Computing the Fubini-Study metric in affine chart
Someone asked this question recently and then deleted it, but I still would like to figure out the answer.
Let us try to compute the Fubini-Study metric in inhomogeneous coordinates on $\mathbb{C} P^n$, using the Hopf fibration. For simplicity, let us do $n=1$.
Start with $\mathbb{C}^2$. In the standard coordinates $(Z_0, Z_1)$, the standard flat metric is given by $g= dZ_0\,d\overline{Z_0} + dZ_1\,d\overline{Z_1}$. This metric restricts to the round metric on the unit sphere $S^3$ seen as $S^3 = \{|Z_0|^2 + |Z_1|^2 = 1\}$.
The group of unit complex numbers $U(1)$ acts on $S^3$ by multiplication, and the quotient is the complex projective line $\mathbb{C}P^1$. (Note that this is the Hopf fibration $S^1 \to S^3 \to S^2$). In homogeneous coordinates $[Z_0 \colon Z_1]$ on $\mathbb{C}P^1$, the projection map $\pi \colon S^3 \to \mathbb{C}P^1$ is just $(Z_0, Z_1) \mapsto [Z_0 \colon Z_1]$. In the inhomogeneous coordinate $z = Z_1/Z_0$ (such that $[Z_0 \colon Z_1] = [1 \colon z]$), the projection map is $(Z_0, Z_1) \mapsto z = Z_1/Z_0$.
Since the metric on $S^3$ is invariant under the action of $U(1)$, it descends to a metric on $\mathbb{C}P^1$ such that the projection map $\pi \colon S^3 \to \mathbb{C}P^1$ is an isometry. This metric on $\mathbb{C}P^1$ is the Fubini-Study metric. In order to compute it, we can use any local section $s \colon \mathbb{C}P^1 \to S^3$, for instance: $$s(z) = \left(Z_0 = \frac{1}{\sqrt{1+z\overline{z}}}, Z_1 = \frac{z}{\sqrt{1+z\overline{z}}}\right)~.$$ The Fubini-Study metric in the coordinate $z$ will simply be given by the pull-back metric $g_{FS} := s^* g = dZ_0\,d\overline{Z_0} + dZ_1\,d\overline{Z_1}$ where $Z_0$ and $Z_1$ denote (somewhat abusively) the functions of $z$ given by the definition of $s(z)$ as above.
A direct computation gives: $$dZ_0 = \frac{-\overline{z}\,dz - z\,d\overline{z}}{2(1+z\overline{z})^{3/2}}$$ $$dZ_1 = \frac{(2+z\overline{z})\,dz - z^2 \,d\overline{z}}{2(1+z\overline{z})^{3/2}}$$ which yields $$dZ_0\,d\overline{Z_0} + dZ_1\,d\overline{Z_1} = \frac{-\overline{z}^2\,dz^2 - z^2\,d\overline{z}^2 + 2(2+z\overline{z})\,dz\,d\overline{z}}{4\,(1+z\overline{z})^2}$$
The problem is that this is not what is expected: instead we expect $$g_{FS} = \frac{dz\,d\overline{z}}{(1+z\overline{z})^{2}}~.$$
Where have I gone wrong?
I found where I went wrong but it took me a while to fix it.
First of all notice that the claim that "the projection map $\pi \colon S^3 \to \mathbb{C}P^1$ is an isometry" cannot be true, since $S^3$ and $\mathbb{C}P^1$ do not have the same dimension. When one does a Riemannian quotient, what is required is that the projection map is isometric for tangent vectors that are orthogonal to the fibers. For this reason, it is also false to say that "in order to compute the metric on $\mathbb{C}P^1$, we can use any local section $s \colon \mathbb{C}P^1 \to S^3$". Only a section which is orthogonally transverse to the orbits of the $U(1)$-action will be isometric.
So in order to fix my calculation I looked for such an orthogonal section. Denote by $s_0 \colon \mathbb{C}P^1 \to S^3$ the section we chose originally, then let us look for another section of the form $s(z) = e^{i\theta(z)}\,s_0(z)$, where $\theta(z)$ is a real-valued unknown function to be determined so that $s(z)$ is orthogonal to the orbits of the $U(1)$-action. We can explicitely find the equation satisfied by $\theta$ by writing that $s_*\left(\frac{\partial}{\partial z}\right)$ must be orthogonal (for the standard flat metric $g$ on $\mathbb{C}^2$) to the vector field $\vec U$ on $\mathbb{C}^2$ generating the $U(1)$-action.
