= Cup square of a Thom class
{title2=$U^2=U\cup\pi^*e(E)$}
For an oriented rank-$r$ <vector bundle>, let $U\in H^r(D(E),S(E);\mathbb Z)$ be its <Thom class>. Forgetting relative supports sends $U$ to $\pi^*e(E)$, where $e(E)$ is the <Euler class>. Compatibility of relative and mixed <cup products> gives
$$
U\cup U=U\cup\pi^*e(E).
$$
For odd $r$, <graded commutativity of the cup product> makes $2U^2=0$. The <Thom isomorphism theorem> is injective, so $2e(E)=0$. This explains why an odd-rank <Euler class> can have only two-torsion.
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