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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 $M$.

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    $\begingroup$ Kosinski has a theorem like this in his differentiable manifolds textbook. But rather than your fibrewise condition he talks about $M$ being a deformation-retract of $E$. I think he uses $h$-cobordism or minimal handle decompositions, though, so it is a little different than your context. $\endgroup$
    – Ryan Budney
    Commented Mar 3, 2020 at 18:21
  • $\begingroup$ One can try to prove this by constructing an exhaustion of $U$ by a sequence of $V_i$, such that 0) $V_0$ is a smooth section of $E$ lying in $U$ 1) For $i>0$ each $V_i$ is a smooth closed submanifold of $U$ with boundary. 2) $V_i$ lies in the interior of $V_{i+1}$ 3) The intersection of $V_i$ with each fiber is a compact convex subset 4) $\cup_i V_i=U$. Note that it is easy to find a smooth section, using the partition of unity (+ convexity). As for constructing these $V_i$, this also looks doable in the same way replacing the sum by the Minkowsky sum. I can try to write this down. $\endgroup$
    – Dmitri Panov
    Commented Mar 4, 2020 at 14:22
  • $\begingroup$ Dear Dmitri, that's a very nice idea. Just so I understand, the idea is then that each $V_i$ is diffeomorphic to a disk bundle of radius $i$ around the zero section for some Riemannian metric on $M$? Is it clear that this is going to be the case? $\endgroup$
    – Dan Petersen
    Commented Mar 4, 2020 at 18:11
  • $\begingroup$ Dear Dan, yes the radius $i$ ball sub-bundle for some Euclidean metric on $E$. I think this is true, if you have a smooth compact convex set $B\subset \mathbb R^n$ (with non-zero interior), containing $0$, then you can construct a smooth map from $B$ to the unit ball, that will be radial, i.e., it will send any straight segment through $0$ to a radius of the unit ball. One can take it isometric close to $0$ and then adjust smoothly, so $\partial B$ goes to $\mathbb S^{n-1}$. This can be also done smoothly in family. $\endgroup$
    – Dmitri Panov
    Commented Mar 4, 2020 at 18:20
  • $\begingroup$ Lovely - yes, that makes sense! $\endgroup$
    – Dan Petersen
    Commented Mar 4, 2020 at 18:24

1 Answer 1

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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.

  1. 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.

  1. 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).
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  • $\begingroup$ Thanks! For completeness, I guess a natural way to ensure that this construction exhausts $U$ is to put a Riemannian metric on the total space of $E$ and then in step 1) note that we can make sure that each $B_x$ contains all vectors whose distance to the complement of $U$ is at least $1/n$. $\endgroup$
    – Dan Petersen
    Commented Mar 4, 2020 at 20:32
  • $\begingroup$ Dan, yes, for example, a good metric would be such that each fiber $\mathbb R^n$ is isometric to an open half of a unite sphere (it is identified with $\mathbb R^n$ by the projection from its centre). Such a metric is good because convex subsets of $\mathbb R^n$ correspond to convex subsets of the half-sphere. $\endgroup$
    – Dmitri Panov
    Commented Mar 4, 2020 at 22:34

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