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We know that (see Hatcher's vector bundles and K-theory Prop. 3.22) the Euler class of an orientable vector bundle or rank $r$, $E\to M$ is the first obstruction to the existence of a never vanishing section of $E$ and thus belongs to $H^r(M,\mathbb{Z})$ .

It follows that if we consider $E= TM$ the tangent bundle $r=\dim (M) =: n$ and thus the Euler class $e(TM)$ is non just the first obstruction but all the obstruction.

Consequently if $e(TM)=0$ there exists a non vanishing section of $TM$

This seems a bit strong as would imply that the Euler characteristic is zero iif we have a never vanishing vector field.

Is my argument correct?

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    $\begingroup$ Note that the Euler class is only defined in the case of an oriented bundle (so you are assuming your manifold to have, and in particular to admit, an orientation). In that case, your argument is correct. As you noted, the Euler class is the one and only obstruction to finding a section of the sphere bundle of the tangent bundle, i.e. a nowhere-zero vector field on the manifold. In the non-orientable case, you could consider the top Stiefel-Whitney class $w_n$, which is the mod 2 reduction of the Euler class in the orientable case, and which integrates to the mod 2 reduction of the (cont'd) $\endgroup$
    – Aleksandar Milivojević
    Dec 2, 2018 at 14:22
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    $\begingroup$ Euler characteristic in both cases. Having $w_n = 0$ does not imply the existence of a nowhere zero vector field, though. For a non-orientable example, consider the connect sum of two Klein bottles, which has even Euler characteristic (thus $w_2 = 0$) but no nowhere-zero vector fields (since the Euler characteristic is $-2 \neq 0$). $\endgroup$
    – Aleksandar Milivojević
    Dec 2, 2018 at 14:25
  • $\begingroup$ Thank you Aleksandar, in the non-orientable case can we also consider the Euler class of the orientable double covering ? P.S. yours is not just a mere comment, if you write it as an answer I will accept it. $\endgroup$
    – Warlock of Firetop Mountain
    Dec 2, 2018 at 14:48
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    $\begingroup$ The orientable double cover has Euler characteristic equal to twice the Euler characteristic of the non-orientable manifold you start with (see this MO post mathoverflow.net/questions/80326/…). This implies that one has Euler char 0 if and only if the other does. A closed connected manifold (no assumption on orientability) with zero Euler char admits a nowhere-zero vector field; see this MO post mathoverflow.net/questions/129752/… $\endgroup$
    – Aleksandar Milivojević
    Dec 2, 2018 at 15:05
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    $\begingroup$ Note that for bundles other than the tangent bundle, the appropriate obstruction is called the twisted Euler class and it lives in cohomology with coefficients twisted by $w_1: \pi_1 \to \pm 1$. A bundle of rank $\dim M$ has a section if and only if the (possibly twisted) Euler class vanishes. For manifolds of dimension larger than the vector bundle you are in trouble: there are more obstructions and they are more complicated. $\endgroup$
    – user98602
    Dec 4, 2018 at 22:38

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(part of comment above converted to answer)

Note that the Euler class is only defined in the case of an oriented bundle, so you are assuming your manifold to have an orientation. In that case, your argument is correct. As you noted, the Euler class is the one and only obstruction to finding a section of the sphere bundle of the tangent bundle, i.e. a nowhere-zero vector field on the manifold.

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