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!
Triviality of Vector bundles.
fiber-bundlesvector-bundles
Related Solutions
Over spheres, you actually have a nice way to distinguish bundles.
Every sphere $S^n$ can be written as a union of two contractible spaces, $U \cup V$, whose intersection is a small neighborhood of the equator, $\mathbb{R} \times S^{n-1}$. For instance if $x,y$ are the north and south poles, respectively, you can take $U = S^n - \{x\}, V = X^n - \{y\}$.
Over $U$ and $V$, any bundle is trivial, as you noted. The bundle on the sphere is glued together from these two trivial bundles.
If you're interested in a vector bundle, the transition function for the bundle is given by a map $U \cap V \to GL_m$, for a vector bundle of rank $m$. If this map is null-homotopic, meaning that it can be homotoped to the constant map, then your bundle is trivial, as any trivialization over $U$ can then be glued to a trivialization over $V$. So in the end you are reduced to the study of maps from $U \cap V \simeq S^{n-1}$ to your structure group -- i.e., $\pi_{n-1}GL_m$.
For $m=1$, this amounts to asking whether any map $S^{n-1} \to GL_1$ is null-homotopic. In your case for $S^3$, since $GL_1(\mathbb{R})$ is a disjoint union of two contractible spaces, and $S^2$ is connected, any map is null-homotopic.
In the fiber bundle case with fiber $S^1$ and base $S^3$, you are looking at maps $S^2 \to S^1$. But $\pi_2(S^1) = *$ so any circle bundle is contractible over $S^3$.
Note that both facts hold for higher spheres as well.
Moishe Kohan's comment contains the answer of the question. This community wiki elaborates it a little.
Let us first observe that if $\xi = (E,p,B)$ is a pre-vector bundle and $f : X \to B$ is a (not necessarily continuous) function on a space $X$, we get a pullback pre-vector bundle $$f^*(\xi) = (f^*(E), p^*,X)$$ where $$f^*(E) = \bigcup_{x \in X} \{x \} \times p^{-1}(f(x)) \subset X \times E$$ and $p^*$ is the restriction of the projection $X \times E \to X$.
Now we generalize the construction of the question (see [1]). Define $$\mathbb K^\infty = \{(x_i)_{i \in \mathbb N} \mid x_i \in \mathbb K, x_i = 0 \text{ for almost all } i \} .$$ This is a vector space with an obvious inner product and we may regard each $\mathbb K^n$ as a genuine subspace of $\mathbb K^\infty$. Doing so, we have $\mathbb K^\infty = \bigcup_{n \in \mathbb N} \mathbb K^n$.
For $0 \le m \le \infty$ and $k \in \mathbb N$ let $G_k(\mathbb K^m)$ denote the set of all $k$-dimensional linear subspaces of $\mathbb K^m$. For $m < \infty$ these are the well-known Grassmann varieties. Each $G_k(\mathbb K^n)$ is a genuine subspace of $G_k(\mathbb K^{n+1})$, and we define $G_k(\mathbb K^\infty) = \bigcup_{n \ge k} G_k(\mathbb K^n)$ as a set, and $U \subset G_k(\mathbb K^\infty)$ to be open iff $U \cap G_k(\mathbb K^n)$ is open in $G_k(\mathbb K^n)$ for all $n$. Thus $G_k(\mathbb K^\infty)$ is the direct limit of the sequence of spaces $G_k(\mathbb K^n)$ bonded by inclusions.
The tautological (or canonical bundle) $\gamma^m_k$ over $G_k(\mathbb K^m)$ has as total space $$E^m_k = \bigcup_{V \in G_k(\mathbb K^m)} \{V\} \times V \subset G_k(\mathbb K^m) \times \mathbb K^m$$ with obvious projection $\pi$ onto the base. The fiber over $V \in G_k(\mathbb K^m)$ is nothing else than $\{V\} \times V$, i.e. a copy of $V \subset \mathbb K^m$. Again we have $E_k^\infty = \bigcup_{n \ge k} E_k^n$. It is well-known that $\gamma^m_k$ is locally trivial.
Now let $f : B \to G_k(\mathbb K^m)$ be any (not necessarily continuous) function. The pullback pre-vector bundle $f^*(\gamma^m_k)$ over $B$ has as total space $$f^*(E_k^m) = \bigcup_{b \in B} \{b \} \times \pi^{-1}(f(b)) = \bigcup_{b \in B} \{b \} \times \{f(b)\} \times f(b) \subset B \times G_k(\mathbb K^m) \times \mathbb K^m .$$ In general this is completely erratic. Let us say that $f^*(\gamma^m_k)$ is continuously parameterized if $f$ is continuous.
Note, however, that $f^*(\gamma^m_k)$ is not the same pre-vector bundle as $\xi(f)$ which was defined in the question. But $f^*(\gamma^m_k)$ is continuously parameterized if and and only $\xi(f)$ is. It seems that we now habe an adequate interpretation of "the spirit of a vector bundle is to continuously parametrize a family of vector spaces by $B$", at least for pre-vector bundles of the form $f^*(\gamma^m_k)$ and $\xi(f)$.
Fact 1. If $f^*(\gamma^m_k)$ is continuously parameterized, then it is locally trivial.
This is well-known. Pullbacks of vector bundles along continuous maps are always vector bundles. This shows that being continuously parameterized is even stronger than locally trivial.
Fact 2. If $f^*(\gamma^m_k)$ is locally trivial, then it is continuously parameterized.
Let $s_0 : B \to f^*(E_k^m) \subset B \times G_k(\mathbb K^m) \times \mathbb K^m$ be the zero-section which is given $s_0(b) = (b,f(b),0)$. Each $b \in B$ has an open neighborhood $U \subset B$ such that the restriction of $f^*(\gamma^m_k)$ to $U$ is trivial. This implies that $s_0 \mid_U$ is continuous. Since the projection $p_2 :B \times G_k(\mathbb K^m) \times \mathbb K^m \to G_k(\mathbb K^m)$ is continuous, we see that $f \mid_U = p_2 \circ s_0 \mid_U$ is continuous. Thus $f$ is continuous.
Fact 3. If $f$ is continuous, then $f^*(\gamma^m_k)$ and $\xi(f)$ are isomorphic. In particular, $\xi(f)$ is locally trivial.
To see this, define $\phi_f : B \times \mathbb K^m \to B \times G_k(\mathbb K^m) \times \mathbb K^m, \phi(b,v) = (b,f(b),v)$. We have $\phi_f(E(f)) = f^*(E_k^m)$ so that we get an induced $\phi'_f : E(f) \to f^*(E_k^m)$ which is bijection which maps the fiber over $b$ in $E(f)$ by a linear isomorphism to the fiber over $b$ in $f^*(E_k^m)$. It is continuous iff $f$ is continuous. Next define $\psi : B \times G_k(\mathbb K^m) \times \mathbb K^m \to B \times \mathbb K^m$ as the projection. Clearly $\psi$ is continuous and $\psi(f^*(E_k^m)) = E(f)$. Hence the restriction $\psi_f : f^*(E_k^m) \to E(f)$ is a morphism of pre-vector bundles which is continuous and fiberwise linearly isomorphic. It is an isomorphism of pre-vector bundles iff $f$ is continuous.
Best Answer
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.