtained is smooth.

In a neighborhood $U$ of any point of $M$, there is a local vector bundle trivialization $\Psi\colon \pi^{-1}(U)\to U\times \mathbb R^k$. Its differential restricts to a smoooth local trivialization $d\Psi|_{(\pi^V)^{-1}(U)}\colon (\pi^V)^{-1}(U) \to U\times \mathbb R^k$. Unwinding the definitions shows that the map $d\Psi\circ\alpha\circ \Psi^{-1}\colon U\times \mathbb R^k\to U\times \mathbb R^k$ has the form $d\Psi\circ\alpha\circ \Psi^{-1}(q,v) =(q,v)$. Since $\Psi$ and $d\Psi|_{(\pi^V)^{-1}(U)}$ are diffeomorphisms, this shows that $\alpha$ is smooth in a neighborhood of each point.

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  • $\begingroup$ Am I correct in saying that this argument cannot be used to rule out the existence of two (necessarily non-trivial) vector bundles which are isomorphic as fiber bundles, but not as vector bundles? $\endgroup$ – Michael Albanese Jul 1 '20 at 14:41
  • $\begingroup$ @MichaelAlbanese: I think it can be adapted to prove that, but it probably needs a slightly different argument. Let me see if I can adapt the proof to cover that case. $\endgroup$ – Jack Lee Jul 5 '20 at 17:52

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