# Triviality of Vector bundles.

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!

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.

• 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? – Michael Albanese Jul 1 '20 at 14:41
• @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. – Jack Lee Jul 5 '20 at 17:52