A form problem between $S^3$ and $S^2$.

Question

Let $\phi: S^3 \rightarrow S^2$ be an smooth map.

a) Suppose that $\omega$ is a 2-form on $S^2$ with $\int_{S^2} \omega =1$. Show that there exists a 1-form $\alpha$ on $S^3$ with $\phi^{*}\omega=d\alpha$.

b) Consider the number $N=\int_{S^3} \alpha \wedge \phi^{*}\omega $. Show that $N$ is independent of the choice of $\omega$ satisfying $\int_{S^2} \omega =1$ and the choice of $\alpha$ satisfying $\phi^{*}\omega=d\alpha$.

For the first part, I use de Rham cohomology, we know that $H_{2}(S^3)=0$ then every closed 2-form is exact. then is enough to show that $\phi^{*}\omega$ is closed. But using the naturality of the operator $d$, we have $d\phi^{*}\omega=\phi^{*}(d\omega)=0$ since $d\omega=0$. So we have that $\phi^{*}\omega$ is exact. My problem is that I never use the hypothesis $\int_{S^2} \omega =1$.

Also, I don't know how to do part b).

Answer 1

1. You don't need $\int \omega=1$ for a); in fact, the statement in a) is scale-invariant, so proving it for $\omega$ with $\int \omega=1$ is precisely equivalent to proving it for all $\omega$. You will, however, need it in order to do b).
2. To show that $N$ is independent of $\alpha$ for a given $\omega$, suppose $\phi^* \omega=d\alpha=d\beta$. Then $$0=\int_{S^3} d(\alpha \wedge \beta)=\int_{S^3} \left(\beta \wedge d\alpha - \alpha \wedge d\beta\right) = \int_{S^3} \beta \wedge \phi^* \omega - \int_{S^3} \alpha \wedge \phi^* \omega$$ via Stokes' theorem and the Leibniz rule.
3. To show that $N$ is independent of $\omega$, suppose $\omega$ and $\eta$ both integrate to $1$. Since $H^2(S^2) \cong \Bbb{R}$ via the integration map, $\omega$ and $\eta$ lie in the same de Rham cohomology class; thus $\eta = \omega+d\theta$ for some 1-form $\theta$. Choose $\alpha$ with $\phi^* \omega=d\alpha$; then $\phi^* \eta=\phi^*(\omega+d\theta)=d(\alpha + \phi^* \theta)$ (since $d$ commutes with pullbacks). Now, if we compute $N$ using $\eta$, we get $$\int_{S^3} (\alpha + \phi^* \theta) \wedge d(\alpha + \phi^* \theta) = \int_{S^3} \alpha \wedge d\alpha + \int_{S^3} \left(\alpha \wedge d(\phi^* \theta) + \phi^* \theta \wedge d\alpha\right) + \int_{S^3} \phi^* \theta \wedge d(\phi^* \theta) \, .$$ The first integral is what we get if we compute $N$ using $\omega$. The second integrand is equal to $d(\phi^* \theta \wedge \alpha)$, so the second integral vanishes by Stokes' theorem. The third integrand may also be written as $\phi^*(\theta \wedge d\theta)$; since $\theta \wedge d\theta$ is a $3$-form on $S^2$, it vanishes (and thus so does its pullback). This completes the proof.