Definition: It is said that a section $F:M\to E$ of a vector bundle $E$ is smooth if it is smooth as a map between manifolds.

Possible Issue: A vector bundle is defined to be, a priori, a smooth manifold, which means that it has some implicit smooth structure $\mathcal{A}$. However, it then has additional structure, i.e. local trivializations

$$\Phi_i:\pi^{-1}(U_i)\to U_i\times \mathbb{R}^k.$$

Since the $\Phi_i$ are defined as diffeomorphisms, $(\Phi_i,\pi^{-1}(U_i))$ is a smooth atlas, which defines another smooth structure $\mathcal{B}$ on $E$.

Two questions: Does the rest of the definition of a vector bundle imply that $\mathcal{A}=\mathcal{B}$? If not, then when we say that vector fields are maps between smooth manifolds, then which manifold: $(E,\mathcal{A}),$ or $(E,\mathcal{B})$?


1 Answer 1


It is part of the (usual) definition of a smooth vector bundle (cf. e.g. http://en.wikipedia.org/wiki/Vector_bundle#Smooth_vector_bundles ) that the local trivializations $\Phi_i$ are required to be smooth (even Diffeomorphisms).

This implies that both smooth structures are the same.

EDIT: Note that if all $\Phi_i$ are smooth w.r.t. $\mathcal{A}$, this implies that $\mathcal{A} \cup \{\Phi_i \mid i \in I\}$ is also an atlas, so that the smooth structures induced by $\mathcal{A}$ and $\{\Phi_i \mid i \in I\}$ are the same.

  • $\begingroup$ I doubt that because the maps $\Phi_i$ take the whole $\mathbb{R}^k$ as domain, so that the topology of that atlas cannot distinguish two points of the same fiber $\pi^{-1}(x)$. In other words, open sets in $(E,\mathcal{B})$ are of the form $U\times \pi^{-1}(U)$ where $U$ is open in $M$. $\endgroup$ Commented Sep 19, 2023 at 4:26

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