Let $M$ be a manifold, $p:E\to M$ a rank $d$ vector bundle. Suppose that $U \subset E$ is an open subset such that $U \cap p^{-1}(x)$ is nonempty and convex for all $x \in M$. Is it true that $U \to M$ is a fiber bundle with fiber $\mathbb R^d$? And that $U \cong E$ as fiber bundles? We may assume with no loss of generality that $U$ contains the zero section.


This seems like a statement that could be a lemma in any number of textbooks (if true), e.g. in connection with the tubular neighborhood theorem, but I haven't seen it anywhere. Lang proves in his differential geometry book that any vector bundle over a manifold is what he calls *compressible*, meaning that any open neighborhood of the zero section of $E$ can be shrunk to a smaller open neighborhood which is diffeomorphic to $E$ as a bundle over $M$.



Shrinking and stretching of vector bundles
Since there are no references so far, let me give a sketch proof along the lines of my comment. I'll assume that $M$ is compact.


0. Let's show first that there is a smooth section of $E$ lying in $U$. Indeed, for any point $x\in M$ there is a neighbourhood $U\_x$ with a section $s\_x$. Take a finite cover $U\_i$ of $M$ by such neighbourhoods and take the corresponding partition $1=\sum f\_i$ of unity. Then by convexity $\sum s\_i f\_i$ is a smooth section lying in $U$.


Clearly we can assume that $s$ is the zero section (by taking an appropriate fiberwise diffeo), we will assume this from now on.


Now we will construct an exhaustion of $U$ by an increasing sequence of fiber-wise compact convex subsets $0\subset {\cal B\_1}\subset ... \subset {\cal B\_i}\subset ...$ so that $U=\cup\_i {\cal B\_i}$.


1. Let me show first how to construct one such subset ${\cal B\_1}\subset U$.


For every point $p\in M$ let us choose some covex compact subset $B\_p$ with smooth boundary in the fiber $U\_p$. Then, since $U$ is open, there is an open neighbourhood $V\_p$ of $p$ in $M$ such that over this neighbourhood there is a smoothly varying family of $B\_x$ ($x\in V\_p$), such that $B\_x\subset U\_x$. Take a finite cover of $M$ by such $V\_i's$, let $\phi\_i$ be the partition of unity. Then the sum


$${\cal B\_1}=\sum\_i \phi\_i B\_i (x)$$


is the desired subset $B\subset U$. Here by sum I mean the Minkowski sum.


2. It is clear that the interior of $\cal B\_1$ is diffeomorphic to the bundle of vectors of length less than $1$ in $E$ (for some fiber-wise Euclidean metric). So the only need to construct a family of $\cal B\_i$ that will exhaust $U$. This can be done as in 1).


