# Trivialization from a smooth frame

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).

• 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? Sep 25, 2018 at 16:15
• @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. Sep 25, 2018 at 16:22

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.

• 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? Sep 26, 2018 at 15:51
• 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. Sep 26, 2018 at 15:59
• Thank you for your help! Your example clearly works so my intuition was just misleading me. Sep 26, 2018 at 16:24