# Why is Fubini-Study metric invariant under $SU(n+1)$ action on $\mathbb{P}^n \mathbb{C}$

In a lot of books on complex geometry, they say that the Fubini-Study metric on $\mathbb{P}^n \mathbb{C}$ is invariant under the action of $SU(n+1)$ on $\mathbb{P}^n \mathbb{C}$, but i can not see why.

The Fubini Study metric $\omega_{FS}$ is defined in coordinates of $\mathbb{P}^n \mathbb{C}$ as follows:

On the open set $U_i = \{[z_0: \dotsb : z_n] \in \mathbb{P}^n \mathbb{C} : z_i\neq 0 \}$ and the coordinate system defined by $$[z_0: \dotsb : z_n] \mapsto \bigg(\frac{z_0}{zi}, \dotsb , \frac{z_{i-1}}{z_i}, \frac{z_{i+1}}{z_i}, \dotsb, \frac{z_n}{z_i}\bigg) = (w_1, \dotsb ,w_n)$$ $\omega_{FS}$ is defined in $U_i\cong \mathbb{C}^n$ by:

$$\displaystyle{\omega_i = \frac{1}{2 i \pi }\partial \bar{\partial} log ( 1 + \sum_{i=1}^{n} \vert w_i \vert^2)}$$

Then it is shown that this expression in coordinates defines a global 2-form.

Doing the calculations we find the formula:

$$\displaystyle{\omega_i = \frac{1}{(1+ \sum_i \vert w_i\vert^2)^2}} \sum h_{ij} dw_i \wedge d\overline{w_j}$$

With $$\displaystyle{h_{ij}= \big(1+ \sum _{i=1}^n \vert w_i\vert^2 \big)\delta_i^j - \overline{w_i}w_j}.$$

Now if $A \in SU(n+1)$ then $A$ descends to a map $$\overline{A} : \mathbb{P}^n\mathbb{C} \to \mathbb{P}^n\mathbb{C}$$ $$[z] \mapsto [A (z)]$$ it is easy verified that this defines a holomorphic map on $\mathbb{P}^n\mathbb{C}$.

How can I show that $\overline{A}^*\omega_{FS} = \omega_{FS}$ ?

I have been working with coordinates since the definition is given in coordinates, the problem I find is that the form of the map $\overline{A}$ in coordinates is not in $U(n)$. I would like to prove something using the commutative property of the $\overline{A}^*$ with $\partial$ and $\bar{\partial}$ in the formula

$$\displaystyle{\omega_i = \frac{1}{2 i \pi }\partial \bar{\partial} log ( 1 + \sum_{i=1}^{n} \vert w_i \vert^2)}$$

fixing a point $[z] \in U_i$ and $[A(z)]$ in $U_j$ using the pullback of both version of the form in corresponding charts. But without luck for the moment.

I have already read a similar post here: Unitary transformation of Fubini-Study metric but i don't think the argument is correct, since as I wrote before, in chats the transformation does not descend in general to an element in $SU(n)$ (I guess).

I would appreciate any help, or suggestion.

• "the transformation does not descend in general to an element in $SU(n)$" - huh? What transformation? – anon Apr 29 '18 at 16:01
• I mean when i take $[z] \in U_i$ and $[A(z)] \in U_j$ if $\Psi_i$ and $\Psi_j$ are the corresponging charts i wrote before, i mean: that the transformation $$\Psi_j \circ \overline{A} \circ \Psi_i^{-1}$$ is not in $SU(n)$. – Alicia Basilio Apr 29 '18 at 16:09
• @AliciaBasilio Can you list a reference where this is mentioned? I will be easier for providing a detailed answer. – Test123 Apr 29 '18 at 16:29
• Hodge Theory and Complex Algebraic Geometry Caire Vousin on pp.78 (there they define the metric a little different, as $\omega_i = \frac{1}{2 i \pi }\partial \bar{\partial} log ( \frac{1}{1 + \sum_{i=1}^{n} \vert w_i \vert^2})$ (but my calculation shows they define the same 2-form. Complex Geometry by Daniel Huybrechts, the define the metric exactly as i did pp 117, and the put like excersice prove this in pp.123 ex 3.1.6. finally encyclopediaofmath.org/index.php/Fubini-Study_metric . But i think there they are working in different coordinates. – Alicia Basilio Apr 29 '18 at 16:38
• @AliciaBasilio View $\mathbb{C}P^n$ as the quotient $S^{2n+1}/S^1$. The unique metric on $\mathbb{C}P^n$ such that the quotient map $S^{2n+1}\rightarrow \mathbb{C}P^n$ is a riemannian submersion, is the FS metric. Note also that the natural action of $U(n+1)$ on $\mathbb{C}^{n+1}\setminus\{0\}$ gives a transitive smooth action of $U(n+1)$ on $\mathbb{C}P^n$. – Test123 Apr 30 '18 at 12:09