I'm trying to solve Exercise 11.19 from Bott & Tu.

Show that the Euler class of an oriented sphere bundle with even-dimensional fibers is zero, at least when the sphere bundle comes from a vector bundle.

I figured that the key to solving this would be to show that since the antipodal map $a: S^{2n} \to S^{2n}$ is orientation reversing, I could do some kind of manipulation with the integrals $\int_M e = \sum_i \int_{\partial \overline{D_i}} s^* \psi$ where $\psi$ is the global angular form on the sphere bundle. We also have $\int_{\partial \overline{D_r}} s^* \psi = \int_{\partial \overline{D_r}} s^* \rho^* \sigma =$ local degree of the section $s$ at $x_i$.

Suppose I have a local section $S: U \to E|_U = U \times S^{2n}$, $S(x) = (x,s(x)v)$, perhaps I could augment it with the antipodal map, $S'(x) = (x,a(s(x)v)) = (x,-s(x)v)$. Since the Euler class is independent of the section, and $S,S'$ both have the same zeros, this would mean that $\int_M e = \sum_i (\text{local degree of $S$ at $x_i$}) = \sum_i (\text{local degree of $S'$ at $x_i$})$.

However I believe that $\sum_i (\text{local degree of $S'$ at $x_i$}) = - \sum_i (\text{local degree of $S$ at $x_i$})$, which would solve the problem. Is this correct? Are there other sources with more information on this way of working with the Euler class?

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    $\begingroup$ This is correct. I guess they put in the comment "at least when the sphere bundle comes from a vector bundle" in order to be sure that you can define a smooth section of $E$ off a finite set of points. $\endgroup$ Jul 31, 2019 at 17:24
  • $\begingroup$ Great, thank you! $\endgroup$ Jul 31, 2019 at 17:24
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    $\begingroup$ But this only shows that the Euler number is zero, instead of the Euler class, am I wrong? The exercise does not assume the dimension of the manifold is the same as the rank of the vector bundle, so in my understanding, the Euler number cannot be identified with the Euler class in this case? $\endgroup$
    – Fan
    Apr 6, 2020 at 10:24

1 Answer 1


For $M$ to be of any (finite) dimension, we could consider the pullback of fibrewise antipodal map acting as an automorphism of the C$\check{e}$ch-de Rham (double) complex of $E\rightarrow M$. Then the pullback of orientation changes sign but the pullback of Euler class does not. That suffices to show the Euler class vanishes.

Let $\pi: E\rightarrow M$ be an oriented sphere bundle with even-dimensional fibres. First of all, we define the fibre-wise antipodal map \begin{align*} A: E&\rightarrow E\\ (p,v) &\mapsto (p,-v)\quad. \end{align*} Note that it is a bundle morphism: $\pi\circ A = \pi$ commutes. Moreover, since $A\circ A = \mathrm{id}_E$, we know $A$ is a bundle isomorphism.

Let $\mathscr{U}=\{U_\alpha\}$ be a good cover of trivializations of M. Since $E$ is orientable, we can find a collection of $(0,n)$-forms $\{\sigma_\alpha\}$ such that $[\sigma_\alpha]=[\sigma_\beta]$ on $U_{\alpha\beta}$ and the restriction of $\sigma$ to each fibre is a generator of $H^n(S^n)$.

Define the pullback of A on the C$\check{e}$ch-de Rham complexes as follows: \begin{align*} A^*: \mathcal{C}(\pi^{-1}\mathscr{U},\Omega^*) &\rightarrow \mathcal{C}(\pi^{-1}\mathscr{U}, \Omega^*)\\ \omega_{\alpha_0\cdots\alpha_p} \in \Omega^q(U_{\alpha_0\cdots\alpha_p})&\mapsto (A|_{U_{\alpha_0\cdots\alpha_p}})^*\omega_{\alpha_0\cdots\alpha_p}\in \Omega^q(U_{\alpha_0\cdots\alpha_p}) \end{align*} This then extends to $C^p(\pi^{-1}\mathscr{U}, \Omega^q) = \prod_{\alpha_0<...<\alpha_p} \Omega^q(U_{\alpha_0\cdots\alpha_p}) \rightarrow C^p(\pi^{-1}\mathscr{U}, \Omega^q) $ natually. Hence the pullback $A^*$ is well-defined in a natural way.

Now, we claim $A$ induces double complex morphism $A^*$ between two C$\check{e}$ch-de Rham complexes $C^*(\pi^{-1}\mathscr{U}, \Omega^*) \rightarrow C^*(\pi^{-1}\mathscr{U},\Omega^*)$. Note that $A^*|_{U_{\alpha_0\cdots\alpha_p}}$ commutes with $d$ locally (properties of pullback operation), so $A^*$ commutes with differentiation $d$. Also, $A^*$ commutes with $\delta$ since $A^*$ is linear. Hence, $A^*$ is a double complex morphism. By $A\circ A = \textrm{id}_E$, we get $(A^*)^{-1} = (A^{-1})^* = A^*$ is a double complex isomorphism.

Consider the "Tic-Tac-Toe" digram which defines the Euler class: \begin{array}{|c|c|c|c|c|c} \beta_0 & & & & & \\ & \beta_1& & & &\\ & & \beta_2 & & &\\ & & & \ddots & &\\ & & & & \beta_n & - \pi^*\epsilon\quad\quad\quad\\ \hline \end{array} where $\beta_0 = (\sigma_\alpha)$ is the orientation of $E$ and $\epsilon$ is a representative of euler class. By applying $A^*$ to every term involved, we get $A^*(-\pi^*\epsilon) = -(\pi\circ A)^* \epsilon = -\pi^*\epsilon $ is the Euler class corresponding to the orientation $A^*\beta_0 = -\beta_0$. Hence, for opposition orientation, we get the same Euler class, i.e. $e(E) = -e(E)$. This shows $e(E) = 0$.


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