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a friend asked me to pose the following problem:

It is known that on an open manifold (connected, not compact and without boundary) there exists a vector field without zero, since its Euler characteristic class is non-zero. I want to construct such a vector field.

In his book, Comprehensive Introduction to Differential Geometry vol I, Mr. Spivak suggests the following steps:

  1. We take a triangulation of the manifold.
  2. In the interior of each n-simplex we construct a vector field with only one zero to obtain a vector field over the whole manifold with an infinite, but discrete, discret zeros (precautions to take in interfaces?)
  3. We join all the zeros of this field with a path.
  4. Take a tubular neighborhood of this path.
  5. Finally, using an isotopy to "hunting" all zeros to infinity to obtain the desired field.

I tried this idea, writing a detailed demonstration without success. Could you help me to do?

One could also try to build on the manifold a function without critical point, then take a Riemannian metric and using the gradient. I do not do well. Is there a demonstration of such a fact?

Thank you for any suggestions.

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    Where did you get stuck in carrying out the outline Spivak suggested? – Sam Lisi Nov 15 '11 at 20:46
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    This appears as an exercise on page 456 in Spivak's book. Sadly, Spivak's suggestion simply does not work. The reason is that for a general discrete subset $Z$ in a connected manifold $M$ it's impossible to find a properly embedded ray containing $Z$. The simplest example is $M=S^1\times {\mathbb R}$ and $Z={p}\times {\mathbb Z}$. One can find a solution which actually works here. – Moishe Kohan Dec 04 '21 at 05:35

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