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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.

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  • $\begingroup$ 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 $M$. $\endgroup$
    – JHF
    Dec 26, 2020 at 21:55
  • $\begingroup$ 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 . $\endgroup$
    – Someone
    Dec 26, 2020 at 21:58

1 Answer 1

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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 $H^n(M) = 0$.

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  • $\begingroup$ 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 $\endgroup$
    – Someone
    Dec 27, 2020 at 8:08
  • $\begingroup$ 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. $\endgroup$
    – Someone
    Dec 27, 2020 at 8:11
  • $\begingroup$ 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 $\mathbb{C}P^1$. $\endgroup$
    – kamills
    Dec 27, 2020 at 15:03
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    $\begingroup$ 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. $\endgroup$
    – JHF
    Dec 27, 2020 at 16:08

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