Local triviality condition in vector bundles [closed]

Question

Let $E$ and $M$ be smooth manifolds (of finite dimension, Hausdorff and second countable). Let $\pi:E\longrightarrow M$ be a surjective submersion such that:

1. $E_p:=\pi^{-1}(p)$ is a real vector space isomorphic to $\mathbb R^n$, $\forall p \in M$.
2. The pointwise sum, multiplication by scalar and zero section are smooth: $$ +:E\times_M E \longrightarrow E $$ $$ \cdot: \mathbb R \times E \longrightarrow E $$ $$ 0:M \longrightarrow E$$

Then, is $E$ a vector bundle with this structure?

I have thought the following:

Let $p\in M$, then exists $\sigma_i:U \longrightarrow E $, local sections such that $\{\sigma_i(p)\}$ is a basis of $E_p$. Define: $$ V:=\{q\in U \; | \; \{\sigma_i(q)\} \text{ is a basis of } E_q \}=\{q\in U \; | \; \sigma_1(q)\wedge \dots \wedge \sigma_n(q)\not=0(q) \} $$ $$ \Xi:V \times \mathbb R^n \longrightarrow \pi^{-1}(V) $$ $$ (q,\lambda_1,...,\lambda_n) \mapsto \lambda_1\sigma_1(q)+\dots+\lambda_n\sigma_n(q) $$

Is $V$ open?

$\Xi$ would be the candidate for a vector bundle chart around $p$: It's an isomorphism on each fiber. If $V$ is open, then $\Xi$ will be smooth (by 2.). Is $\Xi^{-1}$ smooth?

Answer 1

If you see $\Xi$ as a map from $U\times\mathbb R^n$ to $\pi^{-1}(U)$, then it is smooth, and by the inverse mapping theorem it is locally invertible around $(p,0)$ (because of the submersion condition and the condition on the $\sigma_i(p)$). It means that for all $q$ close enough to $p$, the restriction $d\Xi_{(q,0)}:\mathbb R^n\to E_q$ is invertible, so $(\sigma_i(q))_i$ is a basis. So you can take $V$ to be a neighbourhood of $p$, or you can use the same reasoning to show that the $V$ you describe is open. Then $\Xi:V\times\mathbb R^n\to\pi^{-1}(V)$ is a diffeomorphism, because it is a smooth bijection, and the inverse is locally smooth by the inverse mapping theorem.

Note that this is not enough to make $E$ a vector bundle by the usual definition: technically, one needs to check that the change of coordinates are actually vector space isomorphisms. However, in your setting, this is obviously the case since the $\Xi$'s come from the unified vector space structure on $E$.