Existence of specific sections of vector bundles over a manifold

Question

I am trying to do the following exercise from Hirsch, one could say that it's 3 exercises but they are all related so I believe it's best to treat them together:

Let $\xi=(E,M,p)$ be an $n$-plane bundle over a connected $k$-manifold $M$.

a) If $k<n$ then $\xi$ has a non-vanishing section.

Here I belive the idea is just to use the transversality theorem, since we can approximate the zero section map $s$ by mapping the transversal to the zero section $h_k$ and for $h_k$ close enough we will get that $p\circ h_k$ is a diffeomorphism. So we can consider $h_k\circ (p\circ h_k)^{-1}$ , and this will be a section, transversal to the zero section. But then by a dimension argument we have to have that their intersection is empty.

b) If $k=n$ and $x\in M$ then $\xi$ has a section vanishing only at $x$.

Here let's divide this into two cases. First suppose $M$ has a non-vanishing section $s$ , then we can consider a function $f:M\rightarrow \mathbb{R}$ such that $f^{-1}(0)=x$, I believe this is always possible, and just consider the section $fs$. Now from the next question $c)$ and what we just did, we can assume that $\partial M=\emptyset $ and that $M$ is compact. Now if I remove a point from $M$ I get a non-compact manifold with a non-vanishing section, to be proved, and then we can just do an analogous argument to what we just did.

c) If $k=n$, and $\partial M\neq \emptyset $ or $M$ is non-compact, then $\xi$ has a non-vanishing section.

Now let's first assume that $\partial M\neq \emptyset $ and that $M$ is compact . My idea is to try and do something similiar to what was done when we proved that a compact connected manifold with boundary has a non-vanishing vector field. So let's take the double $M'$ of $M$. This will have a section $s$ with a finite number of zeros, by an analogous argument to what we did in $a)$, which we denote by $F$. Here I assume I can create a vector bundle over $M'$ by just taking the fibers of the vector bundle over $M$. I believe this is possible, but would appreciate some input. We will call this $\xi'$. Since $M'$ is connected there is a diffeomorphism $\phi :M'\rightarrow M'$ that takes $F$ into $M-M'$. Then we can consider the map $s\circ \phi^{-1}|M :M\rightarrow \xi'$ that has no zeros. Now I would like to have a vector bundle map that goes from $\xi'\rightarrow \xi $ and covers the identity when we restrict to $M$. But I'm not sure if this is possible since we want this map to not create zeros from the result we get from $s\circ \phi^{-1}$, and I'm not sure how to create a map without using some partitions of unity.

Now for the non-compact case I am kinda lost, have really no ideas on what to do. I thought about using the trivializing charts and then gluing everything together with partitions of unity, but when we glue things together, how can we do it in a way that doesn't create zeros ? I don't think I can have this control. So I am out of ideas and would appreciate some input. Don't really see where I can use the non-compactness hypothesis.

Thanks in advance.

Answer 1

Assume $M$ is a compact $n$-manifold and that $\xi$ is a $n$-plane bundle on $M$. By transversality, there exists a section $s$ of $\xi$ that has finitely many isolated zeros. We shall "push" each zero onto the special point $x$.

For each zero of the section $s$, there is an embedded arc connecting it to $x$ since $M$ is connected. Take a small open neighborhood $U$ of this arc such that $\bar{U}$ does not contain any other zeros of $s$. If $U$ is chosen small enough, it will be contractible and $\xi|_U$ is trivial; we use this to identify all the fibers of $\xi|_U$. We claim that there is a smooth modification $\tilde{s}$ of $s$ such that $\tilde{s} \equiv s$ outside of $U$, and the only zero of $\tilde{s}$ inside $U$ is at $x$.

Without loss of generality, by applying appropriate diffeomorphisms, we may assume that $U$ is the unit ball $B^n$ in $\mathbb{R}^n$ and that $x = 0 \in B^n$. We define $\tilde{s}$ to be identical to $s$ outside of $U$, and inside of $U \cong B^n$, $$\tilde{s}(r, \theta) = \rho(r) \cdot s|_{\partial B^n}(\theta),$$ where $\rho: [0,1] \to [0,1]$ is a smooth function such that $\rho \equiv 1$ in a neighborhood of $1$, $\rho(r) = e^{-1/r}$ in a neighborhood of $0$, and $\rho(r) = 0$ iff $r = 0$. Then the only zero of $\tilde{s}$ inside $U$ is at $x$, as desired.

Repeating these modifications for each zero of $s$, we obtain a section whose only possible zero is located at $x$.

It turns out that this procedure of pushing (and possibly merging) zeros of sections is very useful.

For example, suppose now that $M$ has nonempty boundary $\partial M$. Attach a collar to the boundary (which does not change the topology of $M$), construct a section with isolated zeros, and push all of them onto the collar as above. Then simply detach the collar to get a section with no zeros.

If $M$ is noncompact, take a compact exhaustion $\emptyset = K_0 \subset K_1 \subset K_2 \subset \cdots \subseteq M = \bigcup_i K_i$. Given zeros in $K_i \setminus K_{i-1}$, we push them all to $K_{i+1} \setminus K_i$. Observe that this process leaves the section defined on $K_{i-1}$ unchanged, so as we keep pushing all the zeros of a section off to infinity, we obtain a well-defined nonvanishing section of $\xi$.

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For full disclosure, I'm not sure whether the given construction of $\tilde{s}$ is the most efficient, and checking that it is indeed smooth at the center is a bit of a hassle.

Also, there is a simple argument to show that nonvanishing sections exist on noncompact manifolds if you know a bit about Euler classes. The only obstruction to a nonvanishing section of a rank $n$ bundle on a $n$-manifold $M$ is the Euler class, which is necessarily zero since $H^n(M) = 0$.