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I knew this question was asked before, like here Non-vanishing vector fields on non-compact manifolds .

But it seems there is no detailed satisfying answer I could find. So can anyone please give proof that for this result: Any non-compact smooth manifold $M$ has a non-vanishing vector field?

Thank you in advance!

J. Yang
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  • Do you know what the Euler class of a smooth manifold is? – Michael Albanese Jun 23 '21 at 01:39
  • Yes, but to define that you need orientability; If you are saying Euler characteristic, then I can use Poincare-Hopf to prove this question in the compact case. How about non-compact? – J. Yang Jun 23 '21 at 02:11
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    There is a notion of Euler class for non-orientable bundles too. I am not alluding to the Euler characteristic. The manifold admits a nowhere zero vector field if and only if its Euler class vanishes. For non-compact manifolds, the Euler class belongs to a cohomology group which vanishes, so the class must be zero. – Michael Albanese Jun 23 '21 at 02:13
  • Oh, I'm sorry about that. I saw your post before about twisted Euler class, https://math.stackexchange.com/questions/3481359/twisted-euler-class-for-non-orientable-bundle . Maybe this is what you are trying to explain? But I couldn't find any reference in your post for that. Could you please tell me where to look at this? – J. Yang Jun 23 '21 at 02:24
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    Chapter 39 of Steenrod's Topology of Fiber Bundles is the only reference I know of. – Michael Albanese Jun 23 '21 at 02:30
  • Thank you so much! It seems there are some discussions about my question at the end of Chapter 39 in the book. I will have a look in details – J. Yang Jun 23 '21 at 02:41
  • The best answer on MSE is here. – Moishe Kohan Dec 03 '21 at 19:37

1 Answers1

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I answered the question you mention and found this one afterwards, I suppose I´ll leave the answer here too.

First of all, you can take a vector field $V_0$ in $M$ with isolated zeros as shown here. The idea of the proof is basically "pushing all the zeros to infinity".

Take a covering of the manifold $M$ by compacts $C_n$, $n\geq1$, with $C_n$ contained in the interior of $C_{n+1}$ $\forall n$. We´ll also need $M\setminus C_n$ not to have any component which is relatively compact in $M$ ($*$). You can achieve this by adjoining to $C_n$ all the relatively compact components of $M\setminus C_n$ (the resulting set is compact, see this).

Now we are going to inductively define a field $V_n$ that has no zeros in $C_n$. To do that, we take the vector field $V_n=(\phi_n)_*V_{n-1}$, where:

  • By induction hypothesis, $V_{n-1}$ has isolated zeros, and none of them are in $C_{n-1}$.

  • $\phi_n:M\to M$ is an diffeomorphism fixing $V_{n-1}$ and taking all the (finitely many) zeros of $V_{n-1}$ that were inside $C_n$ outside of it. (To make sure this exists you need the condition ($*$), which implies that every point outside $C_{n-1}$ is connected by a path in $M\setminus C_{n-1}$ to some point outside $C_n$, so you can take it outside $C_n$ by a diffeo fixing $C_{n-1}$).

Now as the sequence $V_n$ of vector fields is eventually constant in any compact subset of $M$, you can consider its limit $V$, which is equal to $V_n$ in $C_n$ so it can´t have any zeros in $C_n$ $\forall n$, meaning it has no zeros.

Saúl RM
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