I am trying to be concrete about the Fubini-Study metric on $\mathbb{C}P^2$, in order to do some Maple calculation for harmonic maps.
On the affine chart $\mathbb{C}^2 = \{[1, z_1, z_1]\}\subset \mathbb{C}P^2$, the Hermitian matrix representing the Fubini-Study metric is
$$
\frac{1}{(1+|z|^2)^2} (\delta_{ij}(1+|z|^2) – \bar {z_i} z_j),
$$
where $z=(z_1, z_2)$ and $|z|^2 = |z_1|^2 + |z_2|^2$.
By the rule that a Hermitian matrix $A+iB$ corresponds to a symmetric
$$
\begin{pmatrix}
A & B\\
-B & A
\end{pmatrix},
$$
the above Fubini-Study metric in the $z_1 = x_1+iy_1, z_2 = x_2 + iy_2$ coordinates should corresponds to the Riemannian metric represented by
$$
\frac{1}{(1+x_1^2+x_2^2+y_1^2+y_2^2)^2}
\begin{pmatrix}
1+x_2^2+y_2^2 & -(x_1 x_2+y_1 y_2) & 0 & -(x_1 y_2 – y_1x_2)\\
-(x_1 x_2+y_1 y_2) & 1+x_1^2 + y_1^2 & (x_1 y_2 – y_1x_2) & 0\\
0 & (x_1 y_2 – y_1x_2) & 1+x_2^2+y_2^2 & -(x_1 x_2+y_1 y_2) \\
-(x_1 y_2 – y_1x_2) & 0 & -(x_1 x_2+y_1 y_2) & 1+x_1^2 + y_1^2
\end{pmatrix}.
$$
However when I compute the curvature of the Riemannian metric using Maple, it is not Einstein: the Ricci curvature is not constant.
I just wonder if someone can point out where my mistake is. Thanks.
Best Answer
It is funny how you tend to find the answer yourself after you post a question. The writing does help with your thoughts.
There is no problem with my matrix above, but the thing that I didn't pay enough attention to is that the matrix represents the Riemannian metric for the coordinates $$ x_1, x_2, y_1, y_2 $$ in this order.
Anyway, it works this way. (I was using $u_1, u_2, u_3, u_4$ and was not aware of the switching of $u_2$ and $u_3$ before.)