Let $S^3$ be the 3-sphere, and $\Sigma$ be a 2-dimension manifold. Let $\omega$ be a 2-form on $\Sigma$. $f:S^3\rightarrow \Sigma$ is a $C^{\infty}$ map.Then there is a 1-form $\alpha$ on $S^3$ such that $f^{\ast}\omega=d\alpha$.For such an $\alpha$, define $I(\alpha)=\int_{S^3}\alpha\wedge d\alpha$.
$(1)$ Prove that $I(\alpha)$is independent of the choice of $\alpha$
$(2)$ In the case $\Sigma$ is the 2-torus $T^2$,prove that there are 1-form $\eta_1,\eta_2$ and $\eta_3$ on $\Sigma$ such that $\omega=\eta_1\wedge \eta_2+d\eta_3, d\eta_1=d\eta_2=0.$
$(3)$ Prove that when $\Sigma$ is the 2-torus $T^2$,$I(\alpha)=0$
I have proved$(1)$:
Since $f^{\ast}\omega$ is a 2-form on $S^3$ and $df^{\ast}\omega=f^{\ast}d\omega=0,$$H_{DR}^2(S^3)=0$ yields that there is a 1-form $\alpha$ on $S^3$ such that $d\alpha=f^{\ast}\omega.$ Suppose another $\hat\alpha$ satisfies $d\hat\alpha=f^{\ast}\omega,$then $d(\alpha-\hat\alpha)=0$,therefore there is a 0-form $g$ such that $\alpha-\hat\alpha=dg.$
Now $I(\alpha)-I(\hat\alpha)=\int_{S^3}(\alpha-\hat\alpha)\wedge d\alpha=\int_{S^3}dg\wedge d\alpha=\int_{S^3}d(g\wedge d\alpha)=\int_{\partial S^3}g\wedge d\alpha=0$ by Stokes's thm.
But for $(2)$ I have no idea how to prove it.I know $H_{DR}^r(M)=0$ means closed r-forms are exact forms but since $H_{DR}^1(T^2)=\Bbb R^2$ and $H_{DR}^2(T^2)=\Bbb R$ , how can I use them to describe the relationship between closed forms and exact forms and how to show the existence of $\eta_i$?For$(3)$ I'm also confused how to prove it,I'll be appreciated if you can give me some hints or help, thank you so much!