Shrinking and stretching of vector bundles

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 $$M$$.

• 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. Mar 3 '20 at 18:21
• 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. Mar 4 '20 at 14:22
• 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? Mar 4 '20 at 18:11
• 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. Mar 4 '20 at 18:20
• Lovely - yes, that makes sense! Mar 4 '20 at 18:24

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).
• 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 $1/n$. Mar 4 '20 at 20:32
• 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. Mar 4 '20 at 22:34