Considering the function $\pi:S^3\to S^2$
$$\pi(x_1,x_2,x_3,x_4):=(2(x_1x_3+x_2x_4),2(x_2x_3-x_1x_4),x_1^2+x_2^2-x_3^2-x_4^2)$$
Let $\omega=x_1\text{d}x_2\wedge\text{d}x_3-x_2\text{d}x_1\wedge\text{d}x_3+x_3\text{d}x_1\wedge\text{d}x_2\in\Omega^2(S^2)$, prove that $\exists\alpha\in\Omega^1(S^3)$ s.t.$\text{d}\alpha=\pi^*\omega$.
My idea:
$\pi^*\omega=4(x_1x_3+x_2x_4)\text{d}(x_2x_3-x_1x_4)\wedge\text{d}(x_1^2+x_2^2-x_3^2-x_4^2)-4(x_2x_3-x_1x_4)\text{d}(x_1x_3+x_2x_4)\wedge\text{d}(x_1^2+x_2^2-x_3^2-x_4^2)+4(x_1^2+x_2^2-x_3^2-x_4^2)\text{d}(x_1x_3+x_2x_4)\wedge\text{d}(x_2x_3-x_1x_4)$
Let $\alpha=f_1\text{d}x_1+f_2\text{d}x_2+f_3\text{d}x_3+f_4\text{d}x_4$, then $\text{d}\alpha=\text{d}f_1\wedge\text{d}x_1+\text{d}f_2\wedge\text{d}x_2+\text{d}f_3\wedge\text{d}x_3+\text{d}f_4\wedge\text{d}x_4=(\frac{\partial f_2}{\partial x_1}-\frac{\partial f_1}{\partial x_2})\text{d}x_1\wedge\text{d}x_2+(\frac{\partial f_3}{\partial x_1}-\frac{\partial f_1}{\partial x_3})\text{d}x_1\wedge\text{d}x_3+(\frac{\partial f_4}{\partial x_1}-\frac{\partial f_1}{\partial x_4})\text{d}x_1\wedge\text{d}x_4+(\frac{\partial f_3}{\partial x_2}-\frac{\partial f_2}{\partial x_3})\text{d}x_2\wedge\text{d}x_3+(\frac{\partial f_4}{\partial x_2}-\frac{\partial f_2}{\partial x_4})\text{d}x_2\wedge\text{d}x_4+(\frac{\partial f_4}{\partial x_3}-\frac{\partial f_3}{\partial x_4})\text{d}x_3\wedge\text{d}x_4$.
But I can't find $f_i,i=1,2,3,4$. Any ideas?
Best Answer
This answer uses some facts about de Rham cohomology. The use of de Rham cohomology is not necessary, but it does simplify matters greatly.
The main facts we will use are
(It's actually not hard to compute $\mathrm{H}^k_\mathrm{dR}(S^n)$ for all $n$ and $k$, but the above is all we will need.)
Another way to state Fact 3 is that, for any $\sigma \in \Omega^2(S^3)$, if $\mathrm{d}\sigma =0$ then there exists $\tau \in \Omega^1(S^3)$ such that $\sigma = \mathrm{d}\tau$.
To apply this to our situation, we will need to know that $\mathrm{d} (\pi^* \omega) = 0$. By Fact 1, we have that $\mathrm{d} (\pi^* \omega) = \pi^*(\mathrm{d} \omega)$. Since $\mathrm{d}\omega \in \Omega^3(S^2) =0$ by Fact 2, we have the desired equality. We can conclude now by Fact 3.