I'm starting to read about smooth vector bundles. My notes suggest that instead of giving a local trivialisation at each point, it is sufficient to give a local frame.

I can see that given a rank $r$ vector bundle $(E, \pi)$ over $M$ and a local frame $\{s_i: 1 \leq i \leq r\}$ for $E$ over $U \subseteq M$, we can define a local trivialisation $t$ by setting for $v_q \in E_q = \pi^{-1}(q)$, $$t(v_q) = (q, \alpha_1, \dots, \alpha_r) \label{eq}\tag{1}$$ where $v_q = \sum_{i=1}^r \alpha_i s_i(q)$. The only non-trivial part of seeing that $t$ is really a trivialisation is seeing that it is smooth. An argument for this assuming that $(E, \pi)$ already has the structure of a vector bundle is given here.

My understanding of what is written in my notes suggests that this is also true without assuming that $(E, \pi)$ is already a vector bundle. That is, we should have something like

Proposition: Let $\pi:E \to M$ be a smooth map of manifolds such that for each $p \in M$, $E_p = \pi^{-1}(p)$ is an $r$-dimensional vector space. Suppose further that for each $p \in M$, there is an open neighbourhood $U$ of $p$ and a smooth frame $s_1, \dots, s_r: U \to E$. Then $(1)$ defines a trivialisation of $E$ over $U$. In particular, $E$ is a rank $r$ vector bundle over $M$.

It is clear that we can use the argument given in the linked question as long as for each $p \in M$, $E_p$ is contained in an open subset of $E$ that is diffeomorphic to $U \times \mathbb{R}^r$ where $U$ is an open neighbourhood of $p$ in $M$ (by looking at a suitable restriction of a local frame at $p$). However it is not clear to me that such a subset of $E$ should exist without assuming that some trivialisation already exists.

I'd like to know if my proposition is true and if so how this attempt at a proof may be fixed (or if there is some entirely different argument that works).

  • $\begingroup$ You need to show that $t: \pi^{-1}(U) \rightarrow U \times \mathbb{R}^n$ is a diffeomorphism. For smoothness it's enough to prove that $q\mapsto \alpha_i(q)$ is smooth on $ \pi^{-1}(U)$, is that what you're struggling with? $\endgroup$
    – Jan Bohr
    Sep 25, 2018 at 16:15
  • $\begingroup$ @JanBohr Exactly that. This is what is done in the linked question under the assumption that we can reduce to the case $E = U \times \mathbb{R}^n$. I can see that that reduction is fine if $(E, \pi)$ is already a vector bundle because then we already have diffeomorphisms on to such sets but I can't see why it's fine in general. $\endgroup$ Sep 25, 2018 at 16:22

1 Answer 1


As it stands, the Proposition is wrong because the vector space structure may not vary smoothly.

Let's start with $\pi:E = U \times \mathbb{R}^n \rightarrow U$ and the trivial sections ($U$ a manifold). Now pick some $p\in U$ and change the vector space structure on $E_p$ such that $0_p$ changes the position and $s_1(p),...,s_n(p)$ remains a basis. Give the newly obtained object (i.e. manifold + projection + new vector space structure at $p$) the name $E'$, then according to the proposition $E'$ should be a vector bundle as well. But then $0':U\rightarrow E'$ (the new zero section) is not smooth, which is impossible for vector bundles.

Example: Consider the projection $\pi:\mathbb{R}^2\rightarrow \mathbb{R}$ onto the first factor as a smooth map between manifolds. For $p\neq 0$ equip the fibre $\pi^{-1}(p)=\{p\}\times \mathbb{R}\subset \mathbb{R}^2$ with the vector space structure inherited from $\mathbb{R}^2$.

In the fibre $V=\pi^{-1}(0)=\{0\}\times \mathbb{R}$ we choose a weird vector space structure such that $0_V = (0,1).$ To this end define addition in $V$ by $$ (0,a) +_V (0,b) = a+b-1$$ and scalar multiplication by $$ \lambda \cdot_V (0,a)= (0,1+\lambda(a-1)).$$

Next consider $s:\mathbb{R}\rightarrow \mathbb{R}^2,p\mapsto(p,2)$. This is a section of $\pi$ (i.e. $\pi\circ s = \mathrm{id}$) and also constitutes a frame, because for each $p\in \mathbb{R}$, the vector $s(p)=(p,2) \in \pi^{-1}(p)$ is different from $ 0_{\pi^{-1}(p)}$ and thus a basis.

Now we are in the situation of your Proposition: $\{s\}$ is a smooth frame (even a global one), nevertheless $\pi$ is not a vector bundle. The reason for this is a again that the zero section $$0_\pi:\mathbb{R}\rightarrow \mathbb{R}^2, p \mapsto \begin{cases} (p,0) & p\neq 0 \\ (0,1) & p= 0 \end{cases}$$ is discontinuous.

  • $\begingroup$ Its possible that Im misunderstanding but something seems to be wrong here. This seems to contradict the argument given in the question I linked that shows that $(1)$ should define a trivialisation in this case. I'm still fuzzy on details but for $E = U \times \mathbb{R}^n$ I expect that the $s_i$ are essentially smooth maps $s_i: U \to \mathbb{R}^n$. Its unclear to me that you can change the vector space structure at $p \in U$ without making the $s_i$ also jump at this point when viewed as maps $U \to \mathbb{R}^n$ (obviously this isn't a rigorous argument, just an intuition). Is this wrong? $\endgroup$ Sep 26, 2018 at 15:51
  • $\begingroup$ The point that the $s_i$ are essentially a smooth maps $U \rightarrow \mathbb{R}^n$ is the tricky one, because they only are, when composing with a trivialisation. Let me add an example to the answer. $\endgroup$
    – Jan Bohr
    Sep 26, 2018 at 15:59
  • $\begingroup$ Thank you for your help! Your example clearly works so my intuition was just misleading me. $\endgroup$ Sep 26, 2018 at 16:24

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