# $\omega_{FS}^n$, the top form of Fubini Study metric

Given $\mathbb{C}P^n$, we look at the open set $U_0 = \{[Z_0, \cdots, Z_n]: Z_0 \neq 0\}$, and a coordinate map $\phi: U_0 \rightarrow \mathbb{C}^n$ by $$[Z_0,\cdots Z_n] \mapsto (Z_1/Z_0, \cdots, Z_n/Z_0).$$

Recall that the Fubini-Study metric $\omega_{FS}$ on $U_0$ is defined as $$(\phi^{-1})^*\omega_{FS} = \frac{i}{2} \sum_{j,k}\frac{(1+|z|^2)\delta_{jk} - \bar z_j z_k}{(1+|z|^2)^2} dz_j \wedge d\bar z_k.$$

I want to show that $$\int_{U_0} \omega_{FS}^n = \int_{\mathbb{C}^n} (\phi^{-1})^*\omega_{FS}^n = \pi^n$$ (I showed this for $\mathbb{C}P^1$) but I am having trouble seeing what the wedge product is for $$(\phi^{-1})^*\omega_{FS}^n= \Big((\phi^{-1})^*\omega_{FS}\Big)^n.$$

Edit: I did the following calculation for $\mathbb{C}P^2$, we have the wedge product with itself \begin{gather} \left(\frac{-z_1 \bar z_2}{(1+|z|^2)^2} dz_2 \wedge d\bar z_1 + \frac{-z_2 \bar z_1}{(1+|z|^2)^2} dz_1 \wedge d\bar z_2 \\ + \frac{((1+|z|^2)-|z_1|^2}{(1+|z|^2)^2} dz_1 \wedge d\bar z_1+\frac{((1+|z|^2)-|z_2|^2}{(1+|z|^2)^2} dz_2 \wedge d\bar z_2\right)^{\wedge 2}\\ = \frac{2}{(1+|z|^2)^3} dz_1\wedge d\bar z_1\wedge dz_2 \wedge d\bar z_2 \end{gather} Now combine the constants from Fubini-Study metric, we have \begin{gather} \int_{U_0} \omega_{FS}^2 = \int_{\mathbb{C}^2} \frac{-1}{4}\frac{2}{(1+|z|^2)^3} dz_1\wedge d\bar z_1\wedge dz_2 \wedge d\bar z_2 \\ = 4\int_{\mathbb{R}^4} \frac{1}{4}\frac{2}{(1+|x|^2)^3} dx_1dx_2dx_3dx_4\\ = 2\left(\frac{2\pi^2}{1!}\right)\int_0^\infty \frac{r^3}{(1+r^2)^3} dr \end{gather} using trig substitution for $[x=\tan \theta]$, I got $\pi^2$ for the final calculation.

## 1 Answer

Denote $$\omega_{FS} = \sum_{i,k=1}^n \frac{i}{2} g_{j\bar k} dz_j \wedge d\bar z_k,$$ we look at the matrix $(dz_j \wedge d\bar z_k)$ for $\mathbb C P^3$. \begin{bmatrix} dz_1\wedge d\bar z_1 \quad & dz_1\wedge d\bar z_2 \quad& dz_1\wedge d\bar z_3 \\ dz_2\wedge d\bar z_1 \quad & dz_2\wedge d\bar z_2 \quad& dz_2\wedge d\bar z_3 \\ dz_3\wedge d\bar z_1 \quad & dz_3\wedge d\bar z_2 \quad& dz_3\wedge d\bar z_3 \end{bmatrix} we see all the terms in $\omega_{FS}^3$ corresponds to picking three entries in the above matrix, and a non-zero term in $\omega_{FS}^3$ means the three entries has to come from different roll and column. This pattern is closely related to computing the determinant using Laplace expansion.

In general for $\mathbb C P^n$, we have $$\boxed{\omega_{FS}^n = \left(\frac{i}{2}\right)^n n! \det(g_{i\bar j}) dz_1\wedge d\bar z_1 \wedge \cdots \wedge dz_n \wedge d\bar z_n.}$$ Recall $g_{j\bar k} = \frac{(1+|z|^2)\delta_{jk} - \bar z_j z_k}{(1+|z|^2)^2}$, from a matrix determinant lemma $$\det(A+uv^T) = (1+ v^TA^{-1}u)\det(A),$$ we have $$\boxed{\det(g_{j\bar{k}}) = \frac{1}{(1+|z|^2)^{n+1}}.}$$ We have \begin{align*} \int_{U_0} \omega_{FS}^n &= \int_{\mathbb C^n} \left(\frac{i}{2}\right)^n n! \det(g_{i\bar j}) dz_1\wedge d\bar z_1 \wedge \cdots \wedge dz_n \wedge d\bar z_n\\ &= \left(\frac{i}{2}\right)^n n! (-2i)^n \text{vol}(S_{2n-1}) \int_0^\infty \frac{r^{2n-1}}{(1+r^2)^{n+1}} dr\\ &= \pi^n 2n \int_0^{\pi/2} \sin^{2n-1}\theta \cos \theta d\theta\\ &=\pi^n \end{align*} Once again, since $U_0^c$ is a null set, we have $\int_{\mathbb C P^n} \omega_{FS}^n = \pi^n$.