# Non-Smooth Vector Bundle

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?

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.
• 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? Jul 9, 2020 at 7:17
• 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? Jul 9, 2020 at 7:37
• 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). Jul 9, 2020 at 7:44
• 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? Jul 9, 2020 at 7:49