# Existence of specific sections of vector bundles over a manifold

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 $$k then $$\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.

• In the second part of your solution to part (b), it's not clear the section on the complement of a point in $M$ can be extended to a section on the whole of $M$.
– JHF
Dec 26, 2020 at 21:55
• My idea would be to kinda normalize that section on $M-\{p\}$ and put it has zero at $p$ and then multiply by that $"bump"$ function so that it is smooth . Dec 26, 2020 at 21:58

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 $$\xi$$.

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 $$H^n(M) = 0$$.

• Thanks for the answer. Yeah I have learned about the euler class but I did not know that characterization of it, I just knew it for being the pull-back of the thom class, and I have learned that for the tangent bundle we get that $\chi(TM)=\chi(M)\mu$, where $\mu \in H^d(M)$ is the canonical generator. Will by any chance we have that $\chi(\xi)=X(\xi)\mu$ for an arbritary rank $n$-vector bundle ? And then we have the result that if $X(\xi)=0$ there exists a non-vanishing section, and so we get that if $H^n(M)=\{0\}$ then $X(\xi)=0$ and hence the existence of such section. @JHF Dec 27, 2020 at 8:08
• Knowing that it get's alot easier since if $M$ is a compact connected manifold we get that $H^d(M-p)=\{0\}$ by using the Mayer-Vietoris sequence, at least for part $b)$, not sure if type arguments work for manifolds with boundary and non-compact manifolds as well. Dec 27, 2020 at 8:11
• When you write $X(\xi)$ do you mean $\chi(X(\xi))$ where $X(\xi)$ is the base space of the bundle $\xi$, and you're asking if $e(\xi) = \chi(X(\xi))\mu$? If so, I don't believe a result like that holds: let $\xi$ be the canonical (complex) line bundle over $\mathbb{C}P^1 \cong S^2$, which is a real $2$-plane bundle. Then $e(\xi)$ generates $H^2$ (it's negative the usual generator for $H^2$) but the Euler characteristic of the base space is $2$, so there is no way that $e(\xi) = \chi(S^2)\mu$ for $\mu$ a fundamental class for $\mathbb{C}P^1$. Dec 27, 2020 at 15:03
• I don't know what you mean by $X(\xi)$. A non-vanishing section of $\xi$ is a section of its sphere bundle $S(\xi)$. Usual obstruction theory tells us that the obstructions to a section of a $S^{n-1}$-bundle on $M$ live in $H^k(M; \pi_{k-1} S^{n-1})$. If $M$ is a $n$-manifold, the only time this group is zero is when $k = n$, and in this case the obstruction class in $H^n(M; \mathbb{Z})$ is the (possibly twisted) Euler class. See e.g., Steenrod's The Topology of Fiber Bundles.
– JHF
Dec 27, 2020 at 16:08