Can vector bundle be thought of as below?

Question

Suppose we have following data.

1. Two smooth manifolds $E,M$ and a surjective submersion $\pi:E\rightarrow M$.
2. A vector space structure on $\pi^{-1}(x)$ for each $x\in M$.
3. A collection of isomorphisms $\{\psi_x:\pi^{-1}(x)\rightarrow \{x\}\times \mathbb{R}^n : x\in M\}$.

I call $\pi:E\rightarrow M$ to be

• a trivial bundle if, there exists a diffeomorphism $\Phi: \pi^{-1}(M)\rightarrow M\times \mathbb{R}^n$ such that $\Phi_{\pi^{-1}(x)}$ is same as $\psi_x:\pi^{-1}(x)\rightarrow \{x\}\times \mathbb{R}^n$ in some sense, I am asking if there I can glue these isomorphisms of vector spaces $\psi_x:\pi^{-1}(x)\rightarrow \{x\}\times \mathbb{R}^n$ to get a diffeomorphism $\Phi: \pi^{-1}(M)\rightarrow M\times \mathbb{R}^n$.

I do not want to restrict my interest to gluing all isomorphisms $\psi_x:\pi^{-1}(x)\rightarrow \{x\}\times \mathbb{R}^n$. I want to consider the case when I can glue isomorphisms locally. This is not new thought or anything. One sees Locally compact more often than compact, Locally connected more than connected and so on. So, I want to consider the case where I can glue isomorphisms locally in the following sense.

• Given $x\in M$, there exists an open set $U$ of $M $containg $x$ such that, I can glue the isomorphisms $\{\psi_x:\pi^{-1}(x)\rightarrow \{x\}\times\mathbb{R}^n:x\in U\}$ i.e., given $x\in M$, there exists a diffeomorphism $\Phi:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^n$ such that $\Phi_{\pi^{-1}(y)}$ is same as $\psi_y:\pi^{-1}(y)\rightarrow \{y\}\times\mathbb{R}^n$ for each $y\in U$.

I call $\pi:E\rightarrow M$ to be

• a locally trivial bundle, if given $x\in M$ there exists an open set $U$ containing $x$ and a diffeomorphism $\Phi:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^n$ such that $\Phi_{\pi^{-1}(y)}$ is same as $\psi_y:\pi^{-1}(y)\rightarrow \{y\}\times\mathbb{R}^n$ for each $y\in U$.

A vector bundle is a locally trivial bundle mentioned above.

Question :

Did I miss anything in the definition?

I used notion of vector bundles very often and may be assuming something and not saying here. Please let me know if I am assuming something and not saying here

Answer 1

Yes, you missed all the non-trivial vector bundles! Your proposed definition has two severe problems:

1. The notion of a vector bundle should generalize the notion of a vector space. In particular, when $M$ is a point, a vector bundle over $M$ should be just a vector space. An arbitrary vector space $V$ doesn't come with a preferred basis (i.e, an isomorphism with $\mathbb{R}^n$) while you propose that a vector bundle over a point is not just a vector space but a vector space together with a choice of a basis. A choice of a basis is not part of a data of a vector space and a choice of isomorphisms $\psi_x$ shouldn't be part of a data of a vector bundle.
2. You write "there exists a diffeomorphism $\Phi$" but this is misleading because you require that $\Phi|_x = \psi_x$. Given $(\psi_x)_{x \in M}$ this requirement completely specifies the map $\Phi$ and the only issue is whether the set theoretic map $\Phi$ is a diffeomorphism or not. If it is a diffeomorphism, your data specifies what is called in the usual terminology a trivial vector bundle with a specific choice of a trivialization (just like in $(1)$ you didn't get a vector space but a vector space together with a choice of a specific isomorphism to $\mathbb{R}^n$).