# Question on the Thom class of an oriented rank 2 vector bundle.

In self studying from Bott & Tu, I came across this problem which asks to find the class $u$ on $M$ such that $\Phi^2 = \Phi\wedge\pi^\ast u$, where $\Phi$ is the Thom class of an oriented rank 2 bundle, $E$, on $M$. Since for a rank 2 bundle, $\Phi\in H^2_{cv}(E)$, we must have $u\in H^2(M)$, in particular, we have that $u=0$ for $M$ a 1-manifold.

My first approach was to try and use the projection formula as follows $$\pi_\ast(\Phi^2) = \pi_\ast(\Phi)^2 = 1^2 = 1 = \pi_\ast (\Phi\wedge\pi^\ast u) = u\wedge\pi_\ast\Phi = u$$ where $\pi_\ast$ is integration along the fiber. However, this oviously makes no sense since $1$ is a $0$-form. After thinking about the defintion of $\pi_\ast$ I realized that while this gives a vector space isomorphism, it doesn't preserve the graded ring structure since, for example, for $\omega = \pi^\ast\phi f(\vec{x},t_1,t_2)dt_1, \eta = dt_2$, $\pi_\ast\omega = \pi_\ast\eta = 0$, but $\pi_\ast(\omega\wedge\eta) = \phi\int_{\mathbb{R}^2}f(\vec{x},t_1,t_2)dt_1 dt_2$ is generally nonzero. So all we really get from the projection formula is that $u = \pi_\ast(\Phi^2)$. Then from the explicit formula for $\Phi$ we can calculate $u$, which I get to be the Euler class $e(E)$. Can anyone verify that this is the case?

Something that supports this result is the fact that for the zero section $s$, $s^\ast\Phi = s^\ast\pi^\ast e = e$, however $s^\ast$ isn't necessarily injective as far as I could tell so the result couldn't be deduced from this observation.

• How does it follow from the explicit formula of $\Phi$ that $u = e$?
– user512346
Apr 22, 2019 at 23:51
• @JuanDiegoRojas I did this about a year ago, but I believe if you use the explicit formula for $\Phi$ derived just before the question is asked and compute $\pi_\ast(\Phi^2)$ you get the definition of the Euler class given earlier. Apr 24, 2019 at 2:52
• Yep, after commenting I figured out that integration along the fiber yields the desired result. Thanks for answering an old post!
– user512346
Apr 24, 2019 at 3:56
• Could you show a little bit of the work you did to show this? I am getting completely lost in the computation of $\pi_*(\Phi^2)$ Jul 16, 2019 at 16:38

First we compute $$\pi_*(\Phi^2) = \pi_*(\Phi \wedge \pi^*u) = \pi_*(\pi^* u \wedge \Phi)$$ since $$\Phi$$ is a 2 form. Then $$\pi_*(\pi^*u \wedge \Phi) = u \wedge \pi_*(\Phi) = u$$ by the projection formula. Thus we have $$u = \pi_*(\Phi^2)$$. Now we must compute this thing explicitly, which at first glance is a horrible calculation, but many things cancel out thanks to $$dx \wedge dx = 0$$ (where $$dx$$ is a basis element of $$\Omega(M)$$ or $$\Omega(E)$$). So lets calculate $$\Phi^2$$ first.
$$\Phi^2 = \left[ d\left( \rho \frac{d \theta}{2\pi} \right) + \frac{1}{2 \pi i} d \left( \rho \, \pi^* \sum_{\gamma} \rho_{\gamma} d \log g_{\gamma \alpha} \right) \right]^2$$ To save my fingers, I will write $$\rho'$$ for $$\rho_{\gamma}$$ and $$d \log$$ for $$d \log g_{\gamma \alpha}$$
First we compute both parts of the sum $$d\left( \rho \frac{d \theta}{2 \pi} \right) = d\rho \frac{d \theta}{2 \pi}$$ $$d \left( \rho \, \pi^* \sum \rho' d \log \right) = d \rho \, \pi^* \sum \rho' d \log + \rho \, \pi^* \sum d (\rho' d \log) =d \rho \, \pi^* \sum \rho' d \log + \rho \, \pi^* \sum d\rho' d \log$$ Now we shall square the whole thing and for the first term we will get $$d \rho \frac{d \theta}{2\pi} \wedge d \rho \frac{d \theta}{2\pi} = 0$$ since it will contain the terms $$d \theta \wedge d \theta = 0$$. The next term will be $$d \rho \frac{d \theta}{2 \pi} \wedge \frac{1}{2 \pi i} d \left( \rho \, \pi^* \sum \rho' d \log \right) =$$ $$= \frac{1}{4\pi^2 i} \left( d \rho d\theta d\rho \pi^* \sum \rho' d \log + d \rho d \theta \rho \pi^* \sum d \rho' d \log \right) = \frac{1}{4 \pi^2 i} \rho d \rho d \theta \, \pi^* \sum d \rho' d \log$$ because $$d \rho \wedge d\rho = 0$$. (Note that $$\rho' = \rho_{\gamma}$$ is not the same function as $$\rho = \rho(r)$$. The first is a function on the manifold, while the second is a function on the bundle, so $$d \rho(r) = \frac{d \rho}{dr} dr \neq d \rho_{\gamma} = \frac{\partial \rho_{\gamma}}{\partial x^i} dx^i$$) Now there are two of these terms, and since both $$d \rho \frac{d \theta}{2\pi}$$ and $$\frac{1}{2 \pi i} d \left( \rho \, \pi^* \sum \rho' d \log \right)$$ are 2 forms, we have that $$d \rho \frac{d \theta}{2 \pi} \wedge \frac{1}{2 \pi i} d \left( \rho \, \pi^* \sum \rho' d \log \right) = \frac{1}{2 \pi i} d \left( \rho \, \pi^* \sum \rho' d \log \right) \wedge d \rho \frac{d \theta}{2 \pi}$$. Thus the middle term will then be $$\frac{1}{2 \pi^2 i} \rho d \rho d \theta \, \pi^* \sum d \rho' d \log$$ Squaring the last term will give us a bunch of terms but won't have any factors of $$d \rho d \theta$$, so when we apply $$\pi_*$$ to it, it will become $$0$$, so we needn't worry about it. Thus we end up with $$\pi_*(\Phi^2) = \frac{1}{2 \pi^2 i} \sum d \rho_{\gamma} d \log g_{\gamma \alpha} \int_{0}^{\infty} \rho(r) \rho'(r) dr \int_0^{2 \pi} d \theta=$$ $$=\frac{1}{2 \pi^2 i} \sum d \rho_{\gamma} d \log g_{\gamma \alpha} \left(\frac{\rho^2(\infty)}{2} - \frac{\rho^2(0)}{2} \right)2 \pi = \frac{1}{\pi i} \sum d \rho_{\gamma} d \log g_{\gamma \alpha} \left( 0 - \frac{1}{2} \right) = - \frac{1}{2 \pi i} \sum_{\gamma} d \rho_{\gamma} d \log g_{\gamma \alpha}$$ and this is the explicit expression for the euler class $$e(E)$$.
Thus $$u = e(E)$$ and $$\Phi^2 = \Phi \wedge \pi^*e$$.