5
$\begingroup$

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 $M$.

$\endgroup$
5
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 3 '20 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. $\endgroup$ Mar 4 '20 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? $\endgroup$ Mar 4 '20 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. $\endgroup$ Mar 4 '20 at 18:20
  • $\begingroup$ Lovely - yes, that makes sense! $\endgroup$ Mar 4 '20 at 18:24
3
$\begingroup$

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).
$\endgroup$
2
  • $\begingroup$ 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 $1/n$. $\endgroup$ Mar 4 '20 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. $\endgroup$ Mar 4 '20 at 22:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.