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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 $\int_{U_1^c} \omega_{FS} = 0?$

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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 $.

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    Thank you very much for the helpful reply! And do you know why people often write $\omega_{FS}=\frac{i}{2}\frac{1}{(1+|z|^2)^2} dz\wedge d\bar z$, or like this for $\mathbb{C}P^n$ with $\omega_{FS} = \sum \cdots dz_j \wedge d\bar z_k$, is this like a standard notation? So I thought $z$ here is a coordinate map from $U_1$ to $\mathbb{C}$. – 2017-02-28
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    It's just shorthand for $(\varphi^{-1})^\star \omega_{FS} = ...dz \wedge d\bar z$. – 2017-02-28
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    And you wrote $(\phi^{-1})^* \omega_{FS}$ is a differential from on $\mathbb{C}$, should it be $(\phi)^* \omega_{FS}$? – 2017-02-28
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    No, I think my way is right. The notation $\phi^\star$ means pullback. – 2017-02-28
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    $\phi$ goes from $U$ to $\mathbb C$. So $\phi^{-1}$ goes from $\mathbb C $ to $U$. So $(\phi^{-1})^\star$ takes pulls $\omega$ from $U$ back to $\mathbb C$. – 2017-02-28
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    I see, but later in the integral, how do we see the term $\phi^* \omega_{FS}$? – 2017-02-28
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    Sorry, that was a typo! – 2017-02-28