confusion in the definition of Euler class

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.

• Um, $TS^2 = \{(p,v)\in S^2\times\Bbb R^3: \langle p,v\rangle = 0\}$. Commented Aug 30, 2021 at 5:53
• @TedShifrin You're right. Thanks for pointing that out. (I feel very, very stupid...) Is the proof that $2e(\xi)=0$ fine?
– Ken
Commented Aug 30, 2021 at 6:08
• No. There can be no orientation-reversing automorphism, as your argument indeed shows. Commented Aug 30, 2021 at 6:10
• @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?"
– Ken
Commented Aug 30, 2021 at 6:13
• 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. Commented Aug 30, 2021 at 6:17

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