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In the Spivak's book on Differential Geometry Euler class $\chi(\xi)$ of an oriented $k-$plane bundle $\xi=\pi:E\to M$ is defined as $$ s^*[\omega]\in H^k(M)$$ where $s:M\to E$ is any section and $[\omega]\in H^k_c(E)$ the Thom class represented by $\omega$ closed $k$-form with compact support on $E$.

Now for a non-zero section $s$, if we pick $c>0$ sufficiently large, some subset of image of $M$ under the section $c\cdot s$ will lie outside of support $\omega$. So by definition $$\chi(\xi)=(c\cdot s)^*[\omega]=0 \in H^k(M).$$ But why is it zero?

For example $\omega_{cs(p)}$ will be zero form for some $p\in M$. Is the form then exact?

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  • $\begingroup$ Perhaps you want a nowhere zero section instead of a nonzero section. If such a section exists, the eular class of the vector bundle is zero. $\endgroup$
    – AG learner
    Mar 31, 2020 at 19:15
  • $\begingroup$ @AGlearner Could you perhaps elaborate on why should it be zero in that case. Thank you! $\endgroup$
    – user3342072
    Mar 31, 2020 at 19:21
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    $\begingroup$ It is exactly what you wrote, if a section $s$ does not vanish, by scaling it will miss the support of the $\omega$, so pullback is a zero form. $\endgroup$
    – AG learner
    Mar 31, 2020 at 19:26
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    $\begingroup$ And note that $M$ is compact? $\endgroup$
    – Ted Shifrin
    Mar 31, 2020 at 22:54
  • $\begingroup$ @TedShifrin Yes :) $\endgroup$
    – user3342072
    Mar 31, 2020 at 23:01

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