Why is the curvature of the connection $\bar{z_1}dz_1 + \bar{z_2}dz_2$ on the Hopf fibration not exact?

Question

Let $\pi : S^3 \to S^2$ be the Hopf fibration, where we take $S^3 \subset \mathbb{C}^2$, $S^2 = \mathbb{C}\mathbb{P}^1$, and $\pi(z_1, z_2) = [z_1 : z_2]$. This is a principal $U(1)$-bundle. The form $\omega = \bar{z_1}dz_1 + \bar{z_2}dz_2$ on $S^3$ defines a connection on this principal bundle. Its curvature is given by $K = d\bar{z_1}\wedge dz_1 + d\bar{z_2}\wedge dz_2$. Since the Lie algebra of $U(1)$ is isomorphic to $\mathbb{R}$ and the adjoint representation is trivial, $K$ descends to a standard $2$-form on $S^2$. I want to show this gives a non-trivial element of $H^2_{dR}(S^2)$.

I have tried to compute the pullback along $\pi$ of the Fubini-Study area form on $\mathbb{C}\mathbb{P}^1$ to see whether this is proportional to $K$, but I didn't succeed. Does someone know how to do this? (A solution using a different approach would also be appreciated.)

Answer 1

HINT: Pull back your $2$-form $K = -(dz_1\wedge d\bar z_1+dz_2\wedge d\bar z_2)$ by the map $\sigma(z) = (z_1,z_2)=\left(\frac1{\sqrt{1+|z|^2}},\frac z{\sqrt{1+|z|^2}}\right)$ (mapping $\Bbb C\subset\Bbb CP^1$ to $S^3$).

Answer 2

If the curvature of the connection is exact, it implies that the bundle is flat, in this case, $S^3$ is the quotient of $S^2\times S^1$ by an action of $\pi_1(S^2)=1$, this is not true since $\pi_1(S^3)$ is trivial.