Could we compute integral of $z_i\bar{z_j}/\sum |z_k|^2$ on $\mathbb{C}P^n$ with the Fubini-Study metric?

Question

Let $\mathbb{C}P^n$ be the complex projective space with homogeneous coordinate $Z=[z_0,\ldots,z_n]$. Let $\omega$ be the Fubini-Study metric $\omega=\frac{i}{2\pi}\partial\bar{\partial}\log||Z||^2$.

From this stackexchange answer we know that the volume of $\mathbb{C}P^n$ with the Fubini-Study metric is $\frac{\pi^n}{n!}$, i.e. $$ \int_{\mathbb{C}P^n}1\cdot \omega^n=\frac{\pi^n}{n!}. $$ Then by symmetry, it is easy to see that $$ \int_{\mathbb{C}P^n}\frac{z_i\bar{z_i}}{||Z||^2}\cdot \omega^n=\frac{\pi^n}{(n+1)!}, \forall~ 0\leq i\leq n. $$ Now for different $i$ and $j$, could we compute $$ \int_{\mathbb{C}P^n}\frac{z_i\bar{z_j}}{||Z||^2}\cdot \omega^n? $$ More generally, for any $k\geq 0$ and any two tuples $(i_0,\ldots i_k)$ and $(j_0,\ldots j_k)$, define a function $$ f(Z)=\frac{z_{i_0}\ldots z_{i_k}\bar{z_{j_0}}\ldots\bar{z_{j_k}}}{||Z||^{2k}}. $$ Could we get a formula for the integral $$ \int_{\mathbb{C}P^n}f\cdot \omega^n? $$

Edit: According to the answer of @AmitaiYuval, we only need to consider the case where $$ f(Z)=\frac{|z_{i_0}|^2\ldots |z_{i_k}|^2}{||Z||^{2k}}. $$

Edit: For $\mathbb{C}P^1$, a iterating computation shows that for $$ f(Z)=\frac{|z_0|^{2k}|z_1|^{2(k-l)}}{||Z||^{2k}} $$ we have $$ \int_{\mathbb{C}P^1}f\cdot \omega=\frac{{k\choose l}\pi}{k+1}. $$ But we are still looking for the answer for higher dimensions.

Answer 1

Let $0\leq i\neq j\leq n$, and let $\varphi:\mathbb{C}P^n\to\mathbb{C}P^n$ be the "reflection" given by $$[z_0:\ldots:z_n]\mapsto[z_0:\ldots:z_{i-1}:-z_i:z_{i+1}:\ldots:z_n].$$ The Fubini-Studi form is preserved by $\varphi$, while your function is anti-preserved by $\varphi$. It follows that the average value is $0$.

Answer 2

(A very late answer, not sure if you are still interested.)

According to the formula at the end of the proof of Proposition 2.1 and Lemma 2.2 in Diverio, Simone: Segre forms and Kobayashi-Lübke inequality. Math. Z. 283 (2016), no. 3-4, 1033–1047, $$\int_{\mathbb{P}^n} |z_{i_1}|^2 \cdots |z_{i_k}|^2 \omega^n = \frac{m_0! \cdots m_n!}{(n+k)!},$$ where $m_j$ is the number of times that $j$ appears among the $i_\ell$.

In the example in your Edit, I imagine you mean $|z_0|^{2\ell}$ to make it homogeneous, then the formula from Diverio matches what you obtained. There is also the difference of the factor $\pi^n$, which I would imagine being due to different normalizations.