I encountered an apparant contradiction while dealing with Euler classes and cannot seem to resolve this issue myself. I feel pretty silly but I need help. Here's the question:

Consider an oriented, real, $n$-dimensional vector bundle $\xi:E\to B$, and let $t\in H^n(E,E\setminus E_0)$ be the Thom class. Here $E_0$ is the image of the zero section $s:B\to E$. If $i:E\to (E,E\setminus E_0)$ denotes the inclusion, then by definition the Euler class $e(\xi )\in H^n(B)$ of $\xi$ is given by $$e(\xi)=s^*i^*t.$$ Now suppose that $\xi$ has an orientation-reversing automorphism, say $\Phi:E\to E.$ Then $\Phi ^*t=-t$, so $$-e(\xi)=s^*i^*\Phi^*t=s^*\Phi^*i^*t=s^*i^*t=e(\xi),$$ whence $2e(\xi)=0$. But the tangent bundle $TS^2=\{(p,v,w)\in S^2\times \mathbb{R}^3\mid\langle v,w\rangle=0\}$ of $S^2$ admits an orientation-reversing automorphism $(p,v,w)\mapsto (p,w,v)$, and its Euler class, which is twice a generator of $H^2(S^2)\cong \mathbb{Z}$, isn't even a torsion! Where did I make a mistake?

Any help is appreciated. Thanks in advance.

  • $\begingroup$ Um, $TS^2 = \{(p,v)\in S^2\times\Bbb R^3: \langle p,v\rangle = 0\}$. $\endgroup$ Aug 30, 2021 at 5:53
  • $\begingroup$ @TedShifrin You're right. Thanks for pointing that out. (I feel very, very stupid...) Is the proof that $2e(\xi)=0$ fine? $\endgroup$
    – Ken
    Aug 30, 2021 at 6:08
  • $\begingroup$ No. There can be no orientation-reversing automorphism, as your argument indeed shows. $\endgroup$ Aug 30, 2021 at 6:10
  • $\begingroup$ @Ted Sorry for not being clear. My question is: "My argument shows that if an oriented vector bundle of rank $n$, which has nothing to do with spheres, admits an orientation-reversing automorphism, then its Euler class $e$ satisfies $2e=0$. Is this argument correct?" $\endgroup$
    – Ken
    Aug 30, 2021 at 6:13
  • $\begingroup$ Ah, yes, that seemed fine to me. From the differential geometry viewpoint, integrating the Pfaffian of curvature should give $0$, which says indeed that the Euler class is torsion. $\endgroup$ Aug 30, 2021 at 6:17

1 Answer 1


As pointed out by @Ted Shifrin, the definition of the tangent bundle of $S^2$ in the question is (blatantly) wrong...


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