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As I understand it the requirements for a vector bundle $p:E \to M$ to be a smooth vector bundle is for $p$ to be a smooth map and for the local trivialization maps $p^{-1}(U) \to U \times \mathbb{R}^k$ to be diffeomorphisms. What is an example where $p:E \to M$ is a vector bundle in the continuous sense, $p$ is still a smooth map, but the trivializations fail to be smooth?

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Let $\tilde{\mathbb{R}}$ denote $\mathbb{R}$ with its standard topology and a smooth structure given by the chart $\phi:\tilde{\mathbb R}\to\mathbb R$, $\phi(x)=x^3$. Note that $\operatorname{id}_{\mathbb R}:\tilde{\mathbb R}\to\mathbb R$ is continuous but not smooth. Then the trivial bundle $p:\tilde{\mathbb R}^2\to\tilde{\mathbb R}$, $p(x,y)=x$ is an example.

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  • $\begingroup$ Thanks, so if I'm understanding this right the trivialization around some $U$ won't be smooth because the map $p^{-1}(U) = U \times \tilde{\mathbb{R}} \to U \times \mathbb{R}$ will be $\text{id}_U \times \text{id}_{\mathbb{R}}$, which isn't differentiable at $(x,0)$ for any $x \in U$, right? $\endgroup$ Commented Jul 9, 2020 at 7:17
  • $\begingroup$ It is slightly more than that. Not only is it true that the "obvious" trivialization for the topological vector bundle is not smooth, no trivialization of it can be smooth. You're right that that's because of problems at 0. $\endgroup$ Commented Jul 9, 2020 at 7:23
  • $\begingroup$ I'm not sure I understand that. Couldn't I take $U \times \tilde{\mathbb{R}} \to U \times \mathbb{R}$ to be $(x,y) \mapsto (x, \phi(y))$ and claim that that's smooth trivialization? $\endgroup$ Commented Jul 9, 2020 at 7:37
  • $\begingroup$ A trivialization is required to preserve the vector space structure of the fibers. We are implicitly using the "natural" real vector space structure on $\{x\}\times\tilde{\mathbb{R}}\simeq\mathbb R$ (this is technically a piece of extra information that I should've specified). $\endgroup$ Commented Jul 9, 2020 at 7:44
  • $\begingroup$ Oh right sorry, $\phi$ isn't linear of course. So any trivialization will have to be something like $(x,y) \mapsto (x,ay)$ for some nonzero $a \in \mathbb{R}$, which will then run into the same differentiability issue at $0$, right? $\endgroup$ Commented Jul 9, 2020 at 7:49

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