Help with showing $\int_{\mathbb{C}P^1} \omega_{FS} = \pi$ Suppose $\mathbb{C}P^1$ is covered by two open sets $U_1, U_2$. 
On 
$U_1 = \{[Z_1,Z_2]\in \mathbb{C}P^1 : Z_1 \neq 0\},$
we have the coordinate map $z:\mathbb{C}P^1 \rightarrow \mathbb{C}$ with 
$$[Z_1, Z_2]\mapsto Z_2/Z_1.$$
To compute $\int_{{U_1}} \omega_{FS} $, I know the Fubini-Study metric on $U_1$ is
$$\omega_{FS}=\frac{i}{2}\frac{1}{(1+|z|^2)^2} dz\wedge d\bar z$$
the next step is really confusing for me, people directly said
$$\int_{U_1} \omega_{FS} = \int_\mathbb{C} \frac{i}{2}\frac{1}{(1+|z|^2)^2} dz\wedge d\bar z.$$
But by definition
$$\int_{U_1} \omega_{FS} = \int_{z(U_1)} \delta z(\omega_{FS})$$
where $\delta z$ is the adjoint of $dz$, so why is 
$$ \delta z(\omega_{FS}) = \frac{i}{2}\frac{1}{(1+|z|^2)^2} dz\wedge d\bar z ?$$
And from 
$$\int_\mathbb{C} \frac{i}{2}\frac{1}{(1+|z|^2)^2} dz\wedge d\bar z = 
-2i  \int_{\mathbb{R}^2} \frac{i}{2}\frac{1}{(1+x^2+y^2)^2}dxdy = \pi,$$ so does this mean $\int_{U_1^c} \omega_{FS} = 0?$  
 A: Let $\varphi : U_1 \to \mathbb C$ mapping $p \mapsto z(p)$ be the first coordinate chart.
$\omega_{FS}$ is a differential form on $\mathbb CP^1$; it can't be written in terms of $z$'s and $\bar z$'s.
However, $(\varphi^{-1})^\star \omega_{FS}$ is a differential form on $\mathbb C$. $z$ and $\bar z$ are coordinates on this $\mathbb C$. We can write
$$
(\varphi^{-1})^\star \omega_{FS} = \frac i 2 \frac 1 {(1 + |z|^2)^2} dz \wedge d\bar z
$$
So
$$
\int_U \omega_{FS} = \int_{\varphi(U)} (\varphi^{-1})^\star \omega_{FS} = \int_{\mathbb C} \frac i 2 \frac 1 {(1 + |z|^2)^2} dz \wedge d\bar z
$$
To address the question about $U_2$, note that $U_1 \cap U_2 $ is non-empty. So
$$
\int_{\mathbb {CP}^1} \omega_{FS} \neq \int_{\mathbb U_1} \omega_{FS} +\int_{\mathbb U_2} \omega_{FS} 
$$
because the expression on the RHS double-counts on the overlap!
To evaulate the integral correctly, you can use a partition of unity subordinate to the cover $U_1, U_2$. Or even better, notice that $U_2 \backslash U_1$ is a single point - the south pole - which has measure zero. so really, $\int_{\mathbb {CP}^1} \omega_{FS} =\int_{\mathbb U_1} \omega_{FS} = \pi $.