If my calculations are correct, with: $$s_*\left(\frac{\partial}{\partial z}\right) = \frac{\partial Z_0}{\partial z} \frac{\partial}{\partial Z_0} + \frac{\partial (\overline{Z_0})}{\partial z} \frac{\partial}{\partial \overline{Z_0}} + \frac{\partial Z_1}{\partial z} \frac{\partial}{\partial Z_1} + \frac{\partial (\overline{Z_1})}{\partial z} \frac{\partial}{\partial \overline{Z_1}}$$ where I have written $$s(z) = \left(Z_0(z), Z_1(z)\right) = \left(\frac{e^{i\theta(z)}}{\sqrt{1+z\overline{z}}}, \frac{z\,e^{i\theta(z)}}{\sqrt{1+z\overline{z}}}\right)$$ and $$\vec U = iZ_0 \frac{\partial}{\partial Z_0} -i\overline{Z_0} \frac{\partial}{\partial \overline{Z_0}} + iZ_1 \frac{\partial}{\partial Z_1} + -i\overline{Z_1} \frac{\partial}{\partial \overline{Z_1}}$$ we find: $$g\left(s_*\left(\frac{\partial}{\partial z}\right), \vec U\right) = 2 \frac{\partial \theta}{\partial z} -\frac{i\overline{z}}{1+z\overline{z}}$$ therefore we get the equation $$\frac{\partial \theta}{\partial z} = \frac{i\overline{z}}{2(1+z\overline{z})}~.$$
This is a very simple PDE on the function $\theta \colon \mathbb{C} \to \mathbb{R}$, unfortunately, it does not have any solutions! Indeed, the equality of the mixed second partial derivatives of $\theta$ is not verified, contradicting Schwarz's theorem. This was very troubling to me at first so I checked my computations again and again. But I am now inclined to think that the conclusion of all this is that there does not exist such orthogonal sections. It is maybe a bit counter-intuitive, but it is entirely possible that the distribution in the tangent space of $S^3$ defined as the orthogonal complement of $\vec U$ is not tangent to submanifolds, in other words is not integrable (i.e. stable under the Lie bracket, by the Frobenius theorem). If someone could please confirm this, that would be great. Maybe there is a nice way to think about this geometrically in terms of the Hopf fibration (where the $U(1)$-orbits are Villarceau circles).
In any case, it is still possible of course to compute the quotient metric $g_{FS}$ on $\mathbb{C}P^1$. We're just going to have to work infinitesimally (in the tangent space) rather than with sections, which is actually easier. Let me do it for completeness: Let $\vec v = \alpha \frac{\partial}{\partial z} + \beta \frac{\partial}{\partial \overline{z}}$ be a (complexified) tangent vector to $\mathbb{C}P^1$ at some point $z$, and let us look for the unique tangent vector $\vec V$ at the point $\left(Z_0 = \frac{1}{\sqrt{1+z\overline{z}}}, Z_1 = \frac{z}{\sqrt{1+z\overline{z}}}\right)$ such that:
- $\vec V$ is tangent to $S^3$ (i.e. orthogonal to the radial vector field $\vec R$).
- $\vec V$ is orthogonal to $\vec U$ (the vector field generating the $U(1)$-action).
- $\pi_* (\vec V) = \vec v$.
It is straightforward to find that $\vec V$ is given by: $$\vec V = (1+z\overline{z})^{-3/2}\left(-\alpha \overline{z}\frac{\partial}{\partial Z_0} -\beta z \frac{\partial}{\partial \overline{Z_0}} + \alpha \frac{\partial}{\partial Z_1} + \beta \frac{\partial}{\partial \overline{Z_1}}\right)~.$$ Since $\pi$ must be isometric on $\vec V$, we can write: $$g_{FS}(\vec v, \vec v) = g(\vec V, \vec V) = \frac{\alpha \beta}{(1 + z \overline{z})^2}$$ and we can finally conclude that $$g_{FS} = \frac{dz\, d\overline{z}}{(1 + z \overline{z})^2}~.$$